## Plus Two Chemistry Notes Chapter 3 Electrochemistry

Students can Download Chapter 3 Electrochemistry Notes, Plus Two Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus Two Chemistry Notes Chapter 3 Electrochemistry

Electrochemistry-
branch of chemistry which deals with the inter-relationship between electrical energy and chemical changes.

Electrolysis – The chemical reaction occuring due to the passage of electric current (i.e., electrical energy is converted into chemical energy).

Electrochemical reaction –
The chemical reaction in which electric current is produced (i.e., chemical energy is converted into electrical energy). Example: Galvanic cell

Electrochemical Cell: – (Galvanic Cell/Voltaic Cell) :
It converts chemical energy into electrical energy during redox reaction, e.g. Daniell Cell
The cell reaction is
Zn(s) + Cu2+ (aq) Zn2+(aq) + Cu(s)
It has a potential equal to 1.1 V.

If an external opposite potential is applied in the Daniell ce|l, the following features are noted:
a) When Eext < 1.1 V,
(i) electrons flow from Zn rod to Cu rod and hence current flows from Cu rod to Zn rod.
(ii) Zn dissolves at anode and Cu deposits at cathode.

b) When Eext= 1.1 V,
(i) No flow of electrons or current,
(ii) No chemical reaction.

c) When Eext > 1.1 V
(i) Electrons flow from Cu to Zn and current flows from Zn to Cu.
(ii) Zn is deposited at the Zn electrode and Cu dissolves at Cu electrode.

Galvanic Cells :
In this device, the Gibbs energy of the spontaneous redox reaction is converted into electrical work.

The cell reaction in Daniell cell is a combination of the following two half reactions:

1. Zn(s) → Zn2+(aq) + 2 $$\overline { e }$$ (oxidation half reaction/ anode reaction)
2. Cu2+(aq) + 2 $$\overline { e }$$ → Cu(s) (reduction half-reaction/ cathode reaction)

These reactions occur in two different vessels of the Daniell cell. The oxidation half reaction takes place at Zn electrode and reduction half reaction takes place at Cu electrode. The two vessels are called half cells or redox couple. Zn electrode is called oxidation half cell and Cu electrode is called reduction half cell. The two half-cells are connected externally by a metallic wire through a voltmeter and switch. The electrolyte of the two half-cells are connected internally through a salt bridge.

Salt Bridge :
It is a U-shaped glass tube filled with agar-agar filled with inert electrolytes like KCl, KNO3, NH4NO3.

Functions of Salt Bridge :

1. It maintains the electrical neutrality of the solution by intermigration of ions into two half-cells.
2. It reduces the liquid-junction potential.
3. It permits electrical contact between the electrode solutions but prevents them from mixing.

Electrode potential –
potential difference developed between the electrode and the electrolyte. According to IUPAC convention, the reduction potential alone is called electrode potential and is represented as $$E_{M^{n+} / M}$$

Standard Electrode Potential :
The electrode potential understandard conditions, (i.e., at 298 K, 1 atm pressure and 1M concentrated solution) is called standard electrode potential. It is represented as EΘ.

Representation of a Galvanic Cell :
A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge.

For example, the Galvanic cell can be represented as,
Zn (s)|Zn2+(aq)||Cu2+(aq)|Cu(s)

Cell Potential or EMF of a Cell :
The potential difference between the two electrodes of a galvanic cell is called cell potential (EMF) and is measured in volts.
EMF = Ecell = Ecathode – Eanode = ERjght – ELeft
Consider a cell, Cu(s) | Cu22+ (aq) || Ag+ (aq) | Ag(s)
Ecell = Ecathode – Eanode = EAg+/Ag – ECu2+/Cu

How to calculate emf of a cell when concentration varies from standard conditions.

Measurement of Electrode Potential using Standard Hydrogen Electrode (SHE)/Normal Hydrogen Electrode :
SHE or NHE consists of a platinum electrode coated with platinum black. The electrode is dipped in an exactly 1 M HCl solution and pure H2 gas at 1 bar is bubbled through it at 298 K. The electrode potential is arbitrarily fixed as zero at all temperatures.

Representation of SHE/NHE :
When SHE acts as anode:
Pt(s), Hsub>2(g, 1 bar) / H+(aq, 1 M)
When SHE acts as cathode:
H+(aq, 1 M)/H2(g, 1 bar), Pt(s)

Electrochemical Series/Activity series :
The arrangement of various elements in the increasing or decreasing order of their standard electrode potentials.

Applications of Electrochemical Series:
1. To calculate the emf of an electrochemical cell – The electrode with higher electrode potential is taken as cathode and the other as anode.
$$E_{\mathrm{cell}}^{\Theta}=E_{\mathrm{cathode}}^{\Theta}-E_{\mathrm{anodo}}^{\Theta}$$

2. To compare the reactivity of elements – Any metal having lower reduction potential (electode potential) can displace the metal having higher reduction potential from the solutions of their salt, e.g. Zn can displace Cu from solution.
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

3. To predict the feasibility of cell reactions -If EMF is positive, the cell reaction is feasible and if it is negative the cell reaction is not feasible.

4. To predict whether H2 gas will be evolved by reaction of metal with acids – All the metals which have lower reduction potentials compared to that of H2 electrode can liberate H2 gas from acids.

5. To predict the products of electrolysis.

Nernst Equation :
It gives a relationship between electrode potential and ionic concentration of the electrolyte. For the electrode reaction,
Mn+ (aq) + n $$\overline { e }$$ → M(s)
the electrode potential at any concentration measured with respect to SHE can be represented by,

R = gas constant (8.314 J K-1 mol-1), T=temperaturein kelvin, n = number of electrons taking part in the electrode reaction, F = Faraday constant (96487 C mol-1)

By converting the natural logarithm to the base 10 and subsitituting the values of R(8.314 J K-1 mol-1),T (298 K) and F (96487 C mol-1) we get,

Nernst Equation for a Galvanic Cell :
In Daniell cell, the electrode potential for any concentration of Cu2+ and Zn2+ ions can be written as,

Converting to natural logarithm to the base 10 and substituting the values of R, F and T=298 K, it. reduces to

Consider a general electrochemical reaction,

Equilibrium Constant and Nernst Equation:

where Kc is the equilibrium constant.

Electrochemical Cell and Gibbs Energy of the Reaction (∆rG):

Conductance of Electrolytic Solutions: .Conductors:
A substance which allows the passage of electricity through it. Conductor are classified as,

Metallic or Electronic Conductors:
In these the conductance is due to the movement of electrons and it depends on:

1. The nature and structure of the metal
2. Number of valence electrons per atom
3. Temperature (it decreases with increase in temperature)
e.g. Ag, Cu, Al etc.

ii. Electrolytic Conductors
Electrolytes – The substances which conduct electricity either in molten state or in solution, e.g. NaCl, NaOH, HCl, H2SO4 etc. The conductance is due to the movement of ions. This is also known as ionic conductance and it depends on:

1. Nature of the electrolyte
2. Size of the ions and their solvation
3. Nature of the solvent and its viscosity
4. Concentration of the electrolyte
5. Temperature (it increases with increase in temperature)

Ohm’s law – It states that the current passing through a conductor (I) is directly proportional to the potential difference (V) applied.
i.e., I ∝ V or I = $$\frac{V}{R}$$
where R – resistance of the conductor- unit ohm. In SI base units it is equal to kg m²/s³ A²

The electrical resistance of any substance/object is directly proportional to its length T, and inversly proportional to its area of cross section ‘A’.
R ∝ $$\frac{\ell}{\mathrm{A}}$$ or R = ρ$$\frac{\ell}{\mathrm{A}}$$ where,

ρ – (Greek, rho) – resistivity/specific resistance – SI unit ohm metre (Ω m) or ohm cm (Ω cm).

Conductance (G):
inverse or reciprocal of resistance (R).
$$G=\frac{1}{R}=\frac{A}{\rho \ell}=\kappa \frac{A}{\ell}$$
where K = $$\frac{1}{\rho}$$ called conductivity or specific conductance (K – Greek, kappa)

SI unit of conductance – S (siemens) or ohm-1.
SI unit of conductivity – S m-1
1 S cm-1 = 100 S m-1

Molar Conductance of a Solution (Λm):
It is the conductance of the solution containing one mole of the electrolyte when placed between two parallel electrodes 1 cm apart. It is the product of specific conductance (K) and volume (V) in cm³ of the solution containing one mole of the electrolyte.

where M is molarity of the solution.
Unit of Λm is ohm’1 cm2 mol’1 Or S cm² mol-1
Λm = $$\frac{K}{C}$$ [C-Concentration of the solution.]

Measurement of the Conductivity of Ionic Solutions :
The measurement of an unknown resistance can be done by Wheatstone bridge. To measure resitance of the electrolyte it is taken in a conductivity cell. The resistance of the conductivity cell is given by the equation.
$$R=\rho \frac{\ell}{A}=\frac{1}{\kappa A}$$

The quantity $$\frac{\ell}{\mathrm{A}}$$ is called cell constant and isdenoted A by G*. It depends on the distance (/) between the electrodes and their area of cross-section (A).

Variation of Conductivity and Molar Conductivity with Concentration :
Conductivity (K) always decreases with decrease in concentration both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity (Λm) increases with decrease in concentration. This is because the total volume, V of the solution containing one mole of electrolyte also increases.

The variation of molar conductance is different for strong and weak electrolytes,

1. Variation of Λm with Concentration for Strong Electrolytes:
The molar conductance increases slowly with decrease in concentration (or increase in dilution) as shown below:

There is a tendency for Λm to approach a certain limiting value when concentration approaches zero i. e., dilution is infinite. The molar conductance of an electrolyte when the concentration approaches zero is called molar conductance at infinite dilution, Λm or Λ°m. The molar conductance of strong electrolytes obeys the relationship.
Λm = Λ°m -AC1/2 where C = Molar concentration, A = constant for a particular type of electrolyte.
This equation is known as Debye-Huckel-Onsagar equation.

2. Variation of Λm with Concentration for Weak Electrolytes :
For weak electrolytes the change in Λm with dilution is due to increase in the degree of dissociation and consequently increase in the number of ions in total volume of solution that contains 1 mol of electrolyte. Here, Λm increases steeply on dilution, especially near lower concentrations.

Thus, the variation of Λm with √c is very large so that we cannot obtain molar conductance at infinite dilution Λ°m by the extrapolation of the graph.

Kohlrausch’s Law:
The law states that, the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the molar ionic conductivities of the cations and anions at infinite dilution.
Λ°m = γ+ λ°+ + γ λ°
λ°+ and λ° are the molar conductivities of cations and anions respectively at infinite dilution, Y+ and V. are number of cations and anions from a formula unit of the electrolyte.

Applications of Kohlaransch’s Law
1) To calculate Λ°m of weak electrolytes

2) To calculate degree of dissociation of weak electrolytes
$$\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}$$

3) To determine the dissociation constant of weak electrolytes

Electrolytic Cell and Electrolysis:
In an electrolytic cell, external source of voltage is used to bring about a chemical reaction. Electrolysis is the phenomenon of chemical decomposition of the electrolyte caused by the passage of electricity through its molten or dissolved state from an external source.

Quantitative Aspects of Electrolysis
Faraday’s Laws of Electrolysis First Law:
The amount of any substance liberated or deposited at an electrode is directly proportional to the quantity of electricity passing through the
electrolyte.
w α Q where ‘Q’ is the quantity of electric charge in coulombs.
w = ZQ .
w = Zlt
(∵ Q = It) where T is the current in amperes , ‘t’ is the time in seconds and ‘Z’ is a constant called electrochemical equivalent.

Second Law:
The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights.

The quantity of electricity required to liberate/deposit 1 gram equivalent of any substance is called Faraday constant ‘F’.
1 F = 96487 C/mol ≈ 96500 C/mol

Products of Electrolysis:
It depend on the nature of the material being electrolysed and the type of electrodes being used.

Electrolysis of Sodium Chloride:
When electricity is passed through molten NaCl, Na is deposited at the cathode and Cl2 is liberated at the anode.
Na+(aq) + $$\overline { e }$$ → Na(s) (Reduction at cathode)
Cl(aq) → ½ Cl2(g) + $$\overline { e }$$ (Oxidation at anode)

When concentrated aqueous solution of NaCl is electrolysed, Cl2 is liberated at anode, but at cathode H2 is liberated instead of Na deposition due to the high reduction potential of hydrogen.

The resultant solution is alkaline due to the formation of NaOH.

Electrolysis of CuSO4 :
When aqueous CuSO4 solution is electrolysed using Pt electrodes, Cu is deposited at the cathode and O2 is liberated at the anode.
Cu2+(aq) + 2 $$\overline { e }$$ → Cu(s) (at cathode)
H2O(l) → 2H+(aq) + 1/2 O2(g) + 2 $$\overline { e }$$ (at anode)

If Cu electrode is used, Cu is deposited at cathode and an equivalent amount of Cu dissolves in solution from the anode (because oxidation potential of Cu is higherthan that of water).
Cu2+(aq) + 2 $$\overline { e }$$ → Cu(s) (at cathode)
Cu(s) → Cu2+(aq) + 2$$\overline { e }$$ (atanode)

Commercial Cells (Batteries)
The electrochemical cells can be used to generate electricity. They are two types:
i) Primary Cells:
Cells in which the electrode reactions cannot be reversed by external energy. These cells cannot be recharged, e.g. Dry cell, Mercury cell.

ii) Secondary Cells :
Cells which can be recharged by passing current through them in the opposite direction so that they can be used again.

Primary Cells
a) Dry Cell:
Anode – Zn container
Cathode – Carbon (graphite) rod surrounded by powdered MnO2 and carbon.
Electrolyte – moist paste of NH4Cl and ZnCl2
The electrode reactions are :
Anode : Zn → Zn2+ + 2 $$\overline { e }$$
Cathode: MnO2 + NH4+ + $$\overline { e }$$ → MnO(OH) + NH3
Dry cell has a potential of nearly 1.5 V.

b) Mercury Cell:
Anode – Zn amalgam (Zn/Hg)
Cathode – paste of HgO and carbon
Eelectrolyte – paste of KOH and ZnO. The electrode reactions are,
Anode : Zn/Hg + 2OH → ZnO(s) + H2O + 2 $$\overline { e }$$
Cathode : HgO + H2O + 2 $$\overline { e }$$ → Hg(l) + 2 OH
Overall reaction : Zn/Hg + HgO(s) → ZnO(s)+ Hg(l)
The cell potential = 1.35 V

2. Secondary Cells
Electrolyte – 38% (by weight) soution of H2SO4.
The cell reactions when the battery is in use are,
Anode: Pb(s) + SO42-(aq) → PbSO4 + 2 $$\overline { e }$$
Cathode: PbO2(s) + SO42-(aq) + 4H+(aq) + 2 $$\overline { e }$$ → PbSO4(s) + 2H2O(I)
The overall cell reaction is,
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

The emf of the cell depends on the concentration of H2SO4. On recharging the battery the reaction is reversed and PbSO4(s) on anode is converted to Pb and PbSO4(s) at cathode is converted into PbO2.

Anode- Cd
Cathode – metal grid containing nickel (IV) oxide. Electrolyte – KOH solution. The overall cell reaction during discharge is,
Cd(s) +2 Ni(OH)3(s) → CdO(s) + 2Ni(OH)2(s) + H2O(l)

3) Fuel Cells :
These are Galvanic cells designed to convert the energy of combustion of fuels directly into electrical energy.

H2 – O2 fuel cell – In this, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous NaOH solution, which acts as the electrolyte.

The electrode reactions are,
Anode : 2H2(g) + 4OH(aq) → 4H2O(l) + 4$$\overline { e }$$
Cathode : O2(g) + 2H2O(l) + 4$$\overline { e }$$ → 4OH(aq)
Overall reaction : 2H2(g) + O2(g) → 2H2O(l)

pollution free, more efficient than conventional methods, Runs continuously as long as the reactants are supplied, electrodes are not affected.

Other examples:
CH4 – O2 fuel cell, CH3OH – O2 fuel cell

Corrosion :
Any process of destruction and consequent loss of a solid metallic material by reaction with moisture and other gases present in the atmosphere. More reactive metals are corroded more easily. Corrosion is enhanced by the presence of impurities, air & moisture, electrolytes and defects in metals.
Examples: Rusting of iron, tarnishing of Ag.

Mechanism:
In corrosion a metal is oxidised by loss of electrons to O2 and form oxides. It is essentially an electro chemical phenomenon. At a particular spot of an object made of iron, oxidation take place and that spot behaves as anode.
2 Fe(s) → 2 Fe2+ + 4$$\overline { e }$$E° = -0.44 V

Electrons released at anodic spot move through metal and go to another spot on the metal and reduce 02 in presence of H+. This spot behaves as cathode.
O2(g) + 4 H+(aq) + 4$$\overline { e }$$ → 2 H2O(l) E° = 1.23 V

The overall reaction is,
2 Fe(s) + O2(g)+ 4H+(aq) → 2 Fe2+ + 2H2O(I) E° = 1,67V

The ferrous ions are further oxidised by atmospheric 02 to ferric ions and form hydrated ferric oxide (rust) Fe2O3.xH2O

Prevention of Corrosion
1) Barrier Protection:
Coating the surface with paints, grease, metals like Ni, Cr, Cu etc.

2) Sacrificial Protection:
Coating the surface of iron with a layer of more active metals like Zn, Mg, Al etc. The process of coating a thin film of Zn on iron is known as galvanisation.

3) Anti-rust Solutions:
Applying alkaline phosphate/ alkaline chromate on iron objects which provide a protectve insoluble film. Also, the alkaline nature of the solutions decreases the availability of H+ ions and thus decreases the rate of corrosion.

## Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds

You can Download Chemical Reactions of Organic Compounds Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemistry Solutions Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 10th Standard Chemistry Solutions Chapter 7 Chemical Reactions of Organic Compounds

### Chemical Reactions of Organic Compounds Text Book Questions and Answers

Sslc Chemistry Chapter 7 Kerala Syllabus Question 1.
Complete stages 2, 3 and 4 in the respective order.

The chemical formula calculator is particularly helpful for establishing the percentage of each element

Chemical Reactions Of Organic Compounds Class 10 Kerala Syllabus Question 2.
What are the compounds formed when CH3-CH3 (ethane) undergoes substitution reaction with chlorine? Write them.
CH3 – CH2 Cl, CH3 – CHCl2,
CH3 – CCl3, CH2Cl – CCl3,
CHCl3 – CCl3, CCl3 – CCl3

Chemistry Class 10 Chapter 7 Kerala Syllabus Question 3.
Write down the structural formulae of ethane and ethene.
CH3 – CH3 – Ethane
CH2 = CH2 – Ethene

Sslc Chemistry Chapter 7 Solutions Kerala Syllabus Question 4.
What is the peculiarity of the Carbon- Carbon bond in ethene?
In ethene, There is carbon – carbon double bond

Text Book Page No: 121

alkyne reactions cheat sheet summary for organic chemistry reactions.

Class 10 Chemistry Chapter 7 Kerala Syllabus Question 5.
What do we get as the product ?
Ethane

Class 10 Chemistry Chapter 7 Notes Kerala Syllabus  Question 6.
Which hydrocarbon is the reactant here? ………..
Unsaturated propene

Chemistry Chapter 7 Class 10 Kerala Syllabus Question 7.
Is the product saturated or unsaturated ?
Saturated Compounds

10th Class Chemistry 7th Chapter Kerala Syllabus Question 8.
Complete table 7.1

Text Book Page No: 123

Hsslive Chemistry 10th Kerala Syllabus Question 9.
Complete table 7.2 Suitably.

 Monomer Polymer Uses Vinyl Chloride PVC Pipe, Helmet Ethene Polyethane Carry bags Isoprene Natural rubber (Poly Isoprene) Tire Tetra Fluro ethene Teflon Nonstick pan

Text Book Page No: 124

Balance the equation calculator allows you to balance chemical equations accurately.

Sslc Chemistry Chapter Wise Questions And Answers Kerala Syllabus Question 10.
Can you write the balanced chemical equation for the combustion of the fuel butane (C4H10) ?
2 C4H10(g) + 13O2(g) → 8CO2 + 10H2O + heat

Text Book Page No: 125

Hsslive Guru 10th Chemistry Kerala Syllabus Question 11.
Complete Table 7.3 and 7.4 contain¬ing chemical reactions of hydrocarbons.

Organic Chemistry Class 10 Kerala Syllabus  Question 12.
Match Columns A, B, and C suitably.

 Reactants (A) Products (B) Name of Reaction (C) CH3 – CH3 + Cl2 CH3 – CH2 Cl + HCl Substitution Reaction C2H6+O2 CO2 + H2O Combustion n CH2 = CH2 [CH2 – CH2]n Polymerisation CH3– CH2 – CH3 CH2 = CH2 + CH4 Thermal Cracking CH = CH + H2 CH2 = CH2 Addition Reaction

Class 10 Chemistry Kerala Syllabus Question 13.
CH2 – OH, CH3 – CH2 – OH
Can you write the IUPAC names of these two compounds?
CH3 – OH – Methanol
CH3 – CH2 – OH – Ethanol

Text Book Page No: 126

Hss Live Guru 10th Chemistry Kerala Syllabus Question 14.
Complete the following word web including more uses of ethanol.

Chemistry Textbook Class 10 Kerala Syllabus Question 15.
List out the uses of ethanoic acid.

• In the manufacture of rayon
• In the rubber and silk industry.
• Vinegar – Impart Sour taste for food item.
• Used as preservative.

Text Book Page No: 129

Hss Live Guru 10 Chemistry Kerala Syllabus Question 16.
Examine the given structural formulae and select the esters. You may also identify the chemicals required for their preparation.
1. CH3 – CH2 – COO – CH3
2. CH3 – CH2 – COOH
3. CH3 – CH2 – CO – CH3
4. CH3 – OH
5. CH3 – CH2 – CH2OH
6. CH3 – COOH
7. CH3 – COO – CH2 – CH2 – CH3
1. CH3 – CH2 – COO – CH3
7. CH3 – COO – CH2 – CH2 – CH3 are esters
1. CH3 – CH2 – COO – CH3
CH3 – CH2 – COOH + OH – CH3 $$\frac{\mathrm{Conc} . \mathrm{H}_{2} \mathrm{SO}_{4}}{ }$$ CH3 – CH2 – COO – CH3+ H2O
7. CH3 – COO – CH2 – CH2 – CH3
CH3 – COOH + OH – CH2 – CH2 – CH3 $$\frac{\mathrm{Conc} . \mathrm{H}_{2} \mathrm{SO}_{4}}{ }$$ CH3 – COO – CH2 – CH2 – CH3 + H2O

Text Book Page No: 130

Chemistry Class 10 Kerala Syllabus Question 17.
Take 10 mL distilled water in a test tube and take the same volume of hard water in another test tube. Add a few drops of soap solution to both the test tubes and shake well. Do both the test tubes contain the same quantity of foam? Which test tube contains more foam? What do you infer?
No. Both the test tube does not contain same quantity of foam. Distilled water taken test tube contains more foam. Soap does not lather well in hard water. The hardness of water is due to dissolved calcium and magnesium salts in it. These salts react with soap to form insoluble compounds resulting in the decrease of lather.

