Category: Class 12

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

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

Time: 3 Hours
Maximum Marks: 70

Instructions:

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

Part – 1

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

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

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

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

Question 4.
The correct order of increasing oxidizing power in the series


Answer:

Hint: VO₂⁺ < Cr₂O₇^2- < MnO₄^– greater the oxidation state, higher is the oxidising power.

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

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

Hint: In diamond carbon forming fee. Carbon occupies comers and face centres and also occupying half of the tetrahedral voids.

Question 7.
In a homogeneous reaction
A → B + C + D, the initial pressure was P₀ and after time t it was P. Expression for rate constant in terms of P₀ , P and t will be ……..

Answer:

Solution:

Question 8.
The solubility of an aqueous solution of Mg(OH)₂ be x then its K_sp is
(a) 4x³
(b) 108x⁵
(c) 27x⁴
(d) 9x
Answer:
(a) 4x³
Solution:
Mg(OH)₂ Mg²⁺(x) + 2OH^– ((2x)²)
K_sp = 4x³

Question 9.
During electrolysis of molten sodium chloride, the time required to produce 0. lmol of chlorine gas using a current of 3 A is ……………..
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Hint: m = ZIt (mass of 1 mole of Cl₂ gas = 71)
Answer:
(b) 107.2 minutes

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

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

Question 11.


Answer:
(a) A- 2, B – 3, C – 4, D – 1

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

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

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

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

Part – II

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

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

Question 17.
Complete the following reactions:
(a) B(OH)₃ + NH₃ →
(b) Na₂B₄O₇ + H₂SO₄ + H₂O
Answer:

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

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

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

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

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

Question 23.
What happens when
Oxidation of acetoneoxime with trifluoroperoxy acetic acid.
Answer:
Oxidation of acetoneoxime with trifluoroperoxy acetic acid gives 2-nitropropane.

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

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

Part – III

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

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

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

During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The positively charged silver cations migrate towards the cathode and get discharged by gaining electrons and deposited on the cathode. Other metals such as copper, zinc etc.,can also be refined by this process in a similar manner.

Question 26.
How will you prepare phosphine and explain the purification of phosphine?
Phosphine’is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.

Phosphine is freed from phosphine dihydride (P2H4) by passing through a freezing mixture. The dihydride condenses while phosphine does not.
Phosphine can also prepared by the hydrolysis of metallic phosphides with water or dilute mineral acids.

Phosphine is prepared in pure form by heating phosphorous acid.

A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda S0lution.

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

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

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

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

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

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

Anisotropy :

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

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

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

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

Question 31.
1mole of Hlis allowed to react with t-butyl methylether. Identify the product and write down the mechanism of the reaction.
Answer:

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

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

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

Part – IV

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

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

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

(b)(i) Reaction between chlorine with cold NaoH

Chlorine reacts with cold NaOH to give sodium chloride and sodium hypochlorite.

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

Question 35.
(a) (i) Discuss the ortho and pyro silicates. (2)
(ii) Compare lanthanides and actinides. (3)
[OR]
(b) (i) Ni²⁺ is identified using alcoholic solution of dimethyl glyoxime. Write the structural formula for the rosy red precipitate of a complex formed in the reaction. (3)
Cu⁺, Zn²⁺, Sc³⁺, Ti⁴⁺ are colourless. Prove this statement. (2)
Answer:
Ortho silicates: The simplest silicates which contain discrete [SiO₄]^4- tetrahedral units are called ortho silicates or neso silicates.

Examples: Phenacite – Be₂SiO₄(Be²⁺ ions are tetrahedrally surrounded by O^2- ions), Olivine – (Fe/Mg)₂ SiO₄ (‘Fe²⁺ and Mg²⁺ cations are octahedrally surrounded by O

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

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

(ii)
Lanthanoids

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

Actinoids

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

[OR]

(b) (i) Ni²⁺ ions present in Nickel chloride solution is estimated accurately for forming an insoluble complex called [Ni(DMG)₂]
Nickel ion reacts with alcoholic solution of DMG in the presence of ammonical medium, to give rosy red precipitate of [Ni(DMG)₂] complex.

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

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

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

(OR)

(b) (i) Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of K_h for that reaction. (3)
(ii) Identify the Lewis acid and the Lewis base in the following reactions. (2)

Answer:
(a) (i) 1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water

NaOH_(aq) + CH₃COOH_(aq) ⇌ CH₃COONa_(aq) + H₂O

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

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

[OR]

(b) (i) 1. Coordination number: The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. In aqueous solution, CH₃COONa is completely dissociated as follows.
CH₃COONa_(aq), →  CH₃COO^–_(aq) + Na⁺_(aq)

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

4.

such cases, the solution is basic due to the hydrolysis and pH is greater than 7.

5. Relationship between equilibrium constant, hydrolysis constant and the dissociation constant of acid is derived as follows:

K_h value in terms of degree of hydrolysis (h) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwald’s dilution law K_h = h²C and \(\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{c}}\)

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

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

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

[OR]

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

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

4. The relationship between molar conductance and equivalent conductance is

(ii) 1. Formic acid contains both an aldehyde as well as an acid group. Hence, like other aldehydes, formic acid can easily be oxidised and therefore acts as a strong reducing agent.

2. Formic acid reduces Tollen’s reagent (ammonical silver nitrate solution) to metallic silver.

3. Formic acid reduces Fehling’s solution. It reduces blue coloured cupric ions to red coloured cuprous ions.

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

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

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

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

5. Structure of α – helix

[OR]

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

Thermoplastic :

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

Thermosetting

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

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

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

Instructions:

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

Time: 3 Hours
Maximum Marks: 90

Part – I

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

Question 1.
If | adj(adj A) | = |A|⁹, then the order of the square matrix A is _______.
(a) 3
(b) 4
(c) 2
(d) 5
Answer:
(b) 4

Question 2.
If |z₁| = 1, |z₂| = 2, |z₃| = 3 and |9z₁z₂ + 4z₁z₂ + z₂z₃| = 12, then the value of |z₁ + z₂+ z₃| is ________.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 3.
The value of \(\left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10}\) is ________.

Answer:
(a) cis \(\frac{2 \pi}{3}\)

Question 4.
If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})=u\), then cos 2u is equal to _____.
(a) tan² α
(b) 0
(c) -1
(d) tan 2α
Answer:
(c) -1

Question 5.
If \(\cot ^{-1} x=\frac{2 \pi}{5}\) for some x∈R, the value of tan^-1 x is _______.

Answer:
(c) \(\frac{\pi}{10}\)

Question 6.
The radius of the circle passing through the point (6, 2) two of whose diameter are x + y = 6 and x + 2y = 4 is ____.
(a) 10
(b) \(2 \sqrt{5}\)
(c) 6
(d) 4
Answer:
(b) \(2 \sqrt{5}\)

Question 7.
The length of the L.R. of x² = -4y is _______.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 8.
Distance from the origin to the plane 3x – 6y + 2z + 7 = 0 is ______.
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 9.
The distance from the origin to the plane \(\vec{r} \cdot(2 \vec{i}-\vec{j}+5 \vec{k})=7\) is _____.

Answer:
(a) \(\frac{7}{\sqrt{30}}\)

Question 10.
The number given by the Mean value theorem for the function \(\frac{1}{x}\), x ∈ [1, 9] is ______.
(a) 2
(b) 2.5
(c) 3
(d) 3.5
Answer:
(c) 3

Question 11.
f is a differentiable function defined on an interval I with positive derivative. Then f is ______.
(a) increasing on I
(b) decreasing on I
(c) strictly increasing on I
(d) strictly decreasing on I
Answer:
(c) strictly increasing on I

Question 12.
If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in our calculation of the volume is ________.
(a) 0.4 cu.cm
(b) 0.45 cu.cm
(c) 2 cu.cm
(d) 4.8 cu.cm
Answer:
(d) 4.8 cu.cm

Question 13.
If u(x, y) = \(e^{x^{2}+y^{2}}\), then \(\frac{\partial u}{\partial x}\) is equal to _______.
(a) \(e^{x^{2}+y^{2}}\)
(b) 2xu
(c) x²u
(d) y²u
Answer:
(b) 2xu

Question 14.
The value of \(\int_{0}^{\infty} e^{-3 x} x^{2} d x\) is _______.
(a) \(\frac{7}{27}\)
(b) \(\frac{5}{27}\)
(c) \(\frac{4}{27}\)
(d) \(\frac{2}{27}\)
Answer:
(d) \(\frac{2}{27}\)

Question 15.
\(\int_{0}^{a} f(x) d x\) is _____.

Answer:
(b) \(\int_{0}^{a} f(a-x) d x\)

Question 16.
The integrating factor of the differential equation \(\frac{d y}{d x}\) + P(x) y = Q (x) is x, then P(x) ______.
(a) x
(b) \(\frac{x^{2}}{2}\)
(c) \(\frac{1}{x}\)
(d) \(\frac{1}{x^{2}}\frac{1}{x^{2}}\)
Answer:
(c) \(\frac{1}{x}\)

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

Question 18 .
Which of the following is a discrete random variable?
I. The number of cars crossing a particular signal in a day.
II. The number of customers in a queue to buy train tickets at a moment.
III. The time taken to complete a telephone call.
(a) I and II
(b) II only
(c) III only
(d) II and III
Answer:
(a) I and II

Question 19.
If p is true and q is false then which of the following is not true?
(a) p → q is false
(b)p ∨ q is true
(c)p ∧ q is false
(d) p ↔ q is true
Answer:
(d) p ↔ q is true

Question 20.
The operation * defined by a*b = \(\frac{a b}{7}\) is not a binary operation on ______.
(a) Q⁺
(b) Z
(c) R
(c) C
Answer:
(b) Z

Part – II

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

Question 21.
Using elementary transformations find the inverse of the following matrix \(\left[\begin{array}{ll}
4 & 7 \\
3 & 0
\end{array}\right]\)
Answer:

Question 22.
If Z₁ = 1 – 3i, z₂ = -4i, and z₃ = 5, show that (z₁ + z₂) + z₃ = Z₁ + (z₂ + z₃)
Answer:
z₁ = 1 – 3i, z₂ = 4i, z₃ = 5
(z₁ + z₂) + z₃ = (1 – 3i – 4i) + 5(1 – 7i) + 5
= 6 – 7i …..(1)
z₁ + (z₂ + z₃) = (1 – 3i) + (-4i + 5)
= 6 – 7i …..(2)
from(1)& (2)we get
∴ (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)

Question 23.
Find a polynomial equation of minimum degree with rational coefficients, having 2+ \(\sqrt{3}\) i as
a root.
Answer:
Given roots is (2 + \(\sqrt{3}\) i)
∴ The other root is (2 – \(\sqrt{3}\) i), since the imaginary roots with real co-efficient occur as conjugate
pairs.
x² – x(S.O.R) + P.O.R = 0 ⇒ x² – x(4) + (4 + 3) = 0
x² – 4x + 7 = 0.

Question 24.
Evaluate: \(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+17 x+29}{x^{4}}\right)\)
Answer:
This is an indeterminate of the form (\(\frac{\infty}{\infty}\)). To evaluate this limit, we apply l’Hôpital Rule.

Question 25.
Let g(x, y) = 2y + x², x = 2r – s, y = r² + 2s, r, s ∈R. Find \(\frac{\partial g}{\partial r}, \frac{\partial g}{\partial s}\).
Answer:

Question 26.
Evaluate: \(\int_{0}^{\frac{\pi}{2}}\left(\sin ^{2} x+\cos ^{4} x\right) d x\)
Answer:

Question 27.
Find the differential equation corresponding to the family of curves represented by the equation y = Ae^8x + Be^-8x, where A and B are arbitrary constants.
Answer:
y = Ae^8x + Be^-8x Where A and B are arbitrary constants.
Differentiate with respect to ‘x’

Question 28.
If F(x) = \(\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right)\) – ∞ < x < ∞ is a distribution function of a continuous variable X, find P (0 ≤ x ≤ 1).
Answer:

Question 29.
Show that p → q and q → p are not equivalent.
Answer:
Truth table for p → q

Truth table for q → p

The entries in the column corresponding to p → q and q → p are not identical, hence they are not equivalent.

Question 30.
Show that the lines \(\frac{x-1}{4}=\frac{2-y}{6}=\frac{z-4}{12}\) and \(\frac{x-3}{-2}=\frac{y-3}{3}=\frac{5-z}{6}\) are parallel.
Answer:
We observe that the straight line \(\frac{x-1}{4}=\frac{2-y}{6}=\frac{z-4}{12}\) is parallel to the vector \(4 \hat{i}-6 \hat{j}+12 \hat{k}\) and the straight line \(\frac{x-3}{-2}=\frac{y-3}{3}=\frac{5-z}{6}\) is parallel to the vector \(-2 \hat{i}+3 \hat{j}-6 \hat{k}\).
Since \(4 \hat{i}-6 \hat{j}+12 \hat{k}=-2(-2 \hat{i}+3 \hat{j}-6 \hat{k})\) the two vectors are parallel, and hence the two straight lines are parallel.

Part – III

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

Question 31.
Using elementary transformations find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]\)

Question 32.
Find the square roots of – 15 – 8i

Question 33.
Find the sum of the squares of the roots of ax⁴ + bx³ + cx² + dx + e = 0, a ≠ 0

Question 34.
For what value of x, the inequality \(\frac{\pi}{2}\) < cos^-1 (3x – 1) < π holds?

Question 35.
Find the foot of the perpendicular drawn from the point (5, 4, 2) to the line \(\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}\). Also, find the equation of the perpendicular.

Question 36.
Evaluate \(\int_{0}^{\infty} \frac{x^{n}}{n^{x}} d x\), where n is a positive integer ≥ 2.

Question 37.
The engine of a motor boat moving at 10 m/s is shut off. Given that the retardation at any subsequent time (after shutting off the engine) equal to die velocity at that time. Find the velocity after 2 seconds of switching off the engine.

Question 38.
The probability that Mr.Q hits a target at any trial is \(\frac{1}{4}\). Suppose he tries at the target 10 times. Find the probability that he hits the target (i) exactly 4 times (ii) at least one time.

Question 39.
Consider the binary operation * defined on the set A = {a, b, c, d} by the following table:

Is it commutative and associative?

Question 40.
Evaluate the following limit, if necessary use l’Hopital Rule. \(\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}\)

Part – IV

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

Question 41.
(a) Find the inverse of A = \(\left[\begin{array}{lll}
2 & 1 & 1 \\
3 & 2 & 1 \\
2 & 1 & 2
\end{array}\right]\) by Gauss-Jordan method
[OR]
(b) Find the point of intersection of the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-4}{5}=\frac{y-1}{2}=z\).

Question 42.
(a) Suppose z₁, z₂ and z₃ are the vertices of an equilateral triangle inscribed in the circle.
|z| = 2. If z₁ = 1 + i\(\sqrt{3}\), then find z₂ and z₃.
[OR]
(b) If A = \(\left[\begin{array}{cc}
5 & 3 \\
-1 & -2
\end{array}\right]\) show that A² – 3A – 7I₂ = O₂. Hence Find A^-1.

