Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium

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

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

Time: 3 Hours
Maximum Marks: 70

Instructions:

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

Part – 1

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

Question 1.
Match items in column -I with the items of column – II and assign the correct code:
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 1
Answer:
(c) A – (iv), B – (ii), C – (iii), D – (i)

Question 2.
Which one of the following is double salt?
(a) Potash alum
(b) Potassium sulphate
(c) Aluminium Sulphate
(d) Ammonium sulphate
Answer:
(a) Potash alum

Question 3.
Most easily liquefiable gas is ……………..
(a) Ar
(b) Ne
(c) He
(d) Kr
Answer:
(c) He

Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium

Question 4.
Which metal is used in manufacturing artificial joints?
(a) Molybdenum
(b) Titanium
(c) Tungsten
(d) Iron
Answer:
(b) Titanium

Question 5.
In which of the following coordination entities the magnitude of A will be maximum?
(a) [CO(CN)6]3-
(b) [CO(C2O4)3]3-
(c) [CO(H2O)6]3+
(d) [CO(NH3)6]3+
Answer:
(a) [CO(CN)6]3-

Question 6.
Which is the coordination number in both hep and ccp arrangements?
(a) 12
(b) 6
(c) 4
(d) 8
Answer:
(a) 12

Question 7.
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200K to 400K? (R = 8.314 Jk-1 mol-1)
(a) 234.65 kJmol-1 K-1
(b) 434.65 kJ mol-1 K-1
(c) 434.65 J mol-1K-1
(d) 334.65 J mol-1K-1
Answer:
(c) 434.65 J mol-1K-1
Solution:
T1 = 200K ; k = k1
T2 = 400K ; k = k2 = 2k1
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 2
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 3

Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium

Question 8.
The following solutions were prepared by mixing different volumes of NaOH of HC1 different concentrations.
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 4
pH of which one of them will be equal to 1?
(a) (iv)
(b) (i)
(c) (ii)
(d) (iii)
Answer:
(d) (iii)
No of moles of HCl = 0.2 × 75 × 10-3 = 15 × 10-3
No of moles of NaOH = 0.2 × 25 × 10-3 = 5 × 10-3
No of moles of HC1 after mixing = 15 × 10-3 – 5 × 10 -3
= 10 × 10-3
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 5
for (iii) solution, pH of 0.1MHCl = -log10(0.1) = 1.

Question 9.
The electrode used in SHE is made of ……….
(a) graphite
(b) copper
(c) platinum
(d) iron
Answer:
(c) platinum

Question 10.
Which one of the following is an example for homogeneous catalysis?
(a) manufacture of ammonia by Haber’s process
(b) manufacture of sulphuric acid by contact process
(c) hydrogenation of oil
(d) Hydrolysis of sucrose in presence of all HCl
Both reactant and catalyst are in same phase. i.e. (1)
Answer:
(d) Hydrolysis of sucrose in presence of all HCl

Question 11.
The alkaline hydrolysis of fats to give glycerol is known as
(a) Esterification
(b) Hydroboration
(c) Hydration
(d) Saponification
Answer:
(d) Saponification

Question 12.
Consider the following statements:
(i) In Rosenmund reduction Barium sulphate act as a Catalyst poison palladium catalyst, so that aldehyde cannot be further reduced to alcohol
(ii) Side chain oxidation of toluene using strong oxidising agent gives benzoic acid.
(iii) Friedle crafts reaction is the best method used to prepare aliphatic ketones.
Which of the above statement is/are correct?
(a) (iii) only
(b) (i) & (ii)
(c) (i) & (iii)
(d) (ii) & (iii)
Answer:
(b) (i) & (ii)

Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium

Question 13.
Assertion : Acetamide on reaction with KOH and bromine gives acetic acid
Reason : Bromine catalyses hydrolysis of acetamide.
(a) if both assertion 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) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(d) both assertion and reason are false

Question 14.
Which is the product formed when fructose undergoes partial reduction with sodium amalgam and water?
(a) Sorbital + mannitol
(b) D-mannose + D-galactose
(c) Gluconic acid + saccharic acid
(d) Aldehyde + ketone
Answer:
(a) Sorbital + mannitol

Question 15.
The polymer used in making blankets (artificial wool) is
(a) polystyrene
(b) PAN
(c) polyester
(d) polythene
Answer:
(b) PAN

Part – II

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

Question 16.
Predict the conditions under which
(а) Aluminium might be expected to reduce magnesia.
(b) Magnesium could reduce alumina.
Answer:
The conditions under which:
(a) Ellingham diagram is used to predict thermodynamic feasibility of reduction of oxides of one metal by another metal. Any metal can reduce the oxides of other metals that are located above it in the Ellingham diagram. In the Ellingham diagram, for the formation of magnesia (magnesium oxide) occupy lower position than aluminium oxide. Therefore aluminium cannot be used to reduce the oxides of magnesium (magnesia). Above 1623K, A1 can reduce MgO to Mg, so that ArG° becomes negative and the process becomes thermodynamically feasible.
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 6

(b)
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 7

At the point of intersection of the Al2O3 and MgO curves in
Ellingham diagram. ∆G0 becomes zero for the reaction.

Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 8

Question 17.
Give the uses of carbon dioxide.
Answer:

  • Carbon dioxide is used to produce an inert atomosphere for chemical processing.
  • Biologically, it is important for photosynthesis.
  • It is also used as fire extinguisher and as a propellent gas.
  • It is used in the production of carbonated beverages and in the production of foam.

Question 18.
What is the coordination entity formed when excess of liquid ammonia is added to an aqueous solution copper sulphate?
Answer:
When excess of liquid ammonia is added to an aqueous solution of copper sulphate to give tetraamminecopper (II) sulphate
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 9
Therefore, the coordination entity is [Cu(NH3)4]2+

Question 19.
Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the cube and Y is at the centre of the cube. What is the formula of the compound?
Answer:
Atoms X and Y form bcc crystalline structure. Atom X is present at the comers of the cube Atom Y is present at die centre of the cube.
No of atoms of X in the unit cell = \(\frac{N_{c}}{8}=\frac{8}{8}=1\)
No of atoms of Y in the’unit cell = \(\frac{N_{b}}{1}=\frac{1}{1}=1\)
Ratio of atoms X : Y = 1 : 1 .
Hence formula of the compound = XY.

Question 20.
Define zero order reaction. Give the unit for its rate constant(k).
Answer:
Zero Order Reaction. The reaction in which die rate of reaction is independent of the concentration of the reactants is called zero order reaction.
Rate = k [A]0
= k
Where k is the rate constant. Its unit is mol L-1 s-1

Question 21.
Distinguish between galvanic cell and electrolytic cell.
Answer:
Galvanic Cell:

  1. It is a device in which a spontaneous chemical reaction generates an electric current.
  2. It converts chemical energy into electrical energy. It is commonly known as Battery.
  3. e.g., Daniel cell, Dry cell.
  4. A salt bridge is used in this.

Electrolytic cell :

  1. It is a device in which an electric current from an external source drives a non spontaneous reaction
  2. It converts electrical energy into chemical energy.
  3. e.g., Electrolysis of molten NaCl.
  4. Na salt bridge is used.

Question 22.
When phenol is treated with propan-2-ol in the presence of HF, Friedel-Craft reaction takes place. Identify the products.
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 10

Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium

Question 23.
Human cannot use cellulose as food – Why?
Answer:
Human cannot use cellulose as food because our digestive systems do not contain the necessary enzymes such as glycosidases (or) cellulases that can hydrolyse the cellulose. But animals contain cellulose enzyme in their digestive system and they can digest cellulose. So cellulose can used as food for animals but not for human.

Question 24.
What are antibiotics?
Answer:
Antibiotics is a chemical substance produced by one microorganism, that selectively inhibits die growth of another micro organism. Example : penicillins and cephalosporins.

Part – III

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

Question 25.
Explain the types of silicones.
Answer:
(i) Linear silicones: They are obtained by the hydrolysis and subsequent condensation of dialkyl or diaryl silicon chlorides.

  • Silicone rubbers: These silicones are bridged together by methylene or similar groups.
  • Silicone resins: They are obtained by blending silicones with organic resins such as acrylic esters.

(ii) Cyclic silicones: These are obtained by the hydrolysis of R2SiCl2
(iii) Cross linked silicones: They are obtained by hydrolysis of RSiCl3

Question 26.
Give the properties of inter halogen compounds.
Answer:
Properties of inter halogen compounds:

  • The central atom will be the larger one.
  • It can be formed only between two halogen and not more than two halogens.
  • Fluorine can’t act as a central metal atom being the smallest one.
  • Due to high electronegativity with small size fluorine helps the central atom to attain high coordination number
  • They can undergo the auto ionization.
  • They are strong oxidizing agents.

Question 27.
Based on VB theory explain why [Cr(NH3 )6]3+ is paramagnetic, while [Ni(CN4)]2- is diamagnetic.
(a) [Cr(NH3 )6]3+
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 11
In this complex Cr is in the +3 oxidation state.
Electronic configuration of Cr atom
Electronic configuration of Cr+3 ion
Hybridisation and formation of [Cr(NH3)6]+3 Complex

  • Due to the presence of three unpaired electrons in [Cr(NH3 )6 ]3+ it behaves as a paramagnetic substance.
  • The spin magnetic moment,
    \(\mu s=\sqrt{3(3+2)}=\sqrt{15}=3.87 \mathrm{BM}\)
  • [Cr(NH3)6]3+ is an inner orbital octahedral complex.

(b) [Ni(CN)4]-2
In this complex Ni is in the +2 oxidation state.
Electronic configuration of Ni atom
Electronic configuration of Ni2+ ion
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 12
Hybridisation and formation of [Ni(CN)4]-2 Complex
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 13

Since CN is strong field ligand, hence the electrons in 3d orbitals are forced to pair up and there is no impaired electron in [Ni(CN)4]2-, hence it’should be diamagnetic substance.

Question 28.
A first order reaction is 20% completed in 10 minutes. Calculate the time taken for the reaction to go to 80% completion.
Answer:
Applying the first order equation,
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 14

Question 29.
Explain common ion effect with an example.
Answer:
Common Ion Effect: When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further.
Acetic acid is a weak acid. It is not completely dissociated in aqueous solution aid hence the following equilibrium exists.
CH3COOH (aq) ⇌ H+ (aq)+ CH3OO(aq)
However, the added salt, sodium acetate, completely dissociates to produce Na+ and CH3COO ion.
CH3COONa (aq) → Na+(aq) + CH3COO (aq) .

Hence, the overall concentration of CH3COO is increased, and the acid dissociation equilibrium is disturbed. We know from Le chatelier’s principle that when a stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, in order to maintain the equilibrium, the excess CH3COO ions combines with H+ ions to produce much more unionized CH3COOH i.e., the equilibrium will shift towards the left. In other words, the dissociation of CH3COOH is suppressed. Thus, the dissociation of a weak acid (CH3COOH) is suppressed in the presence of a salt (CH3COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

Question 30.
Mention the uses of Brownian movement.
Answer:

  • Brownian movement enables us to calculate Avogadro Number.
  • It is used to confirm kinetic theory which considers the ceaseles rapid movement of molecules that increases with increase in temperature.
  • It is used to understand the stability of colloids. As the particles are in continuous rapid movement, they do not come close and hence not get condensed. That is Brownian movement does not allow the particles to be acted on by force of gravity.

Question 31.
How will you convert benzaldehyde into the following compounds?
(i) benzophenone
(ii) benzoic acid
(iii)2 – hydroxyphenylaceticacid.
Answer:
(i) conversion of benzaldehyde into benzophenone.
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 15

(ii) conversion of benzaldehyde into benzoic acid:
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 16

(iii) conversion of benzaldehyde into 2 – hydroxy phenyl acetic acid:
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 17

Question 32.
Complete the following.
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 18
Answer:
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 19

Question 33.
Answer the following questions briefly:
(i) What are reducing sugars?
(ii) What is meant by denaturation of a protein?
(iii) How is oxygen replenished in our atmosphere?
Answer:
(i) Reducing sugar: All those Carbohydrates which reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars.

(ii) Denaturation of a protein: When 2° and 3° structure of a protein is destroyed due to the physical changes like temperature, change in pH, it is called denaturation of a protein. Example: Coagulation of egg white on boiling.

(iii) We take oxygen from atmosphere and release CO2. Plants take up CO2 and H2O from the atmosphere to prepare their food in the presence of sunlight and release O2, thus O2 is replenished in atmosphere.

Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium

Part – IV

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

Question 34.
(a) (i) What is Cementation? (2)
(ii) Write a notes on ionisation enthalpy in p-block elements? (3)
[OR]
(b) (i) Give the uses of sulphuric acid.. (2)
(ii) How alloys are formed in d-block elements? (3)
Answer:
(а) (i) Gold can be recovered by reacting the deoxygenated leached solution with zinc. In this process the gold is reduced to its elemental state (zero oxidation sate) and the process is
called cementation.
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 20

(ii) 1. As we move down a group, generally there is a steady decrease in ionisation enthalpy of elements due to increase in their atomic radius.

2. In p-block elements there are some minor deviations to this general trend. In group 13, from B to A1 the ionisation enthalpy decreases as expected. But from A1 to T1 there is only a marginal difference. This is due to the presence of inner d- and f – elements which has poor shielding effect compared to s andp electrons. As a result, the effective nuclear charge on the valance electrons increase.

3. A similar trend is also observed in group 14. The remaining groups (15-18) follows the general trend, in these groups the ionisation enthalpy decreases as we move down the group. Here poor shielding effect of d- and/electrons are overcome by the increased shielding effect of the additional p-electrons.

4. The ionisation enthalpy of elements in successive groups is higher than the corresponding elements of the previous group as expected.

[OR]

(b) (i)

  1. Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and
    super phosphates and other chemicals such as hydrochloric acid, nitric acid etc.
  2. It is used as a drying agent and also used in the preparation of pigments, explosives etc.

(ii)

  1. An alloy is formed by blending a metal with one or more other elements. The elements may be metals or non-metals or both.
  2. The bulk metal is named as solvent, mid the other elements in smaller portion is called solute.
  3. According to Hume-Rothery rule to form an alloy, the difference between the atomic radii ofthe solvent and solute is less than 15%. Both the solvent and solute must have the same crystal structure and valence and their electro negativity difference must be close to zero.
  4. Since their atomic sizes are similar and one metal atom can be easily replaced by another metal atom from its crystal lattice to form an alloy. The alloys are hard and
    have high melting points. Examples: Gold – copper alloy.

Question 35.
(a) (i) A solution of [CO(NH3) 4I2]Cl when treated with AgNO3, gives a white precipitate . What should be the formula of isomer of the dissolved complex that gives yellow precipitate with AgNO3 What are the above isomers called? (2)
(ii) Write the following in the complex [Cr (en)3 [Cr F6]
(i) Type of complex
(ii) Ligands
(iii) central metal
(rv) Oxidation state of central metal
(v) IUPAC name (3)
[OR]
(b) (i) KF crystallizes in fee structure like sodium chloride, calculate the distance between K+ and F in KF. (given : density of KF is 2.48 g cm-3) (3)
(ii) How do concentrations of the reactant influence the rate of reaction? (2)
Answer:
(a) (i)1. A solution of [CO(NH3)4I2]Cl when treated with AgNO3 gives a white precipitate, because Cl ion is counter ion.
2. Formula of isomer of the dissolved complex that gives yellow precipitate with AgNO3 is, [CO (NH3)4 Cl I] Iθ because Ie is counter ion
3. [CO(NH3)4I2]Cl and [CO(NH3)4 Cl I]I both are ionisation isomers.

(ii)
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 21
[OR]

(b) Density of KF = 2.48cm-3
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 22
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 23

(ii) The rate of a reaction increases with the increase in the concentration of the reactants. The effect of concentration is explained on the basis of collision theory of reaction rates. According to this theory, the rate of a reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.

Question 36.
(a) (i) Point out the differences between ionic product and solubility product. (2)
(ii) The solubility of AgCl in water at 298 K is 1.06 × 10-5 mole per litre. Calculate is solubility product at this temperature. (3)
[OR]
(b) (i) Describe about lithium-ion battery and its uses. (3)
(ii) Give reasons for the following (2)
(1) Rusting of iron is quicker in saline water than in ordinary water.
(2) Aluminium metal cannot be produced by the electrolysis of aqueous solution of aluminium salt.
(a) (i)

Ionic product

  1. It is applicable to all types of solutions.
  2. Its value changes with the change in con-centration of the ions.

Solubility product :

  1. It is applicable to the saturated solutions.
  2. It has a definite value for an electrolyte at a constant temperature.

(ii) The solubility equilibrium in the saturated solution is
AgCl (s) ⇌ Ag+(aq) + Cl (aq)
The solubility of AgCl is 1.06 × 10-5 mole per litre.
[Ag+(aq)] = 1.06 × 10-5molL-1
[Cl (aq)] = 1.06 × 10-5 mol L-1
Ksp = [Ag+(aq)] [Cl(aq)]
= (1.06 × 10-5 mol L-1) × (1.06 × 10-5 mol L-1)
= 1.12 × 10-2 mol2 L-2

[OR]

(b) (i) 1. Lithium-ion battery
Anode : Porous graphite
Cathode : Transition metal oxide as CoO2
Electrolyte : Lithium salt in an organic solvent

2. At the anode oxidation occurs
Li(s) —Li+(aq) + e
At the cathode reduction occurs.
Li+ + CoO2(s) + e → Li CoO2(s)

3. Overall reactions
Li(s) + CoO2 → Li CoO2(s)

4. Both electrodes allow Li+ ions to move in and out of their structures. During discharge the Li+ ions produced at the anode moves towards cathode through the non-aquaeous electrolyte.

5. When a potential greater than the emf produced by the cell is applied across the electrode, the cell reaction is reversed and now the Li+ ions move from cathode to anode where they become embedded on the porous electrode. This is known as intercalation.
6. Uses:- This Li-ion battery is used in cellular phones, Laptop computer and digital camera.

(ii)

  1. It is because in saline water, there are more H+ ions. Greater the number of H+ ions, quicker the rusting will take place.
  2. It is because aluminium metal is more reactive than hydrogen aid it will react with H2O.

Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium

Question 37.
(a) (i) Heat of adsorption is greater for chemisorptions than physisorption. Why? (2)
(ii) Give three examples for heterogeneous catalysis. (3)
[OR]
(b) (i) How is Aniline converted into Phenol? (2)
(ii) What is the action of HCN on
(1) propanone (2) 2,4-dichlorobenzaldehyde. (3)
Answer:
(a) (i) Chemisorption has higher heat of adsorption, because in chemisorption the chemical bonds are much stronger. In adsorbed state the adsorbate is hold on the surface of adsorbent by attractive forces (bond). And chemisorption is irreversible one. Therefore, heat of adsorption is greater for chemisorptions than physisorption.
Chemisorption, heat of adsorption range 40-400kJ/mole.

(ii)
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 24

[OR]

(b) (i) Aniline is diazotized with nitrous acid (NaN02 + HC1) at 273-278K to give benzene diazonium chloride which on further treatment with hot water in the presence of mineral acid gives phenol.
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 25

(ii) 1. Propanone reacts with HCN:
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 26

2. 2,4 – dichlorobenzaldehyde reacts with HCN:
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 27

Question 38.
(a) (i) An aromatic compound ‘A’ of molecular formula C7H7ON undergoes a series of reactions as shown below. Write the structures of A, B, C, D and E in the following reactions. (5)
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 28

[OR]

(b) (i) Write a note on co -polymer. (3)
(ii) What is the difference between elastomers and fibres? Give one example of each.(2)
Answer:
a(i)
Tamil Nadu 12th Chemistry Model Question Paper 1 English Medium - 29

[OR]

(b) (i)

  1. A polymer containing two or more different kinds of monomer units is called a copolymer.
  2. Co-polymers have properties quite different from the homopolymers.
  3. The structural units of co-polymers are derived from the different monomers may be present in regular, alternation or in random order or strings of several units of one kind may alternate with strings of another.
  4.  For example, Buna – S, Buna – N, Nylon – 6,6 etc.
    Buna – S contains styrene and butadiene monomer units.

Elastomers :

  1. These are rubber like solids with elastic properties.
  2. These are held by the weak inter- molecular forces.
    Example: Buna-S and Buna-N.

Fibres :

  1. These are the thread forming solids which possess high tensile strength and high modulus.
    These are held together by strong intermolecular forces like hydrogen bonding. Example: Nylon 6, 6 and
  2. polyesters (terylene)

Tamil Nadu 12th Tamil Model Question Paper 5

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

TN State Board 12th Tamil Model Question Paper 5

நேரம்: 2.30 மணி 
மதிப்பெண்கள் : 90

குறிப்புகள்:

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

பகுதி – I

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

(விடைகள் தடித்த எழுத்தில் தரப்பட்டுள்ளன)

Question 1.
சங்க இலக்கிய மொழியின் அடையாளப் பண்பினை தொகைநிலை என்று எச்சவியலில் பேசும் இலக்கண நூல் _________.
(அ) நன்னூல்
(ஆ) தொல்காப்பியம்
(இ) வீரசோழியம்
(ஈ) தொன்னூல் விளக்கம் கலைவடிவம்
Answer
(ஆ) தொல்காப்பியம்

Question 2.
பொருளும் _______ யும் இணைந்துதான் கலைவடிவம் பெறுகின்றன.
(அ) ஓசை
(ஆ) பாவகை
(இ) இலக்கணம்
(ஈ) எழுத்தும்
Answer:
(அ) ஓசை

Question 3.
உலகச் சுற்றுச் சூழல் தினம் __________ ஆகும்.
(அ) மே 5
(ஆ) ஜுன் 5
(இ) ஜுலை
(ஈ) ஆகஸ்டு 5
Answer:
(ஆ) ஜுன் 5

Tamil Nadu 12th Tamil Model Question Paper 5

Question 4.
செவிலித்தாய் என்பவர் _________ ஆவார்.
(அ) பெற்றதாய்
(ஆ) பெற்றதாயின் தாய்
(இ) வளர்ப்புத்தாய்
(ஈ) தந்தையின் தாய்
Answer:
(இ) வளர்ப்புத்தாய்

Question 5.
சம்பந்தர் தேவாரத்தைத் தொகுத்தவர் _________.
(அ) திருமலை நம்பி
(ஆ) நம்மாழ்வார்
(இ) நம்பியாண்டார் நம்பி
(ஈ) குறுங்குடி நம்பி
Answer:
(இ) நம்பியாண்டார் நம்பி

Question 6.
‘விளியறி ஞமலி’ என்பதில் கூறப்பட்ட விலங்கு எது?
(அ) நாய்
(ஆ) பூனை
(இ) மாடு
(ஈ) மான்
Answer:
(அ) நாய்

Question 7.
கோவில்பட்டி வட்டார எழுத்திற்குக் கரிசல் இலக்கியம் எனப்பெயர் சூட்டியவர் ________ ஆவார்.
(அ) முகமது மீரான்
(ஆ) மு. மேத்தா
(இ) வைரமுத்து
(ஈ) கி. ராஜநாராயணன்
Answer:
(ஈ) கி. ராஜநாராயணன்

Question 8.
வினை, பயன், மெய், உரு ஆகியவற்றின் அடிப்படையில் தோன்றுவது ___________ ஆகும்.
(அ) படிமம்
(ஆ) உவமம்
(இ) இறைச்சி
(ஈ) மரபு
Answer:
(அ) படிமம்

Question 9.
‘தி கிரேட் டிக்டேட்டர்’ திரைப்படத்தைத் தயாரித்தவர் __________ ஆவார்.
(அ) சார்லி
(ஆ) சார்லஸ்
(இ) சார்லி சாப்ளின்
(ஈ) சார்லஸ் டிக்கன்ஸ்
Answer:
(இ) சார்லி சாப்ளின்

Tamil Nadu 12th Tamil Model Question Paper 5

Question 10.
பொருத்துக.
அ) ஆமந்திரிகை – 1. பட்டத்து யானை
(ஆ) அரசு உவா – 2. மூங்கில்
(இ) கழஞ்சி – 3. இடக்கை வாத்தியம்
(ஈ) கழை – 4. எடை அளவு
(ஆ) 2, 1, 3, 4
(ஆ) 3, 1, 4, 2
(இ) 2, 4, 1, 3
(ஈ) 4, 3, 2,1
Answer:
(ஆ) 3, 1, 4, 2

Question 11.
தொல்காப்பியம் நூல் முழுமைக்கும் உரை எழுதியவர் __________ ஆவார்.
(அ) இளம்பூரணர்
(ஆ) நக்கீரர்
(இ) சாத்தனார்
(ஈ) பவணந்தியார்
Answer:
(அ) இளம்பூரணர்

Question 12.
ஆப்பிரிக்க ஆசியத் திரைப்பட விழாவில் சிவாஜிக்கு வழங்கப்பட்ட விருது __________ ஆகும்.
(அ) பத்மஸ்ரீ
(ஆ) பத்ம விபூலின்
(இ) செவாலியர்
(ஈ) கலைமாமணி
Answer:
(இ) செவாலியர்

Question 13.
‘காவ்யதரிசம்’ என்ற வடமொழி நூலைத் தழுவி இலக்கண நூலை இயற்றியவர் ___________.
(அ) பவணந்தி
(ஆ) அகத்தியர்
(இ) தொல்காப்பியர்
(ஈ) தண்டி
Answer:
(ஈ) தண்டி

Question 14.
Tamil Nadu 12th Tamil Model Question Paper 5 - 1
Answer:
(இ) கலிவெண்பா – தூது

பகுதி – II

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

பிரிவு – 1

எவையேனும் மூன்றனுக்கு விடை தருக.

Question 15.
அணியிலக்கணத்தையும் சேர்ந்து கூறும் இலக்கண நூல்கள் யாவை?
Answer:

  • தொல்காப்பியம்
  • வீரசோழியம்
  • இலக்கண விளக்கம்
  • தொன்னூல் விளக்கம்
  • முத்துவீரியம்

Question 16.
எதிர்பாராத நிகழ்வுகளை ஜலாலுத்தீன் ரூமி எவ்வாறு உருவகப்படுத்துகிறார்?
Answer:

  • வரக்கூடிய அனைத்துப் பிரச்சனைகளையும் வரவேற்று விருந்தோம்பல் செய்ய வேண்டும்.
  • துக்கங்கள் ஒரு கூட்டமாக வந்து துப்புரவாக வெறுமைப்படுத்தும் போதும், அவைகளையெல்லாம் நாம் எதிர்கொள்ளும் போது அவை உன்னைத் தூசிதட்டி, தயார்படுத்தி, உன்னைப் பலமிக்கவனாக மாற்றும்.
  • வக்கிரம், அவமானம், வஞ்சனை இவற்றையெல்லாம் வாயிலுக்கே சென்று இன்முகத்தோடு வரவேற்போமேயானால் எல்லாப் பிரச்சனைகளையும் எதிர்கொள்ளும் வலிமையும் நல்வழிகாட்டலும் கிடைக்கும்.

Tamil Nadu 12th Tamil Model Question Paper 5

Question 17.
தேவாரம் குறிப்பு வரைக.
Answer:

  • தேவாரம் = தே + வாரம் என்றும்
    தே + ஆரம் என்றும் பிரிக்கலாம்
  • தே + வாரம் என்றால் தெய்வத் தன்மையை உடைய இசைப் பாடல்கள்
  • தே + ஆரம் எனப்பிரித்தால் தெய்வத்திற்கு சூட்டப் பெற்ற பாமாலை என்று கூறுவர்.
  • சைவத் திருமுறைகள் பன்னிரண்டனுள் முதல் ஏழும் மூவர் தேவாரம் எனப்படும்.

Question 18.
புறநானூற்றில் குறிப்பிடப்பட்ட பல்வேறு பொருண்மைகளை எழுதுக.
Answer:

  • மூவேந்தர்
  • குறுநில மன்னர்
  • வேளிர்
  • மக்கள்
  • போர்ச்செய்திகள்
  • கையறு நிலை
  • நடுகல்

போன்றவற்றைப் பற்றி புறநானூறு குறிப்பிடுகிறது. கையைகாலாக

பிரிவு – 2

எவையேனும் இரண்டனுக்கு விடை தருக

Question 19.
கீழ்த்திசைச் சுவடிகள் குறித்து எழுதுக.
Answer:
காலின் மெக்கன்சியின் தொகுப்புகளை அடிப்படையாகக் கொண்டு 1869 இல் உருவாக்கப்பட்ட இந்நூலகம் அரிய ஓலைச்சுவடிகள், தாள் சுவடிகள், புத்தகங்கள் எனப் பெரும் தொகுப்புகளைக் கொண்டது.

Question 20.
‘ஆளுக்கொரு மரம் வளர்ப்போம்’ என்னும் முழக்கத்தொடர் வாயிலாக எவற்றை வலியுறுத்துவாய்?
Answer:

  • “வீட்டுக்கொரு மரம் வளர்ப்போம்” என்பது அன்றைய வாசகம், ஆளுக்கொரு மரம் வளர்க்க வேண்டிய தேவை இன்று ஏற்பட்டுள்ளது.
  • மரங்கள் இயற்கையின் கொடை இயற்கை அன்னையின் மடியில் மலர்ந்த முதல் குழந்தை தாவரம்.
  • பல வகையில், உதவும் மரங்களை நாம் வளர்க்க வேண்டும்.
  • நிறைய மரங்களை வளர்த்து காடுகளை உருவாக்க வேண்டும்.
  • மரங்கள் நம் நாட்டின் செல்வங்கள் என்பதனை உணர்ந்து இயற்கை வழியில் செல்வோம்! பசுமை பாரதத்தை உருவாக்குவோம்.

Tamil Nadu 12th Tamil Model Question Paper 5

Question 21.
கடலுக்கு வழங்கும் வேறு பெயர்களை எழுதுக.
Answer:
Tamil Nadu 12th Tamil Model Question Paper 5 - 2

பிரிவு – 3

எவையேனும் ஏழனுக்கு விடை தருக.

Question 22.
ஏதேனும் ஒன்றனுக்கு மட்டும் பகுபத உறுப்பிலக்கணம் தருக.
(அ) ஏகுவாய் (ஆ) ஈன்ற
Answer:
அ) ஏகுவாய் – ஏகு + வ் + ஆய்
ஏகு – பகுதி
வ் – எதிர்கால இடைநிலை
ஆய் – முன்னிலை ஒருமை வினைமுற்று விகுதி

(ஆ) ஈன்ற = ஈன் + ற் + அ
ஈன் – பகுதி
ற் – இறந்தகால இடைநிலை
அ – பெயரெச்ச விகுதி

Question 23.
ஏதேனும் ஒன்றனுக்கு மட்டும் இலக்கணக் குறிப்புத் தருக.
(அ) மாதவம்
(ஆ) நுந்தை
Answer:
(அ) மாதவம் = உரிச்சொற்றொடர்
(ஆ) நுந்தை = நும்தந்தை என்பதன் மரூஉ

Question 24.
மரபுப் பிழை நீக்குக.
சிங்கம் பிளிற யானை கர்ஜித்தது
Answer:
யானை பிளிற சிங்கம் முழங்கியது.