Hss Live Guru Chemistry 10 Kerala Syllabus Question 18.
Take 10 mL each of hard water in two test tubes. Add a few drops of soap solution in the first test tube and add the same amount of detergent solution in the second one. Shake both the test tubes well. What do you observe? Which test tube contains more foam?
Soap does not lather well in hard water. The hardness of water is due to dissolved calcium and magnesium salts in it. These salts react with soap to form insoluble compounds resulting in the decrease of lather. But detergents do not give insoluble components on reaction with these salts. Hence detergents are more effective than soaps in hard water.

Hsslive Chemistry Class 10 Kerala Syllabus Question 19.
List out the merits and demerits of detergents, compared to soaps.
Merit:
Detergents are more effective than soaps in hard water.
Detergents are effective in acidic solutions.
Demerits:
excessive use of the detergents causes environmental problems. The microorganisms in water cannot decompose the components of detergents. Hence the detergents released into water lead to the destruction of aquatic life. For example, the detergents which contain phosphate increases the growth of algae and limits the quantity of oxygen. Therefore, it dej creases the quantity of oxygen for the breath of the organisms in water and causes their destruction.

Let Us Assess

Solve using substitution calculator step-by-step solutions.

10th Chemistry Notes Kerala Syllabus Question 1.
Given below are two chemical equations.
a. CH2 = CH2 + H2 → A
b. $$\mathrm{A}+\mathrm{Cl}_{2} \quad \stackrel{\text { sunlight }}{\longrightarrow} \mathrm{B}+\mathrm{HCl}$$
Identify the compounds A and B. Name these reactions.

Question 2.
Name the important chemical reactions of hydrocarbons. Give one example for each.

e. Thermal cracking:
CH3 – CH2 – CH2 -CH3 → CH2 = CH2 + CH3 – CH3

Question 3.
Write chemical formula of propane. Write the names and structural formulae of two compounds, that may be formed during its substitution reaction with chlorine.
CH3 – CH3 – CH3 Propane

Question 4.
Complete the equation for the following chemical reaction. Name this reaction.
$$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}+\ldots \ldots \ldots \ldots \mathrm{O}_{2} \rightarrow–\mathrm{-}+$$
CH3 – CH2 – CH2 – CH3 + 13/2 O2
→ 4CO2 + 5 H2O , Combustion.

Question 5.
Which of the given molecules can form po¬lymers ? Butane, Propane, Propane, Methane, Butene.
Propene, Butene

Extended Activities

Question 1.
You are familiar with different chemical reactions of hydrocarbons. Identify the situations in daily life in which these are used.
a. Substitution Reaction : Chloroform, CCl4 preparation
b. Addition Reaction : Conversion of unsaturated compounds into saturated.
c. Combustion: Preparation of polymers like PVC.
d. Thermal Cracking: Butane (LPG)can be prepared from higher hydrocarbons

Question 2.
List out the different uses of ethanol. Pre-pare an essay on its adverse effects on human body and the related social issues when it is used as a beverage.
Uses of ethanol :

• Fuels
• Medicines
• Preservatives
• Preparation of organic compounds

Health problems :

• Reason for aneamia .
• Increases possibility of cancer
• Problems related with heart

Social problems:

• Reason for spoiling family relationships
• May cause financial crisis

Question 3.
You know how to make soap, don’t you? Try to prepare soaps of different colors and fragrance.Prepare a short note on chemistry of soaps.
Fats and oils are esters. They react with alkalies such as NaOH, KOH to form Sodium/ Potassium Salts of their carboxylic acids and glycerol.ester formed from fatty acids + NaOH / KOH → Soap + glycerol. Fatty acids such as palmitic acid, stearic acid react with alcohol, glycerol to form esters. Oils and fats are esters formed by the reaction between glycerol with fatty acid and stearic acid. Soaps are the salts formed when these react with alkalies.

### Chemical Reactions of Organic Compounds Orukkam Questions and Answers

Question 1.
After completing the chemical reactions write down to which category they belong.
a. CH2Cl + Cl2 → …….. + HCl
b. CH = CH+H2 →…………
c. CH4 + 2O2 → …….. + H2O
d. CH3 – CH2 – CH3 → ……….
a. CH2Cl + Cl2 → CH2Cl2 + HCl Substitution reaction
b. CH = CH + H2 → CH2 = CH2 Addition Reaction
c. CH4 + 2O2 → CO2 + H2O Combustion
d. CH3 – CH2 – CH3 → CH2 = CH2 + CH4 Thermal cracking

Question 2.
Rearrange the table suitably.

Question 3.
Methane is reacting with Cl in presence of sunlight. Complete equation of that reaction.

a. Write down the reaction of C3H8 with chlorine.
b. What type of reaction is this?

C3HCl7+Cl2 → C3Cl8 + HCl
b. Subsititution Reaction

Question 4.
Examples of additon reaction are given below, complete the equation.
a. CH2 = CH2 + H2 → ………
b. CH2 = CH + Cl2 → ……..
c. CH = CH + H2 → ……….
d. CH = CH + Cl2→ ………

Question 5.
a. Examples for combination of Hydrocarbon are given below complete the equation and balance it.
CH4 + O2 → ……. + ……….
C2H6 + O2 → ……. + ………
C3H8 + O2 → …….. + ………
b. Products formed on combustion of Hydrocarbon are ………….
a. CH4 + O2 → CO2 + 2H2O
2C2H6 + 7O2 → 4CO2 + 6H2O
C3H8 + 5O2 → 3CO2 + 4H2O
b. Carbon dioxide and Water

Question 6.
a. Name the product and what type of reaction is this?
nCH2 = CHCl → ………..
b. Write down the names of monomer in it.
c. Give examples for natural polymers

Polyvinyl chloride, polymerization
b. Vinyl chloride
c. Polyisoprene, Protein

Question 7.
Complete the table.

Question 8.
Arrange the points in two separate columns write column heading also.
a. CO and H2 are reacted in presence of a catalyst to form compound.
b. Sugar cane juice is fermented.
c. It is known as wood spirit.
d. It is used to make paint & varnish.
e. Used for drinking.
f. It is used for adding in industrial spirit.

Question 9.
a. Ethanol has very large industrial utility. When it enter into our body it creates large amount of problems in our body as well as in our society. List out the probl-em happening in our body and in the society.

b. In industrial ethanol always Methanol is added to prevent misuse by humans. Name the process and what are the side effects formed after consuming it?
a.

 In human body In society Liver problems Economic problems Cliolestrol Family issues Kidney problems Loses personality

b. Denatured Spirit:

• Loses eyesight permanently
• Vomiting

### Chemical Reactions of Organic Compounds SCERT Questions and Answers

Question 10.
Analyze the reactions and answer the following questions.

a. Identify A, B.
b. Write the name of the compound ‘a’.
c. Write the name of the reaction by which ‘b’ is formed.

b. Polythene

Peptide and Protein Molecular Weight Calculator. Peptide or protein molecular weight is an important parameter for Molecular Biology.

Question 11.
Some reactions of propane are given.
i. Hydrogen atoms are substituted one by one, in presence of sunlight.
ii. When heated in the absence of air, it decomposes to hydrocarbons with lesser molecular mass.
iii. Combines with oxygen to give C02 and H2O.
a. Identify the type of reaction in each case.
b. Write the chemical equation of the reaction (ii).
a. i. Substitutional reaction
ii. Thermal cracking
iii. Combustion
b. CH3 – CH2 – CH3 → CH2 = CH2 + CH4

The chemical formula calculator is particularly helpful for establishing the percentage of each element.

Question 12.
Analyse the reactions and answer the following questions.
$$\text { i. } \mathrm{CH}_{3}-\mathrm{OH}+\mathrm{CO} \stackrel{\text { Catalyst }}{\longrightarrow} \ldots \mathrm{A}$$

$$\text { iii. } \mathrm{A}+\mathrm{B} \stackrel{\mathrm{Con} . \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow} \quad \ldots . \mathrm{C} \ldots .+\mathrm{H}_{2} \mathrm{O}$$
a. Identify A, B, C.
b. What is the general name/ class to which product ‘C’ belongs? Write the IUPAC name.
a. A — CH3 – COOH
B – Methanol/CH3 – OH
C – CH3 – COO – CH3
b. Esters, Methyl ethanoate

Question 13.
Analyze the given reactions and answer the following questions.

a. Identify A and B
b. What is the name of reaction by which ‘B’ is formed?
a. A – CH2 = CH2, B – CH3 Cl
b. Substitution reaction

Question 14.
Some reactions regarding the production of ethanol are given below.

a. Identify A and B.
b. Write the name of the ester formed when the product B reacts with propanoic acid,
c. Write the chemical equation for the formation of the ester.
a. A- C6 H12 O6 B – C2H5 – OH
b. Ethyl Propanoate
c. CH3 – CH2 – COOH + HO – CH2 – CH3
CH3 – CH2 – COO – CH2 – CH3 + H2O

Question 15.
Acetylene (ethyne) is prepared in the laboratory when calcium carbide reacts with water. Write the chemical equations of the reactions for converting it to ethane.
CH = CH + H2 $$\stackrel{N i}{\longrightarrow}$$ CH2 = CH2
CH2 = CH2 + H2 $$\stackrel{N i}{\longrightarrow}$$ CH3 – CH3

Question 16.
Complete the table

Question 17.
a Write the structure of the organic comp¬ound with molecular formula QH.
b. What is the name of the compound formed when one hydrogen atom of benzene is replaced with methyl radical?

b. Methyl benzene (Toluene)

Question 18.
Two equations are given below.
i. CH = CH + HCT → ……… A ………
ii. nA → B
a. Identify A and B.
b. Identify the type of reaction (i)?

Question 19.
Three equations are given below.

a. Identify P, Q, R
b. Identify the name of the chemical reaction (ii) and (iii).
c. Write the IUPAC name of R.
a. P – CH2 = CH2
Q – CH3 – CH3
R – CH3 – CH2Cl
iii. Substitution Reaction
c. Chloroethane

Question 20.
Ethanol is an industrially important compound.
a. What is the name of 8-10% solution of ethanol?
b. How is it converted into rectified spirit?
c. What is denatured spirit?
a. Wash
b. Fractional distillation of wash
c. Product obtained by adding poisonous (methanol, pyridine) substances.

Question 21.
Uses of some important organic compounds are given. Pick out the suitable compounds from the box.
Power alcohol, Teflon, Polythene,
Ethanoic acid, Ethanol
a. For the preparation of rayon,
b. For making the coating of inner surface of non-stick cookware,
c. Solvent in paint industry,
d. As fuel in motor vehicles
a. Ethanoic acid
b. Teflon
c. Methanol
d. Power alcohol

Question 22.
Some reactants, products, and names of reactions are given in the table. Complete it.

a. Substitution reaction
b. CH = CH2
c. HBr
e. H2O
f. Combustion

h. Polymerization

Question 23.
Pick out the suitable compounds from the box for the following reactions.
CH4, C2H4, C3H8, CH3Cl
a Thermal cracking
a. C3H8
b. C2H4

### Chemical Reactions of Organic Compounds Exam Oriented Questions & Answers

Very Short Answer Type Questions (Score 1)

Question 24.
Equation of some chemical Reactions are given below. Complete it.

a. A → CH3 – CH3
B → CH3 – CH2 – Cl
C → HCl
b. A → CH2 = CH2
B → CH3 – CH2 – Cl

Question 25.
Teflon used as nonstick polymer.
a. Write the structure of the monomer of this. Write the IUPAC name.
b. Write the reaction equation for the preparation of the monomer.
a. CF2 = CF2 Tetra fluroethene

Question 26.
Recognize and write A, B, C from following equation.

A. CH3 – CH2 – OH
B. CH3 – COOH – OH
C. H2O

Short Answer Type Questions (Score 2)

Question 27.
Some chemical equations are given below. Write each type of chemical reaction.

CH3 – CH = CH2 + CH3 – CH3
a. Polymerization
b. Substitution reaction
d. Combustion
e. Thermal cracking

Question 28.
Look at the following reactions on heating pentane.
i. In the absence of air
ii. In the presence of air
a. Write the name of the reaction (i), (ii).
b. Write the chemical equation for the reaction.
a. Reaction (i) – Thermal Cracking
Reaction (ii) – Combustion

ii. for writing the combustion equation for hydrocarbons we can use the equation.

Question 29.
Write each of the following.
a. Molasses
b.Wood spirit
c. Vinegar
d. Esters
a. Molasses is mother liquor left after the crystallization of sugar from sugar cane juice.
b. Poisonous chemical methanol (CH3 – OH) is known as wood spirit.
c. 5-8% ethanoic acid (CH3 – COOH) is known as vinegar.
d. esters are salts formed by the reaction ale ‘ whole and organic acids. They have the smell of fruits and flowers.

Short Answer Type Questions (Score 3)

Question 30.
Chloroform can be prepared from Methane.
a. What is the chemical formula of chlor of or m?
b. What is the name of reaction when chloroform is prepared from methane?
c. Write the chemical equation for the reaction.
a. CHCl3
b. Substitution Reaction
$$\mathrm{c.} \mathrm{CH}_{4}+\mathrm{Cl}_{2} \stackrel{\text {sunlight}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}$$
CH2 Cl + Cl2 → CH2Cl2 + HCl
CH2 Cl2 + Cl2 → CHCl3 + HCI

Question 31.
Methyl Ethanoate is an ester,
a. Write the structural formula.
b. Write the structural formula of alcohol and carboxylic acid needed for the preparation.
c. Write the reaction equation for the preparation of this ester.

Question 32.
Fill in the following.

Question 33.
Write one use of each of the following,
a. Methanol
b. Power alcohol
c. Butane
a. Used as solvent in the preparation of paint
b. Fuel in motor vehicle.
c. Cooking gas (LPG)

Question 34.
Write the reason for the following statements.
a. Hydrocarbons like butane are used as fuel.
b. Drinking denatured spirit is harmful
c. Esters are used in perfumes and fruit juice.
a. Combustion of hydrocarbon produces plenty of heat.
b. Ethanol on addition with poisonous material is called denatured spirit.
c. Esters have the pleasant smell of flowers and fruits.

Question 35.
Write structural formula of following com-pounds!
a. Benzene
b. Phenol
c. Toluene

Long Answer Type Questions (Score 4)

Question 6.
Ethyne is a compound that belongs to the class of alkynes.
a. Write chemical formula of ethyne.
b. Write the chemical equation for the preparation of following compounds from ethyne.
i. PVC
ii. 1, 2 – dichloroethane
iii. Chloro ethane
a. CH = CH
b. i. CH = CH + HCl →

Question 37.
Molecular formula of some compounds are given in the box.
C2H4 C6H14 CH3 – CH2 – CI
CH3 – COOH C6H6
a. Which is aromatic compound?
b. Which can be prepared by substitution reaction?
c. Which monomer is used in preparation of polythene?
d. Which compound can be used in food?
a. C6H6
b. CH3 – CH2 – Cl
c. C2H4
d. CH3COOH

Question 38.
a. Name the reaction for the conversion of sugar solution into ethanol.
b. Which enzymes are used in this reaction
c. Chemical involved in grape spirit.
d. Name the monomer of P VC.
e. Products formed during cracking of propane.
a. Fermentation
b. Invertase, Zymase
c. Ethanol (CH3 – CH2 OH)
d. Polyvinyl chloride (CH2 = CH – Cl)
e. Ethene(CH2 = CH2), Methane (CH4)

## Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

Students can Download Chapter 2 Solutions Questions and Answers, Plus Two Chemistry Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 2 Solutions

### Plus Two Chemistry Solutions One Mark Questions and Answers

Question 1.
Which of the following is a liquid in solid type solution?
(a) glucose dissolved in water
(b) Camphor in N2
(c) amalgam of Hg with Na
(d) Cu dissolved in Au
(c) amalgam of Hg with Na

Question 2.
The concentration term used when the solute is present in trace quantities is _______
ppm (parts per million)

Question 3.
A binary solution of ethanol and n-heptane is an example of
(a) ideal solution
(b) Non-ideal solution with +ve deviation
(c) Non-ideal solution with -ve deviation
(d) unpredictable behaviour
(b) Non-ideal solution with +ve deviation

Question 4.
A solution which has higher osmotic pressure as compared to other solution is known as _____
Hypertonic

Question 5.
Which of the following solutions will have the highest boiling point at 1 atm pressure?
(a) 0.1M FeCl3
(b) 0.1MBaCl2
(c) 0.1MNaCl
(d) 0.1Murea
(a) 0.1M FeCl3

Question 6.
1 kilogram of sea water sample contains 6 mg of dissolved O2. The concentration of O2 in the sample in ppm is
6.0

Question 7.
The amount of solute (molar mass 60 g mol-1) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10% is
60 g

Pure Aqua provides users with an osmotic pressure calculator to gain better insight on your osmotic pressure requirements.

Question 8.
The correct equation for the degree of association (a) of an associating solute ‘n‘ molecules of which undergoes association in solution is
a = $$\frac{n(i-1)}{1-n}$$

Question 9.
A solution is prepared by dissolving 10 g NaOH in 1250 ml of a solvent of density 0.8 ml/g. The molarity of the solution is _______
0.25

Question 10.
If the elevation in boiling point of a solution of non volatile, non electrolyte in a solvent (Kb = xk. kg mol-1) is 7 K, then the depression in freezing point would be kf = ZK kg mol-1
$$\frac{Y Z}{x}$$

### Plus Two Chemistry Solutions Two Mark Questions and Answers

Question 1.
Match the terms of list A with those in list B.

 A B Raoult’s Law. Colligative property. Henry’s Law. Ideal solution. Elevation of boiling point. Solution of gases in liquids. Benzene + Toluene. Vapour pressure of solutions.

 A B Raoult’s Law. Vapour pressure of solutions. Henry’s Law. Solution of gases in liquids. Elevation of boiling point. Colligative property. Benzene + Toluene. Ideal solution.

Question 2.
At a particular temperature, the vapour pressure of two liquids A and B are 120 and 180 mm of Hg respectively. Two moles of A and 3 Moles of B are mixed to form an ideal solution. What is the vapour pressure for the solution?
A = 120mm of Hg
B= 180 mm of Hg
χA = 2/5
χB =3/5
PA = P°A × χA
= 120 × 2/5 = 48 mm of Hg
PB = P°B × χB
= 180 × 3/5 = 108 mm of Hg
Ps = PA + PB
= 108 + 48 = 156 mm of Hg

Question 3.
Find the volume of H2O that should be added to 300 mL of 0.5 M NaOH so as to prepare a solution of 0.2 M.
M1V1 = M2V2
300 × 0.5 = V2 × 0.2
V2 $$\frac{300 \times 5}{2}$$ = 750 mL
H2O to be added = 750 mL – 300 mL = 450mL

Question 4.
Calculate the osmotic pressure of 5% solution of urea at 27°C?
Mass of urea, WB = 5 g
R = 0.0821 L atm K-1 mol-1
T = 273 + 27°C = 300 K
Molecular mass of urea, MB = 60 g mol-1
V = 100 mL = 0.1L

= 20.53 atm

Question 5.
Osmotic pressure of 1M solution of NaCl is approximately double than that of 1M sugar solution. Why?
Osmotic pressure is a colligative property and it depends on the number of solute particles present in the solution. In solution, each NaCl unit undergoes dissociation to form two particles (NaCl → Na+ + Cl) and hence osmotic pressure of 1M NaCl is twice that of 1M sugar solution. Sugar molecules does not undergo association or dissociation in solution.

Question 6.
What do you mean by ideal solution?
Ideal solution is a solution which obeys Raoult’s law over the entire range of concentration and temperature, i.e., for an ideal solution having two volatile components A and B.
PA = P°A χA, PB = P°B χB,
Ps = PA + PB =P°A χA + P°B χB

Question 7.
Many countries use desalination plants to meet their potable water requirements. Comment on the phenomenon behind it.
This is based on reverse osmosis. When a pressure more than osmotic pressure is applied, pure water is squezed out of the sea water through a semi-permeable

Question 8.
Movement of solvent molecules through a semipermeable membrane from pure solvent to the solution side is called osmosis.
What are the following

1. Isotonic solution
2. Hypertonic solution?

1. Isotonic solution:
If the osmotic pressure of the two solutions are equal, they are called isotonic solutions.

2. Hypertonic solution:
A solution having higher osmotic pressure than another solution is called hypertonic solution.

Question 9.
What is meant by azeotrope?
Liquid mixtures which distil without change in composition are called azeotropic mixtures or azeotropes.

### Plus Two Chemistry Solutions Three Mark Questions and Answers

Question 1.
A student was asked to define molality. Then he answered that it is the number of gram moles of the solute dissolved per litre of the solution.

1. Is it correct?
2. Can you help the student to define molality?
3. Calculate the molality of a solution containing 20 g of NaOH in 250 g of H2O.

1. It is not correct.
2. It is the number of moles of the solute present in 1000 g of the solvent. Molality can be determined by using the formula.
Molality, m = $$\frac{\text { Mass of the solute in gram } \times 1000}{\text { Molar mass of the solute } \times \text { Mass of the solvent in gram }}$$
3. Molality, m = $$\frac{20 \times 1000}{40 \times 250}$$ = 2 m

Question 2.
The graph of non-ideal solution showing -ve deviation as drawn by a student is given below:

1. Is this diagram correct?
2. Substantiate your argument with a suitable example.

1. No.
2. Because this graph shows the non-ideal solution showing +ve deviation and not -ve deviation. Consider a solution obtained by mixing chloroform and acetone. Here chloroform molecule is able to form hydrogen bond with acetone molecule as shown below:

As a result of this, vapour pressure of the solution decreases. Due to this, boiling point increases. The volume of the solution is less than the expected value. The mixing process is exothermic. So the actual graph is as given below:

Question 3.
Some words are missed in the following paragraph. Add suitable words in the blanks:
If osmotic pressure of 2 solutions are equal they are called ______(a)_____ solution. The solution which is having ______(b)______ osmotic pressure is called hypertonic solution and the solution which is having lower osmotic pressure is called ______(c)_____ solution.