Question 43.
(a) Solve the equation 3x³ – 16x² + 23x – 6 = 0 if the product of two roots is 1.
[OR]
(b) The mean and variance of a binomial variate X are respectively 2 and 1.5.
Find (i) P(X = 0) (ii) P(X = 1) (iii) P(X ≥ 1)

Question 44.
(a) Find the value of \(\cos \left(\sin ^{-1}\left(\frac{4}{5}\right)-\tan ^{-1}\left(\frac{3}{4}\right)\right)\)
[OR]
(b) Find, by integration, the volume of the solid generated by revolving about y-axis the region bounded between the curve y = \(\frac{3}{4} \sqrt{x^{2}-16}\), x ≥ 4, the y-axis, and the lines y = 1 and y = 6.

Question 45.
(a) A semielliptical archway over a one-way road has a height of 3m and a width of 12m. The truck has a width of 3m and a height of 2.7m. Will the truck clear the opening of the archway?
[OR]
(b) For the function f(x, y) = \(\frac{3 x}{y+\sin x}\) find the f_x, f_y, and show that f_xy = f_yx.

Question 46.
(a) Derive the equation of the plane in the intercept form.
[OR]
(b) Let A be Q\{1}. Define * on A by x * y = x +y – xy. Is * binary on A ? If so, examine the existence of identity, existence of inverse properties for the operation * on A .

Question 47.
(a) Find the angle between the rectangular hyperbola xy = 2 and the parabola x² + 4y = 0.
[OR]
(b) A pot of boiling water at 100° C is removed from a stove at time t = 0 and left to cool in the kitchen. After 5 minutes, the water temperature has decreased to 80° C, and another 5 minutes later it has dropped to 65° C. Determine the temperature of the kitchen.

Students can Download Tamil Nadu 12th Maths Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Instructions:

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

Time: 3 Hours
Maximum Marks: 90

Part – I

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

Question 1.
If A^T A^-1 is symmetric, then A² = _______
(a) A^-1
(b) (A^T)²
(c) A^T
(d) (A^-1)²
Answer:
(b) (AT)2

Question 2.
If p + iq = \(\frac{a+i b}{a-i b}\), then p² + q² = ________.
(a) 0
(b) 2
(c) 1
(d) -1
Answer:
(c) 1

Question 3.
If ω ≠ 1 is a cubic root of unity and \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -\omega^{2}-1 & \omega^{2} \\
1 & \omega^{2} & \omega^{7}
\end{array}\right|\) = 3k, then k is equal to _______.
(a) 1
(b) -l
(c) \(\sqrt{3} i\)
(d) \(-\sqrt{3} i\)
Answer:
(d) \(-\sqrt{3} i\)

Question 4.
The value of sin^-1 (cos x), 0 ≤ x ≤ π is _______.
(a) π – x
(b) x – \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{2} – x\)
(d) π – x
Answer:
(c) \(\frac{\pi}{2} – x\)

Question 5.
The radius of the circle 3x² + by² + 4bx – 6by + b² = 0 is ________.
(a) 1
(b) 3
(c) \(\sqrt{10}\)
(d) \(\sqrt{11}\)
Answer:
(c) \(\sqrt{10}\)

Question 6.
The equation of the directrix of the parabola y² = -8x is ______.
(a) y + 2 = 0
(b) x – 2 = 0
(c) y – 2 = 0
(d) x + 2 = 0
Answer:
(b) x – 2 = 0

Question 7.
If \(\vec{a}\) and \(\vec{b}\) are parallel vector, then \([\vec{a}, \vec{c}, \vec{b}]\) is equal to _____
(a) 2
(b) -1
(c) 1
(d) 0
Answer:
(d) 0

Question 8.
The length of the perpendicular from the origin to the plane \(\vec{r} \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})=26\) is _______.
(a) 26
(b) \(\frac{26}{169}\)
(c) 2
(d) \(\frac{1}{2}\)
Answer:
(c) 2

Question 9.
The curve y = ax⁴ + bx² with ab > 0
(a) has no horizontal tangent
(b) is concave up
(c) is concave down
(d) has no points of inflection
Answer:
(d) has no points of inflection

Question 10.
The asymptote to the curve y² (1 + x) = x² (1 – x) is _______.
(a) x = 1
(b) y = 1
(c) y = -1
(d) x = -1
Answer:
(d) x = -1

Question 11.
If f(x, y, z) = xy + yz + zx, then f_x – f_z is equal to ________.
(a) z – x
(b) y – z
(c) x – z
(d) y – x
Answer:
(a) z – x

Question 12.
If f(x, y) = e^xy , then \(\frac{\partial^{2} f}{\partial x \partial y}\) is equal to ________.
(a) xye^xy
(b) (1 + xy) e^xy
(c) (1 + y) e^xy
(d) (1 + x) e^xy
Answer:
(b) (1 + xy) e^xy

Question 13.
The value of \(\int_{0}^{\frac{\pi}{6}} \cos ^{3} 3 x d x\) is _______.
(a) \(\frac{2}{3}\)
(b) \(\frac{2}{9}\)
( c) \(\frac{1}{9}\)
(d) \(\frac{1}{3}\)
Answer:
(b) \(\frac{2}{9}\)

Question 14.
If f(x) is even then \(\int_{-a}^{a} f(x) d x \) _______.

Answer:
(b) \(2 \int_{0}^{a} f(x) d x\)

Question 15.
The order and degree of the differential equation \(\sqrt{\sin x}\)(dx + dy) = \(\sqrt{\sin x}\) (dx- dy) is ________.
(a) 1, 2
(b) 2, 2
(c) 1, 1
(d) 2, 1
Answer:
(c) 1, 1

Question 16.
The solution of the differential equation \(\frac{d y}{d x}\) = 2xy is _______.
(a) y = c e^x2
(b) y = 2x² + c
(c) = ce^-x2 + c
(d) y = x² + c
Answer:
(a) y = c e^x2

Question 17.
If P{X = 0} = 1 – P{X = 1}. If E[X] = 3Var(X), then P{X = 0} ________.
(a) \(\frac{2}{3}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{1}{3}\)
Answer:
(d) \(\frac{1}{3}\)

Question 18.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. Then the possible values of X are _________.
(a) i + 2n, i = 0, 1, 2 … n
(b) 2i – n, i = 0, 1, 2 … n
(c) n – i, i = 0, 1, 2 … n
(d) 2i + 2n, i = 0, 1, 2 … n
Answer:
(b) 2i – n, i = 0, 1, 2 … n

Question 19.
In the set Q define a Θ b= a + b + ab. For what value of y, 3 Θ (y Θ 5) = 7 ?

Answer:
(b) y = \(\frac{-2}{3}\)

Question 20.
If X is a continuous random variable then P(X > a) =
(a) P (X < a)
(b) 1 – P (X > a)
(c) P (X > a)
(d) 1 – P (x ≥ a)
Answer:
(c) P (X > a)

Part – II

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

Question 21.
Reduce the matrix \(\left[\begin{array}{ccc}
3 & -1 & 2 \\
-6 & 2 & 4 \\
-3 & 1 & 2
\end{array}\right]\) to a row-echelon form.
Answer:

Question 22.
Find the least positive integer n such that \(\left(\frac{1+i}{1-i}\right)^{n}=1\)
Answer:

Question 23.
Find the value of \(\sin ^{-1}\left(\sin \left(\frac{5 \pi}{4}\right)\right)\)
Answer:

Question 24.
Identify the type of conic section for the equation 3x² + 3y² – 4x + 3y + 10 = 0
Answer:
Comparing this equation with the general equation of the conic
Ax² + Bxy + cy² + Dx + Ey +F = 0
We get A = C also B = 0
So the given conic is a circle.

Question 25.
If U(x, y, z) = log(x³ + y³ + z³), find \(\frac{\partial \mathrm{U}}{\partial x}+\frac{\partial \mathrm{U}}{\partial y}\) and \(\frac{\partial U}{\partial z}\)
Answer:

Question 26.
Find, by integration, the volume of the solid generated by revolving about the x-axis, the region enclosed by y = 2x², y = 0 and x = 1.
Answer:

Question 27.
Solve the differential equation: \(\frac{d y}{d x}-x \sqrt{25-x^{2}}=0\)
Answer:

Question 28.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Answer:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3
Hence, the probabilities
P(X = 0) = P (No heads) = \(\frac{1}{8}\);
P(X = 1) = P (1 head) = \(\frac{3}{8}\);
P(X = 2) = P (2 heads) = \(\frac{3}{8}\);
P(X = 3) = P (3 heads)= \(\frac{1}{8}\);
∴ The probability mass function is
\(f(x)=\left\{\begin{array}{lll}
1 / 8 & \text { for } & x=0,3 \\
3 / 8 & \text { for } & x=1,2
\end{array}\right.\)

Question 29.
Construct the truth table for the following statements. \(\neg p \wedge \neg q\)
Answer:
Truth table for \(\neg p \wedge \neg q\)

Question 30.
Write the Maclaurin series expansion of the function: ex
Answer:
f (x) = e^x; f (0) = e⁰ = 1
f’ (x) = e^x; f’ (0) = 1
f”(x) = e^x; f”(0) = 1

Part – III

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

Question 31.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)show that A^-1 = \(\frac{1}{2}\) (A² – 3I).

Question 32.
Find the values of the real numbers x and y, if the complex numbers.
(3 – i)x – (2 – i) y + 2i + 5 and 2x + (-1 + 2i) y + 3 + 2i are equal.

Question 33.
It is known that the roots of the equation x³ – 6x² – 4x + 24 = 0 are in arithmetic progression. Find its roots.

Question 34.
Prove that: \(\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}\)

Question 35.
Find the equation of a circle of radius 5 whose centre lies on x-axis and which passes through the point (2, 3).

Question 36.
Using the l’ Hopital Rule prove that, \(\lim _{x \rightarrow 0^{+}}(1+x)^{\frac{1}{x}}=e\)

Question 37.
If v(x, y) = x² – xy + \(\frac{1}{4}\) y² + 7, x, y ∈ R, find the differential dv.

Question 38.
Find the area of the region bounded by 2x – y + 1 =0, y = – 1, y = 3 and y-axis..

Question 39.
Solve: \(\frac{d y}{d x}\) + 2y cot x = 3x² cosec²x

Question 40.
If the straight lines \(\frac{x-5}{5 m+2}=\frac{2-y}{5}=\frac{1-z}{-1}\) and x = \(\frac{2 y+1}{4 m}=\frac{1-z}{-3}\) are perpendicular to each other, find the value of m.

Part – IV

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

Question 41.
(a) Investigate the values of X and p the system of linear equations.
2x + 3y + 5z = 9, 7x + 3y – 5z = 8, 2x + 3y + λz = µ, have
(i) no solution (ii) a unique solution (iii) an infinite number of solutions.
[OR]
(b) If z(x, y) = x tan^-1 (xy), x = t², y = s e^t, s, t ∈ R, Find \(\frac{\partial z}{\partial t}\) and \(\frac{\partial z}{\partial t}\) at s = t = 1.

Question 42.
(a) Form the equation whose roots are the squares of the roots of the cubic equation
x³ + ax² + bx + c = 0.
[OR]
(b) Find the intervals of concavity and the points of inflection of the function.
f(θ) = sin 2θ in (0, π)

Question 43.
(a) If a = cos 2α + i sin 2α, b = cos 2β + i sin 2β and c = cos 2γ + i sin 2γ, prove that.

(b) A closed (cuboid) box with a square base is to have a volume of 2000 c.c. The material for the top and bottom of the box is to cost Rs. 3 per square cm and the material for the sides is to cost Rs. 1.50 per square cm. If the cost of the materials is to be the least, find the dimensions of the box.

Question 44.
(a) Prove that a straight line and parabola cannot intersect at more than two points.
[OR]
(b) Solve \(\left(y-e^{\sin ^{-1} x}\right) \frac{d x}{d y}+\sqrt{1-x^{2}}=0\)

Question 45.
(a) Solve \(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)
[OR]
(b) Show that the lines \(\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\) and \(\frac{x-4}{2}=\frac{y}{0}=\frac{z+1}{3}\) the point of intersection.

Question 46.
(a) A tank initially contains 50 liters of pure water. Starting at time t = 0 a brine containing with 2 grams of dissolved salt per litre flows into the tank at the rate of 3 liters per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time t > 0.
[OR]
(b) If X ~ B(n, p) such that 4P (X = 4) = P (x = 2) and n = 6 . Find the distribution, mean and standard deviation.

Question 47.
(a) Find the centre, foci, and eccentricity of the hyperbola 11x² – 25y² – 44x + 50y – 256 = 0
[OR]
(b) Verify (i) closure property (ii) commutative property (iii) associative property (iv) existence of identity and (v) existence of inverse for the operation +5 on Z5 using table corresponding to addition modulo 5.

Students can Download Tamil Nadu 12th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

General Instructions:

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

Time: 3 Hours
Max Marks: 70

Part – I

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

Question 1.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Question 2.
The work done in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is  ………….
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) Zero
(c)  \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}} \)
(d) infinite
Answer:
(b) Zero
Hint: The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in a electric field.

Question 3.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is……………………..
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Question 4.
An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor when the plates of the capacitor are kept under constant magnetic field of induction \(\overrightarrow{\mathrm{B}}\) is ……….


Answer:
(d)

Question 5.
In a series resonant RLC circuit, the voltage across 100Ω resistor is 40 V. The resonant frequency co is 250 rad/s. If the value of C is 4 pF, then the voltage across L is……………….
(a) 600 V
(b) 4000 V
(c) 400 V
(d) IV
Answer:
(c) 400 V

Question 6.
During the propagation of electromagnetic waves in a medium:
(a) electric energy density is double of the magnetic energy density
(b) electric energy density is half of the magnetic energy density
(c) electric energy density is equal to the magnetic energy density
(d) both electric and magnetic energy densities are zero
Answer:
(c) electric energy density is equal to the magnetic energy density

Question 7.
First diffraction minimum due to a single slit of width 1.0 x 10^-5 cm is at 30°. Then wavelength of light used is, …………………
(a) 400 Å
(b) 500 Å
(c) 600 Å
(d) 700 Å
Answer:
(b) 500 Å
Hint. For diffraction minima, d sin θ = nλ
\(\lambda=\frac{d \sin \theta}{n}=\frac{1 \times 10^{-5} \times 10^{-2} \times \sin 30^{\circ}}{1}=0.5 \times 10^{-7}\)
λ = 500 Å

Question 8.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light

Question 9.
Kinetic energy of emitted electron depends upon
(a) frequency
(b) intensity
(c) nature of atmosphere surrounding the electron
(d) none of these
Answer:
(a) frequency
Hint: Kinetic energy of emitted electron depends on the frequency of incident radiation.