Question 25.
ஏதேனும் ஒன்றிற்குப் பிரித்துப் புணர்ச்சி விதி தருக.
(அ) செல்லிடத்து
(ஆ) வெண்மதி
Answer:
(அ) செல்லிடத்து = செல் + இடத்து
செல் + ல் + இடத்து = செல்லிடத்து

விதிகள் :

  1. தனிக் குறில் முன் ஒற்று உயிர்வரின் இரட்டும்.
  2. உடல்மேல் உயிர் வந்து ஒன்றுவது இயல்பே.

(ஆ) வெண்மதி = வெண்மை + மதி
வெண் + மதி = வெண்மதி
விதி : ஈறுபோதல்

Tamil Nadu 12th Tamil Model Question Paper 5

Question 26.
தனித்தமிழில் எழுதுக.
சீரியல் பார்க்காத லைப் போரடிக்குது என்றார் சீனியர் சிட்டிசன்.
Answer:
தொலைக்காட்சித் தொடர் பார்க்காத வாழ்க்கை வெறுப்பாக உள்ளது என்றார் மூத்த குடிமகன்.

Question 27.
விடைக்கேற்ற வினா எழுதுக.
நாங்கள் இந்த ஆண்டு சுற்றுலாவுக்கு புதுதில்லி சென்று வந்தோம்.
Answer:
நீங்கள் இந்த ஆண்டு சுற்றுலாவுக்கு எங்கே சென்று வந்தீர்கள்?

Question 28.
மயங்கொலிச் சொற்களின் பொருள் வெளிப்படுமாறு ஒரே தொடரில் விடையளி.
மாரி – மாறி
Answer:
மும்மாரிப் பொழிந்த தமிழகத்தில் இந்நாட்களில் மாரி மாறிப் பொழிகிறது.

Question 29.
தேவையற்ற இடங்களில் வல்லின மெய் நீக்கி எழுதுக.
வந்தச்சொல்லே மீண்டும் மீண்டும் வந்துப் பொருள் தந்தால் அதுவும் அதேப் பொருளைத் தந்தால் சொற்பொருள் பின்வரு நிலை அணியாம்.
Answer:
வந்த சொல்லே மீண்டும் மீண்டும் வந்து பொருள் தந்தால். அதுவும் அதே பொருளைத் தந்தால் அது சொற்பொருள் பின்வருநிலை அணியாம்.

Question 30.
LASER என்ற கலைச்சொல்லின் தமிழாக்கம் தருக.
Answer:
சீரொளி

பகுதி – III

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

பிரிவு – 1

எவையேனும் இரண்டனுக்கு மட்டும் விடை தருக

Question 31.
வரிவிதிப்பில் அரசன் எவ்வாறு செயல்பட வேண்டும் எனப் பிசிராந்தையார் விவரிக்கிறார்?
Answer:

  • ஒரு மாவிற்கும் குறைந்த நிலமாயினும் அதன்கண் விளைந்த நெல்லை அறுத்து உணவாக்கிக் கவளமாகக் கொடுத்தால் யானைக்குப் பல நாட்களுக்கு உணவாகும்.
  • நூறு மடங்கு பெரிய வயலாக இருந்தாலும் யானை தனித்துச் சென்று வயலில் புகுந்து உண்ணுமாயின் அதன் வாயில் புகுந்த நெல்லைவிட அதன் கால்களால் மிதிபட்டு அழிந்த நெல்லின் அளவு அதிகமாகும்.
  • அதுபோல அறிவுடைய அரசன், வரி திரட்டும் முறை அறிந்து மக்களிடமிருந்து வரி திரட்டினால் நாடு, கோடிக் கணக்கில் செல்வத்தைப் பெற்றுச் செழிப்படையும்.
  • அரசன் அறிவில் குறைந்தவனாகி, முறை அறியாத சுற்றத்தாரோடு ஆரவாரமாக, குடிமக்களின் அன்பு கெடுமாறு, நாள்தோறும் வரியைத் திரட்ட விரும்புவது. யானை தான் புகுந்த நிலத்தில் தானும் உண்ணாமல் பிறருக்கும் பயன்படாமல் வீணாக்குவது போன்றது.
  • அரசன் தானும் பயனடைய மாட்டான்; நாட்டு மக்களும் துன்புறுவர்.

Tamil Nadu 12th Tamil Model Question Paper 5

Question 32.
இராமலிங்க அடிகள் கந்தவேளிடம் எத்தகையோர் உறவு வேண்டுமெனக் கேட்கிறார்?
Answer:

  • அறம் செய்வார் நிறைந்திருக்கும் சென்னையின் கந்தகோட்டத்துத் திருக்கோவிலில் எழுந்தருளியிருக்கும் கந்தவேளே.
  • குளிர்ந்த முகத்தோற்றத்தையுடைய தூய மாணிக்க மணியே, அம்மணிகளுள் அருள்நிறைந்த சைவமணியே.
  • ஒருநெறிப்பட்ட மனத்துடன் நின்னுடைய மலர்போன்ற திருவடிகளை நினைக்கின்ற உத்தமர்தம் உறவு வேண்டும்.
  • உள்ளத்தில் ஒன்றும் புறத்தில் ஒன்றுமாகப் பேசும் வஞ்சகர்தம் உறவு என்னைப் பற்றாதவாறு காக்க வேண்டும்.
  • பெருமை சான்ற நினது புகழையே நான் பேச வேண்டும்.
  • பொய் பேசாதிருக்க வேண்டும்.
  • சிறந்த வாழ்வியல் நெறியைப் பின்பற்றுமாறு எனக்கு அருள வேண்டும்.
  • மதமான பேய் என்னை அணுகாதிருக்க வேண்டும்.
  • துறவுக்கு எதிரான பெண்ணாசையை என் மனம் மறக்க வேண்டும்.
  • என்றும் உன்னை மறவாதிருக்க வேண்டும். மதியும் நின் கருணையாகிய நிதியும் நோயற்ற வாழ்வும் உடையவனாக நான் இருக்க வேண்டும்.
  • ஆறுமுகங்களை உடைய தெய்வமாகிய மணியே, இத்தகைய சிறப்புகளை நீ எனக்கு அருள்வாயாக என்று கந்தவேளிடம் வள்ளலார் வேண்டுகிறார்.

Question 33.
அதியமான் அவையில் ஒளவையார் செய்த செயலைப் புறநானூறு பாடல்வழி நிறுவுக.
Answer:

  • தாம் எண்ணியதை முடிக்கும் வலிமையுடையவர்கள் புலவர்கள்.
  • அதே வேளையில் அவ்வள்ளல்கள் பற்றித் தாம் எழுதிய கவிதையின் சிறப்பை அறிந்து பரிசளிக்க வேண்டுமே என நினைந்து வருந்தும் தன்மையைக் கொண்டவர்கள்.
  • பரிசிலர்க்கு வாயிலை அடைக்காத காவலனே!விரைந்து ஓடும் குதிரையைக் கொண்ட நெடுமான் அஞ்சி, தன்னுடைய தகுதியை அறியானோ?
  • இவ்வுலகில், அறிவும் புகழும் உடையோர் இன்னும் மாய்ந்துவிடவில்லை.
  • இந்த உலகமும் வெற்றிடமாகிவிடவில்லை
  • ஆகவே, எம் யாழினை எடுத்துக்கொண்டோம். கருவிப்பையையும் சுருக்கிட்டுக் கட்டிக் கொண்டோம்.
  • மரம் வெட்டும் தச்சனின் தொழில் வல்ல பிள்ளைகள், கோடரியுடன் காட்டுக்குச் சென்றால் அவர்கள் வெட்டுவதற்கு ஏதாவது ஒரு மரம் கிடைக்காமலா போகும்? அதுபோல, கலைத்தொழில் வல்ல எங்களுக்கும் இவ்வுலகில் எத்திசையில் சென்றாலும் அத்திசையில் உணவு, தவறாமல்
    கிட்டும். இவையே புறநானூறு பாடலில் ஔவை கூறிய கருத்தாகும்.

Question 34.
இராமன் இலங்கை அரசை யாருக்கு உரிமையாக்கிக் கொடுத்தான்?
Answer:

  • இராவணன் சீதையை கவர்ந்து வந்தது தவறு என்று கூறி இலங்கையை விட்டுவந்தவன் வீடணன். இராமன் இருக்குமிடம் வந்து அடைக்கலம் வேண்டினான்.
  • இராமன் அவனை உடன் பிறந்தவனாக ஏற்று இலங்கை அரசை அவனுக்கே உரிமையாக்கினான்.
  • இலங்கை அரசாட்சி, ஏழேழாகிய பதினான்கு உலகங்களும் எனது பெயரும் இங்கு எவ்வளவு காலம் இருக்குமோ அவ்வளவு காலம் உனக்கே உரிமை எனக் கொடுத்தேன் என்று இராமன் கூறினான்.

பிரிவு – 2

எவையேனும் இரண்டனுக்கு மட்டும் விடை தருக.

Question 35.
தாய முறை (Matrilocal) என்றால் என்ன?
Answer:

  • சங்க காலத்தில் பெண்கள் திருமணம் செய்த பின்னரும் தன் இல்லத்திலேயே தொடர்ந்து வாழ்க்கை நடத்தும் முறையே “தாய முறை” எனப்படும்.
  • திருமணத்திற்குப் பின் மனைவியின் இல்லத்துக்குச் சென்று கணவன் வாழ்வதே நடைமுறையாக சங்க காலத்தில் இருந்துள்ளது.

Tamil Nadu 12th Tamil Model Question Paper 5

Question 36.
சென்னையின் பண்பாட்டு அடையாளங்களில், இன்றும் நிலைத்து இருப்பனவற்றைக் குறிப்பிடுக.
Answer:
சென்னையின் பண்பாட்டு அடையாளங்களில், இன்றும் நிலைத்து இருப்பன:

  • இன்று சென்னையின் புகழுக்குச் சான்றாக நிற்கும் பல்வேறு கல்வி நிறுவனங்கள் ஆங்கிலேயரால் ஏற்படுத்தப்பட்டவை.
  • 8 ஆம் நூற்றாண்டிலேயே சென்னையில் ஐரோப்பிய முறை கல்வி கற்பிக்கும் நிறுவனங்கள் தோன்றின.
  • 1715 இல் உருவான ‘புனித மேரி தேவாலய தர்மப் பள்ளி’ ஆசியாவில் உருவான முதல் ஐரோப்பியக் கல்வி முறையிலான பள்ளியாகும்.
  • 19 ஆம் நூற்றாண்டில் பள்ளிகள் பெருகின.
  • 1812 இல் உருவான சென்னை கோட்டைக் கல்லூரி.
  • 1837 இல் தொடங்கப்பட்ட கிறித்துவக் கல்லூரி.
  • 1840 இல் உருவான பிரசிடென்சி பள்ளி (பின்னாளில் மாநிலக் கல்லூரி) போன்ற பல்வேறு கல்வி நிறுவனங்கள் சென்னையில் கல்வி வளர்ச்சிக்குப் பெரும் பங்காற்றின.

Question 37.
வெ. இறையன்பு பற்றிக் குறிப்பு எழுதுக.
Answer:

  • வெ. இறையன்பு என்பவர் ‘இலக்கியத்தில் மேலாண்மை’ என்னும் நூலை எழுதியுள்ளார். தமிழ்நாடு அரசின் இந்திய ஆட்சிப்பணி அலுவலராகப் பணியாற்றி வரும் இவர், இ. ஆ. ப. தேர்வுக்குத் தமிழை ஒரு விருப்பப் பாடமாகப் படித்து வெற்றி பெற்றவர்.
  • 1990 ஆம் ஆண்டு முதல் பல்வேறு துறைகளில் பதவிகளை வகித்து வருபவர்.
  • தமிழ் இலக்கியப் பற்றுடைய இவர், தமிழில் வாய்க்கால் மீன்கள், ஐ. ஏ. எஸ். வெற்றிப் படிக்கட்டுகள், ஏழாவது அறிவு, உள்ளொளிப் பயணம், மூளைக்குள் சுற்றுலா உள்ளிட்ட பல நூல்களை எழுதியவர்.
  • பட்டிமன்றங்களில் நடுவராகப் பங்கேற்பதுடன் பல்வகைப்பட்ட ஊடகங்களில் பங்களிப்பைச் செய்து வருபவர்.
  • இவர் எழுதிய வாய்க்கால் மீன்கள்’ என்னும் கவிதை நூல், 1995 ஆம் ஆண்டில் தமிழ் வளர்ச்சித் துறையின் சிறந்த நூலுக்கான பரிசினைப் பெற்றது.
  • சிறுகதை, புதினம், தன் முன்னேற்ற நூல், நம்பிக்கை நூல் என இவர் படைப்புக்களம் விரிவானது.

Question 38.
சாப்ளின் உருவாக்கிய ‘தி கிரேட் டிக்டேட்டர்’ என்னும் கதையைப் பற்றி விளக்குக.
Answer:

  • கதையில் ஹிட்லரை உருவகப்படுத்தி ஹென்கோல் என்னும் கதைப்பாத்திரத்தை சாப்ளின் உருவாக்கினார். அதே உருவம் கொண்ட இன்னொரு பாத்திரத்தைக் கடை நடத்தி வரும் யூத இனத்தைச் சார்ந்தவராக அறிமுகப்படுத்தினார்.
  • சர்வாதிகாரி ஹென்கோல், யூதர்களைக் கைது செய்து சிறையில் அடைக்க உத்தரவிடுகிறார். காவலர்களும், யூதரான கடைக்காரரைக் கைதுசெய்து சிறைக்கு அழைத்துச் செல்கின்றனர்.
  • சிறையிலிருந்து தப்பிக்கும் கடைக்காரர், வழியில் ஹென்கோலின் உடையைத் திருடி அணிந்து கொள்கிறார்.
  • அப்போது வழியில் வரும் ஹென்கோலின் படைவீரர்கள் இவர்தான் ஹென்கோல் என தவறாக நினைத்து மரியாதை செய்கின்றனர்.
  • அதேசமயம் தப்பித்த கைதியைத் தேடி வந்த காவலர்கள் சாதாரண உடையில் வந்த ஹென்கோலை, தப்பித்த குற்றவாளி என நினைத்துக் கைது செய்கின்றனர். ஹென்கோல் சிறைக் கைதியாகிறார்.
  • ஒரே நாளில் இருவர் வாழ்க்கையும் தலைகீழாக மாறுகிறது.
  • அதன்பிறகு நடக்கும் காட்சிகள் எல்லாம் கடுமையான அரசியல் விமரிசனங்கள். இறுதிக்காட்சியில் சர்வாதிகாரி வேடத்தில் இருக்கும் யூதர் கைது செய்யப்பட்டிருந்த அனைவரையும் விடுதலை செய்யச் சொல்லி ஆணையிடுகிறார்.
  • மாநாட்டில் மனிதகுல விடுதலை குறித்துப் பேருரை ஆற்றுகிறார்.
  • அந்தப் பேருரைதான் இன்றுவரை திரைப்படங்களின் மிகச் சிறந்த வசனமாகப் போற்றப்படுகிறது.
  • வாழும் காலத்திலேயே ஹிட்லரைக் கடுமையாக விமர்சித்து எடுத்த ஒரே படம் என்ற பெருமையும் இப்படத்திற்கு உண்டு.
  • அதுபோல இரட்டை வேடப்படங்கள் எவ்வளவோ வந்தாலும் அதில் மிகச் சிறந்த திரைப்படம் என்ற பெயரும் இதற்கு உண்டு.

பிரிவு – 3

எவையேனும் மூன்றனுக்கு விடை தருக.

Question 39.
மருதத் திணை அல்லது பாடாண் திணையை விளக்கி எழுதுக.
Answer:
அகத்திணைகள் ஐந்து, அவற்றுள் ஒன்று மருதத்திணை.
முதற்பொருள்
நிலம் – வயலும் வயல் சார்ந்த இடமும்
பொழுது – சிறுபொழுது – காலை
பெரும்பொழுது – ஆறு பெரும்பொழுதுகள்

கருப்பொருள் :
தெய்வம் – வேந்தன்
மக்கள் – உழவர், உழத்தியர்
மரம் – மருதம், காஞ்சி
பறவை – நாரை, அன்னம்
உணவு – செந்நெல், வெண்ணெய்
தொழில் – நெல்லரிதல், களை பறித்தல்

உரிப்பொருள்: ஊடலும் ஊடல் நிமித்தமும்.

(அல்ல து)

பாடாண் திணையைச் சான்றுடன் விளக்குக.
Answer:
திணை விளக்கம்:
ஒரு மன்னனது புகழ், வலிமை, வள்ளன்மை அருள் முதலியவற்றை ஆய்ந்து கூறுவது பாடாண்திணையாகும். பாடப்படுகின்ற ஆண்மகனது ஒழுகலாற்றைக் கூறுவது இத்திணையின் நோக்கமாகும்.

(சான்று) அறுகுளத் துகுத்து மகல்வயிற் பொழிந்தும்
உறுமிடத் துதவா துவர்நில மூட்டியும் – பரணர்

திணைப் பொருத்தம் :
மழையானது நீர் வற்றிய குளத்திலும், அகன்ற விளை நிலத்திலும் பெய்யும். அவ்வாறு பெய்யத்தக்க குளத்திலும் விளைநிலத்திலும் பெய்யாது. பயன்படாத களர் நிலத்திலும் மிகுதியாய் பெய்யும் முறைமையை உடையது. அதுபோல மத மிகுந்த யானையினை உடைய காலில் வீரக் கழல் அணிந்த பேகன் இரவலர்க்கு வரையறையின்றிக் கொடுத்தலில் அறியாமை உள்ளவனாய் இருப்பான். ஆனால் பிறர் படை வந்து கலந்து போரிட்டால் அப்படையினரிடம் தான் அறியாமை இல்லாதவனாய் இருப்பான். பேகனின் படை, கொடை இப்பாடலில் கூறியமையால் பாடாண் திணையாயிற்று.

Tamil Nadu 12th Tamil Model Question Paper 5

Question 40.
உவமையணி அல்லது மடக்கணியை விளக்கி எழுதுக.
Answer:
அணி விளக்கம்:
இரண்டு பொருள்களுக்கு ஒப்புமை தோன்றச் சொல்வது உவமையணி ‘ யாகும். இது தாயணி’ என்று கூறப்படும். ஏனெனில் இந்த அணியிலிருந்தே மற்ற அணிகள் தோன்றின.

(எ.கா.) இழுக்கல் உடையுழி ஊற்றுக்கோல் அற்றே
ஒழுக்கம் உடையார்வாய்ச் சொல்

விளக்கம்
சேற்று நிலத்தில் வழுக்கி விழாமல் இருக்க ஊன்றுகோல் எங்ஙனம் உதவுகிறதோ, அதுபோல நாம் துன்பப்படும் வேளையில் பெரியோர்களின் வாய்ச்சொற்கள் ஊன்றுகோலாக இருந்து நம்மைக் காப்பாற்றும்.

(அல்லது)

மடக்கணி:
Answer:
அணி விளக்கம்:
ஒரு சொற்றொடர் மீண்டும் வந்து வேறு பொருள் தருவது ‘மடக்கணி’ எனப்படும்.
(எ.கா.) ‘அரிவையம்பாகத்தான்’

விளக்கம் :
இச்சொல், அரி + வை + அம்பு + ஆக = திருமால்
கூர்மையான அம்பாக விளங்க, என்னும் பொருளைத் தருகிறது. இதே சொல்லை,
அரிவை + அம் + பாகத்தான் எனப் பிரித்தால், பார்வதி தேவியை அழகிய ஒரு பாகத்தில் உடையவன் சிவபெருமான், என வேறு பொருள் தருகிறது. ஒரே சொற்றொடர் இரு வேறு பொருள் தருகிறது. எனவே, இது ‘மடக்கணி’ எனப்படும்.

Question 41.
உப்பிட்டவரை உள்ளளவும் நினை என்னும் பழமொழியை வாழ்க்கை நிகழ்வுகளோடு தொடர்புபடுத்தி எழுதுக.
Answer:
பழமொழி விளக்கம்: நமக்கு உதவி செய்தவரை நம் உயிர் உள்ளவரை நாம் மறக்கக் கூடாது.

வாழ்க்கை நிகழ்வு:
கல்லூரியில் பயிலும் வேலனும் கந்தனும் சிறு வயதிலிருந்தே நெருங்கிய நண்பர்கள். வேலன் ஏழைக் குடும்பத்தைச் சேர்ந்தவன். கந்தன் தன் நண்பனுக்குத் தன்னால் முடிந்த உதவிகளைச் செய்து வந்தான். ஒருமுறை, கல்லூரியில் தேர்வு தினம் அறிவிக்கப்பட்டது. அனைவரும் அவர்களின் தேர்வுக் கட்டணத்தைச் செலுத்திவிட்டனர். வேலனோ எவ்வளவு முயன்றும் அவனால் தன் கட்டணத்தைச் செலுத்த முடியவில்லை.

பணம் செலுத்தாவிட்டால் தேர்வு எழுத முடியாது என்னும் இக்கட்டான நிலையில் கந்தன் அவனது கட்டணத்தைக் கட்டி உதவினான். இவ்வாறு பல உதவிகளைக் கந்தன் செய்தான். வேலன், ‘இந்த உதவிகளுக்கு உனக்கு நான் என்ன கைம்மாறு செய்யப் போகிறேனோ?’ என்று அடிக்கடி கூறுவான். ஒருநாள் கல்லூரிக்கு வரும்பாதையில் கந்தனுக்கு விபத்து ஏற்பட்டது. அவன் அவசர பிரிவில் சேர்க்கப்பட்டதை அறிந்த வேலன் மருத்துவமனைக்கு விரைந்தான்.

தன் நண்பனின் நிலையைக் கண்டு கண் கலங்கி நின்றான் வேலன். மருத்துவர் அவனிடம் கந்தனுக்கு இரத்தப் பிரிவும் அவனுடைய இரத்தப் பிரிவும் ஒன்று என்பதால் வேலன் தன் உதிரத்தைக் கொடுத்து தன் நண்பனைக் காப்பாற்றினான்.

Tamil Nadu 12th Tamil Model Question Paper 5

Question 42.
தமிழாக்கம் தருக.
1. A guilty conscience needs to accuser.
2. Slow and steady win the race.
3. Do not add fuel to the flame.
4. Honesty is the best policy.
Answer:
1. குற்றமுள்ள நெஞ்சம் குறுகுறுக்கும்.
2. நிதானம் பிரதானம்.
3. எரிகிற நெருப்பில் எண்ணெய் ஊற்றாதே.
4. நேர்மையே சிறந்த கொள்கை.

Question 43.
பின்வரும் தலைப்புகளில் ஏதேனும் ஒன்றில் 8 வரிகளில் கவிதை புனைக.
தூய்மை இந்தியா (அல்லது) புதிய பாரதம்
Answer:
Tamil Nadu 12th Tamil Model Question Paper 5 - 3

(அல்லது)

புதிய பாரதம்
Answer:
புதிய பாரதம் படைத்திடுவோம்
புதிது புதிதாய்ப் படைத்திடுவோம்
பகையில்லா நட்புடனே பழகிடுவோம்
சாதிமத பேதங்கள் தவிர்த்திடுவோம்
சகலருக்கும் நீதியும் சமம்செய்வோம்
சரிநிகராய்ப் பெண்களுக்கு இடமளிப்போம்
சாலையின் விதிகளைக் கடைப்பிடிப்போம்
சத்தமின்றி சட்டமின்றி புதுமையெல்லாம்
சமத்துவமாய் புதிய பாரதம் செதுக்கிடுவோம்

பகுதி – IV

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

Question 44.
(அ) சினத்தைக் காத்தல் வாழ்வை மேன்மைப்படுத்தும் – இக்கூற்றை முப்பால் வழி
விரித்துரைக்க.
Answer:
சினம் கொண்டால் தீமையான விளைவுகள் உண்டாகும். தன்னையும் தன்னை சார்ந்து உள்ளவர்களையும் ஒட்டுமொத்தமாக அழித்து திரும்ப பெற முடியாத நிலையினை சினம் உண்டாக்கிவிடும். எனவே சினத்தை எப்பொழுதும் கைவிட வேண்டும். சினத்தைக் காத்தால் நம் வாழ்வு மேம்படும் என்பது பற்றி காண்போம்.

செல்இடத்துக் காப்பான் சினம்காப்பான் அல்இடத்துக்
காக்கின்என் காவாக்கால் என்?

நம்முடைய சினத்தை நம்மைவிட வலியவரிடத்தில் காட்டினால் என்ன நடக்கும். பாதிப்பு நமக்கே அதிகமாக இருக்கும். எனவே சினத்தை மெலியவரிடத்திலும், வலியவரிடத்திலும் காத்து கொள்ள வேண்டும்.

மறத்தல் வெகுளியை யார்மாட்டும்; தீய
பிறத்தல் அதனான் வரும்.

தீமையான விளைவுகள் அனைத்தும் சினத்தாலேயே ஏற்படும். எனவே யாரிடத்திலும் சினம் கொள்ள நினைக்கக் கூடாது.

நகையும் உவகையும் கொல்லும் சினத்தின்
பகையும் உளவோ பிற?

நம் முகத்தில் உள்ள மகிழ்ச்சியையும், அகத்தில் உள்ள மகிழ்ச்சியையும் அழிப்பது நம் சினமே. எனவே அதைக் கைவிட வேண்டும்.

தன்னைத்தான் காக்கின் சினம்காக்க; காவாக்கால்
தன்னையே கொல்லும் சினம்.

ஒருவர் தன்னைக் காத்துக்கொள்ள விரும்பினால், சினம் வராமல் காத்துக் கொள்ள வேண்டும். அப்படிக் காக்காவிடில் சினமே நம்மை அழித்துவிடும்.

சினம் என்னும் சேர்ந்தாரைக் கொல்லி இனம் என்னும்
ஏமப் புணையைச் சுடும்.

சினம் தன்னையும் அழிக்கும், தன்னைப் பாதுகாக்கும் நம் உறவுகளையும் சுட்டழித்துவிடும். எனவே சினத்தை முழுமையாகக் கைவிட்டு நம் வாழ்வை மேம்படுத்திக்கொள்ள வேண்டும்.

(அல்லது)

Tamil Nadu 12th Tamil Model Question Paper 5

Question 44.
(ஆ) எச். ஏ. கிருட்டிணனார் ‘கிறித்தவக் கம்பரே’ என்பதை நும் பாடப்பகுதி வழி நிறுவுக.
Answer:

  • கிறித்துவக் கம்பர் என்ற ஆல்பிரட் கிருஷ்ணபிள்ளை (H.A. கிருஷ்ணபிள்ளை ) ஏப்ரல் 23, 1827 ஆண்டு திருநெல்வேலி மாவட்டம் கரையிருப்பில் ரெட்டியார் பட்டி என்னும் சிற்றூரில் பிறந்தவர்.
  • இவரது பெற்றோர் சங்கர நாராயண பிள்ளை, தெய்வ நாயகியம்மை. ஹென்றி ஆல்பிரடு என்ற
    பெயர்களின் சுருக்கமே H.A. ஆகும்.
  • தமிழ் ஆசிரியராகப் பணியாற்றிய இவரது படைப்புகள் போற்றித் திருவருகல், இரட்சணிய யாத்திரிகம், இரட்சணிய மனோகரம் ஆகியவை இவர் எழுதியதாகச் சொல்லப்படும் இரட்சணிய குரல், இரட்சணிய பாலா போதனை என்ற நூல்கள் தற்பொழுது கிடைக்கவில்லை.
  • இரட்சணிய மனோகரத்தின் பெரும் பகுதி இரட்சணிய யாத்திரிகத்திலிருந்து எடுத்துத் தொகுக்கப்பட்டது ஆகும்.
  • H.A.கிருஷ்ணபிள்ளை தென்தமிழ் நாடாகிய நெல்லை நாடு, தற்போதைய (திருநெல்வேலி மாவட்டம்) பல மேலை நாட்டு அறிஞர்களை தமிழ்த் தொண்டராக்கிய பெருமை இவரைக் சேரும்
  • இத்தாலிய தேசத்து வித்தகராகிய வீரமாமுனிவரது தமிழ்ப் புலமைக்கு அடிகோலியது நெல்லை நாடும், இவரும் தான். பெருந்தமிழ் தொண்டராகிய போப்பையருக்குத் தமிழ் அறிவு ஊட்டியது நெல்லை நாடு தான்.
  • மொழி நூற்புலமையில் சிறந்து விளங்கிய கால்டுவெல் ஐயர் வாழ்ந்ததும் நெல்லை நாடு தான். இவ்வாறு பிற நாட்டு அறிஞரைப் தமிழ்ப் பணியில் ஈடுபடுத்திய தென்தமிழ் நாட்டில் ரெட்டியார்பட்டி என்ற சிற்றூர் உள்ளது.
  • அந்த ஊரில் வேளாளர் குலத்தில் வைணவ மதத்தில் பிறந்தவர் தான் H.A. கிருஷ்ணபிள்ளை . இளமையிலே தமிழில் உள்ள நீதி நூல்களையும், சமய நூல்களையும் அக்கால முறைப்படி நன்கு கற்றார்.
  • அப்போது நெல்லை நாட்டிலே கிறிஸ்துவ சங்கங்கள் கிளர்ந்து எழுந்தன. சிறந்த சமயத் தொண்டும் செய்யப்பட்டது. சர்ச்சு முறைச் சங்கத்தில் சிறப்பாக சார்சந்தர் என்னும் சிலர் சிறந்த பணி செய்தனர்.

அதன் காரணமாக கிறித்துவனமதத்தின் மீது H.A. கிருட்டிணபிள்ளைக்கு அதிக ஈடுபாடு வந்தது. அந்த மதம் தொடர்பான பல நூல்கள் இவரால் எழுதப்பட்டது.
போற்றித் திருஅகவல்

  • இரட்சணிய யாத்திரிகம்
  • இரட்சணிய மனோகரம்
  • இரட்சணிய குரல்
  • இரட்சணிய பாலா

போன்ற கிறித்துவ தொடர்பான பல படைப்புகள் இவரால் எழுதப்பட்டது. இதன் காரணமாகவே கிறித்துவக் கம்பர் என்று அழைக்கப்படுகிறார்.

