Srinivas

Access to a variety of TS Inter 2nd Year Chemistry Model Papers and TS Inter 2nd Year Chemistry Question Paper March 2020 helps students overcome exam anxiety by fostering familiarity.

TS Inter 2nd Year Chemistry Question Paper March 2020

Note : Read the following instructions carefully.

1. Answer all questions of Section – ‘A’. Answer any six questions in Section – ‘B’ and any two questions in Section – ‘C’.
2. In Section – A, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
3. In Section -”’8′, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
4. In Section – ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
5. Draw labelled diagrams wherever necessary for questions in Section – ‘B’ and ‘C’.

Section – A

Note : Answer all the questions.

Question 1.
State Raoult’s law.
Answer:
Raoult’s law for volatile solute : For a solution of voltile liquids, the partial vapour pressure of each omponent of the solution is directly proportional to its mole fraction present in solution. Raoult’s law for non-volatile solute : The relative lowering of vapour pressure of dilute solution containing non-volatile solute is equal to the mole fraction of solute.

Question 2.
State Faraday’s Law.
Answer:
Faraday’s First Law:
The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of current passing through the electrolyte.
OR
The mass of the substance deposited at an electrode during the electrolysis of electrolyte is directly proportional to quantity of electricity passed through it.

m α Q; m α c × t
m = ect;
m = \(\frac{\text { Ect }}{96,500}\)
e = electrochemical equivalent
t = time in seconds
c = Current in amperes
E = Chemical equivalent

Faraday’s Second Law: The amounts of different substances liberated when the same quantity of electricity is passed through the electrolytic solution are porportional to their chemical equivalent weights, m α E

Question 3.
Write ores with formulae of the following metals :
(i) Aluminium
(ii) Iron
Answer:
i) Ores of Aluminium Bauxite – A1₂O₃.2H₂O
Cryolite – Na₃Al F₆

ii) Ores of Iron Haematite – Fe₂O₃
Magnetite – Fe₃O₄

Question 4.
What is Lanthanoid contraction?
Answer:
Lanthanide contraction : The slow decrease of Atomic in lanthanides with increase in atomic number is called Lanthanide contraction.

Question 5.
What are Polymers? Give example.
Answer:
Polymer : A large molecular weight complex compound which is formed by the repeated combination of simple units (monomers) is called polymer. E.g. : Nylon 6, 6, Buna – S, rubber etc …..

Question 6.
What is vulcanization of rubber?
Answer:
Vulcanization of rubber : The process of heating the raw rubber with sulphur (or) with sulphur compounds to improve its physical properties is called vulcanization of rubber.

Question 7.
What are antacids ? Give example.
Answer:
Antacids : Chemicals that remove the excess of acid in the stomach and maintain the pH to normal level are antacids. E.g. : (Dmeprozole, Lansoprozole etc.,

Question 8.
What is tincture of iodine? What is its use?
Answer:
Tincture of iodine (antiseptic) is a mixture of 2-3% Iodine solution in alcohol – water.

Question 9.
What are Enantiomers?
Answer:
Enantiomers : The stereo isomers related to each other as non-superimposable mirror images are called enantiomers.

1. These have identical physical properties like melting point, boiling points refractive index etc.
2. They differ in rotation of plane polarised right.

Question 10.
What are ambident nucleophiles?
Answer:
Ambident nucleophiles : The nucleophiles which are able to attack at two (or) more different sites are called ambident nucleophiles.
Eg: Alkyl halides react with AgCN to form alkyl cyanide and isocyanides.
Img-1

Section – B

Question 11.
Derive Bragg’s equation.
Answer:
Derivation of Bragg’s equation : When X-rays are incident on the crystal or plane, they are diffracted from the lattice points (lattice points may be atoms or ions or molecules).

In the crystal the lattice points are arranged in regular pattern. When the waves are diffracted from these points, the waves may be constructive or destructive interference.

The 1^st and 21^nd waves reach the crystal surface. They undergo constructive interference. Then from the figure 1st and 2nd rays are parallel waves. So, they travel the same distance till the wave form AD. The second ray travels more than the first by an extra distance (DB + BC) after crossing the grating for it to’interfere with the first ray in a constructive manner. Then only they can be in the same phase with one another. If the two waves are to be in phase, the path difference between the two ways must be equal to the Wavelength (λ) or integral multiple of it (n λ, where n = 1, 2, 3,……)

(i.e.,) nλ = (DB + BC)
[where n = order of diffraction]
DB = BC = d sin θ
θ = angle of incident beam,
(DB + BC) = 2d sin θ
d = distance between the planes
nλ = 2d sin θ This relation is known as Bragg’s equation.

Question 12.
Define molarity. Calculate the molarity of a solution con-taining 5 g of NaOH pi 450 ml solution.
Answer:
Molarity : The number of moles of solute dissolved in one litre of solution is called molarity

Question 13.
What are different types of adsorption ? Give any four differences between characteristics of these different types.
Answer:
Adsorption process is divided into two types :
1) Physiosorption
2) Chemisorption
Distinguishing characteristics of Physiosorption Chemisorption are given in the following table:

Physisorption	Chemisorption
1. This process is weak, due to vander Waall’s forces.	1. This process is strong, due to chemical forces.
2. The process is reversible	2. The process is irreversible
3. This is a quick process i.e. takes place quickly	3. This is a slow process.
4. The process decreases with increase of tempe¬rature	4. The process increases with increase of temperature.
5. This is a multilayered process	5. This is a unilayered process
6. The process depends mainly on the nature of the absorbent.	6. The process depends both on the nature of adsorbent and adsorbate.

Question 14.
Explain the purification of sulphide ore by Froth floatation
method.
Answer:
This method is used to concentrate sulphide ores.

1. In this process a suspension of the powdered ore is made with water.
2. A roating paddle is used to agitate the suspension and air is blown into the suspension in presence of an oil.
3. Froth is formed as a result of blown of air, which carries the mineral particles.
4. To the above slurry froth collectors and stabilizers are added.
5. Collectors like pine oil enhance non-wettability of the mineral particles.
6. Froth stabilizers like cresol stabilize the froth.
7. The mineral particles wet by oil and gangue particles wet by water.
8. The froth is light and is skimmed off. The ore particles are then obtained from the forth.

By using depressants in froth floatation process, it is possible to separate a mixture of two sulphide ores.
Eg: In the ore containing ZnS and PbS, the depressant used in NaCN. It prevents ZnS from coming to the froth, but allows PbS to come with the froth.

Question 15.
Explain the terms :
(a) Ligand
(b) Co-ordination number
(c) Co-ordination entity
(d) Central metal atom/ion
Answer:
i) Ligand:
The ions or molecules bound to the central atom/ ion in the co-ordination entity are called ligands. These may be:
a) simple ions such as Cl^–
b) small molecules such as H₂O or NH3
c) large molecules such as H₂NCH₂CH₂NH₂ or N(CH₂CH₂NH₂)₃ or even
(d) macro – molecules, such as proteins. On the basis of the number of donor atoms available for co-ordination, the ligands can be classified as :

a) Unidentate : One donor atom, Eg : NH₃, :CI: etc.

b) Bidentate : Two donor atoms Eg : H₂NCH₂CH₂NH₂ (ethane-1, 2-diamine or en), C₂O₄^2- (oxalate), etc.

c) Polydentate : More than two donor atoms, Eg:N(CH₂CH₂NH₂)₃ EDTA, etc.

ii) Co-ordination number:
The co-ordination number (CN) of metal ion in a complex can be defined as the number of ligands or donor atoms to which the metal is directly bonded.
Eg : In [PtCl₆]^2-, CN of Pt = 6,
In [Ni(NH₃)₄]²⁺, CN of Ni = 4..

iii) Co-ordination entity:
A central metal atoms or ion bonded to a fixed number of molecules or ions (ligands) is known as co-ordination entity. For example [CoCl₃(NH₃)₃]. Ni(CO)₄], etc.

iv) Central metal atom/ion : In a co-ordination entity, the atom/ion to which a fixed no. of ions/groups are bound in a definite geometrical arrangement around it is called central metal atom or ion.
Eg: K₄[Fe(CN)₆]’Fe’ is central metal.

Question 16.
How are XeF₂ and XeF₄ prepared? Give their structure.
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

a) Structure of XeF₂ :
1) In XeF₂ central atom is Xe’.
2) ‘Xe’ undergoes sp³d hybridisation in it’s 1 st excited state

3) Shape of molecule is linear
4) Xe form two a – bonds with two fluorines (sp³d²-2p_{z(F) overlap)}

b) Structure of XeF₄
1. Central atom in XeF₄ is ‘Xe’
2) ‘Xe’ undergoes sp³d² hybridisation in it’s 2^nd excited state

3) Shape of the molecule is square planar with bond angle 90° and  bond length 1. 95A.

Question 17.
What are hormones ? Give one example for each.
(a) Steroid hormone
(b) Polypeptide hormones and
(c) Amino acid derivatives.
Answer: Hormones : Hormone is defined as an “organic compound synthesised by the ductless glands of the body and carried by the blood stream to another part of the body for its funtion”. Ex: testosterone, estrogen

1. Example for steroid hormones : Testosterone, Estrogen
2. Example for poly peptide hormones : Insulin
3. Example for Amino acid activities: Thyroidal hormones thyroxine.

Question 18.
Write the IUPAC names of the following compounds :
(a) NH₂ – CH₂ – CH = CH₂
(b) C₂H₅ – N – CH₂ – CH₂ – CH₂ – CH₃

Answer:
a) NH₂ – CH₂ – CH = CH₂ = Prop 2 – ene 1 – Amine.
b) C₂H₅ – N – CH₂ – CH₂ – CH₂ – CH₃ = M, N – Di Ethyl Buton amine.

Section – C

Question 19.
a) What are Galvanic Cells ? Explain the working of a gal-vanic cell with a neat sketch taking Daniel cell as ex-ample.
Answer:
Galvanic cell: A device which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell (or) voltaic cell.
Eg: Daniell cell.
Daniell cell: It is a special type of galvanic cell. It contains two half cells in the same vessel.

The vessel is divided into two chambers. Left chamber is filled with ZnSO₄ (aq) solution and Zn – rod is dipped into it. Right chamber is filled with aq. CuSO₄ solution and a copper rod is dipped into it. Process diaphragm acts as Salt bridge. The two half cell’s are connected to external batteiy.

Cell reactions:
Ion Zn/ZnSO₄ half cell, oxidation, reaction occurs
Zn → Zn²⁺ + 2e
Ion Cu/CuS₄ half cell, reduction reaction occurs.
Cu₊₂ + 2e^– → Cu
The net cell reaction is
Zn + Cu⁺² ⇌ Zn⁺² + Cu
Cell is represented as Zn/Zn⁺²//Cu²⁺/Cu

b) What are fuel cells ? Give the construction of H₂, O₂ fuel cell.
Answer:
A fuel cell is a galvanic cell in which the chemical energy of fuel – oxidant system is converted directly into electrical energy.

1. Conventional Galvanic cell converts chemical energy into electrical energy by spontaneous redox reactions.
2. Fhel cell convert energy of combustion of fuels like hydrogen, methane etc., into electrical energy. These cause less pollution.

H₂-O₂ fuel cell: ln this cell, hydrogen and oxygen are bubbled through porous carbon electrodes into Cone. NaOH solution. Electrodes are embedded with suitable catalysts. The electrode reactions are :

Overall reaction : 2H₂(g) + O₂(g) → 2HO₂O(l)
The cell functions as long as the reacting gases are in supply. The heat of‘combustion is directly converted into electrical energy.

Question 20.
a) How is Ammonia manufactured by Haber’s process?
Answer:
In Haber process ammonia is directly synthesised from elements (nitrogen and hydrogen). The principle involved in this is
N₂(g) + 3H₂(g) ⇌ 2NH₃ + 92.4 kJ. This is a reversible exothermic reaction.

According to Le Chatelier’s principle favourable conditions for the better yield of ammonia are low temperature and high pressure. But the optimum conditions are:

Temperature : 720k
Pressure : 200 atmospheres
Catalyst : Finely divided iron in the presence of molybdenum (Promoter).

Procedure: A mixture of nitrogen and hydrogen in the volume ratio 1 : 3 is heated to 720 – 775 K at a pressure of 200 atmospheres is passed over hot finely divided iron mixed with small amount of molybdenum as promotor. The gases coming out of the catalyst chamber consists of 10-20% ammonia gas are cooled and compressed, so that ammonia gas is liquified, and the uncondensed gases are sent for recirculation,

a) Aq.ZnS₄ reacts with ammonia aqueous solution to form white ppt Of zinc hydroxide.
NH_3(ab) + H₂O_(l) ⇌ NH_4(ab)⁺ + OH_(ab)^–

ZnSO_4(ab)^– + 2NH_3(ab) → Zn(OH)_2(s) + (NH₄)₂SO_(ab)

b) Aq.CuSO₄ reacts with ammonia to form a deep blue complex.

c) Solid AgCl reacts with ammonia to form a colourless complex.

b) How does Ozone reacts with the following:
(i) pbS
(ii) KI
(iii) Hg
(iv) Ag
Answer:
Preparation of Ozone :
A slow dry stream of oxygen under silent electric discharge to form ozone (about 10%). The product obtained is known as ozonised oxygen.
3O₂ → 2O₃ ∆H° = 142 kJ/mole

1. The formation of ozone is an endothermic reaction.
2. It is necessary to use silent electric discharge in the preparation of Oa to prevent its decomposition.

i) Reaction with PbS : Black lead sulphide oxidised to white lead sulphate in presence of ozone.
PbS + 4O₃ → PbSO₄ + 4O₂
ii) Reaction with KI: moist KI is oxidised to Iodine in presence of ozone.
2KI + H₂O + O₃ → 2KOH + I₂ + O₂

iii) Reaction with Hg: Mercury loses it’s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.
2Hg + O₃ → Hg₂O + O₂

Question 21.
Explain the following reactions with equations.
a) Hell-Volhard-Zelinsky reaction (HVZ):
Answer:
Carboxylic acids having α – hydrogens are halogenated at the α – position on treatment with chlorine or bromine in presence of small amount of red phosphorous to give α – halo carboxylic acids.
This reaction is named as Hell – volhard – Zelinsky (HvZ) reactions

b) Decarboxylation:
Answer:
Carboxylic acids lose carbon dioxide molecule to produce hydrocarbons on heating their sodium salts with sodalime (a mixture of NaOH & CaO in ratio 3:1). This reaction is called decarboxylation.

c) Aldol condensation
Answer:
When aldol condensation is carried out between two different aldehydes and (or) ketones, it is called cross aldol condensation.

d) Gatterman-Koch reaction
Answer:
Gater man – Koch reactioni:
Benzene or its derivatives treated with CO and HCl in presence of A1Cl₃ gives Benzaldehyde.

Access to a variety of TS Inter 2nd Year Chemistry Model Papers and TS Inter 2nd Year Chemistry Question Paper May 2019 helps students overcome exam anxiety by fostering familiarity.

TS Inter 2nd Year Chemistry Question Paper May 2019

Note : Read the following instructions carefully.

1. Answer all questions of Section – ‘A’. Answer any six questions in Section – ‘B’ and any two questions in Section – ‘C’.
2. In Section – A, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
3. In Section -”’8′, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
4. In Section – ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
5. Draw labelled diagrams wherever necessary for questions in Section – ‘B’ and ‘C’.

Section – A

Note : Answer all the questions.

Question 1.
What is relative lowering of vapour pressure.

Question 2.
What is galvanic cell? Give one example.

Question 3.
Give the composition of the following alloys :
(a) German silver
(b) Brass
(c) Bronze

Question 4.
State Henry’s Law.

Question 5.
What are the neutral oxides of nitrogen?

Question 6.
Why does NH3 act as a Lewis base?

Question 7.
Write the structure of XeF₄.

Question 8.
Write equation for carbylamine reaction.

Question 9.
What are analgesis ? How are they classified?

Question 10.
CuSO₄, 5HO is blue in colour where as unhydroes. CuSO₄ is colourless. Why?

Section – B

Note : Answer any six questions.

Question 11.
Derive Bragg’s equation.

Question 12.
Calculate the molarity of a solution containing 5g of sodium Hydroxide (NaOH) in 450 mZ. solution.

Question 13.
Name any four enzyme catalysed reations.

Question 14.
Explain the following with examples
(i) Calcination
(ii) Roasting.

Question 15.
How are XeF₂ and XeF₆ prepared? Write their structures.

Question 16.
Write the IUPAC names write the systematic name of the following.
i) [Fe(C₂O₄)₃]₄
ii) [Fe(CN⁶)]^-4
iii) [NiCl₄]^-2
iv) [CO(NH₃)₆]Cl₃

Question 17.
Write notes on Antiseptics and Autibiotics.

Question 18.
Explain S¹N and S²N reactions with an examples.

Section – C

Note : Answer any two questions.

Question 19.
How is ozone prepared from Oxygen? Explain its reaction with,
(i) KI
(ii) PbS
(iii) C₂H₄
(iv) Hg

Question 20.
How ammonia manufactured by Haber’s process?

Question 21.
Describe the following :
(i) Cannizaro reaction
(ii) Williamson’s ether synthenis
(iii) Acetylation
(iv) Decorboxylation

Access to a variety of TS Inter 2nd Year Chemistry Model Papers and TS Inter 2nd Year Chemistry Question Paper March 2019 helps students overcome exam anxiety by fostering familiarity.

TS Inter 2nd Year Chemistry Question Paper March 2019

Note : Read the following instructions carefully.

1. Answer all questions of Section – ‘A’. Answer any six questions in Section – ‘B’ and any two questions in Section – ‘C’.
2. In Section – A, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
3. In Section -”’8′, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
4. In Section – ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
5. Draw labelled diagrams wherever necessary for questions in Section – ‘B’ and ‘C’.

Section – A

Note : Answer all the questions.

Question 1.
What are isotonic solutions?
Answer:
Isotonic solutions: Two solutions having same osmotic pressure . ‘ at a given temperature are called Isotonic Solutions.
e.g.: Blood is isotonic with 0.9% \(\left(\frac{W}{V}\right)\) NaCl [Saline]

Question 2.
What is a galvanic cell? Give one example.
Answer:
The electro chemical cell which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell.

Question 3.
Give the composition of the following alloys.
i) Brass
ii) Bronze
Answer:
i) Brass – 60% Cu, 40% Zn.
ii) Bronze – Cu 75 to 90%, Sn-10-25%

Question 4.
What is Lanthanoid contraction?
Answer:
Lanthanide contraction : The slow decrease of Atomic radius (or) atomic size in lanthanides with increase of atomic number is called Lanthanide contraction.

Question 5.
What is Ziegler – Natta catalyst?
Answer:
A mixture of tri alkyl Aluminium and titanium chloride is called Zieglar Natta catalyst.
E.g. : (C₂H₅)₃ Al + TiCl₄.

Question 6.
What is vulcanization of rubber?
Answer:
The process of heating natural rubber with sulphur to improve it’s properties at 373 – 415 K in presence of Zno or Zn – sterate is called vulcanization of rubber.

Question 7.
How Grignard reagent is prepared?
Answer:
Alkyl Magnesium halide is called Grignard reagent. It is prepared by the reaction of Alkyl halides with Magnesium metal in presence of dry ether.

Question 8.
Write equation for carbylamine reaction.
Answer:
Carbylamine reaction: Aniline reacts with chloroform in presence of alcoholic KOH to form foul smelling phenyl Isocyanide.

Question 9.
What are Enantiomers?
Answer:
The optical isomers which are mirror images but non-superior imposable are called Enantiomers.

Question 10.
Arrange the following in decreasing order of the ir basic strength :
C₆H₅NH₂, C₂H₅NH₂, (C₂H₅)₂ NH, NH₃.
Answer:
(C₂H₅)₂NH > C₂H₅NH₂ > NH₃ > C₆H₅NH₂

Section – B

Question 11.
Calculate the molarity of a solution containing 5g of Sodium Hydroxide (NaOH) in 450 ml Solution.
Weight 1000
Answer:

Question 12.
Derive Bragg’s equation.
Answer:
Derivation of Bragg’s equation : When X-rays are incident on the crystal or plane, they are diffracted from the lattice points (lattice points may be atoms or ions or molecules). In the crystal the lattice points are arranged in regular pattern. When the waves are diffracted from these points, the waves may be constructive or destructive interference.

The 1^st and 2_nd waves reach the crystal surface. They undergo constructive interference. Then from the figure 1st and 2nd rays are parallel waves. So, they travel the same distance till the wave form AD.

The second ray travels more than the first by an extra distance (DB + BC) after crossing the grating for it to interfere with the first ray in a constructive manner. Then only they can be in the same phase with one another. If the two waves are to be in phase, the path difference between the two ways must be equal to the wavelength (λ) or integral multiple of it (n λ, where n = 1, 2, 3 )

(i.e.,) nλ = (DB + BC) [where n = order of diffraction]
DB = BC = d sin θ
θ = angle of incident beam,
(DB + BC) = 2d sin θ
d = distance between the planes
nλ. = 2d sin θ
This relation is known as Bragg’s equation.