Question 10.
The ratio between the first three orbits of hydrogen atom is…………….
(a) 1:2:3
(b) 2:4:6
(c) 1:4:9
(d) 1:3:5
Answer:
(c) 1:4:9
Hint :
E_n = \(\frac{-13.6 \times z^{2}}{n^{2}}\)
n = 1; E₁ =- 13.6 eV/ atom
n = 2; E₂ = – 3.4 eV/ atom
n = 3; E₃ = -1.51 eV/atom                             ’
The ratio of three orbits E₁ : E₂ : E₃ = 13.6 : 3.4 : 1.51 = 1 : 4 : 9

Question 11.
Bohr’s theory of hydrogen atom did not explain fully
(a) diameter of H-atom
(b) emission spectra
(c) ionisation energy
(d) the fine structure of even hydrogen spectrum
Answer:
(d) the fine structure of even hydrogen spectrum
Hint: Bohr theory could not explain the five structure of hydrogen spectrum.

Question 12.
If a half-wave rectified voltage is fed to a load resistor, which part of a cycle the load current will flow?
(a) 0° – 90°
(b) 90° – 180°
(c) 0° – 180°
(d) 0° – 360°
Answer:
(c) 0° – 180°

Question 13.
Diamond is very hard because ……………..
(a) it is covalent solid
(b) it has large cohesive energy
(c) high melting point
(d) insoluble in all solvents
Answer:
(b) it has large cohesive energy

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

Question 15.
The blue print for making ultra durable synthetic material is mimicked from
(a) Lotus leaf
(b) Morpho butterfly
(c) Parrot fish
(d) Peacock feather
Answer:
(c) Parrot fish

Part – II

Answer any six questions in which Q. No 17 is compulsory.   [6 x 2 = 12]

Question 16.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines  is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 17.
Determine the number of electrons flowing per second through a conductor, when a current of 32 A flows through it.
Answer:
I = 32A, t= 1 s
Charge of an electron, e = 1.6 x 10^-19 C
The number of electrons flowing per second, n =?

Question 18.
Define magnetic flux.
Answer:
The number of magnetic field lines crossing per unit area is called magnetic flux φ_B
\(\phi_{\mathrm{B}}=\overrightarrow{\mathrm{B}} \cdot\overrightarrow{\mathrm{A}}=\mathrm{B} \mathrm{A} \cos \theta=\mathrm{B} \perp \mathrm{A}\)

Question 19.
Give any one definition of power factor.
Answer:
The power factor is defined as the ratio of true power to the apparent power of an a.c. circuit.
It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit.

Question 20.
State the laws of reflection.
Answer:

• The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
• The angle of incidence i is equal to the angle of reflection r. i = r

Question 21.
Why do metals have a large number of free electrons?
Answer:
In metals, the electrons in the outer most shells are loosely bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving inside the metal in a random manner.

Question 22.
The radius of the 5th orbit of hydrogen atom is 13.25 A. Calculate the wavelength of the electron in the 5th orbit.
Answer:
2πr = nλ
2 x 3.14 x 13.25 Å = 5 x λ
.’. λ = 16.64 Å

Question 23.
Draw the output waveform of a full wave rectifier.
Answer:

Question 24.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Part – III

Answer any six questions in which Q.No. 28 is compulsory. [6 x 3 = 18]

Question 25.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge.
This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Question 26.
What is electric power and electric energy?
Answer:
Electric power: It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
\({ P }=\frac { { W } }{ t } ={ VI }={ I }^{ 2 }{ R }=\frac { { V }^{ 2 } }{ { R } } \)
Electric energy: It is the total workdone in maintaining an electric current in an electric circuit for a given time.
W = Vt = VIr joule = I²R? joule.

Question 27.
A bar magnet having a magnetic moment \(\overrightarrow{\mathrm{M}}\) is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Answer:
Consider a bar magnet of magnetic moment \(\overrightarrow{\mathrm{M}}\). When a bar magnet first cut in two pieces along the axis, their magnetic moment is \(\frac{\overrightarrow{\mathrm{M}}}{2}\)

Question 28.
The current in an inductive circuit is given by 0.3 sin (200t – 40°) A. Write the equation for the voltage across it if the inductance is 40 mH.
Answer:
L = 40 x 10^-3H; i = 0.1 sin (200 t – 40°); X_L = ωL = 200 x 40 x 10^-3 = 8 Ω
V_m = I_m X_L = 0.3 x 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90° Therefore,
υ = V_m sin (ωt + 90°)
υ = 2.4 sin (200f – 40°+ 90°)
υ = 2.4 sin (200f +50°)volt

Question 29.
Writedown the integral form of modified Ampere’s circuital law.
Answer:
This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.

Question 30.
Two light sources have intensity of light as I_o. What is the resultant intensity at a point where the two light waves have a phase difference of π/3?
Answer:
Let the intensities be I_o
The resultant intensity is, I = 4 I_o cos² (φ/2)
Resultant intensity when, ϕ = π/ 3, is I = 4I₀ cos² (π / 6) I
= 4I₀(√3 / 2)² =3I₀

Question 31.
Write the properties of cathode rays.
Answer:

• Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 10⁷m s^-1
• It can be deflected by application of electric and magnetic fields.
• When the cathode rays are allowed to fall on matter, they produce heat.
• They affect the photographic plates and also produce fluorescence when they fall on   certain crystals and minerals.
• When the cathode rays fall on a material of high atomic weight, x-rays are produced.
• Cathode rays ionize the gas through which they pass.
• The speed of cathode rays is up \(\left( \frac { 1 }{ 10 } \right) ^{ th }\) of the speed of light.

Question 32.
Distinguish between wireline and wireless communication.
Answer:

Question 33.
What is the difference between Nano materials and Bulk materials?
Answer:

• The solids are made up of particles. Each of the particle has a definite number of atoms, which might differ from material to material. If the particle of a solid is of size less than 100 nm, it is said to be a ‘nano solid’.
• When the particle size exceeds 100 nm, it is a ‘bulk solid’. It is to be noted that nano and bulk solids may be of the same chemical composition.
• For example, ZnO can be both in bulk and nano form.
• Though chemical composition is the; same, nano form of the material shows strikingly different properties when compared to its bulk counterpart.

Part – IV

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

Question 34.
(a) Calculate the electric field due to a dipole on its equatorial plane.
Answer:
Electric field due to an electric dipole at a point on the equatorial plane. Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure.

Since the point C is equi­distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of \(\overrightarrow{\mathrm{E}}_{+}\) is along BC \(\overrightarrow{\mathrm{E}}_{-} \) and the direction of E is along CA. \(\overrightarrow{\mathrm{E}}_{+} \) and \(\overrightarrow{\mathrm{E}}_{-} \) and E are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(\left|\overrightarrow{\mathrm{E}}_{+}\right| \sin \theta \) and \(\left|\overrightarrow{\mathrm{E}}_{-}\right| \sin \theta \) are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of E₊ and E and its direction is \(\overrightarrow { { E } } _{ + }\quad and\quad \overrightarrow { { E } } _{ – }\) and its direction is along \(-\hat{p}\)

(b) How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer: To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bf and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ₁ and ξ₂ and to be compared are connected to the terminals M₁, N₁ and M₂, N₂, of the DPDT switch. The positive terminals of Bf ξ₁ and ξ₂ should be connected to the same end C.

The DPDT switch is pressed towards M₁, N₁ so that cell ξ₁, is included in the secondary circuit and the balancing length l₁, is found by adjusting the jockey for zero deflection. Then the second cells ξ₂, is included in the circuit and the balancing length l₂, is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have
ξ₁ = Irl₁ ……………….. (1)
ξ₂ = Irl₂ ……………….. (2)
By dividing equation (1) by (2)
\(\frac{\xi_{1}}{\xi_{2}}=\frac{l_{1}}{l_{2}}\)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Question 35.
(a) Discuss the working of cyclotron in detail.
Answer:
Cyclotron : Cyclotron is a device used accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle : When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.

Construction : The particles are allowed to move in between two semicircular metal containers called Dees (hollow D – shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working: Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.

\(\frac { mv^{ 2 } }{ r } =qv{ B }\Rightarrow r=\frac { m }{ q{ B } } v\Rightarrow r\propto v\)
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the Very important condition in cyclotron operation is the resonance condition. It happens when the frequency { at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc From equation
\(f_{\text {osc }}=\frac{q \mathrm{B}}{2 \pi m} \Rightarrow \mathrm{T}=\frac{1}{f_{\text {osc }}}\)
The time Period of oscillation is \(\mathrm{T}=\frac{2 \pi m}{q \mathrm{B}}\)
The kinetic energy of the charged particle is \(\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{q^{2} \mathrm{B}^{2} r^{2}}{2 m}\)

Limitations of cyclotron

• the speed of the ion is limited
• electron cannot be accelerated
• uncharged particles cannot be accelerated

[OR]

(b) Give the uses of Foucault current.
Answer:
Though the production of eddy current is undesirable in some cases, it is useful in some other cases. A few of them are

(1) Induction stove : Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone. When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan. The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.

(2) Eddy current brake : This eddy current braking system is generally used in high speed trains and roller coasters. Strong electromagnets are fixed just above the rails. To stop the train, electromagnets are switched on. The magnetic field of these magnets induces eddy currents in the rails which oppose or resist the movement of the train. This is Eddy current linear brake.

In some cases, the circular disc, connected to the wheel of the train through a common shaft, is made to rotate in between the poles of an electromagnet. When there is a relative motion between the disc and the magnet, eddy currents are induced in the disc which stop the train. This is Eddy current circular brake.

(3) Eddy current testing : It is one of the simple non-destructive testing methods to find defects like surface cracks and air bubbles present in a specimen. A coil of insulated wire is given an alternating electric current so that it produces an alternating magnetic field. When this coil is brought near the test surface, eddy current is induced in the test surface. The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.

(4) Electro magnetic damping : The armature of the galvanometer coil is wound on a soft iron cylinder. Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder. The damping force due to the flow of eddy current brings the armature to rest immediately and then galvanometer shows a steady deflection. This is called electromagnetic damping.

Question 36.
(a) Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:

• Electromagnetic waves are produced by any accelerated charge.
• Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.
• Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.
• Electromagnetic waves travel with speed which is equal to the speed of light in vacuum
  or free space, \(c=\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } } =3\times 10^{ 8 }{ ms }^{ -1 }\)
• The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

[OR]

(b) Explain the Young’s double slit experimental setup and obtain the equation for path difference.
Answer:
Experimental setup
Wavefronts from S₁ and S₂ spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.

S₁ and S₂ travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.

The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.

Equation for path difference
Let d be the distance between the double slits S₁ and S₂ which act as coherent sources of wavelength λ. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S₁ and S₂ is C and the midpoint of the screen O is equidistant from S₁ and S₂. P is any point at a distance y  from O.

The waves from S₁ and S₂ meet at P either inphase or out-of-phase depending upon the path difference between the two waves. The path difference S between the light waves from S₁ and S₂ to the point P is, δ = S₂P and S₁P
A perpendicular is dropped from the point S₁, to the line S₂P at M to find the path difference more precisely.
δ = S₂P – MP = S₂M
The angular position of the point P from C is θ. ∠OCP = θ.
From the geometry, the angles ∠OCP and ∠S₂S₁ M are equal.
∠OCP = ∠S₂S₁ M = θ
In right angle triangle ΔS₁S₂M, the path difference, S₂M = d sin θ
δ = d sin θ
If the angle θ is small, sin θ ≈ tan θ ≈ θ From the right angle triangle ΔOCP, tan θ = y/D. The path difference, δ =dy/D

Question  37.
(a) Give the construction and working of photo emissive cell.
Answer:
Photo emissive cell:
Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.

• It consists of an evacuated glass or quartz bulb in which two metallic electrodes – that is, a cathode and an anode are fixed.
• The cathode C is semi-cylindrical in shape and is coated with a photo sensitive material. The anode A is a thin rod or wire kept along the axis of the semi-cylindrical cathode.
• A potential difference is applied between the anode and the cathode through a galvanometer G.

Working:

• When cathode is illuminated, electrons are emitted from it. These electrons are attracted by anode and hence a current is produced which is measured by the galvanometer
• For a given cathode, the magnitude of
  the current depends on the intensity to incident radiation and
  the potential difference between anode and cathode.

[OR]

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

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

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

(ii) Determination of specific charge
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
\(e \mathrm{V}=\frac{1}{2} m v^{2} \Rightarrow \frac{e}{m}=\frac{v^{2}}{2 \mathrm{V}}\)
Substituting the value of velocity from equation (1) , we get
\(\frac{e}{m}=\frac{1}{2 \mathrm{V}} \frac{\mathrm{E}^{2}}{\mathrm{B}^{2}}\) ………..(2)
Substituting the value of E ,B and V, the specific charge vam be determined as
\(\frac{e}{m}=1.7 \times 10^{11} \mathrm{Ckg}^{-1}\)

(iii) Deflection of charge only due to uniform electric field
When the magnetic field is turned off, the deflection is only due to electric field. The
deflection in vertical direction is due to the electric force.
F_e = eE ………………….. (3)
Let m be the mass of the electron and by applying Newton’s second law of motion,
acceleration of the electron is
\(a_{e}=\frac{1}{m} \mathrm{F}_{e}\) ………………… (4)
Substituting equation (4) in equation (3),
\(a_{e}=\frac{1}{m} e E=\frac{e}{m} E\)
Let y be the deviation produced from original position on the screen. Let the initial upward velocity of cathode ray be μ = 0 before entering the parallel electric plates. Let l be the time taken by the cathode rays to travel in electric field. Let l be the length of one of the plates, then the time taken is
\(t=\frac{l}{v}\) …………(5)

Hence, the deflection y’ of cathod rays is( note : u = 0 and \(a_{e}=\frac{e}{m} \mathrm{E}\))

Therefore, the deflection y on the screen is
y α y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get

Substituting the values on RHS, the value of specific charge is calculated as
e/m = 1.7 x 10¹¹Ckg^-1

Question 38.
(a) What is an LED? Give the principle of operation with a diagram.
Answer:
Light Emitting Diode (LED): LED is a p-n junction diode which emits visible or invisible light when it is forward biased. Since, electrical energy is converted into light energy, this process is also called electroluminescence. The cross-sectional view of a commercial LED is shown in figure (b). It consists of a p-layer, n-layer and a substrate. A transparent window is used to allow light to travel in the desired direction. An external resistance in series with the biasing source is required to limit the forward current through the LED. In addition, it has two leads; anode and cathode.

When the p-n junction is forward biased, the conduction band electrons on n-side and valence band holes on p-side diffuse across the junction. When they cross the junction, they become excess minority carriers (electrons in p-side and holes in n-side). These excess minority carriers recombine with oppositely charged majority carriers in the respective regions, i.e. the. electrons in the conduction band recombine with holes in the valence band as shown in the figure (c). During recombination process, energy is released in the form of light (radiative) or heat (non-radiative). For radiative recombination, a photon of energy hv is emitted. For non- radiative recombination, energy is liberated in the form of heat.

The colour of the light is determined by the energy band gap of the material. Therefore, LEDs are available in a wide range of colours such as blue (SiC), green (AlGaP) and red (GaAsP). Now a days, LED which emits white light (GalnN) is also available.

[OR]

(b) Give the applications of ICT in mining and agriculture sectors.
Answer:
(i) Agriculture
The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the Challenges and risk factors.