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

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

தொடக்ககால ஆய்வுகள்:

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

வரலாற்று ஆய்வு:

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

கலையியல் ஆய்வு:

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

கல்வெட்டு ஆய்வுகள்:

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

பன்மொழிப் புலமை:

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

ஆராய்ச்சிப் பேரறிஞர்:

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

முடிவுரை:

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

(அல்லது)

Question 45.
(ஆ) குடும்பம் என்னும் சிறிய அமைப்பிலிருந்தே மனித சமூகம் என்னும் பரந்த அமைப்பு கட்டமைக்கப்படுகிறது – எவ்வாறு? விளக்குக.
Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

Tamil Nadu 12th Tamil Model Question Paper 5

Question 46.
(அ) தோப்பில் முகமது மீரான் எழுதிய தலைக்குளம் கதையை கருப்பொருள் மாறாமல் எழுதுக.
Answer:
சிங்காரச் சென்னையில் மக்கள் நெருக்கம் மிகுந்த பகுதியில் தன் பிள்ளைகளோடு, வேண்டா வெறுப்பாக ஒரு பெரியவர் வாழ்ந்து வந்தார்.

ஊரில் பழைய ஒரு ஓட்டு வீடும், சுத்தமான தண்ணீர் கிடைக்கும் வறண்டே போகாத கிணறும் சும்மா கிடக்குடா, அங்கே போய்த் தங்குவோம்? என பெரியவர் கேட்கும் போது – அவரது பிள்ளைகள் அதைக் காதில் வாங்கிக் கொள்வதுமில்லை. அவரது எண்ணங்களுக்கு மதிப்பளிப்பதும் இல்லை.

மாறாக. சொந்த ஊர்ல யாரையும் எங்களுக்குத் தெரியாது கிணற்று நீரைவிட மினரல் வாட்டர்தான் உயர்வானது என்றும் பிள்ளைகள் வாப்பாவிடம் பதிலுக்குக் கூறியதுண்டு.

பெரியவர் ‘நான் பிறந்த மண்ணுடா, என்னோட அப்பன் வச்ச வீடுடா, அது அங்கே சும்மா கிடக்குடா.. அந்தத் தண்ணீர் குடிச்சித்தான் நானே வளர்ந்திருக்கேன் என்றார். அத வித்திடுங்க வாப்பா, இந்த நாகரீக காலத்துல ஓடு போட்ட மர வீட்ல யாரு தங்குவா? என பிள்ளைகள் ஏளனம் செய்வதைக் கேட்கக் பரிதாபமாக இருந்தது.

பெரியவர் தன் வீட்டைப் பற்றியும், தலைக்குளம் கிராமத்தைப் பற்றியும் எவ்வளவு உயர்வாகப் பேசினாலும், அவரது பிள்ளைகள் ஏற்றுக் கொள்ளாவிட்டாலும் பரவாயில்லை. அவர்கள் பெரியவரின் பேச்சைக் காதால் கேட்கக்கூடத் தயாராக இல்லை.

தலைக்குளத்தின் கதைகள், கன்னியார் கோண ஏலாவைப் பற்றிய கதைகள், ஞானியார் நீரோடை, புன்னமூட்டடி இதைப் பற்றியெல்லாம் பெரியவர் சொன்னாலும் பிள்ளைகள் அறிய விரும்புவதில்லை. வேகாத வெயிலில் மொட்டை மாடியில் காற்றுவாங்கும் நகரத்து நரக வேதனையை அனுபவிக்கும்போது, பெரியவர் தனது கிராமத்தின் வயல்வெளியில் வீசும் தலைக்குளத்துக் குளிர்காற்றை அனுபவித்ததை எண்ணிப் பார்க்க மறக்கவில்லை.

கிராமத்து இயற்கையை ரசிக்கத் தெரியாத தலைமுறையாக நமது பிள்ளைகள் ஆகிவிட்டனரே என்று பெரியவர் கவலை கொண்டார். ஒரு நாள் மூத்த மகன் பெரியவரிடம், வாப்பா நம்ம தலைக்குளத்து ஓட்டு வீடு இனி வேண்டாம். நல்ல விலைக்கு வித்திடுங்க வாப்பா. இனிமே நாங்க யாரும் அங்கே போய் இருக்கப்போறது இல்லை. உன்னையும் அங்கே தங்கவிடப் போறதில்லை. அதனை பேப்பர்ல விளம்பரம் போடுங்க… வீட்டை நல்ல விலைக்கு வித்திடுங்க.

ஓடு வேய்ந்த வீடு… நாலு கட்டு வீடு – ஈட்டி மரம் – எந்த வறட்சியிலும் வற்றாத கிணறு, ருசிமிகுந்த தண்ணீர் – பக்கத்தில் எப்போதும் தண்ணீர் ஓடும் வாய்க்கால் – வீடு விற்பனைக்கு – அணுகவும் என்று மூத்த மகன் பத்திரிகையில் விளம்பரம் செய்திருந்தான்.

விளம்பரத்தைக் கண்டு ஓய்வு பெற்ற நீதிபதி ஒருவர் வீட்டை வாங்கும் எண்ணத்தில் பெரியவரை அணுகியிருந்தார். ஓய்வுக்காலத்தைக் கிராமத்தில் கழிக்க எண்ணியிருந்த நீதிபதியை அழைத்துக் கொண்டு தலைக்குளம் பற்றியும், புன்னமூட்டி குளிர்ச்சி, புன்னப்பூ வாசம், தாழை வாசம் என பயணத்தின் போது பெரியவர் சொல்லச் சொல்ல; நீதிபதி வீட்டை உடனே விலைபேசி வாங்கும் நிலைக்கு வந்துவிட்டிருந்தார்.

பெரியவர், முன்னாள் நீதிபதியை அழைத்துச் சென்றார். நீண்ட நாட்களாக பூட்டப்பட்டிருந்த தனது வீட்டைத் திறந்துகாட்டினார். அடைபட்ட வீட்டிலிருந்து வௌவால்கள் சத்தமிட்டபடி அங்குமிங்கும் பறந்தன. வீட்டினுள் பழமைநெடி மூக்கைத் துளைத்தது. மூக்கைப் பொத்தியபடியே நீதிபதி வீட்டைப்பார்த்தார். அவருக்கு வீட்டின் அமைப்புமுறை மிகவும் பிடித்திருந்தது. பெரியவர் பின்பக்க கதவைத் திறந்தார். இருவரும் கிணற்றுப் பக்கம் சென்றார். நீதிபதிதான் முதலில் கிணற்றை எட்டிப் பார்த்தார். கிணற்றுக்குள் வறண்ட பூமியின் வெடிப்பைக் கண்டார். உடனே மனம் மாறினார். வீடு வேண்டாமென்று கூறிவிட்டுத் தனது காரில் கிளம்பத் தயாரானார். உடன் வருமாறு பெரியவரை அழைத்தும் வராததால் நீதிபதி சென்றுவிட்டார்.

பெரியவர் தனது ஊரின் மீதுள்ள ஆசை மிகுதியால் தலைக்குளத்துத் தண்ணீரில் மூழ்க ஆசையோடு தலைக்குளம் நோக்கி நடந்தார். அங்கே குளம் இருந்த சுவடே மறைந்து போயிருந்தது. குளம் இருந்த இடத்திலும், கன்னியார்கோணம் ஏலா, அரசகுளம் ஆகிய எல்லா இடங்களிலும் புதிய புதிய குடியிருப்புகள்.

பெரியவருக்குத் தலைசுற்றியது. நீர்ப்பிடிப்புப் பகுதியெல்லாம் குடியிருப்புகள் ஆனதினால் நிலத்தடி நீர் மட்டம் குறைந்து போகுமே என ஏக்கப் பெருமூச்சு விட்டபடி, தன்மகன் வாழும் நகரத்தை நோக்கிப் பயணப்பட்டார் பெரியவர்.

(அல்லது)

Question 46.
(ஆ) சங்ககால வரலாற்றை அறிந்து கொள்ள, புகளூர்க் கல்வெட்டு எவ்வகையில் துணைபுரிகிறது – விளக்குக.
Answer:
புகளூர் கரூர் மாவட்டத்தில் அமைந்துள்ளது. இந்த ஊரில் ஆறுநாட்டான் என்று ஒரு மலை உள்ளது. மலையடி வாரத்து ஊரை வேலாயுதம்பாளையம் என்பர். இம்மலைப்பகுதியில் இக்கல்வெட்டுகள் காணப்படுகின்றன. மொத்தம் 12 கல்வெட்டுகள் காணப்படுகின்றன. இவை அனைத்தும் சங்ககாலத் தமிழ் எழுத்தில் எழுதப் பெற்றுள்ள மொழி தமிழாகும். இவற்றுள் இரு கல்வெட்டுகள் சேர மன்னர்கள் வழங்கிய கொடை பற்றிக் கூறுகின்றன. எனவே, இக்கல்வெட்டுகள் சங்க வரலாற்றைத் தெரிந்து கொள்ளப் பெரிதும் துணைபுரிகின்றன.

கல்வெட்டு -1
மூதா அமண்ணன் யாற்றூர்
செங்காயபன் உறைய்
ஆதன் செல்லிரும் பொறை மகன்
பெருங்கடுங்கோன் மகன் (இளங்
கடுங்கோ (இளங்கோ ஆக அறுத்த கல்

யாற்றூரைச் சேர்ந்த செங்காயபன் என்னும் துறவிக்குச் சேர மன்னர் செல்லிரும் பொறை மகனான பெருங்கடுங்கோவின் மகன் இளங்கடுங்கோ இளவரசர் ஆவதை முன்னிட்டு வழங்கப்பட்ட கொடை. சேர அரசின் மூன்று தலைமுறை இதன் மூலம் தெரிய வந்துள்ளது. இதிலுள்ள அரசர்கள் பதிற்றுப்பத்தில் 7, 8, 9 ஆம் பத்திற்குரிய தலைவர்களாக அடையாளப்படுத்தப் பெற்றுள்ளனர்.

கல்வெட்டு -2, 3
யாற்றூர் செங்காயப்பன்
(தா)வன் பின்னம் கொற்றன்
அறுபித்த அதிட்டானம்

யாற்றூரில் செங்காயபனுக்குத் தாவன் பின்னன் கொற்றன் என்பவர் அதிட்டானம் கொடுத்தது பற்றிக் கூறுகிறது. அதிட்டானம் என்றால் தரைப்பகுதி என்று பொருள்.

கல்வெட்டு – 4
நலிய்) ஊர் பிடன் குறும்மகள்கீரள்
கொற்றி செய்பிதபளி

நலி ஊரை சேர்ந்த பிடனுடைய இள மகளான கீரன் கொற்றி செய்பித்த பாளியைப் பற்றி கூறுகிறது.

கல்வெட்டு – 5
நலி(ய்) ஊர் பிடந்தை மகள் கீரன்
கொறி அதிட்டான்

நலி ஊரை சேர்ந்த பிடந்தையின் மகளான கீரன் கொற்றி கொடுத்த படுக்கை. மேற்கண்டவாறு மொத்தம் 10 கல்வெட்டுகள் கிடைத்துள்ளன. அவை அனைத்தும் தமிழின் தமிழரின் பெருமையையும் அரசனின் பெருமையையும் எடுத்தியம்புகின்றன.

சிறப்புகள் :

  • சங்ககால சேர அரசர்களின் கல்வெட்டு.
  • கோ ஆதன் செல்லிரும் பொறை பெருங்கடுங்கோ, இளங்கடுங்கோ ஆகிய அரசர்களின் பெயர்கள் குறிப்பிடப் பெற்றுள்ளன.
  • சேர அரசின் மூன்று தலைமுறை இதன் மூலம் தெரியவந்துள்ளது.
  • பதிற்றுப்பத்தில் 7, 8, 9 ஆம் பத்திற்குரிய தலைவர்களாக அடையாளப்படுத்த பெற்றுள்ளனர்.

பகுதி – V

அடிமாறாமல் செய்யுள் வடிவில் எழுதுக. [1 × 4 = 4]

Question 47.
(அ) ‘அறிவும்’ என்று தொடங்கும் புறநானூற்றுப் பாடல்.
Answer:
அறிவும் புகழும் உடையோர் மாய்ந்தென,
வறுந்தலை உலகமும் அன்றே; அதனால்
காவினெம் கலனே; சுருக்கினெம் கலப்பை;
மரங்கொல் தச்சன் கைவல் சிறாஅர்
மழுவுடைக் காட்டகத்து அற்றே
எத்திசைச் செலினும், அத்திசைச் சோறே. – ஒளவையார்

(ஆ) ‘வையத்துள்’ என்று துவங்கும் திருக்குறளை எழுதுக. [1x 2 = 2]
வையத்துள் வாழ்வாங்கு வாழ்பவன் வான்உறையும் – தெய்வத்துள் வைக்கப் படும்.
– திருவள்ளுவர்

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 6 Random Variable and Mathematical Expectation Ex 6.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 1.
Value which is obtained by multiplying possible values of random variable with probability of occurrence and is equal to weighted average is called
(a) Discrete value
(b) Weighted value
(c) Expected value
(d) Cumulative value
Solution:
(c) Expected value

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 2.
Demand of products per day for three days are 21, 19, 22 units and their respective probabilities are 0.29, 0.40, 0.35. Profit per unit is 0.50 paisa then expected profits for three days are
(a) 21, 19, 22
(b) 21.5, 19.5, 22.5
(c) 0.29, 0.40, 0.35
(d) 3.045, 3.8, 3.85
Solution:
(d) 3.045, 3.8, 3,85
Hint:

x 21 19 22
P(x) 0.29 0.40 0.35

E(X) = \(\sum_{ x }\) xP(x)
For Day 1
E(X) = 21 × 0.29
= 6.09
Expected profit = 6.09 × 0,50 = 3.054
For Day 2
E(X) = 19 × 0.40
= 7.6
Expected Profit = 7.6 × 0.50 = 3.8
For Day 3
E(X) = 22 × 0.35
= 7.7
Expected Profit = 7.7 × 0.50 = 3.85

Question 3.
Probability which explains x is equal to or less than particular value is classified as
(a) discrete probability
(b) cumulative probability
(c) marginal probability
(d) continuous probability
Solution:
(b) Cumulative Probability

Question 4.
Given E(X) = 5 and E(Y) = -2, then E(X – Y) is
(a) 3
(b) 5
(c) 7
(d) -2
Solution:
(c) 7
Hint:
E(x – 7) = E(x) – E(y) = 5 – (-2) = 7

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 5.
A variable that can assume any possible value between two points is called
(a) discrete random variable
(b) continuous random variable
(c) discrete sample space
(d) random variable
Solution:
(b) Continuous random Variable

Question 6.
A formula or equation used to represent the probability distribution of a continuous random variable is called
(a) probability distribution
(b) distribution function
(c) probability density function
(d) mathematical expectation
Solution:
(c) Probability density function

Question 7.
If X is a discrete random variable and p(x) is the probability of X, then the expected value of this random variable is equal to
(a) Σf(x)
(b) Σ[x + f(x)]
(c) Σf(x) + x
(d) Σxp(x)
Solution:
(d) Σxp(x)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 8.
Which of the following is not possible in probability distribution?
(a) Σp(x) > 0
(b) Σp(x) = 1
(c) Σxp(x) = 2
(d) p(x) = -0.5
Solution:
(d) P(x) = -0.5

Question 9.
If c is a constant, then E(c) is
(a) 0
(b) 1
(c) c f(c)
(d) c
Solution:
(d) c

Question 10.
A discrete probability distribution may be represented by
(a) table
(b) graph
(c) mathematical equation
(d) all of these
Solution:
(d) all of these

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 11.
A probability density function may be represented by:
(a) table
(b) graph
(c) mathematical equation
(d) both (b) and (c)
Solution:
(d) both (b) and(c)

Question 12.
If c is a constant in a continuous probability distribution, then p(x = c) is always equal to
(a) zero
(b) one
(c) negative
(d) does not exist
Solution:
(a) Zero

Question 13.
E[X – E(X)] is equal to
(a) E(X)
(b) V(X)
(c) 0
(d) E(X) – X
Solution:
(c) 0

Question 14.
E[X – E(X)]² is
(a) E(X)
(b) E(X)²
(c) V(X)
(d) S.D(X)
Solution:
(c) V(X)

Question 15.
If the random variable takes negative values, then the negative values will have
(a) positive probabilities
(b) negative probabilities
(c) constant probabilities
(d) difficult to tell
Solution:
(a) Positive probabilities

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 16.
If we have f(x) = 2x, 0 ≤ x ≤ 1, then f(x) is a
(a) probability distribution
(b) probability density function
(c) distribution function
(d) continuous random variable
Solution:
(b) Probability density function

Question 17.
\(\int_{ -∞ }^{∞}\) f(x) dx is always equal to
(a) zero
(b) one
(c) E(X)
(d) f(x) + 1
Solution:
(b) one

Question 18.
A listing of all the outcomes of an experiment and the probability associated with each outcome is called
(a) probability distribution
(b) probability density function
(c) attributes
(d) distribution function
Solution:
(a) Probability distribution

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 19.
Which one is not an example of random experiment ?
(a) A coin is tossed and the outcome is either a head or a tail
(b) A six-sided die is rolled
(c) Some number of persons will be admitted to a hospital emergency room during any hour.
(d) All medical insurance claims received by a company in a given year.
Solution:
(d) All medical insurance claims received by a company in a given year

Question 20.
A set of numerical values assigned to a sample space is called
(a) random sample
(b) random variable
(c) random numbers
(d) random experiment
Solution:
(b) random variable

Question 21.
A variable which can assume finite or countably infinite number of values is known as
(a) continuous
(b) discrete
(c) qualitative
(d) none of them
Solution:
(b) Discrete

Question 22.
The probability function of a random variable is defined as
Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3 1
Then k is equal to
(a) zero
(b) \(\frac { 1 }{4}\)
(c) \(\frac { 1 }{15}\)
(d) one
Solution:
(c) k = 1/15
Hint:
W.K.T Σi=1 P(xi) = 1
p(x= -1)+ p(x= -2) + p(x = 0) + p(x = 1)
+ p(x = 2) = V
k + 2k + 3k + 4k + 5k = 1
5k = 1 ⇒ k = 1/5

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 23.
If p(x) = \(\frac { 1 }{10}\), x = 10, then E(X) is
(a) zero
(b) \(\frac { 6 }{8}\)
(c) 1
(d) -1
Solution:
(c) 1
Hint:
P(x) + 1/10 and x = 10
E(x) = Xp(x) = 10(1/10) = 1

Question 24.
A discrete probability function p(x) is always
(a) non-negative
(b) negative
(c) one
(d) zero
Solution:
(a) non-negative

Question 25.
In a discrete probability distribution the sum of all the probabilities is always equal to
(a) zero
(b) one
(c) minimum
(d) maximum
Solution:
(b) one

Question 26.
An expected value of a random variable is equal to it’s
(a) variance
(b) standard deviation
(c) mean
(d) con variance
Solution:
(c) mean

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Question 27.
A discrete probability function p(x) is always non-negative and always lies between
(a) 0 and ∞
(b) 0 and 1
(c) -1 and +1
(d) -∞ and ∞
Solution:
(b) 0 and 1

Question 28.
The probability density function p(x) cannot exceed
(a) zero
(b) one
(c) mean
(d) infinity
Solution:
(b) One

Question 29.
The height of persons in a country is a random variable of the type
(a) discrete random variable
(b) continuous random variable
(c) both (a) and (b)
(d) neither (a) not (b)
Solution:
(b) Continuous random variable

Question 30.
The distribution function F(x) is equal to
(a) P(X = x)
(b) P(X ≤ x)
(c) P (X ≥ x)
(d) all of these
Solution:
(b) p(x ≤ x)

Samacheer Kalvi 12th Business Maths Guide Chapter 6 Random Variable and Mathematical Expectation Ex 6.3

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Chemistry Guide Pdf Chapter 8 Physical and Chemical Equilibrium Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium

11th Chemistry Guide Physical and Chemical Equilibrium Text Book Back Questions and Answers

Textbook Evaluation:

I. Choose the best answer:

Question 1.
If Kb and Kf for a reversible reaction are 0.8 × 10-5 and 1.6 × 10-4 respectively, the value of the equilibrium constant is,
a) 20
b) 0.2 × 10-1
c) 0.05
d) None of these
Answer:
a) 20

Question 2.
At a given temperature and pressure, the equilibrium constant values for the equilibria
3A2 + B2 + 2Cr Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 1 2A3BC and
A3BC Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 2 \(\frac{3}{2}\) [A2] + \(\frac{1}{2}\) B2 +C
The relation between K1 and K2 is
a) K1 = \(\frac{1}{\sqrt{\mathrm{K}_{2}}}\)

b)K2 = \(\mathrm{K}_{1}^{-1 / 2}\)

C) K12 = 2K2

d) \(\frac{\mathrm{K}_{1}}{2}\) = K2
Answer:
b)K2 = \(\mathrm{K}_{1}^{-1 / 2}\)

Question 3.
The equilibrium constant for a reaction at room temperature is K1 and that at 700 K is K2 If K1 > K2, then
a) The forward reaction is exothermic
b) The forward reaction is endothermic
c) The reaction does not attain equilibrium
d) The reverse reaction is exothermic
Answer:
a) The forward reaction is exothermic

Question 4.
The formation of ammonia from N2(g) and H2(g) is a reversible reaction 2NO(g) + O2(g) Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 2 2NO2(g) + Heat What is the effect of increase of temperature on this equilibrium reaction
a) equilibrium is unaltered
b) formation of ammonia is favoured
c) equilibrium is shifted to the left
d) reaction rate does not change
Answer:
c) equilibrium is shifted to the left

Question 5.
Solubility of carbon dioxide gas in cold water can be increased by
a) increase in pressure
b) decrease in pressure
c) increase in volume
d) none of these
Answer:
a) increase in pressure

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
Which one of the following is incorrect statement?
a) for a system at equilibrium, Q is always less than the equilibrium constant
b) equilibrium can be attained from either side of the reaction
c) the presence of catalyst affects both the forward reaction and reverse reaction to the same extent
d) Equilibrium constant varied with temperature
Answer:
a) for a system at equilibrium, Q is always less than the equilibrium constant

Question 7.
K1 and K2 are the equilibrium constants for the reactions respectively.
N2(g) + O2(g)Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 1 2NO(g)
2NO(g) + O2(g) Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 22NO2 (g)
What is the equilibrium constant for the reaction NO2(g) ⇌ 1/2 N2(g) + O2(g)
a) \(\frac{1}{\sqrt{\mathrm{K}_{1} \mathrm{~K}_{2}}}\)

b) \(\left(\mathrm{K}_{1}=\mathrm{K}_{2}\right)^{1 / 2}\)

c) \(\frac{1}{2 \mathrm{~K}_{1} \mathrm{~K}_{2}}\)

d) \(\left(\frac{1}{\mathrm{~K}_{1} \mathrm{~K}_{2}}\right)^{3 / 2}\)
Answer:
a) \(\frac{1}{\sqrt{\mathrm{K}_{1} \mathrm{~K}_{2}}}\)

Question 8.
In the equilibrium, 2A(g) ⇌ 2B(g) + C2(g) the equilibrium concentrations of A, B and C2 at 400K are 1 × 10-4M, 2.0 × 10-3 M, 1.5 × 10-4 M respectively. The value of Kc for the equilibrium at 400 K is
a) 0.06
b) 0.09
c) 0.62
d) 3 × 10-2
Answer:
a) 0.06

Question 9.
An equilibrium constant of 3.2 × 10-6 for a reaction means, the equilibrium is
a) largely towards forward direction
b) largely towards reverse direction
c) never established
d) none of these
Answer:
b) largely towards reverse direction

Question 10.
\(\frac{\mathrm{K}_{\mathrm{C}}}{\mathrm{K}_{\mathrm{P}}}\) for the reaction,
N2(g) + 3H2(g) ⇌ 2NH3 (g) is
a) \(\frac{1}{\mathrm{RT}}\)
b) \(\sqrt{\mathrm{RT}}\)
c) RT
d) (RT)2
Answer:
d) (RT)2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 11.
For the reaction AB (g) ⇌ A(g) + B(g), at equilibrium, AB is 20 % dissociated at a total pressure of P, the equilibrium constant Kp is related to the total pressure by the expression
a) P = 24 Kp
b) P = 8 Kp
c) 24 P = Kp
d) none of these
Answer:
a) P = 24 Kp

Question 12.
In which of the following equilibrium, Kp and Kc are not equal?
a) 2N0(g) ⇌ N2(g) + O2(g)
b) SO2(g) + NO2 ⇌ SO3(g) + NO(g)
c) H2(g) + I2(g) ⇌ 2HI(g)
d) PCl5 ⇌ PCl3(g) + Cl2(g)
Answer:
b) SO2(g) + NO2 ⇌ SO3(g) + NO(g)

Question 13.
If x is the fraction of PCl5 dissociated at equilibrium in the reaction PCl5 ⇌ PCl3 + Cl2 then starting with 0.5 mole of PCl5, the total number of moles of reactants and products at equilibrium is
a) 0.5 – x
b) x + 0.5
c) 2x + 0.5
d) x + 1
Answer:
b) x + 0.5

Question 14.
The values of Kp1 and Kp2; for the reactions, X ⇌ Y + Z, A ⇌ 2B are in the ratio 9 : 1 if degree of dissociation of X and A be equal then total pressure at equilibrium P1, and P2 are in the ratio
a) 36 : 1
b) 1 : 1
c) 3 : 1
d) 1 : 9
Answer:
a) 36 : 1

Question 15.
In the reaction
Fe(OH)3(S) ⇌ Fe3+ (aq) + 3OH (aq),
if the concentration of OH ions is decreased by 1/4 times, then the equilibrium concentration of Fe3+ will
a) not changed
b) also decreased by 1/4 times
c) increase by 4 times
d) increase by 64 times
Answer:
d) increase by 64 times

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 16.
Consider the reaction where Kp = 0.5 at a particular temperature PCl5(g) ⇌ PCl3 (g) + Cl2 (g) if the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true.
a) more PCl3 will be produced
b) more Cl2 will be produced
c) more PCl5 will be produced
d) none of these
Answer:
c) more PCl5 will be produced

Question 17.
Equimolar concentrations of H2 and I2 are heated to equilibrium in a 1 litre flask. What percentage of the initial concentration of H2 has reacted at equilibrium if rate constant for both forward and reverse reactions are equal
a) 33%
b) 66%
c) (33)2%
d) 16.5 %
Answer:
a) 33%

Question 18.
In a chemical equilibrium, the rate constant for the forward reaction is 2.5 × 10-2, and the equilibrium constant is 50. The rate constant for the reverse reaction is,
a) 11.5
b) 50
c) 2 × 102
d) 2 × 10-3
Answer:
b) 50

Question 19.
Which of the following is not a general characteristic of equilibrium involving physical process
a) Equilibrium is possible only in a closed system at a given temperature
b) The opposing processes occur at the same rate and there is a dynamic but stable condition
c) All the physical processes stop at equilibrium
d) All measurable properties of the system remains constant
Answer:
c) All the physical processes stop at equilibrium

Question 20.
For the formation of Two moles of SO3(g) from SO2 and O2, the equilibrium constant is K1. The equilibrium constant for the dissociation of one mole of SO3 into SO2 and O2 is
a) \(1 / \mathrm{K}_{1}\)
b) K12
c) \(\left(\frac{1}{\mathrm{~K}_{1}}\right)^{1 / 2}\)
d) \(\frac{\mathrm{K}_{1}}{2}\)
Answer:
c) \(\left(\frac{1}{\mathrm{~K}_{1}}\right)^{1 / 2}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 21.
Match the equilibria with the corresponding conditions:
i) Liquid ⇌ Vapour
ii) Solid ⇌ Liquid
iii) Solid ⇌ Vapour
iv) Solute(s) ⇌ Solute (Solution)
1) Melting point
2) Saturated solution
3) Boiling point
4) Sublimation point
5) Unsaturated solution

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 3
Answer:
b) 3 1 4 2

Question 22.
Consider the following reversible reaction at equilibrium, A + B ⇌ C, If the concentration of the reactants A and B are doubled, then the equilibrium constant will
a) be doubled
b) become one fourth
c) be halved
d) remain the same
Answer:
d) remain the same

Question 23.
[Co(H2O)6]2+ (aq) (pink) + 4Cl (aq) ⇌ [CoCl4]2-(aq) (blue) + 6 H2O (l)
In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in color. On the basis of this information, which one of the following is true?
a) ∆H > 0 for the forward reaction
b) ∆H = 0 for the reverse reaction
c) ∆H < 0 for the forward reaction
d) Sign of the ∆H cannot be predicted based on this information
Answer:
a) ∆H > 0 for the forward reaction

Question 24.
The equilibrium constants of the following reactions are:
N2 + 3H2 ⇌ 2NH3; K1
N2 + O2 ⇌ 2NO; K2
H2 + 1/2O2 ⇌ H2O; K3
The equilbrium constant (K) for the reaction ;
2NH3 + 5/2 O2 ⇌ 2NO + 3H2O, will be
a) K23 K3/K1

b) K1 K33/K2

c) K2K33/K1

d) K2K3/K1
Answer:
c) K2K33/K1

Question 25.
A 20 litre container at 400 K contains CO2 (g) at pressure 0.4 atm and an excess (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when the pressure of CO2 attains its maximum value will be: Given that: SrCO3(S) ≅ SrO + CO2 (g) [Kp = 1.6 atm]
a) 2 litre
b) 5 litre
c) 10 litre
d) 4 litre
Answer:
b) 5 litre

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

II. Write brief answer to the following questions:

Question 26.
If there is no change in concentration, why is the equilibrium state considered dynamic?
Answer:
Rate of forward reaction = Rate of backward reaction

Question 27.
For a given reaction at a particular temperature, the equilibrium constant has constant value. Is the value of Q also constant? Explain.
Answer:
Kc and Qc are constant at equilibrium, both are temperature dependent. When Kc is constant at given temperature, Qc also constant.

Question 28.
What is the relation between Kp and Kc? Given one example for which Kp is equal to Kc.
Answer:
Kp = Kc(RT)∆ng
H2(g) + I2(g) ≅ 2HI (g) ; ∆ng = 0 ;
Kp = Kc.