1. Isotonic
2. Higher
3. Hypotonic

Question 4.
The solubility of gases depends upon some factors.

1. Can you suggest the factors?
2. Which is the law behind it?
3. What are the limitations of this law?

1. Nature of the gas, Nature of the solvent, Pressure, Temperature
2. Henry’s law. It states that the mass of a gas dissolved per unit volume of the solvent at a given temperature is proportional to the pressure applied on it.
3. Henry’s law is valid only under the following conditions:
• The pressure of the gas is not too high.
• The temperature is not too low.
• The gas is not highly soluble.

Question 5.
A solution of 12.5 g of an organic solute in 170g of H2O a boiling point elevation of 0.63 K. Calculate the molecular mass of the solute (K2=0.52 K/m).
Kb for water = 0.52 K Kg mol-1
WA = 170g
ΔTb = 0.63 K
WB = 12.5 g
MB = ?

Question 6.
Find the freezing point of the solution containing 3.6 g of glucose dissolved in 50 g of H20. (Kf for H2O = 1.86 K/m).
Mass of glucose, WB = 3.6 g
Molecular mass of glucose = 180 g mol-1
Mass of solvent, WA = 50 g
Kf for H2O = 1.86 K/m = 1.86 K kg-1 mol-1

i.e., ΔTf = T°f – Tf= 0.744 K
∴ Freezing point of the solution, Tf = T°f – ΔT
= 273 K – 0.744 K = 272.3 K

Question 7.
Raw mangoes shrivel when pickled in brine solution.

1. Name the process behind this.
2. Define that process.

1. Osmosis
2. When a solution is separated from its solvent by a semipermeable membrane, the solvent flows into the solution through the semipermeable membrane. This process is called osmosis.

Question 8.
200 cm3 of an aqueous of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein.

Question 9.
What type of deviation from Rauolt’s law is exhibited by a mixture of phenol and aniline? Explain with the help of graph.
Negative deviation.

In this case the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules. Thus, escaping tendency of the particles decreases in solution and hence the liquid mixture shows negative deviation.

Question 10.
Wilted flowers revive when placed in fresh water.

1. Identify and define the phenomenon.
2. Mixing of acetone and CHCl3 results in a reduction in volume. What type of deviation from Raoult’s law is observed here? Give reason.
3. Benzoic acid dissolved in benzene shows double of its molecular mass. Explain.

1. Osmosis. It is the phenomenon of flow of solvent from pure solvent into a solution or from a solution of lower concentration into a solution of higher concentration through a semi-permeable membrane.

2. Negative deviation. Here chloroform molecule is able to form hydrogen bond with acetone molecule.

Thus escaping tendency of the particles decreases in solution and hence the liquid mixture shows negative deviation.

3. Benzoic acid undergoes association in solution.
2C6H5COOH $$\rightleftharpoons$$ (C6H5COOH)2 Thus, the number of particles as well as colligative properties decreases. So molecular mass increases.

Question 11.
A solution is obtained by mixing 300 g of 25 % solution and 400 g of 40 % solution by mass. Calculate the mass percentage of the resulting solution.
Mass of solute in 300 g of 25 % solution
$$\frac{300 \times 25}{100}$$ = 75 g
Mass of solute in 400 g of 40 % solution
$$\frac{400 \times 40}{100}$$ = 160 g
Total mass of solute = (75 + 160) g = 235 g
Total mass of solution = (300 + 400) g = 700 g
Mass % of solute in resulting solution = $$\frac{235 \times 100}{700}$$ = 33.57%
Mass % of solvent (water) in resulting solution
= 100 – 33.57 = 66.43%

Question 12.
A graph showing vapour pressure against mole fraction of an ideal solution with volatile components A and B are shown below:

1. Which law is supported by the graph?
2. State the law.
3. Resketch the graph if the attraction between A and B is greater than that between A-A and B-B.

1. Raoult’s law.
2. Raoult’s law states that vapour pressure of a volatile component in a solution is the product of vapour pressure of that component in the pure form and mole fraction of that component in the solution.
3. If the A-B attraction is greater than A-A and B-B attractions the liquid mixture behaves as a non-ideal solution with negative deviation.

Question 13.
Concentrated nitric acid used in the laboratory work is 68% nitric by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL-1?

Question 14.
Why does gases always tend to be less solube in liquids as the temperature is raised?
Dissolution of gases is exothermic process. It is because of the fact this process involves decrease of entropy (ΔS < 0). Thus, increase of temperature tends to push the equilibrium,
Gas + Solvent $$\rightleftharpoons$$ Solution; ΔH = -ve
in the backward direction, thereby, supressing the dissolution.

Question 15.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
H2O= 12.3 kPa
1000
In 1 molal solution, nsolute = 1; nH2O= $$\frac{1000}{18}$$ = 55.5
∴ χH2O = $$\frac{55.5}{55.5+1}$$
Vapour pressure of the solution, Ps = P°H2O × χH2O
= 0.982 × 12.3 = 12.08 kPa

Question 16.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated.
Since K2SO4 is completely dissociated as K2SO4 → 2K+ + SO42- Thus, i = 3
Osmotic pressure of the solution, π = i CRT
$$\frac{3 \times 25 \times 10^{-3} \mathrm{g} \times 0.0821 \mathrm{L} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 298.15 \mathrm{K}}{174 \mathrm{g} \mathrm{mol}^{-1} \times 2 \mathrm{L}}$$
= 5.27 × 10-3 atm

Question 17.
Solution of sucrose is prepared by dissolving 34.2 g of it in 1000 g of water. Find out the freezing point of the solution, if Kf of water is 1.86 K/kg/mol? (Molecular mass of sucrose is 342 g/mol).
ΔTf = kf × m
=1.86 K kg mol-1 × $$\frac{34.2 \mathrm{g} \times 1000 \mathrm{g} \mathrm{kg}^{-1}}{342 \mathrm{g} \mathrm{mol}^{-1} \times 1000 \mathrm{g}}$$ = 0.186K
Tf = OK – 0.186 K
Freezing point of the solution = – 0.186 K

Question 18.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
In the first case, π1 = C1RT
i.e., 4.98 bar = $$\frac{36 \mathrm{g} \times \mathrm{R} \times 300 \mathrm{K}}{180 \mathrm{g} \mathrm{mol}^{-1}}$$ ——– (1)
In second case, π2 = C2RT
i.e., 1.52 bar = C2R × 300 K ———- (2)

Question 19.
A solution containing 12.5 g of non-electrolytic substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molecular mass of the substance? (Kb for water = 0.52 K kg mol-1)
Molecular mass of the solute, MB = $$\frac{1000 \mathrm{K}_{\mathrm{b}} \mathrm{W}_{\mathrm{B}}}{\mathrm{W}_{\mathrm{A}} \Delta \mathrm{T}_{\mathrm{b}}}$$
$$\frac{1000 \mathrm{g} \mathrm{kg}^{-1} \times 0.52 \mathrm{K} \mathrm{kg}^{-1} \mathrm{mol}^{-1} \times 12.5 \mathrm{g}}{175 \mathrm{g} \times 0.70 \mathrm{K}}$$
= 53.06 mol-1

Question 20.

1. “For any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction. Name the law stated above.
2. Study the graph. What phenomenon it denotes? Based on your observation predict the reason for the greater volatility of a mixture of carbon disulphide and acetone?

1. Raoult’slaw
2. Positive deviation from Raoult’s law. In a solution formed by adding CS2 to acetone, the dipolar interactions between solute-solvent molecules are weaker than the respective interactions among the solute-solute and solvent-solvent molecules. Thus, the escaping tendency of the particles increases and the solution shows +ve deviation.

Question 21.

1. To get hard boiled eggs, common salt is added to water during boiling. Give reason.
2. Which colligative property is more suitable for the determination of molecular mass of polymers? Give the expression to determine molecular mass by this method.

1. The egg which is placed in sodium chloride solution will shrink due to osmosis of water out of the egg and thus get hardened.
2. Osmotic pressure method.

Question 22.
18 g of glucose is dissolved in 1kg of water in a beaker. At what temperature will water boil at 1.013 bar? (Kb for water is 0.52 K kg mol-1)
Number of moles of glucose = 18/180 = 0.1 mol.
Mass of solvent = 1 kg.
Morality of glucose , m = $$\frac{n_{8}}{W_{A}}$$
= $$\frac{0.1 \mathrm{mol}}{1 \mathrm{kg}}$$ = 0.1 mol/Kg
Elevation of boiling point ΔTb = Kb × m
= 0.52 K kg/mol × 0.1 mol/kg = 0.052 K
Since water boils at 373.15 K at 1.013 bar pressure, the boiling point of solution will be 373.15 K + 0.052 K = 373.202 K

### Plus Two Chemistry Solutions Four Mark Questions and Answers

Question 1.
Colligative properties are exhibited by dilute solutions.

1. What do you mean by colligative properties?
2. Which are the four colligative properties?

1. Colligative properties are those properties of dilute solutions of non-volatile solutes whose value depend upon the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.
2. The four colligative properties
• Relative lowering of vapour pressure
• Elevation of boiling point
• Depression of freezing point
• Osmotic pressure

Question 2.
PA = P°A χA
PB = P°B χB
ΔmixV = O

1. Arun said to Subin that it is the condition for a type of solutions.
2. Identify the type of solutions.
3. What are the differences between ideal and non-ideal solutions?

1. Ideal solutions.

2. Vapour pressure of a volatile component in the solution is the product of vapour pressure of pure component and mole fraction of that component in the solution.

3.

Question 3.

1. What are the characteristics of a non-ideal solution?
2. Explain +ve and -ve deviations from Raoult’s law with suitable examples.

1. The characteristics of a non-ideal solution

• Does not obey Raoult’s law over the entire range or concentration. Fora non-ideal solution having two volatile components A and B,
PA ≠ P°A χA
PB ≠ P°B χB
• Volume of mixing not equal to zero, DmixV10
• Enthalpy of mixing not equal to zero, DmixH10

2.

Question 4.
Osmosis and osmotic pressure are two important terms related to solutions.

1. Explain these terms.
2. How will he determine the molar mass of a substance by this method?

1. The phenomenon of the spontaneous flow of a solvent from a solution of lower concentration to higher concentration, separated by a semipermeable membrane is called osmosis.

The excess hydrostatic pressure that builds up when the solution is separated from the solvent by a semipermeable membrane is called osmotic pressure.

2. Osmotic pressure (p) is proportional to the molar concentration/molarity (C) of the solution at a given temperature (T K).
π = CRT, where R is the gas constant.
π = $$\frac{n_{\mathrm{B}}}{V}$$ RT, where nB is the number of moles of the solute and V is the volume of the solution in litres.
π V = nBRT
π V = $$\frac{w_{B}}{M_{B}}$$RT, where WB is the mass of the solute and MB is the molar mass of the solute.
Or MB = $$\frac{\mathrm{W}_{\mathrm{B}} \mathrm{RT}}{\pi \mathrm{V}}$$
Osmotic pressure measurement is widely used to determine molar mass of proteins, polymers and other macro molecules.

Question 5.
The value of molecular mass determined by colligative property measurement is sometimes abnormal.

1. Explain these abnormalities in the case of benzoic acid in benzene and KCl in water.
2. What is van’t Hoff factor?

1. This is caused by dissociation in the case of KCl and association in the case of acetic acid. KCl in aqueous solution undergoes dissociation as KCl → K+ + Cl
Molecules of ethanoic acid (acetic acid) dimerises in benzene due to hydrogen bonding. As a result of dimerisation the actual number of solute particles in solution is decreased. As colligative property decreases molecular mass increases.

Question 6.
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution. This is the commonly used law for expressing the solubility of gas in liquid.

1. Name the law. Write its mathematical expression.
2. What are the factors affecting the solubility of a gas in a liquid? Explain.

1. Roult’s law.
For an ideal solution containing two volatile components A and B,
PA = P°A χA,
PB = P°B χB and
P[Total] = PA + PB = P°A χA + P°B χB

2. The factors affecting the solubility of a gas in a liquid:

• Nature of the gas and the liquid – Each gas has a characteristic solubility in a particular liquid at a particular temperature and pressure.
• Temperature – solubility of a gas in a liquid is an exothermic process. Hence according to Le Chatelier’s principle solubility of gases in liquids decreases with rise in temperature.
• Pressure – According to Henry’s law, solubility of a gas in a liquid is directly proportional to the pressure of the gas.

Question 7.
Concentration of solution may be expressed in different ways.

1. Mention any one of the concentration terms.
2. What are colligative properties?
3. Show that elevation in boiling point is a colligative property.

1. Molarity – It is the number of moles of the solute present in one litre of the solution.

2. Colligative properties are those properties which depends only on the number of solute particles.

3. ΔTb = Kbm
= $$\mathrm{k}_{\mathrm{b}} \frac{\mathrm{n}_{\mathrm{B}} \times 1000}{\mathrm{W}_{\mathrm{A}}}$$
i.e., ΔTb α nB i.e., elevation of boiling point depends on number of moles of solute. Hence, it is a colligative property.

### Plus Two Chemistry Solutions NCERT Questions and Answers

Question 1.
Concentrated nitric acid used in the laboratory work is 68% nitric by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL-1?

Question 2.
Why does gases always tend to be less solube in liquids as the temperature is raised?
Dissolution of gases is an exothermic process. It is because of the fact this process involves decrease of entropy (ΔS < 0). Thus, increase of temperature tends to push the equilibrium,
Gas + Solvent $$\rightleftharpoons$$ Solution; ΔH = -ve
in the backward direction, thereby, supressing the dissolution

Question 3.
The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.
H2O= 12.3 kPa
1000
In 1 molal solution, nsolute = 1; nH2O= $$\frac{1000}{18}$$ = 55.5
∴ χH2O = $$\frac{55.5}{55.5+1}$$
Vapour pressure of the solution, Ps = P°H2O × χH2O
= 0.982 × 12.3 = 12.08 kPa

Question 4.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2 litre of water at 25°C, assuming that it is completely dissociated.
Since K2SO4 is completely dissociated as K2SO4 → 2K+ + SO42- Thus, i = 3
Osmotic pressure of the solution, π = i CRT
$$\frac{3 \times 25 \times 10^{-3} \mathrm{g} \times 0.0821 \mathrm{L} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 298.15 \mathrm{K}}{174 \mathrm{g} \mathrm{mol}^{-1} \times 2 \mathrm{L}}$$
= 5.27 × 10-3 atm

## Plus Two Chemistry Notes Chapter 2 Solutions

Students can Download Chapter 2 Solutions Notes, Plus Two Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus Two Chemistry Notes Chapter 2 Solutions

Solutions:
homogeneous mixtures of two or more pure substances, having uniform composition and properties throughout. The substances forming a solution are called components.

Solvent and Solute:
The component that is present in the largest quantity is known as solvent.

One or more components present in the solution other than solvent are called solutes. e.g. In sugar solution, water is the solvent and sugar is the solute.

Binary solution:
A solution containing only two components.

Aqueous solutions:
solutions in which the solvent is water.

Types of Solutions

Percent Concentration. One way to describe the concentration of a solution is by the percent of the solution that is composed of the solute.

Expressing Concentration of Solutions :
The concentration of a solution is defined as the amount of solute present in the given quantity of the solution.

1. Mass percentage (w/w) :
The mass % of a component in a given solution is the mass of the component (solute) per 100 g of solution.

e.g. 10% glucose solution means 10 g of glucose dissolved in 90 g of water resulting in a 100 g solution.

2. Volume percentage (v/v) :
The volume % of a component in a given solution is the volume of the component per 100 volume of solution.

Example:
10% ethanol solution means 10 mL of ethanol dissolved in 90 mL of water.

3. Mass by volume percentage (w/v):
It is the mass of solute dissolved in 100 mL of the solution. Used in medicine and pharmacy.

4. Parts per million (ppm):
It is the parts of a solute (component) per million parts of the solution. When a solute is present in very minute amounts, parts per million (ppm) is used.

Use of Mole Fraction Equation … This formula is easy to use if you know the number of moles of all the solutes and solvents.

5. Mole fraction (X):
ratio of number of moles of one component to the total number of moles of all the components present in the solution.

For a binary solution, nA be the number of moles of A and nB be the number of moles of B.

The sum of mole fractions of all the components present in the solution is always equal to 1.
i.e., χA + χB = 1
Fora solution containing ‘i’ number of components,
χ1 + χ2 +……………… + χi = 1
Mole fraction is independent of temperature.

6. Molarity (M):
number of moles of solute dissolved in one litre of the solution.

7. Molality (m):
number of moles of solute per kilogram of the solvent.

Solubility:
Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a particular temperature.

Factors affecting solubility
Nature of the solute, nature of the solvent, temperature ‘ and pressure

Solubility of Solids in Liquids :
Like dissolves like:
Polar solutes are soluble in polar solvents and non-polar solutes are soluble in non-polar solvents.

Unsaturated solution:
Solution in which more solute can be dissolved at the same temperature.

Saturated solution:
Solution in which no more solute can be dissolved at the same temperature and pressure.

Effect of temperature :
Solubility increases with temperature if the reaction is endothermic. Solubility decreases with temperature if the reation is exothermic.

Effect of pressure :
Pressure does not have any significant effect on solubility of solids in liquids because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

Solubility of a Gas in a Liquid :
It is greately affected by pressure and temperature.

Effect of pressure
Henry’s law :
The law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.
The most commonly used form of Henry’s law states that the partial pressure of the gas in the vapour phase (p) is proportional to the molefraction of the gas (χ) in the solution.
P = KH
where KH is the Henry’s law constant.
Different gases have different KH values at the same temperature. Thus, KH is a function of the nature of the gas.

Higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid.

The solubility of gases increase with decrease of temperature. Therefore, aquatic species are more comfortable in cold waters rather than in hot waters.

Applications of Henry’s law
1. To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.
2. To avoid bends (a medical condition which is painful and dangerous to life caused by the formation of bubbles of N2 in the blood) the tanks used by scuba divers are filled with air diluted with He (11.7% He, 56.2% N2 and 32.1% O2).
3. At high altitudes, low pressure leads to low concentrations of O2 in blood. It causes climbers to become weak and unable to think clearly (anoxia).

Effect of temperature :
Dissoloution of gases in liquids is an exothermic process. Hence, according to Le Chatelier’s principle solubility of gases in liquids decreases with rise in temperature.

Vapour Pressure of Liquid Solutions

Vapour Pressure of Liquid-Liquid Solutions:
Consider the two volatile liquids denoted as ‘A’ and ‘B’. When both liquids are taken in a closed vessel, both components would evaporate and an equilibrium would be established between liquid and vapour phase.
Let, PA– Partial vapour pressure of component A’
PB – Partial vapour pressure of component ‘B’
χA Mole fraction of A
χB Moiefraction of B

Raoult’s Law :
The law states that fora solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction.
For component ‘A’
PA ∝ χA.
PA= P°A χA
where P°A is the vapour pressure of pure component ‘A’ at the same temperature.
Similarly, for component ‘B’
PB ∝ χB
PB= P°B χB
where PB° is the vapour pressure of pure component ‘B’. Rauolt’s law also states that, at a given temperature for a solution of volatile liquids, the partial vapour pressure of each component is equal to the product of the vapour pressure of pure component and its mole fraction.

According to Dalton’s law of partial pressures,
Total pressure, P[Total] = PA + PB

A plot of PA or PB versus the mole fractions χA and χB for a solution gives a linear plot as shown in the figure.

Raoult’s Law as a special case of Henry’s Law:
According to Raoult’s law, the vapour pressure of volatile liquid in a solution is proportional to its mole fraction, i.e., Pi = Pi° χi

According to Henry’s law, the vapour pressure of a gas in a liquid is proportional to its mole fraction, i. e., p=KHχ

Thus, Raoult’s law becomes a special case of Henry ’s law in which KH becomes equal to Pi°.

Vapour Pressure of Solution of Solids in Liquids:
If a non-volatile solute is added to a solvent to give a solution, the surface of solution has both solute and solvent molecules; thereby the fraction of surface covered by the solvent molecules gets reduced. Consequently, the number of solvent molecules escaping from the surface is reduced. Hence, the vapour pressure of solution is lower than vapour pressure of pure solvent.

General form of Raoult’s Law:
For any solution, the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.

In a binary solution, let us denote the solvent by ‘A’ and solute by ‘B’.
According to Raoult’s law,
PA ∝ χA
PA = PA° χA
Total pressure, P = PA Here, PB = 0
(∵ solute is non-volatile)
P = PA° χA
For binary solution,
χA + χB = 1
χA = 1 – χB
Thus, the above equation becomes,

lowering of vapour pressure.

Ideal and Non-ideal Solutions :
Ideal Solutions:
The solutions which obey Raoult’s law over the entire range of concentrations.

Important properties of Ideal Solutions
i. PA = P°A χA ; PB = P°B χB
ii. Enthalpy of mixing is zero (∆mixH = 0)
iii. Volume of mixing is zero (∆mixV = 0)

If the intermolecular attractive forces between A – A and B – Bare nearly equal to those between A – B, it leads to the formation of ideal solution.

Examples:

1. Solution of n-hexane and n-heptane
2. Solution of bromoethane and chloroethane
3. Solution of benzene and toluene

Non-ideal Solutions :
solutions which do not obey Raoult’s law overthe entire range of concentration. The vapour pressure of such solutions is either higher or lower than that predicted by Raoult’s law.

If the vapour pressure is higher, it exhibits positive deviation and if the vapour pressure is lower it exhibits negative deviation from the Raoult’s law.

Solutions showing positive deviation :
the intermolecular attractive forces between the solute- solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. Thus, in such solutions molecules will find it easier to escape than in pure state. This will increase the vapour pressure and results in the positive deviation.

(dotted line represents graph for ideal solution).
Examples:
Ethanol + Water, Ethanol + Acetone, CCl4 + Chloroform, C6H6 + Acetone , n-Hexane + Ethanol

Solution showing negative deviation:
In the case of negative deviation, the intermolecular attractive forces between solvent-solute molecules are greater than those between solvent-solvent and solute-solute molecules and leads to decrease in the vapour pressure.

Examples:
1. Mixture of phenol and aniline – In this case the intermolecular hydrogen bonding between phenolic proton and lone pair on nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules.
2. Mixture of acetone and chloroform – Here chloroform molecule is able to form hydrogen bond with acetone molecule.