Question 13.
Write any four differences between physisorption and chemisorption.
Answer:
Distinguishing characteristics of Physisorption and Chemisorption are given in the following table :

Physisorption	Chemisorption
1. This process is weak, due to vander Waal’s forces.	1. This process is strong, due to chemical forces.
2. The process is rever sible.	2. The process is irreversible.
3. This is a quick process. i.e., takes place quickly.	3. This is a slow process.
4. The process decreases with increase of tempe rature.	4. The process increases with increase of temperature.
5. This is a multilayered process.	5. This is a unilayered process.
6. The process depends mainly on the nature of the adsorbent.	6. The process depends both on the nature of adsorbent and adsorbate.

Question 14.
Explain roasting and calcination.
Answer:
Roasting :
Removal of the volatile components of a mineral by heating mineral either alone (or) mixed with some other substances to a high temperature in the presence of air is called Roasting.

1. It is applied to the sulphide ores.
2. SO₂ gas is producted along with metal oxide.
   

Calcination :
Removal of the volatile components of a mineral by heating in the absence of air is called calcination.

1. It is applied to carbonates and bicarbonates.
2. CO₂ gas is produced along with metal oxide.
   

Question 15.
How is nitric acid manufactured by Ostwald’s process?
Answer:
Step – 1 : In step -1 Ammonia is oxidised to form Nitric oxide in presence of catalyst.

Step – 2 : Nitric oxide is oxidised to form Nitrogen dioxide.
2 NO + O₂ → 2 NO₂

Step – 3 : Nitrogen dioxide dissolved in water to form Nitric acid.
3 NO₂ + H₂O → 2 HNO₃ + NO
Here to get 98% HNO₃ it undergo distillation followed by dehydration with H₂SO₄.

Question 16.
Write the IUPAC names of the following co-ordination compounds :
a) [CO(NH₃)₄ (H₂O) Cl]Cl₂
b) [Ni(CO)₄
c) K₃(Fe(CN)₆]
d) [Cr(NH₃)₄(H₂O)₃]Cl₃
Answer:
a) [CO(NH₃)₄ (H₂O) Cl]Cl₂ – Mono aquo monochloro – Tetra amine cobalt (III) chloride
b) [Ni(CO)₄ – Tetra carbonyl Nickel (O)
c) K₃(Fe(CN)₆] – Potassium Hexa Cyano ferrate (III)
d) [Cr(NH₃)₄(H₂O)₃]Cl₃ – Tri amine Tri aquo Chromium Chloride (III)

Question 17.
What are harmones? Give one example for steroid hormones and polypeptide hormones.
Answer:
Hormones : Hormone is defined as an “organic compound synthesised by the ductless glands of the body and carried by the blood stream to another part of the body for its function”. Eg : testosterone, estrogen.

1. Example for steroid hormones :  Testosterone, Estrogen
2. Example for poly peptide hormones : Insulin
3. Example for Amino acid derivative : Thyroidal hormones thyroxine.

Question 18.
Write notes on Antiseptics and Antibiotics.
Answer:
Antiseptics :
The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics. Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
E.g. : Dettol, Bithional etc.

Antibiotics :
The chemical substances produced by micro organisms and inhibit the growth or destroy microorganisms are called antibiotics.
(Or)
The substance produced totally or partly by chemical synthesis which in low concentration inhibits the growth (or) destroy micro¬organism by intervening in their metabolic process are called antibiotics. E.g. : Penicillin, Chloramphenicol etc.

Section – C

Question 19.
How is chlorine prepared by Deacon’s process? Explain its reactions with Hot Cone. NaOH. H₂S and Na₂S₂O₃
Answer:
Deacon’s process : In Deacon’s process chlorine is obtained by the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCZ2 (catalyst) at 723 K.

Cl₂ reacts with hot Cone. NaOH to form sodium chloride and sodium chlorate
3Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O
Cl₂ reacts with H₂S and forms HCl and ‘S’
Cl₂ + H₂S → 2 HCl + S
‘S’ is precipitated by the reaction of Cl₂ with Na₂S₂O₃.
Na₂S₂O₃ + Cl₂ + H₂O → Na₂SO₄ + 2 HCl + S

Question 20.
Give a detailed account of the Collision theory of reaction rates of bimolecular reaction.
Answer:
Collision theory of reaction rate bimolecular reactions salient features.

1. The reaction molecules are assumed to be hard spheres.
2. The reaction is postulated to occur when molecules collide with each other.
3. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).
4. For a bimolecular elementary reaction A + B products
   Rate = Z_AB. e^-E_a/RT; Z_AB = collision frequency.
5. All collisions do not lead to product formation.
6. The collisions with sufficient kinetic energy (Threshold energy) are responsible for product formation. These are called as effective collisions.
7. To account for effective collisions a factor p called to proba¬bility factor or steric factor is introduced.
   Rate = P Z_AB . e^-E_a/RT
8. The proper orientation of reactant molecular lead to bond formation where as improper orientation makes them back and no product are formed.

Question 21.
With a suitable example write equations for the following:
i) Kolbe’s reaction
ii) Reimer – Tiemann reaction
iii) Williamson s ether synthesis
iv) Cannizaro reaction.
Answer:
i) Kolbe’s reaction :
Phenol reacts with sodium hydroxide to form sodium phenoxide. This undergoes electrophillic substitution with CO₂ to form salicylic acid.

ii) Reimer-Tiemann reaction :
Phenol reacts with chloroform in presence of NaOH to form salicylaldehyde (O – Hydroxy benzaldehyde). This reaction is known as Reimer-Tiemann reaction.

iii) Williamsons ether synthesis :
This method is used for the preparation of symmetrical and unsymmetrical ethers. The reaction of an alkyl halide with sodium alkoxide to form • ethers is known as Williamson’s Synthesis.

iv) Cannizaro’s reaction :
Aldehydes without a-hydrogen reacts with strong base to form an alcohol and salt of carboxylic acid.

Access to a variety of TS Inter 2nd Year Chemistry Model Papers and TS Inter 2nd Year Chemistry Question Paper May 2018 helps students overcome exam anxiety by fostering familiarity.

TS Inter 2nd Year Chemistry Question Paper May 2018

Note : Read the following instructions carefully.

1. Answer all questions of Section – ‘A’. Answer any six questions in Section – ‘B’ and any two questions in Section – ‘C’.
2. In Section – A, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
3. In Section -”’8′, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
4. In Section – ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
5. Draw labelled diagrams wherever necessary for questions in Section – ‘B’ and ‘C’.

Section – A

Note : Answer all the questions.

Question 1.
Give the units of rate constants for zero and first order reactions.
Answer:

Order	Units
Zero	mole.lit-1
First	Sec-1

Question 2.
What is tailing of mercury? How is it removed?
Answer:
Mercury loses it’s lustreness, meniscus and consequently sticks. to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.
2Hg + O₃ → Hg₂O + O₂
It is removed by shaking it with water which dissolves Hg₂O.

Question 3.
PH₃ has lower boiling point than NH₃ Why?
Answer:
In NH₃ hydrogen bonding is present. Where as in case of PH₃ hydrogen bonding is not present. So PH₃ has lower boiling point than NH₃.

Question 4.
State Henry’s law.
Answer:
At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid.
(Or)
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.
P = K_H × x K_H – Henry’s constant
P = Partial pressure
x = Mole fraction of gas

Question 5.
What is the role of cryolite in the metallurgy of aluminium?
Answer:
By adding the cryolite to the pure Alumina, the melting point of pure Alumina is lowered (which is very high 2324K) and eletrical conductivity of pure alumina is increased.

Question 6.
Arrange the following bases in decreasing order of pK_b values :
C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂ NH and C₆H₅NH₂.
Answer:
The decreasing order of pKb values of gives amines is
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂ NH

Question 7.
What are Enantiomers?
Answer:
Enantiomers: The stereo isomers related to each other as non- superimposable mirror images are called enantiomers.
These have identical physical properties like melting point, boiling points refractive index etc.  They differ in rotation of plane polarised right.

Question 8.
What is PDI (Poly Dispersity Index)?
Answer:
Poly Dispersity Index (PDI): The ratio between weight average molecular mass (Mw) and the number average molecular mass (Mn) of a polymer is called Poly Dispersity Index (PDI).

Question 9.
Write the name of the monomers present in Buna-S and Bakelite.
Answer:
→ In Buna – S monomers are 1,3- Butadiene
(CH₂ = CH – CH = CH₂)
and Styrene (C₆H₅ – CH = CH₂)
→ In Bakelite monomers are phenol (C₆H₅ OH) Formaldehyde (HCHO)

Question 10.
CuSO₄. 5H₂O is blue in colour whereas anhydrous CuSO₄ in colourless. Why?
Answer:
CuSO₄.5H₂O is blue in colour whereas anhydrous CuSO₄ is colourless because in the absence of ligand, crystal field splitting doesnot occurs.

Section – B

Question 11.
Derive Bragg’s equation.
Answer:
Derivation of Bragg’s equation : When X-rays are incident on the crystal or plane, they are diffracted from the lattice points (lattice points may be atoms or ions of molecules) . In the crystal the lattice points are arranged in regular pattern. When the waves are diffracted from these points, the waves may be constructive or destructive interference.

The 1^st and 2^nd waves reach the crystal surface. They undergo constructive interference. Then from the figure 1st and 2nd rays are parallel waves. So, they travel the same distance till the wave form AD. The second ray travels more than the first by an extra distance(DB + BC) after crossing the grating for it to interfere with the first ray in a constructive manner. Then only they can be in the same phase with one another. If the two waves are to be in phase, the path difference between the two ways must be equal to the wavelength (λ) or integral

multiple of it (n λ, where n = 1,2, 3 )
(i.e.,) nλ = (DB + BC)
[where n = order of diffraction]
DB = BC = d sin θ
θ = angle of incident beam,
(DB + BC) = 2d sin θ
d = distance between the planes
nλ = 2d sin θ
This relation is known as Bragg’s equation.

Question 12.
Calculate the molefraction of Ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass.
Answer:
Given 20% of C₂H₆O₂ solution by mass

Question 13.
What is catalysis? How is catalysis classified? Give two examples for each type of catalysis.
Answer:
Catalysis :
A substance which alters the rate of a chemical reaction without itself being consumed in the process, is called a catalyst. The action of catalyst in altering the rate of a chemical reaction is called catalysis.

Types of catalysis :
Catalysis is classified into two types as a) Homogeneous catalysis and b) Heterogeneous catalysis.

Homogeneous catalysis :
The catalytic process in which the catalyst is present in the same phase as that of reactants, is known as homogeneous catalysis.

Heterogeneous catalysis :
The catalytic process in which the catalyst is present in a phase different from that of the reactants is known as heterogeneous catalysis.

Question 14.
Write the IUPAC names of the following co-ordination compounds :
a) [PtCl₂ (NH₃)₂]
b) [Ni(CO)₄]
c) K₃[Al(C₂O₄)₃]
d) [CoCl₂(en)₂]Cl
Answer:
a) [PtCl₂ (NH₃)₂] – Diamine Diamine Platinum (II)
b) [Ni(CO)₄] – Tetra Carbonyl Nickel (O).
c) K₃[Al(C₂O₄)₃] – Potassium Trioxalate Aluminate (III)
d) [CoCl₂(en)₂]Cl – Bis Ethylene diamine dichloro Cobalt (III) Chloride.

Question 15.
Explain the purification of Sulphide ore by Froth floatation method.
Answer:
Froth floatation method :

1. This method is used to concentrate sulphide ores.
2. In this process a suspension of the powdered ore is made with water.
3. A rotating paddle is used to agitate the suspension and air is blown into the suspension in presence of an oil.
4. Froth is formed as a result of blown of air, which carries the mineral particles.
5. To the above slurry froth collectors and stabilizers are added. Collectors like pine oil enhance non-wettability of the mineral particles.
6. Froth stabilizers like cresol stabilize the froth.
7. The mineral particles wet by oil and gangue particles wet by water.
   
8. The froth is light and is skimmed off. The ore particles are then obtained from the froth.

By using depressants in froth floatation process, it is possible to separate a mixture of two sulphide ores. Eg : In the ore containing ZnS and PbS, the depressant used is NaCN. It prevents ZnS from coming to the froth but allows PbS to come with the froth.

Question 16.
Write short notes on :
a) Carbylamine reaction
b) Racemisation
Answer:
a) Carbylamine reaction : Aniline reacts with chloroform in present of alc.KOH to form foul smelling phenyl isocyanide. This reaction is called carbylamine reaction.
C₆H₅NH₂ + CHCl₃ + 3 KOH (ale) → C₆H₅NC + 3 KCZ + 3H₂O

b) Racemisation :
Racemic mixture: Equal portions of Enantiomers combined to form an optically inactive mixture. This mixture is called racemic mixture.
1) Here rotation due to one isomer will be exactly cancelled by the rotation of due to other isomer.
2) The process of conversion of enantiomer into a racemic mixture is called as racemisation.

Question 17.
What are Hormones? Give one example for each:
i) Steroid hormones.
ii) Polypeptide hormones.
iii) Amino acid derivatives.
Answer:
Hormones : Hormone is defined as an “organic compound synthesised by the ductless glands of the body and carried by the blood stream to another part of the body for its function”. Eg : testosterone, estrogen.

1. Example for steroid hormones : Testosterone, Estrogen
2. Example for polypeptide hormones : Insulin
3. Example for Amino acid derivative: Thyroidal hormones thyroxine.

Question 18.
Write short notes on :
a) Artificial sweetening agents.
b) Food preservatives.
Answer:
a) Artificial sweetening agents:
The chemical substances which are used instead of sucrose (or) sugar are called artificial sweetening agents. E.g. : Aspartame, Alitame, saccharin. These decrease the Calorific in take and at the same time several times sweeter than sucrose.

b) Food preservatives:
The chemical substances which prevent the spoilage of food due to microbial growth are called food preservatives.E.g. : Sodium benzoate, Salt of sorbic acid etc.

Section – C

Question 19.
a) How is XeF₂ and XeF₄ prepared? Give their structures.
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

Structure of XeF₂ :
1) In XeF₂ central atom is ‘Xe’. ,
2) Xe’ undergoes sp³d hybridisation in it’s 1^st excited state.

3) Shape of molecule is linear.
4) Xe form two σ.bonds with two fluorines.

b) Structure of XeF₄ :
1) Central atom in XeF₄ is Xe’.
2) Xe undergoes sp³d² hybridisation in it’s 2^ndexcited, state.

3) Shape of the molecule is square planar with bond angle 90° and bond length 1.95A.

4) Xe – forms four o-bonds by the overlap of sp³d² – 2p_Z (F) orbitals.

b) How is chlorine prepared by Deacon’s method? Explain its reaction with the following:
i) Iron
ii) Na₂S₂O₃
Answer:
Deacon’s process : In Deacon’s process chlorine is obtained by the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl₂ (catalyst) at 723 K.

i) Iron : Cl₂ reacts with Iron to form FeCl₃
2 Fe + 3Cl₂ → 2 FeCl₃

ii) Na₂S₂O₃ : ‘S’ is precipitated by the reaction of Cl₂ with Na₂S₂O₃.
Na₂S₂O₃ + Cl₂ + H₂O → Na₂SO₄ + 2 HCl + S

Question 20.
a) Describe the salient features of the collision theory of reaction rates of bimolecular reaction.
Answer:
Collision theory of reaction rate bimolecular reactions salient features.

1. The reaction molecules are assumed to be hard spheres
2. The reaction is postulated to occur when molecules collide with each other.
3. The number of collisions per second per unit volume of the reaction mixture is known as collision
   frequency (Z).
4. For a bimolecular elementary reaction A + B products
   K= Z_AB.e^-E_a/RT;Z_AB = collision frequency.
5. All collisions do not lead to product formation.
6. The collisions with sufficient kinetic energy (Threshold energy) are responsible for product formation. These are called as effective collisions.
7. To account for effective collisions a factor p called to probability factor or steric factor is introduced. K = PZ_AB.e^-E_a/RT
8. The proper orientation of reactant molecular lead to bond formation where as improper orientation makes them back and no product are formed.
   
9. In this theory activation energy and proper orientation of the molecules to gather determine the creteria for an effective collision and hence the rate of a chemical reaction.

b) State Kohlrausch’s law of independent migration of ions. Give its applications.
Answer:
Kohlrausch’s law of independent migration of ions : The limiting molar conductivity of an electrolytes can be represented as the sum of the individual contributions of the anion and the cation of the electrolytes.

λm0 = λA+0 + λB–0
λA+0 = Limiting molar conductivity
λA+0 = Limiting molar conductivity of cation
λB–0 = Limiting molar conductivity of anion

Applications :
1) Kohlrausch’s law is used in the calculation of the limiting molar conductivity of weak electrolytes.
Eg: λ_CH3COOH = λ_CH3COONa + λ_HCl – λ_NaCl
= λ_CH3COO^– + λ_Na⁺ + λ_H⁺ + λ_Cl^– – λ_Na⁺ – λ_Cl^–
= λ_CH3COO^– + λ_H⁺
2) This law is used in the calculation of degree of dissociation of a weak electrolyte.
3) This law is used in the calculation of solubility of sparingly soluble salts like AgCl, BaSO₄

Question 21.
Explain the following reactions :
i) Williamson’s synthesis.
ii) Hell-Volhard-Zelinsky reaction.
iii) Cannizaro reaction.
iv) Aldol condensation.
Answer:
i) Williamson’s synthesis :
Alkyl halides reacts with sodium j alkoxides to form ethers. This reaction is called Williamson’s
synthesis.
R-X + R’-ONa → R-O-R’ + NaX
E.g. : Ethyl chloride reacts with sodium ethoxide to form. diethyl ether.
C₂H₅Cl + C₂H₅ONa → C₂H₂OC₂H₅ + NaCl

ii) Hell-Volhard-Zelinsky reaction :
Carboxylic acids having a – hydrogen reacts with halogen and red phosphorus to form a-halo carboxylic acids. This reaction is called H.V. Z. reaction.

Eg : Acetic acid reacts with Cl₂ and red ‘P’ to form finally trichloro acetic acid.

iii) Carinizaro reaction:
Aldehydes without a-hydrogen undergo reaction with strong base to form an alcohol and salt of carboxylic acid. This reaction is called Cannizaro reaction.

iv) Aldol condensation:
Aldehydes having a-hydrogen undergo reaction in presence of base to form the products Aldol. This resction is called aldol condensation

Access to a variety of AP Inter 2nd Year English Model Papers and AP Inter 2nd Year English Question Paper March 2023 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year English Question Paper March 2023

Time : 3 Hours
Max. Marks : 100

SECTION – A

I. Annotate ANY TWO of the following in about 10-15 lines each. (2 × 4 = 8)

a) Every work must necessarily be a mixture of good and evil yet we are commanded to work incessantly.

b) A mortal dose of chloroform was dropped into the water. 1 The graph became the record of a death agony.

c) People that value its privileges above its principles soon lose both.

d) Studies serve for delight, for ornament and for ability.

n. Annotate ANY TWO of the following in about 10-15 lines each. (2 × 4 = 8)

a) All are architects of fate
Working in these walls of time.

b) All is, If I have grace to use it so
As ever in my great Task-Master’s eye.

c) Yet will I slake my individual sorrow
At the deep source of universal joy.

d) Books ! T is a dull and endless strife
Come, hear the woodland linnet.

III. Answer ANY TWO of the following questions in about 10-15 lines each. (2 × 4 = 8)

a) How does studies cure the diseases of the mind ?

b) What was the accident that Dr. Barnard Hord ? How did he react to the accident ?

c) What does Vivekananda say about a man of character ?

d) Give a list of the lessons that Narayana Murthy feels we should learn from the west.

IV. Answer ANY TWO of the following in about 10-15 lines each. (2 × 4 = 8)

a) Attempt a critical appreciation of the poem “On His Having Arrived At The Age of Twenty Three”.

b) What is the central idea of the poem, The Builders’ ?

c) Why are people generally afraid of fate ? How did Sarojini Naidu challenge such a fearful fate ?

d) Why does Wordsworth consider Nature to be a good teacher?

V. Answer ANY ONE of the following questions in about 25 lines. (1 × 8 = 8)

a) Give a character analysis of Huckleberry Finn.

b) What are the turning points in the novel “The Adventures of Tom Sawyer” ? Discuss the main events in the plot.

c) Write a character sketch of Tom Sawyer.

SECTION – B

VI. Read the following passage carefully and answer the ques-tions that follow. (5 × 1 = 8)

Fifty feet away, three male lions lay by the road. They didn’t appear to have a hair on their heads. Noting the colour of their noses (leonine noses darken as they age, from pink to black). Craig estimated that they were six years old-young adults. “This is what we came to see. They are really mane-less”. Craig, a professor at the University of Minnesota, is arguably the leading expert on the majestic Serengeti lion, whose head is mantled, in long, thick hair. He and Peyton West, a doctoral student who has been working with him in Tanzania, had never seen the Tsavo lions that live some 200 miles east of Serengeti. The scientists had partly suspected that the maneless male lions were adolescents mistaken for adults by amateur observers. Now they know better.
The Tsavo research expedition was mostly Peyton’s show. She had spent several years in Tanzania, compiling the data she needed to answer a question that ought to have been answered long ago. Why do lions have manes ? It’s the only cat wild or domestic, that displays such ornamentation.

Question 1.
Name the two scientists in the passage.

Question 2.
State true or false :
Serengeti lions are maneless.