Question 29.
For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is Kc is larger or smaller than Kp.
Answer:
Kp >Kc
np > nR

Question 30.
When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant, in which direction does the reaction proceed to reach equilibrium?
Answer:
Q > Kc
Reverse reaction is favoured.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 31.
For the reaction, A2(g) + B2(g) ⇌ 2AB(g); ∆H is -ve.
the following molecular scenes represent differenr reaction mixture.(A-green, B-blue)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 4

i) Calculate the equilibrium constant Kp and (Kc).
ii) For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
iii) What is the effect of increase in pressure for the mixture at equilibrium.
Answer:
Kc = \(\frac{[\mathrm{AB}]^{2}}{\left[\mathrm{~A}_{2}\right]\left[\mathrm{B}_{2}\right]}\) ; A – green, B – blue
Given that ‘V’ is constant(closed system)
At equilbrium,
Kc = \(\frac{(-)}{\left(\frac{2}{V}\right)\left(\frac{2}{V}\right)}\) = \(\frac{16}{4}\) = 4
Kp = Kc(RT)∆n = 4 (RT)0 = 4
At stage ‘x’,
Q = \(\frac{\left(\frac{6}{\mathrm{~V}}\right)^{2}}{\left(\frac{2}{\mathrm{~V}}\right)\left(\frac{1}{\mathrm{~V}}\right)}=\frac{36}{2}\) = 18
Q > Kc i.e., reverse reaction is favoured.
At stage ‘y’,
Q = \(\frac{\left(\frac{3}{V}\right)^{2}}{\left(\frac{3}{V}\right)\left(\frac{3}{V}\right)}=\frac{9}{3 \times 3}\) = 1
Kc > Q i.e., forward reaction is favoured.

Question 32.
State Le – Chateller principle.
Answer:
“If a system at equilibrium is distributed, then the system shifts itself in a direction that nullifies the effect of that disturbance”.

Question 33.
Consider the following reactions,
a) H2(g) + I2(g) ⇌ 2HI(g)
b) CaCO3(s) ⇌ CaO (s) + CO2(g)
c) S(s) + 3 F2(g) ⇌ SF6(g)
In each of the above reactions find out whether you have to increase (or) decrease the volume to increase the yield of the product.
Answer:
a) H2(g) + I2(g) ⇌ 2HI(g)
In this reaction, there is no effect on changing the volume. ∆ng = 0

b) CaCO3(s) ⇌ CaO (s) + CO2(g)
In this reaction, increases in volume favours forward reaction.

c) S(s) + 3 F2(g) ⇌ SF6(g)
In this reaction, decreases in volume favours forward reaction.

Question 34.
State law of mass action.
Answer:
The law states that, “At any instant, the rate of a chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant”.
Rate ∝ [Reactant]x
Active mass = \(\left(\frac{\mathrm{n}}{\mathrm{V}}\right)\) mol dm-3 (or) mol L-1

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 35.
Explain how will you predict the direction of an equilibrium reaction.
Answer:
From the knowledge of equilibrium constant, it is possible to predict the di¬rection in which the net reaction is taking place for a given concentration or partial pressure of reactants and products.
Consider a general homogeneous reversible reaction,

xA + yB ⇌ lC + mD

For the above reaction under non-equilibrium conditions, reaction quotient ‘Q’ is defined as the ratio of the product of active masses of reaction products raised to the respective stiochiometric coefficients in the balanced chemical equation to that of the reactants.

Under non – equilibrium conditions, the reaction quotient Q can be calculated using the following expression.
Q = \(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\)

As the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the Q value until the reaction reaches equilibrium. At equilibrium, Q is equal to Kc at a particular temperature. Once the equilibrium is attained, there is no change in the Q value. By knowing the Q value, we can predict the direction of the reaction by comparing it with Kc.
If Q = Kc, the reaction is in equilibrium state.
If Q > Kc, the reaction will proceed in the reverse direction, i.e., formation of reactants.
If Q < Kc, the reaction will proceed in the forward direction i.e., formation of products.
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 5
Predicting the direction of a reaction

Question 36.
Derive a general expression for the equilibrium constant Kp and Kc for the reaction, 3H2(g) + N2(g) ⇌ 2NH3(g).
Answer:
Let us consider the formation of ammonia in which, ‘a’ moles nitrogen and h’ moles hydrogen gas are allowed to react in a container of volume V. Let ‘x’ moles of nitrogen react with 3 x moles of hydrogen to give 2x moles of ammonia.
3H2(g) + N2(g) ⇌ 2NH3(g)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 6

Applying law of mass action,
K2 = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}\)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 7
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 8

Question 37.
Write the balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression.
Answer:
Kc = \(\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{7}}{\left.[\mathrm{NO}]^{4} \mid \mathrm{H}_{2} \mathrm{O}\right]^{6}}\)
Balanced equation is:
4NO + 6H2O ⇌ 4NH3 + 7O2

Question 38.
What is the effect of added Inert gas on the reaction at equilibrium?
Answer:
When an inert gas (i.e., a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles of gases present in the container increases, that is, the total pressure of gases increases. The partial pressure of the reactants and the products are unchanged. Hence at constant volume, the addition of inert gas has no effect on the equilibrium.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 39.
Derive the relation between Kp and Kc.
Answer:
Let us consider the general reaction in which all reactants and products are ideal gases.
xA + yB ⇌ lC + mD
The equilibrium constant, Kc is,
Kc = \(\frac{\left[\mathrm{Cl}^{l}[\mathrm{D}]^{\mathrm{m}}\right.}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\) ……………..(1)

and Kp is,
Kp = \(\frac{p_{C}^{l} \times p_{D}^{m}}{p_{A}^{x} \times p_{B}^{y}}\) ………….(2)

The ideal gas equation is,
PV = nRT
P = \(\frac{n}{V}\) RT
since, Active mass = molar concentration = n/V
P = active mass × RT
Based on the above expression the partial pressure of the reactants and products can be expresssed as,
PXA = [A]x[RT]x
PYB = [A]y[RT]y
PlC = [A]l[RT]l
PmD = [A]m[RT]m
On substituting in Eqn.(2).,
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 9
By comparing equation (1) and (4),
we get
Kp∆ng =Kc (RT) …………..(5)
Where,
∆ng is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
The following relations become immediately obvious.
when ∆ng = 0.
Kp = Kc (RT)0 = Kc
Example :
H2(g) + I2(g) ⇌ 2HI(g)
N2 + O2 ⇌ 2NO(g)
when ∆ng = +ve
Kp = Kc(RT)+ve
Kp > Kc
Example:
2NH3(g) ⇌ N2(g) + 3H2(g)
PCl5(g) ⇌ PCl3(g) + Cl2(g)
When ∆ng = -ve
Kp = Kc(RT)-ve
Kp < Kc
Example:
2H2(g) + O2(g) ⇌ 2H2O(g)
2SO2(g)+ O2(g) ⇌ 2SO3(g)

Question 40.
One mole of PCl5 is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, Calculate the value of equilibrium constant.
Solution:
Given that,
[PCl5]initial = \(\frac{1 \text { mole }}{1 \mathrm{dm}^{3}}\)
[Cl2]eq = 0.6 mole dm-3
PCl5 ⇌ PCl3 + Cl2
[PCl3]eq = 0.6 mole dm-3
[PCl5]eq = 0.4 mole dm-3
∴ Kc = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{0.6 \times 0.6}{0.4}\)
= 0.9 mole dm-3

Question 41.
For the reaction SrCO3 (s) ⇌ SrO(s) + CO2(g) the value of equilibrium constant Kp = 2.2 × 10-4 at 1002 K. Calculate Kc for the reaction.
Solution :
For the reaction,
SrCO3(s) ⇌ SrO(s) + CO2(g)
∆ng = 1 – 0 = 1
∴ Kp = Kc (RT)
2.2 × 10-4 = Kc (0.0821) (1002)
Kc = \(\frac{2.2 \times 10^{-4}}{0.0821 \times 1002}\) = 2.674 × 10-6

Question 42.
To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at 500°C. At equilibrium he found the concentration of HI which is equal to 0.05 M. Calculate Kc and Kp.
Solution :
V = 3 L
[HI]initial = \(\frac{0.3 \mathrm{~mol}}{3 \mathrm{~L}}\) 0.1 M
[HI]eq = 0.05 M
2 HI(g) ⇌ H2(g) + I2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 10

Kc = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}=\frac{0.025 \times 0.025}{0.05 \times 0.05}\) = 0.25
Kp = Kc(RT)(∆ng)
∆ng = 2 – 2 = 0
Kp = 0.25(RT)0
Kp = 0.25.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 43.
Oxidation of nitrogen monoxide was studied at 200 with initial pressures of 1 atm NO and 1 atm of O2. At equilibrium partial pressure of oxygen is found to be 0. 5 atm calculate Kp value.
Solution:
2 NO(g) + O2(g) ⇌ 2NO2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 11
Kp = \(\frac{\left(P_{N O_{2}}\right)^{2}}{\left(P_{N O}\right)^{2}\left(P_{O}\right)}\)

= \(\frac{0.96 \times 0.96}{0.04 \times 0.04 \times 0.52}\)
Kp = 1.017 × 103.

Question 44.
1 mol of CH4, 1 mole of CS2 and 2 mol of H2S are 2 mol of H2 are mixed in a 500 ml flask. The equilibrium constant for the reaction Kc = 4 x 10-2 mol2 lit-2. In which direction will the reaction proceed to reach equilibrium ?
Solution :
CH4(g) + 2 H2S (g) ⇌ CS2(g) + 4H2(g)
Kc = 4 x 10-2 mol2 lit-2
Volume = 500 ml = \(\frac{1}{2}\) L
[CH4]in = \(\frac{1 \mathrm{~mol}}{\frac{1}{2} L}\) = 2 mol L-1;

[CH2]in = \(\frac{1 \mathrm{~mol}}{\frac{1}{2} L}\) = 2 mol L-1
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 12

Question 45.
At particular temperature Kc = 4 × 10-2 for the reaction, H2S ⇌ 2H2(g) + S2(g). Calculate the Kc for each of the following reaction.
i) 2H2S(g) ⇌ 2H2 + S2(g)
ii) 3H2S(g) ⇌ 3H2(g) + 3/2 S2(g)
Solution :
Kc = 4 × 10-2 for the reaction,
H2S ⇌ 2H2(g) + S2(g)
Kc = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]^{1 / 2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]}\)

⇒ 4 × 10-2 = \(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{S}_{2}\right]^{1 / 2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]}\)

(i) 2H2S(g) ⇌ 2H2 + S2(g)
For the reaction,
2H2S(g) ⇌ 2H2 + S2(g)
Kc = \(\frac{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{~S}_{2}\right]}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}}\)
= (4 × 10-2)2 = 16 × 10-4

(ii) 3H2S(g) ⇌ 3H2(g) + 3/2 S2(g)
For the reaction,
3H2S(g) ⇌ 3H2(g) + 3/2 S2(g)
Kc = \(\frac{\left[\mathrm{H}_{2}\right]^{3}\left[\mathrm{~S}_{2}\right]^{3 / 2}}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{3}}\)
= (4 × 10-2)3 = 64 × 10-6

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 46.
28 g of Nitrogen and 6 g of hydrogen were mIed In a 1 litre closed container. At equIlibrium 17 g NH3 was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.
Solution:
Given mN2 =28g;
mH2 = 6g;
V = 1 L.
(nN2)initial = \(\frac{28}{28}\) = 1

(nH2)initial = \(\frac{6}{2}\) = 3
N2(g) + 3 H2(g) ⇌ 2NH3(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 13
[NH3] = \(\frac{17}{17}\) = 1 mol
Weight of N2 = (no. of moles of N2) × (molar mass of N2)
= 0.5 × 28 = 14g
Weight of H2 = (no. of moles of H2) × (molar mass of H2)
= 1.5 × 2 = 3g

Question 47.
The equilibrium for the dissociation of XY2 is given as,
2 XY2(g) ⇌ 2 XY(g) + Y2(g)
if the degree of dissociation x is so small compared to one. Show that 2 Kp = PX3 where P is the total pressure and Kp is the dissociation equilibrium constant of XY2.
Answer:
2 XY2(g) ⇌ 2 XY(g) + Y2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 14

Question 48.
A sealed container was filled with 0.3 mol H2(g), 0.4 mol I2(g) and 0.2 mol HI(g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 870 for the reaction, H2(g) + I2(g) ⇌ 2 HI (g).
Solution:
A2(g) + B2(g) ⇌ 2 AB (g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 15

Given that, Kp = 1;
\(\frac{4 x^{2}}{(1-x)^{2}}\) = 1
⇒ 4x2 = (1 – x)2 = 1
⇒ 4x2 = 1 + x2 – 2x
3x2 + 2x – 1 = 0
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 16
x = 0.33 – 1(not possible)
∴ [A2]eq = 1 – x = 1 – 0.33 = 0.67
[B2]eq = 1 – x = 1 – 0.33 = 0.67
[AB2]eq = 2x × 0.33 = 0.66

Question 49.
Deduce the Vant Hoff equation.
Answer:
This equation gives that quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change (∆G°) and equilibrium constant is
∆G° = – RT In K ………………..(1)
We know that, ∆G° = ∆H° – T∆S° …………(2)
Substituting (1) in equation (2)
– RT In K = ∆H° – T∆S°
Rearranging, In K = \(\frac{\Delta \mathrm{H}^{0}}{\mathrm{RT}}+\frac{\Delta \mathrm{S}^{0}}{\mathrm{RT}}\) ……………(3)
Differentiating equation (3) with respect to temperature,
\(\frac{\mathrm{d}(\ln \mathrm{K})}{\mathrm{dT}}=\frac{\Delta \mathrm{H}^{0}}{\mathrm{RT}^{2}}\) ……………….(4)
Equation (4) is known as differential form of van,t Hoff equation.
On integrating the equation (4), between T1 and T2 with their respective equilbrium consatnts K1 and K2.
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 17
Equation 5 is known as integrated form of Van’t Hoff equation.

Question 50.
The equilibrium constant Kp for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is 8.19 × 102 at 298 K and 4.6 × 10-1 at 498 K. Calculate ∆H° for the reaction.
Solution :
Kp1 = 8.19 × 102;
T1 = 298 K
Kp1 = 8.19 × 102;
T1 = 298 K
Kp2 = 4.16 × 10-1;
T2 = 498 K
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 18

Question 51.
The partial pressure of carbon dioxide in the reaction CaCO3(s) ⇌ CaO(s) + CO2(g) is 1.017 × 10-3 atm at 500°C. Calculate Kp at 600°C for the reaction. H for the reaction is 181 KJ mol-1 and does not change in the given range of temperature.
Solution:
PCO2 = 1.017 × 10-3 atm
T = 500°C;
Kp = PCO2
∴ Kp1 = 1.017 × 10-3;
T = 500 + 273 = 773 K
Kp2 = ?
T = 600 + 273 = 873 K
∆H° = 181 KJ mol-1
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 19

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

11th Chemistry Guide Physical and Chemical Equilibrium Additional Questions and Answers

I.Choose the best answer:

Question 1.
For which of the following Kp is less than Kc?
a) N2O4 ⇌ 2NO2
b) N2 + 3H2 ⇌ 2NH3
c) H2 + I2 ⇌ 2HI
d) CO + H2O ⇌ CO2 + H2
Answer:
b) N2 + 3H2 ⇌ 2NH3

Question 2.
In which of the following reaction, the value of Kp will be equal to Kc?
a) H2 + I2 ⇌ 2HI
b) PCl5 ⇌ PCl3 + Cl2
c) 2NH3 ⇌ N2 + 3H2
d) 2SO2 +O2 ⇌ 2SO3
Answer:
a) H2 + I2 ⇌ 2HI

Question 3.
For homogeneous gas reaction 4NH3 + 5O2 ⇌ 4NO + 6H2O. The equilibrium constant Kc has the unit
a) (concentration)1
b) (concentration)1
c) (concentration)9
d) (concentration)10
Answer:
a) (concentration)1

Question 4.
The reaction, 2SO2(g) + O2(g) ⇌ 2SO3(g) is carried out in a 1 dm3 vessel and 2 dm3 vessel separately. The ratio of the reaction velocities will be
a) 1 : 8
b) 1 : 4
c) 4 : 1
d) 8 : 1
Answer:
d) 8 : 1

Question 5.
Kp for the following reaction at 700 K is 1.3 × 10-3 atm-1. The Kc same temperature for the reaction. 2SO2 + O2 ⇌ 2SO3 will be
a) 1.1 × 10-2
b) 3.1 × 10-2
c) 5.2 × 10-2
d) 7.4 × 10-2
Answer:
d) 7.4 × 10-2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
The equilibrium constant expression for the equilibrium :
2NH2(g) + 2O2(g) ⇌ N2O(g) + 3H2O(g) is
a) Kc = \(\frac{\left[\mathrm{N}_{2} \mathrm{O}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}}{\left[\mathrm{NH}_{3}\right]\left[\mathrm{O}_{2}\right]}\)

b) Kc = \(\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}\left[\mathrm{~N}_{2} \mathrm{O}\right]}{\left[\mathrm{NH}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}}\)

c) Kc = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}}{\left[\mathrm{~N}_{2} \mathrm{O}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}}\)

d) Kc = \(\frac{\left[\mathrm{NH}_{3}\right]\left[\mathrm{O}_{2}\right]}{\left[\mathrm{N}_{2} \mathrm{O}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]}\)
Answer:
b) Kc = \(\frac{\left[\mathrm{H}_{2} \mathrm{O}\right]^{3}\left[\mathrm{~N}_{2} \mathrm{O}\right]}{\left[\mathrm{NH}_{3}\right]^{2}\left[\mathrm{O}_{2}\right]^{2}}\)

Question 7.
For the system 3A + 2B ⇌ C, the expression for equilibrium constant is
a) \(\frac{[3 \mathrm{~A}][2 \mathrm{~B}]}{[\mathrm{C}]}\)

b) \(\frac{[\mathrm{C}]}{[3 \mathrm{~A}][2 \mathrm{~B}]}\)

c) \(\frac{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}{[\mathrm{C}]}\)

d) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)
Answer:
d) \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)

Question 8.
For the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) at 250°C, the value of Kc is 26, then the value of Kp at the same temperature will be
a) 0.61
b) 0.57
c) 0.83
d) 0.46
Answer:
a) 0.61

Question 9.
The equilibrium constant for the reaction N2(g) + O2(g) ⇌ 2NO(g) at temperature T is 4 × 10-4. The value of Kc for the reaction NO(g) ⇌ 1/2 N2(g) + 1/2 O2(g) at the same temperature is ___________.
a) 4 × 10-4
b) 50
c) 2.5 × 102
d) 0.02
Answer:
b) 50

Question 10.
In which of the following equilibria, the value of Kp is less than Kc?
a) H2 + I2 ⇌ 2HI
b) N2 + 3 H2 ⇌ 2NH3
c) N2 + O2 ⇌ 2NO
d) CO + H2O ⇌ CO2 + H2
Answer:
b) N2 + 3 H2 ⇌ 2NH3

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 11.
In the reversible reaction A + B ⇌ C + D. The concentration of each C and D at equilibrium was 0.8-mole lit, then the equilibrium constant will be
a) 6.4
b) 0.64
c) 1.6
d) 16.0
Answer:
d) 16.0

Question 12.
The Kc for H2(g) + I2(g) ⇌ 2 HI(g), is 64. If the volume of the container is reduced to one – half of its original volume, the value of the equilibrium constant will be
a) +28
b) 64
c) 32
d) 16
Answer:
b) 64

Question 13.
For the reaction C(s) + CO2(g) ⇌ 2CO (g), the partial pressure of CO2 and CO are 2.0 and 4.0 atm respectively at equilibrium. The Kp for the reaction is
a) 0.5
b) 4.0
c) 8.0
d) 32.0
Answer:
c) 8.0

Question 14.
When the rate of forward reaction becomes equal to backward reaction, this state is termed as
a) chemical equilibrium
b) Reversible state
c) Equilibrium
d) All of these
Answer:
d) All of these

Question 15.
The equilibrium constant of the reaction H2(g) + I2(g) ⇌ 2 HI(g) is 64. If the volume of the container is reduced to one – fourth of its original volume, the value of the equilibrium constant will be
a) 16
b) 32
c) 64
d) 128
Answer:
c) 64

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 16.
For the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g), the position of equilibrium can be
shifted to the right by _______.
a) increasing the temperature
b) Doubling the volume
c) Addition of Cl2 at constant volume
d) Addition of equimolar quantities of PCl3 and PCl5
Answer:
b) Doubling the volume

Question 17.
For the reaction 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g), Kc at 427 is 3 × 10-6 L mol-1. The value of Kp is nearly
a) 7.50 × 10-5
b) 2.50 × 10-5
c) 2.50 × 10-4
d) 1.72 × 10-4
Answer:
d) 1.72 × 10-4

Question 18.
Rate of reaction curve for equilibrium can be like [rf = forward rate, rb = backward rate]
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 20
Answer:
a) Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 21

Question 19.
For a reaction, N2 + 3H2 ⇌ 2NH3(g), the value of Kc does not depend upon
A) initial concentration of the reactants
B) pressure
C) temperature
D) Catalyst
a) only C
b) A, B, C
c) A, B, C
d) A, B, C, D
Answer:
c) A, B, C

Question 20.
2NO2 ⇌ 2NO + O2, K = 1.6 × 10-12 NO + \(\frac{1}{2}\)O2 ⇌ NO2; K’ = ?
a) K’ = \(\frac{1}{\mathrm{~K}^{2}}\)
b) K’ = \(\frac{1}{\mathrm{~K}}\)
c) K’ = \(\frac{1}{\sqrt{\mathrm{K}}}\)
d) None of these
Answer:
c) K’ = \(\frac{1}{\sqrt{\mathrm{K}}}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 21.
At 1000 K, the value of Kp for the reaction, A(g) + 2B(g) ⇌ 3C (g) + D(g) is 0.05 atm. The value of Kc in terms of R would be
a) 20000 R
b) 0.02 R
c) 5 × 10-5 R
d) 5 × 10-5 × R-1
Answer:
d) 5 × 10-5 × R-1

Question 22.
In a chemical equilibrium, the rate constant for the backward reaction is 7.5 x 10-4 and the equilibrium constant is 1.5. The rate constant for the forward reaction
a) 2 × 10-3
b) 5 × 10-4
c) 1.12 × 10-3
d) 9.0 × 10-4
Answer:
c) 1.12 × 10-3

Question 23.
A reversible reaction is one which
a) Proceeds in one direction
b) proceeds in both direction
c) proceeds spontaneously
d) All the statements are wrong
Answer:
b) proceeds in both direction

Question 24.
The equilibrium constant in a reversible reaction at a given temperature
a) depends on the initial concentration of the reactants
b) depends on the concentration of the products at equilibrium
c) does not depend on the initial concentrations
d) It is not characteristic of the reaction
Answer:
c) does not depend on the initial concentrations

Question 25.
For the system A(g) + 2B(g) ⇌C (g), the equilibrium concentrations are (A) 0.06 mole / lit (B) 0.12 mole / lit and (C) 0.216 mole / lit. The Keq for the reaction is
a) 250
b) 416
c) 4 × 10-3
d) 125
Answer:
a) 250

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 26.
A chemical reaction is at equilibrium when
a) Reactants are completely transformed into products
b) The rates of forward and backward reactions are equal
c) Formation of products is minimized
d) Equal amounts of reactants and products are present
Answer:
b) The rates of forward and backward reactions are equal

Question 27.
Partial pressures of A, B, C and D on the basis of gaseous system A + 2B ⇌ C + 3D are A = 0.20, B = 0.10, C = 0.30 and D = 0.50 atm. The numerical value of equilibrium constant is _______.
a) 11.25
b) 18.75
c) 5
d) 3.75
Answer:
b) 18.75

Question 28.
Molar concentration of 96 g of O2 contained in a 2 L vessel is ______.
a) 16 mol / L
b) 1.5 mol / L
c) 4 mol / L
d) 24 mol / L
Answer:
b) 1.5 mol / L

Question 29.
According to law of mass action rate of a chemical reaction is proportional to
a) concentration of reactants
b) molar concentration of reactants
c) concentration of products
d) molar concentration of products
Answer:
b) molar concentration of reactants

Question 30.
The equilibrium constant of the reaction SO2(g) + \(\frac{1}{2}\) O2(g) ⇌ SO3 (g) is 4 × 10-3 atm-1/2. The equilibrium constant of the reaction 2SO3(g) ⇌ 2SO2(g) + O2(g) would be
a) 250 atm
b) 4 × 103 atm
c) 0.25 × 104 atm
d) 6.25 × 104 atm
Answer:
d) 6.25 × 104 atm

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 31.
The rate constant for forward and backward reactions of the hydrolysis of ester are 1.1 × 10-2 and 1.5 × 10-3 per minute respectively. The equilibrium constant for the reaction is ___________.
a) 4.33
b) 5.33
c) 6.33
d) 7.33
Answer:
d) 7.33

Question 32.
The active mass of 64 g of HI is a two litre of flask would be
a) 2
b) 1
c) 5
d) 0.25
Answer:
d) 0.25

Question 33.
At a certain temperature, 2HI ⇌ H2 + I2. Only 50 % HI is dissociated at equilibrium. The equilibrium constant is
a) 0.25
b) 1.0
c) 3.0
d) 0.50
Answer:
a) 0.25

Question 34.
Unit of equilibrium constant for the reversible reaction, H2 + I2 ⇌ 2HI is ________.
a) mol-1 litre
b) mol-2 litre
c) mol litre-1
d) none of these
Answer:
d) none of these

Question 35.
For the reaction
CO(g) + \(\frac{1}{2}\)O2(g) ⇌ CO2(g), Kp/Kc is _______.
a) RT
b) (RT)-1
c) (RT)-1/2
d) (RT)1/2
Answer:
c) (RT)-1/2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 36.
The following equilibrium are given
N2 +3H2 ⇌ 2NH2 ………K1
N2 + O2 ⇌ 2N0 ………….K2
H2 + 1/2 O2 ⇌ H2O …….. K3
The equilibrium constant of the reaction.
2 NH3 + 5/2 O2 ⇌ 2NO + 3H2O, in terms of K1, K2 and K3 is _________.
a) \(\frac{\mathbf{K}_{1} \mathbf{K}_{2}}{\mathbf{K}_{3}}\)

b) \(\frac{\mathrm{K}_{1} \mathrm{~K}_{3}^{2}}{\mathrm{~K}_{2}}\)

c) \(\frac{\mathrm{K}_{2} \mathrm{~K}_{3}^{2}}{\mathrm{~K}_{1}}\)

d) K1 K2 K3
Answer:
c) \(\frac{\mathrm{K}_{2} \mathrm{~K}_{3}^{2}}{\mathrm{~K}_{1}}\)

Question 37.
For the reaction
2NO2(g) ⇌ 2NO(g) + O2(g) Kc = 1.8 × 10-6 at 184°C and R = 0.083 Jk-1 mol-1. When Kp and are compared at 184C, it
is found that
a) Kp > Kc
b) Kp < Kc
c) Kp – Kc
d) Kp × Kc( depends upon pressure of gases)
Answer:
a) Kp > Kc

Question 38.
The rate of forward reaction is two times that of reverse reaction at a given temperature and identical concentration. K equilibrium is
a) 2.5
b) 2.0
c) 0.5
d) 1.5
Answer:
b) 2.0

Question 39.
In a reaction A+B ⇌ C+D,the concentrations of A, B, C and D [in mole/lit] are 0.5, 0.8, 0.4 & 1.0 respectively. Th. equilibrium constant is
a) 0.1
b) 1.0
c) 10
d) ∞
Answer:
b) 1.0

Question 40.
For the following thre. reactions 1, 2 and 3 equilibrium constants are given:
1. CO(g) + H2O(g) ⇌ CO(g) + H2(g); K1
2.CH4(g) + H2O(g) ⇌ CO + 3H(g); K2
3. CH4(g) + 2H2O(g) ⇌ CO(g) + 4H2(g); K3
Which of the following relations is correct?
a) K1√K2 = K3
b) K2K3 = K1
c) K3 = K1K2
d) K3 = K23 K12
Answer:
c) K3 = K1K2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 41.
4 moles of A is mixed with 4 moles of B. At equilibrium for the reaction A + B ⇌ C + D. 2 moles of C and D is formed. The equilibrium constant for the reaction will be
a) \(\frac{1}{4}\)
b) \(\frac{1}{2}\)
c) 1
d) 4
Answer:
c) 1

Question 42.
In a reaction PCl5 ⇌ PCl3 + Cl2 degree of dissociation is 30%. If Initial moles of PCl5 is one then total moles at equilibrium is
a) 1.3
b) 0.7
c) 1.6
d) 1.0
Answer:
a) 1.3

Question 43.
When 3 moles of A and 1 mole of B are mixed in 1 lit vessel, the following reaction takes place A(g) + B(g) ⇌ 2C(g), 1.5 moles of ‘C’ are formed. The equilibrium constant for the reaction is
a) 0.12
b) 0.25
c) 0.50
d) 4.0
Answer:
d) 4.0

Question 44.
In which of the following, the reaction proceeds towards completion
a) K = 103
b) K = 10-2
c) K = 10
d) K = 1
Answer:
a) K = 103

Question 45.
Two moles of NH3 when put into a previously evacuated vessel (1L) partially dissociate into N2 and H2. If at equilibrium one mole of NH3 is present, the equilibrium constant is ________.
a) \(\frac{3}{4}\) mol2 lit-2

b) \(\frac{27}{64}\) mol2 lit-2

c) \(\frac{27}{32}\) mol2 lit-2

d) \(\frac{27}{16}\) mol2 lit-2
Answer:
d) \(\frac{27}{16}\) mol2 lit-2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 46.
For the reaction N2(g) + O2(g) ⇌ 2NO(g) the value of Kc at 800°C is 0.1. When the equilibrium concentration of both the reactions is 0.5 mol, what is the value of Kp at the same temperature?
a) 0.5
b) 0.1
c) 0.01
d) 0.025
Answer:
b) 0.1

Question 47.
28 g of N2(g) and 6g of H2(g) were mixed at equilibrium 17 g NH3 was produced. The weight of N2 and H2 at equilibrium are respectively.
a) 11g, 0g
b) 1g, 3g
c) 14g, 3g
d) 11g, 3g
Answer:
c) 14g, 3g

Question 48.
2SO3 ⇌ 2SO2 + O2 If Kc = 100, α = 1, half of the reaction is completed, the concentration of SO3 and SO2 are equal, the concentration of O2 is _________.
a) 0.001 M
b) \(\frac{1}{2}\) SO2
c) 2 times of SO2
d) Data in complete
Answer:
d) Data in complete