3. H2O + HCl, (4) H2O + HNO3, (5) CHCl3 + (C2H5)2O

Azeotropes:
binary mixtures having same composition in liquid and vapour phase and boil at a constant temperature. It is not possible to separate the components of azeotropes by fractional distillation.

Solutions which show large positve deviation from Raoult’s law form minimum boiling azeotrope at a specific composition. For example, ethanol-water mixture forms a minimum boiling azeotrope (b.p. 351.1 K) when approximately 95% by volume of ethanol is reached.

The solutions that show large negative deviation from Raoult’s law form maximum boiling azeotrope at a specific composition. For example, nitric acid and water form a maximum boiling azeotrope (b.p. 393.5 K) at the approximate composition, 68% nitric acid and 32% water by mass.

Colligative Properties :
properties which depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution. These are,
i. Relative lowering of vapour pressure of the solvent $$\left(\frac{\Delta p_{1}}{p_{1}^{0}}\right)$$
ii. Elevation of boiling point of the solvent (∆Tb)
iii. Depression of freezing point of the solvent (∆Tf)
iv. Osmotic pressure of the solution (π)

Relative Lowering of Vapour Pressure:
When a non-volatile solute (B) is dissolved in a liquid solvent (A), the vapour pressure of the solvent is lowered. This phenomenon is called lowering of vapour pressure. It depends only on the concentration of the solute particles and it is independent of their identity. The relation between vapour pressure of solution, mole fraction and vapour pressure of the solvent is given as,
PA = χAA ……………(1)
The lowering of vapour pressure of solvent ∆ PA is given as,
∆ PA = P°A – PA ……………(2)
Substitute the equation (1) in (2)
∆ PA = P°A – P°AχA
= P°A(1 – χA)
∆ PA = P°AχB …………..(3) ∵ (1 – χA) = χB
The relative lowering of vapour pressure is given as,

of vapour pressure and is equal to the mole fraction of solute.
From equation (4),

For dilute solutions nB < < nA, hence neglecting nB In the denominator, the above equation becomes,

where wA and wB are the masses and MA and MB are the molar masses of solvent and solute respectively.

Elevation of boiling point (∆Tb):
The boiling point of a solution is higher than that of the pure solvent. The elevation in the boiling point depends ‘ on the number of solute molecules rather than on their nature.

Let T°b be the boiling point of pure solvent and Tb be the boiling point of solution. The increase in the boiling point ∆Tb = Tb – T°b is known as elevation of boiling point.

For a dilute solution, the elevation of boiling point ( ∆Tb) is directly proportional to the molal concentration of the solute in a solution (i.e., molality).
∆Tb ∝ m
∆Tb = Kbm …………(1)

where, m → molality and Kb → Boiling Point Elevation Constant/Molal Elevation Constant/ Ebullioscopic Constant.
Unit of Kb is K kg mol-1 Or K m-1

Substituting the value of‘m’ in equation (1),

Depression of Freezing point (∆Tf) :
The lowering of vapour pressure of a solution causes a lowering of the freezing point compared to that of the pure solvent.

Let T°f be the freezing point of pure solvent and Tf be the freezing point of solution.
Depression in freezing point ∆Tf= T°f – Tf
For a dilute solution, depression of freezing point (∆Tf) is directly proportioned to molality (m) of the solution. Thus,
∆Tf ∝ m
∆Tf = Kfm ………………(1)
where, Kf – Freezing Point Depression Constant/ Molal Depression Constant/Cryoscopic Constant.
Unit of Kf is K kg mol-1 Or K m-1

[Note: The values of Kb and Kf, depend upon the nature of the solvent. They Can be ascertained from the following equations:

where,
R → Gas constant, MA → Molar mass of solvent
Tb → Boiling point of pure solvent of kelvin
Tf → Freezing point of pure solvent in kelvin
fusH → Enthalpy of fusion, ∆vapH → enthalpy of vapourisation.
For water, Kb = 0.52 K kg mol-1 and Kf = 1.86 K kg mol-1]

Osmosis and Osmotic Pressure:
The process of flow of the solvent molecules from pure solvent to the solution through semipermeable membrane (SPM) is called osmosis.

Semi Permeable Membrane :
The membrane which allows the passage of solvent molecules but ’ not the solute molecule is called SPM.

Example:
Parchment paper, Pig’s bladder, Cell wall, Film of cupric ferrocyanide.

Osmotic Pressure (π):
the excess pressure which must be applied to a solution to prevent osmosis or the pressure that just stops the flow of solvent.

Osmotic pressure (π) is proportional to the molarity (C) of the solution at a given temperature (T K).
π = CRT, where R is the gas constant.
π = $$\frac{n_{B}}{V}$$RT, where nc is the number of moles of the solute and V is the volume of the solution in litres.
π = nBRT
π V= $$\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}}$$RT , where wB is the mass of the solute and MB is the molar mass of the solute.
Or MB = $$\frac{\mathbf{w}_{\mathrm{B}} \mathrm{RT}}{\pi \mathrm{V}}$$

Osmotic pressure measurement is widely used to determine molar mass of proteins, polymers and other macro molecules.

i) pressure measurement is around the room temperature
ii) molarity of the solution is used instead of molality
iii) the magnitude of osmotic pressure is large compared to other colligative properties even for very dilute solutions.

Isotonic Solution :
Two solutions having same (equal) osmotic pressure at a given temperature. A 0.9% solution of NaCI (normal saline solution) is isotonic with human blood, and it is safe to inject intravenously.

Hypertonic Solution :
A solution having higher osmotic pressure than another solution.
Hypotonic Solution :
A solution having lower osmotic pressure than another solution.

Reverse Osmosis:
flow of the pure solvent from solution side to solvent side through semipermeable membrane when a pressure larger than the osmotic pressure is applied to the solution side.

Uses of reverse osmosis:
Desalination of sea water, Purification of water.

Abnormal Molar Mass :
In some cases, the molar mass determined by colligative properties do not agree with the theoretical values. This is due to association ordissociation of the solute particles in the solution.

Association of Solute Particles :
When solute particles undergo association the number of the solute particles in the solution decreases. Consequently, the experimental values of colligative properties are less than the expected values, e.g. Molecules of ethanoic acid (acetic acid) dimerise in benzene due to intermolecular hydrogen bonding.

Similarly, benzoic acid undergo dimerisation when dissolved in benzene.

Dissociation of Solute Particles :
When the solute particles dissociate or ionise in the solvent, the number of particles in solution increases and so the experimental values of the colligative properties are higher than the calculated values.
e.g. KCl in water ionises as
KCl → K+ + C
Molar mass either lower or higher than the expected or normal value is called as abnormal molar mass.

van’t Hoff factor (i):
It accounts for the extent of association or dissociation.

Significance of van’t Hoff factor.
i > 1 ⇒ there is dissociation of solute particles.
i < 1 ⇒ there is association of solute particles.
i < 1 ⇒ there is no dissociation and association of solute particles.

Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows:
Relative lowering of vapour pressure of solvent,

## Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Students can Download Chapter 5 Surface Chemistry Notes, Plus Two Chemistry Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus Two Chemistry Notes Chapter 5 Surface Chemistry

Surface chemistry deals with phenomena that occurs at the surfaces or interfaces.

accumulation of molecular species at the surface rather than in the bulk of a solid or liquid, it is a surface phenomenon, e.g. Moisture gets adsorbed on silica gel.

molecular species or substance, which accumulates at the surface.

material on the surface of which adsorption takes place, e.g. Charcoal, Silica gel, etc.

Desorption:

the substance is concentrated only at the surface and does not penetrate to the bulk of the adsorbent.

Absorption –
the substance is uniformly distributed throughout the bulk of the solid, e.g. Moisture gets absorbed on anhydrous CaCl2 while adsorbed on silical gel.

Sorption:
term used when both adsorption and absorption take place simultaneously.

The unbalanced or residual attractive forces are responsible for attracting the adsorbate particle on adsorbent surface. During adsorption energy decreases, therefore adsorption is exothermic process, i.e., ∆H of adsorption (heat of adsorption) is always negative. The entropy of the system also decreases (∆S = – ve).

Here the adsorbed molecules are held on the surface of the adsorbent by physical forces such as van der Waals’ forces. It is reversed by reducing pressure or by heating.

Characteristics:
Lack of specificity, easily liquifiable gases readily adsorbed, reversible in nature, extent of adsorption increases with increase in surface area of adsorbent, enthalpy of adsorption quite low (20 – 40 kJ mol’ ).

the forces of interaction between the adsorbent and adsorbate are chemical in nature. It cannot be easily reversed.

Characteristics:
High specificity, irreversibility, increases with increase in surface area, enthalpy of adsorption is high (80 -240 kJ mol”1).

Sometimes physisorption and chemisorption occur simultaneously and it is not easy to ascertain the type of adsorption. A physisorption at low temperature may pass into chemisorption as the temperature is increased. For example, dihydrogen is first adsorbed on Ni by van der Waals’forces. Molecules of hydrogen then dissociate to form hydrogen atoms which are held on the surface by chemisorption.

Comparison of Physisorption and Chemisorption

 Physisorption Chemisorption 1) Arises because of van der Waals’ force 1) Caused by chemical bond formation 2) Not specific 2) Highly specific 3) Reversable 3) Irreversible 4) More easily liquefiable gases are adsorbed readily. 4) Gases which can react with the adsorbent show chemisorption. 5) Enthalpy of adsorption is low (20-40 kJ mol’1) 5) Enthalpy of adsorption is high (80-240 kJ mol-1) 6) Low temperature is favourable. It decreases with increase of temperature 6) Hig temperature is favourable. It increases with increase of temperature 7) No appreciable activation energy is needed. 7) High activation energy is sometimes needed. 8) Increases with an increase of surface area. 8) Increases with an increase of surface area. 9) Results into multimolecular layers on adsorbent surface under high pressure. 9) Results into unimolecular layer

The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve termed as adsorption isotherm.

empirical relation between the quantity of gas adsorbed by unit mass of the solid adsorbant and pressure at a particular temperature.
x/m = k.P1/n (n > 1)
x → mass of the gas adsorbed
‘k’ and ‘n’ are constants which depend on the nature of the adsorbent and the gas at a particular temperature.
OR log x/m = log k + $$\frac{1}{n}$$ log P

Freundlich’s equation approximately describes the behaviour of adsorption from solution.
$$\frac{x}{m}$$ = k.C1/n m
C – equilibrium concentration
log x/m = log k + $$\frac{1}{n}$$ log C
Plotting log x/m vs log C a straight line is obtained

Production of high vacuum, in Gas masks – activated charcoal is filled in gas mask to adsorb poisonous gases, for removal of colouring matter from solution in heterogeneous catalysis, in chromatographies analysis, in froth floatation process.

How to find Kp with the entropy and enthalpy amounts and with Gibb’s Free Energy.

Catalysis :
The process of altering the rate of chemical reaction by the addition of a foreign substance (catalyst) is called catalysis, e.g. MnO2 acts as a catalyst in the thermal decomposition of KClO3.

Promoters:
substances that enhance the activity of a catalyst, e.g. In Haber’s process, iron is used as catalyst and molybdenum acts as a promoter.

Poisons:
substances which decrease the activity of a catalyst.

Homogeneous and Heterogeneous Catalysis
a) Homogeneous Catalysis:
When the reactants and catalyst are in the same phase, the process is said to be homogeneous catalysis.

Here both the reactants and the catalyst are in the liquid phase.

Heterogeneous Catalysis:
If the reactants and the catalyst are in different phase, the catalysis known as heterogeneous catalysis.

Here reactants are gaseous state while the catalysts are in the solid state.

Important Features of Solid Catalysts
a) Activity:
ability of catalysts to accelerate a chemical reaction.

But pure mixture of H2 and O2 does not react at all in the absence of a catalyst.

b) Selectivity:
ability of a catalyst to direct a reaction to yield a particular product.

e.g. CO and H2 combine to form different products by using different catalysts.

Shape Selective Catalysis by Zeolites:
The catalytic reaction that depends upon the pore structure of the catalyst and size of the reactant and the product molecules.

Zeolites are good shape-selective catalysts because of their honey comb-like structures. Zeolites are widely used in petrochemical industries for cracking and isomerisation of hydrocarbon.
e.g. ZSM – 5 – which convert alcohols into petrol.

Enzyme Catalysis:
Enzymes are biological catalysts. They catalyse biological reaction in animals and plants to maintain life. e.g.

1. Invertase – Cane sugar into glucose and fructose
2. Zymase – Glucose into alcohols
3. Maltase – Maltose into glucose
4. Diastase – Starch into maltose
5. Cellulase – Cellulose into glucose
6. Urease – Urea into NH3 and CO2

Characteristics:
Highly efficient, highly specific in nature, highly active under optimum temperature, highly active under optimum pH

Mechanism (Lock and key model)
The molecules of the reactant (substrate), which have complementary shape, fit into the cavities on the surface of enzyme particles just like a key fits into a lock. The enzyme catalysed reactions proceeds in two steps:
Step -1 :
Binding of enzyme to sutbstrate to form an activated complex.
E + S → ES*
Step-2 :
Decomposition of the activated complex to form product.
ES* → E + P

Catalysts in Industry

1. Finely divided iron with molybdenum as promoter in Haber’s process. (New catalyst: a mixture of iron oxide, potassium oxide and alumina)
2. Platinised asbestos in Ostwald’s process
3. Platinised asbestos or V205 in Contact process

Colloids:
Heterogeneous system in which one substance is dispersed (dispersed phase) as very fine particles in another substance called dispersion medium, e.g. Starch, Gelatin. In colloids the particle size (diameter) is between 1nm and 1000 nm.

Classification of Colloids:
i) Based on physical state of dispersed phase and dispersion medium:

ii) Based on Nature of Interaction between Dispersed Phase and Dispersion Medium:
1. Lyophilic (solvent attracting) Colloids:
there is strong interaction between the dispersed phase and dispersion medium. They are reversible sols. e.g. Starch, gelatin, albumin etc.

2. Lyophobic (solvent repelling) Colloids:
there is little or no interaction between the dispersed phase and dispersion medium. They are also irreversible colloids and are not stable.

iii) Based on Types of Particle of the Dispersed Phase
a) Multimolecular Colloids :
the individual particles consist of an aggregate of atoms or small molecules with molecular size less than 1 nm, the particles are held together by van der Waals’ forces, e.g. Sulphur sol, Gold sol etc.

b) Macromolecular Colloids :
the particles of dispersed phase are sufficiently big in size, maybe in the colloidal range, e.g. Starch, cellulose, proteins.

c) Associated Colloids (Micelles):
colloids which behave as normal strong electrolytes at low concentration but get associated at higher concentrations and behaves as colloidal solutions. The associated particle formed are called micelles.
e.g. Soap, detergents etc.

The formation of micelles take place only above a particular temperature called Kraft temperature (Tk.) and above a particular concentration called Critical Micelle Concentration(CMC).

Mechanism of micelle formation –
In soaps, the RCOO ions are present on the surface with their COO groups in water and R staying away from it and remain at the surface. At CMC, the anions are pulled into the bulk of the solution and aggregate to form ‘ionic micelle’ having spherical shape with R pointing towards the centre of the sphere and COO part remaining outward on the surface of the sphere.

Preparation of Colloids
a) Chemical Methods
Some examples:

b) Electrical Disintegration or Bredig’s Arc Method
Metallic sols can be prepared by striking an arc between two electrodes of the metal, immersed in the dispersion medium. The metal is vapourised by the arc which then condenses to form particles of colloidal size. e.g. Gold sol, Platinum sol, Silver sol etc.

c) Peptization:
process of converting a precipitate into colloidal sol by shaking it with dispersion medium in the presence of small amount of electrolyte (peptizing agent), e.g. Freshly prepared Fe(OH)3 is peptized by adding small quantity of FeCI3 solution (peptizing agent).

Mechanism of peptization –
During peptization, the precipitate adsorbs one of the ions of the electrolyte on its surface. This causes the development of positive or negative charge on precipitates, which ultimately break up into smaller particles of the size of a colloid.

Purification of Colloids:
process of reducing the amount of impurities to a requisite minimum from the colloids.
i) Dialysis:
process of removing a dissolved substance from a colloid by means of diffusion through a suitable membrane.

ii) Electro-dialysis:
process of dialysis in presence of an applied electric field. It is faster and is applicable if the dissolved substance in the impure colloid is only an electrolyte. The ions present in the colloid migrate out to the oppositely charged electrodes.

iii) Ultrafilteration:
process of separating the colloidal particles from the solvent and soluble solutes present in the colloid by ultra filters. The ultra filter paper is prepared by soaking the filter paper in a colloidion solution (4% solution of nitro cellulose in a mixture of alcohol and ether). It is then hardened by formaldehyde and finally dried.

Properties of Colloids
1) Colligative Properties:
values of colligative properties as smaller due to smaller number of particles.

2) Tyndall Effect (Optical Property):
phenomenon of the scattering of light by colloidal particles.

Conditions for observing Tyndall effect:
1. The diameter of the dispersed particles is not much smaller than the wavelength of the light used; and

2. The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude. The ultramicroscope used to observe the light scattered by colloidal particles is based on Tyndall effect.

The colour of the sky can be explained by Tyndall effect. The dust and other colloids present in the atmosphere scatter the light. Only blue light reaches to our eyes.

3) Colour:
It depends on the wavelength of lighty scattered by the dispersed particles which in turn depends on the size and nature of the particles and changes with the manner in which the observer receives the light, e.g. a mixture of milk and water appears blue when viewed by the reflected light and red when viewed by transmitted light.

4) Brownian Movement:
The constant zig-zag movement of the colloidal particles. It is due to the unbalanced bombardment of the particles by the molecules of the dispersion medium. It does not permit the particles to settle and is responsible for the stability of sols. ,

5) Charge on Colloidal Particles:
Colloidal particles carry an electric charge.
Positive charged sols: Al2O3. xH2O, CrO3.xH20, basic dye stuffs, blood (Haemoglobin) etc.

Negatively charged sols:
Metal sols (Cu, Ag, Au), metallic sulphides, acid dyes stuffs, starch, gelatin.

Reason for charge:
It is due to
i) electron capture by sol particles during electrodispersion of metals,
ii) preferential adsorption of ions from solution and/ or
iii) formulation of electrical double layer.

Helmholtz Electrical Double Layer:
combination of two layers of opposite charges around the colloidal particle. The first layer of ions is firmly held and is termed fixed layer while the second layer is mobile which is termed as diffused layer.

Electrokinetic Potential or Zeta Potential:
It is the potential difference between the fixed layer and the diffused layer of opposite changes in the electrical ‘ double layer.

Significance of Charge on Colloidal Particles:
provides stability to the colloid because the repulsive forces between charged particles having same charge prevent them from coalescing or aggregating when they come closer to one another.

6) Electrophoresis:
lled anaphoresis and that of cathode is called cataphoresis.

Coagulation/Flocculation/Precipitation:
process of settling of colloidal particles by the addition of electrolyte.

Coagulation of lyophobic sols can be carried out by the following ways:
Electrophoresis, mutual coagulation (mixing two oppositely charged sols), boiling, persistent dialysis, addition of electrolytes, etc.

Colloids interact with ion carrying charge opposite to that present on themselves. This causes neutralisation leading to their coagulation.

Hardy – Schulze Rule:
the greater the valence of the flocculating ion added, the greater is its power to cause precipitation.

The ion having opposite charge to sol particles (coagulating ion) cause coagulation.

In the coagulation of negative sol, the flocculating power is in the order: Al3+ > Ba2+ > Na+

In the coagulation of positive sol, the flocculating power in the order:
[Fe(CN)6]4- > PO43- > SO42-> Cl

Protective Colloids:
the lyophilic sol used for protection of lyophobic sol.

Emulsions:
liquidin liquid colloidal systems i.e., the dispersion of finely divided droplets in another liquid. There are two types of emulsions.
1) Oil dispersed in water (O/W type):
water acts as dispersion medium, e.g. Milk, Vanishing cream.

2) Water dispersed in oil (W/O type):
oil, acts as dispersion medium.e.g. Butter, Creams, Cod liveroil Emulsification – process of making an emulsion. Emulsion may be obtained by vigourously agitating a mixture of both liquids.

Emulsifying agent or emulsifier –
substance used to stabilise an emulsion. It forms an interfacial film between suspended particles and the medium, e.g.

Emulsifying agents for O/W emulsions :
Proteins, gums, natural and synthetic soaps etc.

Emulsifying agents for W/O emulsions:
Heavy metal salts of fatty acids, long chain alcohols, lampblack etc.

Colloids Around Us :
Fog, mist and rain; food materials, blood, soils, formation of delta.

Application of Colloids
I) In Medicine:
Colloidal medicines are more effective because they have large surface area and are, therefore, easily assimilated, e.g. Colloidal silver (Argyrol) – as eye lotion, Colloidal antimony – in curing Kalaazar, Colloidal gold – for intramuscular injection. Milk of magnesia – in stomach disorder.

II) In industries :
Electrical precipitation of smoke – by Cottrell smoke precipitator, purification of water, tanning, cleansing action of soaps and detergents (micelle formation), photographic plates and films, rubber industry and Industrial products.

## Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting

Students can Download Chapter 2 National Income Accounting Questions and Answers, Plus Two Economics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations

## Kerala Plus Two Economics Chapter Wise Questions and Answers Chapter 2 National Income Accounting

### Plus Two Economics National Income Accounting One Mark Questions and Answers

Question 1.
GNP – depreciation is called
(a) GDP
(b) NNP
(c) PCI
(d) PI
(b) NNP

Question 2.
The GDP deflator is equal to
i) Real GDP-Nominal GDP

iii) $$\frac{{ No minal GDP }}{\text { Real GDP }} \times 100$$

Question 3.
NFIA is included in:
(a) NNPFC
(b) NDPFC
(c) GDPFC
(d) All the above
(a) NNPFC

Question 4.
Which among the following in a flow concept?
(a) export
(b) wealth
(c) capital
(d) foreign exchange reserve
(a) export

Question 5.
When does net factor income from abroad become negative?
(a) NDP < NNP
(b) NNP < NDP
(c) NDP = NNP
(d) none of the above
(b) NNP < NDP

Question 6.
When does GDP and GNP of an economy become equal?
(a) When net factor income from abroad is positive
(b) When net factor income from abroad is negative
(c) When net factor income from abroad is zero
(d) None ofthe above.
(c) When net factor income from abroad is zero

### Plus Two Economics National Income Accounting Two Mark Questions and Answers

Question 1.
Same job is done by a servant and housewife, whose service is included in the national income calculation? Why?
Service of a servant is included in the national income calculation, whereas, the service of housewife is not included in the national income. This is because the housewife is not paid for the service she does.