• ICT is widely used in increasing food productivity and farm management.
• It helps to optimize the use of water, seeds and fertilizers etc.
• Sophisticated technologies that include robots, temperature and moisture sensors, aerial images, and GPS technology can be used.
• Geographic information systems are extensively used in farming to decide the suitable place for the species to be planted.

(ii) Mining

• ICT in mining improves operational efficiency, remote monitoring and disaster locating system.
• Information and communication technology provides audio-visual warning to the trapped underground miners.
• It helps to connect remote sites.

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

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

Instructions:

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

Time: 3 Hours
Maximum Marks: 90

Part – I

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

Question 1.
If A is a 3 × 3 non-singular matrix such that AA^T = A^T A and B = A^-1 A^T, then BB^T = ________.
(a) A
(b) B
(c) I₃
(d) B^T
Answer:
(c) I₃

Question 2.
The rank of the matrix \(\left[\begin{array}{cc}
7 & -1 \\
2 & 1
\end{array}\right]\) is ________.
(a) 9
(b) 2
(c) 1
(d) 5
Answer:
(b) 2

Question 3.
The value of \(\sum_{i=1}^{13}\left(i^{n}+i^{n-1}\right)\) is ________.
(a) 1 + i
(b) i
(c) 1
(d) 0
Answer:
(a) 1 + i

Question 4.
Which of the following is incorrect?
(d) Re(z) ≤ |z|
(b) Im (z) ≤ |z|
(c) \(z \bar{z}=|z|^{2}\)
(d) Re(z) ≥ |z|
Answer:
(d) Re(z) ≥ |z|

Question 5.
According to the rational root theorem, which number is not possible rational zero of 4x⁷ + 2x⁴ – 10x³ – 5?
(a) -1
(b) \(\frac{5}{4}\)
(c) \(\frac{4}{5}\)
(d)5
Answer:
(c) \(\frac{4}{5}\)

Question 6.
If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})=u\), then cos 2u is equal to ______.
(a) tan² α
(b) o
(c) -1
(d) tan 2α
Answer:
(c) -1

Question 7.
The domain of the function defined by f(x) = sin^-1 \(\sqrt{x-1}\) is _______.
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Answer:
(a) [1, 2]

Question 8.
The area of quadrilateral formed with foci of the hyperbolas \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\) is ________.
(a) 4(a² + b²)
(b) 2(a² + b²)
(c) a² + b²
(d) \(\frac{1}{2}\) (a² + b²)
Answer:
(b) 2(a² + b²)

Question 9.
The directrix of the parabola x² = -4y is ________.
(a) x = 1
(b) x = 0
(c) y = 1
(d) y = 0
Answer:
(c) y = 1

Question 10.
If the line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane x + 3y – αz + b = β, then (α, β) is ______
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Answer:
(b) (-6, 7)

Question 11.
If \(\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \times \vec{b}) \times \vec{c}\) for non-coplanar vectors \(\vec{a}, \vec{b}, \vec{c}\) then ________.
(a) \(\vec{a}\) parallel to \(\vec{b}\)
(b) \(\vec{b}\) parallel to \(\vec{c}\)
(c) \(\vec{c}\) parallel to \(\vec{a}\)
(d) \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)
Answer:
(c) \(\vec{c}\) parallel to \(\vec{a}\)

Question 12.
The maximum product of two positive numbers, when their sum of the squares is 200, is ________.
(a) 100
(b) \(25 \sqrt{7}\)
(c) 28
(d) \(24 \sqrt{14}\)
Answer:
(a) 100

Question 13.
If w(x,y, z) = x² (y – z) + y² (z – x) + z² (x – y), then \(\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}+\frac{\partial w}{\partial z}\) is ________.
(a) xy + yz + zx
(b) x (y + z)
(c) y (z + x)
(d) 0
Answer:
(d) 0

Question 14.
If u (x, y) = x² + 3xy + y – 2019, then \(\left.\frac{\partial u}{\partial x}\right|_{(4,-5)}\) is equal to ______.
(a) -4
(b) -3
(c) -7
(d) 13
Answer:
(c) -7

Question 15.
The volume of solid of revolution of the region bounded by y² = x (a – x) about x-axis is ________.

Answer:
(d) \(\frac{\pi a^{3}}{6}\)

Question 16.
\(\int_{0}^{a} f(x) d x+\int_{0}^{a} f(2 a-x) d x\) = _________.

Answer:
(c) \(\int_{0}^{2 a} f(x) d x\)

Question 17.
The differential equation of the family of curves y = Ae^x + Be^-x, where A and B are arbitrary constants is _______.

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

Question 18.
The differential equation corresponding to xy = c² where c is an arbitrary constant, is ______.
(a) xy”+ x = 0
(b) y” = 0
(c) xy’ + y = 0
(d) xy”- x = 0
Answer:
(c) xy’ + y = 0

Question 19.
If \(f(x)=\left\{\begin{array}{ll}
2 x, & 0 \leq x \leq a \\
0 & , \text { otherwise }
\end{array}\right.\)
is a probability density function of a random variable, then the value of a is _________.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 20.
The proposition \(p \wedge(\neg p \vee q)\) is
(a) a tautology
(b) a contradiction
(c) logically equivalent to p ∧ q
(d) logically equivalent to p ∨ q
Answer:
(c) logically equivalent to p ∧ q

Part – II

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

Question 21.
A 12 metre tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away.
Answer:
Let the two parts be x and (12 – x)
Given that x = \(\sqrt[3]{12-x}\)
Cubing on both side, x³ = 12 – x
x³ + x – 12 = 0

Question 22.
Find the value of \(\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} \sin \frac{\pi}{9}\right)\)
Answer:

Question 23.
Obtain the equation of the circles with radius 5 cm and touching x-axis at the origin in general form.
Answer:

Given radius = 5 cm and the circle is touching x axis
So centre will be (0, ± 5) and radius = 5
The equation of the circle with centre (0, ± 5) and radius 5 units is
(x – 0)² + (y ± 5)² = 5²
(i.e) x² + y² ± 10y + 25 – 25 = 0
(i.e) x² + y² ± 10y = 0

Question 24.
Find the length of the perpendicular from the origin to the plane.
\(\bar{r} \cdot(3 \vec{i}+4 \bar{j}+12 \vec{k})=26\).
Answer:
Taking the equation of the plane in cartesian form we get,
\((x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})=26\)
i.e. 3x + 4y + 12z – 26 = 0
The length of the perpendicular from (0, 0, 0) to the above plane is

Question 25.
Evaluate \(\lim _{x \rightarrow \pi / 2} \frac{\log (\sin x)}{(\pi-2 x)^{2}}\)
Answer:

Note that here l’ Hospitals rule, applied twice yields the result.

Question 26.
Evaluate \(\int_{-1}^{1} e^{-\lambda x}\left(1-x^{2}\right) d x\)
Answer:
Taking u = 1 – x² and v= e^-λx, and applying the Bernoulli’s formula, we get

Question 27.
Solve: \(\frac{d y}{d x}+2 y=e^{-x}\)
Answer:
Given that \(\frac{d y}{d x}+2 y=e^{-x}\)
This is a linear differential equation.
Here P = 2; Q = e
∫P dx = ∫2 dx = 2x
Thus, I.F = e^∫Pdx = e^2x
Hence the solution of (1) is \(y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c\)
That is, ye^2x = ∫e^-x e^2x dx + c (or) ye^2x = e^x + c (or) y = e^-x + ce^-2x is the required solution.

Question 28.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Answer:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3
Hence, the probabilities
P(X = 0) = P (No heads) = \(\frac{1}{8}\);
P(X = 1) = P (1 head) = \(\frac{3}{8}\);
P(X = 2) = P (2 heads) = \(\frac{3}{8}\);
P(X = 3) = P (3 heads)= \(\frac{1}{8}\);
∴ The probability mass function is
\(f(x)=\left\{\begin{array}{lll}
1 / 8 & \text { for } & x=0,3 \\
3 / 8 & \text { for } & x=1,2
\end{array}\right.\)

Question 29.
Construct the truth table for \((p \vee q) \vee \neg q\)
Answer:
truth table for \((p \vee q) \vee \neg q\)

Question 30.
Show that f(x, y) = \(\frac{x^{2}-y^{2}}{y^{2}+1}\) is continuous at every (x, y) ∈ R².
Answer:

Here, f satisfies all the three conditions of continuity at (a, b). Hence, f is continuous at every point of R² as (a, b) ∈ R².

Part – III

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

Question 31.
Form a polynomial equation with integer coefficients with \(\sqrt{\frac{\sqrt{2}}{\sqrt{3}}}\) as a root.

Question 32.
Find the equation of the tangents from the point (2, -3) to the parabola y² = 4x

Question 33.
Find the vector and cartesian equations of the straight line passing through the points (-5, 2, 3) and (4,-3, 6).

Question 34.
Find the points on the curve y = x³ – 6x² + x + 3 where the normal is parallel to the line x + y =1729.

Question 35.
Assuming log₁₀ e = 0.4343, find an approximate value of log₁₀ 1003.

Question 36.
Evaluate: \(\int_{0}^{\frac{\pi}{2}} \sin ^{2} x \cos ^{4} x d x\)

Question 37.
Solve the differential equation: \(\frac{d y}{d x}\) = e^x+y + x³ e^y

Question 38.
Using binomial distribution find the mean and variance of X for the following experiments
(i) A fair coin is tossed 100 times, and X denote the number of heads.
(ii) A fair die is tossed 240 times, and X denote the number of times that four appeared.

Question 39.
Let

be any three Boolean matrices of the same type. Find (A ∨ B) ∧C

Question 40.
Sketch the graph of y = sin\(\left(\frac{1}{3} x\right)\) for 0 ≤ x < 6π.

Part – IV

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

Question 41.
(a) A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself ? (Use Cramer’s rule to solve the problem).
[OR]
(b) Using elementary transformations find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)

Question 42.
42. (a) Let z₁, z₂, and z₃ be complex numbers such that |z₁| = |z₂| = |z₃| = r > 0 and z₁ +z₂ + z₃ ≠ 0.
Prove that \(\left|\frac{z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}}{z_{1}+z_{2}+z_{3}}\right|=r\)
[OR]
(b) Find all cube roots of \(\sqrt{3}\) + i.

Question 43.
(a) Find all zeros of the polynomial x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135, if it is known that 1 + 2i and \(\sqrt{3}\) are two of its zeros.
[OR]
(b) Let W(x, y, z) = x² – xy + 3 sin z, x, y, z∈R. Find the linear approximation at (2, -1, 0)

Question 44.
(a) Prove p → (q → r) ≡ (p ∧ q) → r without using truth table.
[OR]
(b) Evaluate \(\int_{0}^{\frac{\pi}{4}} \frac{1}{\sin x+\cos x} d x\)

Question 45.
(a) Prove that \(\text { (i) } \tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2} \text { (ii) } \sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{16}{65}\)
[OR]
(b) Find two positive numbers whose product is 100 and whose sum is minimum.

Question 46.
(a) If X is the random variable with distribution function F (x) given by,
\(\mathrm{F}(x)=\left\{\begin{array}{ll}
0, & x<0 \\
\frac{1}{2}\left(x^{2}+x\right) & 0 \leq x<1 \\
1, & x \geq 1
\end{array}\right.\)
then find (i) the probability density function f(x) (ii) P(0.3 ≤ X ≤ 0.6)
[OR]
(b) Find the vertex, focus, equation of directrix and length of the latus rectum of the following: y² – 4y – 8x + 12 = 0

Question 47.
(a) The rate at which the population of a city increases at any time is proportional to the population at that time. If there were 1,30,000 people in the city in 1960 and 1,60,000 in 1990, what population may be anticipated in 2020? [log_e \(\left(\frac{16}{3}\right)\) = 0.2070; e^-0.42 = 1.52]
[OR]
(b) Find the parametric vector, non-parametric vector and Cartesian form of the equation of the plane passing through the point (3, 6, -2), (-1, -2, 6) and (6, 4, -2).

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

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

General Instructions:

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

Time: 3 Hours
Max Marks: 70

Part – I

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

Question 1.
If voltage applied on a capacitor is increased from V to 2V, Choose the correct conclusion,
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Question 2.
The electric field in the region between two concentric charged spherical shells
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Question 3.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be …………
(a) R
(b) 2R
(c) \(\frac{\mathbf{R}}{4}\)
(d) \(\frac{\mathbf{R}}{2}\)
Answer:
(c) \(\frac{\mathbf{R}}{4}\)

Question 4.
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(b) 45°

Question 5.
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is
(a) 1.2
(b) 0.83
(c) 0.12
(d) 0.9
Answer:
(b) 0.83

Question 6.
Faraday’s law of electromagnetic induction is related to the …………….
(a) Law of conservation of charge
(b) Law of conservation of energy
(c) Third law of motion
(d) Law of conservation of angular momentum
Answer:
(b) Law of conservation of energy

Question 7.
The dimension of \(\frac{1}{\mu_{0} \varepsilon_{0}}\) is ……….
(a) [L T^-1]
(b) [L² T^-2]
(c) [L^-1 T]
(d) [L^-2 T²]
Answer:
(b) [L² T^-2]

Question 8.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,…………..
(a) 2D
(b) \(\frac{\mathrm{D}}{2}\)
(c) \(\sqrt{2} \mathrm{D}\)
(d) \(\frac{\mathrm{D}}{\sqrt{2}}\)
Answer:
(a) 2D

Question 9.
The speed of light in an isotropic medium depends on,
(a) its intensity
(b) its wavelength
(c) the nature of propagation
(d) the motion of the source w.r.to medium
Answer:
(b) its wavelength

Question 10.
The mean wavelength of light from sun is taken to be 550 nm and its mean power is 3.8 x 10²⁶ W. The number of photons received by the human eye per second on the average from sunlight is of the order of ……………..
(a) 10⁴⁵
(b) 10⁴²
(c) 10⁵⁴
(d) 10⁵¹
Answer:
(a) 10⁴⁵
Hint:

Question 11.
As the intensity of incident light increases
(a) kinetic energy of emitted photoelectrons increases
(b) photoelectric current decreases
(c) photoelectric current increases
(d) kinetic energy of emitted photoelectrons decreases
Answer:
(c) photoelectric current increases
Hint: As the intensity of incident light increases, photoelectric current increases.

Question 12.
In J.J. Thomson e/m experiment, a beam of electron is replaced by that of muons (particle with same charge as that of electrons but mass 208 times that of electrons). No deflection condition is achieved only if
(a) B is increased by 208 times
(b) B is decreased by 208 times
(c) B is increased by 14.4 times
(d) B is decreased by 14.4 times
Answer:
(c) B is increased by 14.4 times
Hint: In the condition of no deflection \(\frac{e}{m}=\frac{\mathrm{E}^{2}}{2 v \mathrm{B}^{2}}\)
If m is increased by 208 times then B should be increased \(\sqrt{208}\) = 14.4 times

Question 13.
A forward biased diode is treated as
(a) An open switch with infinite resistance
(b) A closed switch with a voltage drop of 0V
(c) A closed switch in series with a battery voltage of 0.7V
(d) A closed switch in series with a small resistance and a battery.
Answer:
(d) A closed switch in series with a small resistance and a battery.