Question 49.
For reaction HI ⇌ \(\frac{1}{2}\)H2 + \(\frac{1}{2}\)I2 value of Kc is \(\frac{1}{8}\), then value of K is \(\frac{1}{8}\) for H2 + I2 ⇌ 2HI.
a) \(\frac{1}{64}\)
b) 64
c) \(\frac{1}{8}\)
d) 8
Answer:
b) 64

Question 50.
2 moles of PCl5 were heated in a closed vessel of 2 liter capacity. At equilibrium, 40% of PCl5 is dissociated into PCl3 and Cl2. The value of equilibrium constant is
a) 0.266
b) 0.53
c) 2.66
d) 5.3
Answer:
a) 0.266

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 51.
For a reaction H2 + I2 ⇌ 2HI at 721 k, the value of equilibrium constant is 50. If 0.5 mol each of H2 and I2 is added to the system the value of equilibrium constant will be
a) 40
b) 60
c) 50
d) 30
Answer:
c) 50

Question 52.
In a reaction the rate of reaction is proportional to its active mass, this statement is known as
a) Law of mass action
b) Le – Chatelier principle
c) Faradays’s law of electrolysis
d) law of constant proportion
Answer:
b) Le – Chatelier principle

Question 53.
In a chemical equilibrium A + B ⇌ C + D, when one mole each of the two reactants are mixed, 0.6 mole each of the products are formed. The equilibrium constant calculated is
a) 1
b) 0.36
c) 2.25
d) \(\frac{4}{9}\)
Answer:
c) 2.25

Question 54.
Under a given set of experimental conditions, with an increase in the concentration of the reactants, the rate of a chemical reaction.
a) Decreases
b) Increases
c) Remains unaltered
d) First decreases and then increases
Answer:
c) Remains unaltered

Question 55.
A + B ⇌ C + D, If finally the concentrations of A and B are both equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of reaction.
a) \(\frac{4}{9}\)
b) \(\frac{9}{4}\)
c) \(\frac{1}{9}\)
d) 4
Answer:
d) 4

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 56.
Theory of ‘active mass’ indicates that the rate of chemical reaction is directly proportional to the ___________.
a) Equilibrium constant
b) Properties of reactants
c) Volume of apparatus
d) Concentration of reactants
Answer:
d) Concentration of reactants

Question 57.
In the reaction, A + B ⇌ 2C, at equilibrium the concentration of A and B is 0.20 mol L-1 each and that of C was found to be 0.60 mol L-1. The equilibrium constant of the reaction is
a) 2.4
b) 18
c) 4.8
d) 9
Answer:
d) 9

Question 58.
The rate at which substances react depends on their _______.
a) Atomic weight
b) Molecular weight
c) Equivalent weight
d) Active mass
Answer:
c) Equivalent weight

Question 59.
If in the reaction N2O4 ⇌ 2NO2, α is that part of N2O4 which dissociates, then the number of moles at equilibrium will be
a) 3
b) 1
c) (1- α)2
d) (1 + α)
Answer:
d) (1 + α)

Question 60.
On decomposition of NH4HS, the following equilibrium is established NH4HS(s) ⇌ NH3(g) + H2S(g). If the total pressure is P atm, then the equilibrium constant Kp is equal to _______.
a) P atm
b) P2 atm2
c) \(\frac{\mathrm{P}^{2}}{4}\) atm2
d) 2P atm
Answer:
c) \(\frac{\mathrm{P}^{2}}{4}\) atm2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 61.
Some gaseous equilibrium are given below:
i) CO + H2O ⇌ CO2 + H2
ii) 2CO + O2 ⇌ 2CO
iii) 2H2 + O2 ⇌ 2H3O
find out the relation between equilibrium constants.
a) K = K1K2
b) K = (K1K2)2
c) K = (K1K2)–\(\frac{1}{2}\)
d) K = \(\left(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\right)^{\frac{1}{2}}\)
Answer:
d) K = \(\left(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\right)^{\frac{1}{2}}\)

Question 62.
Kc = 9 for the reaction A + B ⇌ C + D. If A and B are taken In equal amounts, then amount of ‘C’ in equilibrium is
a) 1
b) 0.25
c) 0.75
d) none of these
Answer:
c) 0.75

Question 63.
In the reaction C(S) + CO2(g) ⇌ 2CO(g) the equilibrium pressure is 12 atm. If 50 % of CO2 reacts, then Kp will be ________.
a) 12 atm
b) 16 atm
c) 20 atm
d) 24 atm
Answer:
b) 16 atm

Question 64.
For the following gases equilibrium, N2O4(g) ⇌ 2NO2(g) Kp is found to be equal to Kc. This is attained when,
a) 0°C
b) 273 k
c) 1 k
d) 12.19 k
Answer:
d) 12.19 k

Question 65.
Consider the following reversible gaseous reactions {At 298 K)
a) N2O4 ⇌ 2NO2
b) 2SO2 + O2 ⇌ 2SO3
c) 2HI ⇌ H2 + I2
d) X + Y ⇌ 4Z
Answer:
a) N2O4 ⇌ 2NO2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 66.
For the reaction A + 2B ⇌ 2C at equilibrium [C] = 1.4 M, [A]0 = 1 M, [B]0 = 2M, [C]0 = 3 M. The value of Kc is .
a) 0.084
b) 8.4
c) 84
d) 840
Answer:
a) 0.084

Question 67.
For the reaction H2(g) + I2(g) ⇌ 2HI(g). Kc = 66.9 at 350°C and Kc = 50.0 at 448°C. The reaction has
a) ∆H = +ve
b) ∆H = -ve
c) ∆H = zero
d) ∆H = not found the signs.
Answer:
b) ∆H = -ve

Question 68.
In an equilibrium reaction H2(g) + I2(g) ⇌ 2HI(g), ∆H = -3000 calories, which factor favours dissociation of HI
a) Low temperature
b) High pressure
c) High temperature
d) Low pressure
Answer:
c) High temperature

Question 69.
In an equilibrium reaction for which ∆G°= 0, the equilibrium constant K Is
a) 0
b) 1
c) 2
d) 10
Answer:
b) 1

Question 70.
Consider the following reversible reaction at equilibrium, 2H2O(g) ⇌ 2H2(g) + O2(g). Which one of the following ∆H = 241. 7 kJ changes in a conditions will lead to maximum decomposition of H2O(g)?
a) Increasing both temperature and pressure
b) Decreasing temperature and increasing pressure
c) Increasing temperature and decreasing pressure
d) Increasing temperature at constant pressure.
Answer:
c) Increasing temperature and decreasing pressure

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 71.
In which of the following system doubling the volume of the container causes a shift to the right.
a) H2(g) + Cl2(g) ⇌ 2HCl(g)
b) 2CO(g) + O2(g) ⇌ 2CO2(g)
c) N2(g) + 3H2(g) ⇌ 2NH3(g)
d) PCl5(g) ⇌ PCl3(g) + Cl2(g)
Answer:
d) PCl5(g) ⇌ PCl3(g) + Cl2(g)

Question 72.
∆G° (HI, g) ≈ 1.7 kJ. What is the equilibrium constant at 25° C for 2HI(g) ⇌ H2(g) + I2(g)
a) 24.0
b) 3.9
c) 2.0
d) 0.5
Answer:
d) 0.5

Question 73.
In the reaction A2(g) + 4B2(g) ⇌ 2AB4 ∆H < 0 the formation of AB4 is will, be favoured at
a) Low temperature, high pressure
b) High temperature, low pressure
c) Low temperature, low pressure
d) High temperature, high pressure
Answer:
a) Low temperature, high pressure

Question 74.
N2 + 3H2 ⇌ 2NH3. If temperature of following equilibrium reaction increases then the reaction.
a) Shifts Right side
b) Shifts left side
C) Remains unchanged
d) No change
Answer:
b) Shifts left side

Question 75.
Consider the equilibrium N2(g) + 3H2(g) ⇌ 2NH3 ∆H = -93.6 kJ. The maximum yield of ammonia is obtained by
a) Decrease of temperature and increase of pressure.
b) Increase of temperature and decreases of pressure.
c) Decrease of both the temperature and pressure
d) Increase of both the temperature and pressure
Answer:
a) Decrease of temperature and increase of pressure.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 76.
H2(g) + I2(g) ⇌ 2HI(g) ∆H = +q cal, then formation of HI
a) Is favoured by lowering the temperature
b) Is favoured by increasing the pressure
c) Is unaffected by change in pressure
d) Is unaffected by change in temperature
Answer:
c) Is unaffected by change in pressure

Question 77.
The formation of SO3 takes place according to the following reaction, 2SO2 + O2 ⇌ 2SO3; ∆H = 45.2 k cal. The formation of SO3 is favoured by
a) Increasing in temperature
b) Removal of oxygen
c) Increase of volume
d) Increase of pressure
Answer:
d) Increase of pressure

Question 78.
Which of the following equilibrium is not shifted by an increase in the pressure?
a) H2(g) + 3H2(g) ⇌ 2HI(g)
b) N2(g) + 3H2(g) ⇌ 2NH3(g)
c) 2CO(g) + O2(g) ⇌ 2CO2(g)
d) 2C(g) + O2(g) ⇌ 2CO(g)
Answer:
a) H2(g) + 3H2(g) ⇌ 2HI(g)

Question 79.
Consider the heterogeneous equilibrium in a closed container. NH4HS(s) ⇌ NH3(g) + H2S(g) if more NH4HS is added to the equilibrium
a) Partial pressure of NH3 increases
b) Partial pressure of H2S increases
c) Total pressure in the container increases
d) No effect on partial pressure of NH3 and H2S
Answer:
d) No effect on partial pressure of NH3 and H2S

Question 80.
According to Le – Chatelier’s principle adding heat to a solid and liquid in equilibrium with endothermic nature will cause the ____________.
a) Temperature to rise
b) Temperature to fall
c) Amount of solid to decrease
d) Amount of liquid to decrease
Answer:
c) Amount of solid to decrease

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 81.
According to Le-chatelier’s principle, adding heat to a solid to liquid in equilibrium will causes the
a) temperature to increase
b) temperature to decrease
c) amount of liquid to increase
d) amount of solid to increase
Answer:
c) amount of liquid to increase

Question 82.
For the reaction A + B + Q ⇌ C + D, if the temperature is increased, then concentration of the products will
a) increase
b) decrease
c) Remain same
d) become zero
Answer:
a) increase

Question 83.
H2(g) + I2(g) ⇌ 2HI(g). In this reaction when pressure increases, the reaction direction
a) does not change
b) is forward
c) is backward
d) decreases
Answer:
a) does not change

Question 84.
Which of the following factor is shifted the reaction PCl3 + Cl2 ⇌ PCl5 at the left side?
a) Adding PCl5
b) increase pressure
c) constant temperature
d) catalyst
Answer:
a) Adding PCl5

Question 85.
The rate of reaction of which of the following is not affected by pressure?
a) PCl3 + Cl2 ⇌ PCl5
b) N2 + 3 H2 ⇌ 2NH3
c) N2 + O2 ⇌ 2NO
d) 2 SO2 + O2 ⇌ 2 SO3
Answer:
c) N2 + O2 ⇌ 2NO

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 86.
Which reaction is not affected by change in pressure?
a) H2 + I2 ⇌ 2HI
b) 2 C + O2 ⇌ 2CO
c) N2 + 3 H2 ⇌ 2NH3
d) PCl5 ⇌ PCl3 + Cl2
Answer:
a) H2 + I2 ⇌ 2HI

Question 87.
Which of the following reactions proceeds at low pressure?
a) N2 + 3H2 ⇌ 2NH3
b) H2 + I2 ⇌ 2HI
c) PCl5 ⇌ PCl3 + Cl2
d) N2 + O2 ⇌ 2NO
Answer:
c) PCl5 ⇌ PCl3 + Cl2

Question 88.
In the following reaction PCl5 ⇌ PCl3(g) + Cl2(g) at constant temperature, rate of backward reaction, is increased by
a) inert gas mixed at constant volume
b) Cl2 gas mixed at constant volume
c) inert gas mixed at constant pressure
d) PCl5 mixed in constant volume
Answer:
b) Cl2 gas mixed at constant volume

Question 89.
On cooling of following system at equilibrium CO2(s) ⇌ CO2(g)
a) There is no effect on the equilibrium state
b) more gas is formed
c) more gas solidifies
d) None
Answer:
c) more gas solidifies

Question 90.
On the velocity in a reversible reaction the correct explanation of the effect of catalyst is _______.
a) it provides a new reaction path of low activation energy
b It increases a kinetic energy of the reacting molecules
c) it displaces the equilibrium state on right side
d) it decreases the velocity of backward reaction
Answer:
a) it provides a new reaction path of low activation energy

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 91.
According to Le – Chatelier’s principle, if heat is given to solid – liquid system, then
a) Quantity of solid will reduce
b) Quantity of liquid will reduce
c) increase in temperature
d) decrease in temperature
Answer:
a) Quantity of solid will reduce

Question 92.
In a gives system, water and ice are in equilibrium, if pressure is applied to the system then .
a) more of ice is formed
b) amount of ice and water will remain same
c) more of ice is melted
d) either (a) or (c)
Answer:
c) more of ice is melted

Question 93.
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 22
Answer:
b) Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 23

Question 94.
The value of ∆G° for a reaction in aqueous phase having Kc = 1, would be
a) – RT
b) -1
c) 0
d) + RT
Answer:
c) 0

Question 95.
The equilibrium constants for the reaction, Br2 ⇌ 2 Br at 500 K and 700 K are 1 × 10-10 and 1 × 10-5 respectively. The reaction is
a) endothermic
b) exothermic
c) fast
d) slow
Answer:
a) endothermic

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 96.
In the reaction A2(g) + 4 B2(g) ⇌ 2AB4(g), ∆H < 0, the decomposition of AB4(g) will be favoured at
a) low temperature and high pressure
b) high temperature and low pressure
c) low temperature and low pressure
d) high temperature and high pressure
Answer:
c) low temperature and low pressure

Question 97.
The equilibrium constant for the reaction N2(g) + O2(g) ⇌ 2NO (g) is 4 × 10-4 at 200 K. In the presence of a catalyst the equilibrium is attained 10 times faster. Therefore the equilibrium constant in pressure of the catalyst at 200 K is
a) 4 × 10-3
b) 4 × 10-4
c) 4 × 10-5
d) None
Answer:
b) 4 × 10-4

Question 98.
Change in volume of the system does not alter the number of moles in which of the following equilibrium.
a) N2(g) + O2(g) ⇌ 2 NO (g)
b) PCl5(g) ⇌ PCl3(g) + Cl2(g)
c) N2(g) + 3H2(g) ⇌ 2 NH3(g)
d) SO2Cl2(g) ⇌ SO2(g) + Cl2(g)
Answer:
a) N2(g) + O2(g) ⇌ 2NO2 (g)

Question 99.
In which of the following equilibrium reactions the equilibrium would shift to the right, if total pressure is increased
a) N2 + 3H2 ⇌ 2 NH3
b) H2 + I2 ⇌ 2HI
c) H2 + Cl2 ⇌ 2 HCl
d) N2O4 ⇌ 2NO
Answer:
a) N2 + 3H2 ⇌ 2 NH3

Question 100.
The graph in K (eq) vs \(\frac{1}{T}\) relates for a reaction must be
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 24
a) Exothermic
b) Endothermic
c) ∆H is negligible
d) High spontaneous at ordinary temperature
Answer:
a) Exothermic

Question 101.
Which of the following statements is cor¬rect for a reversible process in state of equilibrium?
a) ∆G° = -2.30 RT log K
b) ∆G° = 2.30 RT log K
c) ∆G = -2.30 RT log K
d) ∆G = 2.30 RT log K
Answer:
a) ∆G° = -2.30 RT log K

Question 102.
Which of the following is a characteristic of a reversible reaction?
a) Number of moles of reactants and products are equal
b) it can be influenced by a catalyst
c) it can never proceed to completion
d) none of the above
Answer:
c) it can never proceed to completion

Question 103.
The equilibrium constant in a reversible reaction at a given temperature
a) depends on the initial concentration of the reactants
b) depends on the concentration of the products at equilibrium
c) does not depend on the initial concentration
d) it is not characteristic of the reaction
Answer:
c) does not depend on the initial concentration

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

II. Very short question and answers(2 Marks):

Question 1.
What are reversible reactions?
Answer:
Reactions in which the products can react to give back reactants under suitable conditions are called reversible reactions.

Question 2.
What are irreversible reactions?
Answer:
Reactions in which the products formed do not react to give back reactants under normal conditions are called irreversible reactions.

Question 3.
Why the chemical equilibrium is referred to as Dynamic Equilibrium?
Answer:
At equilibrium the forward and the backward reactions are proceeding at the same rate and no macroscopic change is observed. So chemical equilibrium is often called as dynamic equilibrium.

Question 4.
What is equilibrium constant?
Answer:
The equilibrium constant Kc may be defined as the ratio of product of the active masses of the products to that of the reactants.

Question 5.
Write the Applications of equilibrium constant?
Answer:
The knowledge of equilibrium constant helps us in

  • predicting the extent of a reaction, predicting the direction in which the net reaction is taking place and
  • calculation of equilibrium constants and concentration of the reactants and products.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
What is Le – Chatelier’s principle?
Answer:
It states that “If a system at equilibrium is subjected to a change of concentration, pressure or temperature, then the equilibrium is shifted in such a ways to nullify the effect of that disturbance.

Question 7.
For the H2(g) + I2(g) ⇌ 2HI(g) reaction the value of Kc at 717 K is 48. If at a particular instant, the concentration of H2, I2, and HI is 0.2 mol lit-1 and 0.6 mol lit-1 respectively, then calculate Q and predict the direction of reaction.
Solution:
Q = \(\frac{\left[\mathrm{H}_{1}\right]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}=\frac{0.6 \times 0.6}{0.2 \times 0.2}\) = 9
Q < Kc, the reaction will proceed in the forward direction.

Question 8.
Define Degree of dissociation.
Answer:
The fraction of the total number of moles of the reactants are dissociated.
Degree of dissociation ‘x’ =
‘x’ = \(\frac{\text { no.of moles dissociated }}{\text { Total no. of moles present initially }}\)

Question 9.
Dissociation of PCl5 decreases in presence of increase in Cl2, Why?
Answer:

  1. The dissociation equilibrium of PCl5 in gaseous state is written as PCl5(g) ⇌ PCl3(g) + Cl2(g)
  2. In a chemical equilibrium increasing the concentration of the products results in shifting the equilibrium in favour of the reactants. So in the presence of increased Cl2 reversed reaction only favoured.
  3. Hence dissociation of PCl5 decreases in presence of increase in Cl2.

Question 10.
The equilibrium constant for Kc for A(g) ⇌ B(g) is 2.5 × 10-2 . The rate constant of the forward reaction is 0.05 sec-1. Calculate the rate constant of the reverse reaction.
Answer:
Kc = 2.5 × 10-2;
Kf = 0.05 sec-1
Kc = \(\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{r}}}\)
= \(\frac{0.05}{2.5 \times 10^{-2}}\) = 2
Rate constant of the reverse reaction,
Kr = 2 sec-1.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 11.
The equilibrium constant for the reaction 2SO3(g) ⇌ 2SO2(g) + O2(g) is 0.15 mol dm-3 at 900K. Calculate the equilibrium constant for the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) at the same temperature. 2SO3(g) ⇌ 2SO2(g) + O2(g).
Answer:
Kc = \(\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}}\) = 0.15
K’ = \(\frac{1}{\mathrm{~K}_{\mathrm{c}}}=\frac{1}{0.15}\)
= 6.6 mole dm-3
K’ = 6.6 mole dm-3

III. Short question and answers(3 Marks):

Question 1.
What is solid – liquid equilibrium? Give example.
Answer:

  • The equilibrium between solid and liquid at its melting point (or freezing point) at normal atmospheric pressure is called solid-liquid equilibrium.
  • If some ice-cubes along with water are placed in a thermos flask (at 00 C and 1 atm), then there will be no change in the mass of ice and water.
  • At equilibrium,
    Rate of melting of ice = Rate of freezing of water
    H2O(s) ⇌ H2O(l)

Question 2.
What is liquid – vapour equilibrium?
Answer:
The equilibrium of a liquid with its gas at its boiling point at 1 atm pressure is called liquid – vapour equilibrium.
Rate of evaporation = Rate of condensation
H2O(l) ⇌ H2O(g).

Question 3.
What is Solid – vapour equilibrium?
Answer:

  • When solid iodine is placed in a closed transparent vessel, after sometime the vessel gets filled up with violet vapour.
  • The intensity of the colour increases and finally becomes constant, i.e., equilibrium is attained.
    I2(s) ⇌ I2(g)

Question 4.
What are Homogeneous and Heterogeneous equilibriums?
Answer:
Homogeneous equilibrium:
In a homogeneous equilibrium, all the reactants and products are in the same phase.
Example:
H2(g) + I2(g) ⇌ 2HI (g)
In the above equilibrium, H2, I2, and HI are in the gaseous state.
Similarly, for the following reactions,
CH3COOCH3(aq) + H2O(aq) ⇌ CH3COOH (aq) + CH3OH (aq) all the reactants and products are in homogenous solution phase.

Heterogeneous equilibrium:
If the reactants and products are present in different phases, it is known as heterogeneous equilibrium.
Example:
H2O(l) ⇌ H2O (g)
CaCO3(s) ⇌ CaO(s) + CO2(g).

Question 5.
What is Law of mass action?
Answer:
“At any instant, the rate of a chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants”.
Rate ∝ [Reactants]x
[ ] = \(\frac{\mathrm{n}}{\mathrm{v}}\), {Unit is mol.dm-3 or mol.L-1}
where,
n is the number of moles and
V is the volume of the container (dm3 or L).

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
What is the reaction quotient? Write the relation of K & Q.
Answer:
Reaction quotient ‘Q’ is defined as the ratio of the product of active mass of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants under non equilibrium.
Reaction quotient Q.
Q = \(\frac{[\mathrm{C}]^{1}[\mathrm{D}]^{\mathrm{m}}}{[\mathrm{A}]^{\mathrm{x}}[\mathrm{B}]^{\mathrm{y}}}\)
If Q = Kc, the reaction is in equilibrium
If Q > Kc, the reaction will proceed in the reverse direction i.e., formation of reactants.
If Q < Kc, the reaction will proceed in the forward direction i.e., formation of products.

Question 7.
Explain effect of pressure of formation of HI.
Answer:
Pressure has no effect on the equilibrium when the total number of the moles of the gaseous reactants and the gaseous products are equal. ( ∆ng = 0).
Therefore, pressure has effect only on the equilibrium with ∆ng = 0.
Let us consider the following reactions
H2(g) + I2(g) ⇌ 2HI(g)
N2(g) + O2(g) ⇌ 2NO(g)
Here,
The number of moles on reactants and products are equal. So pressure has no effect on them.

Question 8.
The equilibrium concentrations of NH3, N2 and H2 are 1.8 × 10-2 M, 1.2 × 10-2 M and 3 × 10-2 M respectively. Calculate the equilibrium constant for the formation of NH3 from N2 and H2. [Hint: M= mol lit-1]
Solution:
Given data:
[NH3] = 1.8 × 10-2 M ;
[N2] = 1.2 × 10-2 M ;
[H2] = 3 × 10-2 M ;
Kc = ?
N2(g) + 3H2(g) ⇌ 2NH3(g)
Kc = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}\)
Kc = \(\frac{1.8 \times 10^{-2} \times 1.8 \times 10^{-2}}{1.2 \times 10^{-2} \times 3 \times 10^{-2} \times 3 \times 10^{-2} \times 3 \times 10^{-2}}\)

Kc = 1 × 10-2 l2 mol-2

Question 9.
One mole of H2 and one mole of I2 are allowed to attain equilibrium. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.
Answer:
Given data:
[H2] = 1 mole;
[I2] = 1 mole
At equilibrium, [HI] = 0.4 mole; Kc = ?
H2(g) + I2 ⇌ 2HI(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 25

Question 10.
Relate the three equilibrium constants.
i) N2 + O2 ⇌ 2NO; K1
ii) 2NO + O2 ⇌ 2NO2; K2
iii) N2 + 2O2 ⇌ 2NO2; K3
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 26

Question 11.
Write the equilibrium constant for the following:
i)H2O2(g) ⇌ H2O + 1/2O2(g)
ii) CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
iii) N2O4(g) ⇌ 2NO2(g)
Answer:
i)H2O2(g) ⇌ H2O + 1/2O2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 27

ii) CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 28

iii) N2O4(g) ⇌ 2NO2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 29

Question 12.
Calculate the ∆ng, for the following reactions:
i) H2(g) + I2(g) ⇌ 2HI
ii) 2H2O(g) + 2Cl2(g) ⇌ 4HCl(g) + O2(g)
∆ng = (Total number of moles of gaseous products) – (Total number of moles of gaseous reactants)
Answer:
i) H2(g) + I2(g) ⇌ 2HI
∆ng = 2 – 2 = 0
∆ng = 0

ii) 2H2O(g) + 2Cl2(g) ⇌ 4HCl(g) + O2(g)
∆ng = 5 – 4 = 1
∆ng = 1

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 13.
For the reaction A+B ⇌ 3C at 25°C, a 3 litre volume reaction vessel contains 1, 2, and 4 moles of A, B, and C respectively at equilibrium, calculate the equilibrium constant Kc of the reaction at 25°C.
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 30

Question 14.
At temperature T1 the equilibrium constant of reaction is K1 At a higher temperature T2, K2 is 10% of K1. Predict whether the equilibrium is endothermic or exothermic.
Answer:
At temperature, T1, the equilibrium constant is K1
At a high temperature, T2, the equilibrium constant K2 = 10%KX
K2 = \(\frac{10}{100}\) K1
K2 = 0.1 K1
K1 > K2
Hence, the equilibrium is exothermic.

Question 15.
Calculate the Kc when a mixture containing 8.07 moles of H2 and 9.08 moles of I2 are reacted at 448°C until 13.38 moles of HI was formed at the equilibrIum.
Answer:
The equilibrium reaction is,
H2(g) + I2(g) ⇌ 2HI(g)
Kc = \(\frac{(2 x)^{2}}{(a-x)(b-x)}\)
a = 8.07 moles;
b = 9.08 moles
2x = 13.38 moles
x = 6.69 moles
Kc = \(\frac{4(6.69)^{2}}{(8.07-6.69)(9.08-6.69)}\)

= \(\frac{4 \times 44.75}{1.38 \times 2.39}\)
Kc = 54.29

Question 16.
For the equilibrium 2NOCl(g) ⇌ 2NO(g) + Cl2(g) the value of the equilibrium constant Kc is 3.75 × 10-6 at 790°C. Calculate Kp for this equilibrium at the same temperature. [Hint: Kp = Kc(RT)∆ng]
Answer:
Kc = 3.75 × 10-6;
T = 790 + 273 = 1063 K
Kp = Kc × (RT)∆ng
For the reaction,
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
∆ng = 3 – 2 = 1
Kp = 3.75 × 10-6 × (8.314 × 1063)
= 33141.6 × 10-6
= 3.31 × 104 × 10-6
Kp = 3.31 × 10-2

Question 17.
How much PCl5 must be added to one litre volume reaction vessel at 250°C in order to obtain a concentration of 0.1 mole of Cl2, Kc for PCl5 ⇌ PCl3 + Cl2 is 0.0414 mol dm-3 at 250°C.
Answer:
PCl5 ⇌ PCl3 + Cl2
Number of moles of PCl5 at equilibrium = a – 0.1
Number of moles PCl3 at equilibrium = 0.1
Number of moles Cl2 at equilibrium = 0.1
Kc = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
= 0.0414
= \(\frac{0.1 \times 0.1}{a-0.1}\)
0.01 = (a – 0.1) 0.0414
0.01 = 0.0414a – 0.00414
0.0414a = 0.01414
a = 0.3415 mole

Question 18.
At 540 K, the equlllbrium constant Kp for PCl5 dissociation equilibrium at 1.0 atm is 1.77 atm. Calculate equilibrium constant in molar concentration (Kc) at same temperature and pressure.
Answer:
Kp = 1.77 atm;
T = 540 K;
a = 0.082
PCl5 ⇌ PCl3 + Cl2
∆ng = 2 – 1 = 1
Kp = Kc(RT)∆ng
1.77 = Kc × (0.0821 × 450)1
Kc = \(\frac{1.77}{(0.0821 \times 540)}\)
= 4 × 10-2 moles/litres.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

III. Long question and answers (5 Marks):

Question 1.
Derive the Equilibrium constants Kp and K for the homogeneous reactions.
Answer:
Let us consider a reversible reaction,
xA + yB ⇌ lC + mD
where A and B are the reactants, C and D are the products, and x, y, l, and m are the coefficients of A, B, C, and D respectively.
Applying Law of mass action, the rate of the forward reaction,
rf α [A]x [B]y (or)
rf = kf [A]x [B]y
Similarly, the rate of the backward reaction,
rb a [C]l [D]m (or)
rb = k, [C]l [D]m
where kf and kb are proportionality constants
At equilibrium, Rate of forward reaction(rf) = Rate of backward reaction(rb)
kf α [A]x [B]y = kb a [C]l [D]m (or)
\(\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{b}}}=\frac{[\mathrm{C}]^{l}[\mathrm{D}]^{m}}{[\mathrm{~A}]^{x}[\mathrm{~B}]^{y}}\) = Kc
where, Kc is the equilibrium constant in terms of concentration.

At a given temperature, the ratio of the product of active masses of products to that the reactants is a constant, known as equilibrium constant. Each concentration term is raised to the power which is equal to the respective stoichiometric coefficients in the balanced chemical equation. Later when we study chemical kinetics we will learn that this is only approximately true. If the reactants and products behave as gases, then the equilibrium constant in terms of partial pressures can be written as

\(\frac{\mathrm{P}_{\mathrm{C}}^{l} \times \mathrm{P}_{\mathrm{D}}^{m}}{\mathrm{P}_{\mathrm{A}}^{x} \times \mathrm{P}_{\mathrm{B}}^{y}}\)

where, Pc is the partial pressure of the gas, C and so on.