Question 3.
Where do Tsavo lions live ?

Question 4.
Pick the adjective in the passage which means ‘pertaining to or characteristic of a lion’.

Question 5.
Write the noun form of the verb ‘estimated’.

VII. Read the following passage carefully and answer the questions that follow. (5 × 1 = 5)

In the evening the other children returned to the boat. They laughed and talked, they were very tired. They did not see that Tom and Becky were not there. The boat took them back to St. Petersburg.
Huck saw the boat but he did not know about the picnic. The mothers of St. Petrsburg did not like him. They never invited him to birthday picnics. But tonight Huck was not interested in the birthday picnics. He was interested in Injun Joe’s treasure. He hid behind a tree and watched an old house ‘Injun Joe’s in that house’, he thought. To stay here and wait. When he comes out I’lll find the treasure’. It was late and very dark. Soon two men came out. It was Injun Joe and his friend. Huck followed them quietly. ‘They’re going to Widow Dougals’s House, thought Huck, ’But why ?’ Suddenly the ’ two men stopped. Injun Joe said, ‘Many years ago Widow Dougals’s husband was very cruel to me. Now I want to hurt the widow. I want to cut her face, her nose and her ears. And you must help me’.

Question 1.
Why were Tom and Becky not with the other children ?

Question 2.
Why did Huck not know about the picnic ?

Question 3.
What was Huck interested in that night ?

Question 4.
Why were the two men going to Widow Dougals’s house ?

Question 5.
State true or false :
Huck was waiting for Injun Joe behind the tree because he knew that Injun Joe was going to hurt Widow Dougals.

VIII. Study the advertisement and answer the questions that follow. (5 × 1 = 5)

Question 1.
What is the advertisement about ?

Question 2.
Write down two major benefits of the new frame work of India.

Question 3.
What is the TRAI website ?

Question 4.
Where can we find ‘consumer information’ ?

Question 5.
Write a synonym of ‘disruption’.

IX. Study the Pie-Chart given below and answer the questions that follow. (5×1 = 5)

THE MAJOR SOURCES OF ENERGY IN INDIA

Question 1.
What does the Pie-chart show ?

Question 2.
State True or False.
Natural gas and coal put together is almost equal to our major source of energy.

Question 3.
What is the second major source of energy in our country ?

Question 4.
How many sources of energy are taken into consideration ?

Question 5.
What is hydroelectric power ?

SECTION – C

X. Write a letter to your friend describing the college annual day celebrations conducted recently in your college. (1 × 5 = 5)
Hints : Describe college annual day-sports and games-com- petitions-literary and cultural:Chief guest-principal-annual report-prize distribution-cultural programmes.

(Or)

Write a leter to the principal of your college requesting him/ her to issue your transfer, conduct and bona fide certificates.
Hints : Completed intermediate – group – admn. No. – passed in distinction – year of study – dues paid – going to further studies – certificates required – requested.

XI. Write a paragraph of about 8 lines describing the process of how you get ready for college regularly. (1 × 5 = 5)
(Or)
Describe the process of planting a sapling.

XII. Prepare a Curriculum Vitae based on the information given below. (1 × 5 = 5)

N. Sunitha-aged 24 years-B.Ed., 78% Govt., College of Edn., Rajahmundry-B.Sc. (Biology) 85% marks-Govt College for Women, Rajahmundry-Intermediate (Bi.RC.) 78% Govt. Junior College, Palakollu, S.S.C.-88%-Govt., High School, Palakollu-applying for the post of Biology Teacher.

XIII. You are Mr. Mohan. Your father Mr. Narasimha has asked you to deposit Rs. 10,600/- (in Rs. 2,000 and Rs. 100 denomination) in his Post Office Savings Account in Antarvedi Post Office. His account No. is 20231803. Fill in the form with the information given. (It is not necessary to draw the form. Write the numbers 1 to 10 and the corresponding answers). (1 × 5 = 5)

XIV. Write a dialogue between you and your friend inviting him to your birthday party. (1 × 5 = 5)
(Or)
Imagine you are selected for an interview to get a job in a reputed company. Write a dialogue between yourself and the interviewer.

XV. Read the following passage and make notes. (1 × 5 = 5)

Earthworms are the answer for every garden problem, according to Harold John Weigel. They can increase crop production, turn and freshen soil, and produce faster growth. Simply take care of the earthworms, and the earthworms will take care of the garden. Weigel is extremely enthusiastic about earthworms. They are tremendous creatures, “the intestines of the Earth”, Weigel says quoting Charles Darwin. Weigel is so excited about the benefits of Worms that he is writing a book about them. He gardens using thousands of earthworms. He has persuaded his wife to put worms in her house plant pots. He even suggests eating worms, which he claims, are 70 per cent protein. He has dreams of armies of earthworms helping to replace topsoil in the 74 countries. It is a fact, he says, that topsoil is disappearing every year through erosion. Wind and water carry away the soil and nature need’s centuries to replace it.

XVI. Match the words in Column-A with their meanings/definitions in Column-B. (5 × 1 = 5)

XVII. Mark the stress for any five of the following words. (5 × 1 = 5)
1) country
2) failure
3) polite
4) student
5) among
6) discuss
7) engineer
8) beautiful
9) delight
10) intermediate

Access to a variety of TS Inter 2nd Year Maths 2B Model Papers and TS Inter 2nd Year Maths 2B Question Paper March 2019 allows students to familiarize themselves with different question patterns.

TS Inter 2nd Year Maths 2B Question Paper March 2019

Time : 3 Hours
Max. Marks : 75

Section – A
(10 × 2 = 20)

I. Very Short Answer type questions.

1. Attempt all questions.
2. Each question carries two marks.

Question 1.
Write the parametric equations of the circle 2x² + 2y² = 7.
Solution:
Given circle equation is 2x² + 2y² = 7
⇒ x² + y² = \(\frac{7}{2}\)
Here centre (h, k) = (0, 0) and
radius r = \(\sqrt{\frac{7}{2}}\)
Circle Parametric equations are

Question 2.
Find the value of k, if the points (1, 3) and (2, k) are conjugated with respect to the circle x² + y² = 35.
Solution:
Let S = x² + y² – 35= 0
Since the points (1, 3) and (2, k) are conjugated with respect to the circle S = 0
∴ S₁₂ = 0
⇒ x₁x₂ + y₁y₂ – 35 = 0
⇒ 2 + 3k – 35 = 0 ⇒ 3k = 33 ⇒ k = 11

Question 3.
Find the equation of radical axis of the circles x² + y² + 4x + 6y – 7 = 0, 4(x² + y²)+ 8x + 12y – 9 = 0.
Solution:
Let S ≡ x² + y² + 4x + 6y – 7 = 0,
S¹ ≡ x² + y² + 2x + 3y – \(\frac{9}{4}\) = 0
The equation of the radical axis of the circles S = 0, S¹ = 0 is S – S¹ = 0
⇒ (x² + y² + 4x + 6y – 7) – (x² + y² + 2x + 3y – \(\frac{9}{4}\)) = 0
⇒ x² + y² + 4x + 6y – 7 – x² – y² – 2x – 3y + \(\frac{9}{4}\) = 0
⇒ 8x + 12y – 19 = 0.

Question 4.
Find the equation of the normal to the parabola y² = 4x which is parallel to y – 2x + 5 = 0.
Solution:
Given parabola equation is y² = 4x
Here 4a = 4 ⇒ a = 1
Given line equation is y – 2x + 5 = 0
⇒ y = 2x – 5
∴ Slope m = 2
Since required normal is parallel to y = 2x – 5
∴ Slope of normal = 2
⇒ -t = 2 ⇒ t = -2
Equation of the normal at t is
y + xt = 2at + t³
⇒ y + x(-2) = 2(1) (-2) + (-2)³
⇒ y – 2x = -4 – 8
⇒ 2x – y – 12 = 0.

Question 5.
If the eccentricity of the hyperbola is \(\frac{5}{4}\), then find the eccentricity of its conjugate hyperbola.
Solution:
If e₁, e₂ are the eccentricities of hyperbola and conjugate hyperbola then \(\frac{1}{e_1^2}+\frac{1}{e_2^2}\) = 1

Question 6.
Evaluate \(\int \frac{1+\cos ^2 x}{1-\cos 2 x} d x\), on I ⊂ R\{nπ : n ∈ z}.
Solution:

Question 7.
Evaluate \(\int \frac{1}{x \log x[\log (\log x)]}\)dx, on (1, ∝).
Solution:

Question 8.
Evaluate \(\int_0^a(\sqrt{a}-\sqrt{x})^2\) dx
Solution:

Question 9.
Find \(\int_0^{\pi / 2}\)cos¹¹ xdx.
Solution:

Question 10.
Find the general solution of : \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2 y}{x}\)
Solution:
Given \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2 y}{x}\) ……… (1)
⇒ \(\frac{\mathrm{dy}}{\mathrm{y}}\) = 2\(\frac{d x}{x}\)
2\(\frac{d x}{x}\) = \(\frac{d y}{y}\) = 0
Integrating

Section – B
(5 × 4 = 20)

II. Short Answer type questions.

1. Answer ANY FIVE questions.
2. Each question carries FOUR marks.

Question 11.
Find the equation of the circle which cuts orthogonally the circle x² + y² – 4x + 2y – 7 = 0 and having the centre at (2, 3).
Solution:
Let x² + y² + 2gx + 2fy + c = 0 be the required circle equation ………. (1)
Given centre (-g, -f) = (2, 3)
∴ -g = 2,-f = 3
⇒ g = -2, f = -3.
Given circle equation in x² + y² – 4x + 2y – 7 = 0 …….. (2)
Since (1), (2) cut orthogonally
∴ 2g₁g₂ + 2f₁f₂ = c₁ + c₂
⇒ 2(-2)(-2)+2(-3)(1) = -7 + c
⇒ 8 – 6 = -7 + c
⇒ 2 = -7 + c ⇒ c = 9
Hence required circle equation is
x² + y² + 2(-2)x + 2(-3)y + 9 = 0
⇒ x² + y² – 4x – 6y + 9 = 0.

Question 12.
The line y = mx + c and the circle x² + y² = a² intersect at A and B. If AB = 2λ, then show that:
c² = (1 + m²)(a² – λ²).
Solution:
Given circle equation is x² + y² = a²
Centre c = (0, 0) and radius r = a
d = The ⊥ distance from c to line
y = mx + c

Question 13.
Find the equation of the ellipse with focus at (1, -1), e = \(\frac{2}{3}\) and directrix as x + y + 2 = 0.
Solution:
Given focus s = (1, -1)
Eccentricity e = \(\frac{2}{3}\)
Equation of directrix is x + y + 2 = 0
Let P(x, y) be any point on the ellipse
\(\frac{S P}{P M}\) = e
⇒ SP = e.PM
⇒ SP² = e².PM²

⇒ 9x + 9y² – 18x + 18y + 18 = 2x² + 2y² + 8 + 4xy + 8x + 8y
⇒ 7x² – 4xy + 7y² – 26x + 10y + 10 = 0
Hence required ellipse equation is
7x² – 4xy + 7y² – 26x + 10y + 10 = 0.

Question 14.
Find the equations of tangents to the ellipse 2x² + 3y² = 11 at the points whose ordinate is 1.
Solution:
Given ellipse equation is 2x² + 3y² = 11
Given ordinate y = 1
⇒ 2x² + 3 = 11
⇒ 2x² = 8
⇒ x² = 8
⇒ x = ±2
∴ Points on the ellipse are P(1, 2) and Q(-2, 1)

Case (i): P(2, 1)
∴ Equation of the tangent at P is S₁ = 0
⇒ 2x(2) + 3y(1) – 11 = 0
⇒ 4x + 3y – 11 = 0
∴ Equation of the normal at P is
3(x – 2) – 4(y – 1) = 0
⇒ 3x – 6 – 4y + 4 = 0
⇒ 3x – 4y – 2 = 0

Case (ii): Q (-2,1)
Equation of the tangent at Q is S₁ = 0
⇒ 2x(-2) + 3y(1) – 11 = 0
⇒ – 4x + 3y – 11 = 0
⇒ 4x – 3y + 11 = 0
Equation of the normal at Q is
3(x + 2) + 4(y – 1) = 0
⇒ 3x + 6 + 4y – 4 = 0
⇒ 3x + 4y + 2 = 0.

Question 15.
Find the foci, eccentricity, equations of the directrix, length of latus rectum of the hyperbola x² – y² = 4.
Solution:
Given equation of the hyperbola is x² – 4y² = 4
Here a² = 4, b² = 1
Centre = (0, 0)
Eccentricity e = \(\sqrt{\frac{a^2+b^2}{a^2}}\)
= \(\sqrt{\frac{4+1}{4}}\)
= \(\sqrt{\frac{5}{4}}\)
= \(\frac{\sqrt{5}}{2}\)

Question 16.
Find: \(\int_0^{2 \pi}\) sin⁴ x cos⁶ x dx.
Solution:

Question 17.
Solve the differential equation : cos x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + y sin x = sec² x.
Solution:
Given cos x\(\frac{d y}{d x}\) + y sin x = sec² x
⇒ \(\frac{d y}{d x}\) + y tanx = sec³x …….. (1)
Here P = tan x, Q = Sec³ x

Section – C
(5 × 7 = 35)

III. Long Answer type questions.

1. Answer ANY FIVE questions.
2. Each question carries SEVEN marks.

Question 18.
Show that the points (9, 1), (7, 9), (-2, 12), (6, 10) are concyclic and find the equation of the circle on which they lie.
Solution:
Let A = (9, 1), B = (7, 9), C = (-2, 12), D = (6, 10)
The equation of the circle which having \(\overline{\mathrm{AB}}\) as a diameter is
(x – x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
⇒ (x – 9)(x – 7) + (y – 1 )(y – 9) = 0
⇒ x² – 7x – 9x + 63 + y² – 9y – y + 9 = 0
⇒ x² + y² – 16x – 10y + 72 = 0 ……… (1)
Slope of \(\overline{\mathrm{AB}}\) = \(\frac{y_2-y_1}{x_2-x_1}\) = \(\frac{9-1}{7-9}\) = \(\frac{8}{-2}\) = -4
Equation of \(\overline{\mathrm{AB}}\) is y – 1 = – 4(x – 9)
⇒ y – 1 = – 4x + 36
⇒ 4x + y – 37 = 0
The equation of circle passing through A, B is
(x² + y² – 16x – 10y + 72) + C(4x + y – 37) = 0
If this circle passes through C(-2, 12) then
4 + 144 + 32 – 120 + 72 + C(-8 + 12 – 37) = 0
⇒ 132 – 33C = 0
⇒ 33C = 132 ⇒ C = 4
∴ Equation of the circle passes through A, B, C is
(x² + y² – 16x – 10y + 72) + 4(4x + y – 37) = 0
⇒ x² + y² – 16x – 10y + 72 + 16x + 4y – 148 = 0
⇒ x² + y² – 6y – 76 = 0 ……….. (2)
x² + y² – 6y – 76 = (6)² + (10)² – 6(10) – 76
= 36 + 100 – 60 – 76 = 0
∴ D(6, 10) lies on (2)
∴ A, B, C, D are concydic and circle equation is x² + y² – 6y – 76 = 0.

Question 19.
Show that, four common tangents can be drawn for the circles given by x² + y² – 14x + 6y + 33 = 0 and x² + y² + 30x – 2y + 1 = 0 and find the internal and external centres of similitude.
Solution:
Given circle equations are x² + y² – 14x + 6y + 33 = 0
x² + y² + 30x – 2y + 1 = 0
Centres A = (7, -3), B = (-15, 1)
r₁ = \(\sqrt{49+9-33}\) = \(\sqrt{25}\) = 5

∴ Four tangents can be drawn for the given circles.
Let P, Q be the internal and external centres of similitude.
∴ P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 3 internally

Question 20.
From an external point ‘P’ tangents are drawn to the parabola y² = 4ax and these tangents make angles θ₁, θ₂ with its axis, such that cot θ₁ + cot θ₂ is a constant ‘d’. Then show that all such P lie on a horizontal line.
Solution:
Equation of a tangent to the Parabola y² = 4ax is a
y = mx + \(\frac{a}{m}\)
This tangent passes through P(x₁y,)
∴ y₁ = mx₁ + \(\frac{\mathrm{a}}{\mathrm{m}}\)
⇒ my₁ = m²x₁ + a
⇒ m²x₁ – my₁ + a = 0 which is a quadratic eq. in m
If m₁, m₂ are the slopes of the tangents drawn from P then m₁, m₂ are the roots of the equation m²x₁ – my₁ + a = 0

Question 21.
Evaluate : ∫e^ax sin(bx + c)dx; (a, b, c ∈ R; b ≠ 0) on R.
Solution:

Question 22.
Obtain reduction formula for I_n = ∫cotⁿ x dx; n being a positive integer; n ≥ 2 and deduce the value of ∫cot⁴ x dx.
Solution:

Question 23.
Find : \(\int_0^\pi x \sin ^7 x \cos ^6 x d x .\)
Solution:

Question 24.
Solve the differential equation : \(\frac{d y}{d x}\) = \(\frac{2 y+x+1}{2 x+4 y+3}\)
Solution:


Integrating
\(\int \frac{1}{2} d z\) + \(\frac{1}{2} \int \frac{1}{4 z+5}\)dz = \(\int d x\) + \(\frac{c}{8}\)
⇒ \(\frac{1}{2}\)z + \(\frac{1}{2} \cdot \frac{1}{4}\)log|4z + 5| = x + \(\frac{c}{8}\)
⇒ 4z – log|4z + 5| = 8x + c
⇒ 4(x + 2y) + log|4(x + 2y) + 5| = 8x + c
⇒ 8y – 4x + log|4x + 8y + 5| = c
∴ The general solution of (1) is
8y – 4x + log|4x + 8y + 5| = c.

Access to a variety of AP Inter 2nd Year English Model Papers and AP Inter 2nd Year English Question Paper April 2022 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year English Question Paper April 2022

Time : 3 Hours
Max. Marks : 100

SECTION – A

I. Annotate ANY TWO of the following. (2 × 4 = 8)

Question 1.
Some books are tasted, others to be swallowed and some few to be chewed and digested.

Question 2.
The only solution of the problem is to make mankind pure.

Question 3.
You don’t become a better person because you are suffering but you become a better person because you have experienced suffering.

Question 4.
People that value its privileges above its principles soon lose both.

II. Annotate ANY TWO of he following. (2 × 4 = 8)

Question 1.
Enough Science and a Arts
Close up those barren leaves.
Answer:

Question 2.
Time is with materials filled
Our todays and yesterdays
Are the blocks with which build.

Question 3.
Take me away and roof and wall
Would fall to ruin me utterly

Question 4.
For all the cruel folly you pursure
I will not cry with suppliant hands to you.

III. Answer ANY TWO of the following questions in 10-15 lines each. (2 × 4 = 8)

Question 1.
How do studies cure the diseases of the mind ?

Question 2.
Explain the things that guide the conduct of human beings.

Question 3.
What lessons did Bernard learn from the two children ?

Question 4.
Give a list of lessons that Narayana Murthy feels we should learn from the West.

IV. Answer ANY TWO of the following questions in 10-15 lines each. (2 × 4 = 8)

Question 1.
Why does Wordsworth consider Nature to he a good teacher?

Question 2.
What is the central idea of the poem – The Builders’?

Question 3.
The poem ‘Any Woman’ is a celebration of the glory of ‘womanhood’. Illustrate.

V. Answer ANY ONE of the following questions in about 25-30 lines each. (1 × 8 = 8)

Question 1.
Sketch the character of Tom Sawyer

Question 2.
What are turning points in the story ? Discuss the main events in the story.

Question 3.
Write at least five lines about the following characters.
a) Aunt Polly
b) Muff Potter
c) Becky Thatcher

SECTION – B

VI. Read the following passage carefully and answer he questions that follow. (5 × 1=5)

A man came to Hyderabad to attend a conference. He was visiting the city for the first time. He wanted to see the Charminar. So, the next day after he attended the conference, he set out to visit the Charminar. He asked a police officer for the directions to reach that place. The officer said, ‘Wait at this bus stop for the number 65 bus. It will take you right there”. He thanked the police officer and the officer drove off.

Three hours later the police officer returned to the same area to find the man still waiting at the same bus stop. Surprised, he got out of his, car and asked the man, “Why are you still waiting ? Did not you get the bus”?
The man replied “Don’t worry officer, it won’t be late now. The 58th bus just went by!”

Question 1.
Why did the man come to Hyderabad ?

Question 2.
Write a synonym for ‘surprised’.

Question 3.
Why was the man waiting at the bus stop ?

Question 4.
Do you think the police officer gave the right directions to reach the Charminar?

Question 5.
State True of False.
The man was waiting to visit the Charminar for the first time.

VII. Read the following passage carefully and answer the questions that follow. (5 × 1 = 5)

They went to the small jail and they saw Muff. He was happy to see them. ‘No one remembers old Muff anymore. But you are my friends and you remember me. Thank you boys!’ said Muff smiling. Now, Tom felt terrible. He was worried about the trial. He could not sleep at night. Everyone in the village went to muff power’s trial. Muff looked old, tired and unhappy. Injun joe was at the trial too. During the trial there were many questions and answers. All the answers were against old Muff. Then, the lawyer said ‘Call Homas Sawyer!’ Everyone was surprised and looked at Tom.