Question 2.
From the following, classify the material into final goods and intermediary goods. Wheat, Bench, Bread, Wood, Rubber, Tyre.

 Final Goods Intermediary goods Bench Wheat Bread Wood Tyre Rubber

Question 3.
Distinguish between real flow and money flow?
Flow of goods and services from firms to households is called real flow. Factors of production receive reward for their services in the form of money. Households use this money to buy goods and services produced by firms. This flow of money from firms to households and back to firms is called money flow.

Question 4.
Some variables are given below. Classify them into Stock and Flow

1. Wealth
2. Income of a household
3. Consumption
4. Capital
5. Money Supply
6. Capital formation
7. Inventories
8. Saving of a household

a. Stock

• Wealth
• Inventories
• Capital
• Money supply

b. Flow

• Income of a household
• Consumption
• Capital formation
• Saving of a household

Question 5.
GDP = C + I + G + (X – M) = C + S + T Derive the Budget Deficit and Trade Deficit equations from the above identity.
GDP = C + I + G + (X – M) = C + S + T
Budget deficit = G – T
Trade deficit = M – X

### Plus Two Economics National Income Accounting Three Mark Questions and Answers

Question 1.
“Transfer payments are not included in the national income calculation”. Do you agree? Justify your answer.
Yes. Transfer payments like pension, old age pension, etc. are not included in the national income. This is because they are transfer earnings not generated by any economic activity. These payments are usually made by the government out of tax revenue collected from the public. Since these generated incomes are already included in national income calculation there is no need to include transfer payment in the national income calculation again.

Question 2.
State whether the following are included or excluded in the national income.

1. purchase of second hand goods
2. operating surplus
3. production for self-consumption
4. interest
5. windfall gains and loses

1. Purchase of second hand goods – excluded
2. operating surplus – included
3. old age pension – excluded
4. Production for self consumption – excluded
5. interest – included
6. windfall gains and loses – excluded

Question 3.
Provide appropriate term.

2. GNP
3. NNP
4. NNPFC

Question 4.
Point out any 3 uses of national income accounting.
The uses of national income accounting are given below.

1. It shows the distribution of national income among the various factors of production.
2. National income statistics indicate the contribution of different sectors in the economy.
3. Structural changes in the economy can be assessed by the national income accounting.

Uniform Distribution Calculator is an online tool that helps to calculate the probability distribution for the given values.

Question 5.
Classify the following under proper heads.
Flow of teacher services, Flow of subsidies and taxes, Flow of factor rewards, flow of finished goods, Flow of consumption expenditure, Flow of import goods.

 Real Flow Money Flow Flow of teacher services Flow of subsidies and taxes Flow of finished goods Flow of factor rewards Flow of import goods Flow of consumption

Question 6.

• Does not includes prices of imported goods
• Weights are different
• It includes all goods and services
• Includes prices of imported goods
• Weights are constant
• Does not include all goods and services

a. Consumer price index

• Includes prices of imported goods
• Weights are constant
• Does not include all goods and services

b. GDP deflator

• Does not include prices of imported goods
• Weights are different
• It includes all goods and services

Question 7.
Assume that there are three goods produced in an economy and they are sold at different prices in dif-ferent years. Calculate GDP Deflator.

Question 8.
Calculate Depreciation, Net Indirect Tax and NNPFC from the below data.
GDPMP = 11300
NDPMP = 10300
NDPFC = 10000
NFIA = 1500
1. Depreciation = GDPMP – NDPMP
= 11300 – 10300
= 1000

2. Net Indirect tax = NDPMP – NDPFC
= 10300 – 10000 = 300

3. NNPFC = NDPFC + NFIA
= 10000 + 1500
= 11500

### Plus Two Economics National Income Accounting Five Mark Questions and Answers

Question 1.

1. GNP, NNP, CSO, GDP
2. Salary, bonus, GPF, free housing, saving
3. Smuggling, production of wheat, sale of second-hand goods, services of housewives
4. Services of teacher, services of engineer, services of lawyer, services of housewife
5. Unemployment allowances, scholarships, old age pension, support price.

1. C.S.O. Others are national income concepts.
2. Saving. Others come under compensation to employees
3. Production of wheat. Others are excluded from national income
4. Services of housewife. Others are included in the national income calculation.
5. Support price. Others are transfer payments.

Question 2.
Match the following.

 A B NNP GDP – net factor income from abroad GNP Personal income – direct taxes Value added GNP-depreciation GDP at market prices value of output – intermediate consumption Disposable income GDP at factor cost – net indirect tax

 A B NNP GNP – depreciation GNP GDP – net factor income from abroad Value added Value of output- intermediate consumption GDP at market prices GDP at factor cost – net indirect tax Disposable income Personal income – direct taxes

Question 3.
Categorize the following into stocks and flows, wealth, salary, food grain stock, foreign exchange reserves, export, gross domestic saving, capital, change in money supply, quantity of money, capital formation.

 Stock Flow Wealth Export Foreign exchange reserves Salary Food grain stock Gross domestic saving Capital Change in money supply Quantity of money Capital formation

Question 4.
The phase of circular flow of income in a two sector economy is given below.

1. Complete the diagram.
2. Explain the process of circular flow

1.

2. Circular flow of income:
The concept that the aggregate value of goods and services produced in an economy is going around in a circular way. Either as factor payments, or as expenditures on goods and services, or as the value of aggregate production.

Question 5.
Suppose that in a two sector economy the value of finished goods is equal to ₹100 crore and the income generated as factor rewards is also equal to ₹100 crore. The households spend only ₹80 crore.

1. What will happen to the circular flow?
2. Which system can be introduced to correct the circular flow?
3. Name the leakages and injections.

1. There will be a mismatch between the real flow and money flow in the circular flow. In other words, the flow will be broken.
2. As a corrective measure, the financial system can be introduced.
3. The leakages is the difference between the income generates and household spending.

This is saving. The injection are the savings that the households, firms and the government take from the financial institutions as borrowings.

Question 6.
1. Estimate the NI of India and Pakistan from the data given below.

2. Which method is used here?
3. What are the other methods of measuring national income?

1. National income of India = ₹2885 crore
National income of Pakistan = ₹1860 crore
2. The method used here is the product method or value added method.
3. Income method and expenditure method are the other two method of measuring national income.

Question 7.
What do you mean by GDP deflator? How far GDP deflator differs from Consumer Price Index?
The ratio of nominal to real GDP is a well known index of prices. This is called GDP Deflator. GDP deflator differs from Consumer Price Index. The major points of difference are given below.

1. The goods purchased by consumers do not represent all the goods which are produced in a country. GDP deflator takes into account all such goods and services.

2. CPI includes prices of goods consumed by the representative consumer; hence it includes prices of imported goods. GDP deflator does not include prices of imported goods.

3. The weights are constant in CPI – but they differ according to production level of each good in GDP deflator.

Question 8.
Write down some of the limitations of using GDP as an index of welfare of a country.
GDP is the sum total of value of goods and services created within the geographical boundary of a country in a particular year. It gets distributed among the people as incomes. So we may be tempted to treat higher level of GDP of a country as an index of greater well-being of the people of that country. But there are at least three reasons why this may not be correct. They are discussed below.

1. Distribution of GDP – how uniform is it:
If the GDP of the country is rising, the welfare may not rise as a consequence. This is because the rise in GDP may be concentrated in the hands of very few individuals or firms. For the rest, the income may, in fact, have fallen.

In such a case the welfare of the entire country cannot be said to have increased. If we relate welfare improvement in the country to the percentage of people who are better off, then surely GDP is not a good index.

2. Non-monetary exchanges:
Many activities in an economy are not evaluated in monetary terms. For example, the domestic services women perform at home are not paid for. The exchanges which take place in the informal sector without the help of money are called barter exchanges.

This is a case of underestimation of GDP. Hence GDP calculated in the standard manner may not give us a clear indication of the productive activity and well-being of a country.

3. Externalities:
Externalities refer to the benefits (or harms) a firm or an individual causes to another for which they are not paid (or penalized). Externalities do not have any market in which they can be bought and sold. Therefore, if we take GDP as a measure of welfare of the economy we shall be overestimating the actual welfare.

This was an example of negative externality. There can be cases of positive externalities as well. In such cases, GDP will underestimate the actual welfare of the economy.

Question 9.
Assume that GDP in the year 2007 was ₹1,200 which rose to ₹1,800 in 2008. Calculate GDP deflator.
GDP deflator = Current year GDP / Base year GDP x 100
= 1800/1200 × 100
= 1.5 × 100
= 1.5 (in percentage terms 150)

Question 10.
Relate and complete the identities/equations in column A with column B.

Question 11.
Estimate the Gross National Product at market price and GNP at factor cost through the expenditure method.

 Item Amount (in Crores) Inventory investment 15 Net factor income from abroad 10 Personal consumption expenditure 475 Gross residential construction investment 48 Exports 25 Government purchase of goods and services 175 Gross public investment 15 Gross business fixed investment 38 Imports 12 Net indirect tax 8

GNPMP = private consumption expenditure + govt, final consumption expenditure( gross fixed capital formation + change in stock or inventory investment) + net export + net factor income from abroad
= 475 + 175 + 101 (i.e., 48 + 15 + 38) + 15 + 13
= ₹779 crores.
GNPC = GNPUD – net indirect taxes
= 779 – 8 = ₹771 crores

Question 12.
Suppose that in a two sector economy, the value o finished goods is equal to ₹200 crore and the income generated as factor rewards is equal to ₹200 crore. The households spend only ₹180 crore. The remaing 20 crore economy saved then.

1. Is ₹20 (saving) included in the circular flow?
2. Which system can be introduced to correct the circular flow?
3. Is saving leakage or injection.

1. No, saving (₹20) is excluded in the circular flow.
2. Financial system can be introduced to correct the circular flow.
3. Yes, saving is a leakage.

Question 13.
Fill in the blanks

1. GNPMP – ……….. = NNPMP
2. NNPMP – ………… = NNPFC
3. GDPFC+ – ………… = GDPMP
4. GDP + -………….. = GNP

1. GNPMP – depreciation = NNPMP
2. NNPMP – net indirect tax = NNPFC
3. GDPFC + net indirect tax = GDPMP
4. GDP + net factor income from aborad = GNP

Question 14.
Write down the 3 identities of calculating the GDP of a country by the 3 methods. Also briefly explain why each of those should give us the same value of GDP.
Gross National Product (GNP) equals Gross National Income equals Gross National Expenditure, i.e.
GNP = GNI = GNE
These are equal because national income is a circular flow of income. Aggregate expenditure is equal to aggregate output which in turn, is equal to aggregate income. However each method has some different items, yet they show exactly identical results.

Their identity can be shown in the following manner:
Reconciling Three Methods of Measuring Gross

Question 15.
The economic recession of 2008 affected the market economics in general and the US in particular. Thou-sands of Indians working abroad lost their job especially in IT and banking sectors and they returned to India. Evaluate its consequences on Indian economy with regard to the following macro variables.

1. The value of GNP
2. Gneral unemployment level
3. Foreign exchange rate

1. The value of GNP decreases due to reduction in NFIA.
2. General unemployment level increases.
3. Foreign exchange rate increases.

### Plus Two Economics National Income Accounting Eight Mark Questions and Answers

Question 1.
Given below some macro economic indicators. Derive the equations of the following terms:

1. GNP
2. NNP
3. NNP at factor cost
4. Personal income
5. Personal disposable income
6. Private Income
7. National Disposable Income

1. GNP = GDP + Factor income earned by the domestic factors of production employed in the rest of the world – Factor income earned by the factors of production of the rest of the world employed in the domestic economy

2. NNP = GNP – Depreciation

3. NNP at factor cost = National Income (NI) = NNP at market prices – (Indirect taxes – Subsidies)

4. Personal income (PI) = NI – Undistributed profits – Net interest payments made by households – Corporate tax + Transfer payments to the households from the government and firms.

5. Personal Disposable Income (PDI) = PI – Personal tax payments – Non-tax payments.

6. Private Income = Factor income from net domestic product accruing to the private sector + National debt interest + Net factor income from abroad + Current transfers from government + Other net transfers from the rest of the world

7. National Disposable Income = Net National Product at market prices + other current transfers from the rest of the world

Question 2.
Prepare a seminar report on the topic ‘Measurement of National Income’.
Measurement of National Income Respected teachers and dear friends,
The topic of my seminar paper is ‘measurement of national income or the methods of measuring national income’. The concept of national income occupies an important place in economic theory.

National income is the aggregate money value of all goods and services produced in a country during an accounting year. In this seminar paper, I would like to present various methods of measuring national income.

Content:
National income can be measured in different ways. Generally there are three methods for measuring national income. They are

2. Expenditure method
3. Income method

The term that is used to denote the net contribution made by a firm is called its value-added. We have seen that the raw materials that a firm buys from another firm which are completely used up in the process of production are called ‘intermediate goods’.

Therefore the value-added of a firm is the value of production of the firm – value of intermediate goods used by the firm. The value-added of a firm is distributed among its four factors of production, namely, labor, capital, entrepreneurship, and land.

Therefore wages, interest, profits, and rents paid out by the firm must add up to the value-added of the firm. Value-added is a flow variable.

2. Expenditure Method:
An alternative way to calculate the GDP is by looking at the demand side of the products. This method is referred to as the expenditure method. The aggregate value of the output in the economy by expenditure method will be calculated.

In this method we add the final expenditures that each firm makes. Final expenditure is that part of expenditure which is undertaken not for intermediate purposes.

3. Income Method:
As we mentioned in the beginning, the sum of final expenditures in the economy must be equal to the incomes received by all the factors of production taken together (final expenditure is the spending on final goods, it does not include spending on intermediate goods).

This follows from the simple idea that the revenues earned by all the firms put together must be distributed among the factors of production as salaries, wages, profits, interest earnings, and rents.
That is GDP = W + P + In + R

Conclusion:
Thus it can be concluded that there are three methods for measuring national income. These methods are value-added method, income method and expenditure method. Usually in estimating national income, different methods are employed for different sectors and sub sectors.

Question 3.
From the following data, calculate personal income and personal disposable income (₹in Crores).

1. NDPFC – 8,000
2. net factor income from abroad – 200
3. Undistributed profit – 1,000
4. Corporate tax – 500
5. Interest received by households – 1,500
6. Interest paid by households – 1,200
7. Transfer income – 300
8. Personal Tax – 500

Personal income = NDPfc + Net factor income from abroad – undistributed profits – corporate taxes + transfer payments + net interest received from households.
= 8000 + 200-1000 – 500 + 300 (1500 -1200)
= 7,300 crores
Personal disposable income = Personal income – personal tax
= 7,300 – 500 = 6,800 crores

Question 4.
Production generates income. Prove this statement with the help of a simple two sector model of circular flow of income.
circular flow of income:
It is a pictorial representation of interdependence or interrelationship between the various sectors of the economy. It is a concept associated with income earning and spending. The circular flow of income in a simple economy works on the basis of certain assumptions.
They are as follows:

1. Households and firms are the only two sectors in an economy (2 sector model)
2. Households supply factor services to firms.
3. Firms hire factor services households
4. Household spends their entire income on consumption and thereby no savings are left with them.
5. Firms sell their entire products to the households
6. There is no government in the economy.
7. The economy is not related to any other economies or the economy is a ‘closed’ system. As a result, there is no export or imports from the economy.

In such an economy, there would be two types of markets.
They are:

1. product-market for goods and services
2. factor markets for buying and selling various factor services.

The relationship between the sectors of an economy can be explained with the help of a diagram.

The households own the factors of production such as land, labour, capital, and organization. The households sell these factors of production to the firms for producing goods and services are known as real flow. The rewards for factors of production are rent to land, interest to capital, wage to the labour and profit to the entrepreneur is known as the money flow.

## Plus Two Computer Application Chapter Wise Questions and Answers Chapter 1 Review of C++ Programming

Students can Download Chapter 1 Review of C++ Programming Questions and Answers, Plus Two Computer Application Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

## Kerala Plus Two Computer Application Chapter Wise Questions and Answers Chapter 1 Review of C++ Programming

### Plus Two Computer Application Review of C++ Programming One Mark Questions and Answers

Question 1.
IDE means ______
IDE _______.
Integrated Development Environment.

Question 2.
We know that C++ is a high level language. From the following which statement is true.
(a) C++ contains English like statements.
(b) C++ contains mnemonics
(c) C++ contains only 0 and 1
(d) None of these
(a) C++ contains English like statements.

Question 3.
C++ is a_____language.
(a) High level
(b) Low level
(c) Middle level
(d) None of these
(a) High level.

Question 4.
C++ was developed at_______.
(a) AT & T Bell Laboratory
(b) Sanjose Laboratory
(c) Kansas University Lab
(d) None of these
(a) AT & T Bell Laboratory.

Question 5.
C++ is a successor of______language.
(a) C#
(b) C
(c) Java
(d) None of these
(b) C.

Question 6.
The most adopted and popular approach to write programs is_______.
Structured programming.

Question 7.
From the following which uses OOP concept.
(a) C
(b) C++
(c) Pascal
(d) Fortran
(b) C++.

Question 8.
_____is the smallest individual unit.
Token

Question 9.
Pick the odd one out
(a) float
(b) void
(c) break
(d) Alvis
(d) Alvis, the others are keywords.

Question 10.
Reserved words for the compiler is_______.
(a) Literals
(b) Identifier
(c) Keywords
(d) None of these
(c) Key words.

Question 11.
Pick an identifier from the following.
(а) auto
(b) age
(c) float
(d) double
(b) age.

Question 12.
Pick the invalid identifier
(a) name
(b) Date of birth
(c) age
(d) joining_time
(b) Date of birth, because it contains space.

Question 13.
Pick the octal integer from the following.
(a) 217
(b) 0 × 217
(c) 0217
(d) None of these
(c) 0217, an octal integer precedes 0.

Question 14.
Pick the hexa decimal integer from the following.
(а) 217
(b) 0 × 217
(c) 0217
(d) None of these
(b) 0 × 217, an hexa decimal integer precedes Ox.

Question 15.
From the following pick a character constant.
(a) ’A’
(b) ‘ALL’
(c) ‘AIM’
(d) None of these
(a) ‘A’, a character enclosed between single quote.

Question 16.
Non graphic symbol can be represented by using______
Escape Sequence.

Question 17.
Manish wants to write a program to produce a beep sound. Which escape sequence is used to get an alert (sound).
(a) \a
(b) \d
(c) \s
(d) None of these
(a) \a.

Question 18.
Ajo wants to print a matter in a new line. Which es-cape sequence is used for this?
(a) \a
(b) \n
(c) \s
(d) None of these
(b) \n.

Question 19.
To represent null character______is used
(a) \n
(b) \0
(c) \f
(d) \s
(b) \0.

Question 20.
State True/False a string is automatically appended by a null character.
True.

Question 21.
From the following pick a string constant.
(a)  ‘a’
(b) “abc”
(c) ‘abc’
(d) None of these
(b) “abc”, a character constant must be enclosed between double quotes.

Question 22.
C++ was developed by______
(a) Bjarne Stroustrup
(b) James Gosling
(c) Pascal
(d) None of these
(a) Bjarne Stroustrup.

Question 23.
From the following which is not a character constant.
а) ‘c’
b) ‘e’
c) ‘d’
d) “c”
(d) “c”, It is a string constant the others are character constant.

Question 24.
From the following which is a valid declaration.
(a) int 91;
(b) int x;
(c) int 9x;
(d) int “x”;
(b) int x;

Question 25.
Symbols used to perform an operation is called_______
(a) Operand
(b) Operator
(c) Variable
(d) None of these
(b) Operator.

Question 26.
Consider the following. C = A + B. Here A and Bare called______
(a) Operand
(b) Operator
(c) Variable
(d) None of these
(b) Operand.

Question 27.
The. execution of a program starts at_______function.
main()

Question 28.
The execution of a program ends with______function
main()

Question 29.
______USed to write single line comment
(a) //
(b) /*
(c) */
(d) None of these
(a) //

Question 30.
const k = 100 means const k = 100
(a) const float k = 100
(b) const double k = 100
(c) const int k = 100
(d) const chark = 100
(c) const int k = 100

Question 31.
Each and every statement in C++ must be end with_____
(а) Semi colon
(b) Colon
(c) full stop
(d) None of these
(a) Semi colon

Question 32.
From the following select the input operator.
(a) >>
(b) <<
(c) >
(d) <
(a) >>

Question 33.
From the following select the output operator.
(a) >>
(b) <<
(c) >
(d) <
(b) <<

Question 34.
In while loop, the loop variable should be updated?
(a) along with while statement
(b) after the while statement
(c) before the while statement
(d) inside the body of while
(d) Inside the body of while

Question 35.
Adeline wrote a C++ program namely sum.cpp and she compiled the program successfully with no error. Some files are generated. From the following which file is a must to run the program
(a) sum.exe
(b) sum.obj
(c) sum.vbp
(d) sum.htm
(a) sum.exe

Question 36.
Adeline wrote a C++ program namely sum.cpp and she compiled the program successfully with no error. Some files are generated namely sum.obj and sum.exe. From this which file is not needed to run the program
sum.obj is not needed and can be deleted.

Question 37.
To terminate a program, from the following which is used.
(a) break
(b) continue
(c) end()
(d) exit()
(d) exit()

Question 38.
To write a C++ program, from the following which statement is a must.
(a) sum()
(b) main()
(c) #include
(d) #include
(b) main(). A C++ program must contains at least one main() function.

Question 39.
State True / False. Comment statements are ignored by the compiler
True.

Question 40.
More than one input / output operator in a single statement is called______.

Question 41.
From the following which is ignored by the compiler.
(a) statement
(c) loops
(d) None of these

Question 42.
From the following which is known as string terminator
(a) ‘\0’
(b) ‘\a’
(c) *\s*
(d) ‘\t’
(a) ‘\0’

Question 43.
______is the main activity carried out in computers.
Data processing.

Question 44.
The data used in computers are different. To differentiate the nature and size of data______is used.
Data types.

Question 45.
Classify the following data types.

1. int
2. array
3. function
4. char
5. pointer
6. void
7. float
8. double
9. structure

 Fundamental data types Derived data types int array float function double pointer void structure char

Question 46.
Sheela wants to store her age. From the following which is the exact data type.
(a) void
(b) char
(c) int
(d) double
(c) int

Question 47.
Integer data type uses_____bytes of memory
(a) 5
(b) 2
(c) 3
(d) 4
(b) 2

Question 48.
char data type uses______bytes of memory
(a) 1
(b) 3
(c) 7
(d) 8
(a) 1

Question 49.
From the following which data type uses 4 bytes of memory
(a) float
(b) short
(c) char
(d) double
(a) float

Question 50.
Full form of ASCII is______.
American Standard Code for Information Interchange.