Question 2.
Derive Kp & Kc relation for the formation of HI.
Answer:
Synthesis of HI:
Consider that ‘a’ moles and ‘b’ moles are the initial concentration of H2 and I2 respectively, taken in a container of volume V. Let ‘x’ moles of each of H2 and I2 react together to form 2x moles of HI.
H2(g) + I2(g) ⇌ 2HI(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 31
The equilibrium constant Kp can also be calculated as follows:
Total number of moles at equilibrium = a – x + b – x + 2x = a + b.
If the total pressure of the system at equilibrium be P, then
Partial pressure of H2,
pH2 = \(\left(\frac{a-x}{a+b}\right)\)P

Partial pressure of I2,
pI2 = \(\left(\frac{b-x}{a+b}\right)\)P

Partial pressure of HI,
pHI = \(\left(\frac{2x}{a+b}\right)\)P

Kp = \(\frac{\left(\frac{2 x}{a+b}\right)^{2} \mathrm{P}^{2}}{\left(\frac{a-x}{a+b}\right) \mathrm{P}\left(\frac{b-x}{a+b}\right) \mathrm{P}}=\frac{4 x^{2}}{(a-x)(b-x)}\)
In the above equilibrium, Kc and Kp are thus shown to be equal.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 3.
Derive the equilibrium constant Kp & Kc for the Dissociation of PCl5.
Answer:
Dissociation of PCl5:
Consider that ‘a’ moles is the initial concentration of PCl5 taken in a container of volume V. Let ‘x’ moles of PCl5 be dissociated into x moles each of PCl3 and Cl2.
PCl5(g) ⇌ PCl3(g) + PCl2(g)
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 32
The equilibrium constant Kp can also be calculated as follows:
Total number of moles at equilibrium = a – x + x + x = a + x
If the total pressure of the system at equilibrium be P, then
Partial pressure of PCl5,
pPCl5 = \(\left(\frac{a-x}{a+x}\right)\)P;

Partial pressure of PCl3,
pPCl3 = \(\left(\frac{x}{a+x}\right)\)P;

Partial pressure of Cl2
pCl2 = \(\left(\frac{x}{a+x}\right)\)P

Kp = \(\frac{\left(\frac{x}{a+x}\right) \mathrm{P}\left(\frac{x}{a+x}\right) \mathrm{P}}{\left(\frac{a-x}{a+x}\right) \mathrm{P}}=\frac{x^{2} \mathrm{P}}{(a-x)(a+x)}\)

Question 4.
Explain the effect of Concentration of formation of HI?
Answer:
Effect of concentration:
At equilibrium, the concentration of the reactants and the products does not change. The addition of more reactants or products to the reacting system at equilibrium causes an increase in their respective concentrations.

According to Le Chatelier’s principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance. Suppose at equilibrium if one of the reactants is added, then the equilibrium will shift in the forward direction and if one of the products is added, then the equilibrium will shift in backward direction.
Let us consider the reaction,
H2(g) + I2(g) ⇌ 2HI(g)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 33

On addition of H2 or I2 to the equilibrium mixture product HI will tend to increase. In order to minimize the stress, the system shifts the reaction in a direction where H2 and I2 are consumed i.e., the formation of HI would relieve the stress. Hence, the equilibrium shifts to the right (forward direction).

Similarly, removal of HI (product) also boosts the forward reaction and increases the concentration of products.
If HI is added to the equilibrium mixture, the concentration HI is increased, then the system proceeds in the reverse direction to nullify the stress.

Question 5.
The equilibrium constant at 298 K for a reaction is 100. A + B ⇌ C + D. If the initial concentration of all the four species were 1 M each what will be equilibrium concentration of D (in mol lit-1)
Solution:
Given data:
[A] = [B] = [C] = 1 M;
Kc = 100; [D] = ?
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 34
10(1 – x) = 1 + x
10 – 10x – 1 – x = 0
9 – 11x = 0
11x = 9
x = 9/11 = 0.818
So, [D] at equilibrium = 1 + 0.818 = 1.818 M

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 6.
What happens when ∆ng = 0, ∆ng = -ve, ∆ng= +ve In a gaseous reaction?
Answer:
(i) When ∆ng = 0, the total number of moles of gaseous products are equal to the total number of moles of gaseous reactants.
Eg:
H2(g) + I2(g) ⇌ 2HI(g)
∆ng = 2 – 2 = 0
Kp = Kc(RT)∆ng
If ∆ng = 0,
Kp = Kc(RT)0
∴ Kp = Kc

(ii) When ∆ng = +ve, the total number of moles of gaseous products are greater than the total number of moles of gaseous reactants.
Eg:
2H2O + 2Cl2(g) ⇌ 4HCl(g) + O2(g)
∆ng = 5 – 4 = 1
Kp = Kc(RT)1
Kp = Kc(RT)
∴ Kp > Kc

(iii) When ∆ng = -ve, the total number of moles of gaseous products are greater than the total number of moles of gaseous reactants.
N2(g) + 3H2(g) ⇌ 2NH3(g)
∆ng = 2 – 4 = -2
Kp = Kc(RT)-2

Kp = \(\frac{K_{c}}{(R T)^{2}}\)

∴ Kp < Kc

Question 7.
What is the relationship between the dissociation and formation equilibrium constant?
Answer:
In a chemical equilibrium reaction, the equilibrium constant for the dissociation equilibrium reaction which is also known as dissociation constant is found to be the reciprocal value of the equilibrium constant for the formation equilibrium reaction.
Formation of equilibrium reaction,
2SO2(g) + O2 ⇌ 2SO3
Kc = \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}\) dm3 /mole.

Dissociation equilibrium reaction of SO3 the reactants became products and vice versa.
2SO3(g) ⇌ 2SO2(g) + O2(g)
Kc‘ = \(\frac{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{SO}_{3}\right]^{2}}\) mole/dm3
= \(\frac{1}{\mathrm{~K}_{\mathrm{c}}}\)

Kc‘ is considered as the dissociation constant of SO3 gas. Usually, the equilibrium constant of the dissociation equilibrium is the reciprocal of the equilibrium constant of the formation equilibrium reaction.

Question 8.
In the equilibrium H2 + I2 ⇌ 2HI, the number of moles of H2, I2, and HI are 1, 2, 3 moles respectively. Total pressure of the reaction mixture is 60 atm. Calculate the partial pressures of H2, I2, and HI in the mixture.
Solution:
H2 + I2 ⇌ 2HI
Number of moles of H2 at equilibrium = 1 mole
Number of moles of I2 at equilibrium = 2 mole
Number of moles of HI at equilibrium = 3 mole
Total number of moles of equilibrium = 1 + 2 + 3 = 6 moles
Mole fraction of H2 = 1/6;
Mole fraction of I2 = 2/6;
Mole fraction of HI = 3/6
Partial pressure = Mole fraction × Total pressure
PH2 = \(\frac{1}{6}\) × 60 = 10 atm;
PI2 = \(\frac{2}{6}\) × 60 = 20 atm;
PHI = \(\frac{3}{6}\) × 60 = 30 atm.

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 9.
In 1 litre volume reaction vessel, the equilibrium constant K of the reaction PCl5 ⇌ PCl3 + Cl2 is 2 × 10-4 What will be the degree of dissociation assuming only a small extent of 1 mole of PCl5 has dissociated?
Answer:
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 35
since ‘x’ is very small compared to 1,
then K2 = x2
2 × 10-4 = x2
Degree of dissociation x = \(\sqrt{2 \times 10^{-4}}\)
= 1.414 × 10-2

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium

Question 10.
At 35°C, the value of K for the equilibrium reaction N2O4 ⇌ 2NO2 is 0.3174, Calculate the degree of dissociation when P is 0.2382 atm.
Answer:

N2O4 ⇌ 2NO2 N2O4 NO2
Initial number of moles 1 0
Number of moles dissociated x 2x
Number of moles  equilibrium 1 – x 2x

Total number of moles at equilibrium = 1 – x + 2x = 1 + x
Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 36

Question 11.
Degree of dissociation of PCl5 at 1 atm and 25°C is 0.2. Calculate the pressure at which PCl5 is half dissociated at 25°C.
Answer:
For PCl5 dissociation equilibrium,
Kp = \(\frac{x^{2} p}{1-x^{2}}\) [ P = total pressure = 1 atm]
x = 0.2
Kp = \(\frac{(0.2)^{2}(1.0)}{1-(0.2)^{2}}\)
= \(\frac{0.04}{1-0.04}\)

Samacheer Kalvi 11th Chemistry Guide Chapter 8 Physical and Chemical Equilibrium 37

Question 12.
Initially, 0.1 moles each of H2 and I2 gases and 0.02 moles of HI gas are mixed in a reaction vessel of constant volume at 300K. Predict the direction towards which the reaction proceeds [K = 3.5 x 10-2].
Answer:
The formation of equilibrium is
H2(g) + I2(g) ⇌ 2HI(g)
Kc = \(\frac{[\mathrm{HI}]_{\mathrm{e}}^{2}}{\left[\mathrm{H}_{2}\right]_{\mathrm{e}}\left[\mathrm{I}_{2}\right]_{\mathrm{e}}}\)
= 3.5 × 10-2 at 300 K.

Under non equilibrium conditions;
Q = \(\frac{[\mathrm{HI}]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}\)

= \(\frac{\left(2 \times 10^{-2}\right)^{2}}{0.1 \times 0.1}\)

= 4 × 10-2
Thus Q > Kc.
Therefore, the reaction will proceed initially before attaining the equilibrium in a direction such that the Q value is reduced. That is, concentrations of H2 and I2 should be increases, which is a reverse reaction. The Traction proceeds in the left side of fomiation equilibrium of HI and HI decomposes initially to H2 and I2 until Q = Kc.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Bio Botany Guide Pdf Chapter 8 Biomolecules Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 8 Biomolecules

11th Bio Botany Guide Biomolecules Text Book Back Questions and Answers

Part I

Question 1.
The most basic amino acid is
a. Arginine
b. Histidene
c. Glycine
d. Glutamine
Answer:
c. GIycine

Question 2.
An example of feed back inhibition is
a. cyanide action on cytochrome
b. Sulpha drug on folic acid
c. Allosteric inhibition of hexokinase by glucose- 6- phosphate
d. The inhibition of succinic dehydrogenase by malonate
Answer:
c. Allosteric inhibition of hexokinase by glucose-6-phosphate

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Enzymes the catalyse interconversion of optical, geometrical or positional isomers are
a. Ligases
b. Lyases
c. Hydrolases
d. Isomerases
Answer:
d. Isomerases

Question 4.
Proteins perform many physiological functions, for example, some functions as enzymes one of the following represents an additional function that some proteins discharge
a. Antibiotics
b. Pigment conferring colour to skin
c. Pigments making colours of flowers
d. Hormones
Answer:
d.Hormones

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Given below is the diagrammatic representation of one of the categories of small molecular weight organic compounds in the living tissues. Identify the category shown &one Blank component ‘X’ in it.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 1

Category   Compound
I. Cholesterol A. Guanine
II. Amino acid B. IVH2
III. Nucleotide C. Adenine
IV. Nucleoside D. Uracil

Answer:
IV

Question 6.
Distinguish  between Nitrogen base and a base found in inorganic chemistry
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 2
Question 7.
What are the factors affecting the rate of enzyme reactions?
Answer:
Enzymes being bio-molecules sensitive to environmental condition

(i) Temperature

  • Heating increases molecular motion-quicken enzyme reaction
  • Optimum temperature is the temperature that promotes maximum activity

(ii) pH

  • Change in the pH – leads to an alteration of enzyme shape (active site)
  • Extremes of pH’ denatures enzymes
  • Optimum pH’ is that at which the maximum rate of reaction occurs

(iii) Substrate concentration
For a given enzyme concentration, the rate of reaction increase with increasing substrate concentration

(iv) Enzyme concentration
The rate of reaction is directly proportional to enzyme concentration.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 8.
Briefly outline the classification of enzymes?
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 3

Question 9.
Write down the characteristic features of DNA?
Answer:
The characteristic feature of DNA.

  1. If one strand runs in the 5′ – 3′ direction, the other runs in 3′ – 5′ direction and thus are antiparallel (they run in the opposite direction). The 5′ end has the phosphate group and 3’end has the OH group.
  2. The angle at which the two sugars protrude from the base pairs is about 120°, for the narrow-angle and 240° for the wide-angle. The narrow-angle between the sugars generates a minor groove and the large angle on the other edge generates major groove.
  3. Each base is 0.34 nm apart and a complete turn of the helix comprises 3.4 nm or 10 base pairs per turn in the predominant B form of DNA.
  4. DNA helical structure has a diameter of 20 Å and a pitch of about 3 Å. X-ray crystal study of DNA takes a stack of about 10 bp to go completely around the helix (360°).
  5. Thermodynamic stability of the helix and specificity of base pairing includes
    • The hydrogen bonds between the complementary bases of the double helix
    • stacking interaction between bases tend to stack about each other perpendicular to the direction of the helical axis.
  6. Electron cloud interactions (\({ \Pi -{ \Pi } }\)) between the bases in the helical stacks contribute to the stability of the double helix.
  7. The phosphodiester linkages give an inherent polarity to the DNA helix. They form strong covalent bonds, gives strength and stability to the polynucleotide chain.
  8. Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure. Whereas in paranemic coiling the two strands simply lie alongside one another, making them easier to pull apart.
  9. Based on the helix and the distance between each turn, the DNA is of three forms – A DNA, B DNA and Z DNA.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 10.
Explain the structure and function of different types of RNA?
Answer:
I. mRNA (messenger RNA)

  • single-stranded
  • carries a copy of instructions to carry out amino acid assembling &protein synthesis
  • unstable
  • 5% of total RNA
  • In Prokaryotes – it is (polycistronic carrying coding sequence for many polypeptides
  • Eukaryotes – (monocistronic) contain information for only one polypeptide

II. tRNA (transfer RNA)

  • single-stranded clover-shaped with 4 arms highly folded -3 D structure
  • translates the code from mRNA and transfers amino acid to ribosomes (to built proteins)
  • unstable (also known as soluble RNA)
  • 15% of total RNA

III. rRNA (ribosomal RNA)

  • single-stranded
  • make up the 2 subunits of ribosomes
  • metabolically stable
  • 80% total RNA
  • A polymer with varied length from 120 – 3000 nucleotides & give ribosomes their shape
  • Genes of rRNA employed for phylogenetic studies

Part II

11th Bio Botany Guide Biomolecules Additional Important Questions and Answers

I Choose the right answer.

Question 1.
Who invented the electron microscope? (2010 AIIMS, 2008 JIPMER)
(a) Janssen
(b) Edison
(c) Knoll and Ruska
(d) Landsteiner
Answer:
(c) Knoll and Ruska

Question 2.
Polysaccharides also called
a. Polymers
b. Glycans
c. Glycosidic compounds
d. Glycones
Answer:
b. Glycans

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Omnis – cellula – e – cellula was given by ……………. (2007 AIIMS)
(a) Virchow
(b) Hooke
(c) Leeuwenhoek
(d) Robert Brown
Answer:
(a) Virchow

Question 4.
Nitrocellulose is used in making
a. cellophane
b. drapers
c. explosives
d. pain balms
Answer:
c. explosives

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Genes present in the cytoplasm of eukaryotic cells are found in ……………. (2006 AIIMS)
(a) mitochondria and inherited via egg cytoplasm
(b) lysosomes and peroxisomes
(c) Golgi bodies and smooth endoplasmic reticulum
(d) Plastids inherited via male gametes
Answer:
(a) mitochondria and inherited via egg cytoplasm

Question 6.
Chitin when added with amino acid becomes
a.myeopolysaccharide
b. amylopolysaceharide
c .mucopolysaccharide
d. peptidopolysaccharide
Answer:
c. mucopolysaccharide

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 7.
A quantosome is present in …………… . (JIPMER 2012)
(a) Mitochondria
(b) Chloroplast
(c) Golgi bodies
(d) ER
Answer:
(b) Chloroplast

Question 8.
Among the following one is not a non-polar solvent
a. benzene
b. sulphuric acid
c. ether
d. chloroform
Answer:
b. sulphuric acid

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 9.
One of the given below is a complex found in the cell membrane of animal cell
a. cholesterol
b. myelin
c. proline
d. Ieeithin
Answer:
a. cholesterol

Question 10.
A major site for the synthesis of lipids ……………. (2013 NEET)
(a) Rough ER
(b) smooth ER
(c) Centriole
(d) Lysosome
Answer:
(b) smooth ER

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 11.
Principle information molecules of the cell are known as
a. Nucleus
b. DNA
c. RNA
d. Nucleic acids
Answer:
d. Nucleic acids

Question 12.
(I) Cellulose – A most abundant organic compound
(II) Morphine – Pain relieving alkaloid
(III) Aldose – reducing sugar & Ketose
(IV) Glycogen – mucopolysaccharide
Answer:
(IV) Glycogen – mucopolysaccharide

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 13.
The following is a general formula of
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 4
a. Amino acid
b. Fatty acid
c. Nucleotide
d. Monosaccharide
Answer:
a. Amino acid

Question 14.
Lactose is a disaccharide of
a. Glucose – Glucose
b. Fructose – Fructose
c. Glucose – Galactose
d. Fructose – Galactose
Answer:
c. Glucose – Galactose

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 15.
Number of fatty acids in triglyceride is …………… .
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 16.
Heparin the anti-coagulant is got from
a. D – glucuronic acid
b. Polymer of fructose
c. Mucopolysaccharide from red algae
d. Glucosaminoglycan
Answer:
d. Glycosaminoglycan

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 17.
The pH at which Zwitterion is formed is known as
a. Iso ionic balance
b. Isoelectric potential
c. Isoelectric point
d. Iso ionic point
Answer:
c. Isoelectric point

Question 18.
Aspartate and Glutamate are amino acids of
a. Negatively charged ‘R’ groups
b. Positively charged ‘R’ groups
c. Non-polar aliphatic ‘W groups
d. Non-polar aromatic ‘R’ groups
Answer:
a. Negatively charged ‘R’ groups

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 19.
The test for protein ¡s
a. iodine test
b. Biuret test
c. Benedict’s test
d. Hydrolysis test
Answer:
b. Biuret test

Question 20.
The competitive inhibitor is …………… for succinic dehydrogenase.
(a) malonate
(b) succinate
(c) oxalate
(d) citrate
Answer:
(a) malonate

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 21.
Formation of new chemical bonds using ATP as a source of energy
a. Lyase
b. Hydrolase
c. Telomerase
d. Ligase
Answer:
d. Ligase

Question 22.
Uridylic acid is an
a. Dinucleotide
b. Nucleoside
c. Nucleotide
d. Ribo nucleotide
Answer:
d. Ribonucleotide

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 23.
Phosphate forming linkage with sugar is known as
a. diester linkage
b. peptide linkage
c. phosphodiester linkage
d. Ionic linkage
Answer:
c. phosphodiester linkage

Question 24.
…………… is a catalytic RNA.
(a) mRNA
(b) Ribozyme
(c) Ribonuclease
(d) rRNA
Answer:
(b) Ribozyme

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 25.
A class of lipid that serves as a major component of the cell membrane is
a. triglyceride
b. glycerol
c. phospho lipid
d. lipoprotein
Answer:
c. phospholipid

Question 26.
One molecule of sucrose on hydrolysis give
a. 2 molecules of glucose
b. 1 molecule glucose & 1 molecule fructose
c. 2 molecules of glucose & 1 molecule of fructose
d. 2 molecules of fructose
Answer:
b. 1 molecule glucose & 1 molecule fructose

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 27.
In fibrous proteins polypeptide chai are held together by
a. Vander Waals forces
b. disulphide linkage
c. electrostatic forces
d. hydrogen bonds
Answer:
a. Vander Waals forces

Question 28.
According to Chargaff’s rule, the hydrogen bonding between Adenine and Thymine is …………….
(a) 2
(b) 3
(c) 4
(d) Nil
Answer:
(a) 2

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 29.
Which polymer is stored in liver
a. Amylose
b. Amylo pectin
c. Cellulose
d. Glycogen
Answer:
d. Glycogen

Question 30.
The bond that is not needed for protein formation is
a. Hydrogen bond
b. Peptide bond
c. Ionic bond
d. glucosidic bond
Answer:
d. glucosidic bond

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 31.
A complete turn of the helix comprises …………….
(a) 34 nm
(b) 3.4 nm
(c) 20 nm
(d) 2nm
Answer:
(b) 3.4 nm

Question 32.
The acid is also known as vitamin C
a. Aspartic acid
b. Tartaric acid
c. Ascorbic acid
d. Adipic acid
Answer:
c. Ascorbic acid

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 33.
Which is the left-handed DNA?
(a) B – DNA
(b) A – DNA
(c) Z – DNA
(d) dsDNA
Answer:
(c) Z – DNA

II. Choose the wrong answer.

Question 1.
(a) Hydrolase – Amylase
(b) Oxidoreductase – Dehydrogenase
(c) Transferase – Transaminase
(d) Isomerase – Hexokinase
Answer:
(d) Isomerase – Hexokinase

Question 2.
This has nothing to do with the structure of
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 5
a. Cytosine
b. Pyrimidine
c. Adenine
d. Thyamine
Answer:
c. Adenine

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Choose the right answer
a) Amylose – linear unbranched polymer of with 20% starch
b) Amylopectin – a polymer with some 1,6 linkages that give it a linear structure
c) Inulin – Polymer of galactose
d) Amino acid – Here a basic structure of carbon linked to a basic amino group
Answer:
d)  Amino acid – Here a basic structure of carbon linked to a basic amino group

III. Match The Following And Find The Correct Answer.

Question 1.
(I) Morphine – A. Hectins
(II) Concanavalin A – B. Drug
(III) Vinblastin – C. Pigment
(IV) Anthocyanin – D. Toxin
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 6
Answer:
(a) B – A – D – C

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 2.
(I) Lactose – A. Penta saccharide
(II) Ramnose – B. Tetra saccharide
(III) Stachyose – C. Disaccharide
(IV) Verbascose – D. Tri saccharide
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 7
Answer:
(b) C-D-B-A

Question 3.
(I) Fred Sanger 1st sequenced
(II) Di sulphide bridges formed between sulphur & amino acids
(III) non-protein enzyme
(IV) homo polysaccharide with amino acid
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 8
Answer:
(b) D-A-B-C

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 4.
(I) Amino acid chain is twisted into coiled configuration call a helix – A. Tertiary Protein
(II) Protein fold into a globular structure called domains – B. Quaternary protein
(III) Linear arrangement of amino acids in a Polypeptide chain – C. Secondary protein
(IV) more than one polypeptide forms a large multiunit multimer – D. Primary Protein
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 9
Answer:
(c) C – A – D – B

Question 5.
(I)) Esters formed between long-chain alcohol another negative. – A. a molecule with two or more & saturated fatty acid function group one +ve and
(II) lipids have both hydrophobic & hydrophilic end known for permeability – B. fluid nature & selective
(III) The amino acids are both acidic & basic exoskeleton of insects – C. waxy substance coating
(IV) Zwitter is also called dipolar – D. amophoteric in nature
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 10
Answer:
(c) C-B-D-A

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

IV. Find Out The True And False Statements From The Following And That Basis Find Out The Right Answer.

Question 1.
(I) Esters are formed between long-chain alcohol & saturated fatty acid.
(II) Lecithin is a food additive & dietary supplement. ‘
(III) Lipids in their structure have two hydrophilic ends
(IV) Solid fats are usually unsaturated
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 11
Answer:
(a) True – True – False – False

Question 2.
(I) In saturated fatty acids, the hydrocarbon chain is single-bonded
(II) Triglycerides are composed of a single molecule of glycerol bound to 2 fatty acids
(III) Palmitic acid is an example of saturated fatty acid.
(IV) Oleic acid is an example of unsaturated fatty acid.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 12
Answer:
(d) Deoxyribose sugar – Phosphate – Nitrogen base – Nucleotide

V.

Question 1.
Label the diagram parts correctly by choosing the right option.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 13

A

B C

D

a Deoxyribose sugar Nitrogen base Nucleotide Phosphate
b Deoxyribose sugar Phosphate Nucleotide Nitrogen base
c Deoxyribose sugar Nitrogen base Nucleotide Phosphate
d Deoxyribose sugar Phosphate Nitrogen base Nucleotide

Answer:
(d) Deoxyribose sugar – Phosphate Nitrogen base – Nucleotide

Question 2.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 14

A

B C

D

a Q arm Centromere Sister Chromatids Q arm
b P arm Centromere Sister Chromatids Q arm
c Sister Chromatids Centromere Q arm P arm
d Q arm Centromere P arm Sister Chromatids

Answer:
(b) P arm – Centromere – Sister chromatids – Q arm

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

VI. Assertion & Reason – Find Out The Correct Answer.

Question 1.
Assertion (A): Adhesion refers to the tendency of water molecules to cling together
Reason (R): Because of hydrogen bonding, water molecules interact with one another continuous column of water is raised in xylem vessels.
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion.
Answer:
(a) Assertion & Reason correct Reason Explaining Assertion.

Question 2.
Assertion (A): Glycine is a non-essential amino acid
Reason (R): It must be taken through diet
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion.
Answer:
(c) Assertion is true but Reason is wrong.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Assertion (A): In the presence of enzyme substance molecules can be attached by the reagent.
Reason (R): Active sites of enzymes hold the substance in a suitable position.
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion
Answer:
(a) Assertion & Reason correct Reason Explaining Assertion.

Question 4.
Assertion (A): Aminoacids behave like salt rather than simple amines or carboxylic acid.
Reason (R): In aqueous solution, the COOH group of amino acid loses a protein and the NH2 group accepts a proton to form zwitterion (salt).
(a) Assertion & Reason correct Reason Explaining Assertion
(b) Assertion & Reason correct- Reason not explaining Assertion
(c) Assertion is true but Reason is wrong
(d) Assertion is true but Reason is not explaining Assertion
Answer:
(a) Assertion & Reason correct Reason Explaining Assertion.

2 Marks

Question 1.
Define Micronutrients.
Answer:
Nutrients which are required in trace amounts is called micronutrients.
Cobalt, zinc, boron, copper, molybdenum, and manganese – essential for enzyme action.
Eg – Molybdenum is necessary for the fixation of nitrogen by enzyme nitrogenase.

Question 2.
Write down the properties of Water.
Answer:

  • It has Adhesion & cohesion property
  • High latent heat of vaporisation
  • High melting and boiling point
  • Universal solvent
  • Has specific heat capacity.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 3.
Differentiate between Primary and Secondary Metabolites.
Answer:

Primary metabolites

Secondary Metabolites

Required for the basic metabolic processes, like Photosynthesis, Respiration, Protein & lipid metabolism. No direct function in the growth and development of organisms.

Question 4.
Define Polymerisation.
Answer:
A process in which repeating subunits termed monomers is bound into chains of different lengths called polymers.
Eg – Starch – Polynucleotide.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Distinguish between Glycogen and Cellulose.
Answer:

Glycogen Cellulose
Storage polysaccharide Structural polysaccharide
Animal starch Plant starch
made up of glucose with ( α- 1-6) linked branches Made up of 1000s of glucose units held by β  glucose units held by 1,4 glucosidic linkage
Seen in liver cells skeletal muscle fibre throughout the human body except brain. Occur in cotton. In the form of nitrocellulose used as explosives.

Question 6.
Distinguish between Dinucleotide & Polynucleotide.
Answer:

Dinucleotide

Polynucleotide

2 nucleotide joined to form Dinucleotide
They are linked through 3′- 5′- Phospho – diester linkage by condensation between phosphate group of one with the sugar of other.
Like dinucleotide,when many nucleotides then it leads to the formation of polynucleotides Eg-DNA-RNA

Question 7.
Differentiate between Nucleoside & Nucleotide.
Answer:

Nucleoside Nucleotide
Nitrogen + Sugar → Nucleoside
Eg – Adenine + Ribose → Adenosine
Nucleoside +→ Nucleotide Phosphoric acid (N + S) + P
Adenine + Ribose → Adenosine
Adenosine + Phosphoric  Acid → Adenylic acid

Question 8.
State Chargaff’s Law.
Answer:
Chargaffs Law in 1949
I. A = T & G = C
Between A & T double bond Between G & C triple bond
II. A + G number equal to T + C
III. But A : T, need not be equal to G : C

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 9.
Differentiate Between DNA & RNA.
Answer:

DNA

RNA

Double-stranded Single-stranded
The genetic material in almost all living
organism except for RNA virus
Not genetic material except RNA virus
2 types
Prokaryotic DNA is circular
Eukaryotic DNA is linear
3 types
m RNA
t RNA
r RNA
Controls all aspects of a cell plays important role in protein synthesis

Question 10.
Distinguish between Cation, Zwitterion & Anion.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 15

Question 11.
How will you test reducing sugar?
Answer:

Substrate

Reagent

Result

1. Glucose is taken in a test tube (Aldehyde) An alkaline solution of copper di sulfate (Benedicts’ reagent) added & heated So brick-red precipitate of copper oxide is formed (i.e.) (Cu + is reduced to Cu+) Aldehyde is oxidised to COOH group.

Question 12.
Draw the structure of a basic amino acid.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 16

3 Marks

Question 1.
Classify Polysaccharides.
Answer:
Polysaccharides have more than 10 monosaccharides
They can be divided in to

  • Homopolysaccharides
  • Hetero polysaccharides
1. Homopolysaccharides 2. Heteropolysaccharides
a. Starch a. Peptidoglycan
b. Glycogen b. Hyaluronic acid
c. Cellulose c. chondroitin
d. Chitin d. keratan sulphate
e. Inulin e. Agar Agar

Question 2.
Why do we call Glucose and Fructose Isomers -Discuss.
Answer:
Glucose and Fructose have a same molecular formula, C6H1206 but different structural formulas- so they are known as Isomers.
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 17

Question 3.
Explain the formation of Disaccharide Lactose.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 18

Question 4.
Draw the structure of Fatty acid.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 19

Question 5.
Distinguish between Waxes & Steroids.
Answer:

Waxes

Steroids

Esters formed between long-chain alcohol and saturated fatty acids
Fur feathers, fruits, leaves, skin, and insect exoskeleton are waterproofed with a coating of wax.
Complex compounds found in cell membrane & animal hormones. Eg – Cholesterol
It reinforces the structure of the life cell membrane in animal cells also in Mycoplasma.

Question 6.
Draw the structural formula of 3 simple amino acids – Glycine, Alanine & Valine.
Answer:
The non-polar aliphatic R group has 6 amino acids Glycine, Alanine, Valine, Leucine, Methionine & Isoleucine.
Structure of Glvcine Alanine Valine
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 20

Question 7.
Distinguish between Macronutrients & Micronutrients.
Answer:
Macronutrients:

  • Nutrients required in larger quantities for plant growth are called Macronutrients.
  • e.g. Potassium and Calcium

Micronutrients:

  • Nutrients required in trace amount for plant growth are called Micronutrients
  • e.g. Zinc and Bora

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 8.
Define Activation energy?
Answer:

  • The minimum quantity of energy the reactants must possess in order to undergo a specified reaction is known as Activation energy.
  • Energy being the biocatalysts reduce the activation energy, thereby help the reaction occurs.
  • The rate of reaction increases if activation energy decreases.