Question 1.
Who went to see Muff in the jail ?

Question 2.
Why was Muff in the jail ?

Question 3.
All answers were against old Muff. That means
a) Muff is cornered at a criminal.
b) Muff is innocent.
c) Muff really planned the crime.

Question 4.
Was Muff happy to see Tom and Huck ?

Question 5.
Why did the lawyer call Tom ?

VIII. Study the advertisement and answer the questions that follow. (5 × 1 = 5)

Question 1.
Who is the national leader that can be seen in the advertisement ?

Question 2.
Who has issued the advertisement ?

Question 3.
Write the antonym of rural.

Question 4.
Write the verb form of ‘Development’.

Question 5.
What is being guaranteed in the advertisement?

IX. Study the following bar graph carefully and answer the questions that fellow. (5 × 1 = 5)

The following bar graph depicts the favourite sports of a group of students.

Question 1.
What does the graph depict ?

Question 2.
Which is the most liked game ?

Question 3.
State True of False.
The number of students who like Basketball and other games is equal to the number of students who like Soccer.

Question 4.
Which sport is liked by only 6 students ?

Question 5. How many students like Softball ?

SECTION – C

X. Write a letter to the principal of your college asking your T.C., study and conduct certificates.
Hints :Year of study – group – adm No. – pursue higher studies – certificates – required. (5 × 1 = 5)

OR

Write a letter to the editor of a Newspaper complaining about the bad conditions of roads and drains in your area.
Hints : Resident of colony – roads damaged – pot holes are formed – manholes are displaced – difficult for travellers – during night times – request for publish.

XI. Write a paragraph of 8 lines describing the process. Describe the process of paying electricity bill on-line. (5 × 1 = 5)

OR

Describe the process of creating Gmail account.

XII. Prepare a Curriculum Vitae based on the information given below. (5×1 = 5)

Pocha Sumalatha – aged 22 years – B.Sc.(CS) – 79% marks – Govt. Degree College, Visakhapatnam – Intermediate (MPC) – 88% – Govt. Jr.College, Anakapalli – SSC – 90% – Govt. High School, Anakapalli – applying for the post or System Analyst – Mozaic Software Company, Visakhapatnam.

XIII. You are Mr. K. Venu Gopal. You have an SB A/c in Bank of India and A/c No: 11164927471. Draw personally Rs. 10,500/- today by using the following cheque. (5 × 1 = 5)

XIV. Construct a dialogue between a teacher and a student who came to class late. (1 × 5 = 5)
OR
Construct a dialogue between a lecturer and a student about the prospects after the completion of Intermediate.

XV. Read the following passage and make notes. (1 × 5 = 5)

There are many different kinds of musical instruments. They are divided into three main classes according to the way that they are played. For example, some instruments are played by blowing air into them. These are called wind instruments. In some of these the air is made to vibrate inside a wooden tube and these are said to be of the woodwind family. Examples of woodwind instruments are the flute, the clarinet and the bassoon. Other instruments are made of brass, the trumpet and the hom, for example. There are also various other wind instruments like mouth organ and the bagpipes. Some instruments are played by banging or striking them. One obvious example is the drum. Such instruments are called percussion instruments.
The last big group of musical instruments are the ones which have strings. There are two main kinds of stringed instrument, those in which the music is made by plucking the strings and those where the player draws a bow across strings. Examples of the former are the harp and the guitar, examples of the latter are the violin and the cello.

XVI. Match the words with their meanings : (5 × 1 = 5)

XVII. Mark the stress for any five of the following words.

1) student
2) college
3) examination
4) valuation
5) project
6) prolong
7) present
8) ourselves
9) ability
10) call centre

Access to a variety of AP Inter 2nd Year English Model Papers Set 10 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year English Model Paper Set 10 with Solutions

Time : 3 Hours
Max. Marks : 100

SECTION – A

I.Annotate ANY TWO of the following in about 10-15 lines each : (2 × 4 = 8)

Question 1.
Studies serve for delight, for ornament and for ability.
Answer:
Context: This line is taken from the essay “Of Studies” written by Francis Bacon. He was called the Father of English Essay. This line is about the benefits of studies.

Explanation : We get delight when we read a book. When we are at leisure, we read and enjoy. Reading enriches our capacity of communication and judgement. Thus it makes us efficient and it is like an ornament. Effective thinking and perfect knowledge are possible by studies. Many people take the advice of a man who studies well.

General Relevance: The short sentence, has got great meaning. Bacon’s sentences are aphoristic and are filled with sensitive views.

Question 2.
Every work must necessarily be a mixture of good and evil yet we are commanded to work incessantly.
Answer:
Context: This passage is taken from the essay “The Secret of Work” written by the great thinker Swami Vivekananda. In this sentence he writes about the need of regular work. The secret of work is discussed here.

Explanation : Vivekananda gives prominence to work. Every work has got the two faces, the good and the bad. We have to pick up good and work without any stop. Through good work, good effect will come into being. The good things coming out of one’s character belong to our Samskara. The senses are controlled and there is no attachment with the evil. The secret of work is doing the things without any delay.

General Relevance : According to Vivekananda, the secret of work, is doing it without any stop. Love drives the man into action and the secret of work lies in continuity.

Question 3.
My gloomy thoughts probably stem from an accident I had few years ago.
Answer:
Context: This sentence is ta^en from the essay “In Celebration of Being Alive” written by Dr. Christian Barnard. When the writer thought about the suffering prevailing in the world, he remembered an accident in his life. He described the accident in the essay.

Explanation :.His thoughts were gloomy but his suffering was alive in his mind. As he and his wife were crossing the street, they were caught in the accident. They got treatment at the hospital. But it was strange for him, why he should get such a punishment, by God. His anguish was great, the unexpected suffering, according to the narrator, restricted him from the service he *had to render to the patients.

General Relevance : The suffering stopped the services of the doctor. He had broken ribs. So, the narrator was angry with God, for this situation.

Question 4.
Meritocracy by definition means that we cannot let personal prejudices affect our evaluation of an individual’s performance.
Answer:
Context: This passage is taken from the essay “Learning from the West” written by N.R.Narayanh Murthy. The writer was a great Industrialist and a Writer. Here he writes about the virtues, one should possess in life.

Explanation : In the West people evaluate based on merit. No k personal prejudices or interests come in the way. In the West right from the young age, they are accustomed to this way of living. So, the Indian,-has to give respect to this clean thinking. It enhances our respect in the society. Meritocracy is a great virtue.

General Relevance : Personal interests should not come in the way, one gives respect to the merits. This will add to the character of an individual.

II. Annotate ANY TWO of the following in about 10-15 lines each : (2 × 4 = 8)

Question 1.
It shall be still in strictest measure ev’n,
To that same lot, however mean or high.
Answer:
Context: These lines are taken from the poem “On His Having Arrived at the Age of Twenty Three”. It was written by the epic poet John Milton. These lines explain the introspection of John Milton.

Explanation : The poet finds displeasure with his development. He attains twenty three years of age. There is no proper development according to the poet. He thinks that God has not provided the talents due to him. He wants to get maturity in writing. But when he comes to introspection, he understands the will and authority of God. Whatever is to be given by God, is given. Hie has a plan for Milton and that will be adopted. Whether the development is slow or fast it is to be accepted by the individual. So, the poet has to wait upon God and get his lot.

General Relevance : Every body in this world has to obey God and his plans. Though there is dissatisfactionin the beginning in course of time the individual will satisfy himself with the blessings of God.

Question 2.
Enough of Science and of Arts,
Close up those barren leaves.
Answer:
Context: These lines are extracted from the poem “The Tables Turned” written by William Wordsworth. These lines suggest how the poet creates interest in the study of nature.

Explanation : These are the lines from the concluding stanza. The central idea of the poem is knowing the importance of nature in education. The poet wants the tables of books to be turned. There is no need of the books. Laborious study is to be shun. There are many books.of science or of arts. They have plenty of knowledge in each subject. But This study is not equivalent with the study of nature which gives a lot of knowledge. Those books do not think of the natural messages available in the nature. So, the poet stresses the need of the study of nature.

General relevance : All the knowledge, including that, which belongs to science and art, is available in the study of nature. So one should not hesitate to go in for the study of nature.

Question 3.
I am their wall against all danger,
Their door against the wind and snow.
Answer:
Context: These are the lines taken from the poem “Any Woman” written by the British writer, Katharine Tynan. It is about the sacred place of a mother in the family.

Explanation: The mother is like a pillar in the family. She takes care of the children every now and then. She gives warmth to the children. She keeps them united. She shows her love and keeps them safe in her lap. She is like a wall when there is a danger for the children. She saves them from the winds and rain. Just like a bird guarding her chicks, the mother saves her children from the outwards dangers. In thfe world she refines her children, to be fit for future life.

General Relevance : A mother is like a mother bird, to the children. Natural calamities create problems but the mother does not leave her children to their fate. She is always at their rescue.

Question 4.
Time is with materials filled;
Our todays and yesterdays,
Are the blocks with which we build.
Answer:
Context: These lines are taken from the poem “The Builders” written by H.W. Longfellow. Here people are compared to builders. The life is the structure and time controls it.

Explanation; The poet aims at constructing the building of life, based on time. There is one experience today and it is like a block for the construction. Today’s block shall become a base for the next day’s construction. The blocks make the building qualitative. So one should look to the blocks and utilise them as time passes. The life of an individual should become great by changing according to time.

General Relevance : Whatever be the experience of today, is like a base for the next day. Time will decide the changes and the life style can be improved, accordingly.

III. Answer ANY TWO of the following questions in about 10-15 lines:(2 × 4 = 8)

Question 1.
What according to bacon is the theme of’of studies’?
Answer:
Francis Bacon was a great english essayist. ‘Of Studies’ is an interesting essay depicting the benefits and disadvantages of studies. , Reading books gives us happiness. It helps us to become more intelligent in communication. Natural abilities can be developed by reading books. Sometimes it becomes harmful if too-much time is spent on it. Showy tendency should be avoided. There are books to be skipped through, books to be read completely and some others to be read with great care and to be digested. Through studies an individual becomes a full man. The author tells that the study of different subjects helps us to cure the weaknesses of the mind.

Question 2.
Explain the views of Huxley about the instruments for experimentation.
Answer:
Aldous Huxley was anEnglish Writer. He wrote a number of books. ‘j.C.Bose’ is an essay taken from the book “Testing Pilate”. It is about the experimenting on Science and Technology. In the beginning of the essay he writes that experimentation does not need expensive instruments. Michael Faraday invented the powers of electricity using very simple things like the tea kettle, silk wires, wax and jam pots only. Similarly J.C.Bose invented the machine, to measure the growth of plant. Here also a little clockwork, needles, filaments, a sheet of smoked glass etc., are used in the experiment. So, the writer thinks that there is no need of costly instruments, if the instrumenter is with curiousity and talent. Even the suffering of a plant or death of a poisoned plant could be seen on the smoked glass. The things are be arranged properly.

Question 3.
What lessons did Dr. Barnard learn from the two children ?
Answer:
Dr. Barnard was a heart surgeon and a writer. He wrote the ” essay “In Celebrations of Being Alive” giving his experience at the time of an accident. There was an incident at the Cape Town Red Cross Children’s Hospital.. The trolly of breakfast was moved by two physically handicapped children. One was blind and the other was a heart patient. They drove it and made it worthy scene for all the patients and nurses around. The boys did not think of their deformities but did whatever they could do with all that they possessed. So, the writer learnt that one should try to utilise the possessions carefully. One should not aspire for that which was not available. The business of living is the celebration of being alive. Thus the poem teaches us the quality satisfaction.

Question 4.
Indians become infinate even without being friendly.
Answer:
N.R.Narayana Murthy was a great Industrialist and Writer. In the essay “Learning from the West” he gives a number of examples from the life of European countries and asks us to emulate. One of the qualities is that the people of the West become friendly even without being intimate. This statement can be seen in the behaviour of Rudyard Kipling. He said that an easterner tries to become intimate without being friendly. In this example we find how an Indian differs from the Western person in behaviour.

IV. Answer ANY TWO of the following questions in about 10-15 lines :(2 × 4 = 8)

Question 1.
‘Time and Tide wait for no man” is an old saying. Discuss this in the context of the poem “On His Having Arrived at the Age Of Twenty Three”.
Answer:
John Milton was an epic poet. He writes in a grand style. “On His Having Arrived at the Age of Twenty Three” is a sonnet, written by this poet. Here the saying of “Time and Tide wait for no man’ is suitable to be discussed. It is a fact that the tides in the sea move on to the shores at a high speed and they cannot be stopped by anything. Similarly time doesnot stay Or stop at any place. It is a natural phenomena. The consumption of time is a continuous process. Milton looks to himself and wonders how his years have passed away. He cannot find a reason in the beginning. So he complains against God.
But after introspection, he finds a solace. He thinks that time does npt wait for him or for his maturity. Thus the saying can be applied to the views of John Milton.

Question 2.
Why does Wordsworth consider nature to be a good teacher ?
Answer:
‘The Tables Turned’ is a famous poem written by William Wordsworth. He was a romantic poet and his love towards nature is of great value. The poet believes that nature is the main source for education. From books, we learn only a little but nature gives us complete knowledge. The beautiful scenery, the lustrous plants and trees, the birds and all that is there give us complete knowledge of the creation. The birds do riot read books but they know the secrets of this world. For research, we dissect and kill certain creatures, But if we have an affectionate heart, we can acquire a lot of knowledge from the nature. So, the poet thinks that the nature is a good teacher.

Question 3.
What is the appropriateness of the title “The Builders” ? Do r you agree with the poet about building one’s own life ?
Answer:
‘The Builders’ is an interesting pqem written by the great American poet H.W Longfellow. The poet compares men to builders. An individual’s life is controlled by time and he is the one who constructs the building of life. According to the poet, everybody plans his life according to his will. He experiences something and advances based on the previous experience. Everyday’s work is taken as a block in the construction of thp building. There are people who build their character on good behaviour, with good blocks. Those lives have become buildings suitable for the dwelling of God. Thus people have to build huge buildings of character. The poet suggests the title, ‘The Builders’ and if is quite apt.

Question 4.
The poem “Any Woman” is a celebration of the glory of womanhood. Illustrate.
Answer:
The poem “Any Woman” was written by Katharine Tynan, a British writer. She produced a number of novels and poems and became a noted writer. Any woman is a poem depicting the supremacy of a woman in the family. She loves her children and guards them carefully. Without a woman, there is no house of recognition. A mother takes care of the children providing food and shelter. She keeps the house neat and tidy. She protects them at the time of danger. If the mother leaves a Child, he will be spoiled. She is like a pillar in the house and keeps the house graceful. A good Woman brings respect to the family.

V. Answer ANY ONE of the following questions in about 25 lines : (1 × 8 = 8)

Question 1.
What are the turning points in the story ? Discuss the main events in the plot.
Answer:
Mark Twain, wrote the novel “The adventures of Tom Sawyer”. The story of Tom Sawyer is narrated by the writer, with interesting turning points at different places- The story moves round the tender boy Tom Sawyer. He was being brought up by Aunt Polly. At the school, there was an interesting episode. He got punishment but he drew a beautiful picture. He showed love towards Becky, a girl in the class. There is,a turning point at the graveyard. The murder of the doctor, by Injun Joe, created frustration in the minds of Tom and Huck. It continued throughout the stoiy until the person died in the cave.

There is an important point, when they attended the funeral function. It adds to the humour of the novel. Muff Potter was jailed but Tom gave witness before the judge and he was set free. Becky’s parents arranged a picnic. Tom and Becky were caught in the cave but were saved. The judge arranged an iron door. This door became the cause for the death of the villain Injiin Joe at the end. Huck was adopted by Widow Dougals. He became a person with culture.

Afterwards Tom wanted to form a pirate gang and Huck would become the bold pirate. Thus, there are some turning points which make the story interesting.

Question 2.
Write at least five lines about the following characters.
a) Auntpolly
b) Muff Potter
c) Becky Thatcher.
Answer:
a) Aunt Polly : Aunt pollywas the care taker of Tom Sawyer. As his mother died, she was carefully looking to the needs of Tom, When she was misguided by Tom, she showed anger but blended with .motherly affection. She wanted the boy to be in a discplined way. When we look to her character, She was kind hearted, cool and considerate. N She was loved by Tom and was deceived at times showing some love towards her. When she saw the boy Tom and his friend Huck at the mourning cermony she was astonished. She was filled with happiness.
She show ther love and affection to both of them. On the whole, the character should have been given based on the life of Twain’s another.

b) Muff Potter : One of the small characters is Muff Potter. He was the father of Huckleberry Finn. He was an aweful drunkard. He ‘ could not provide a place of residence for his son. Muff Potter thought that Injun Joe was his friend. He did not know that he was cheated by him. When the people believed that he was a murderer, his position still degraded. Muff Potter was saved from the plight, when Tom gave . his witness in the. court. He was kind hearted and was not harmful.
But the circumstances drove him into troubles.

c) Becky Thatcher : Becky is an interesting character in the novel. Though she was the beautiful girl at the school along with Tom, she was taken as the heroine. She was at a tenderage and her pretty face was attracted by Tom. The writer did not want to make Tom a hero having maturity. Becky’s love towards Tom was a puppy love. We find her to be sacrificing her life also showing her bravery. Becky was a splendid girl, exhibiting virtues suitable for a tender and pretty girl.

Question 3.
Write a character sketch of Tom Sawyer.
Answer:
Tom Sawyer was the young hero in the novel “The Adventures of Tom Sawyer” written by Mark Twaim, an American writer. Tom was a boy of a tender age. He was a misfit in the school. His behaviour with Becky Thatcher is interesting because he had a liking for her. He expressed it boldy. He was equally like by the girl, Tom had another ” friend, Huck- leberry Finn, who was emtidy, and spent his time fishing and swimming. Tom liked this boy because he wanted to have adventures in life.

Tom went to the graveyard one day along with Huck, to see the ghosts. It was a mysterious experience for him. The three men who came there to rob a dead body quarrelled among themselves. One of them, a doctor was killed by Injun Joe. Tom was frightened to watch Injun killing the doctor. Tom had a liking to wander and his stay at Jackson’s island, was wonderful.

The parents arranged a funeral, thinking that Tom and Huck were dead, which indeed created a humorous occasion for Tom’s jolly mind. Tom’s presence at the church created a hilarious situation. Tom s determination to save Muff Potter and his brave witness at the court, adds to his virtuous behaviour. His relationship with Becky Thatcher was very much appreciative.

Tom wanted to save Widow Bougals. His determination to punish Injun Joe and his pursuasion in the Search of the treasure in the cave, were note worthy. Toni got the treasure and changed Huck, the dirty, lazy and unworthy boy to a neat and tidy boy. Thus Tom was typical of the days of Mark Twainer in America.

SECTION – B

VI. Read the following passage carefully and answer the questions kthat follow: (5 × 1 = 5)

The maneless Lion

Fifty feet away, three male lions lay by the road. They didn’t appear to have a hair on their heads. Noting the colour of their noses (leonine noses darken as they age, from pink to black),
Craig estimated that they were six years old-young adults. “This is what we came to see. They are really maneless.” Craig, a professor at the University of Minnesota, is arguably the leading expert on the majestic Serengeti lion, whose head is mantled, in long, thick hair. He and Peyton West, a doctoral student who has been working with him in Tanzania, had never seen the Tsavo lions that live some 200 miles east of Serengeti. The scientists had partly suspected that the maneless male lions were adolescents mistaken for adults by amateur observers. Now they know better.

The Tsavo research expedition was mostly Peyton’s show. She had spent several years in Tanzania, compiling the data she needed to answer a question that ought to have been answered long ago : Why do lions have manes ? It’s the only cat wild or domestic, that displays such ornamentation.

Questions:
1.Name the two scientists in the passage.
Answer:
Craig and Peyton West.

2.State true or false :
Serengiti lions are maneless.
Answer:
False.

3. Where do Tsavo lions live ?
Answer:
Some 200 miles away from Serengeti.

4.Pick the adjective in the passage which means ‘pertaining to or characteristic of a lion’.
Answer:
Leonine.

5.Write the noun form of the verb ‘estimated’.
Answer:
Estimation.

VII. Read the passage and answer the questions that follow : (5 × 1 = 5)

‘I’m going swimming but you can’t come with me. You’re working,’ said Ben. ‘Do you calllhis work ?’ asked Tom. ‘Of course it’s work. You’re painting a fence,’ said Ben. May be it’s work but may be it isn’t. I like it!’• said Tom, ‘I can swim everyday, but I can’t paint a fence everyday.’
Ben watched Tom. He painted slowly and carefully. He often stopped to and moved back from the fence. He looked at his work and smiled. Ben was suddenly interested in the fence and said, ‘Let me paint a little, Tom.’ Tom thought for a moment. Tm sorry, Ben, Aunt Polly wants me to do it because I’m very good at painting. My brother Sid wanted to do it, but he’s not good at painting.’ ‘Oh please, Tom! Please can I paint ? I’m good at painting too. Here, you can have some of my apple.’ ‘No, Ben I can’t’ – ‘Then take all of my apple !’

Questions :
1.What is the work Tom was doing ?
Answer:
Painting the fence.