Question 51.
Ramu wants to store the value of From the following which is correct declaration
(а) char pi = 3.14157
(b) int pi = 3.14157
(c) float pi = 3.14157
(d) long pi = 3.14157
(c) float pi = 3.14157.

Question 52.
From the following which is not true, to give a variable name.
(a) Starting letter must be an alphabet
(b) contains digits
(c) Cannot be a key word
(d) special characters can be used
(d) special characters can be used

Question 53.
Pick a valid variable name from the following
(а) 9a
(b) float
(c) age
(d) date of birth
(c) age

Question 54.
To perform a unary operation how many number of operands needed?
(a) 2
(b) 3
(c) 1
(d) None of these.
(c) 1 (Unary means one)

Question 55.
To perform a binary operation how many number of operands needed?
(a) 2
(b) 3
(c) 1
(d) None of these.
(a) 2 (binary means two)

Question 56.
To perform a ternary operation how many number of operands needed?
(a) 2
(b) 3
(c) 1
(d) None of these.
(b) 3(eg: ternary means three)

Question 57.
In C++ 13 % 26 =_______
(a) 26
(b) 13
(c) 0
(d) None of these
13. % is a mod operator i.e. it gives the remainder. Here the remainder is 13.

Question 58.
In C++ 41/2 =______
(a) 20.5
(b) 20
(c) 1
(d) None of these
(b) 20. (The actual result is 20.5 but both 41 and 2 are integers so .5 must be truncated).

Question 59.
++ is a_____operator
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
(a) Unary.

Question 60.
Conditional operator is______operator
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
(c) Ternary

Question 61.
% is a_____operator
(a) Unary
(b) Binary
(c) Ternary
(d) None of these
(b) Binary

Question 62.
State True/False

1. Multiplication, division, modulus have equal priority
2. Logical and (&&) has less priority than logical or ()

1. True
2. False

Question 63.
______is composed of operators and operands
(a) expression
(b) Key words
(c) Identifier
(d) Punctuators
(a) expression

Question 64.
Supply value to a variable at the time of declaration is known as______.
Initialisation.

Question 65.
From the following which is initialisation
(a) int k;
(b) int k = 100;
(c) int k[10];
(d) None of these
(b) int k= 100;

Question 66.
State True/False
In an expression, aH the operands having lower size are converted(promoted) to the data type of the highest sized operand.
True

Question 67.
Classify the following as arithmetic / Logical expression.
(a) x + y * z
(b) x < y && y > z
(c) x/y
(d) x > 89 || y < 80
(a) and (c) are Arithmetic
(b) and (d) are Logical

Question 68.
Suppose x = 5 and y = 2 then what will be count << (float) x/y.
2.5 The integer x is converted to float hence the result.

Question 69.
Considerthe following ,
a = 10; a* = 10;
Then a =______
(a) a = 100
(b) a = 50
(c) a = 10
(d) a = 20
(a) a = 100, This short hand means a = a * 10

Question 70.
Consider the following a = 10; a+ = 10; Then a =_____
(a) a = 30
(b) a = 50
(c) a = 10
(d) a = 20
(d) a = 20. This short hand means a = a + 10.

Question 71.
Pick the odd one out
(a) structure
(b) Array
(c) Pointer
(d) int
(d) int, it is fundamental data type the others are derived data types.

Question 72.
From the following select not a character of C++ language
(a) A
(b) 9
(c) \
(d) @
(d) @

Question 73.
Consider the following float x = 25.56; cout << (int)x;
Here the data type of the variable is converted. What type of conversion is this?
(a) type promotion
(b) type casting
(c) implicit conversion
(d) None of these
(b) type casting (explicit conversion);

Question 74.
From the following which is ignored by the compiler
(a) statement
(c) loops
(d) None of these

Question 75.
Multi line comment starts with____and ends with_____
(a) /’ and ‘/
(b) */ and /*
(c) /* and */
(d) ‘/ and /’
(c) /* and */

Question 76.
Single line comment starts with_____
(a) **
(b) @@
(c) */
(d) //
(d) //

Question 77.
Alvin wants to store the value of π From the following which is correct declaration
(a) char pi = 3.14157
(b) const int pi = 3.14157
(c) const float pi = 3.14157
(d) long pi = 3.14157
(c) const float pi = 3.14157

Question 78.
To store 70000 which modifier is used with int.
(a) long
(b) short
(c) big
(d) none of these
(a) long

Question 79.
To store 60000 which modifier is used with int.
(a) unsigned
(b) short
(c) big
(d) none of these
(a) unsigned

Question 80.
Consider x++(post fix form). Select the correct definition from the following
(a) The operation is performed after the value is used
(b) The operation is performed before the value is used
(c) First change then use
(d) None of these
(a) The operation is performed after the value is used

Question 81.
Consider ++x(pre fix form). Select the correct definition from the following
(a) The operation is performed after the value is used
(b) The operation is performed before the value is used
(c) First use then change
(d) None of these
(b) The operation is performed before the value is used

Question 82.
Consider the following int a = 10; float b = 4; cout << a/b; We know that the result is 2.5 a float. What type of conversion is this?
(a) type promotion
(b) type casting
(c) explicit coversion
(d) None of these
(a) type promotion (implicit conversion);

Question 83.
From the following which has the major priority?
(а) ++
(b) =
(c) ==
(d) &&
(a) ++

Question 84.
One of your friend told you that post increment (eg:x++) has more priority than pre increment (eg: ++x). State True/False
It is true.

Question 85.
Raju declared a variable as follows. unsigned number;
So he can a store a number in the range______
(a) 0 to 65535
(b) -32768 to 32767
(c) 0 to 65536
(d) 0 to 95536
(a) 0 to 65535. Unsigned uses only 2 bytes of memory but no negative numbers can store.

Question 86.
Considerthe following declaration. signed number;
So we can a store a number in the range______
(a) 0 to 65535
(b) -32768 to 32767
(c) 0 to 65536
(d) 0 to 95536
(b) -32768 to 32767

Question 87.
Pick the odd one out
(a) long
(b) short
(c) unsigned
(d) int
(d) int. It is fundamental type modifiers.

Question 88.
Memory size and sign can be changed using______with fundamental data types.
Type modifiers.

Question 89.
“Its value does not change during execution”. What is it?
Constant.

Question 90.
“BVMHSS” is called______
(a) integer constant
(b) float constant
(c) string constant
(d) None of these
(c) string constant

Question 91.
The address of a variable is called______
L-value (Location value) of a variable.

Question 92.
The content of a variable is called_______
R-value (Read value) of a variable.

Question 93.
Suppose the address of a variable age is 1001 and the content i.e. age = 33. Then what is R-value and L-value?
R-value is 33 and L-value is 1001.

Question 94.
A total of 65535 single window +1 application forms are sold in a district. To store the application form, from the following which is valid?
(a) unsigned app_no;
(b) intapp_no;
(c) signed app_no;
(d) none of these
(a) unsigned app_no;

Question 95.
is it possible to declare a variable as and when a need arise. What kind of declaration is this?
Yes. It is known as Dynamic declaration.

Question 96.
Emerin wants to store a constant value. Which key word is used for this?
constant.

Question 97.
Suppose x = 5. Then cout << x++ displays_____
5. Here post increment first use the value then incremented.

Question 98.
Suppose x = 5. Then cout << ++x displays_____
6. Here pre increment first incremented and then use the value.

Question 99.
An if statement contains another if statement completely. Then it is known as______
Nested if

Question 100.
From the following which is not optional with switch statement.
(a) break
(b) default
(c) case
(d) none of these
(c) case.

Question 101.
To exit from a switch statement______is used
(a) quit
(b) exit
(c) break
(d) none of these
(c) break

Question 102.
From the following which statement is true for switch statement
(a) switch is used to test the equality
(b) switch is used to test relational or logical expression
(c) switch can handle real numbers case data
(d) none of these
(a) switch is used to test the equality

Question 103.
Sonet wants to execute a statement more than once. From the following which is exactly suitable.
(a) if
(b) loop
(c) switch
(b) loop

Question 104.
Odd one out
(a) for
(b) if
(c) switch
(a) for. It is a loop the others are branching statement

Question 105.
Odd one out
(a) for
(b) if
(c) while
(d) do while
(b) if. It is a branching statement and the others are loops.

Question 106.
From the following which loop does the three things, initialisation, checking, and updation.
(a) while
(b) do while
(c) for
(d) none of these
(c) for

Question 107.
Predict the output for(i=1;i<=10;i++); cout<<i;
(a) 10
(b) 1 to 10.
(c) 11
(d) None of these
(c) 11.

Question 108.
From the following which is exit controlled loop
(a) for
(b) while
(c) do while
(d) None of these
(c) do while

Question 109.
_____statement is used for unconditional jump from one location to another.
goto.

Question 110.
Sunitha wants to skip one iteration. From the following which will help her?
(a) continue
(b) break
(c) for
(d) case
(a) continue.

Question 111.
Pick the odd one out from the following. Give reason
1. (a) for
(b) while
(c) do____while
2. (a) if
(b) switch
(c) for

1. do_____while . It is an exit controlled loop others are entry controlled loop
2. for. It is a loop while others are branching statements.

Question 112.
Identify the error in the following C++ statement and correct it. short population = 68000;
The maximum number that can store in short type is less than 32767. So to store 68000 we have to use long data type.

Question 113.
What would be the appropriate data type to store the following?

1. Number of students in a classroom
2. Age of a student
3. Average mark of a student
4. A question mark (?)

1. short or int
2. short or int
3. float
4. char

Question 114.
Pick odd one out from the following loops. Give the reasons.
(a) for
(b) while
(c) do___while
(c) do while. This is an exit controlled loop others are entry controlled loop.

Question 115.
Which of the following data types of C++ has no type modifier?
(a) void
(b) int
(c) char
(d) short
(a) void

### Plus Two Computer Application Review of C++ Programming Two Mark Questions and Answers

Question 1.
Mr. Dixon declared a variable as follows
int 9 age. Is it a valid identifier. If not briefly explain the rules for naming an identifier.
It is not a valid identifier because it violates the rule 1. The rules for naming an identifier is as follows.

• Underscore can be considered as a letter
• White spaces and special characters cannot be used.
• Key words cannot be considered as an identifier

Question 2.
Rose wants to print as follows \n is used for New Line
Write down the C++ statement for the same.
# include
using namespace std;
{
cout<<“\\n is used for New Line”;
}

Question 3.
Alvis wants to give some space using escape sequence as follows
Welcome to C++
Write down the C++ statement for the same
# include
using namespace std;
int main()
{
cout<<“welcome to \t C++”;
}

Question 4.
How many bytes used to store ‘\a’.
To store ‘\a’ one byte is used because it is an escape sequence. An escape sequence is treated as one character. To store one character one byte is used.

Question 5.
How many bytes used to store “\abc”.
A string is automatically appended by a null character.
Here one byte for \a(escape sequence).
One byte for character b.
One byte for character c.
And one byte for null character.
So a total of 4 bytes needed to store this string.

Question 6.
How many bytes used to store “abc”.
A string is automatically appended by a null character.
Here one byte for a.
One byte for character b.
One byte for character c.
And one byte for null character.
So a total of 4 bytes needed to store this string.

Question 7.
We know that the value of pi = 3.14157, a constant (literal). What is a constant? Explain it?
A constant ora literal is a data item its value doe not change during execution.

1. Integer literals:
Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal, and hexadecimal.
Eg. For decimal 100,150,etc
For octal 0100,0240, etc
For hexadecimal 0 × 100, 0 × 1 A, etc

2. Float literals:
A number with fractional parts and its value does not change during execution is called floating-point literals.
Eg. 3.14157,79.78,etc.

3. Character literal:
A valid C++ character enclosed in single.

Question 8.
Write a program to print the message “TOBACCO CAUSES CANCER” on screen.
#include
using namespace std;
int main()
{
cout<<“TOBACCO CAUSES CANCER”;
}

Question 9.
Consider the following code
int main()
{
cout<<“welcome to C++”;
}
After you compile this program there is an error called prototype error. Why it is happened? Explain Answer:
Here we used the output operator cout<<. It is used to display a message “welcome to C++” to use this operator the corresponding header file must be included and using namespace std; is also include. We didn’t included the header file hence the error.

Question 10.
You are supplied with a list of tokens in C++ program, Classify and Categorise them under proper headings. Explain each category with its features. tot_mark, age, M5,_____break,(), int, _pay, ; , cin

Question 11.
In C++ the size of the string “book” is 5 and that of “book\n” is 6. Check the validity of the above statement. Justify your answer.
A string is automatically added by a null character(\0). The null character is treated as one character. So the size of string “book” is 5. Similarly, a null character (\0) is also added to “book\n”. \n and \0 is treated as single characters. Hence the size of the string “book\n” is 6.

Question 12.
Is 0 × 85B a valid integer constant in C++? If yes why?
Yes. It is a hexa decimal number.

Question 13.
Pick the odd man out. Justify
TOTSAL, TOT_SAL, totsal5, Tot5_sal, SALTOT, tot.sal
tot.sal. Because it contains a special character dot(.). An identifier cannot contain a special character. So it is not an identifier. The remaining satisfies the rules of naming identifier. So they are valid identifier.

Question 14.
Write a C++ statement to print the following sentence. Justify “\ is a special character”

1. cout<<“\\ is a special character”
2. \\ is treated as an escape sequence.

Question 15.
A student type a C++ program and saves it in his personal folder as Sample.cpp. After getting the output of the program, he checks the folder and finds three files namely Sample.cpp, Sample.obj and Sample.exe. Write the reasons for the generation of the two files in the folder.
After the compilation of the program sample.cpp, the operating system creates two files if there is no error. The files are one object file (sample.obj) and one executable file(sample.exe). Now the source file(sample.cpp) and object file(sample.obj) are not needed and can be deleted. To run the program sample.exe is only needed.

Question 16.
Write a program to print the message “SMOKING IS INJURIOUS TO HEALTH” on screen.
#include
using namespace std;
int main()
{
cout<<” SMOKING IS INJURIOUS TO HEALTH”;
}

Question 17.
Consider the following
short number;
number = 76543;
Is it valid. Explain?
It is not valid. Because the data type int uses only two bytes(16 bits) of memory. That is we can store a total of 216 = 65536 integers. There are 2 types of integers negative integers and positive integers, i.e. 32768 each.

So we can store a number in between -32768 to +32767 (0 included in the positive section). The number 76543 is bigger than this range. Hence there is an error overflow. To store this number declare the variable is as follows.
long number;

Question 18.
Consider the following declaration.
const int bp;
bp = 100;
Is it valid? Explain it?
This is not valid. This is an error. A constant variable cannot be modified. That is the error and a constant variable must be initialised. So the correct declaration is as follows, const int bp = 100;

Question 19.
Consider the following statements in C++

• cout<<41/2;
• cout<<41/2.0;

Are this two statements give same result? Explain?
This two statements do not give same results. The first statement 41/2 gives 20 instead of 20.5. The reason is 41 and 2 are integers. If two operands are integers the result must be integer, the real part must be truncated.

To get floating result either one of the operand must be float. So the second statement gives 20.5. The reason is 41 is integer but 2.0 is a float.

Question 20.
If mark = 70 then what will be the value of variable result in the following result = mark > 50? ‘P’: ‘F’;
The syntax of the conditional operator is given below Condition? Value if true: Value if false; Here the conditional operator first checks the condition i.e.,70 > 50 it is true. So ‘P’ is assigned to the variable result. So the result is d ‘P’;

Question 21.
Is it possible to initialise a variable at the time of execution. What kind of initialisation is this? Give an example.
Yes it is possible. This is known as Dynamic initialisation. The example is given below
Eg: int a=10, b=5;
int c = a*b;
here the variable c is declared and initialised with the value 10*5.

Question 22.
Boolean data type is used to store True / False in C++. Is it true? Is there any data type called Boolean in C++?
No there is no data type for storing boolean value true /false. But in C++ non -zero (either negative or positive) is treated as true and zero is treated as false.

Question 23.
Consider the following
n=-15;
if (n)
cout<<“Hello”;
else
cout<<“hai”;
What will be the output of the above code?
The output is Hello, because n = -15 a non zero number and it is treated as true hence the result.

Question 24.
Is it possible to declare a variable in between the program as and when the need arise? Then what is it?
Yes it is possible to declare a variable in between the program as and when the need arise. It is known as dynamic initialisation.
Eg. int x=10, y=20;
_____
_____
int z=x*y;

Question 25.
charch;
cout<<“Enter a character”; cin>>ch;
Considerthe above code, a user gives 9 to the variable ‘ch’. Is there any problem? Is it valid?
There is no problem and it is valid since 9 is a character. Any symbol from the key board is treated as a character.

Question 26.
“With the same size we can change the sign and range of data”. Comment on this statement.
With the help of type modifiers we can change the sign and range of data with same size. The important modifiers are signed, unsigned, long and short.

Question 27.
Write short notes about C++ short hands?
x = x + 10 can be represented as x+=10, It is called short hands in C++. It is faster. This is used with all the arithmetic operators as follows.

 Arithmetic Assignment Expression Equivalent Arithmetic Expression x+ = 10 x = x + 10 x- = 10 x = x -10 x* = 10 x = x * 10 x/ = 10 x = x /10 x% = 10 x = x % 10

Question 28.
What is the role of ‘const’ modifier?
This ‘const’ key word is used to declare a constant. Eg. const int bp = 100; By this the variable bp is treated as constant and cannot be possible to change its value during execution.

Question 29.
Specify the most appropriate data type for handling the following data.

1. Rollno. of a student.
2. Name of an employee.
3. Price of an article.
4. Marks of 12 subjects

1. short Rollno;
2. charname[20];
3. float price;
4. short marks[12];

Question 30.
Write C++ statement for the following.

1. The result obtained when 5 is divided by 2.
2. The remainder obtained when 5 is divided by 2.

1. 5/2
2. 5%2

Question 31.
Predict the output of the following code. Justify.
int k = 5, b = 0;
b = k++ + ++k;
cout<<b; (2 Scores)
Output is 12. In this statement first it take the value of k in 5 then increment it K++. So first operand for + is 5. Then it becomes 6. Then ++k makes it 7. This is the second operand. Hence the result is 12.

Question 32.
Predict the output.

1. int sum = 10, ctr= 5;
sum = sum + ctr __;
cout<<sum;
2. int sum = 10, ctr = 5;
sum = sum + ++ctr; court<<sum;

1. 15
2. 16

Question 33.
Predict the output
int a;
float b;
a = 5;
cout<<sizeof(a + b/2);
Output is 4. Result will be the memory size of floating point number.

Question 34.
Predict the output.
int a, b, c;
a = 5; b = 2;
c = a/b;
cout<<c;
Output is 2. Both operands are integers. So the result will be an integer.

Question 35.
The multiple use of input or output operators in a single statement is called cascading of i/o operators. Eg: To take three numbers by using one statement is as follows cin>>x>>y>>z; To print three numbers by using one statement is as follows. cout<<x<<y<<z;

Question 36.
Trace out and correct the errors in the following code fragments

1. cout<<“Mark=”45;
2. cin<<“HellowWorld!”;
3. cout>>”X + Y;
4. Cout<<‘Good'<<‘Moming’

1. cout<<“Mark=45”;
2. cout<<“HellowWorld!”;
3. cout<<X + Y
4. Cout<<“Good Morning”;

Question 37.
What do you mean by preprocessor directive?
A C++ program starts with the preprocessor directive i.e., #include, #define, #undef, etc, are such a preprocessor directives. By using #include we can link the header files that are needed to use the functions. By using #define we can define some constants.
Eg. #definex100. Here the value of x becomes 100 and cannot be changed in the program.

Question 38
Write a program to print a message as ” Hello, Welcome to C++”.
#include
using namespace std;
int main()
{
cout<<” Hello, Welcome to C++”;
}

Question 39.
Write a program to generate the following table.

 2013 100% 2012 99.9% 2011 95.5% 2010 90.81% 2009 85%

Use a single cout statement for output
# include
using namespace std;
int main ()
{
cout<<“2013\t 100%\n2012\t99.9%\n2011\t95.5%\n2010\t90.81%\n2009\t85%”;
}

Question 40.
Determine the data type of the following expression If a is an int, b is a float, c is a long int and d is a double

In type promotion the operands with tower data type will be converted to the highest data type in the expression. So consider the following,

= double + long
= double (Which is the highest data type).

Question 41.
White writing a program Geo uses while loop but forgets to update the loop variable. What will happen?
The loop variable inside the while loop must be updated otherwise the loop will not be terminated. The loop will be work infinitely.

Question 42.
Draw the flow chart of if statement.

Question 43.
Draw the flow chart of if else statement.

Question 44.
Write a while loop that display numbers from 500 to 550.
int i=500
while (i<=550)
{
cout<<i;
i=i+1;
}

Question 45.
Compare if else and conditional operator?
We can use conditional operator as an alternative of if-else statement. The conditional operator is a ternary operator.
The syntax of if-else
if (expression 1)
expression 2;
else
expression 3;
First expression 1 is evaluated if it is true expression 2 will be executed otherwise expression 3 will be executed. Instead of this, we can be written as follows using conditional operator Expression 1? expression 2: expressions;

Question 46.
Draw the flow chart of for loop.

Question 47.
Find out the error in syntax if any and correct it? esm<oi!Bl

a) No need of semi colon. The corrected loop is given below
while (test condition)
{
}

b) In do____white loop the while must be end with; semicoion.
do (condition)
{
}while;

c) switch contains expression instead of condition switch(expression)
{
Case 1:
Case 2:
Case 3:
Case 4:
}

Question 48.
State whether the following statements are True or False. In either case Justify your answer

1. Break statements essential in switch
2. For loop is an entry controlled loop
3. Do____white loop is an entry controlled loop
4. Switch is a selection statement

1. False. It is not essential in single case statement
2. True. Because it will first check the condition. If it is true then only the body will be executed.
3. False. It is an exit controlled loop.
4. True.