Question 9.
Distinguish between Primary metabolite & Secondary metabolite.
Answer:
Between Primary metabolite & Secondary metabolite:

  • Primary metabolites are those that are required for the basic metabolic processes like photosynthesis, respiration, etc Example: Lipase, a protein.
  • Secondary metabolites do not show any direct function in growth and development of organisms. Example: Ricin, gums.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 10.
Give examples for Secondary metabolites.
Answer:
Secondary metabolites :

  • Pigments – Carotenoids/Anthocyanins
  • Alkaloids – Morphine, codeine
  • Essential oil – Lemongrass oil, Rose oil
  • Toxins – Abrin & ricin
  • Lectins – Concanavalin. A
  • Drugs – Vinbiastine, curcumin
  • Polymeric substances – Rubber, gums, cellulose

Question 11.
Draw the various structures of Protein.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 21

Question 12.
Draw the structure of Purine – Adenine & Guanine Nucleotides.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 22

Question 13.
Why do some people have curly hair?
Answer:
Human hair is made of protein. The more the distance between the sulphur atoms, the more the proteins bend; the more the hair curls.

Question 14.
Deoxyribose (C5H10O4) is not a carbohydrate – Discuss.
Answer:
Carbohydrates are hydrates of carbon Deoxy ribose is a carbohydrate, but its formula C5H10O4 -does not apply the general formula of Carbohydrate (C2H20)X formula has Carbohydrates formula.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

5 Marks Questions

Question 1.
How will you identify the presence of glucose in a given food sample?
Answer:
Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called Benedict’s solution), the aldehyde or ketone group reduces Cu2+ ions to Cu+ ions forming brick red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (-COOH).

This reaction is used as test for reducing sugar and is known as Benedict’s test. The results of Benedict’s test depends on concentration of the sugar. If there is no reducing sugar it remains blue. Sucrose is not a reducing sugar The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 2.
What is protein denaturation.
Answer:
Exposure to heat causes atoms to vibrate violently distrust hydrogen and ionic bonds. There is the loss of 3D structure protein become elongated, disorganised strands. Soaps, detergents, acid, alcohol and some disinfectants disrupt the interchain bond cause the molecule to be non-functional.

Question 3.
Tabulate other sugar compounds
Answer:

Other Polysaccharides

Structure

Functions

Inulin Polymer of fructose It is not metabolised in the human body and is readily filtered through the kidney
Hyaluronic acid Hcteropolymcr of d glucuronic acid and D-N acetyl glucosamine 11 accounts for the toughness and flexibility of cartilage and tendon
Agar Mucopolysaccharide from red algae Used as solidifying agent in culture medium in laboratory
Heparin Glycosamino glycan contains variably sulphated disaccharide unit Drcscnt in liver Used as an anticoagulant
Chondroitin sulphate Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid Dietary supplement for treatment of osteoarthritis
Kcratan sulphate Sulphated glycosaminoglycan and is a structural carbohydrate Acts as cushion to absorb mechanical shock

Question 4.
Classify enzyme reactions:
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 23
II. a. Extracellular enzymes
Enzymes secreted outside & work externally Eg digestive enzymes. b. Intracellular Enzymes
Remain within cells & work there Eg Insulin.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 5.
Explain the three types of Co-Factors.
Answer:
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 24

II. Prosthetic groups:
Eg. Vit B2(Riboflavin) – & Flavin adenine dinucleotide (FAD) Kreb’s cycle
III. Co- Enzymes:
These are co-factors but don’t remain attached to enzymes Eg. NAD, NADP, Co-enzyme A, ATP etc.

Question 6.
Tabulate the uses of enzymes
Answer:

Enzyme Source Application
Bacterial protease Bacillus Biological detergents
Bacterial glucose isomerase Bacillus Fructose- Syrup manufacture
Fungal lactose Kluyveromyces Breaking down of lactose
glucose + glactose
Amylases Aspergillus Removal of starch in woven cloth production

Question 7.
Enumerate the properties of Enzyme.
Answer:
The properties of Enzyme:

  • Enzymes are globular proteins.
  • They act as catalysts and effective even in small quantities.
  • They remain unchanged at the end of the reaction.
  • They are highly specific.
  • They have an active site where the reaction takes place.
  • Enzymes lower the activation energy of the reaction they catalyse.

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Question 8.
Explain lock and key mechanism of enzymes.
Answer:
The substrate Enzyme product
The substrate binds to a specific pocket in an enzyme known as the Active site.
Active site = Lock
Substrate = Key
Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules 25

  • The substrate binds to the active site of the enzyme
  • As enzyme and substrate form ES- Complex, the substrate is raised in Energy —
  • This was explained by Fischer
  • transition stage break down in to, products, Enzyme remain unchanged.

Question 9.
What are the various types of inhibitors of enzymes.
Answer:
Definition:
Substances present in the cells may react with enzyme and lower the rate of reactions Inhibitors
I. Competitive Inhibitors:
Substances resemble the shape of substrate & compete to occupy active sites
Eg. 1. Enzyme RUBISCO – is competitively inhibited by oxygen/carbon dioxide in the chloroplast
2. Succinic dehydrogenase – Inhibited by malonate.

II. Non-Competitive Inhibitors
Unlike substrates, blocks by binding on active sites, change its shape so enzyme unable to accept substrate.
Enzyme – pyruvate kinase- inhibited by amino acid Alanine.

III. Non-reversible/ Irreversible Inhibitors
They bind to an enzyme tightly & permanently destroying their catalytic nature Enzyme cytochrome oxidase inhibited by cyanide ions Neurotransmitter – blocked by nerve gas sarin.

Question 10.
Distinguish between feedback allosteric inhibition negative feedback (end product) inhibition.
Answer:

Feedback Allosteric Inhibition

Negative feedback Inhibition

Allosteric inhibitors modify enzyme active sites (reversible)
E.g. Glucose  Hexokinase G-6 Phosphate
G.6. Phosphate – Inhibit Hexokinase
When end products accumulate they cause negative feedback or end product inhibition
After products get used up the enzyme reaction is switched on once again.

Question 11.
Tabulate other sugar compounds.
Answer:

Other polysaccharides

Structure

Functions

Inulin Polymer of fructose It is not metabolized in the human body and is readily filtered through the kidney
Hyaluronic acid Heteropolymer of d glucuronic acid and D-N acetyl glucosamine It accounts for the toughness and flexibility of cartilage and tendon
Agar Mucopolysaccharide from red algae Used as a solidifying agent in culture medium in a laboratory
Heparin Glucosamine glycan contains variably sulphated disaccharide unit present in liver Used as anticoagulant
Used as anticoagulant Sulphated glycosaminoglycan composed of altering sugars (N-acetylglucosamine and glucuronic acid) Dietary supplement for treatment of osteoarthritis
Keratan sulphate Sulphated glycosaminoglycan and is a structural carbohydrate Acts as cushion to absorb mechanical shock

Samacheer Kalvi 11th Bio Botany Guide Chapter 8 Biomolecules

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Commerce Guide Pdf Chapter 30 Performance of Contract Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Commerce Solutions Chapter 30 Performance of Contract

11th Commerce Guide Performance of Contract Text Book Back Questions and Answers

I. Choose the Correct Answer

Question 1.
On the valid performance of the contractual obligations by the parties, the contract
a. Is discharged
b. Become enforceable
c. Becomes void
d. Becomes legal
Answer:
a. Is discharged

Question 2.
Which of the following persons can perform the contract?
a. Promisor alone
b. Legal representatives of promisor
c. Agent of the promisor
d. All the above
Answer:
d. All the above

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 3.
A, B, C jointly promised to pay Rs.50,000 to D. Before performnce of the contract, C dies. Here, the contract
a. Becomes void on C’s death
b. Should be performed by A and B along with C’s legal representatives.
c. Should be performed by A and B alone.
d. Should be renewed between A, B and D.
Answer:
b. Should be performed by A and B along with C’s legal representatives.

Question 4.
Which of these parties cannot demand performance of promise?
a. Promisee
b. Any of the Joint Promisees
c. On the death of a Promisee, his Legal Representative.
d. Stranger to the Contract
Answer:
d. Stranger to the Contract

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 5.
A person is said to be a third person if he is not a
a. promisor
b. promise
c. agent
d. Legal Representative
Answer:
c. agent

II Very Short Answer Questions

Question 1.
State the ways of Performing a Contract.
Answer:
There are mainly two ways of performing a contract such as:

  1. Actual Performance
  2. Attempted Performance

Question 2.
Who is a Legal Representative?
Answer:
A person who oversees the legal affairs of another. Examples: The executor or administrator of an estate and a court-appointed guardian of a minor or incompetent person.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 3.
Who is an agent?
Answer:
According to Para 2 of Section 40, the promisor may employ a competent person such as an agent to perform the promise if the contract is not formed on personal condition.

Question 4.
Define Reciprocal Promise.
Answer:
Promises which form consideration or part of the consideration for each other are called ‘reciprocal promise’.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 5.
By whom must contracts be performed?
Answer:

  1. Promisor himself
  2. Agent
  3. Representations
  4. Third-person
  5. Joint Promisors

III. Short Answer Questions

Question 1.
What is a Valid tender?
Answer:
The tender which is in proper form, which contains the whole obligation and must not be in installments, which is unconditional is said to be a valid tender.

Question 2.
Who can execute and Perform a Contract?
Answer:
Promisor himself – Under Para 1 to Section 40, it is laid down that where it appears from the nature of the contract, the intention of the parties that any promise contained in it must be performed by the promisor himself or by his legal representatives or by any other competent person employed by him.

Question 3.
Who can demand performance?
Answer:
The following are those who can demand the performance of a valid contract:

  • Promise
  • Legal Representative
  • Third-Party

Question 4.
Write a note on the benefits of Reciprocal Promise.
Answer:
Meaning:
Promises which form the consideration or part of the consideration for each other are called reciprocal promises.
Benefits of Reciprocal Promise:

  • Since both the parties are working simultaneously, there is no chance of losses.
  • Each party are performing their promises independently to make the promise a successful one.
  • There is no loss raised due to lack of time.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 5.
Who is a Joint Promisors?
Answer:
Joint Promisors:
(a) Devolution of Joint Liabilities (Section 42)
Section 42 of Indian Contract Act lays down that “When two or more persons have made a joint promise, then unless a contrary intention appears in the contract, all such persons, during their joint lives and after the death of the last survivor, representatives of all, jointly must fulfil the promise”

(b) Devolution of Joint Rights (Section 45)
“When a person has made a promise to two or more persons jointly, then unless there is a contract to the contrary, the right to claim performance rests as between him and them, with them during their joint lives and after the death of them with representatives of such deceased person jointly with survivors, and after the death of last survivor, with the representatives of all jointly”.

IV Long Answer Questions

Question 1.
Explain rules relating to place of performance of promise:
Under Section 47, specified time and place for performance: If the promise is to be performed on a certain day, the promisor may undertake to perform it without application of the promisee.

According to Section 47, In such a case the promisor may perform the promise at any time during the usual hours of business on such day and at the place at which the promise ought to be performed.

Question 2.
Elucidate the provision regarding time as a factor in performance.
Answer:
1. Under Section 46, performance within a reasonable time:
According to Section 46, a promisor is to perform his promise within a reasonable time. On the other hand, a reasonable time will depend upon the circumstance of the case, the usage of trade or on the intention of the parties entering into the contract.
Example: A has given an order of supply of books in July which should be performed within 4 to 5 days of the month of July.

2. Under Section 47, specified time and place for performance:
If the promise is to be performed on a certain day, the promisor may undertake to perform it without application of the promisee. According to Section 47, In such a case the promisor may perform the promise at any time during the usual hours of business on such day and at the place at which the promise ought to be performed.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 3.
How do you think appropriation of payments takes place?
Answer:
Appropriation means application of payments The question of appropriation of payments arises when a debtor owes several debts to the same creditor and makes a payment that is not sufficient to discharge the whole indebtedness.
Appropriation of Payments:
Sometimes, a debtor owes several distinct debts to the same creditor and he makes a payment which is insufficient to satisfy all the debts. In such a case, a question arises as to which particular debt the payment is to be appropriated. Section 59 to 61 of the Act lay down the following rules as to appropriation of payments which provide an answer to this question. Therefore, where the debtor expressly states that the payment is to be applied to the discharge of a particular debt, the payment must be applied accordingly.

Application of payment:
Where debt to be discharged is not indicated [60] If section 60 is attracted, the creditor shall have the discretion to apply such payment for any lawful debt which is due to him from the person making the payment.

Application of payment:
Where neither party appropriates [61] The payment shall be applied in discharge of the debts in order of time whether they are or are not based by the limitation Act 1963 if the debt is of equal standing (i.e. payable on the same date) the payment shall be applied in discharge of each of this debt proportionately.

11th Commerce Guide Performance of Contract Additional Important Questions and Answers

I. Choose the Correct Answer:

Question 1.
Every promise and every set of promises, forming the consideration for each other is an
a. agreement
b. contract
c. offer
d. acceptance
Answer:
a. agreement

Question 2.
When, at the desire of the promisor, the promisee or any other person has done or abstained from doing or, does or abstain from doing or promises to do or to abstain from doing something, such act or abstinence or promise under section 2(d) is called
(a) Reciprocal promise
(b) consideration for the promise
(c) counteroffer
(d) acceptance
Answer:
(b) consideration for the promise

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 3.
Promises which form the consideration or part thereof, for each other under section 2(F) are called
(a) acceptances for different proposals
(b) agreements
(c) reciprocal promises
(d) consideration
Answer:
(c) reciprocal promises

Question 4.
In a valid contract, what comes first
a. enforceability
b. acceptance
c. promise
d. proposal
Answer:
a. enforceability

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Question 5.
The void agreement signifies ………..
a. agreement illegal in nature
b. agreement not enforceable by law
c. agreement violating legal procedure
d. agreement against public policy
Answer:
b. agreement not enforceable by law

II. Very Short Answer Questions:

Question 1.
What do you mean by the appropriation of payments?
Answer:
Appropriation means the application of payments. It arises when a debtor owes several debts to the same creditor and makes a payment that is not sufficient to discharge the whole indebtedness.

Question 2.
What is the nature of the third person with regard to the performance of the contract?
Answer:
According to Section 41, if a promisee accepts the performance of the promise by a third person he cannot afterward enforce it against the promisor.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

III. Short Answer Questions:

Question 1.
Explain the ways of performing a contract:
Answer:
There are mainly two ways of performing a contract such as:

  1. Actual Performance: When the party has done what he had undertaken to do, it is called actual performance. In actual performance, the party is to fulfill all his obligations under the contract.
  2. Attempted Performance: When the party offers to perform his obligation, it is not accepted by the promisee. A valid tender of performance is considered to be the performance of a promise.

Question 2.
Explain the term Devolution of Joint Rights:
Answer:
According to section 45 of Indian Contract, Act Devolution means, “When a person has made a promise to two or more persons jointly, then unless there is a contract to the contrary. the right to claim performánce rests as between him and them, with them during their joint lives: and after the death of them with representatives of such deceased person jointly with survivors, and after the death of’ last survivor, with the representatives of all jointly”.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

IV. Long Answer Questions

Question 1.
State the essentials of a valid tender of performance:
Answer:
A tender, to be valid, must satisfy the following essential requirements

  • It must be unconditional
  • It must be for the whole obligation and must not be in installments if the contract requires in full. It must be by a person who is in a position and willing to perform the promise.
  • It must be at the proper time and place.
  • It must be in proper form.
  • It must be made to a proper person i.e. to the promisee or his authorized agent.
  • In case of the tender of goods, the promisee must be given a reasonable opportunity to inspect the goods.
  • It may be made to one of the several joint promisees.

Question 2.
Explain the types of Reciprocal Promises:
Answer:

  1. Mutual and Indepéndent: Where each party must perform his promise independently without the performance of the other, the promise is mutual and independent, For example, Ramu agrees to pay Somu the amount for the rice supplied on 10th June. Some promises to deliver rice on 18th June.
  2. Mutual and Dependent: Where the performance of the promise by one party depends upon the prior performance of the promise by the other party, the promises are conditional and dependent.
  3. For example, A agrees to construct a building for B. B agrees to supply cement for the construction. Hence A’s promise to perform depends on B’s promise.
  4. Mutual and Concurrent: Where the two promises are said to be performed simultaneously, they are said to be mutual and concurrent.

Samacheer Kalvi 11th Commerce Guide Chapter 30 Performance of Contract

Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 9 Applications of Integration Ex 9.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.5

Question 1.
Evaluate the following:
(i) \(\int_{0}^{π/2}\) \(\frac{dx}{1+5cos^2x}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5 1

Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5

(ii) \(\int_{0}^{π/2}\) \(\frac{dx}{5+4sin^2x}\)
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5 2

Samacheer Kalvi 12th Maths Guide Chapter 9 Applications of Integration Ex 9.5

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.4

Question 1.
If sin x = \(\frac{15}{17}\) and cos y = \(\frac{12}{13}\), 0 < x < \(\frac{\pi}{2}\), 0 < y <\(\frac{\pi}{2}\), find the value of
(i) sin(x + y)
(ii) cos(x – y)
(iii) tan(x + y)
Answer:
Given sin x = \(\frac{15}{17}\), 0 < x < \(\frac{\pi}{2}\)
we have cos2x + sin2x = 1
∴ cos2x = 1 – sin2x
= 1 – \(\left(\frac{15}{17}\right)^{2}\)
= 1 – \(\frac{225}{289}\)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 1

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Given that 0 < x < \(\frac{\pi}{2}\), that is x lies in the first quadrant ∴ cos x is positive.
cos x = \(\frac{8}{17}\)
Also given cos y = \(\frac{12}{13}\), 0 < x < \(\frac{\pi}{2}\)
we have cos2y + sin2y = 1
sin2y = 1 – cos2y
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 2
Since 0 < y < \(\frac{\pi}{2}\), y lies in the first quadrant
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 3

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

(i) sin (x + y)
sin (x + y) = sin x cos y + cos x sin y
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 4

(ii) cos (x – y)
cos (x – y) = cos x cos y + sin x sin y
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 5

(iii) tan (x + y)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 6

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 2.
If sin A = \(\frac{3}{5}\) and cos B = \(\frac{9}{41}\) 0 < A < \(\frac{\pi}{2}\), 0 < B < \(\frac{\pi}{2}\), find the value of
(i) sin (A + B)
(ii) cos (A – B)
Answer:
Given sin A = \(\frac{3}{5}\) 0 < A < \(\frac{\pi}{2}\)
we have sin2A + cos2A = 1
cos2A = 1 – sin2A
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 7
Since 0 < A < \(\frac{\pi}{2}\), A lies in the first quadrant cos A is positive. ∴ cos A = \(\frac{4}{5}\)
Also given cos B = \(\frac{9}{41}\), 0 < B < \(\frac{\pi}{2}\)
We have cos2B + sin2B = 1
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 8

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

(i) sin (A + B)
sin (A + B) = sin A cos B + cos A sin B
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 9

(ii) cos (A – B)
cos (A – B) = cos A cos B + sin A sin B
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 10

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 3.
Find cos(x – y),given that cos x = –\(\frac{4}{5}\) with π < x < \(\frac{3 \pi}{2}\) and sin y = –\(\frac{24}{25}\) with π < y < \(\frac{3 \pi}{2}\)
Answer:
Given cos x = –\(\frac{4}{5}\), π < x < \(\frac{3 \pi}{2}\)
we have cos2 x + sin2 x = 1
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 11
Since π < x < \(\frac{3 \pi}{2}\), x lies in the third quadrant.
Since x is negative in the third quadrant. ∴ sin x = –\(\frac{3}{5}\)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 12

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 4.
Find sin(x – y) , given that sin x = \(\frac{8}{17}\) with 0< x < \(\frac{\pi}{2}\), and cos y = –\(\frac{24}{25}\), x < y < \(\frac{3 \pi}{2}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 13
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 14
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 15

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 5.
Find the value of
(i) cos 105°
Answer:
cos 105° = cos(90° + 150)
= -sin 15°
= -sin(45°- 30°)
= -[sin 45° cos 30° – cos 45° sin 30°]
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 16

(ii) sin 105°
Answer:
sin 105° = sin (90° + 15°)
= cos 15°
= cos (45° – 30°)
= cos 45° cos 30° + sin 45° sin 30°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 17

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

(iii) tan \(\frac{7 \pi}{12}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 18
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 19

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 6.
prove that
(i) cos (30° + x) = \(\frac{\sqrt{3} \cos x-\sin x}{2}\)
(ii) cos (π + θ) = – cos θ
(iii) sin (π + θ) = – sin θ
Answer:
(i) cos (30° + x) = \(\frac{\sqrt{3} \cos x-\sin x}{2}\)
cos ( 30° + x) = cos 30°. cos x – sin 30° sin x
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 20

(ii) L.HS = cos(π + θ) = cos(180° + θ)
= cos 180° cos θ – sin 180° sin θ
= (-1) cos θ – (0) sin θ
= – cos θ = RHS

(iii) LHS = sin (π + θ) = sin π cos θ + cos π sin θ
= (0) cos θ + (-1) sin θ
= -sin θ = RHS

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 7.
Find a quadratic equation whose roots are sin 15° and cos 15°.
Answer:
sin 15° = sin (45° – 30°)
= sin 45°. cos 30° – cos 45°. sin 30°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 21
cos 15° = cos(45° – 30°)
= cos 45° . cos 30° + sin 45° . sin 30°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 22
The quadratic whose roots cos 15° and sin 15° is
x2 – (cos 15° + sin 15°)x + (cos 15°) (sin 15°) = 0 ——— (3)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 23
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 24
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 25
Substituting in equation (3) we have
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 26

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 8.
Expand cos(A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B if A + B + C = \(\frac{\pi}{2}\)
Answer:
Taking A + B = X and C = Y
We get cos (X + Y) = cos X cos Y – sin X sin Y
(i.e) cos (A + B + C) = cos (A + B) cos C – sin (A + B) sin C
= (cos A cos B – sin A sin B) cos C – [sin A cos B + cos A sin B] sin C
cos (A + B + C) = cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C If (A + B + C) = π/2 then cos (A + B + C) = 0
⇒ cos A cos B cos C – sin A sin B cos C – sin A cos B sin C – cos A sin B sin C = 0
⇒ cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B

Question 9.
Prove that
(i) sin (45° + θ) – sin (45° – θ) = √2 sin θ
(ii) sin (30° + θ) + cos (60° + θ) = cos θ
Answer:
(i) sin (45° + θ) – sin (45° – θ) = √2 sin θ
sin(45° + θ) – sin(45° – θ) = (sin 45° cos θ + cos 45° sin θ) – (sin 45° cos θ + cos 45° sin θ)
= sin 45° cos θ + cos 45° sin θ – sin 45° cos θ + cos 45° sin θ
= 2 cos 45° sin θ
= 2 × \(\frac{1}{\sqrt{2}}\) sin θ
= \(\frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\) × sin θ
sin (45° + θ) – sin (45° – θ) = \(\frac{2 \sqrt{2}}{2}\) sin θ
= √2 sin θ

(ii) sin (30° + θ) + cos (60° + θ) = cos θ
sin (30° + θ) + cos (60° + θ)
= sin 30° cos θ + cos 30° sin θ + cos 60° cos θ – sin 60° sin θ
= \(\frac{1}{2}\) cos θ + \(\frac{\sqrt{3}}{2}\) sin θ + \(\frac{1}{2}\) cos θ – \(\frac{\sqrt{3}}{2}\) sin θ
= cos θ

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 10.
If a cos (x + y) = b cos(x – y), show that (a + b) tan x = (a – b) cot y.
Answer:
a cos (x + y) = b cos (x – y)
a [cos x cos y – sin x sin y] = b [cos x cos y + sin x sin y]
a cos x cos y – a sin x sin y = b cos x cos y + b sin x sin y
a cos x cos y – b cos x cos y = a sin x sin y + b sin x sin y
(a – b) cos x cos y = (a + b) sin x sin y
(a – b) \(\frac{\cos y}{\sin y}\) = (a + b) \(\frac{\sin x}{\cos x}\)
(a – b) cot y = (a + b) tan x
(a + b) tan x = (a – b) cot y .

Question 11.
Prove that sin 105° + cos 105° = cos 45°
Answer:
sin 105° + cos 105° = sin (90° + 15°) + cos ( 90° + 15°)
= cos 15° – sin 15°
= cos (45° – 30°) sin (45° – 30°)
= (cos 45° . cos30° + sin 45° sin 30°) – (sin 45° cos 30° – cos 45° sin 30°)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 27

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 12.
Prove that sin 75° – sin 15° = cos 105° + cos 15°
Answer:
RHS = cos 105° + cos 15° = cos (90° + 15°) + cos (90° – 75°)
= -sin 15° + sin 75°
= sin 75° – sin 15°
= LHS

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 13.
Show that tan 75° + cot 75° = 4
Answer:
tan 75° = tan (45° + 30°)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 29
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 30

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 14.
Prove that cos(A + B). cos C – cos(B + C) cos A = sin B sin (C – A)
Answer:
LHS = (cos A cos B – sin A sin B) cos C – (cos B cos C – sin B sin C) cos A
= cos A cos B cos C – sin A sin B cos C – cos A cos B cos C + cos A sin B sin C
= cos A sin B sin C – sin A sin B cos C
= sin B [sin C cos A – cos C sin A]
= sin B [sin (C – A)] = RHS

Question 15.
Prove that sin (n + 1)θ . sin(n – 1)θ + cos(n + 1)θ . cos(n – 1)θ = cos 2θ, n ∈ Z
Answer:
Taking (n + 1) θ = A and (n – 1) θ = B
LHS = sin A sin B + cos A cos B
= cos (A – B)
= cos[(n + 1) – (n – 1)]θ
= cos (n + 1 – n + 1)θ = cos 2θ = RHS

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 16.
If x cos θ = y cos\(\left(\theta+\frac{2 \pi}{3}\right)\) = z cos \(\left(\theta+\frac{4 \pi}{3}\right)\) find the value of xy + yz + zx.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 31
= cos θ + cos (θ + 120°) + cos (θ + 240°)
= cos θ + cos θ cos 120° – sin θ sin 120° + cos θ . cos 240° – sin θ sin 240°
= cos θ + cos θ cos (180° – 60°) – sin θ sin( 180°- 60°) + cos θ cos (180°+ 60°) – sin θ sin (180° + 60°)
= cos θ + cos θ × – cos 60° – sin θ × sin 60° + cos θ × – cos 60°- sin θ (- sin 60°)
= cos θ – cos θ cos 60° – sin θ sin 60° – cos θ cos60° + sin θ sin 60°
= cos θ – 2 cos θ cos 60°
= cos θ – 2 cos θ × \(\frac { 1 }{ 2 }\) = cos θ – cos θ = 0
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 32

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 17.
Prove that
(i) sin(A + B) . sin(A – B) = sin2A – sin2B
Answer:
LHS = sin (A + B) sin (A – B)
= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)
= sin2A cos2 B – cos2 A sin2 B
= sin2 A (1 – sin2 B) – (1 – sin2 A) sin2 B
= sin2 A – sin2 A sin2 B – sin2 B + sin2 A sin2 B
= sin2 A – sin2 B = RHS

(ii) cos (A + B) . cos (A – B) = cos2A – sin2B = cos2B – sinA
Answer:
cos(A + B) . cos(A – B) = (cos A cos B – sin A sin B) (cos A cos B + sin A sin B)
= (cos A cos B )2 – (sin A sin B )2 = cos2 A cos2 B – sin2 A sin2 B
= cos2A(1 – sin2B) – (1 – cos2A) sin2B
= cos2A – cos2A sin2B – sin2B + cos2A sin2B
cos(A + B) . cos(A – B) = cos2A – sin2B
Also cos(A + B) . cos(A – B) = cos2A cos2B – sin2A sin2B
= (1 – sin2A)cos2B – sin2A(1 – cos2B)
= cos2B – sin2A cos2B – sin2A + sin2A cos2B
cos(A + B) . cos(A – B) = cos2B – sin2A

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

(iii) sin2 (A + B) – sin2(A – B) = sin 2 A sin 2 B
Answer:
sin2 A – sin2 B = sin (A + B) sin (A – B)
LHS = sin2 (A + B) – sin2 (A – B) = sin [(A + B) + (A – B)] [sin (A + B) – (A – B)]
= sin 2A sin 2B = RHS

(iv) cos 8θ . cos 2θ = cos2 5θ – sin2
Answer:
cos8θ . cos2θ = cos(5θ + 3θ) cos(5θ – 3θ)
= (cos 5θ . cos 3θ – sin 5θ sin 3θ) (cos 5θ . cos3θ + sin 5θ sin 3θ)
= (cos 5θ . cos 3θ)2 – (sin 5θ sin 3θ)2
= cos2 5θ cos23θ – sin2 5θ sin2
= cos2 5θ (1 – sin2 3θ) – (1 – cos2 5θ) sin2
= cos2 5θ – cos2 5θ sin2 3θ – sin2 3θ + cos2 5θ sin2
cos 8θ . cos 2θ = cos2 5θ – sin2

Question 18.
Show that cos2A + cos2B – 2 cos A cos B cos (A + B) = sin2(A + B)
Answer:
LHS = cos2 A + cos2 B – 2 cos A cos B [cos A cos B – sin A sin B]
= cos2 A + cos2 B – 2 cos2 A cos2 B + 2 sin A cos A sin B cos B
= (cos2 A – cos2 A cos2 B) + (cos2 B – cos2 A cos2 B) + 2 sin A cos A sin B cos B
= cos2 A (1 – cos2 B) + cos2 B (1 – cos2 A) + 2 sin A cos A sin B cos B
= cos2 A sin2 B + cos2 B sin2 A + 2 sin A cos B sin B cos A
= (sin A cos B + cos A sin B)2
= sin2 (A + B) = RHS

Question 19.
If cos (α – β) + cos(β – γ) + cos(γ – α) = [lαtex]-\frac{3}{2}[/lαtex],then prove thαt
cos α + cos β + cos γ = sin α + sin β + sin γ = 0
Answer:
Given cos( α – β) + cos (β – γ) + cos (γ – α) = [lαtex]-\frac{3}{2}[/lαtex]
cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α = [lαtex]-\frac{3}{2}[/lαtex]
2 [cos α cos β + sin α sin β + cos β cos γ + sin β sin γ + cos γ cos α + sin γ sin α] = – 3
(2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (2 sin α sin β + 2sin β sin γ + 2 sin γ sin α) + 3 = 0
(2 cos α cos β + 2 cos β cos γ + 2cos γ cos α) + (2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α)+
(cos2 α + sin2 α) + (cos2β + sin2 β) + (cos2 γ + sin2 γ) = 0
(cos2 α + cos2β + cos2γ + 2 cos α cos β + 2 cos β cos γ + 2 cos γ cos α) + (sin2α + sin2β + sin2γ + 2 sin α sin β + 2 sin β sin γ + 2 sin γ sin α) = 0
(cos α + cos + cos γ)2 + (sin α + sin β + sin )2 = 0
cos α + cos β + cos γ = 0 and sin α + sin β + sin γ = 0
cos α + cos β + cos γ = sin α + sin + sin γ = 0

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 20.
Show that
(i) Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 33
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 35

(ii) Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 34
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 36

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 21.
Prove that cot (A + B) = \(\frac{\cot A \cot B-1}{\cot A+\cot B}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 37
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 38

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 22.
If tan x = \(\frac{n}{n+1}\) and tan y = \(\frac{1}{2 n+1}\) find tan (x + y).
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 39

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 23.
Prove that tan \(\left(\frac{\pi}{4}+\theta\right)\) tan \(\left(\frac{3 \pi}{4}+\theta\right)\) = -1
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 40

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 24.
Find the value of tan (α + β), given that cot α = \(\frac{1}{2}\), α ∈ \(\left(\pi, \frac{3 \pi}{2}\right)\) and sec β = –\(\frac{5}{3}\) β ∈ \(\left(\frac{\pi}{2}, \pi\right)\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 42

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4

Question 25.
If θ + Φ = α and tan θ = k tan Φ, then prove that sin (θ – Φ ) = \(\frac{k-1}{k+1}\) sin α
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 43
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.4 44

Samacheer Kalvi 8th Social Science Guide Civics Chapter 5 Road Safety Rules and Regulations

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Social Science Guide Pdf Civics Chapter 5 Road Safety Rules and Regulations Text Book Back Questions and Answers, Important Questions, Notes.