2.Was Tom really enjoying his work ?
Answer:
No.

3.Who was suddenly interested in painting the fence ?
Answer:
Ben.

4.What did Ben offer Tom to give him a chance to paint the fence ?
Answer:
An apple.

5.What do you think of Tom from the above incident ?
a.Tom is hard working.
b.Tom is plain natured.
c.Tom is cunning.
Answer:
c. Tom is cunning.

VIII. Study the advertisement given below and answer the questions that follow : (5 × 1 = 5)

1.Where should we cross the road ?
Answer:
At Zebra crossing.

2.Why should we call the National Consumer Help Line ac-cording to the advertisement ?
Answer:
For any help / clarification.

3.What is the toll free number for the National Consumer Help Line ?
Answer:
1800-11-4000.

4.What should we do before we assert our right ?
Answer:
We should be responsible to follow rules / It’s our responsibility to follow rules.

5.Write the noun form of ‘assert’.
Answer:
Assertion.

IX. Study the Pie Chart carefully and answer the questions that follow. (5 × 1 = 5)

The following bar graph shows the number of rhinoceros poached in South Africa from 2007 to 2015.
Recorded number of rhinos poached in South Africa

Questions :
1.In which year were the maximum rhinos poached ?
Answer:
2014.

2.Explain if poaching decreased or increased from 2007 to 2015.
Answer:
Poaching increased from 2007 to 2014 and decreased a bit in 2015.

3.During which years can we observe the maximum difference in the number of rhinos poached ?
Answer:
2012 to 2013.

4.A person who poaches is called a.
Answer:
Poacher.

5.State true or false.
The number of rhinos poached in 2014 is almost double to the number, poached in 2012..
Answer:
True.

SECTION – C

X. Write a letter to the editor of a newspaper about the bad condition of roads in a locality.(1 × 5 = 5)
You are interested in doing a short term course in Spoken English during your summer vacation. Write a letter to the director, NEO Spoken English and Grammar Institute, Annamaiah circle, Tirupati, enquiring about the terms and conditions for the course.
Answer:
K.Amaranath
2-3-40 Gandhinagar
Vijayawada.

15-4-2019

The Director
NEO Spoken English Institute
Annamalai circle
Tirupathi.
Sir,
Sub: Information about short term course-request-regarding. Having come to know that the institute is running a short term course in spoken English. I wish to know the particulars about it/1 wish to know the duration of the course and the date of starting. The fee particulars, the time of running the classes, may also be sent. Is there any hostel facility ? If there is hostel, the fee there of may also’be informed.
Thanking you
Yours faithfully
K.Amaranath.

OR

You have been invited to attend your friend’s marriage. You are not able to attend the marriage due to personal problems. Write a letter to your friend congratulating him and expressing your inability to attend.his marriage.
Answer:
10-25, Srinagar colony
Guntur.

10-6-2019

My dear Ramana,
I am doing well and hope the same with you. Here is a piece of obligation to you. First of all accept my hearty congratulations, for the happy occasion of your wedding recently. At the same time I request you to pardon me, for I could not be there at that time. I had an interview in Delhi two days before the marriage.
I could not reach Vijayawada by that date. Hope you would, understand my inability. Convey my best wishes to your better half.
With Love
Yours lovingly
D. Ravindra

Address:
B.VRamana
2-3-11.0 Narasimhanagar
Vishakhapatnam.

XI. Write a short paragraph describing the process of how you get ready for college regularly.(1 × 5 = 5)
Answer:
These days study to a college, needs so much care and preparation. Everyday I have to attend my college at 8 A.M’. So, I get up at 6 A.M and finish off the daily chores in time. I should not be lazy in getting up from bed. I will have an eye upon anytime table, to sort out the books necessary. I have to verify whether all the Home work given has been finished or not. As, I go to college by bus, I must start from my house early. I will take the breakfast, sufficiently early. My mother keeps the box of meals ready. 1 will take water in a bottle. The bag of books is kept ready and so, I go to the main road taking the bag, the bottle and others. I wait there until the college bus arrives the point of halt. Then I reach the college and go on with the classwork. It is a routine work.

OR

Describe the process of making tea for two people.
Answer:
It would be better if’we learn certain things, needed for our smooth living. How to make tea for two people. The following are needed to prepare tea.
Tea Kettle
Spoons
Filter
Cups and Saucers
Tea powder
Sugar
Milk.
First of all take some water in the kettle. Mix some tea powder in it and boil it for a few minutes. Pour some milk into it and boil it for another few minutes. Take the kettle out of the stove and filter the tea. Add some sugar into the tea. Serve the tea in two cups to the people waiting.

XII.Prepare a curriculum vitae in response to the following advertisement.(1 × 5 = 5)

Now prepare a Resume based on the above format. In this Resume, Dandi Deepa is applying for the position of Area Sales Manager in the reputed NCL Company based on the newspaper advertisement given at the beginning of this section.
Answer:
DANDI DEEPA
45/24, Teachers Colony
Venkata Ramana Colony Road No. 4
KURNOOL – 518003
Mobile : 8500012340
kdeepa@gmail.com

OBJECTIVE
A position as Area Sales Manager in a reputed company.

EDUCATION
2009-2011 – MBA (Sales & Marketing)
Sri Padmavati Mahila Visvavidyalayam, Tirupati, 81% aggregate.
2006-2009 – B.Com (General) Silver Jubilee College (Autonomous), Kumool, 76% aggregate.
2004-2006 – Intermediate (C.E.C), Board of Intermediate Education, AP. Govt. Junior College (Town), Kumool, 82.5% aggregate.
2004 – SSC, Board of Secondary Education, AE Govt. Girls High School, Kumool, 80% aggregate.

WORK EXPERIENCE
2011-2016 Worked as Sales Officer for five years in XYZ Company, Vijayawada.
2016 – Present Working as Area Sales Manager for three years in Abhiram Industries, Ananthapuram.

STRENGTHS
Good communication skills, Problem solving and Managerial skills, Good motivator and coordinator.

ADDITIONAL INFORMATION
Languages known : English, Telugu and Hindi
Hobbies : Reading books and listening to music
Father’s Name : Sri Dandi Gopal Rao
References : Available on request

DECLARATION
I hereby declare that the details furnished above are true and correct to the best of my knowledge and belief and I undertake to inform you of any changes therein, immediately.

D. Deepa
SIGNATURE

Place : Kumool.
Date : 23 April, 2019.

A cover, letter which is to be sent with the CV/Resume of Dandi Deepa is given below.
DANDI DEEPA
45/24, Teachers Colony
Venkata Ramana Colony Road No. 4
KURNOOL – 518003
Mobile : 8500012340

23 April, 2019

kdeepa@gmail.com.
The Manager
M/s NCL Industries Ltd.,
Gayathri Nursing Home, Near BSNL Tel. Exchange
TUDA Plots, RC Road, Tirupati-517501

Dear Sir
APPLICATION FOR AREA SALES MANAGER
In response to your advertisement in the Eenadu newspaper of April 23, 2019 for the post of Area Sales Manager, I am enclosing my resume for your consideration. I feel I have the required qualifications and skills for the position in your industry.
I completed my academics with good percentage. Moreover, my work experience in the area you have asked for is very effective and always I worked for the benefit of the companies and proved myself worthy in the positions I held.
I am excited to see your job listed in the newspaper. I have been looking for exactly this position, and I think that my work experience and related skills will be an excellent match for your needs.
You can call me for an interview on any day convenient to you. With best regards
Yours faithfully
D. Deepa .
(DANDI DEEPA)

XIII.Fill in the form (10 × 1/2 = 5)
Mr. Rajesh Kumar wants to subscribe to the Textile Magazine for a period of two years. He is enclosing a Cheque for the mentioned amount bearing No. 244261 dated 12.06.2019. He lives at H.No. 124/2-B, Prakash Nagar, Narasaraopet. – 522601. His rpobile No. is 944164901 and his Mail Id is rkumar1995@gmail.com.

Answer:
1) The Textile Magazine 2 years (24 issues) ? 800
2) Mr. Rajesh Kumar
3) Personal
4) No
5) H.No. 124/2-B Prakash Nagar, Narasaraopet
6) 944164901 .
7) rkumar1995@gmail.com
8) 244261
9) 12-06-2019
10) 800

XIV.Construct adialogue between the doctor and the patient about treatment for fever and pain. (1 × 5 = 5)
Answer:
Patient : Good morning, doctor.
Doctor : Good morning, What’s your problem ?
Patient : Sir, I’m feeling weak and I’m unable to walk even a 5 little distance.
Doctor : What’s your appetite like ?
Patient : . Not at all good. I don’t feel like eating anything.
Doctor : Have you had any fever ?
Patient : Well, I do feel feverish all the time.
Doctor : All right, let me check your temperature first. There’s nothing wrong with the pulse. You please lie down on that table, I will examine.
Patient : Yes, doctor.
Doctor : Do you feel pain here ?
Patient : Yes, some.
Doctor : And here ?
Patient : Oh! That’s quite painful.
Doctor : Nothing to worry. I’m prescribing two types of tablets. Take one before meals and the other after meals ‘ – for three days. Don’t eat any fried or spicy food. Drink
milk and follow your regular diet.
Patient : Thank you doctor.
Doctor : It’s ok.

OR

Write a dialogue between a customer and the waiter of a hotel (enquire about room for two days)
Answer:
Customer : Waiter!
Waiter : Yes sir, what can I do for you ?
Customer : Bring the Menu card.
Waiter : Sir, here it is.
Customer : Is this veg or non-veg ?
Waiter : It is purely veg sir.
Customer : OK, what do you have, hot ?
Waiter : Idly, vada and dosa are hot.
Customer : Bring me a plate of Idly and tell the master to prepare a dosa also.
Waiter : Yes sir.
Customer : Waiter!
Waiter : Yes, please.
Customer : This jug is empty. Pour some water. Is it aqua water ?
Waiter : Yes sir.
Customer : Where’s the wash basin ?
Waiter : That side, sir.
Customer : Waiter! Get me a cup of coffee.
Waiter : Here you are, sir.
Customer : Get me the bill.
Waiter : Yes sir, here’s the change.
Customer : Keep it for you.
Waiter : Thank you sir.

XV.Read the following and make notes. (1 × 5 = 5)

Women empowerment has becorpe the buzzword today with women working alongside men in all spheres. They profess an independent outlook, whether they are living inside their home or working outside. They are increasingly gaining control over their lives and taking their own decisions with regard to their education, career, profession and lifestyle.

With steady increase in the number of working women, they have gained financial independence, which has given them con-fidence to lead their own lives and build their own identity. They are successfully taking up diverse professions to prove that they are second to none is any respect.

But while doing so, women also take care to strike a balance between their commitment to their profession as well as their home and family. They are playing multiple roles of a mother, daughter, sister, wife and a working professional with remarkable harmony and ease. With equal opportunities to work, they are functioning with a spirit of team work to render all possible co-operation to their male counter parts in meeting the deadlines and targets set in their respective professions!

Women empowerment is not limited to urban, working women but women in even remote towns and villages are now increasingly making their voices heard loud and clear in society. They are no longer willing to play a second fiddle to their male counter parts. Educated or not, they are asserting their social and political rights and making their presence felt, regardless pf their socioeconomic backgrounds.

While it is true that women, by and large, do not face discrimi-nation in society today, unfortunately, many of them face exploita-tion and harassment which can be of diverse types : emotional, physical, mental and sexual. They are often subjected to rape, abuse and other forms of physical and intellectual violence.

Women empowerment, in the trust sense, will be achieved only when there is attitudinal change in society with regard to womenfolk, treating them with proper respect, dignity, fairness and equality. The rural areas of the country are, by and large, steeped in a feudal and medieval outlook, refusing to grant women equal say in the matters of their education, marriage, dress code, profession and social interactions.
Let us hope, women empowerment spreads to progressive as well as backward areas of our vast country.
Answer:
Women empowerment : Now-a-days women are trying to be independent-own decisions – Financial independence for women – second to none in every respect.
Woman in society : Mother, Daughter, Sister-tearti,work-co-operation from male partners – respect in professions.
Place: Both urban and rural areas to develop – social and political rights – No exploitation of ladies.
Prosperity : Women to prosper – respect, dignity and equality wrought.

XVI.Match the wards in Column A with their meanings/definitions in Column ‘B’.

Column ‘A’	Column ‘B’
1.incredible	a.call on for inspiration
2.contradict	b.throwout
3.invoke	c.say against
4.eject	d.a system where there is no rule
5.anarchy	e.that which cannot be believe unbelievable
	f.shaped like a cross
	g.mathematical shape

Answer:

Column A	Column ‘B’
1. interact	e.that which cannot be believe unbelievable
2. animate.	c.say against
3. calligraphy	a.call on for inspiration
4. cacophony	b.throwout
5. egoity	d.a system where there is no rule

XVII. Mark the stress for ANY FIVE of the following words : (5 × 1 = 5)
1) minister
2) parliament
3)nation
4) confidential
5)kilometer
6) tourism
7)commandeer
8) telegraph
9) ornament
10) architect
Answer:
1) minister = ‘minister
2) parliament = ‘parliament
3) nation = ‘nation
4) confidential = confi’dential
5) kilometer = ‘kilometer
6), tourisom = ‘tourisom
7) commander = comman’der
8) telegraph = te’legraph
9) ornament = ‘ornament
10) architect = ‘architect

Access to a variety of TS Inter 2nd Year Maths 2B Model Papers and TS Inter 2nd Year Maths 2B Question Paper May 2018 allows students to familiarize themselves with different question patterns.

TS Inter 2nd Year Maths 2B Question Paper May 2018

Time : 3 Hours
Max. Marks : 75

Section – A
(10 × 2 = 20)

I. Very Short Answer type questions.

1. Attempt all questions.
2. Each question carries two marks.

Question 1.
If x² + y² – 4x + 6y + c = 0 represents a circle with radius 6, then find the value of C.
Solution:
Given circle equation is x² + y² – 4x + 6y + c = 0 .
Here 2g = -4 ⇒ g = -2
2f = 6 ⇒ f = 3 and c = c
Given radius = 6
⇒ \(\sqrt{g^2+f^2-c}\) = 6
⇒ \(\sqrt{4+9-c}\) = 6
⇒ 13 – c = 36
⇒ c = 13 – 36
⇒ c = -23.

Question 2.
If the length of the tangent from (2, 5) to the circle x² + y² – 5x + 4y + k = 0 is \(\sqrt{37}\) then find k.
Solution:
Let S = x² + y² – 5x + 4y + k = 0
Given length of the tangent from (2, 5) to the circle S = 0 is \(\sqrt{37}\)
\(\sqrt{S_{11}}\) = \(\sqrt{37}\)
⇒ S₁₁ = 37
⇒ 4 + 25 – 5(2) + 4(5) + k = 37
⇒ 29 – 10 + 20 + k = 37
⇒ k = 37 – 39
⇒ k = -2

Question 3.
If pair of circles x² + y² – 5x – 14y – 34 = 0,
x² + y² + 2x + 4y + k = 0 are orthogonal, then find the value of k.
Solution:
Given circle equations are
x² + y² – 5x – 14y – 34 = 0 ……. (1)
x² + y² + 2x + 4y + k = 0 ……. (2)
Since (1), (2) are orthogonal
∴ 2g₁g₂ + 2f₁f₂ = c₁ + c₂
⇒ (-5) (1) + (-14) (2) = -34 + k
⇒ -5 – 28 = -34 + k
⇒ k = – 33 + 34
⇒ k = 1.

Question 4.
Find the value of k, if the line 2y = 5x + k is a tangent to the parabola y² = 6x.
Solution:
Given line equation is 2y = 5x + k
⇒ y = \(\frac{5}{2}\)x + \(\frac{k}{2}\) ….. (1)
Here m = \(\frac{5}{2}\), c = \(\frac{k}{2}\)
Given parabola equation is y² = 6x …….. (2)
Here 4a = 6 ⇒ a = \(\frac{6}{4}\)
⇒ a = \(\frac{3}{2}\)
If (1) is a tangent to the Parabola (2) then
c = \(\frac{a}{m}\)
⇒ \(\frac{k}{2}\) = \(\frac{\frac{3}{2}}{\frac{5}{2}}\)
⇒ \(\frac{k}{2}\) = \(\frac{3}{5}\)
⇒ k = \(\frac{6}{5}\)

Question 5.
If e„ e₁, are the eccentricities of a hyperbola and its conjugate hyperbola, then prove that \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}_1^2}\) = 1.
Solution:

Question 6.
Evaluate : ∫Sec²x Cosec² x dx, x ∈ I ⊂ R \
({nπ : n ∈ Z} ∪ {(2n + 1)\(\frac{\pi}{2}\):n∈Z})
Solution:
∫Sec²x Cosec² xdx = ∫\(\frac{1}{\cos ^2 x \sin ^2 x}\)dx
= ∫\(\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}\)dx
= ∫\(\frac{1}{\cos ^2 x}\) + ∫\(\frac{1}{\sin ^2 x}\)
= ∫sec²x dx + ∫cosec²xdx
= Tanx – Cotx + c
∴ ∫Sec²x Cosec² x dx = Tanx – cotx + c

Question 7.
Evaluate ∫(Tanx + Log Sec x)e^x dx on
((2n – \(\frac{1}{2}\))π, (2n + \(\frac{1}{2}\))π), n∈Z
Solution:
∫(Tanx + Log Secx) e^x dx = ∫e^xTanx dx + ∫e^x.log sec x dx
= ∫e^xTanx dx + log sec x. e² – ∫\(\frac{1}{\sec x}\).Secx Tanx. e^x dx
= ∫e^x Tanx dx + ∫e^xlog Secx – ∫e^x Tanx dx
= e^x log Secx + c
∴ ∫(Tanx + Log Secx) e^x dx = e^x log Secx + c

Question 8.
Evaluate \(\int_0^1\left(\frac{x^2}{x^2+1}\right) d x\)
Solution:

Question 9.
Find the area bounded by the parabola y = x², the X-axis and the lines x = -1, x = 2.
Solution:

Question 10.
Find the order and degree of the differential equation
\(\frac{d^2 y}{d x^2}\) = \(\left[1+\left(\frac{d y}{d x}\right)^2\right]^{5 / 3}\)
Solution:
Given differential equation is

Section – B

II. Short Answer Type questions.

1. Attempt any five questions.
2. Each question carries four marks.

Question 11.
Find the value of k, if x + y – 5 = 0 and 2x + ky – 8 = 0 are conjugate with respect to the circle x² + y² – 2x – 2y – 1 =0.
Solution:
Given line equations are
x + y – 5 = 0 …….. (1)
2x + ky – 8 = 0 ………. (2)
Here l₁ = 1, m₁ = 1, n₁ = – 5
l₂ – 2, m₂ = k, n₂ = – 8
Given circle equation is S ≡ x² + y² – 2x – 2y – 1 = 0 ……. (3)
Here 2g = -2 ⇒ g = -1
2f = -2 ⇒ f = -1 and c = -1
Radius r = \(\sqrt{g^2+f^2-c}\) = \(\sqrt{1+1-(-1)}\) = \(\sqrt{3}\)
If (1), (2) are conjugate w.r. to the circle S = 0 then

Question 12.
Find the radical centre of the circles x² + y² – 4x – 6y + 5 = 0, x² + y² – 2x – 4y – 1 = 0 and x² + y² – 6x – 2y = 0.
Solution:
Let S ≡ x² + y² – 4x – 6y + 5 = 0
S’ ≡ x² + y² – 2x – 4y – 1 = 0
S” ≡ x² + y² – 6x – 2y = 0
The radical axis of S = 0, S’ = 0 is S – S’ = 0
⇒ -2x – 2y + 6 = 0
⇒ x + y – 3 = 0 ………. (1)
The radical axis or S = 0, S” = 0 is S – S” = 0
⇒ 2x – 4y + 5 = 0 ……… (2)
Solving (1) and (2)

Question 13.
If a tangent to the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (a > b) meets its major axis and minor axis at M and N respectively, then prove that \(\frac{a^2}{(C M)^2}+\frac{b^2}{(C N)^2}\) = 1 where C is the centre of the ellipse.
Solution:
Given ellipse equation is \(\frac{x^2}{a^2}\) + \(\frac{y^2}{b^2}\) = 1 (a > b)
Let P(θ) (a cosθ, b sinθ) is any point on the ellipse.
Equation of the tangent at P(θ) is S₁ = 0

Question 14.
Find the eccentricity, co-ordinates of foci, length of latus rectum and the equations of directrices of the ellipse 9x² + 16y² – 36x + 32y – 92 = 0.
Solution:
Given ellipse equation is 9x² + 16y² – 36x + 32y – 92 = 0
⇒ 9(x² – 4x + 4) + 16 (y² + 2y + 1) = 92 + 36 + 16
⇒ 9(x – 2)² + 16 (y + 1)² = 144
⇒ \(\frac{(x-2)^2}{16}\) + \(\frac{(y+1)^2}{9}\) = 1
Here a² = 16, b² = 9
Also h = 2, k = -1

Question 15.
Find the equations of the tangents to the hyperbola 3x² – 4y² = 12 which are
(i) Parallel and
(ii) Perpendicular to the line y = x – 7.
Solution:
Given hyperbola equation is 3x² – 4y² = 12
⇒ \(\frac{x^2}{4}-\frac{y^2}{3}\) = 1
Here a² = 4, b² = 3

i) The tangent is parallel to y = x – 7
∴ m = slope of the tangent = 1
Equation of the parallel tangents are
y = mx ± \(\sqrt{a^2 m^2-b^2}\)
⇒ y = x± \(\sqrt{4-3}\)
⇒ y = x ± 1
⇒ y = x + 1, y = x – 1
⇒ x – y + 1 = 0, x – y – 1 = 0

ii) The tangents perpendicular to y = x – 7
∴ m = slope of the tangent = -1
∴ Equation of the perpendicular tangents are
y = (-1)x ± \(\sqrt{4-3}\)
y = -x ± 1
⇒ y = -x + 1, y = -x – 1
⇒ x + y – 1 = 0, x + y + 1 = 0

Question 16.
Evaluate \(\int_0^{\pi / 2} \frac{d x}{4+5 \cos x}\)
Solution:

Question 17.
Solve \(\frac{d y}{d x}\) + yTanx = Cos³x. dx
Solution:
Given differential equation is
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) + yTanx = Co³x ……. (1)
Here P = Tan x and Q = Cos³x.
I.F = \(\text { e } \int^{P d x}\) = e∫Tanx dx
= e^log|secx|
= Secx.