Question 14.
The variation of frequency of carrier wave with respect to the amplitude of the modulating
signal is called …………
(a) Amplitude modulation
(b) Frequency modulation
(c) Phase modulation
(d) Pulse width modulation
Answer:
(b) Frequency modulation

Question 15.
Who is the father of the modem robotics industry formed the world’s first robotic company in 1956?
(a) Joliot
(b) Cormark
(c) Engelberger
(d) Edward purcell
Answer:
(c) Engelberger

Part – II

Answer any six questions in which Q. No 17 is compulsory.

Question 16.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from  – q to +q. The SI unit of dipole moment is coulomb meter (cm).
\(\vec{p}=q a \hat{i}-q a(-\hat{i})=2 q a \hat{i}\)

Question 17.
Calculate the electric flux through the rectangle of sides 5 cm and 10 cm kept in the region of a uniform electric field 100 NC^-1. The angle 0 is 60°. Suppose 0 becomes zero, what is the electric flux?
Answer:

Question 18.
What is meant by hysteresis?
Answer:
The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. Hysteresis means Tagging behind’.

Question 19.
What is meant by wattles current?
Answer:
The component of current (I_pMS sin φ), which has a phase angle of \(\frac{\pi}{2}\) with the voltage is called reactive component. The power consumed is zero. So that it is also known as ‘Wattless’ current.

Question 20.
What is Snell’s window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle i_c. The restricted illuminated circular area is called Snell’s window.

Question 21.
Define work function of a metal. Give its unit.
Answer:
The minimum energy needed for an electron to escape from the metal surface is called work function of that metal. It’s unit is electron volt (eV).

Question 22.
What is meant by radioactivity?
Answer:
The phenomenon of spontaneous emission of highly penetrating radiations such as α, β and γ rays by an element is called radioactivity.

Question 23.
What is the angular momentum of an electron in the third orbit of an atom?
Answer:
Here n = 3; h = 6.6 x 10^-34 Js
Angular momentum.

Question 24.
Explain the current flow in a NPN transistor
Answer:

• The conventional flow of current is based on the direction of the motion of holes
• In NPN transistor, current enters from the base into the emitter.

Part – III

Answer any six questions in which Q.No. 27 is compulsory. [6×3 = 18]

Question 25.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
\(C=\frac{Q}{V} \text { or } Q \propto V\)
The SI unit of capacitance is coulomb per volt or farad (F).

Question 26.
Distinguish between drift velocity and mobility
Answer:

Question 27.
A coil of a tangent galvanometer of diametre 0.24 m has 100 turns. If the horizontal component of Earth’s magnetic field is 25 x 10^-6 T then, calculate the current which gives a deflection of 60°.
Answer:
The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T

Question 28.
Mention the ways of producing induced emf.
Answer:
Induced emf can be produced by changing magnetic flux in any of the following ways:

• By changing the magnetic field B
• By changing the area A of the coil and
• By changing the relative orientation 0 of the coil with magnetic field

Question 29.
A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.
Answer:
L = 400 x 10^-3 H; I_eff = 6 x 10^-3A; f= 1000 Hz
Inductive reactance, X_L = Lω = L x 2πf = 2 x 3.14 x 1000 x 0.4 = 2512Ω
Voltage across L, V – IX_L = 6 x 10^-3 x 2512 =15.072 V (RMS)

Question 30.
What are the sign conventions followed for lenses?
Answer:
The sign conventions for thin lenses differ only in the signs followed for focal lengths.
(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).

(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.

Question 31.
Write down the draw backs of Bohr atom’model.
Answer:
Limitations of Bohr atom model
The following are the drawbacks of Bohr atom model

• Bohr atom model is valid only for hydrogen atom or hydrogen like-atoms but not for
  complex atoms.
• When the spectral lines are closely examined, individual lines of hydrogen spectrum is
  accompanied by a number of faint lines. These are often called fine structure. This is not explained by Bohr atom model.
• Bohr atom model fails to explain the intensity variations in the spectral lines.
• The distribution of electrons in atoms is not completely explained by Bohr atom model.

Question 32.
What do you mean by leakage current in a diode?
Answer:
The leakage current in a diode is the current that the diode will leak when a reverse voltage is applied to it. Under the reverse bias, a very small current in μA, flows across the junction. This is due to the flow of the minority charge carriers called the leakage current or reverse saturation current.

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

Part – IV

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

Question 34.
(a) Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field: Consider an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\) whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec{E}\)  in the direction of the field and charge -q will experience a force -q \(\vec{E}\) in a direction opposite to the field. Since the external field \(\vec{E}\) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O.

Using right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.
The magnitude of the total torque


where θ is the angle made by \(\vec{p} \text { with } \overrightarrow{\mathrm{E}}\) .
Since p = 2aq, the torque is written in terms of the vector product as \(\vec{\tau}=\vec{p} \times \vec{E}\)
The magnitude of this torque is τ = pE sin θ and is maximum
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field [/latex]\overrightarrow{\mathrm{E}}[/latex] . Once \(\overrightarrow{\mathrm{p}}\) is aligned with \(\overrightarrow{\mathrm{E}}\), the total torque on the dipole becomes zero.

[OR]

(b) Explain the equivalent resistance of a parallel resistor network.
Answer:
Resistors in parallel: Resistors are in parallel when they are connected across the same potential difference as shown in fig. (a).
In this case, the total current I that leaves the battery in split into three separate paths. Let I₁, I₂ and I₃ be the current through the resistors R_{1 ,}R₂ and R₃ respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I_{1 +} I₂ + I₃ ………………… (1)

Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have

Substituting these values in equation (1) we get.

Here R_p is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).
Note: The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Question 35.
(a) Obtain a relation for the magnetic induction at a point along the axis of a circular coil ‘ carrying current.
Answer:
Magnetic field produced along the axis of the current carrying circular coil: Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O. It is computed by taking two diametrically opposite line elements of the coil each of length \(d \bar{l} \) at C and D Let r be the vector joining the current element (I \(d \bar{l}\)) at C to the point P.
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
angle ∠CPO = ∠DPO = θ
According to Biot-Savart’s law, the magnetic field at P due to the current element \(d \bar{l} \) is


The magnitude of magnetic field due to current element I \(d \bar{l} \) at C and D are equal because of equal distance from the coil. The magnetic field dB due to each current element I dl is resolved into two components; dB sin θ along y-direction and dB cos θ along the z-direction. Horizontal components of each current element cancels out while the vertical components (dB cos θ \(\hat{k}\) ) alone contribute to total magnetic field at the point P.
If we integrate \(d \bar{l} \) around the loop, \(d \bar{B} \) sweeps out a cone, then the net magnetic field \(\overrightarrow{\mathrm{B}}\) at point P is

Using Pythagorous theorem r² = R² + Z² and integrating line element from 0 to 2πR, we get

Note that the magnetic field \(\bar{B} \) points along the direction from the point O to P. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point P to O.

[OR]

(b) Explain the working of a single-phase AC generator with necessary diagram.
Answer:
Single phase AC generator: In a single phase AC generator, the armature conductors are connected in series so as to form a single circuit which generates a single-phase alternating emf and hence it is called single-phase alternator.

The simplified version of a AC generator is discussed hire. Consider a stator core consisting of 2 slots in which 2 armature conductors PQ and RS are mounted to form single-turn rectangular loop PQRS. Rotor has 2 salient poles with field windings which can be magnetized by means of DC source.

Working: The loop PQRS is stationary and is perpendicular to the plane of the paper. When field windings are excited, magnetic field is produced around it. The direction of magnetic field passing through the armature core. Let the field magnet be rotated in clockwise direction by the prime mover. The axis of rotation is perpendicular to the plane of the paper.

Assume that initial position of the field magnet is horizontal. At that instant, the direction of magnetic field is perpendicular to the plane of the loop PQRS. The induced emf is zero. This is represented by origin O in the graph between induced emf and time angle.

When field magnet rotates through 90°, magnetic field becomes parallel to PQRS. The induced emfs across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Fleming’s right hand rule.

Care has to be taken while applying this rule; the thumb indicates the direction of the motion of the conductor with respect to field. For clockwise rotating poles, the conductor appears to be rotating anti-clockwise. Flence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper. For PQ, it is downwards and for RS upwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.

For the rotation of 180° from the initial position, the field is again perpendicular to PQRS and the induced emf becomes zero. This is represented by point B. The field magnet becomes again parallel to PQRS for 270° rotation of field magnet. The induced emf is maximum but the direction is reversed. Thus the current flows along SRQP. This is represented by point C.

On completion of 360°, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature.

Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle. For this construction, the frequency of the induced emf depends on the speed at which the field magnet rotates.

Question 36.
(a) What is emission spectra? Give their types.
Answer:
Emission spectra: When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum _ can be divided into three types:

(i) Continuous emission spectra (or continuous spectra): If the light from incandescent lamp filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours. Thus, it consists of wavelengths containing all the visible colours ranging from violet to red. Examples: spectrum obtained from carbon arc, incandescent solids, liquids gives continuous spectra.

(ii) Line emission spectrum (or line spectrum): Suppose light from hot gas is allowed to pass through prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra are sharp lines of definite wavelengths or frequencies. Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element which means it is different for different elements. Examples: spectra of atomic hydrogen, helium, etc.

(iii) Band emission spectrum (or band spectrum): Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces, known as band spectra. This spectrum has a sharp edge at one end and fades out at the other end. Such spectra arise when the molecules are excited. Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube etc.

[Or]

(b) Describe the Fizeau’s method to determine speed of light.
Answer:
Fizeau’s method to determine speed of light:
Apparatus: The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism.

The light passing through one cut in the wheel will, get reflected by a mirror M kept at a long distance d, about 8 km from the toothed wheel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.

Working: The angular speed c rotation of the toothed wheel was increased from zero to a value co until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.

Expression for speed of light: The speed of light in air v is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back 2d to the time taken t.
\(v=\frac{2 d}{t}\) ………………… (1)

The distance d is a known value from the arrangement. The time taken t for the light to travel the distance to and fro is calculated from the angular speed ω of the toothed wheel. The angular speed ω of the toothed wheel when the light disappeared for the first time is,
\(\omega=\frac{\theta}{t}\) ………………….. (2)

Here, θ is the angle between the tooth and the slot which is rotated by the toothed wheel within that time t

Rewriting the above equation for t
\(t=\frac{\pi}{\mathrm{N} \omega}\) ……………….. (3)
Substituting t from equation (3) in equation (1)
\(v=\frac{2 d}{\pi / \mathrm{N} \omega}\)
\(v=\frac{2 d \mathrm{N} \omega}{\pi}\) …………….(4)
Fizeau had some difficulty to visually estimate the minimum intensity blocked by the adjacent tooth, and his value for speed of light was very value.

Question 37.
(a) List out the laws of photoelectric effect.
Answer:
Laws of photoelectric effect:

• For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
• Maximum kinetic energy of the photo electrons is independent of intensity of the incident light.
• Maximum kinetic energy of the photo electrons from a given metal is directly proportional to the frequency of incident light.
• For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
• There is no time lag between incidence of light and ejection of photoelectrons.

[OR]

(b) Describe the working of nuclear reactor with a block diagram.
Answer:
Nuclear reactor:

• Nuclear reactor is a system in which the nuclear fission takes place in a self-sustained controlled manner and the energy produced is used either for research purpose or for power generation.
• The main parts of a nuclear reactor are fuel, moderator and control rods. In addition to this, there is a cooling system which is connected with power generation set up.

Fuel:

• The fuel is fissionable material, usually uranium or plutonium. Naturally occurring uranium contains only \(0.7 \% \text { of }_{92}^{235} \mathrm{U} \text { and } 99.3 \% \text { are only }^{238} \mathrm{g}_{2} \mathrm{U}\) . So the \(_{ 92 }^{ 235 }{ U }\) must be enriched such that it contains at least 2 to 4% of \(_{ 92 }^{ 235 }{ U }\) .
• In addition to this, a neutron source is required to initiate the chain reaction for the first time. A mixture of beryllium with plutonium or polonium is used as the neutron source. During fission of \(_{ 92 }^{ 235 }{ U }\), only fast neutrons are emitted but the probability of initiating fission by it in another nucleus is very low. Therefore, slow neutrons are preferred for sustained nuclear reactions.

Moderators:

• The moderator is a material used to convert fast neutrons into slow neutrons. Usually the moderators are chosen in such a way that it must be very light nucleus having mass comparable to that of neutrons. Hence, these light nuclei undergo collision with fast neutrons and the speed of the neutron is reduced
• Most of the reactors use water, heavy water (D₂0) and graphite as moderators. The blocks of uranium stacked together with blocks of graphite (the moderator) to form a large pile.

Control rods:

• The control rods are used to adjust the reaction rate. During each fission, on an average 2.5 neutrons are emitted and in order to have the controlled chain reactions, only one neutron is allowed to cause another fission and the remaining neutrons are absorbed by the control rods.
• Usually cadmium or boron acts as control rod material and these rods are inserted into the uranium blocks. Depending on the insertion depth of control rod into the uranium, the average number of neutrons produced per fission is set to be equal to one or greater than one.
• If the average number of neutrons produced per fission is equal to one, then reactor is said to be in critical state. In fact, all the nuclear reactors are maintained in critical state by suitable adjustment of control rods. If it is greater than one, then reactor is said to be in super-critical and it may explode sooner or may cause massive destruction.

Shielding:

• For a protection against harmful radiations, the nuclear reactor is surrounded by a concrete wall of thickness of about 2 to 2.5 m.

Cooling system:

• The cooling system removes the heat generated in the reactor core. Ordinary water, heavy water and liquid sodium are used as coolant since they have very high specific heat capacity and have large boiling point under high pressure.
• This coolant passes through the fuel block and carries away the heat to the steam generator through heat exchanger. The steam runs the turbines which produces electricity in power reactors.

Question 38.
(a) Explain the working principle of a solar cell. Mention its applications.
Answer:
Solar cell:
A solar cell, also known as photovoltaic cell, converts light energy directly into electricity or electric potential difference by photovoltaic effect. It is basically a p-n junction which generates emf when solar radiation falls on the p-n junction. A solar cell is of two types: p-type and n-type.

Both types use a combination of p-type and n-type Silicon which together forms the p-n junction of the solar cell. The difference is that p-type solar cells use p-type Silicon as the base with an ultra-thin layer of n-type Silicon as shown in Figure, while n-type solar cell uses the opposite combination. The other side of the p-Silicon is coated with metal which forms the back electrical contact. On top of the n-type Silicon, metal grid is deposited which acts as the front electrical contact. The top of the solar cell is coated with anti-reflection coating and toughened glass.

In a solar cell, electron-hole pairs are generated due to the absorption of light near the junction. Then the charge carriers are separated due to the electric field of the depletion region. Electrons move towards n-type Silicon and holes move towards p-type Silicon layer.

The electrons reaching the n-side are collected by the front contact and holes reaching p-side are collected by the back electrical contact. Thus a potential difference is developed across solar cell. When an external load is connected to the solar cell, photocurrent flows through the load.