Question 49.
Given a code segment for(i=1; i<10; i++) cout<<i;

1. Rewrite the code using do____while loop
2. What will be the output when i=0? Give reason.

1. i=1;
do{
cout<<i;
i++;
}while(i<10);

2. When i=0, it will execute one more time. ie. the for loop execute 9 times but here this loop executes 10 times.

Question 50.
Write the equivalent code for the following statement R=(P<Q?P: Q)
if(P<Q)
R=P;
else
R=Q;

Question 51.
Examine the following code snippet and find out the output? What will happen if the statement int ch; is replaced char ch;
int ch;
for(ch=’A’;ch<=’Z’;++ch)
cout<<ch<<”;
This code snippet will print 65, 66, 67, ______,90. If the statement int ch; is replaced by char ch; it prints A, B, C,_____,Z.

Question 52.
Your friend usee the following identifiers in a program. Find out the invalid identifiers with reason if not valid, basic pay area, data-of-birth, B3, 9A, switch
basic pay – Invalid because white space is not allowed.
area – valid
date-of-birth – Invalid because hyphen(-) is not allowed.
B3 – Valid
9A – Invalid because should not begin with number.
switch – Invalid because keywords not allowed.

Question 53.
int x=5;
int y=10;
cout<<(x+y)%2;
(x+y) % 2 = (5+10)%2 = 15% 2 =1 It prints 1.

Question 54.
Classify the following C++ tokens in accordance to the table given below.

keyword: int, do
Identifier: digit, cin(pre defined identifier, eg. int cin; is possible);
Literal: “break”(string literal), 25.6(float literal).
Operator: %(Mod operator), = (assignment operator).

Question 55.
Rewrite the following C++ code using conditional operator.
if (a>b)
max=a;
else
max=b;
max=(a>b)?a:b;

Question 56.
Write the C++ expression to calculate the value of the following expression.

1. x = $$\frac{\left(b^{2}-4 a c\right)}{2 a}$$
2. y = a2 + 2ab + b2

1. x= (b*b – 4*a*c)/(2*a);
2. y= a*a + 2*a*b + b*b;

Question 57.
for(int i=i;i<10;++i)

The output is 1 2 3 4 5 6 7. This loop is used to print 1 to 10 but this loop will terminate when the value of i becomes 8. Hence it prints 1 to 7.

Question 58.
Predict the output for the following program code.
for(i=1;i<=10;++i)
{
if (i==7)
coutinue;
cout<<“\t”;
cout<< i;
}
The output is 1 2 3 4 5 6 8 9 10.
This loop is used to print 1 to 10 but this loop will by pass one iteration when the value of i becomes 7. Hence it prints 1 to 10 except 7.

Question 59.
How many times the following loop will be executed?
int s = 0, i = 0;
do
{
S+= i;
i++;
} while(i < = 5);
This loop will be executed 6 times. The value of i becomes 1 to 6.

### Plus Two Computer Application Review of C++ Programming Three Mark Questions and Answers

Question 1.
In a panchayath or municipality, all the houses have a house number, house name, and members. Similar situation is in the case of memory. Explain.
The named memory locations are called variable. A variable has three important things

1. variable name: A variable should have a name
2. Memory address: Each and every byte of memory has an address. It is also called location (L) value
3. Content: The value stored in a variable is called content.lt is also called Read(R) value.

Question 2.
Briefly explain constants
A constant or a literal is a data item its value doe not change during execution. The keyword const is used to declare a constant. Its declaration is as follows
const data type variable name=value;
eg.const int bp = 100;
const float pi = 3.14157;
const char ch = ‘a’;
const char[] = “Alvis”;

1. Integer literals:
Whole numbers without fractional parts are known as integer literals, its value does not change during execution. There are 3 types decimal, octal, and hexadecimal.
Eg. For decimal 100,150,etc
For octal 0100,0240, etc
For hexadecimal 0 × 100, 0 × 1A,etc.

2. Float literals:
A number with fractional parts and its value does not change during execution is called floating point literals.
Eg. 3.14157,79.78,etc.

3. Character literal:
A valid C++ character enclosed in single quotes, its value does not change during execution.
Eg. ‘m’, ‘f, etc.

4. String literal:
One or more characters enclosed in double quotes is called string constant. A string is automatically appended by a null charater(‘\0’)
Eg. “Mary’s”,”India”,etc.

Question 3.
Considerthe following statements
int a=10, x=20;
float b=49000.34, y=56.78;

1. a=b;
2. y=x;

Is there any problem for the above statements? What do you mean by type compatibility?
Assignment operator is used to assign the value of RHS to LHS. Following are the two chances

1. The size of RHS is less than LHS. So there is no problem and RHS data type is promoted to LHS. Here it is compatible.
2. The size of RHS is higher than LHS. Here comes the problem sometimes LHS cannot possible to assign RHS. There may be a chance of wrong answer. Here it is not compatible.
Here

• a=b; There is an error since the size of LHS is 2 but the size of RHS is 4.
• y=x; There is no problem because the size of LHS is 4 and RHS is 2.

Question 4.
A company has decided to give incentives to their salesman as per the sales. The criteria is given below.
If the total sales exceeds 10,000 the incentive is 10%

1. If the total sales >=5,000 and total sales <10,000, the incentive is 6 %
2. If the total sales >=1,000 and total sales <5,000, the incentive is 3 %

Write a C++ program to solve the above problem and print the incentive after accepting the total sales of a salesman. The program code should not make use of if statement.
# include
using namespace std;
int main()

Question 5.
A C++ program code is given below to find the value of X using the expression
X = $$\frac{a^{2}+b^{2}}{2 a}$$ where a and b are variables 2a
# include
using namespace std;
int main()

Pre did the type of errors during compilation, execution, and verification of the output. Also write the output of two sets of input values

1. a=4, b=8
2. a=0, b=2

This program contains some errors and the correct program is as follows.
#include
using namespace std;
int main()

The output is as follows

1. a=4 and b= 8 then the output is 10
2. a=0 and b= 2 then the output is an error divide by zero error(run time error).

Question 6.
We know that a program has a structure. Explain the structure of C++ program.
A typical C++ program would contain four sections as shown below.
Include files
Function declarations
Function definitions
Main function programs
Eg.
# include
using namespace std;
int sum(int x, int y)
{return (x+y);}
int main()
{
cout<<sum(2,3);
}
}

Question 7.
Write a program to read two numbers and find its sum.
#include
using namespace std;
int main()

Question 8.
Write a program to read three scores and find the average.
#include
using namespace std;
int main()

Question 9.
Write a program to find the area and perimeter of a circle.
#include
using namespace std;
int main()

Question 10.
Write a program to find the simple interest
# include
using namespace std;
int main()

Question 11.
Write a program to convert temperature from Celsius to Fahrenheit.
# include
using namespace std;
int main()

Question 12.
Write a program to read weight in grams and convert it into Kilogram.
# include
using namespace std;
int main()

Question 13.
Write a program to read your height in meter and cm convert it into Feet and inches
# include
using namespace std;
int main()

Question 14.
Write a program to find the area of a triangle.
# include
using namespace std;
int main()

Question 15.
Write a program to

1. print ASCII for a given digit.
2. print ASCII for backspace.

# include
using namespace std;
int main()

Question 16.
Write a program to read time in seconds and convert it into hours, minutes and seconds.
# include
using namespace std;
int main()

Question 17.
Two pairs C++ expressions are given below.

1. a=10, a==10
2. b=a++, b=++a
• How do they differ?
• What will be the effect of the expression

1. = is an assignment operator that assigns a value 10 to the LHS (Left Hand Side)variable a But == is equality operator that checks whether the LHS and RHS are equal or not. If it is equal it returns a true value otherwise false

2. In a++,++is a post(means after the operand) increment operator and in ++a, ++ is a pre(means before the operand) increment operator. They are entirely different.

Post increment:
Here first use the value of ‘a’ and then change the value of ‘a’.
Eg: if a= 10 then b=a++. After this statement b= 10 and a=11

Pre increment:
Here first change the value of a and then use the value of a.
Eg: if a=10 then b=++a. Afterthis statement b=11 and a=11.

Question 18.
Consider the following
int a=45.65;
cout<<a;
What is the output of the above. Is it possible to convert a data type to another type? Explain.
The output of the code is 45, the floating-point number is converted into integer. It is possible to convert a data type into another data type. Type conversions are two types.

1. Implicit type conversion:
This is performed by C++ compiler internally. C++ converts all the lower sized data type to the highest sized operand. It is known as type promotion. Data types are arranged lower size to higher size is as follows, unsigned int(2 bytes), int(4 bytes), long(4 bytes), unsigned long(4 bytes), float(4 bytes), double(8 bytes), long double(10 bytes).

2. Explicit type conversion:
It is known as type casting. This is done by the programmer. The syntax is given below.
(data type to be converted) expression
Eg.int x=10;
(float) x; This expression converts the data type of the variable from integer to float.

Question 19.
Your friend Arun asked you that is there any loop that will do three things, initialization, testing, and updation. What is your answer? Explain?
Yes. There is only one loop namely for loop that will do this three things. The other loops will do the checking only, initialisation must be do before the loop and updation must be inside the loop.
The syntax of for loop is given below
For(initialisation; testing; updation)
{
Body of the for loop;
}

Question 20.
Distinguish between exit(0) function and return statement
Both are used to terminate the program but both are different. Return is a keyword and exit(O) is a function. The difference is, we can use more than one exit(0) function but we can use only one return statement in a scope. To use exit(0), the header file cstdlib should be used.

Question 21.
Rewrite the program following program using if else
#include
using namespace std;
int main()

# include
using namespace std;
int main()

Question 22.
Rewrite the following using nested switch construct.
#include
using namespace std;
int main()

#include
using namespace std;
int main()

Question 23.
Consider the following output and write down the code for the same.

include
using namespace std;
int main()

Question 24.
Consider the following output and write down the code for the same.

# include
using namespace std;
int main()

Question 25.
Consider the following output and write down the code for the same.

# include
using namespace std;
int main()

Question 26.
Consider the following output and write down the code for the same.

# include
using namespace std;
int main()

Question 27.
Consider the following output and write down the code for the same.

#include
using namespace std;
int main()

Question 28.
Consider the following output and write dawn the code for the same.

#include
using namespace std;
int main()

Question 29.
Write a program to print the sum of first n natural numbers
# include
using namespace std;
int main()
int n,i,sum=0;
cout<<“Entera value torn”; cin>>n;
for(i=1;i<=n;i++)

Question 30.
Write a program to read a number and check whether it is palindrome or not
# include
using namespace std;
int main()

Question 31.
Write a program to print the factorial of a number.
#include
using namespace std;
int main()

This Fibonacci calculator will generate a list of Fibonacci numbers from start and end values of n.

Question 32.
Write a program to print the Fibonacci series Fibonacci
# include
using namespace std;
int main()

Question 33.
Write a program to read a number and check whether the given number is Armstrong or not
# include
using namespace std;
int main()

Question 34.
Distinguish between entry controlled loop and exit controlled loop
An entry controlled loop first checks the condition and execute(or enters in to) the body of loop only if it is true. But exit control loop first execute the body of the loop once even if the condition is false then check the condition. The for loop and while loop are entry controlled loops but do-while loop is an exit controtted loop.

Question 35.
Write a program to find the largest of 3 numbers
#include
using namespace std;
int main()

Question 36.
Check whether a given number is prime or not
#include
using namespace std;
int main()

Question 37.
Write a program to read number and display its factors.
# include
using namespace std;
int main()

Question 38.
You are given the heights of 3 students. Write the relevant code segment to find the maximum height?
# include
using namespace std;
int main()

Question 39.
Write the easiest code snippet for printing your name 1000 times. Explain
# include
using namespace std;
int main()

Question 40.
Whenever a string is entered the inverse of that string is displayed( eg: if we enter ‘CAR’ the output is ‘RAC’). Write a suitable programme for the output.
#include
#include
using namespace std;
int main()

Question 41.
Write a C++ program to display as follows

#include
using namespace std;
int main()

OR

#include
using namespace std;
int main()

Question 42.
Consider the following code The new line.character is \n. The output of the following code does not contain the \n. Why it is happened? Explain.
\n is a character constant and it is also known as escape sequence. This is used to represent the non graphic symbols such as carriage return key(enter key), tab key, back space, space bar, etc. It con¬sists of a back slash symbol and one more characters.

 Escape sequence Non-graphic character \a Audible bell \b back space \n for new line \r carnage return \t horizontal tab \v vertical tab \\ to print \ \’ to print ‘ \” to print ” \? To print? \0 null character

Question 43.
The Maths teacher gives the following problem to Riya and Raju.
x= 5 + 3 * 6. Riya got x = 48 and Raju got x = 23. Who is right and why it is happened? Write down the operator precedence in detail?
Here the answer is x = 23. It is because of precedence of operators. The order of precedence of operators are given below.

Here multiplication has more priority than addition.

Question 44.
Rewrite the following code using while and do..while loop.
for(i=1;i<5;++i)
{
cout<<“\n”<<i;
}
1. using while loop
i=1;
while(i<=5)
{
cout<<“\n”<<i;
i++;
}

2. using do while loop
i=1;
do
cout<<“\n”<<i;
i++;
}while(i<=5);

Question 45.
Rewrite the following C++ code using switch
statement.

switch(choice)

Question 46.
Considerthe following code and predict the output
int a=5, b=6, c=7;
if (a<b) || b>c)
cout<<“\nOne”;
if (a<b&&b>c)
cout<<“\nTwo”; if (! *a>b)
cout<<“\nThree”;
case 1:
if(ac)
cout<<“\nOne”;
here ac
5<6 || 6>7
True || False = True.
This condition returns True
Hence it prints One.

case 2:
if(ac)
cout<<“\nTwo”;
here ac
5<6 && 6>7
True && False = False
This condition returns False
Hence it prints nothing.

Case 3:
if(! (a>b))
cout<<“\nThree”); here !(a>b)
! (5 >6)
! (False)
This condition returns True
Hence it prints Three.

Question 47.
Match the following

 Name Symbol Modulus operator + + Logical operator = = Relational operator = Assignment operator ?= Increment operator % Conditional operator &&

 Name Symbol Modulus operator % Logical operator && Relational operator = = Assignment operator = Increment operator + + Conditional operator ?=

Question 48.
Write a C++ program to calculate the simple interest SI, by accepting the value principal amount P, rate of interest R and number of years N using the equation SI = PNR/100.
# include
using namespace std;
int main()

Question 49.
Rewrite the following conditional statement with ‘if statement in C++.
min=(a<b) ? (a<c? a:c) : (b<c? b:c);
if(a<b && a<c)
min=a;
else if(b<c) min=b; else min=c;

### Plus Two Computer Application Review of C++ Programming Five Mark Questions and Answers

Question 1.
Define operator and explain operators in detail. Operators
An operator is a symbol that performs an operation. The data on which operations are carried out are called operands. Following are the operators.
1. Input(>>) and output(<<) operators are used to perform input and output operation. Eg. cin>>n; cout<<n;

2. Arithmetic operators:
It is a binary operator. It is used to perform addition(+), subtraction (-), division (/), multiplication (*) and modulus (%-gives the remainder) operations.
Eg. If x=10 and y=3 then

x/y = 3, because both operands are integer. To get the floating point result one of the operand must be float.

3. Relational operator:
It is also a binary operator. It is used to perform comparison or relational operation between two values and it gives either true(1) or false(O). The operators are <,<=,>,>=,== (equality)and !=(not equal to)
Eg. If x=10 and y=3 then

4. Logical operators:
Here AND(&&), OR (||) are binary operators and NOT (!) is a unary operator. It is used to combine relational operations and it gives either true (1) or false (0).
If x=1 and y=0 then

Both operands must be true to get a true value in the case of AND (&&) operation
If x=1 and y=0 then

Either one of the operands must be true to get a true value in the case of OR(||) operation
If x=1 and y=0 then

5. Conditional operator:
It is a ternary operator hence it needs three operands. The operator is?: Syntax: expression? value if true: value if false. First evaluates the expression if it is true the second part will be executed otherwise the third part will be executed.
Eg. If x=10 and y=3 then
x>y? cout<< Here the output is 10.

6. sizeof():
This operator is used to find the size used by each data type.
Eg. sizeof(int) gives 2.

7. Increment and decrement operator:
These are unary operators.

• Increment operator (++): It is used to increment the value of a variable by one i.e., x++ is equivalent to x=x+1;
• Decrement operator (–): It is used to decrement the value of a variable by one i.e., x- is equivalent to x = x-1.

8. Assignment operator (=):
It is used to assign the value of a right side to the left side variable.eg. x=5; Here the value 5 is assigned to the variable x.

Question 2.
A list of data items are given below
45, 8.432, M, 0.124, 8, 0, 8.1X 1031, 1010, a, 0.00025, 9.2 × 10120, 0471, -846, 342.123E03

1. Categorise the given data under proper headings of fundamental data types in C++
2. Explain the specific features of each data type. Also mention any other fundamental data type for which sample data is not given

1.

2.
(i) int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 4 bytes (32 bits) of memory. i.e. 232 numbers. That is 231 negative numbers and 231 positive numbers (0 is considered as +ve) So a total of 232 numbers. We can store a number in between -231 to + 231-1.

(ii) char data type:
Any symbol from the key board, eg. ‘A’,’?’, ‘9’,___It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCI I code 97 and zero is having ASCII code 48.

(iii) float data type:
It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory. Eg. 67.89, 89.9 E-15.

(iv) double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.
void data type:- void means nothing. It is used to represent a function returns nothing.

Question 3.
Write valid reasons after reading the following statements in C++ and comment on their correctness by give reasons.

1. char num = 66;
2. char num – B’;
3. 35 and 35L are different
4. The number 14,016 and OxE are one and the same
5. Char data type is often said to be an integer type
6. To store the value 4.15 float data type is preferred over double

1. The ASCII number of B is 66. So it is equivalent.
2. 35 is of integer type but 35L is Long
3. The decimal number 14 is represented in octal is 016 and in hexadecimal is OxE.
4. Internally char data type stores ASCII numbers.
5. To store the value 4.15 float data type is better because float requires only 4 bytes while double needs 8 bytes hence we can save the memory.

Question 4.
Suggest most suitable derived data types in C++ for storing the following data items or statements

1. 0 Age of 50 students in a class
2. Address of a memory variable
3. A set of instructions to find out the factorial of a number
4. An alternate name of a previously defined variable
5. Price of 100 products in a consumer store
6. Name of a student

1. Integer array of size 50
2. Pointer variable
3. Function
4. Reference
5. Float array of size 100
6. Character array

Question 5.
Considering the following C++ statements. Fill up the blanks

1. lf p=5 and q=3 then q%p is _____
2. If E1 is true and E2 is False then E1 && E2 will be_____
3. If k=8, ++k <= 8 will be______
4. If x=2 then (10* ++x) % 7 will be_____
5. If t=8 and m=(n=3,t-n), the value of m will be______
6. If i=12 the value i after execution of the expres¬sion i+=i– + –i will be______

1. 3
2. False
3. False(++k makes k=9. So 9<=8 is false)
4. 2(++x becomes 3, so 10 * 3 =30%7 =2)
5. 5( here m=(n=3,8-3)=(n=3,5), so m=5, The maximum value will take)
6. Here i=12

i + = i– + –i
here post decrement has more priority than pre decrement. So i — will be evaluated first. Here first uses the value then change so it uses the value 12 and i becomes 11
i + =12 + –i
now i =11.
Here the value of i will be changed and used so i– becomes 10
i + = 12 + 10
= 22
So i = 22 + 10
i = 32
So the result is 32.

Question 6.
Analyses the following statements and write Time or False. Justify

1. There is an Operator in C++ having no special character in it
2. An operator cannot have more than 2 operands
3. Comma operator has the lowest precedence
4. All logical operators are binary in nature
5. It is not possible to assign the constant 5 to 10 different variables using a single C++ expression
6. In type promotion the operands with lower data type will be converted to the highest data type in expression

1. True (sizeof operator)
2. False( conditional operator can have 3 operands
3. True
4. False
5. False(Multiple assignment is possible, eg: a=b=c=___=5)
6. True

Question 7.
Match the following numbers and data types in C++ to form the most suitable pairs.

 1. 142789 a. Signed 2. 240 b. Double 3. -150 c. Long int 4. 8.4 × 10-4000 d. Float 5. 0 e. Long double 6. 0.0008 f. Unsigned short 7. -127 g. Short int 8. 2.8 × 10308 h. Signed char

 1. 142789 a.  Long int 2. 240 b. Short int 3. -150 c.  Signed 4. 8.4 × 10-4000 d.  Long double 5. 0 e.  Unsigned short 6. 0.0008 f.   Float 7. -127 g.  Signed char 8. 2.8 × 10308 h.  Double

Question 8.
#include
using namespace std;
int main()

Rewrite the above code using if else if ladder.
#include
using namespace std;
int main()

Question 9.
Consider the following code
# include
using namespace std;
int main()

Is it possible to rewrite the above program using switch statement? Distinguish between switch and if else if ladder. Answer:
No. It is not possible to write the above code using switch statement. Following are the difference between switch and if else if ladder.

1. Switch can test only for equality but if can evaluate a relational or logical expression
2. If else is more versatile
3. If else can handle floating values but switch can not
4. If the test expression contains more variable if else is used
5. Testing a value against a set of constants switch is more efficient than if else

Question 10.
Write down the code for the following output using while loop.

# include
using namespace std;
int main()

Question 11.
“We know that the execution of a program is sequential”. Is it possible to change this sequential manner and explain different jump statements in detail.
The execution of a program is sequential but we can change this sequential manner by using jump statements. The jump statements are
1. goto statement:
By using goto we can transfer the control anywhere in the program without any condition. The syntax is goto label; Eg. # include
using namespace std;
int main()

2. break statement:
It is used to skip over a part of the code i.e. we can premature exit from a loop such as while, do-while, for or switch. Syntax:
while (expression)

The output is
1
2
3
4
5.

3. continue statement:
It bypasses one iteration of the loop.
Syntax:
while (expression)

The output is
1
2
3
4
5
6
7
8
9
10.

4. exit(O) function:
It is used to terminate the program. For this the header file cstdlib must be included.

Question 12.
Write a program to print the prime numbers less than 100
# include
using namespace std;
int main()

Question 13.
Write a program to print the Armstrong numbers less than 1000
# include
using namespace std;
int main()

Question 14.
Given the total mark of each student in SSLC examination. Write a C++ code fragment to find the grades.
# include
using namespace std;
int main()

Question 15.
You are about to study the fundamentals of C++ programming Language. Do a comparative study of the basics of the new language with that of a formal language like English or Malayalam to familiarize C++?. Provide sufficient explanations for the compared items in C++ Language.
Character set:
To study a language first we have to familiarize the character set. For example to study English language first we have to study the alphabets. Similarly here the characterset includes letters(A to Z & a to z), digits(0 to 9), special characters(+,-,?,*,/,___) white spaces(non printable) etc. Token: It is the smallest individual units similar to a word in English or Malayalam language. C++ has 5 tokens
1. Keywords:
These are reserved words for the compiler. We can’t use for any other purposes
Eg: float is used to declare variable to store numbers with decimal point. We can’t use this for any other purpose

2. Identifier:
These are user defined words.
Eg: variable name, function name, class name, object name etc….

3. Literals (Constants):
Its value does not change during execution
Eg: In maths pi = 3.14157 and boiling point of water is 100.