Section – C
(5 × 7 = 35)

III. Long Answer typecfoestions,

1. Attempt any five questions.
2. Each question carries seven marks.

Question 18.
Find the equation of a circle passing through the points (1, 2), (3, -A), (5, -6).
Solution:
Let A = (1, 2), B = (3, -4), C = (5, -6) and D = (19, 8)
The equation of the circle which having \(\overline{\mathrm{AB}}\) as a diameter is
(x – 1) (x – 3) + (y – 2) (y + 4) = 0
⇒ x² – 4x + 3 + y² + 2y – 8 = 0
⇒ x² + y² – 4x + 2y – 5 = 0 …….. (1)
Equation of \(\overline{\mathrm{AB}}\) is
y – 2 = \(\frac{-4-2}{3-1}\)(x – 1)
y – 2 = -3(x – 1)
y – 2 = -3x + 3
3x + y – 5 = 0
The equation of circle passing through A, B is (x² + y² – 4x + 2y – 5) + λ(3x + y – 5) = 0 If this circle through C (5, – 6) then
(25 + 36 – 20 – 12 – 5) + λ(15 – 6 – 5) = 0
⇒ 24 + 4λ = 0
⇒ 6 + λ = 0
⇒ λ = – 6
Equation to the circle through A, B, C is
(x² + y² – 4x + 2y – 5) – 6 (3x + y – 5) = 0
⇒ x² + y² – 4x + 2y – 5 – 18x – 6y + 30 = 0
⇒ x² + y² – 22x – 4y + 25 = 0 …….. (2)
Substituting D (19, 8)
361 + 64 – 22(19) – 4(8) + 25 = 361 + 64 – 418 – 32 + 2
= 450 – 450 = 0
∴ D lies on (2)
Hence A, B, C, D are concyclic.

Question 19.
Show that x² + y² – 6x – 9y + 13 = 0, x² + y² – 2x – 16y = 0 touch each other. Find the point of contact and the equation of common tangent at their point of contact.
Solution:
Equations of the circles are

AB = \(\sqrt{(3-1)^2+\left(\frac{9}{2}-8\right)^2}\) = \(\sqrt{4+\frac{49}{4}}\) = \(\frac{\sqrt{65}}{2}\)
AB = |r₁ – r₂|
∴ The circles touch each other internally. The point of contact ‘P’ divides AB
externally in the ratio r₁ : r₂ = \(\frac{\sqrt{65}}{2}\) : \(\sqrt{65}\)
= 1 : 2 Co-ordinates of P are

∴ Equation of the common tangent is
S₁ – S₂ = 0
-4x + 7y + 13 = 0
4x – 7y – 13 = 0

Question 20.
Derive the equation of a parabola in the standard form y² = 4ax with a diagram.
Solution:
To study the nature of the curve, we prefer its equation in the simplest possible form we proceed us follows to derive such an equation.

Let S be the focus, I be the directrix as shown in fig. Let Z be the projection of ‘S’ on I and ‘A’ be the midpoint of SZ. A lies on the parabola because SA = AZ. A is called the vertex of the parabola. Let YAY be the straight line through A and parallel to the directrix. Now take ZX as the -axis and YY as the Y-axis.

Then A is the origin (0, 0). Let S = (a, 0), (a > 0). Then Z = (-a, 0) and the equation of the directrix is x + a = 0.

If P(x, y) is a point on the parabola and PM is the perpendicular distance from P to the directrix l, then \(\frac{S P}{P M}\) = e = 1.
∴ (SP)² = (PM)²
⇒ (x – a)² + y² = (x + a)²
∴ y² = 4ax.
Conversely if P (x, y) is any point such that y² = 4ax then
SP = \(\sqrt{(x-a)^2+y^2}\) = \(\sqrt{x^2+a^2-2 a x+4 a x}\)
= \(\sqrt{(x+a)^2}\) = |x + a | = PM.
Hence P (x, y) is on the locus. In other words, a necessary and sufficient condition for the point P(x, y) to be on the parabola is that y² = 4ax.
Thus the equation of the parabola is y² = 4ax.

Question 21.
Evaluate ∫(6x + 5) (\(\sqrt{6-2 x^2+x}\)) dx.
Solution:
Let 6x + 5 = A (1 – 4x) + B
Equating the co-efficients of x
6 = -4 A ⇒ A = \(\frac{-3}{2}\)
Equating the constants
A + B = 5
B = 5 – A = 5 + \(\frac{3}{2}\) = \(\frac{13}{2}\)

Question 22.
If I_n = ∫sinⁿ x dx, then show that

for n ≥ 2 and deduce the value of ∫sin⁴x x dx.
Solution:
I_n = ∫sinⁿ x dx
= ∫sin^n-1x sin xdx
= sin^n-1 x (-cos x) – ∫(n – 1) sin^(n-2) x cos x (-cos x) dx
= -sin^n-1 xcosx + (n- 1) ∫sin^n-2x cos² x dx
= – sin^n-1 xcosx + (n – 1) ∫sin^n-2x(1 – sin²x)dx
= -sin^n-1 xcosx + (n- 1) ∫sin^n-2xdx – (n – 1) ∫sin,sup>n xdx
= -sin^n-1 xcosx + (n – 1) I_n-2 – (n – 1)I_n

Question 23.
Evaluate \(\int_0^\pi\left(\frac{x \sin ^3 x}{1+\cos ^2 x}\right)\)dx
Solution:

Question 24.
Solve xy²dy – (x³ + y³) dx = 0.
Solution:
Given differential equation is
xy²dy – (x³ + y³) dx = 0
⇒ xy²dy = (x³ + y³) dx
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^3+y^3}{x y^2}\) ……… (1)
This is a homogeneous differential equation
Put y = Vx
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = V + x\(\frac{\mathrm{dv}}{\mathrm{dx}}\)
Substituting these values in (1)

Access to a variety of AP Inter 2nd Year English Model Papers Set 9 allows students to familiarize themselves with different question patterns.

AP Inter 2nd Year English Model Paper Set 9 with Solutions

Time : 3 Hours
Max. Marks : 100

SECTION – A

I. Annotate ANY TWO of the following in about 10-15 lines each : (2 × 4 = 8)

Question 1.
Unfortunately, our attitude towards family life is not reflected in our attitude towards community behaviour.
Answer:
Context: This passage is taken from the essay “Learning from the West” written by N.R.Narayana Murthy. The writer was a great Industrialist and a Writer. Here he writes about the virtues, one should possess in life.

Explanation: The writer exposes himself against the limitations in the behaviour of the Indian. An Indian is interested in his family. He wants to develop it in a healthy way. One should be loyal to the community, in addition to the family relations. It is unfortunate that we lack it in India.
Therefore, the writer suggests to think about the behaviour in the West and follow. It is our responsibility to act, accordingly.

General Relevance : In the beginning of the essay, Narayana Murthy explains as to why he has entitled the essay “Learning from the West”. He thinks that it is obvious to have a broad outlook.

Question 2.
My gloomy thoughts probably stem from an accident I had a few years ago.
Answer:
Context: This sentence is taken from the essay “In Celebration of Being Alive” written by Dr. Christian Barnard. When the writer thought about the suffering prevailing in the world, he remembered an accident in his life. He described the accident in the essay.

Explanation : His thoughts were gloomy but his suffering was alive in his mind. As he and his wife were crossing the street, they were caught in the accident. They got treatment at the hospital. But it was strange for him, why he should get such a punishment, by God. His anguish was great, the unexpected suffering, according to the narrator, restricted him from the service he had to render to the patients.

General Relevance : The suffering stopped the services of the doctor. He had broken ribs. So, the narrator was angry with God, for this situation.

Question 3.
Some books are to be tasted, others to be swallowed and some few to.be chewed and digested.
Answer:
Context : This sentence is taken from the lesson “Of Studies” written by Francis Bacon. He was called the Father of English Essay. This passage is about the three types of books available. Different books are to be treated differently.

Explanation : The essayist differentiates the books according to the subject matter. Actually one should not read much. If any body reads, he should know the content of the book. Some books are to be read formally. We have to taste it and leave it. Some books are to be read completely. We read them because we are curious. Some of the other books are to be studied carefully. They are useful for us in real life experiences. The knowledge acquired shall become our own and so they have to be chewed and digested

General Relevance : Bacon’s style is aphoristic. The words tasted, swallowed and chewed give an interesting experience. Thus the language of the essay is appreciated.

Question 4.
The only solution of the problem is to make manking pure.
Answer:
Context: This line is taken from the lesson “The Secret of Work” written by the great thinker Swami Vivekananda. In this sentence, he tells about the problems in the world and leads us to a solution.

Explanation: Vivekananda takes up the solution for the problem of the miseries in this world. Ignorance is the cause for this problem. Purity of behaviour, incessant work and understanding of the problem will solve the problem. We have to pick up good and follow it. The impressions of goodness should come out from the heart. The secret of work lies in self denial.

General Relevance : It is a known thing that continuous work yields good results. Work with love, work without selfishness shall bring a change; in the society. Vivekananda hopes ’Work is Worship”.

II. Annotate ANY TWO of the following in about 10-15 lines each : (2 × 4 = 8)

Question 1.
And inward ripeness doth much less appear
that some more timely happy spirits end with
Answer:
Context: These lines are taken from the poem “On His Having Arrived at the Age of Twenty Three”. It was written by the epic poet John Milton. It is about his dissatisfaction towards God. Later he reconcil.

Explanation : John Milton regrets that he has no proper development in life. He gets 23 years of age but his talents are not ripened . He thinks that he is growing rapidly. His age grows but his talent of writing has not developed. His capability is much less than what is expected to be. He is unhappy that his appearance, his age and his achievements are quite disagree one with the other. But afterwards he turns to introspection and consoles himself.

General Relevance : These lines give a vivid description of the . originality of the poet. It is aspiration to become a poet is given here.

Question 2.
Thus alone can we attain
To those turrents, where the eye
sees the world as one vast plan
Answer:
Context: These lines’are taken from the poem “The Builders” written by H.W Longfellow. Here people are compared to builders. The life is the structure and time controls it.

Explanation : The poet wants to find the builder quite worthy. One’s life is built based on one’s behaviour. In this world every minute is valuable. The construction of the building of life is based on time. Time, gives experience and experience is life. People should not be fricted to one place. They should become great citizens of the world.
If there is magnanimity the individual has a pleasing construction of life and he becomes great in life.

General Relevance : The poet wishes that an individual should not be limited to a country but they should be world citizens.

Question 3.
Enough of science and of Arts
close up those barren leaves
Answer:
Context: These lines are extracted from the poem “The Tables Turned” written by William Wordsworth. These lines suggest how the poet creates interest in the study of nature.

Explanation : These are the lines from the concluding stanza. The central idea of the poem is knowing the importance of nature in education. The poet wants the tables of books to be turned. There is no need of the books. Laborious study is to be shun. There are many books of science or of arts. They have plenty of knowledge in each subject. But This study is not equivalent with the study of nature which gives a lot of knowledge. Those books do not think of the natural messages available in the nature. So, the poet stresses the need of the study of nature.

General relevance : All the knowledge, including that, which belongs to science and art, is available in the study of nature. So one should not hesitate to go in for the study of nature.

Question 4.
How will you daunt any free, far journeying fancy
that rides upon the pinions of rain ?
Answer:
Context: These lines are taken from the poem “A Challenge to Fate” written by Sarojini Naidu. It is about the courageous behaviour of the writer. It is a challenge against Fate.

Explanation : The writer is an individual having many talents. She can speak well, she can freely travel in the world, she has gather own freedom of thought. Fate cannot dominate her movements. She has a challenging career before her. How can all this be removed by ‘Fate’. She thinks that the efforts of Fate shall become unsuccessful. Her dreams cannot be spoiled.

General Relevance : These two lines depict the free movement. The weakness of Fate is also touched here. Happy experience of the writer cannot be altered by Fate.

III. Answer ANY TWO of the following questions in about 10-15 lines: (2 × 4 = 8)

Question 1.
Who were the driver and the mechanic in the Grand prix held at the Cape Town Red Cross Children’s Hospital ? In what way was the choice of their roles suitable ?
Answer:
Dr.Christian Barnard was a heart sargeon and a writer. He wrote the essay “In Celebrations of Being Alive” giving his experience at the time of an accident, The writer recollected an incident when he had dissatisfaction with God. At the Cape Town Red Cross Children’s Hospital a strange thing happened. There was a breakfast trolly at the hospital. Two boys drove it in the hospital. One acted as a mechanic and the other as the driver. Actually both were physically handicapped. One lost his eyes and the other had a hole in the heart. The trolley was run at a high speed and all the spectators were filled with joy. The blind one acted as mechanic galloping along, with the trolley. The other seated at the lower deck, drove the trolley, with one leg on the floor. The choice of the roles was suitable. Though blind he could push it on and the other sat at a place and steered it with one hand.

Question 2.
Give a list of lessons that Narayana Murthy feels we should learn from the west.
Answer:
N.R.Narayana Murthy is a brilliant Industrialist and Writer. In the essay “Learning from the West”, he gives a list of things that the Indians should learn from the west.
Narayana Murthy likes the tendency of community thinking, seen in the West. There is no corruption in the Western society. The Indians want to get bribe at every stage. Problems of the society, are preferred. The people of the West are accountable to the work they do.
Dignity of labour, merit in professionalism, keeping up the vows are the merits of the people in the West. Similarly, we should learn the lessons given by Narayana Murthy.

Question 3.
What does Vivekananda say about a man of character ?
Answer:
Swami Vivekananda was a great thinker and an orator. He became the disciple of Sri Ramakrishna Paramahamsa. He said “Work is Worship”. In this world, according to Vivekananda, ignorance is seen every where. It is the cause for all the miseries. In the Gita, the importance of continuous work, is explained. In every work there is good and evil. If an individual does good deeds, his character is great. He will never allow the bad in his deeds to come out, Good thoughts and good deeds will establish his character. He continues following i good action only. The secret of work lies in non-attachment. It can be ’ called Samskara. Thus good character is needed for a good citizen.

Question 4.
What according toBacon is the theme of “of studies” ?
Answer:
Francis Bacon was a great english essayist. ‘Of Studies’ is an interesting essay depicting the benefits and disadvantages of studies. Reading books gives us happiness. It helps us to become more intelligent in communication. Natural abilities can be developed by reading books. Sometimes it becomes harmful if too-much time is spent on it. Showy tendency should be avoided. There are books to be skipped through, books to be read completely and some others to be read with great care and to be digested. Through studies an individual becomes a full rnan. The author tells that the study of different subjects helps us to cure the weaknesses of the mind.

IV. Answer ANY TWO of the following questions in about 10-15 lines : (2 × 4 = 8)

Question 1.
The poem ‘Any woman’ is a celebration of the glory of womanhood. Illustrate.
Answer:
The poem “Any Woman” was written by Katharine Tynan, a British writer. She produced a number of novels and poems and became a noted writer. Any woman is a poem depicting the supremacy of a woman in the family. She loves her children and guards them carefully. Without a woman, there is no house of recognition. A mother takes ; care of the children providing food and shelter. She keeps the house neat and tidy. She protects them at the time of danger. If the mother leaves a child, fie will be spoiled. She is like a pillar in the house and keeps the house graceful. A good woman brings respect to the family.

Question 2.
What is the appropriateness of the title “The Builders” ? Do you agree with the poet about building one’s own life ?
Answer:
‘The Builders’ is an interesting poem written by the great American poet H.W. Longfellow. The poet compares men to builders.
An individual’s life is controlled by time and he is the one who constructs the building of life. According to the poet, everybody plans his life according to his will. He experiences something and advances based on the previous experience. Everyday’s work is taken as a block in the construction of the building. There are people who build their character on good behaviour, with good blocks. Those lives have become buildings suitable for the dwelling of God. Thus people have to build huge buildings of character. The poet suggests the title, The Builders’ and it is quite apt.

Question 3.
Do you prefer to gain knowledge through books or become wise through exprience of nature ? Give reasons in support of your answer, with reference to the poem “The tables Turned”.
Answer:
‘The Tables Turned’ is a famous poem written by William Wordsworth. He was a romantic poet and his love towards nature is of great value. In the poem, the poet suggests two ways of learning. A common view is reading books and get knowledge. But in the view of the poet learning is possible and complete by the study of nature. The secret of creation, the dynamic changes, the natural powers, the happiness of the birds and creatures and others are found in nature.

The bookish knowledge is not enough for the ripening of mind. The good and bad can be seen by practical study only. Kind behaviour j towards the things of nature, fills our heart with happiness. So along with the study of books, the study of nature should also be given importance. Everything is possible when the study of nature takes importance in our life.

Question 4.
“Time and Tide want for no; man”, is an old saying, discuss this in the context of the poem” On his having arrived at the age of twenty three.
Answer:
John Milton was an epic poet. He writes in a grand style. “On , His Having Arrived at the Age of Twenty Three” is a sonnet, written by this poet. Here the saying of ‘Time and Tide wait for no man’ is suitable to be discussed. It is a fact that the tides in the sea move on to k the shores at a high speed and they cannot be stopped by anything.
Similarly time doesnot stay or stop at any place. It is a natural phenomena. The consumption of time is a continuous process. Milton looks to himself and wonders how his years have passed away. He cannot find a reason in the beginning. So he complains against God. But after introspection, he finds a solace. He thinks that time does not wait for him or for his maturity. Thus the saying can be applied to the views of John Milton.

V. Answer ANY ONE of the following questions in about 25 lines : (1 × 8 = 8)

Question 1.
Write a character sketch of Tom-Sawyer.
Answer:
Tom Sawyer was the young hero in the novel “The Adventures of Tom Sawyer” written by Mark Twaim, an American writer. Tom ; was a boy of a tender age. He was a misfit in the school. His behaviour with Becky Thatcher is interesting because he had a liking for her. He 1 expressed it boldy. He was equally like by the girl. Tom had another friend, Huck leberry Finn, who was emtidy, and spent his time fishing and swimming. Tom liked this boy because he wanted to have adventures in life.

Tom went to the graveyard one day along with Huck, to see the ghosts. It was a mysterious experience for him. The three men who came there to rob a dead body quarrelled among themselves. One of them, a doctor was killed by Injun Joe. Tom was frightened to watch Injun killing the doctor. Tom had a liking to wander and his stay at Jackson’s island, was wonderful.

The parents arranged a funeral, thinking that Tom and Huck , were dead, which indeed created a humorous occasion for Tom’s jolly mind. Tom’s presence at the church created a hilarious situation. Tom’s determination to save Muff Potter and his brave witness at the court, adds to his virtuous behaviour. His relationship with Becky Thatcher was very much appreciative.

Tom wanted to save Widow Bougals. His determination to punish Injun Joe and his pursuasion in the search of the treasure in the cave, were note worthy. Tom got the treasure and changed Huck, the dirty, lazy and unworthy boy to a neat and tidy boy. Thus Tom was typical of the days of Mark Twainer in America.

Question 2.
What is the part played by Injun Joe in the life of Tom sawyer.
Answer:
Mark Twain carved interesting characters in the novel “The Adventures of Tom Sawyer”. The main character is Tom Sawyer but there are other important characters, fulfilling the background to make, Tom an impressive one. One such character is Injun Joe, the individual with villainy.

Injun Joe was seen by Tom and HuCk at the burial ground. He was one among the three who came to steal a dead body. As there was a tussle among themselves Injun did not hesitate to kill the doctor. He made Muff the other person of the three believe that he had killed the doctor, in his drunken stage. This behaviour of Injun was a frightening experience to Tom.

In addition to this at the court also Injun escaped through the window, when Tom gave his witness against him. Tom’s impression still strengthened. He wanted to punish the villain some how or the other. Injun was after the hunt of the treasure for which he entered the cave. The judge closed the door. So he could not come out and , died of hunger. Now Tom had great satisfaction because the world could get rid of a dangerous man.
Tom sawyer was of a tender age. His character could become complete with the activities and their effect upon it.