Many solar cells are connected together either in series or in parallel combination to form solar panel or module. Many solar panels are connected with each other to form solar arrays. For  high power applications, solar panels and solar arrays are used.

Applications:

• Solar cells are widely used in calculators, watches, toys, portable power supplies, etc.
• Solar cells are used in satellites and space applications
• Solar panels are used to generate electricity.

(b) Elaborate on the basic elements of communication system with the necessary block diagram.
Answer:
Elements of an electronic communication system
Information (Baseband or input signal): Information can be in the form of a sound signal like speech, music, pictures, or computer data which is given as input to the input transducer.

Input transducer
A transducer is a device that converts variations in a physical quantity (pressure, temperature, sound) into an equivalent electrical signal or vice versa. In communication system, the transducer converts the information which is in the form of sound, music, pictures or computer data into corresponding electrical signals. The electrical equivalent of the original information is called the baseband signal. The best example for the transducer is the microphone that converts sound energy into electrical energy.

Transmitter:
It feeds the electrical signal from the transducer to the communication channel. It consists of circuits such as amplifier, oscillator, modulator, and power amplifier. The transmitter is located at the broadcasting station.

Amplifier:
The transducer output is very weak and is amplified by the amplifier. Oscillator: It generates high-frequency carrier wave (a sinusoidal wave) for long distance transmission into space. As the energy of a wave is proportional to its frequency, the carrier wave has very high energy.

Modulator:
It superimposes the baseband signal onto the carrier signal and generates the modulated signal.

Power amplifier:
It increases the power level of the electrical signal in order to cover a large distance.

Transmitting antenna:
It radiates the radio signal into space in all directions. It travels in the form of electromagnetic waves with the velocity of light (3 ^x 10⁸ ms^-1).

Communication channel:
Communication channel is used to carry the electrical signal from transmitter to receiver with less noise or distortion. The communication medium is basically of two types: wireline communication and wireless communication.

Noise:
It is the undesirable electrical signal that interfaces with the transmitted signal. Noise attenuates or reduces the quality of the transmitted signal. It may be man-made (automobiles, welding machines, electric motors etc .) or natural (lightening, radiation from sun and stars and environmental effects). Noise cannot be completely eliminated. However, it can be reduced using various techniques.

Receiver:
The signals that are transmitted through the communication medium are received with the help of a receiving antenna and are fed into the receiver. The receiver consists of electronic circuits like demodulator, amplifier, detector etc. The demodulator extracts the baseband signal from the carrier signal. Then the baseband signal is detected and amplified using amplifiers. Finally, it is fed to the output transducer.

Repeaters:
Repeaters are used to increase the range or distance through which the signals are sent. It is a combination of transmitter and receiver. The signals are received, amplified, and retransmitted with a carrier signal of different frequency to the destination. The best example is the communication satellite in space.

Output transducer:
It converts the electrical signal back to its original form such as sound, music, pictures or data. Examples of output transducers are loudspeakers, picture tubes, computer monitor, etc.

Attenuation:
The loss of strength of a signal while propagating through a medium is known as attenuation.

Range:
It is the maximum distance between the source and the destination up to which the signal is received with sufficient strength.

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

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

General Instructions:

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

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

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

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

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

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

Question 4.
Match the following column I with column II


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

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

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

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

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

Part – II

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

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

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

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

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

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

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

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

Part – III

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

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

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

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

Applications:

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

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

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

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

Question 19.
Mention the name of man-made cereal. How it is developed?
Answer:

Triticale, the successful first man made cereal. Depending on the ploidy level Triticale can be divided into three main groups:

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

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

Part – IV

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

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

Each seed has a thick outer covering called seed coat. The seed coat is developed from integuments of the ovule. The outer coat is called testa and is hard whereas the inner coat is thin, membranous and is called tegmen.

In Pea plant the tegmen and testa are fused. Two cotyledons laterally attached to the embryonic axis are present. It stores the food materials in pea whereas in other seeds like castor the endosperm contains reserve food and the cotyledons are thin. The portion of embryonal axis projecting beyond the cotyledons is called radicle or embryonic root.

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

[OR]

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

(ii)

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

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

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

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

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

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

[OR]

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

Bio-Zoology [Maximum Marks: 35]

Part – I

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

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

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

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

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

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

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

Question 7.
Match column I with column II

(a) A – 4, B – 5, C – 2, D – 3, E – 1
(b) A – 3, B – 1, C – 4, D – 2, E – 5
(c) A – 2, B – 3,C – 1, D – 5, E – 4
(d) A – 5, B – 4, C – 2, D – 3, E – 1
Answer:
(a) A – 4, B – 5, C – 2, D – 3, E – 1

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

Part – II

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

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

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

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

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

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

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

Part – III

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

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

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

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

Question 17.
State Fisher and Race hypothesis.
Answer:
Fisher and Race hypothesis: Rh factor involves three different pairs of alleles located on three different closely linked loci on the chromosome pair. This system is more commonly in use today, and uses the ‘Cde’ nomenclature.

In the given figure, three pairs of Rh alleles (Cc, Dd and Ee) occur at 3 different loci on homologous chromosome pair-1. The possible genotypes will be one C or c, one D or d, one E or e from each chromosome. For e.g. CDE/cde; CdE/cDe; cde/cde; CDe/CdE etc. All genotypes carrying a dominant ‘D’ allele will produce Rh⁺positive phenotype and double recessive genotype ‘dd’ will give rise to Rh negative phenotype.

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

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

Part – IV

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

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

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

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

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

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

[OR]

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

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

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

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

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

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

Passive Immunity:

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

[OR]

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

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

At the end of incubation both single template strands will be made partially double stranded. The new strand of each double stranded DNA extends to a variable distance downstream. These steps are repeated again and again to generate multiple forms of the desired DNA, This process is also called DNA amplification.

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

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

Instructions:

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

Time: 3 Hours
Maximum Marks: 90

Part – I

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

Question 1.
If A = \(\left[\begin{array}{cc}
2 & 3 \\
5 & -2
\end{array}\right]\) be such that λA^-1 = A, then λ is _______.
(a) 17
(b) 14
(c) 19
(d) 21
Answer:
(c) 19

Question 2.
If ω ≠ 1 is a cubic root of unity and (1+ ω)⁷ = A+ B ω, then (A, B) equals to _______.
(a) (1,0)
(b) (-1, 1)
(c) (0, 1)
(d) (1, 1)
Answer:
(d) (1, 1)

Question 3.
The value of z – \(\bar{Z}\) is ______.
(a) 2 Im (z)
(b) 2 i Im (z)
(c) Im (z)
(d) i Im (z)
Answer:
(b) 2 i Im (z)

Question 4.
If x³ + 12x² + 10ax + 1999 definitely has a positive zero, if and only if ________.
(a) a ≥ 0
(b) a > 0
(c) a < 0
(d) a < 0
Answer:
(c) a < 0

Question 5.
sin(tan^-1 x), |x| < 1 is equal to _______.

Answer:
(d) \(\frac{x}{\sqrt{1+x^{2}}}\)

Question 6.
The centre of the circle inscribed in a square formed by the lines x² – 8x – 12 = 0 and y² – 14y + 45 = 0 is _____.
(a) (4, 7)
(b) (7, 4)
(c) (9, 4)
(d) (4, 9)
Answer:
(a) (4, 7)

Question 7.
The axis of the parabola x² = – 4y is ______.
(a) y= 1
(b) x = 0
(c) y = 0
(d) x = 1
Answer:
(b) x = 0

Question 8.
The coordinates of the point where the line \(\vec{r}=(6 \hat{i}-\hat{j}-3 \hat{k})+t(-\hat{i}+4 \hat{k})\) meets the plane \(\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=3\) are _______.
(a) (2, 1, 0)
(b) (7, -1, -7)
(c) (1, 2, -6)
(d) (5, -1, 1)
Answer:
(d) (5, -1, 1)

Question 9.
If the vectors \(\vec{a}=3 \vec{i}+2 \vec{j}+9 \vec{k}\) and \(\vec{b}=\vec{i}+m \vec{j}+3 \vec{k}\) are parallel then m is _________.

Answer:
(b) \(\frac{2}{3}\)

Question 10.
The minimum value of the function |3 – x | + 9 is ________.
(a) 0
(b) 3
(c) 6
(d) 9
Answer:
(d) 9

Question 11.
The curve y² = x² (1 – x²) has ______.
(a) an asymptote x = -1
(b) an asymptote x = 1
(c) two asymptotes x = 1 and x = -1
(d) no asymptote
Answer:
(d) no asymptote

Question 12.
If /(x, y, z) = xy + yz + zx, then f_x – f_z is equal to _______.
(a) z – x
(b) y – z
(c) x – z
(d) y – x
Answer:
(a) z – x

Question 13.
A circular template has a radius of 10 cm. The measurement of radius has an approximate error of 0.02 cm. Then the percentage error in calculating area of this template is _______.
(a) 0.2%
(b) 0.4%
(c) 0.04%
(d) 0.08%
Answer:
(b) 0.4%

Question 14.
The value of \(\int_{0}^{\pi} \sin ^{4} x d x\) is _______.

Answer:
(b) \(\frac{3 \pi}{8}\)

Question 15.
\(\int_{a}^{b} f(x) d x\) is _______.

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

Question 16.
The degree of the differential equation \(y(x)=1+\frac{d y}{d x}+\frac{1}{1.2}\left(\frac{d y}{d x}\right)^{2}+\frac{1}{1.2 .3}\left(\frac{d y}{d x}\right)^{3}+\ldots\) is ________.
(a) 2
(b) 3
(c) 1
(d) 4
Answer:
(c) 1

Question 17.
In finding the differential equation corresponding toy = e^mx where m is the arbitrary constant, then m is ____.
(a) \(\frac{y}{y^{\prime}}\)
(b) \(\frac{y^{\prime}}{y}\)
(c) y’
(d) y
Answer:
(b) \(\frac{y^{\prime}}{y}\)

Question 18.
Let X be random variable with probability density function f(x) = \(\left\{\begin{array}{ll}
2 / x^{3} & x \geq 1 \\
0 & x<1
\end{array}\right.\)
Which of the following statement is correct
(a) both mean and variance exist
(b) mean exists but variance does not exist
(c) both mean and variance do not exist
(d) variance exists but mean does not exist
Answer:
(b) mean exists but variance does not exist

Question 19.
The random variable X has the probability density function f(x) = \(\left\{\begin{array}{cc}
a x+b, & 0<x<1 \\
0, & \text { otherwise }
\end{array}\right.\)
and E(X) = \(\frac{7}{12}\), then a and b are respectively _______.
(a) 1 and \(\frac{1}{2}\)
(b) \(\frac{1}{2}\) and 1
(c) 2 and 1
(d) 1 and 2
Answer:
(a) 1 and \(\frac{1}{2}\)

Question 20.
A binary operation on a set S is a function from ________.
(a) S → S
(b)(S x S) → S
(c) S → (S x S)
(d) (S x S) → (S x S)
Answer:
(b)( S x S) → S

Part – II

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

Question 21.
Solve the following system of homogeneous equations.
3x + 2y + 7z = 0, 4x – 3y – 2z = 0, 5x + 9y + 23z = 0
Answer:
The matrix form of the above equation is

The augmented matrix [A, B] is

The above matrix is in echelon form. Here ρ(A, B) = ρ( A) < number of unknowns.
⇒ The system is consistent with infinite number of solutions. To find the solutions.
Writing the equivalent equations.
We get 3x + 2y + 7z = 0 ……..(1)
-17y – 34z = 0 …….(2)
Taking z = t in (2) we get -17y – 34t = 0
⇒ -17y = 34t
⇒ y= \(\frac{34 t}{-17}\) = -2t
Taking z = t; y = -2t in (1) we get
3x + 2 (-2t) + 7t = 0
3x – 4t + 7t = 0
⇒ 3x = -3t ⇒ x = -t
So the solution is x = -t; y = -2t; and z = t, t∈R

Question 22.
Show that |3z – 5 + i| = 4 represents a circle, and, find its centre and radius.
Answer:
The given equation |3z – 5 + i| = 4 can be written as

It is of the form |z – z| = r and so it represents a circle, whose center and radius are \(\left(\frac{5}{3},-\frac{1}{3}\right)\) and 4/3 respectively.

Question 23.
Find the equation of the circle whose centre is (2, -3) and passing through the intersection of the line 3x – 2y = 1 and 4x + y = 27.
Answer:
Solving 3x – 2y = 1 and 4x + y = 27
Simultaneously, we get x = 5 and y = 7
∴ The point of intersection of the lines is (5, 7)
Now we have to find the equation of a circle whose centre is
(2, -3) and which passes through (5, 7)

∴ Required equation of the circle is
(x – 2)² + (y + 3)² = \((\sqrt{109})^{2}\)
⇒ x² + y² – 4x + 6y – 96 = 0

Question 24.
Find the intercepts cut off by the plane \(\vec{r} \cdot(6 \hat{i}+4 \hat{j}-3 \hat{k})=12\) on the coordinate axes.
Answer:
\(\vec{r} \cdot(6 \vec{i}+4 \vec{j}-3 \vec{k})=12\)
Compare the above equations into \(\vec{r} \cdot \vec{n}=q\) so q = 12
Let a, b, c are intercepts of x-axis, y-axis and z-axis respectively.
Clearly

x – intercept = 2; y – intercept = 3; z – intercept = -4

Question 25.
Find the values in the interval (1, 2) of the mean value theorem satisfied by the function f(x) = x – x² for 1 ≤ x ≤ 2.
Answer:
f(1) = 0 and f(2) = -2. Clearly f(x) is defined and differentiable in 1 < x < 2. Therefore, by the Mean Value Theorem, there exists a c ∈(1, 2) such that
f'(c) = \(\frac{f(2)-f(1)}{2-1}\) = 1 – 2c
That is, 1 – 2c = -2 ⇒ c = \(\frac{3}{2}\)

Question 26.
Show that the percentage error in the n^th root of a number is approximately \(\frac{1}{n}\) times the percentage error in the number.
Answer:

Question 27.
Solve the differential equation: tany \(\frac{d y}{d x}\) = cos (x + y) + cos (x -y)
Answer:
tan y \(\frac{d y}{d x}\) = cos (x + y) + cos(x – y)
tan y \(\frac{d y}{d x}\) = cos x cos y – sin x sin y + cos x cos y + sin x sin y
tan y \(\frac{d y}{d x}\) = 2 cos x cos y
seperating the variables
\(\int \frac{\tan y}{\cos y}\) dy = 2∫cos x dx ⇒ ∫sec y tan y dy = 2∫cos x dx
sec y = 2 sin x + c

Question 28.
The probability density function of X is given by \(f(x)=\left\{\begin{array}{cc}
k e^{-\frac{x}{3}} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array}\right.\)
Find the value of k.
Answer:

Question 29.
Construct the truth table for the following statement. \(\neg(p \wedge \neg q)\).
Answer:

Question 30.
Find an approximate value of \(\int_{1}^{1.5} x^{2} d x\) by applying the right-hand rule with the partition {1.1, 1.2, 1.3, 1.4, 1.5}.
Answer:
Here a = 1; b = 1.5; n = 5; f(x) = x²
So, the width of each subinterval is

x₀ = 1; x₁ = 1.1; x₂ = 1.2; x₃ = 1.3; x₄ = 1.4; x₅ = 1.5
The Right hand rule for Riemann sum,
S = [f(x₁) + f(x₂) + f(x₃) + f(x₄) + f(x₅)] Δx
= [f(1.1) + f(1.2) + f(1.3) + f(1.4) + f(1.5)] (0.1)
= [1.21 + 1.44 + 1.69 + 1.96 + 2.25] (0.1)
= [8.55] (0.1)
= 0.855.