Tamilnadu Samacheer Kalvi 8th Social Science Solutions Civics Chapter 5 Road Safety Rules and Regulations

Samacheer Kalvi 8th Social Science Road Safety Rules and Regulations Text Book Back Questions and Answers

I. Choose the correct answer:

1. At a red light
a) You can keep going if the path looks clear.
b) You must stop and wait for it turn green.
c) Quickly you can reply your friend’s text message.
d) You can attend call.
Answer:
b) You must stop and wait for it turn green

2. Pedestrians can cross the road only …………….
a) at anywhere
b) near the signals
c) at Zebra crossing
d) none
Answer:
c) at Zebra crossing

3. Road Safety Week is celebrated in the month of ……………. every year
a) December
b) January
c) March
d) May
Answer:
b) January

4. For emergency, call ……………. for ambulance service.
a) 108
b) 100
c) 106
d) 101
Arts: a) 108

5. What are the causes for the road accidents?
a) Over Speeding
b) Drunken Driving
c) Distraction to Drivers d) All of these Answer:
d) AH of these

6. The first category of traffic signs is …………….
a) Mandatory Signs
b) Cautionary Signs
c) Informatory Signs
d) None of these
Answer:
a) Mandatory Signs

7. ‘Setu Bharatam’, a program was launched in …………….
a) 2014
b) 2015
c) 2016
d) 2017
Answer:
c) 2016

8. Expand ABS:
a) Anti Brake start
b) Annual Base System
c) Anti – locking Brake System
d) None of these
Answer:
c) Anti-locking Brake System

9. Overtaking when approaching a bend is
a) permissible
b) not permissible
c) permissible with care
d) our wish
Answer:
b) not permissible

10. When the ambulance is approaching
a) allow passage if there are no vehicles from the front side
b) no preference need be given
c) allow free passage by drawing to the side of the road
d) drive behind the ambulance with great speed
Answer:
c) allow free passage by drawing to the side of the road

II. Fill in the Blanks.

1. Always keep ……………. while driving.
Answer:
to the left

2. Mandatory signs are exhibited in ……………. shape.
Answer:
circular

3. ……………. controls the speed of the vehicle.
Answer:
Brake

4. Higher the speed; ……………. the risk.
Answer:
greater

5. Use of ……………. in four-wheelers and ……………. for two-wheelers has been brought under the law.
Answer:
seat belt, helmet

III. Match the Following.

1. Pollution under control certificate a) Zebra crossing
2. One-time tax for a new car b) Comic book on road safety
3. Pedestrian c) 6 months
4. Brasilia Declaration d) 15 years
5. Swachha safer e) Global conference

Answer:

1. Pollution under control certificate c) 6 months
2. One-time tax for a new car d) 15 years
3. Pedestrian a) Zebra crossing
4. Brasilia Declaration e) Global conference
5. Swachha safer b) Comic book on road safety

IV. State true or false.

1. The problem of accidents lies with roads only.
Answer:
False

2. Check mirrors before changing lanes.
Ans.’True

3. Flashing yellow signal indicates to slow down and proceed with caution.
Answer:
True

4. On a two-wheeler only one pillion rider is allowed.
Answer:
True

5. The roads are one of the worst inventions of man.
Answer:
False

V. Consider the following statements and tick the appropriate answer.

1. Which of the following statement is/are not correct?
i) Maintain the right distance to the vehicle in the front.
ii) Maintain speed limit, never exceed the speed limit.
iii) Wearing seat belt is not necessary while driving.
iv) Don’t slow down on bend and turn in the road.
a) i, iii only
b) ii, iv only
c) i, ii only
d) iii, iv only
Answer:
d) iii, iv only

2. Assertion: Drunken driving causes accidents.
Reason: It hampers vision due to dizziness.
a) A is correct and R is the correct explanation of A.
b) A is correct and R is not the correct explanation of A.
c) A is wrong and R is correct.
d) Both are wrong.
Answer:
a) A is correct and R is the correct explanation of A.

What is the aim of observing Road Safety Week?
Answer:
The event aimed at raising public awareness about traffic rules and ultimately reducing casualties due to road accidents. It is organised by the National Safety Council of India (NSC).

7. Write any four traffic rules.
Answer:
i) On a two-way road, the driver must drive on the left side of the road.
ii) It is mandatory for the driver to slow down at all inter junctions and pedestrian crossing.
iii) Drivers should not use the horns in prohibited areas like hospital zones, school zones, etc.,
iv) It is our responsibility to give way to emergency vehicles such as Army convoy fire engine and ambulance.

8. How does alcohol affect driving?
Answer:
Because it reduces concentration, it hampers vision due to dizziness. This may cause an accident.

VII. Answer the following in detail.

1. Explain the factors that contributed to road accidents.
Answer:

  1. Drivers – Over speeding, rash driving, violation of rules, failure, to understand signs, fatigue and consumption of alcohol.
  2. Pedestrian – Carelessness, illiteracy, crossing at wrong places, moving on roads and jaywalkers.
  3. Passengers – Projecting their body outside tie vehicle by talking to drivers, travelling on footboards, catching a running bus etc.
  4. Vehicles – Failure of brakes or steering, tyre burst, insufficient headlights, overloading and projecting loads.
  5. Road Condition – Damaged road, potholes, eroded road merging of rural road with highways, diversion and illegal speed breakers.
  6. Weather Conditions – Fog, snow, heavy rainfall, wind, storms and hail storms.

2. Describe the steps taken by the Government of India to prevent road accidents.
Answer:

  • Always keep to the left – while driving, keep to the left, and allow vehicles to pass from the opposite direction.
  • Slow down on bends and turn. A very important thing to keep in mind is to become cautious and slow down on the bends.
  • Use helmets – Make it a habit of strapping the helmet before mounting the bikes.
  • Follow the road signs – Road signs are mostly pictorial so it is not hard to comprehend.
  • For pedestrians – Cross only at Zebra crossing.

3. What are the preventive measures for road accidents?
Answer:
Safety Measures:

  1. Always keep to the left-While driving, keep to the left, and allow vehicles to pass from the opposite direction.
  2. Use helmets – Make it a habit of strapping the helmet before mounting the bikes.
  3. Never exceed the speed limit – The speed limit is related to the traffic condition of the area. To maintain the speed limit.
  4. Maintain the right distance – Collisions occur because we do not maintain adequate distance from the vehicle in the front. To maintain a safe distance from the bigger vehicles.
  5. Park the vehicles only along the designated parking bays/zones, not on the sides of the highways.
  6. Follow the road signs – Road signs are mostly pictorial so it is not hard to comprehend.
  7. Some other safety tips are – Never ever drive when drunk, never use cell phone while driving, avoid listening to too loud music, check your mirrors before changing lanes.
  8. Never cross on a red and yellow light. Take a sidewalk and avoid walking on the roads.

VIII HOTS

1. What is the requirement for a two-wheeler rider during the night?
Answer:
Few tips for Riding (Motorcycle) two-wheeler during the night:

  1. Upgrade your bike’s lightings.
  2. Increase your visibility to others on the road.
  3. Be ensured that your vision is clear.
  4. Use the lights of other vehicles equally.
  5. Ride slowly.
  6. Keep controlling your speed.
  7. Take adequate rest.
  8. Proper planning.

2. Tabulate tips for avoiding driver fatigue.

  • Many people think fatigue is only a problem for long-distance drivers, however, it is just as relevant for short-distance drives.
  • People generally don’t become fatigued from driving. Usually, they are already tired when they get behind the wheel from long hours, lack of sleep apnoea, or physically demanding roles.
  • Take a 15 minutes power nap if you feel yourself becoming drowsy.

IX. Project and Activity.

1.  Tabulate a few basic road safety rules for children.
Answer:

  1. Know your signals
  2. Stop, look and cross
  3. Pay attention – Listen
  4. Don’t Run on Roads
  5. Always use sidewalks
  6. Crossroads and Pedestrian crossing
  7. Never crossroad at bends
  8. Staying safe on a bicycle

2. Demonstrate road safety techniques.
Answer:
Road Safety Techniques

  • A day keeps the accident away
  • A constant glance at the mirrors.
  • Properly hold the steering wheel.
  • Regularly check the speed control.
  • Use Signals
  • Keep your feed on the breakers
  • Wrapping up

3. Prepare a Road Safety Awareness pocket guide and circulate in school.
Answer:
Slogan:

  • Start Early
  • Drive Slowly
  • Reach Safely.

Samacheer Kalvi 8th Social Science Road Safety Rules and Regulations Additional Important Questions and Answers

I. Choose the Correct answer.

1. For emergency calls ………….. for traffic accidents.
a) 103
b) 104
c) 102
d) 100
Answer:
a) 103

2. Totally ………….. cautionary traffic signs in our Roadways and Transport Department
a) 30
b) 40
c) 50
d) 20
Answer:
b) 40

3. In which year, the motor vehicle act came into force?
a) 1988
b) 1989
c) 1990
d) 1987
Answer:
b) 1989

4. The police will file a criminal case under this section
a) 300A
b) 302 A
c) 303A
d) 303 A
Answer:
d) 304 A

5. A multi – Pronged strategy has been adopted based on …………..
a) E
b) Three E’S
c) Four E’S
d) Two E’S
Answer:
c) Four E’S

6. The government had made of mandatory for a person who wants to obtain ………….. to be well versed in a traffic signal.
a) Bank Passbook
b) Aadhar number
c) Voters ID
d) Driving License
Answer:
c) Driving License

7. ………….. colour number plate used in the vehicle for the president of India and
Governor of states
a) Red
b) blue c) White
d) yellow
Answer:
a) Red

8. Quick response ambulances are parked in every ………….. on the National Highways.
a) 40 km
b) 50 km
c) 10 Km
d) 20 km
Answer:
b) 50 km

9. Decade of action for road safety from the year 2011 to ………….. proclaimed by the Un General Assembly.
a) 2015
b) 2020
c) 2000
d) 2010
Answer:
b) 2020

10. In which year Indian signed in
a) 2014
b) 2016
c) 2017
d) 2015
Answer:
d) 2015

II. Fill in the Blanks.

1. ……………………… is Random break test illegal and dangerous habit
Answer:
Tailgating

2. ……………………… is done to detect the consumption of alcohol.
Answer:
Random breath Test

3. The major distraction now a day is talking on ……………………… while driving.
Answer:
mobile phone

4. We are not allowed to overtake and to stay with in the lane mentioned by ………………………
Answer:
Solid white line

5. ……………………… ministry is responsible to prevent Road accidents.
Answer:
Road Transport and Highways

6. ……………………… made mandatory for two wheelers to make them more conspicuous..
Answer:
Automatic headline.

7. A set of comic books on road safety released for children are ………………………, ………………………
Answer:
Swacha, suvarshit

8. A public charitable trust that in workers to improve road safety is ………………………
Answer:
Save life foundation

9. ……………………… suggested 10 points a few Road safety.
Answer:
IPRC

10. ……………………… is for Commercial Vehicle.
Answer:
Yellow

11. The main distribution of road accident is ………………………
Answer:
adjusting mirror

12. ……………………… are mostly pictorial. so it’s easy to comprehend.
Answer:
Essential documents

13. Never ever drive when ………………………
Answer:
drunk

14. We never cross the roads when the ……………………… and lights are flashed.
Answer:
Red, yellow

15. ……………………… lines are used in areas where visibility is low.
Answer:
Single solid yellow

III. Match the Following.

1. Mandatory signs a) Triangle Shape
2. Cautionary signs b) Signalling device
3. Informatory signs c) Circular in shape
4. Traffic digits d) Complete Stop
5. Flashing red signal e) Rectangular shape

Answer:

1. Mandatory signs c) Circular in shape
2. Cautionary signs a) Triangle Shape
3. Informatory signs e) Rectangular shape
4. Traffic digits b) Signalling device
5. Flashing red signal d) Complete Stop

IV. State true or false.

1. Violation of any mandatory traffic sign is an offence punishable by law by the transport Department.
Answer:
True

2. Informatory signs never provide information to the drivers in board
Answer:
False

3. On a two-wheeler only one pillion ride is allowed.
Answer:
True

4. Never exceed the speed limit
Answer:
True

5. Pedastrinas cross only at a zebra crossing
Answer:
True

V. Consider the following statements and tick the appropriate answer.

1. i) Road accidents are desired events.
ii) Wearing a helmet while drivers is not mandatory.
iii) Road safety primarily means the protection and security of all road users.
iv) The ministry of road transport and highways has taken a number of steps to prevent road accidents.
a) i and ii correct
b) i and iii correct
c) ii and iv correct
d) iii and iv correct
Answer:
d) iii and iv correct

VI. Answer the following in one or two sentences.

1. Write a Short note on -108 Emergency Response Service.
Answer:

  • It is a free emergency service providing integrated medical (ambulance), police, and fire services.
  • If you find any victims on road don’t panic.
  • Call 108 for help and 103 for traffic accidents.

2. What is the importance of Indian Penal code 304A?
Answer:
The police will file a criminal case under this section, which deals with offences relating to death due to rashness and negligence of the driver.

3. Different colour number plates used in India – explain.
Answer:
In India, four different colour number plates are used. They are

  • Red Number Plate – is used in the vehicle for the President of India and Governor of States.
  • Blue Number Plate – is given to a vehicle that is used by foreign delegates/ ambassadors.
  • White Number Plate – it means that the car belongs to a common citizen.
  • Yellow Number Plate – is for commercial vehicles.

4. What are the essential Documents Possessed by a driver?
Answer:
Driving License, Registration Certificate of the Vehicle, Taxation Certificate, Insurance Certificate, Fitness Certificate, and Permit.

5. Write Short note on Setu Bharatam.
Answer:

  • A program was launched in 2016 for building bridges for safe and seamless travel on the National Highways.
  • It aims to make all National Highways free of railway level crossing by 2019.

6. Explain – Save LIFE Foundation.
Answer:
Save LIFE Foundation is an independent, non-profit, non-governmental, and public charitable trust that is working to improve road safety and emergency medical care across India.

VII. Answer the following in Detail.

1. What are the Significances of the lines marked on Indian roads?
Answer:

  • Broken white line – basic marking on roads, you may change lanes, and are allowed to overtake a vehicle or take U-turn of it is safe to do so.
  • Solid white line – seen on areas of strategic importance. This implies that you are not allowed to overtake and to stay within the lane.
  • Single solid Yellow lines – used in areas where visibility is low. It implies that you can overtake And should drive on your side.
  • Double solid yellow lines – used on dangerous roads or for two-way traffic. It strictly prohibits anybody from crossing over into the lane. You can overtake inside your own lane.
  • Stop line – This is marked before the pedestrian crossing and sets the deadline where the car should stop before the traffic signal.
  • Solid and Broken lines – if you are driving on the side with the broken line you are allowed to overtake and if you are driving on the side of the solid line you are not.

2. What are Traffic Signals?
Answer:
Traffic signals are there to regulate traffic, warn about hazards, and to guide the road users. Importance of Traffic Signals:

  • Red: This signal indicates to stop behind the stop line.
  • Yellow: This signal indicates to stop. Do not pass through or start until green shows.
  • Green: This signal indicates you may go if the way is clear.
  • Steady Green Arrow Signal: This signal may be provided in addition to the full green signal. This indicates to proceed with caution in the direction the arrow points.
  • Flashing Red Signal: It means to come to a complete stop. Proceed only when the way is clear.
  • Flashing Yellow Signal: It indicates to slow down and proceed with caution.

Samacheer Kalvi 8th Social Science Guide History Chapter 6 Development of Industries in India

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Social Science Guide Pdf History Chapter 6 Development of Industries in India Text Book Back Questions and Answers, Important Questions, Notes.

Tamilnadu Samacheer Kalvi 8th Social Science Solutions History Chapter 6 Development of Industries in India

Samacheer Kalvi 8th Social Science Development of Industries in India Text Book Back Questions and Answers

I. Choose the correct answer.
1. Which of the following activities of the people will not come under handicraft?
a) Carving statues out of stone
b) Making bangles with glass
c) Weaving silk sarees
d) Smelting of iron
Answer:
d) Smelting of iron

2. The oldest industry in India was industry.
a) Textile
b) Steel
c) Electrical
d) Fertilizers
Answer:
a) Textile

3. The woollen and leather factories became prominent in
a) Bombay
b) Ahmadabad
c) Kanpur
d) Dacca
Answer:
c) Kanpur

4. What was the aim of first Three Five year Plans of India?
a) To control population growth
b) To reduce illiteracy rate
c) To built a strong industrial base
d) To empower the women
Answer:
c) To built a strong industrial base

5. What was not the reason for the decline of Indian Industries?
a) Loss of royal patronage
b) Competition of machine made goods
c) Industrial policy of India
d) Trading policy of British
Answer:
c) Industrial policy of India

II. Fill in the blanks.
1. ………………. was the integral part in the life of the people.
Answer:
crafts

2. Industrial revolution took place in ………………. .
Answer:
England

3. The Assam Tea Company was founded in ………………. .
Answer:
1839

4. Jute industry was started in the Hoogly Valley at ………………. near Calcutta.
Answer:
Rishra

5. ………………. shortened the distance between Europe and India.
Answer:
Suez canal

III. Match the following.

1. Tavernier a) Drain Theory
2. Dacca b) Paper mill
3. Dadabai Naoroji c) Artisan
4. Ballygunj d) Muslin
5. Smiths e) French traveller

Answer:

1. Tavernier e) French traveller
2. Dacca d) Muslin
3. Dadabai Naoroji a) Drain Theory
4. Ballygunj b) Paper mill
5. Smiths c) Artisan

IV. State True or False.
1. India was famous for cotton and silk cloths.
Answer:
True

2. The railway was introduced in India by the British.
Answer:
True

3. Steel was first manufactured by modem methods at Jamshedpur.
Answer:
False

4. The Industrial policy of 1948, brought mixed economy in industrial sector.
Answer:
True

5. The tenth and eleventh five-year plans witnessed a high growth rate of Agricultural production.
Answer:
False

V. Consider the following statements and tick the appropriate answer.
1. Which of the following statements are correct?
i) According to Edward Baines, ‘The birthplace of cotton manufacture is in England’.
ii) Before mechanised industry handicrafts were the second-largest source of employment in rural India.
iii) Saurashtra was known for the tin industry.
iv) Construction of the Suez Canal made the British goods cheaper in India,
a) i and ii are correct
b) ii and iv are correct
c) iii and iv are correct
d) i, ii, and iii are correct
Answer:
b) ii and iv are correct

2. Assertion (A): Indian handicrafts collapsed under colonial rule.
Reason (R): British made India the producer of raw materials and markets for their finished
products.
a) A is correct R is the correct explanation of A
b) A is correct and R is not the correct explanation of A
c) Both A and R are correct
d) Both A and R is wrong
Answer:
A is correct R is the correct explanation of A

3. Which one of the following is wrongly matched?
a) Bernier – Shajahan
b) Cotton mill – Ahmadabad
c) TISCO – Jamshedpur
d) Economic Liberalisation – 1980
Answer:
d) Economic Liberalisation -1980

VI. Answer the following in one or two sentences.
1. What are the traditional handicraft industries of India?
Answer:

  1. The traditional handicrafts industries of India are textiles, woodwork, ivory, stone cutting, leather, fragrance wood, metalwork, and jewellery.
  2. The village artisans such as potters, weavers, smiths produced articles and utensils.

2. Write about the drain theory.
Answer:
Dadabai Naoroji was the first to acknowledge that the poverty of the Indian people was due to the British exploitation of India’s resources and the drain of India’s wealth to Britain.

3. Name the inventions which made the production of textiles on a large scale
Answer:
The invention of cotton gin, flying shuttle, spinning jenny and steam engine in England, which made the production of textiles on large scale.

4. Write a short note on the Confederation of Indian Industry.
Answer:

  • It is a business association in India. CII is a non-government, not for profit, industry-led, and industry- managed organisation.
  • It was founded in 1985.

5. What is de-industrialization?
Answer:
The process of disruption of traditional Indian crafts and a decline in national income has been referred to as de-industrialisation.

VII. Answer the following.
1. How was the trading policy of the British caused the decline of Indian Industries?
Answer:
The decline of Indian Industries:

  • Loss of Royal Patronage.
  • Transition from producer to exporter of raw materials.
  • Competition of Machine-Made goods.
  • The trading policy of the British.
  • De-industrialization.

2. Write in detail about the plantation industries.
Answer:
Plantation industries:

  1. The plantation industry was the first to attract Europeans. This provides jobs on a large scale.
  2. In reality, it could meet the increasing demands for tea, coffee and indigo by the British Society.
  3. The Assam Tea Company was founded in 1839.
  4. The coffee plantations also started simultaneously.
  5. As the tea plantation was the most important industry of Eastern India, the coffee plantation became the center of activities in South India.
  6. The Third important plantation, which gave birth to the factory was jute.
  7. All these Industries were controlled by many former employees of the British East India Company.

3. Explain Industrial development after the 1991 reforms.
Answer:
Industrial development after the 1991 reforms:

  • The year 1991 ushered a new era of economic liberalisation.
  • India took a major decision to improve the performance of the industrial sector.
  • The tenth and Eleventh Five-Year plans witnessed a high growth rate of industrial production.
  • The abolition of industrial licensing, dismantling of price controls, dilution of reservation of small- scale industries, and the virtual abolition of monopoly law enabled the Indian industry to flourish.
  • The new policy welcomes foreign investments.

VIII. HOTs.
1. How do handicraft products differ from machine-made products?
Answer:
Handicraft:
Something you make with your own hands, especially an ornament or decoration, is a handicraft. Instead, items made by artisans like pottery, handwoven blankets, handmade jewellery, and quilts stitched by hand are all examples of handicrafts.

Machine-made Products:
Machine-made products are produced faster and all are exactly the same. Machine manufacturing is faster and more economical. Also, machine-made goods are cheaper than hand made goods.

X. Project and Activity.
1. Name the industries in your state and divide them into agro-based metal-based, and forest-based. Agro-based industries:
Answer:
Cotton textile industries, Jute industry, Sugar industries, etc are agro-based industries.
Eg: Coimbatore

Metal-based industries:

  • Mineral-based industries use both metallic and nonmetallic as raw material. Eg: Chennai.
  • Forest-based industries: India has a rich diversity of forest resources. The most important industry is the paper industry. Eg: Chennai.

Samacheer Kalvi 8th Social Science Development of Industries in India Additional Important Questions and Answers

I. Choose the Correct answer.
1. The established of cotton textile industry ……………. in 1854
a) Calcutta
b) Hyderabad
c) Gujarat
d) Bombay
Answer:
d) Bombay

2. The first paper mill was started in Ballyguni in the year
a) 1870
b) 1871
c) 1872
d) 1950
Answer:
19478

3. The industrial policy Resolution act introduced in the year
a) 1947
b) 1948
c) 1949
d) 1950
Answer:
b) 1948

4. India has emerged as the ……………. largest produces of electricity in Asia
a) First
b) Second
c) Third
d) fourth
Answer:
c) Third

5. Introduction of ……………. plan the most important innovations in the industrial field
a) Annual
b) Four – year
c) Five – Year
d) Six-year
Answer:
c) Five – Year

6. On the basis of raw material used industries can be classified into two types of sectors are ……………., ……………..
a) Public and private
b) Mineral and private
c) basic and important
d) agro and mineral
Answer:
d) agro and mineral

7. The period can be considered as the period of the industrial recovery
a) 1980
b) 1981
c) 1982
d) 1983
Answer:
a) 1980

8. Steel was the first manufactured by modern methods in 1874
a) Jamshedpur
b) Kulti
c) Bally guni
d) None of these
Answer:
b) Kulti

9. The openings of Suez canal also shortened the distance between Europe and India by
about
a) 4800 km
b) 4890 Km
c) 4830 km
d) 4000 Km
Answer:
c) 4830 km

10. The consideration of Indian Industry was founded in
a) 1980
b) 1982
c) 1985
d) 1985
Answer:
c) 1985

II. Fill in the blanks.
1. “The birth place of cotton manufacture is India,” told by, ………………..
Answer:
Baines

2. ………………. cloth was used to Preserved Mummies in Egyptian.
Answer:
The Muslin of Dacca

3. ………………. is was identified with muslin cloths.
Answer:
Dacca

4. ………………. was known for tin Industry
Answer:
Bengal

5. The drain of India’s wealth to Britain acknowledge by ………………..
Answer:
Dadabai Naoroji

6. The period of the 1980s can be considered as the period of the ………………. recovery
Answer:
Industrial

7. The year ………………. where a new era of the economic liberalisation.
Answer:
1991

8. India ………………. has become one of the largest in the world.
Answer:
Road network

9. The industrial expansion over the plan period presents a ………………. picture
Answer:
mixed economic

10. In India a fourth sector industries are ……………….
Answer:
Information related industries

III. Match the following.

1. 1839 a) industrial recovery
2. 1854 b) Steel industry
3. 1874 c) Tata Iron and steel company
4. 1907 d) Tea company
5. 1980 e) Cotton textile industry

Answer:

1. 1839 d) Tea company
2. 1854 e) Cotton textile industry
3. 1-874 b) Steel industry
4. 1907 c) Tata Iron and steel company
5. 1980 a) industrial recovery

IV. State True or False.
1. Dacca was identified with Muslim clothes.
Answer
True

2. The plantation industry did not attract the European in the beginning.
Answer
False

3. Textile was the oldest industries in India.
Answer
True

4. The woolen and leather factories became prominent in Kanpur
Answer
True

5. The credit for the development of large – scale manufacture of steel in India goes to Jansheji Tata
Answer
True

V. Consider the following statements and tick the appropriate answer.
1. i) Textile was the oldest industry in India.
ii) India became the market for the finished products of Britain.
iii) The Indian goods made with primitive techniques could not compute with Industrial goods made in England
iv) The English good enterious India were charged.
a) i and ii are correct
b) ii and iv are correct
c) iii and iv are correct
d) i, ii and iii are correct
Answer:
b) ii and iv are correct

2. Assertion: The Indian road network has been one of the largest in the world
Reason: The Indian road network has directly contributed to industrial growth
a) A is correct R is the correct explanation of A
b) A is correct and R is not the correct explanation of A
c) Both A and R are correct
d) Both A and R is wrong
Answer:
c) Both A and R are correct

VI. Answer the following in one or two sentences.
1. Give an account of Muslin Cloth.
Answer:

  • Muslin was famous under the Hindu kings.
  • Early these cloths are light in nature. And also so costliest in the early period.
  • After the Muslim invasion, the muslin preparation was unknown by the other rulers.

2. What are the classification Industries?
Answer:

  • On the basis of raw materials used, industries can be classified into agro-based and mineral-based.
  • According to their role, it can be classified into basic and key industries.
  • On the basis of ownership, it can be classified into the public sector, private sector, joint sector, and co-operative sector.

3. How the Indian road network used in industrial growth?
Answer:

  • The Indian road network has become one of the largest in the world.
  • Government efforts led to the expansion of the network of National Highways, State highways, and major district roads, which in turn has directly contributed to industrial growth.

VII. Answer the following.
1. Explain the growth of Heavy industries in the beginning period.
Answer:

  • The heavy industries included the iron and steel industry, Steel was first manufactured by modem methods at Kulti in 1874.
  • Iron and steel industries began rooted in the Indian soil at the beginning of the 20th century.
  • However, the credit for the development of large-scale manufacture of steel in India goes to Jamshedji Tata.
  • The Tata Iron and Steel Company (TISCO) was set up in 1907 at Jamshedpur.
  • It started producing pig iron in 1911 and steel ingots in 1912.

2. Explain the classification of industries per the industrial policy Resolution 1956.
Answer:
As per the Industrial Policy Resolution 1956, industries were classified into three categories:
Schedule A:
Only the Government can handle these industries. Some of these are atomic energy, electrical, iron and steel, and others.

Schedule B:
These comprise road and sea transportation, machine tools, aluminium, chemicals including plastics and fertilisers, ferroalloys, and certain types of mining.

Schedule C:
Under this category, the remaining industries and left to the private sector.

3. What are the phases of Industrial development in India?
Answer:
Industrial development during 1950 – 1965:

  • During this phase, a majority of consumer goods were produced in India.
  • As a result, this phase witnessed a strong acceleration in the growth rate of production.

Industrial development during 1965 – 1980:

  • It mostly focused on the development of the capital goods sector, the consumer goods sector was neglected.
  • So this period is marked as the period of structural retrogression.

Industrial development during 1980 – 1991:
This period witnessed quite a healthy industrial growth.

Industrial development post-1991 Reforms:

  • The year 1991 a new era of economic liberalisation.
  • India took a major decision to improve the performance of the industrial sector.
  • The new policy welcomes foreign investments.