3. Whpt are the turning points in the story ? Discuss the main events In the plot.
Answer:
Mark Twain, wrote the novel “The adventures of Tom Sawyer”.
The story of Tom Sawyer is narrated by the writer, with interesting turning points at different places. The story moves round the tender boy Tom Sawyer. He was being brought up by Aunt Polly. At the school, there was an interesting episode. He got punishment but he drew a beautiful picture. He showed love towards Becky, a girl in the class. There is a turning point at the graveyard. The murder of the doctor, by Injun Joe, created frustration in the minds of Tom and Huck. It continued throughout the story until the person died in the cave.

There is an important point, when they attended the funeral function. It adds to the humour of the novel. Muff Potter was jailed but Tom gave witness before the judge arid he was set free. Becky’s parents arranged a picnic. Tom and Becky were caught in the cave but were saved. The judge arranged an iron door. This door became the cause for the death of the villain Injun Joe at the end. Huck was adopted by Widow Dougals. He became a person with culture.

Afterwards Tom wanted to form a pirate gang and Huck would become the bold pirate. Thus, there are some turning points which make the story interesting.

VI. Read the following passage carefully and answer the questions that follow: (5 × 1 = 5)

Light enters the eye by refracting, or bending, as it passes through the cornea. Light rays then pass through the opening into the eye known as the pupil. The pupil size is controlled by a muscle known as the iris. The pupil becomes smaller when in bright area and larger in dark area. After leaving the pupil, light rays are refracted once again as they pass through the convex lens of eye. Light rays continue travelling through a jelly-like material called the vitreous humor. An upside down image is formed on the back of the eye known as the retina. Cone cells on the retina interpret the colour of the image and rod cells interpret the black and white colours. Lastly the image is taken to the brain for the image to be seen correctly.

Questions:
1. What is the opening into the eye called ?
Answer:
The pupil

2. What is the muscle that controls the size of the pupil ?
Answer:
The iris

3. State true or false :
The pupil becomes smaller in dark areas.
Answer:
False

4. Write the antonym of ‘convex’.
Answer:
Concave

5. Where does an upside down image form in the eye ?
Answer:
The retina

VII. Read the passage and answer the questions that follow : (5 × 1 = 5)

‘Mr. Welsh, help, help!’ Mr. Welsh opened the door.’Mr. Welsh, please help me ! Two men want to kill Widow Dougals !’ Mr. Wfelsh and his sons took their rifles. They rah to the widow’s house. Suddenly there was a loud shot. Injun Joe and his friend escaped, and the widow was not hurt.
The next morning Huck returned to see Mr. Welsh. ‘You’re a courageous boy, Huck,’ said Mr. Welsh, You saved the widow’s life. The ,two men escaped but we’ll find them. Sit down and have breakfast with us !’ Huck was happy because he saved the widow’s life. And now he had new friends, Mr. Welsh and his family. That morning all people of St. Petersburg knew that Tom and Becky were lost in Mc.Dougal’s Cave and they were wor¬ried.

Questions :
1. Why is Huck asking for Mr. Welsh’s help ?
Answer:
Because he knew that two men wanted to kill Widow Douglas.

2. Did Mr. Welsh catch Injun Joe and his friend ? What happened to them ?
Answer:
No. They both escaped.

3. What are the two reasons for which Huck was happy ?
Answer:
a. He saved the widow’s life.
b. He had new friends, Mr. Welsh and his family.
c. Because they came to know that Tom and Becky were lost.

4. Why were the people of St. Petersburg worried ?
a. because the Widow was attacked.
b. because they knew Injun Joe was involved in the attack.
c. because they came to know that Tom and Becky were lost.
Answer:
c. because they came to know that Tom and Becky were lost.

5. Write the noun form of ‘courageous’ ?
Answer:
Courage.

VIII. Study the advertisement given below and answer the questions that follow: (5 × 1 = 5)

Questions :
1. Mention any two offensive acts of ragging.
Answer:
Disorderly behaviour or treatment with fellow student by words spoken or written / or by an act which has effect of teasing / treating and handling with rudeness with any other students / indulging in rowdy or indecent activities.

2. Who has defined ragging in the above advertisement ?
Answer:
The Honourable Supreme Court of India.

3. Pickout the word that means encourage from the passage.
Answer:
Abetment.

4. Write verb forms of the following words.
a) Prohibition
b) apprehension.
Answer:
a) Prohibit
b) Apprehend.

5. Write down any two punishments for students involved directly or indirectly in ragging.
Answer:
1) Cancellation of admission
2) Suspension from classes.

IX. Study the following Pie Chart is about the percentage of people in the U.S. who speak different languages of the Indian subcontinent. (5 × 1 = 5)
Questions :
1. What does the pie chart show ?
Answer:

2. How many languages have been taken into consideration ?
Answer:

3. What is the major Indian language spoken in the U.S. ?
Answer:

4. Which position does Telugu stand in the pie chart from the highest to the lowest order ?
Answer:

5. Which of our neighbouring countries speak Sinhalese ?
Answer:

SECTION – C

X. Write a letter to the chairman of your local municipality, complaining about the poor sanitary condition and mosquito menance in your locality. (1 × 5 = 5)
(Hints : Nehru Nagar – sanitary conditions – mosquitoes – contaminated water.)
Answer:
17-461 B2
Nehru Nagar
Madanapalle.

11 March, 2019

The Chairman
Madanapalle Municipality
Madanapalle.
Sir,
Sub : Complaining about the poor sanitary conditions and mosquito menace in Nehru Nagar area of Madanapalle Municipality – Submitted – Reg.
I would like to bring the following few lines to your kind notice. I am a resident of Nehru Nagar in Madanapalle Municipality. In our area, sanitary conditions are very poor and garbage is dumped everywhere. The municipal workers are not cleaning the streets and garbage bins daily. Due to this, pedestrians have to actually cover their nose while walking down the roads.
The public toilets also stink and are not cleaned by the sanitation workers every day. This is a major cause of breeding mosquitoes. Contaminated water and waste material are the reason for the diseases such as dengue, diarrhea and others.
For the interest of public health, I request that immediate action be taken against the poor sanitary conditions and to arrest the mosquito menace that has been growing at a shocking rate over the weeks.
Thanking you sir
Yours faithfully
Signature.

OR

You are interested in doing a .short term course in Spoken English during your summer vacation. Write a letter ,to the director, NEO Spoken English and Grammar Institute, Annamaiah circle, Tirupati, enquiring about the terms and conditions for the course.
(Hints: spoken english – course duration – terms and conditions.)
Answer:
K.Amaranath
2-3-40 Gandhinagar
Vijayawada.

15-4-2019

The Director
NEO Spoken English Institute
Annamalai circle
Tirupathi.
Sir,
Sub: Information about short term course-request-regarding. Having come to know that the institute is running a short term course in spoken English. I wish to know the particulars about it. I wish to know the duration of the course and the date of starting. The fee particulars, the time of running the classes, may also be sent. Is’ there any hostel facility ? If there is hostel, the fee there of may also be informed.
Thanking you
Yours faithfully
K. Amaranath.

XI. Write a short paragraph of about 8 lines describing the process df an idea daily routine. (1 × 5 = 5)
Answer:
Some people are very particular to maintain a dairy of them-selves. Generally, the dairy should be ideal. It adds to our happiness , at a later date, giving a ghimpse of our experiences. The daily routine from the moment you get up from bed, to the time you go to bed at night shquld be written, in the dairy. We should not hide any of our experiences of the day. Whether it is happy or unhappy, whether it is good or not it is a part of our routine and so it should .find a place in the narration. Very minute things of our life shall provide great happy occasions. Thus we have to maintain an ideal routine, in our life.

OR

Write a short paragraph narrating in a sequence the steps involved in making tea for two people.
Answer:
It would be better if we learn certain things, needed for our smooth living. How to make tea for two people. The following are needed to prepare tea.
Tea Kettle
Spoons
Filter
Cups and Saucers
Tea powder
Sugar
Milk.
First of all take some water in the kettle. Mix some tea powder in it and boil it for a few minutes. Pour some milk into if and boil it for another few minutes. Take the kettle out of the stove and filter the tea. Add some sugar into the tea. Serve the tea in-two cups to the people waiting.

XII. Prepare a curriculum vitae. (1 × 5 = 5)

Pocha Sumalatha – aged 22 years – B.Sc.(CS)-79% marks – Govt.Degree College, Visakhapatnam – Intermediate(MPC) – 88% – GovtJunior College, Anakapalli – SSC -90% – Govt, high School, Anakapalli – applying for the post of System Analyst – Mozaic Software Company, Visakhapatnam.
Answer:

COVERING LETTER

Pocha Sumalatha
20 June, 2019
10/151, Market Street,
Anakapalli.

The Director
Mozaic Software Company
Visakhapatnam.
Sir,
Sub : Application for the Post of Analyst – Regarding. Having given to understand that there is a vacancy for the post of Analyst at your company, I wish to apply as a candidate for the post. As per the qualifications, I passed B.Sc (C.S.) with 79% of marks, intermediate with 88% and S.S.C with 90% of marks. I have got a good academic record.
I request you to appoint me as analyst and oblige.
Yours Faithfully
R Sumalatha. .
Enclosures : Resume, Copies of Certificates.

RESUME

Pocha Sumalatha
10/151, Market Street,
Anakapalli.

OBJECTIVE
Post of an analyst at Mozaic Software Company, Visakhapatnam.

EDUCATION
2015-2019 – B.Sc (C.S.) Govt. Degree College, Visakhapatnam 79% of marks.
2014-2016 – Intermediate Govt. Junior College, Anakapalli 88% of marks.
2013-2014 – S.S.C Govt. High School, Anakapalli, 90% of marks.

WORK EXPERIENCE
Nil.

STRENGTHS
Good Communication skills in English. Good co-ordinater.

ADDITIONAL INFORMATION
Languages Known : English, Telugu and Hindi
Hobbies : Reading Books and Listening to Music
Father’s name : Pocha Subba Rao
Reference : Will be given on request.

DECLARATION
I here by undertake that the details furnished above are true and correct to the best of my knowledge and belief.
P Sumalatha
Place : Anakapalli

XIII. Fill in the Bank forms writing your answers against the number given two blanks. (10 × 1/2 = 5)

Mr. Kiran went to the Treasury Branch of SBI to deposit an amount of Rs. 22,500/- (10 notes of 2,000/- and 5 notes of 500/-) in Chandrasekhar’s account on 2 Jan, 2019. Mr. Chandrasekhar holds an SB account in that branch with account No. 10878827556 and his mobile number is 9441844435.

Answer:
1) 2-1-2019
2) Treasury
3) 10878827556
4) Mr. Kiran
5) 9441844435
6) Twenty two thousand five hundred
7) 22,500/-
8) Signature
9) 2000 × 10, 500 × 5
10) 22,500/-

XIV. Construct a dialogue between the teacher and a student who has come to class late. (1 × 5 = 5)
Answer:
Student : Good morning sir, May I come in ?
Teacher : Good morning Naveen, Why are you late ?
Student : Sorry sir, my bus waslate.
Teacher : Naveen, don’t blame on the bus. You missed the bus. At what time did you come to the bus stop ?
Student : Sir, I came to the bus stop at 8.30 a.m. in the morning.
Teacher : Naveen, you have to come to the bus stop a little early, then only you can catch the early bus and come to class in time.
Student : Yes, Sir, I will start early from home from tomorrow onwards.
Teacher : Naveen, at what time do you get up in the morning ?
Student : Sir, I get up at 6 a.m. daily.
Teacher : Naveen, it is not right and you have to get up between 4:00 and 4:30 a.m daily in the morning. Then only, you will be active.
Student : Yes sir, from tomorrow onwards, I will try to get up early in the morning.
Teacher : Naveen, it is my last warning. If you come late from tomorrow, you will be strictly punished.
Student : Yes sir, from tomorrow, I will not he late.
Teacher : OK, Now you may come in.
Student : Thank you sir.

OR

A dialogue at the bank between a customer and the clerk about opening an account.
Answer:
Customer : Excuse me sir.
Clerk : Yes, what can I do for you ?
Customer : Sir, I would like to open an account in your bank.
Clerk : you are most welcome. Where do you live ?
Customer : I live in Lakshmi Nagar, Srikakulam.
Clerk : OK, please read this form carefully. Do you want to open savings or current account ?
Customer : Savings account.
Clerk : You have to bring a document for address proof such as an electricity bill or phone bill.
Customer : How much will I have to deposit for opening the account ?
Clerk : Do you want a cheque book ?
Customer : Yes.
Clerk : You have to open the account with one thousand rupees. You’ll get the cheque book by post after 10 or 15 days.
Customer : Sir can I have an ATM Card ?
Clerk : Yes, just tick at the relevant column. You’ll get it after 6 or 7 days by post.
Customer : Sir, how many photos will be required ?
Clerk : Two photos will be needed, one you have to affix on the form and another will be pasted on your pass book.
Customer : Thank you sir, I will bring photos and deposit the amount tomorrow.
Clerk : It’s ok.

XV. Read the passage and make notes. (1 × 5 = 5)

The process of study involves four operations – perception, comprehension, retention and retrieval. In other words, you should first perceive what is relevant to your needs and select only those areas which are important. You cannot study everything available in every book you can lay your hands on. Once you have decided on areas, which are important, you have to read and understand or comprehend the material that you have selected for no learning can take place without comprehension. What is notunderstood is not learnt. Comprehension is thus imperative in the process of learning. However, mere comprehension is not enough. What one understands now, may also be easily forgotten later. Hence, you, as a student, have to make special efforts to retain what you comprehend. You also have to retrieve all that you learn throughout the year at the time of Examination.
Answer:
Four operations – perception, comprehension, retention and retrieval. Select the area of study – understand the material and comprehend. Sometimes forgetting is possible.
Try to retain the subject you studied.
You have to retrieve all that you learnt, at the time of examination.

XVI. Match the words in Column A with their meanings/definitions in Column ‘B’. (5 × 1 = 5)

Column A	Column ‘B’
1. interact	a. a harsh, unpleasant sound
2. animate.	b. beautiful handwriting
3. calligraphy	c. act between
4. cacophony	d. the essence of ego, personality
5. egoity	e. having life
	f. to see something in a way that may or may not be factual, to think
	g. the way in which a person sees the world, a view

Answer:

Column A	Column ‘B’
1. interact	c. act between
2. animate.	e. having life
3. calligraphy	b. beautiful handwriting
4. cacophony	a. a harsh, unpleasant sound
5. egoity	d. the essence of ego, personality

XVII. Mark the stress for ANY FIVE of the following words : (5 × 1=5)

1) perform
2) library
3) polite
4) question
5) romantic
6) conscious
7) grammar
8) intermediate
9) student
10) trophy
Answer:
1) perform = per’form
2) library = ‘library
3) polite = po’lite
4) question – = ‘question
5) romantic = ro’mantic
6) conscious = ‘conscious
7) grammar = ‘grammar
8) intermediate = inter’mediate
9) student = ‘student
10) trophy = ‘trophy

Access to a variety of TS Inter 2nd Year Chemistry Model Papers and TS Inter 2nd Year Chemistry Question Paper March 2018 helps students overcome exam anxiety by fostering familiarity.

TS Inter 2nd Year Chemistry Question Paper March 2018

Note : Read the following instructions carefully.

1. Answer all questions of Section – ‘A’. Answer any six questions in Section – ‘B’ and any two questions in Section – ‘C’.
2. In Section – A, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
3. In Section -”’8′, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
4. In Section – ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
5. Draw labelled diagrams wherever necessary for questions in Section – ‘B’ and ‘C’.

Section – A

Note : Answer all the questions.

Question 1.
State Raoult’s law.
Answer:
Raoult’s law for non-volatile solute : The relative lowering of vapour pressure of dilute solution containing non-volatile solute is equal to the mole fraction of solute. 1

Question 2.
State Faraday’s second law of electrolysis.
Answer:
The amounts of different substances liberated when the same quantity of electricity is passed through the electrolytic solution are proportional to their chemical equivalent weights.

Question 3.
What is PHBV? How is it useful to man?
Answer:
Poly p – hydroxy butyrate – CO – P – hydroxy Valerate (PHBV) : It is a Copolymer of 3 -hydroxy butanoic acid and 3 – hydroxy pentanoic acid.

It is used in medicine for making capsules. PHBV also undergoes degradation by bacteria.

Question 4.
Write the names of the monomers for the following polymers.
a) Bakelite
b) Nylon 6, 6
Answer:
a) Bakelite : In Bakelite monomers are phenol.
(C₆H₅OH) and formaldehyde (HCHO)
Monomers : Phenol, Formaldehyde

b) Nylon 6, 6 :
The repeating monomeric units of Nylon 6, 6 are hexamethylene diamine and Adipic acid.

Question 5.
What is a ligand ? Give one example for unidentate ligand.
Answer:
Ligand : A co-ordinating entity which is bound to the central atom by donating electron pairs is called a ligand.
Eg. : Cl^–, NH₃, CN^– etc.
A unidentate ligand containing two possible donor atoms can co-ordinate through either of donor atoms. Such ligands are called ambidentate ligands.
Eg : NO₂^–

Question 6.
What are the neutral oxides of nitrogen?
Answer:
Nitrous oxide (N₂O) and Nitric oxide (NO) are neutral oxides of nitrogen.

Question 7.
What happens when Cl₂ reacts with dry slaked lime?
Answer:
Chlorine reacts with dry slaked lime and forms bleaching powder.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O

Question 8.
What is blister copper? Why is it so called?
Answer:
During the extraction of ‘Cu’ from copper pyrites when the matte is charged into a Bessemer converter then cuprous oxide combine with cuprous sulphide and forms Cu metal.
2CU₂o + Cu₂S → 6Cu + SO₂ The ‘Cu’ metal is cooled.

The obtained copper metal is impure and is known as “Blister copper” (98% pure). The solid fied copper has blistered appearance due to evolution of SOz Hence it is called blister copper.

Question 9.
Explain carbylatnine reaction with an example.
Answer:
Aniline reacts with chloroform in presence of ale. KOH to form phenyl isocyanide.
C₆H₅ NH₂ + CHCl₃ + 3KOH → C₆H₅ NC + 3KCl + 3H₂O

Question 10.
Explain the reaction of Aniline with Nitrous acid.
Answer:
Reaction of Aniline with nitrous acid : Aniline reacts with nitrous acid at low temperatures (0 – 5°C) to form diazonium salts. (Benzene diazonium salt)

Section – B

Question 11.
Explain Schottky and Frenkel defects.
Answer:
Schottky defect :

1. It is a point defect in which an atom or ion is missing from its normal site in the lattice”.
2. In order to maintain electrical neutrality, the number of missing cations and anions are equal.
3. This sort of defect occurs mainly in highly ionic compounds, where cationic and anionic sizes are similar. In such compounds the co-ordination number in high. Ex. : NaCl, CsCl etc.
4. Illustration :
   
5. This defect decreases the density of the substance.

Frenkel defect :

1. “It is a point defect in which an atom or ion is shifted from its normal lattice position”. The ion or the atom now occupies an interstitial position in the lattice.
2. This type of a defect is favoured by a large difference in sizes between the cation and anion. In these compounds co-ordination number is low. E.g.: Ag – halides, ZnS etc.
3. Illustration :
   
4. Frenkel defect do not change the density of the solids significantly.

Question 12.
Define mole fraction. Calculate the mole fraction of H₂SO₄ in a solution containing 98% (w/w) H₂SO₄ by mass.
Answer:
Mole fraction : The ratio of number of moles of the one component of the solution to the total number of moles of all the components of the solution is called mole fraction.

Mole fraction of solute X_s = \(\frac{n_s}{n_0+n_s}\)
Mole fraction of solvent X₀ = \(\frac{n_s}{n_0+n_s}\)
n_s = number of moles of solute
n₀ = number of moles of solvent
→ It has no units.

Given a solution containing – 98% H₂SO₄ by mass. It means 98 gms of H₂SO₄ and 2 gms of H₂O mixed to form a solution.

Question 13.
What are ’emulsions? How are they classified ? Give one example for each.
Answer:
Emulsion: The colloidal system in which a dispersion of finely divided droplets of a liquid in another liquid medium is called Emulsion. Ex : Milk.
In Milk, the droplets of liquid fat are dispersed in water. This is an example for oil in water type emulsion.

Classification of emulsions : Emulsions are classified into two classes. These are
a) Oil in Water (O/W) and
b) Water in Oil (W/O), (O = Oil; W = Water).
These emulsions are classified as such depending on which is dispersed phase and which is dispersion medium.
a) Oil in Water (O/W) type emulsions:
In this type of emulsions, the dispersed phase is oil and the dispersion medium is water.
Ex : Milk, liquid, fat (oil) in water. Vanishing cream; fat in water.

b) Water in Oil (W/O) type emulsions:
In this type of emulsions the dispersed phase is water and the dispersion medium is oil.
Ex : Stiff greases : water in lubrication oils
Cod liver oil : water in cod liver oil
Cold cream : water in fat.

Question 14.
Explain calcination and roasting with suitable examples.
Answer:
Roasting : Removal of the volatile components of a mineral by heating mineral either alone (or) mixed with some other substances to a high temperature in the presence of air is called Roasting.

1.  It is applied to the sulphide ores. .
2. SO₂ gas is producted along with metal oxide.

Calcination :
Removal of the volatile components of a mineral by heating in the absence of air is called calcination.