Part – III

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

Question 31.
Find a matrix A if adj (A) = \(\left[\begin{array}{ccc}
7 & 7 & -7 \\
-1 & 11 & 7 \\
11 & 5 & 7
\end{array}\right]\)

Question 32.
Obtain the Cartesian form of the locus of z = x + iy in the following case Im[(1 – i)z +1] = 0

Question 33.
If \(\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}, \vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}\), show that \([\vec{a} \vec{b} \vec{c}]\) depends on neither x nor y.

Question 34.
The Taylor’s series expansion of f(x) = sin x about x = \(\frac{\pi}{2}\) is obtained by the following way.

Question 35.
The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error in computing (i) the volume of the cube and (ii) the surface area of cube.

Question 36.
Evaluate \(\int_{0}^{1} \frac{\sin \left(3 \tan ^{-1} x\right) \tan ^{-1} x}{1+x^{2}} d x\)

Question 37.
Find the particular solution of (1 + x³) dy – x² ydx = 0 satisfying the condition y(1) = 2.

Question 38.
If X is the random variable with distribution function F(x) given by,
\(\mathrm{F}(x)=\left\{\begin{array}{ll}
0, & x<0 \\
x, & 0 \leq x<1 \\
1, & 1 \leq x
\end{array}\right.\)
then find (z) the Probability density function f(x)

Question 39.
Show that \(((\neg q) \wedge p) \wedge q\) is a contradiction.

Question 40.
Show that the absolute value of difference of the focal distances of any point P on the hyperbola is the length of its transverse axis.

Part – IV

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

Question 41.
(a) By using Gaussian elimination method, balance the chemical reaction equation:
C₂H₆ + O₂ → H₂O + CO₂.
[OR]
(b) \(\frac{d y}{d x}+\frac{3 y}{x}=\frac{1}{x^{2}}\), given that y = 2 when x = 1

Question 42.
(a) Find the real values of x and y for the equation \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}=i\)
[OR]
(b) Find the area between the line y = x + 1 and the curve y = x² – 1.

Question 43.
(a) Determine k and solve the equation 2x³ – 6x² + 3x + k = 0 if one of its roots is twice the sum of the other two roots.
[OR]
(b) Evaluate: \(\int_{0}^{\frac{\pi}{2}} \frac{d x}{5+4 \sin ^{2} x}\)

Question 44.
(a) A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16 m, and the height at the edge of the road must be sufficient for a truck 4 m high to clear if the highest point of the opening is to be 5 m approximately. How wide must the opening be?
[OR]
(b) Using truth table check whether the statements \(\neg(p \vee q) \vee(\neg p \wedge q)\) and \(\neg p\) are logically equivalent.

Question 45.
(a) Find the value of cot^-1 x – cot^-1 (x + 2) = \(\frac{\pi}{12}\), x > 0
[OR]
(b) Verify Euler’s theorem for f(x, y) = \(\frac{1}{\sqrt{x^{2}+y^{2}}}\)

Question 46.
(a) Find the points where the straight line passes through (6, 7, 4) and (8, 4, 9) cuts the xz and yz planes.
[OR]
(b) If X is the random variable with probability density function f(x) given by,
\(f(x)=\left\{\begin{array}{rc}
x+1, & -1 \leq x<0 \\
-x+1, & 0 \leq x<1 \\
0, & \text { otherwise }
\end{array}\right.\)
then find (z) the distribution function f(x) (ii) P (-0.5 ≤ X ≤ 0.5)

Question 47.
(a) Sketch the graph of the function: y = \(x \sqrt{4-x}\)
(b) The velocity v, of a parachute falling vertically satisfies the equation, \(v \frac{d v}{d x}=g\left(1-\frac{v^{2}}{k^{2}}\right)\)
where g and k are constants. If v and x are both initially zero, find v in terms of x.

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

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

Time: 3 Hours
Maximum Marks: 70

Instructions:

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

Part – 1

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

Question 1.
In the extraction of aluminium from alumina by electrolysis, cryolite is added to …………..
(a) Lower the melting point of alumina
(b) Remove impurities from alumina
(c) Decrease the electrical conductivity
(d) Increase the rate of reduction
Answer:
(a) Lower the melting point of alumina

Question 2.
Compound used for propellant is
(a) BN
(b) H₂B₄O₇
(c) B₂H₆
(d) Borax
Answer:
(c) B₂H₆

Question 3.
P₄O₆ reacts with cold water to give …………..
(a) H₃PO₃
(b) H₄P₂O₇
(c) HPO₃
(d) H₃PO₄
Answer:
(a) H₃PO₃

Question 4.
Which one of the following elements show high positive electrode potential?
(a) Ti²⁺
(b) Mn²⁺
(c) CO²⁺
(d) Cr²⁺
Answer:
(c) CO²⁺

Question 5.
As per IUPAC guidelines, the name of the complex [Co(en)₂(ONO)Cl]Cl is …………..
(a) chlorobisethylenediaminenitritocobalt(III) chloride
(b) chloridobis(ethane-l,2-diamine)nitro k-Ocobaltate(III) chloride
(c) chloridobis(ethane-l,2-diammine)nitrito k-Ocobalt(II) chloride
(d) chloridobis(ethane-l,2-diamine)nitro k-Ocobalt(III) chloride
Answer:
(d) chloridobis(ethane-l,2-diamine)nitro k-Ocobalt(III) chloride

Question 6.
Solid NH₃ solid CO₂ are examples of …………………..
(a) Covalent solids
(b) polar molecular solids
(c) molecular solids
(d) ionic solids
Answer:
(b) polar molecular solids

Question 7.
After 2 hours, a radioactive substance becomes\(\left(\frac{1}{16}\right)^{4}\) of original amount. Then the half life ( in min ) is ………
(a) 60 minutes
(b) 120 minutes
(c) 30 minutes
(d) 15 minutes
Answer:
(c) 30 minutes
Solution:

Question 8.
What is the decreasing order of strength of bases?

Answer:

Question 9.
Which electrolyte is used in Leclanche cell?
(a) ZnSO₄ + CuSO₄
(6) NH₄Cl + ZnCl₂
(e) NaCl + CuSO₄
(d) MnSO₄ + MnO₂
Answer:
(6) NH₄Cl + ZnCl₂

Question 10.
The phenomenon observed when a beam of light is passed through a colloidal solution is ………………
(a) Cataphoresis
(b) Eleætrophoresis
(c) Coagulation
(d) Tyndall effect
Answer:
(d) Tyndall effect

Question 11.
In the following sequence of reactions,

(a) Butanal
(b) n-butyl alcohol
(c) propan-1-ol
(d) Propanal
Answer:
(c) propan-1-ol

Question 12.
Of the following, which is the product formed when cyclohexanone undergoe’s aldol condensation followed by heating?

Answer:
(a)

Question 13.

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

Question 14.
Insulin, a hormone chemically is ………………..
(a) Fat
(b) Steroid
(c) Protein
(d) Carbohydrates
Answer:
(c) Protein

Question 15.
The role of phosphate in detergent powder is
(a) control pH level of the detergent water mixture
(b) remove Ca²⁺ and Mg²⁺ ions from water that causes hardness of water
(c) provide whiteness to the fabric
(d) more soluble in soft water
Answer:
(b) remove Ca²⁺ and Mg²⁺ ions from water that causes hardness of water

Part – II

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

Question 16.
What are all the steps involved in metallurgical process?
Answer:
The extraction of a metal from its ore consists the following metallurgical process.

• Concentration of the ore
• Extraction of crude metal
• Refining of crude metal

Question 17.
Give the uses of Borax.
Answer:

• Borax is used for the identification of coloured metal ions.
• In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
• It is also used as a flux in metallurgy and also acts as a good preservative.

Question 18.
Differentiate primary valency and secondary valency.
Answer:
Primary Valency :

1. The primary valence of a metal ion positive in most of the cases and zero in certain cases.
2. The primary valence is always satisfied by negative ions.
3. The primary valences are non directional
4. Example: In COCl₃.6NH₃, the primary valence of cobalt is +3 Example: In CoCl3.6NH3, the secondary valence of cobalt +3

Secondary Valency :

1. The secondary valence as the coordination number.
2. The secondary valence is satisfied by negative ions, neutral molecular or positive ions.
3. The secondary valences are directional
4. Example: In COCl₃.6NH₃, the secondary valence of cobalt is 6

Question 19.
Write a note about molecular solids.
Answer:

• In molecular solids, the constituents are neutral molecules. They are held together by weak vander waals forces.
• Molecular solids are soft and they do not conduct electricity. Eg., Solid CO₂

Question 20.
What are the limitations of Arrhenius concept?
Answer:

• Arrhenius theory does not explain the behaviour of acids and base in non-aqueous solvents such as acetone, tetrahydro furan.
• This theory does not account for the basicity of the substances like ammonia which do not possess hydroxyl group.

Question 21.
Write a note on catalytic poison.
Answer:
Catalytic poison: Certain substances when added to a catalysed reaction, decreases or completely destroys the activity of catalyst and they are often known as catalytic poisons.
For example, in the reaction, 2SO₂ + O₂ → 2SO₃ with a Pt catalyst, the poison is AS₂O₃. i. e., AS₂O₃ destroys the activity of pt. AS₂O₃ blocks the activity of the catalyst. So, the activity is lost.

Question 22.
Convert phenyl magnesium bromide to phenyl methanol (or) How would you prepare phenyl methanol from Grignard reagent?
Answer:

Question 23.
Identify compounds A,B and C in the following sequence of reactions.

Answer:

Question 24.
Why cannot Vitamin C be stored in our body?
Answer:
Vitamin C is water soluble, therefore it is readily excreted in urine and hence cannot be stored in the body.

Part – III

Answer any six questions. Question No. 29 is compulsory. [6 x 3 = 18]

Question 25.
All ores are minerals, but all minerals are not ores. Explain.
Answer:
A naturally occurring substance obtained by mining which contains the metal in free state or in the form of compounds is called a mineral. In most of the minerals, the metal of interest is present only in small amounts and some of them contains a reasonable percentage of metal. Such minerals that – contains a high percentage of metal, from which it can be extracted conveniently and economically are called ores. Hence all ores are minerals but all minerals are not ores.

Question 26.
Discuss the Commercial method to prepare Nitric acid.
[OR]
How will you prepare nitric acid by Ostwald’s process?
Answer:
Nitric acid prepared in large scales using Ostwald’s process. In this method ammonia from Haber’s process is mixed about 10 times of air. This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze. The temperature rises to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO, which then oxidised to nitrogen dioxide.

4NH₃ + 5O₂ → 4NO + 6H₂O + 120 kJ
2NO + O₂ → 2NO₂

The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.

6NO₂ + 3H₂O → 4HNO₃ + 2NO + H₂O

Question 27.
Actinoid contraction is greater from element to element than the lanthanoid contraction, why?
Answer:

• Actinoid contraction is greater from element to element than lanthanoid cintraction. The 5f orbitals in Actinoids have a very poorer shielding effect than 4f orbitals in lanthanoids.
• Thus, the effective nuclear charge experienced by electron in valence shells in case of actinoids is much more than that experienced by lanthanoids.
• In actinoids, electrons are shielded by 5d, 4f, 4d and 3d whereas in lanthanoids, electrons are shielded by 4d, 4f only. ,
• Hence, the size contraction in actinoids is greater as compared to that in lanthanoids.

Question 28.
What are the examples of first order reaction?
Answer:
(i) Decompostion of dinitrogen pentoxide
2N₂O_5(g) → 2NO_2(g) + 1/2 O_2(g)

(ii) Decomposition of thionylchloride
SO₂Cl_2(g) → SO_2(g) + Cl_2(g)

(iii) Decompostion of H₂O₂ in aqueous solution.
H₂O_2(aq) → H₂O_(l) + 1/2 O_2(g)

(iv) Isomerisation of cyclopropane to propene

Question 29.
Complete the following reaction.

Answer:

Question 30.
How will you calculate degree of dissociation of weak electrolytes and dissociation constant using Kohlrausch’s law?
Answer:
(i) The degree of dissociation of weak electrolyte can be calculated from the molar conductivity at a given concentration and the molar conductivity in infinite dilution using the formula \(\alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\)

(ii) According to Ostwald’s dilution law \(\mathrm{K}_{\mathrm{a}}=\frac{\alpha^{2} \mathrm{C}}{1-\alpha}\)
Substituting α value in the above equation

Question 31.
What are the characteristics of adsorption?
Answer:

• Adsorption can occur in all interfacial faces i.e., the adsorption can occur in between gas – solid, liquid – solid, liquid – liquid, solid – solid and gas – liquid. .
• Adsorption is always accompanied by decrease in free energy. When AG reaches zero, the equilibrium is attained.
• Adsorption is a spontaneous process.
• When molecules are getting adsorbed, there is always decrease in randomness of the molecules.
  ∆G = ∆H – T ∆S where ∆G = change in free energy
  ∆H = change in enthalpy
  ∆S = change in entropy
  .’. ∆H = ∆G + T∆S
• Adsorption is exothermic and it is a quick process.
• If simultaneous adsorption and absorption take place, it is termed as ‘sorption’ and sorption of gases on metal surface is called occlusion.

Question 32.
What are the uses of cellulose?
Answer:

• Cellulose is used extensively in manufacturing paper, cellulose fibres and rayon explosive.
• Gun cotton – nitrated ester of cellulose an explosive is prepared from cellulose.
• Cellulose act as food for animals.

Question 33.
How will you prepare PHBV? Give its use?
Answer:
(i) The biodegrable polymer PHBV (Poly hydroxy butyrate-co hydroxyl valerate) is prepared by the polymerisation of monomers 3 – hydroxy butanoic acid and 3 – hydroxy pentanoic acid.

(ii) It is used in orthopaedic devices and in controlled release of drugs.

Part – IV

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

Question 34.
(a) (i) Which type of ores can be concentrated by froth floatation method?
Give two examples for such ores. (2)
(ii) Explain the variation In E°_M3+/M2+ series. (3)
[OR]
(b) (i) Mention the uses of silicon tetrachtoride. (2)
(ii) What are all the conditions that are necessary for catenation? (3)
Answer:
(a) (i) Suiphide ores can be concentrated by froth floatation method.
e.g., (i) Copper pyrites (CuFeS₂) (ii) Zinc blende (ZnS) (iii) Galena (PbS)

(ii)

• In transition series, as we move down from Ti to Zn, the standard reduction potential E° _M3+/M2+value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than Cu²⁺
• E°_M3+/M2+ value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺.
• The standard electrode potential for the M³⁺ /M²⁺ half cell gives’the relative stability between M³⁺ and M²⁺.
• The high reduction potential of Mn³⁺ / Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺.
• For Fe³⁺/Fe²⁺ the reduction potential is 0.77 V, and this low value indicates that both Fe³⁺ and Fe²⁺ can exist under normal condition.
• Mn³⁺ has a 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible [E° = +1.5 IV]

[OR]

(b) (i) Silicon tetrachloride is used in the production of semiconducting silicon.
It is used as a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

(ii) Essential condition for catenation:

• The valency of elements is greater than or equal to two.
• Element should have an ability to bond with itself.
• The self bond must be as strong as its bond with other elements.
• Kinetic inertness of catenated compound towards other molecules.

Question 35.
(a) (i) Discuss the manufacture of chlorine. (3)
(ii) What is inert pair effect? (2)
[OR]
(b) (i) Calculate the magnetic moment of Ti³⁺ and V⁴⁺. (2)
(ii) Draw all possible geometrical isomers of the complex [CO(en)₂Cl₂]⁺ and identify the optically active isomer. (3)
Answer:
(a) (i) Electrolytic process: When a solution of brine (NaCl) is electrolysed, Na⁺ and Cl^– ions are formed. Na⁺ ion reacts with OH^– ions of water and forms sodium hydroxide.

Hydrogen and chlorine are liberated as gases.

Deacon’s process: In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed. Hot gases at about 723 K are passed through a jacket that surrounds the chamber.

The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is given below,

(ii) In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

[OR]

(b) (i) Ti (Z = 22) Ti³⁺3d¹
V (Z = 23) V⁴⁺ 3d¹
.’. \(\mu=\sqrt{1(1+2)}=\sqrt{3}=1.73 \mu_{\mathrm{B}}\) So they are paramagnetic.

(ii) 1. Cis – [Co(en)₂Cl₂]⁺

The coordination complex [Co(en)₂Cl₂]⁺ has three isõmers two optically active cis forms and the optically inactive trans form.

Question 36.
(a) (i) Calculate the number of atoms in a fee unit cell.
(ii) How do nature of the reactant influence rate of reaction?
[OR]
(b) (i) Account for the acidic nature of HClO₄
In terms of Bronsted – Lowry theory, identify its conjugate base.
(ii) IS it possible to store copper sulphate in an iron vessel for a long time?
Given \(\mathbf{E}_{\mathrm{Cu}^{2+} \mathrm{Cu}}^{\circ}=\mathbf{0 . 3 4 V}\) and \(\mathbf{E}_{\mathbf{F e}^{2+} \mathbf{F e}}^{s}=+\mathbf{0 . 4 4 V}\)
Answer:
(a) (i) Number of atoms in a fcc unit cell,

(ii) Nature and state of the reactant:
We know that a chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.
Let us compare the following two reactions that we carried out in volumetric analysis.

1. Redox reaction between ferrous ammonium sulphate (FAS) and KMnO₄
2. Redox reaction between oxalic acid and KMnO₄

The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe²⁺ . In fact heating is required for the reaction between KMnO₄ and Oxalate ion and is carried out at around 60°C.

The physical state of the reactant also plays an important role to influence the rate of reactions. Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants. For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine. Let us consider another example that we carried out in inorganic qualitative analysis of lead salts. If we mix the aqueous solution of colorless potassium iodide with the colorless solution of lead nitrate, precipitation of yellow lead iodide take place instantaneously, whereas if we mix the solid lead nitrate with solid potassium iodide, yellow coloration will appear slowly.

[OR]

(b) (i) HClO₄ ⇌ H⁺ + ClO₄^–

According to Lowry – Bronsted concept, a strong acid has weak conjugate base and a weak acid has a strong conjugate base.

Let us consider the stabilities of the conjugate bases ClO₄^– , CIO₃^– , ClO₂^– and CIO^– formed from these acid HClO₄, HClO₃ , HClO₂, HOCl respectively. These anions are stabilized to greater extent, it has lesser attraction for proton and therefore, will behave as weak base. Consequently the corresponding acid will be strongest because weak conjugate base has strong acid and strong conjugate base has weak acid.

The charge stabilization increases in the order, ClO^– < ClO^–₂ < ClO^–₃ < CIO^–₄ . This means ClO^–₄ will have maximum stability and therefore will have minimum attraction for H⁺. Thus ClO^–₄ will be weakest base and its conjugate acid HClO₄ is the strongest acid.

ClO^–₄ is the conjugate base of the acid HClO₄.

(ii) E⁰_cell = E⁰_ox + E⁰_red = 0.44 V +0.34 V = 0.78 V
These +ve E⁰_cell values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.

Question 37.
(a) (i) Explain the formation of water with copper catalyst by intermediate compound formation theory. (3)
(ii) O-nitro phenol is slightly soluble in water where as P-nitro phenol is more soluble. Give reason. (2)
[OR]
(b) (i) What happens when the following alkenes are subjected to reductive ozonolysis.
1. propene
2.1-Butene
3. Isobutylene (3)
(ii) What are reducing and non – reducing sugars? (2)
Answer:
(a) (i) Formation of water due to the reaction of H₂ and O₂ in the presence of Cu can be given as

(ii) O-nitro phenol is slightly soluble in water and more volatile due to intra molecular hydrogen bonding, whereas P-nitro phenol is more soluble in water and less volatile due to intermolecular hydrogen bonding.

[OR]

(b)

(ii) Reducing sugars: Those carbonhydrates which contain free aldehyde or ketonic group and reduces Fehling’s solution and Tollen’s reagent are called reducing sugars. All monosacchaides whether aldose or ketone are reducing sugars.

Non-reducing sugars: Cabohydrates which do not reduce Tollen’s reagent and Fehling’s solution are called non-reducing sugars. Example Sucrose. They do not have free aldehyde group.

Question 38.
(a) A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO₂ (B) on heating with liquid ammonia followed by treating with Br₂ /KOH gives (C) which on treating with NaNO₂ and HCl at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D. (5)
[OR]

(b) Explain the mechanism of cleansing action of soaps and detergents.(5)
Answer:

[OR]

(b) Mechanism of cleansing action of soaps and detergents:

• The cleansing action of both soaps and detergents from their ability to lower the surface tension of water, to emulsify oil or grease and to hold them in a suspension in water.
• This ability is due to the structure of soaps and detergents.
• In water a sodium soap dissolves to form soap anions and sodium cations. For example, the following chemical equation shows the ionisation of sodium palmitate.
  
• A soap anion consists of a long hydrocarbon chain with a carboxylate group on one end. The hydrocarbon chain, which is hydrophobic, is soluble in oils or grease. The ionic part is the carboxylate group which is hydrophilic, is soluble in water.
  
• In water, detergent dissolves to form detergent anions and sodium cations. For example the following chemical equations show the ionisation of sodium alkyl sulphate and sodium alkyl benzene sulphate.
  

6. The following explains the cleansing action of a soap or detergent on a piece of cloth with a greasy stain:

• A soap or detergent anion consists of a hydrophobic part and a hydrophilic part.
• Soap or detergent reduces the surface tension of water. Therefore the surface of the cloth is wetted thoroughly.
• The hydrophobic parts of the soap or detergents anions are soluble in grease.
• The hydrophilic parts of the anions are soluble in water.
• Scrubbing or mechanical agitation helps to pull the grease away from the cloth and the
• grease is broken into smaller droplets.
• Repulsion between the droplets causes the droplets to be suspended in water, forming an emulsion.
• Thus the droplets do not coagulate or redeposit on the cloth. Rinsing washes away the droplets.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Time: 3 Hours
Maximum Marks: 70

Instructions:

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

Part – 1

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

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

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

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

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

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

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

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

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

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

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

dispersion medium-gas
dispersed phase-liquid

Question 11.
Match the following Column-I with Column-II using the code given below.
c
Answer:
(a) A – 2,B – 3, C – 4, D – 1

Question 12.

(a) anilinium chloride
(b) O – nitro aniline
(c) benzene diazonium chloride
(d) m – nitro benzoic acid

Answer:
(c) benzene diazonium chloride

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

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

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

Part – II

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

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

Question 17.
What happens when PCI5 is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.

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

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

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

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

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

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

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

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

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

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

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

Question 22.
Draw the major product formed when 1-ethoxyprop-l-ene is heated with one equivalent
Answer:

This reaction follows SN¹ mechanism because in this reaction the more stable carbocation is formed that is double bonded carbocation. Therefore, the given molecule reacts with HI to form ethanol and 1-iodo prop-1-ene.

Question 23.
Define Tautomerism. Give example. Why tertiary nitro alkanes do not exhibit tautomerism?
Answer:
(i) Tautomerism is an isomerism in which the isomers change into one another with great ease of shifting of proton so that they exist together in equilibrium.

(ii) Tertiary nitro alkanes do not exhibit tautomerism due to absence of a-H atom.

Question 24.
Why do soaps not work in hard water?
Answer:
Soaps are sodium or potassium salts of long-chain falty acids. Hard water contains calcium and magnesium ions. When soaps are dissolved in hard water, these ions displace sodium or potassium from insoluble calcium or magnesium salts of fatty acids. These insoluble salts separate as scum.

This is the reason why soaps do not work in hard water.

Part – III

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

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

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

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

Question 26.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono (H₂SO_4.H₂O) and di (H₂SO_4.H₂O) hydrates and the reaction is exothermic.
The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.

Question 27.
Give evidence that [CO(NH₃)gCl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they will give different ions in the solution which can be
tested by adding AgNO₃ solution and BaCl₂ solution. When Cl^– ions are the counter ions, a white precipitate will be obtained with AgNO₃ solution. If SO₄^2- ions are the counter ions, a white precipitate will be obtained with BaCl₂ solution.

Question 28.
For a reaction, X + Y → Product; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall .order of the reaction?
Answer:
z = k [x]^m [y]ⁿ …… (1)
8z = k [4x]^m [y]ⁿ…… (2)
I 6z = k [4x]^m [4y]ⁿ…… (3)
Dividing Eq (2) by Eq (1) we get,

1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,

16 = 4^m . 4ⁿ
16 = 4².4ⁿ
16/16 = 4ⁿ
1 = 4n
∴ n = 0 [Zero order with respect toy]
Overall order of the reaction, ‘
k[x]^m[y]ⁿ
k [x]^1.5 [y]⁰
Order = (1.5 + 0) = 1.5

Question 29.
Identify the conjugate acid base pair for the following reaction in aqueous solution
Answer:

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

(ii)

• HPO₄^2- and PO₄^3- , SO₃^-2 and HSO₃^– are two conjugate acid – base pairs.
• PO₄^3- is the conjugate base of the acid HPO₄^2- (or)
  HPO₄^2- is the conjugate acid of PO₄^3-.
• HSO₃^– is the conjugate acid of SO₃^-2 (or) SO₃^-2 is the conjugate base of HSO₃^–

(iii)

• NH₄⁺ and NH₃, CO₃^2- and HCO₃^– are two conjugate acid – base pairs.
• HCO₃^– is the conjugate of acid CO₃^2- (or) CO₃^2- is the conjugate bases of HCO₃^–
• NH₃ is the conjugate base of NH₄⁺ (or) NH₄⁺ is the conjugate acid of NH₃.

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

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

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

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

Question 31.
Identify A, B, C, and D

Answer:

Question 32.
Draw the structural formula and write the IUPAC name of
(i) N, N-dimethyl aniline (ii) Benzyl amine (iii) N-methyl benzylamine
Answer:

Question 33.
Differentiate soap and detergents?
Soap :

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

Detergent :

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

Part – IV

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

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

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

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

[OR]

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

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

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

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

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

Therefore, tetrahedral complexes do not exhibit geometrical isomerism.

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

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

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

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

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

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

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

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

[OR]

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

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

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

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

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

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

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

Octahedral void:
A hole or void formed by three spheres of a hexagonal layer and another three spheres of the adjacent layer is a hole between six spheres. The spheres are arranged at the vertices of a regular octahedron. Such a hole or void is abc arrangement – ccp structure called octahedral void.

Question 36.
(a) (i) The rate constant for a first order reaction is 1.54 × 10^-3s^-1 (2)
Calculate its half life time.
(ii) Explain about protective action of colloid. (3)
[OR]
(b) (i) What is buffer solution? Give an example for an acidic buffer and a basic buffer. (2)
(ii The value of k_spof two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 × 10^-15 and 6 × 10^-17 respectively. Which salt is more soluble? Explain. (3)
Answer:
(a) (i) We know that, t_1/2 = 0.693/ k
t_1/2= O.693/1.54 ×10^-3s ^-1 = 450s

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

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

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

[OR]

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

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

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

(ii)

Ni(OH)₂ is more soluble than AgCN.

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

2. Oxidation occurs at the anode

Pb_(s) → Pb²⁺ _(aq) + 2e^– …………(1)

The Pb²⁺ ions combine with SO₄^2- to form PbSO₄ precipitate

Pb²⁺_(aq) + SO₄^2- → PbS_4(s) …………….(2)

3. Reduction occurs at the cathode

PbO_2(s) + 4H⁺_(aq) + 2e^– – → Pb²⁺_(aq) + 2H₂O ………….(3)

The Pb²⁺ ions also combines with SO₄^2- ions to form sulphuric acid to form PbSO₄ Precipitate

Pb²⁺_(aq) + SO₄^2-_(aq) → PbSO₄ …………..(4)

4. The overall reaction is,
(1) + (2) + (3) + (4)
pb_(s) + PbO_2(s) + 4H⁺ + 2SO₄^2-_(aq) → 2PbSO_4(s) + 2H₂O(l)

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

6. The emf of the cell depends on the concentration of H₂SO₄. As the cell reaction uses SO₄^2- ions, the concentration H₂SO₄ decreases. When the cell potential falls to about 1,8V, the cell has to be recharged.

7. Recharge of the cell: During recharge process, the role of anode and cathode is reversed and H₂SO₄ is regenerated.

The above reaction is exactly the reverse of redox reaction which takes place while discharging.

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

[OR]

(b) (i) When m-cresol is treated with acidic solution of sodium dichromate it gives 4-hydroxy benzoic acid.

(ii) The electron releasing groups (+1 groups) increase the relative charge on the carboxylate ion and destabilise it and hence the loss of proton becomes difficult.
+I groups are CH₃, – C₂H₅, – C₃H₇

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

2. Since ‘C’ is aniline, therefore, the amide from which it is formed must be benzamide (C₆H₅CONH₂). Thus, compound ‘B’ is benzamide:
The chemical equation showing the conversion of ‘B’ to ‘C’ is

Step-2: To find out the structure of compound ‘A’. Since compound ‘B’ is formed from compound ‘A’ by treatment with aqueous ammonia and heating. Therefore, compound ‘A’ must be benzoic acid or benzenecarboxylic acid.

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

[OR]

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

In a typical reaction, enzyme E binds with the substrate molecule leversity to produce an enzyme substrate complex. During this stage the substrate is converted into product and the enzyme becomes free and is ready to bind to another substrate molecule.

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