1. It is applied to carbonates and bicarbonates.
2. CO₂ gas is produced along with metal oxide.

Question 15.
Explain Werner’s theory of coordination compounds with suitable examples.
Answer:
Werner’s theory :
Postulates :

1. Every complex compound has a central metal atom (or) s ion.
2. The central metal shows two types of valencies namely primary valency and secondary valency.

A) Primary valency:
The primary valency is numerically equal to the oxidation state of the metal. Species or groups bound by primary valencies undergo complete ionization. These valencies are identical with ionic bonds and are non-directional. These valencies are represented by discontinuous lines (……)
Eg. : C0Cl₃ contains CO³⁺ and 3Cl^– ions. There are three Primary Valencies or three ionic bonds.

B) Secondary Valency:
Each metal has a characteristic number of Secondary Valencies. They are directed in space around the central metal. The number of Secondary Valencies is called Coordination number (C.N.) of the metal. These valencies are directional in Nature.For example in COCl₃ 6NH₃

Three Cl^– ions are held by primary Valencies and 6NH₃ molecules are held by Secondary Valencies. In CuSO₄. 4NH₃ complex SO₄^2- ion is held by two Primary Valencies and 4NH₃ molecules are held by Secondary Valencies.

3) Some negative ligands, depending upon the complex, may satisfy both primary and secondary valencies. Such ligands, in a complex, which satisfy both primary as well as secondary valencies do not ionize.

4) The primary valency of a metal is known as its outer sphere of attraction or ionizable valency while the Secondary Valencies are known as the inner sphere of attraction or coordination sphere. Groups bound by secondary valencies do not undergo ionization in the complex.

Question 16.
What are hormones? Give one example for eaph of the following.
a) Steroid hormones.
b) Polypeptide hormones.
c) Amino acid derivations.
Answer:
Hormones : Hormone is defined as an “organic compound synthesised by the ductless glands of the body and carried by the blood stream to another part of the body for its function”.
Eg : testosterone, estrogen.

1. Example for steroid hormones : Testosterone, Estrogen.
2. Example for poly peptide hormones : Insulin
3. Example for Amino acid derivative: Thyroidal hormones thyroxine.

Question 17.
What are analgesics ? How are they classified? Give one example for each.
Answer:
Analgesics: These are to, reduce or totally abolish pain without. ‘causing impairment of consciousness, mental confusion, incoordination, paralysis, disturbances of nervous system etc.
Analgesics are classified as

i) Narcotic analgesics :
These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics. E.g. : Morphine, Codeine etc.

ii) Non-narcotic analgesics : These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains. E.g. : Aspirin, Ibuprofen etc.

Question 18.
Explain and reactions with an example.
Answer:
i) S_N¹ : Substitution nucleophilic unimolecular reaction :
In this reaction the first step is the formation of stable carbonium ion. This step is slow and is the rate determining step. The alkyl halides which form stable carbonium ion follow S_N¹ reaction.

The order of reactivity of alkyl halides towards S_N¹ reaction.
Tertiary halide > Secondary halide > Primary halide > CH₃ – X.

Benzyl halide and allyl halides are primary halides but participate in SN1 mechanism due to the formation of stable benzyl and allyl carbonium ions (Benzyl and allyl carbonium ions are stabilised due to resonance).

ii) SN² : Substitution nucleophilic bimolecular reaction :
In this reaction the rate of reaction depends on the concentration of alkyl halide and also on the concentration of nucleophile hence it is a bimolecular reaction.

The order of reactivity of alkyl halides towards SN² reaction.
CH₃ – X > Primary halide > Secondary halide > Tertiary halide For a given alkyl group, the reaction of the alkyl halide, R – X follows the same’ order in both the mechanisms R-I > R – Br > R – Cl > R – F

Question 19.
a) What are Galvanic cells? Explain the working of a Galvanic cell by taking Daniel cell as example.
Answer:
Galvanic cell : A device which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell (dr) voltaic cell.
Eg : Daniel cell.

Daniel cell: It is a special type of galvanic cell. It contains two half cells in the same vessel.
The vessel is devided into two chambers. Left chamber is filled with ZnSO₄ (aq) solution and Zn – rod is dipped into it. Right chamber is filled with aq. CuSO₂ solution and a copper rod is dipped into it. Process diaphragm acts as Salt bridge. The two half cell’s are connected to external battery.

Cell reactions : Ion Zn/ZnSO⁴ half cell, oxidation, reaction occurs
Zn → Zn²⁺ + 2e^–
Ion Cu/CuSO₄ half cell, reduction reaction occurs.
Cu⁺² + 2e^– → Cu
The net cell reaction is
Zn + Cu⁺² ⇌ Zn⁺² + Cu
Cell is represented as Zn / Zn⁺² || Cu²⁺ / Cu

b) Describe the salient features of the collision theory of reaction rates.
Answer:
Collision theory of reaction rate bimolecular reactions salient features :

1. The reaction molecules are assumed to be hard spheres.
2. The reaction is postulated to occur when molecules collide with each other.
3. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).
4. For a bimolecular elementary reaction A + B —> products
   K = Z_AB. e^E_a/RT ; Z_AB = Collision frequency.
5. All collisions, do not lead to product formation.
6. The collisions with sufficient kinetic energy (Threshold energy) are responsible for product formation. These are called as effective collisions.
7. To account for effective collisions a factor p called to probability factor or steric factor is introduced.
   K = P Z_AB. e^E_a/RT

Question 20.
a) Write the important reactions involved in the manufacture of sulphuric acid by ‘contact process’.
Answer:
Manufacture of H₂SO₄.by contact process : Manufacturing of H₂SO₄ involves three main steps.

Step-1 :
SO₂ production : The required SO₂ for this process is obtained by burning S (or) Iron pyrites in oxygen.
S + O₂ → SO₂
4FeS₂ + 15O₂ > 2Fe₂O₃ + 8SO₃

Step-2 :
SO₃ formation : SO₂ is oxidised in presence of catalyst with atmosphric air to form SO₃.

The equation reveals the following points :

1. 3 volumes of the reactants convert into 2 volumes of SO₃. i.e., a decrease of volume accompanies the reaction.
2. The reaction is an exothermic change.
3. The catalyst may be present to increase the SO₃ yields.

Step-3 :
Formation of H₂SO₄: SO₃ formed in the above step absorbed in 98% H₂SO₄ to get oleum (H₂S₂O₇). This oleum is diluted to get H₂SO₄
SO₃ + H₂SO₄ → H₂S₂O₇
H₂S₂O₇ + H₂O → 2H₂SO₄

b) Write the preparation of the compounds XeF₂ and XeF₄. Give their structures.
Answer:
Xenon forms the binary fluorides XeF₂, XeF₄, XeF₆ as follows. These are formed by direct combination of Xe and F₂.

structures of XeF₂ :
1) In XeF₂ central atom is ‘Xe’
2) ‘Xe’ undergoes sp³d hybridisation in it’s 1^st excited state

3) Shape of molecule is linear.
4) Xe form two σ.bonds with two fluorines.

b) structures of XeF₄:
1) Central atom in XeF4 is ‘Xe’.
2) Xe undergoes sp³d² hybridisation in it’s 2^nd excited, state.

3) Shape of the molecule is square planar with bond angle 90° and bond length 1.95A.

4) Xe – forms four o-bonds by the overlap of sp³d² – 2p_Z (F) orbitals.

Question 21.
Explain the following reactions with equations.
i) Kolbe’s reaction.
ii) Riemer – Tiemann reaction,
iii) Williamson’s ether synthesis,
iv) HVZ reaction.
Answer:
i) Kolbe’s reaction :
Phenol reacts with sodium hydroxide to form sodium phenoxide. This undergoes electrophillic substitution with CO² to form salicylic acid.

ii) Reimer-Tiemann reaction :
Phenol reacts with chloroform in presence of NaOH to form salicylaldehyde (O-Hydroxy benzal- dehyde). This reaction is known as ii) Reimer-Tiemann reaction : Phenol reacts with chloroform in presence of NaOH to form salicylaldehyde (O-Hydroxy benzal- dehyde). This reaction is known as Reimer-Tiemann reaction.

iii) Williamsons ether synthesis :

1. This method is used for the preparation of symmetrical and unsymmetrical ethers.
2. The reaction of an alkyl halide with sodium alkoxide to form ethers is known as Williamson’s Synthesis.

E.g. C₂H₅Cl + CH₃ONa → CH₃ – O – CH₂CH₃ + NaCl
(CH₃)₃ – C -ONa + CH₃ – Br → CH₃ – O – C – (CH₃)₃ + NaBr

iv) HVZ reaction :
Carboxylic acids having α-hydrogens are halogenated at the α-position on treatment with chlorine or bromine in presence of small amount of red phosphorus to give α-halo carbo¬xylic acids. This reaction is named as Hell-Volhard – Zelinsky (HVZ) reactions.

Access to a variety of TS Inter 2nd Year Chemistry Model Papers and TS Inter 2nd Year Chemistry Question Paper May 2017 helps students overcome exam anxiety by fostering familiarity.

TS Inter 2nd Year Chemistry Question Paper May 2017

Note : Read the following instructions carefully.

1. Answer all questions of Section – ‘A’. Answer any six questions in Section – ‘B’ and any two questions in Section – ‘C’.
2. In Section – A, questions from Sr. Nos. 1 to 10 are of “Very Short Answer Type”. Each question carries two marks. Every answer may be limited to 2 or 3 sentences. Answer all these questions at one place in the same order.
3. In Section -”’8′, questions from Sr. Nos. 11 to 18 are of “Short Answer Type”. Each question carries four marks. Every answer may be limited to 75 words.
4. In Section – ‘C’, questions from Sr. Nos. 19 to 21 are of “Long Answer Type”. Each question carries eight marks. Every answer may be limited to 300 words.
5. Draw labelled diagrams wherever necessary for questions in Section – ‘B’ and ‘C’.

Section – A

Note : Answer all the questions.

Question 1.
What is ideal solution?
Answer:
The solution which obeys Raoult’s law over the entire range of concentration is known as ideal solution.

Question 2.
Give two examples of zero order reactions.
Answer:
Examples for zero order reactions :

Question 3.
Write any two ores with formulas of the following metals.
(a) Iron
(b) Copper
Answer:
a) Ores of Iron :
Haematite (Fe₂O₃)
Magnetite (Fe₃O₄)

b) Ores of Copper:
Copper pyrites (CuFeS₂)
Copper glance (Cu₂S)

Question 4.
What is tailing of mercury? How is it removed?

Question 5.
Noble gases are inert. Explain.
Answer:
Noble gases are chemically inert:
1) Noble gases have stable electronic configuration octet configuration except He.
2) Noble gases have high Ionisation energy values and have large positive values of electron gain enthalpy.

Question 6.
Why Zn⁺² is diamagnetic where as Mn+2 is paramagnetic?

Question 7.
What is zwitter iron 7 Give anexample.
Answer:
Zwitter ion : In aqueous solution of amino acids, the carboxyl group can lose a proton and amino acid can accept that proton to form a dipolar ion. This ion is called as zwitter ion.

Question 8.
Write any two biological functions of Nucleic acids.
Answer:
Biological functions of Nucleic acids:

1. DNA is the chemical basis of heredity.
2. DNA is the reserve of genetic information.
3. DNA is capable of self duplication during cell division and identical DNA strands are transfered to daughter cells.
4. RNA molecules are used in the protein synthesis in the cell.
5. The message for the synthesis of a protein is present in DNA.

Question 9.
What are antibiotics? Give an example.
Answer:
Antibiotics : The chemical substances produced by micro organisms and inhibit the growth or destroy micro-organisms are called antibiotics.
(or)
The substance produced totally or partly by chemical synthesis which in low concentration inhibits the growth (or) destroy micro-organism by intervening in their metabolic process are called antibiotics. E.g. : Penicillin, Chloramphenicol etc.

Question 10.
What are antifertility drugs? Give an example.
Answer:
Antifertility drugs : These are birth control pills and contain a mixture of synthetic oestrogen and progesterone derivatives. E.g. : Norethindrone, Ethynylestradiol (novestrol).

Section – B

Question 11.
What is relative lowering of vapour pressure ? How is it useful to determine the molar mass of a solute?
Answer:
Raoult’s law for volatile solute:
For a solution of volatile liquids,the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.

Raoult’s law for non-volatile solute :
The relative lowering of vapour pressure of dilute solution containing non-volatile solute is equal
to the mole fraction of solute.
Relative lowering of vapour pressure \(\frac{P_0-P_5}{P_0}\) = X_S (mole fraction of solute)
\(\frac{P_0-P_s^{\prime}}{P_0}\) = \(\frac{n_s}{n_0+n_s}\)
For very much dilute solutions n_s <<< …..n₀
∴ \(\frac{P_0-P_5}{P_0}=\frac{n_5}{n_0}=\frac{w}{m} \times \frac{M}{W}\)
W – Weight of solute
m = Molar mass of solute
w = Weight of solvent
M = Molar mass of solvent
Molar mass of solute m = \(\frac{w \times M}{W} \times \frac{P_0}{P_0-P_s}\)

Question 12.
Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer:
The solids which are^havfng~Thoderate conductivity between insulators and conductors are called semi conductors.

1. These have the conductivity range from 10^-6 to 10⁴ – Ohm^-1m^-1.
2. By doping process the conductivity of semi conductors increases. E.g. : Si, Ge, crystal.

Semi conductors are of two types. They are :
1. Intrinsic semi-conductors:
In case of semi-conductors, the gap between the valence band and conduction band is small. Therefore, some electrons may jump to conduction band and show some conductivity, electrical conductivity of semi-conductors increases with rise in “temperature”, since more electrons can jump to the conduction band. Substancesjike silicon and germanium show this type of behaviour and are called intrinsic semi-concuctors.

2. Extrinsic semi-conductors:
Their conductivity is due to the presence of impurities. They are formed by “doping”.

Doping : Conductivity of semi-conductors is too low to be of practical use. Their conductivity is increased by adding an appufiate amount of suitable impurity. This process is called “doping”. ‘
Doping can be done with an impurity which is electron rich of electron deficient. Extrinsic semi-conductors are of two types.

a) n-type semi-conductors:
It is obtained by adding trace amount of V group element (P, As, Sb) to pure Si or Ge by doping.
When P, As, Sb (or) Bi is added to Si or Ge some of the Si or Ge in the crystal are replaced by P or As atoms and four out of five electrons of P or As atom will be used for bonding with Si or Ge atoms while the fifth electron serve to conduct electricity.

b) p – type semi-conductors:
It is obtained by doping with impurity atoms containing less electrons i.e.. Ill group elements (B, Al, Ga or In).

When B or Al is added to pure Si or Ge some of the Si or Ge in the crystal are replaced by B or aZ atoms and four our of three electrons of B or AZ atom will be used for bonding with “Si” or Ge atoms while the fourth valence electron is missing is called electron hole (or) electron vacancy. This vacancy on an atom in the structure migrates from on atom to another. Hence it facilitates the electrical conductivity.

Question 13.
What are lyophilic and lyophobic sols ? Compare the two terms in terms of stability and reversibility.
Answer:
Lyophilic colloid:
The colloidal solution in which the dispersed phase has great affinity to the dispersion, medium is called a Lyophilic colloid or Lyophilic solution. Ex : Starch solution. The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid :
The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution. Ex : Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution. Gold’particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 14.
Out line the principles of metals by the following methods
(a) Zone refining
(b) Poling
Answer:
a) Zone refining :

1. Zone refining is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
2. A circular mobile heater is fixed at one end of a rod of impure metal.
3. The molten zone moves along with the heater moves forward the pure metal crystallises out of the melt and the impurities pass into the adjacent molten zone.
4. The above process is repeated several times and the heater is moved in the same direction from one end to the other end. At one end impurities get concentrated. This end is cut off.
5. This method is very useful for producing semiconductor grade metals’of very high purity. Eg : Ge, Si, B, Ga etc.

(b) Poling :
When the metals are having the metal oxides as impurities this method is employed. The impure metal is melted and is then covered by carbon powder. Then it is stirred with green wood poles. The reducing gases formed from the green wood and the carbon, reduce the oxides to metal. Eg : Cu & Sn metals are refined by this method.

Question 15.
Write IUPAC names of the following.
(a) [CU(NH₃)₄]SO₄
(b) K₃[Cr(C₂O₄)₃]
(c) [co(scN)₄^-2
(d) [pt cl₂ (NH₃)₂]
Answer:
(a) [CU(NH₃)₄]SO₄ – Tetra amine copper (11) sulphate
(b) K3[Cr(C₂O₄)₃] – Potassium trioxalato chromate (11)
(c) [CO(SCN)₄]^-2 – Tetra thiocyano cobalt (11) ion
(d) [Pt Cl₂ (NH₃)₂] – Diamine dichloro platinum (11)

Question 16.
Write the names of the monomers in the polymers.
(a) Bakelite
(b) Polyvinyl chloride
(c) polystyrene
(d) Teflon
Answer:
(a) Bakelite

(b) Polyvinyl chloride
Monomers : Vinyl chloide
Structure : CH₂ = CH – Cl

(d) Teflon:
Monomers : Tetrafluoro erthylene
Structure : CF₂ = CF₂

Question 17.
Define the following :
(a) Racemic mixture
(b) Enantioaers
Answer:
(a) Racemic mixture:
Equal portions of Enantiomers combined to form an optically inactive mixture. This mixture is called racemic mixture.

1. Here rotation due to one isomer will be exactly cancelled by the rotation of due to other isomer.
2. The process of conversion of enantiomer into a racemic mixture is called as racemisation.

(b) Enantiomers : The stereo isomers related to each other as non-superimposable mirror images are called enantiomers.

1. These have identical physical properties like melting point, boiling point, refractive index etc.
2. They differ in rotation of plane polarised light.

Question 18.
Explain the following.
(a) Carbyl amine reaction
(b) Sand meyer reaction
Answer:
(a) Carbyl amine reaction : Aniline reacts with chloroform in presence of alc.kOH to form phenyl iso cyanide. This reaction is called carbylamine reaction.

(b) Sandmeyer reaction : Formation of chloro benzene, Bromo benzene (or) cyano benzene from benzene diazonium salts with reagents Cu₂Cl₂/HCl, Cu₂Br₂/HBr, CuCN/KCN is called sandmeyer’s reaction.

Question 19.
(a) How is nitric acid manufactured by Ostwald’s process ?
(b) How does chlorine reacts with the following and write equations,
(i) H₂S
(ii) l₂
(iii) Ca(OH)₂
(iv) H₂0
Answer:
(a) Ostwald’s process ; Ammonia, mixed with air in 1:7 or 1:8, when passed over a hot platinum gauze catalyst, is oxidised to NO mostly (about 95%). The reaction is,

(b) i) H₂S : Cl₂ reacts with HsS and forms H_sS and ‘S’. Cl₂ + H₂S → 2 HCl + S

ii) I₂ : Cl₂ reacts with Iodine to form ICl.
I₂ + Cl₂ → 2 ICl(Equi molar)

iii) Ca(OH)₂ : Chlorine reacts with dry slaked lime and forms bleaching powder.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂0

iv) H₂O : Chlorine reacts with water and forms chlorine water.
Cl₂+ H₂O → HCl + HOCl
HOCl is unstable and decompose to give nascent oxygen.
HOCl → HaCl + (O)

Question 20.
(a) Give detailed account of the collision theory of reaction
rates of bimolecular gaseous reactions.
(b) State Faraday’s first law. A solution of CuSO₄ is electrolysed for 10 min with a current of 1.5 amp. What is the mass of copper deposited at the cathode?
Answer:
(a) The reaction molecules are assumed to be hard spheres.

1. The reaction is postulated to occur when molecules collide with each other.
2. The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).
3. For a bimolecular elementary reaction A + B → products
   Rate = Z_AB.e^-E<_a/RT Z_AB = collision frequency.
4. All collisions do not lead to product formation.
5. The collisions with sufficient kinetic energy (Threshold energy) are responsible for product formation. These are called as effective collisions.
6. To account for effective collisions a factor p called to probability factor or steric factor is introduced.
   Rate = P Z_AB.e^-Ea/RT
7. The proper orientation, of reactant molecular lead to bond formation where as improper orientation makes them back and no products are formed.

(b) The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of current passing through the electrolyte.
(Or)
The mass of the substance deposited at an electrode during the electrolysis of electrolyte is directly proportional to quantity of electricity passed through it.

m ∝ Q; m ∝ = c × t
m = ect:  m = \(\frac{\text { Ect }}{96,500}\)
e = electrochemical equivalent
t = time in seconds
c = Current in amperes
E = Chemical equivalent

Question 21.
Explain the following,
(a) Acylation
(b) Esterification
(c) Gatterman – koch reaction
(d) Decarboxylation.
Answer:
(a) Acylation :
Aniline reacts with acetyl chloride/acetic acid/acetic anhydride to form acetanilide.

(b) Esterification:
Alcohol reacts with carboxylic acid in presence of acid to form ester. This reaction is called esterification.

(c) Gatterman – koch reaction:
Benzene reacts with Co/HCl in presence of anhydrous AlCl₃ to form benzaldehyde.

(d) Decarboxylation : Sodium salt of carboxylic acid undergo reaction with NaOH + CaO and removes CO₂ from the compound. This reaction is called decarboxylation.