4. Punctuators:
In English or Malayalam language punctuation mark are used to increase the read ability but here it is used to separate the tokens.
Eg:{.}.(.)…..

5. Operators:
These are symbols used to perform an operation(Arithmetic, relational, logical,etc…)code, the middle button displays the object and the right button toggles the folder.

Question 16.

Consider the above data, we know that there are different types of data are used in the computer. Explain different data types used in C++.
1. int data type:
It is used to store whole numbers without fractional (decimal point) part. It can be either negative or positive. It consumes 2 bytes(16 bits) of memory.i.e. 216 = 65536 numbers. We can store a number in between -32768 to + 32767.

2. char data type:
Any symbol from the keyboard, eg. ‘A’,’9′,…. It consumes one byte( 8 bits) of memory. It is internally treated as integers, i.e. 28 = 256 characters. Each character is having a ASCII code, ‘a’ is having ASCII code 97 and zero is having ASCII code 48.

3. float data type:
It is used to store real numbers i.e. the numbers with decimal point. It uses 4 bytes(32 bits) of memory. Eg. 67.89, 89.9 E-15.

4. double data type:
It is used to store very large real numbers. It uses 8 bytes(64 bits) of memory.

5. void data type:
void means nothing. It is used to represent a function returns nothing.

Question 17.
Explain conditional statements in detail?
1. Simple if:
The syntax is given below
if(expression)
statement;
or
if(expression)
{
Block of statements
}
First expression evaluates if it is true then only statement will be executed.
Eg. if (n>0)
cout<<n<<” is positive”;

2. if else:
The syntax is given below.
if (expression)

First expression evaluates if it is true statement block 1 will be executed otherwise statement block 2 will be executed. Only one block will be executed at a time so it is called branching statement.
Eg.
if (n>0)
cout<<n<<” is positive”;
else
cout<<n<<” is negative”;

The syntax will be given below
if (expression 1)

Here first expression 1 will be evaluated if it is true only the statement block 1 will be executed otherwise expression 2 will be executed if it is true only the statement block2 will be executed and soon. If all the expression evaluated is false then only statement block n will be evaluated.
Eg.
If(mark>=90)

4. conditional operator:
It is a ternary operator and it is an alternative for if else construct. The syntax is given below. expression 1? expression 2: expression 3; or expression 1? Value if true: value if false; Here expression 1 will be evaluated if it true ex¬pression 2 will be executed otherwise expression 3 will be executed. Eg. n>0?cout<<n<<” is positive”:cout<<n<<” is negative”;

5. Switch:
It is a multiple branch statement. Its syntax is given below.
switch(expression)

Question 18.
Explain different loops in detail?
1. For loop:
The syntax of for loop is
for(initialization; checking ; update loop variable)
{
Body of loop;
}
First part, initialization is executed once, then checking is carried out if it is true the body of the for loop is executed. Then loop variable is updated and again checking is carried out this process continues until the checking becomes false. It is an entry controlled loop.
Eg. for(i=1 ,j=1 ;i<=10;i++,j++)
cout<<i<<” * “<<j<<” = “<<i*j;

2. While loop:
It is also an entry controlled loop
The syntax is given below
Loop variable initialised
while(expression)
{
Body of the loop;
Update loop variable;
}
Here the loop variable must be initialised out side the while loop. Then the expression is evaluated if it is true then only the body of the loop will be executed and the loop variable must be updated inside the body. The body of the loop will be executed until the expression becomes false.
Eg.
i=1;
j=1;
while(i<=10)
{
cout<<i<<” * “<<j<<” = “<<i*j;
i++;
j++;
}

3. do While loop:
It is an exit controlled loop.
The syntax is given below
do
{
Statements
}while(expression);
Here the body executes atleast once even if the condition is false. After executing the body it checks the expression if it false it quits the body otherwise the process will be continue.

Question 19.
Write a program to find simple interest and compound interest.
#include<iostream>
#include<cmath>
using namespace std;

Question 20.
Mr. X wants to get an output 9 when inputting 342 and he also wants to get 12 when inputting 651. Write the program and draw a suitable flowchart for X?
Flow chart

#include
using namespace std;
int main()

Question 21.
Differentiate break and continue statements with suitable examples.
1. break statement:
It is used to skip over a part of the code i.e. we can premature exit from a loop such as while, do-while, for or switch.
Syntax:
while (expression)

The output is
1
2
3
4
5.

2. continue statement:
It bypasses one iteration of the loop.
Syntax :
while (expression)

The output is
1
2
3
4
5
6
7
8
9
10.

Question 22.
Write a C++ program to print to get the following output.

#include
using namespace std;
int main()

## HSSLive Plus Two Study Material / Question Bank Kerala

HSSLive.Guru providing HSE Kerala Board Syllabus HSSLive Higher Secondary SCERT Plus Two Study Material, Question Bank, Quick Revision Notes, Chapter Wise Notes, Chapter Wise Previous Year Important Questions and Answers, Previous Year Sample Question Papers with Answers based on CBSE NCERT syllabus in both English Medium and Malayalam Medium for Class 12. Students can also read NCERT Solutions.

## HSSLive Plus Two Study Material / Question Bank Kerala

### HSSLive Plus Two Previous Year Question Papers and Answers Kerala

Here HSSLive.Guru have given HSSLive Higher Secondary Kerala Plus Two Previous Year Sample Question Papers with Answers.

### HSSLive Kerala Plus Two Notes Chapter Wise Kerala

Here HSSLive.Guru have given HSSLive Higher Secondary Kerala Plus Two Chapter Wise Quick Revision Notes.

### HSSLive Plus Two Chapter Wise Questions and Answers Kerala

Here HSSLive.Guru have given HSSLive Higher Secondary Kerala Plus Two Chapter Wise Questions and Answers.

### HSSLive Plus Two Chapter Wise Previous Year Questions and Answers Kerala

Here HSSLive.Guru have given HSSLive Higher Secondary Kerala Plus Two Chapter Wise Previous Year Important Questions and Answers.

## Plus Two History Notes Chapter Wise HSSLive Kerala

HSE Kerala Board Syllabus HSSLive Plus Two History Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala HSSLive Plus Two Notes. Here HSSLive.Guru has given Higher Secondary Kerala Plus Two History Chapter Wise Quick Revision Notes based on CBSE NCERT syllabus.

 Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject History Chapter All Chapters Category Kerala Plus Two

## Kerala Plus Two History Notes Chapter Wise

We hope the given HSE Kerala Board Syllabus HSSLive Plus Two History Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus Two History Chapter Wise Quick Revision Notes based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

## Plus Two English Textbook Answers Unit 3 Challenges of Life

Kerala State Board New Syllabus Plus Two English Textbook Answers Unit 3 Challenges of Life Text Book Questions and Answers, Summary, Notes.

## Kerala Plus Two English Textbook Answers Unit 3 Challenges of Life

“Be the change you want to see in the world.”

– Mahatma Gandhi

Challenges of Life About The Unit

Teenage is the period of dreams and hopes, enthusiasm and adventure. It is the most appropriate time to develop different views of life and to understand the challenges of the world. In this unit we will see the lives and attitudes of people who have made contributions to the society in their own ways. Definitions of success have changed. Success is not about just making money. It is about pursuing excellence to make a difference in the society. The unit introduces the concept of ‘entrepreneurship’ and emphasizes the dignity of labour.

The Unit has an interview with a young entrepreneur, a personal story and a poem. Irfan Alam tells us how concepts of business can be used for a social cause successfully. ‘Didi’ tells us how a 19-year old girl mad a difference in the slums of Mumbai. The ‘Stammer’tells us how something looks different when viewed from a different point.

Let’s Begin

Look at the following cartoon strip, Can you build a story on them?

Question 1.
Use your imagination to develop a story in the above comic strip. Make sure to give your story a beginning, a climax (turning point) and an ending. You can add details about the setting of the story (time and place) and the characters (name, profession, personality traits etc.) Discuss your ideas with your partner and modify your story. Now narrate the story to your class. Have fun!
Gopal worked as a chef in a restaurant. He prepared delicious meals for the customers and people came to the restaurant to enjoy Gopal’s food. He was a happy with a loving wife and a darling daughter.

It was Monday the 10th of May 2014. As usual Gopal went to his restaurant to do his work. He was summoned to the office of the Manager. Gopal got the shock of his life when his manager told him that he was fired from his job as he got another trained chef in his place. Gopal did not wait to argue with the rude manager, who dismissed him from service without any notice or any valid reason.

With a sad heart Gopal was walking back home thinking what he would to make a living. He was passing by the local bus stop. There were many people waiting for the bus. Suddenly he saw a bird sitting on an electric wire with a straw in her beaks. It was trying to build a nest. An idea suddenly occurred to him: “Why don’t I begin a restaurant here?” There was an empty shop nearby which he could get on rent. The problem was money.

He went home and told his wife what happened. She was an understanding wife. She immediately took out her bangles and told him to pawn them to get money to start a small restaurant. He was reluctant to take the bangles. But she insisted. He went to the local financiers and got a loan. With the loan, he started “Gopal’s Food Comer” near the bus stop. Gopal’s Food Comer became very popular within days. Soon he got the bangles back and bought a couple more for his wife. The Food Corner soon developed into a big restaurant with a lot tables and chairs for customers. Even as Gopal is sitting on the Manager’s chair, he has not forgotten his past.

Question 2.
Comment on the following quotes:
a) “If plan A didn’t work, the alphabet has 25 more letters! Stay cool!”
This is a very positive and optimistic quote. Sometimes people feel very disappointed when their plan fails. They feel sad and disappointed as if the whole world has crashed. The quote tells us that if a plan has failed there are other plans you can pursue. There can be plan A, plan B, plan C, etc. A is just one letter in the alphabet. There are 25 more! So be cool and don’t get worried.

b) If you don’t build your dream, someone will hire you to help build theirs. ”
All people, especially the youths, have dreams. But often we are afraid to build them for fear of failure. That is why many educated people look for government jobs because they feel secure there. But if you are intelligent and adventurous try to build your own dream. Bill Gates and Mukesh Ambani had big dreams. They hired people to build their dream. So instead of working for others to realize their dreams, be courageous and be an entrepreneur, building your own dreams.

Question 3.
Can you find more quotes on success? Collect them and write them on the chart to display in the classroom.

• Where there is a will there is away.
• Rome was not built in a day.
• No pain, no gain.
• Nothing ventured, nothing gained.
• Success favours the lion-hearted.
• Success is not final, failure is not fatal; it is the courage to continue that matters.

Question 1.
Most of the teenagers dream big. Some of them pursue their dreams, choosing a different path from others. They plan their own route of life. Success is for those who think big and act differently.

Question 2.
Have you heard about ‘entrepreneurship’. Meet an entrepreneur who changed the lives of many people through his innovative enterprise. Here is an interview with him.

## Plus Two English Textbook Answers Unit 4 Chapter 2 Rice (Poem)

Kerala State Board New Syllabus Plus Two English Textbook Answers Unit 4 Chapter 2 Rice Text Book Questions and Answers, Summary, Notes.

## Kerala Plus Two English Textbook Answers Unit 4 Chapter 2 Rice (Poem)

Rice Poem Appreciation Hsslive Chapter 2 Question 1.
Of all the memories of his homeland, the narrator thinks of rice’ first. What does this show?
It shows his extreme love for rice. He is a rice eater. For the last 4 years he has been eating chapattis. Now he is craving for rice.

Rice Poem Summary In Malayalam Chapter 2 Question 2.
What are the memories of the narrator about the paddy cultivating season?
He remembers his father in the fields below his house. He is wearing a handloom dhoti stained with yellow mud. He is excited about the water of the Varanganal canal. He also remembers his little brother carrying the tender saplings to be planted where the ploughing is done. On the dyke baskets full of seeds are kept. There is the noise of shouting as the oxen draw the plough in the field.

Plus Two English Rice Appreciation Chapter 2 Question 3.
The narrator wants the train to move a little faster. What does this tell us about his feeling for his native village?
It tells that he loves his village dearly. He has been missing it for long. He wants to reach it as quickly as possible. We see his nostalgic feelings here.

Appreciation Of The Poem Rice Plus Two English Question 4.
What changes in the native village does the narrator notice on his return.
There are many changes in his native village. The palm-thatched houses are gone. There are only rubber plants there. There are no rice fields any more. There is no noise of people below. No shouts of ploughing. The whole field is planted with areca nut palms. In the corner, along the canal, there are the dealwood trees which were not there before.

Plus Two English Chapter Rice Appreciation Chapter 2 Question 5.
Rubber plants have taken the place of paddy. What does this imply?
It implies that food crops are replaced by cash/commercial crops. People don’t any more cultivate their favourite varies of rice like athikira, modan and vellaran. Now they survive on their rations which consist of wheat.

Plus Two English Chapter Rice Summary Chapter 2 Question 6.
“Only fools turn to rice farming for gain.” Why does the father say so?
The father says so because nobody promotes the farming of rice. Rice farming was quite inconvenient and the farmer gained nothing. Rubber brings better money. The government gives rice to those who have no paddy fields.

Plus Two English Textbook Malayalam Translation Question 7.
What does the ship of the sky represent?
It represents the aeroplane, as the ship of the desert represents the camel.

Plus Two Rice Poem Appreciation Chapter 2 Question 8.
“Can we get some husk from the Centre, too, to make toys with it?” Bring out the satire in these lines.
The poet had gone to North India and did a 4-year research on making toys with husk. He got a doctoral degree at the end of his research. Now he comes back to Kerala to find out that there is no more rice cultivation here and consequently no husk to make toys with. To make toys with husk, he has to get husk from the Centre! He did the research to help the State to find employment and. income. What use the State will have now with his doctorate?

Think And Write

Plus Two English Rice Poem Appreciation Chapter 2 Question 1.
Why does the narrator feel confused as he walks home?
The narrator feels confused as he walks home because things were quite different from the state he had left them before he went to North India for his research that took him 4 years. The palm-thatched houses that were in the distance had gone. All around him he sees rows of rubber plants on the ridge. They have grown twice his height. He used to see modan and vellaran there in the past. Now nothing. So he is confused.

Plus Two English Rice Summary Chapter 2 Question 2.
Why does the father wear a contented look?
The father wears a contented look because he is excited about the water of the Varanganal canal. He is getting enough water from the canal so that he can start his planting. He is happy with his work.

Rice Poem Questions And Answers Chapter 2 Question 3.
What changes have occurred in the lifestyle of the farmers when they shifted from food crops to cash crops?
Their life has become comparatively easy. They don’t have to struggle in the field wearing wet and muddy clothes. They are happy that they have stopped producing food crops. It was quite inconvenient. The farmer gained nothing. Cash crops bring more money and life is better. They feel good times have come. But they have to eat wheat instead of their favourite varieties of rice.

Plus Two English Rice Notes Chapter 2 Question 4.
Read the lines “My father says ……….. quite inconvenient” (Page 118).
Cite other instances of satire in the poem.

• “Only fools turn to rice-farming for gain.”
• “The government gives rice to those who don’t have paddy fields.”
• The narrator wants to eat athirika rice but his younger brother is bringing the ration for the household which consists of only wheat!
• The Chief Minister flying like an arrow to the Centre to clamour for more gains.
• Can we get some husk from the Centre, too, to makes toys with it?” This is the bitterest satire because he has a doctorate on making toys with husk.

Plus Two English Poem Rice Summary Chapter 2 Question 5.
The poem is a contrast between expectations and reality. Prepare a write-up substantiating this.
The poem is definitely a contrast between expectations and reality. The narrator goes to North India to do a research on making toys with husk as there is a lot of husk in Kerala because of the huge amount of paddy produced at that time. He spends 4 years and gets a doctorate. Naturally he has plans to use his knowledge and expertise to make toys from husk and thus generate employment opportunities for others. But when he comes back with his doctorate what does he see? The paddy fields have been converted to rubber plantations. No husk!

He is tired of eating chapatti day after day and he longs to eat his favourite rice when he comes home. What does he see? His younger brother brings wheat as the ration for the whole family. Now he has to continue eating chapatti!

Plus Two English Rice Summary In Malayalam Chapter 2 Question 6.
Comment on the style of writing of the poem.
The poem is written in a colloquial style. The poem is a translation from the Malayalam original. A number of Malayalam words are used in the poem. These words will not be understood by foreigners even with some explanations as they would not even know the various varieties of rice that we cultivate here. Such words are ‘athirika’, ‘modan’ and ‘vellaran’.

The imagery is quite good. The picture of the father wearing a mud-stained dhoti and working in his field is very touching. We see the ploughing scene. We can see the rubber trees on the ridge and dealwood trees along the canal. Then we see another picture of the father sitting and watching workers fixing the machine for making rubber sheets. We see the little brother of the narrator running in with the ration and tripping and falling down scattering the wheat he has brought. We see the aeroplane carrying the Chief Minister, flying to the Centre, to clamour for more grain. He is flying over the cash crops.

The stanzas are irregular – sometimes 4 lines, sometimes 5 lines and there are three 6-line stanzas. There is also a 2-line stanza in the end. It is a prose- poem. Being a satirical poem, the style is simple, direct and lucid and it suits the theme.

#### Rice (Poem) Edumate Questions and Answers

Plus Two English Rice Short Summary Chapter 2 Question 1.
“… only fools turn to rice-farming for gains”
Do you agree with this idea expressed by the father in the poem Rice? Express your opinion/suggestion in three or four sentences. You may use expressions like ‘I think…, I feel…, I suggest…, In my opinion…’
I think the father is wrong here. I feel that saying that only fools turn to rice farming is an insult to the rice farmers who work hard to give us rice to eat. I suggest that the father should have told his son that any job is good provided one does it well. He should have given importance to the principle of dignity of labour. In my opinion rice farming is a good way of earning one’s livelihood and also a way of making good profits. My neighbour is a rice farmer and he is very rich and he lives in style in a palatial bungalow, with a pair of cars and a few servants working for him every day.

Question 2.
a) The poet in Rice finds his place totally changed in four years. What are the major changes mentioned by the poet?
b) Describe a place in your locality elaborating the changes it has undergone within a short period of time.
a) The poet in “Rice” finds the place totally changed in four years. The palm thatched houses are gone. There are only rubber plants now. There are no rice fields any more. There is no noise of people ploughing. The whole field is planted with areca nut palms. In the corner along the canal there are the dealwood trees which were not there four years ago.

b) My locality has undergone many changes in a short time. Mine was a village. In the junction there were a few shops. A grocery shop, a tea shop, a shop that sold tobacco, beedi, cigarette and ‘paan’, a stationery shop where one bought his pen, pencil, notebooks etc. There was also a barbershop. In the tea shop there would be some people all the time discussing all the things under the sun – from the foreign policy of Donald Trump to the love affair of a local boy and girl. But all this is gone.

Now there is a huge shopping mall there. If one left the place a couple of years ago and came back only now, he would not even recognize the place. So many changes have taken place in my locality.

Question 3.
Much of our water bodies are polluted by industrial waste and toxic chemicals and fertilizers from farmlands. Prepare an essay describing the increasing rate of water pollution in our state.
(Hints: reason for pollution – impact of pollution – remedial measures etc.)
Water covers two-thirds of the Earth’s surface, with over 97% present in the oceans and less than 1% in freshwater streams and lakes. Water is also present in the atmosphere in solid form in the polar icecaps and as groundwater in water-bearing rocks deep underground. Water is called a universal solvent because many things get dissolved in it.

Water pollution may be defined as any chemical or physical change in water, harmful to living organisms. It can occur through natural processes. For example, water can be polluted by sediments produced by natural erosion. Water bodies get polluted as they receive a lot of waste produced by human activity. This waste is discharged directly into the water bodies by sewers or pipes from factories and washed down from agricultural or urban areas, especially after heavy rains.

Sources of pollution may be domestic, agricultural or industrial. In the underdeveloped and developing countries human and animal waste and sediments from defective agricultural and forestry practices are the main pollutants. In developed countries, industrial pollutants such as toxic metals and organic chemicals add to the water pollution. This is more dangerous than the pollution caused by human and animal waste.

Water pollution is caused by different things. They include excessive plant nutrients, acidification by acid rain and acid mine drainage, organic compounds containing chlorine like DDT and other pesticides, oil getting into water sources through drilling or accidental spillage from oil tankers, the discharge of huge quantities hot water into water bodies causing thermal pollution, fluoride and arsenal pollutants. Now the question comes how can we prevent water pollution? If the water is polluted people and animals easily become sick and die. But we can do certain things to prevent or reduce pollution. First of all industries should not be allowed to discharge untreated chemicals into water bodies, especially rivers and lakes.

Secondly, the domestic sewage system should be designed in a scientific manner. Thirdly, Underground water can be saved from pollution if the land is not exposed to pesticides and other industrial chemicals. Farmers should be encouraged to do organic farming instead of using a lot of artificial fertilizers, pesticides and other such things that pollute the soil and later get washed into the water bodies.

Question 4.
In connection with the activities of the Haritha Keralam project, the Nature Club of your school has organized a programme for planting trees in the campus. You are asked to deliver a speech on the importance of preserving nature and natural resources. Draft the speech you would like to present there.
Respected Principal, dear teachers and friends,
As you are aware in connection with the activities of the Haritha Keralam Project, the Nature Club of our
School has organized this programme for planting trees in the campus. Ralph Waldo Emerson once said, “Nature never hurries, atom by atom, little by little, she achieves’ fier work.” He is very right when we think how the trees grow taking their time to reach their full status as trees. Before a tree becomes a full-fledged tree it passes through many stages – seed, seedling, sapling and then tree. Some trees take years to become real trees. But look at man! How cruelly he cuts them down in just a few minutes to use it as fuel, for furniture work, construction works and even to make paper. Sometimes man destroys entire forests to convert them into farmlands or to make factories and residential areas. Large scale deforestation brings about climatic changes. Even in Kerala there are climatic changes because of the large scale destruction of forests.

Trees absorb the carbon dioxide in the atmosphere and supply us oxygen. They prevent the soil from eroding. They help in blocking the clouds and bring rain. They supply us a lot of forest products like honey, wax and different kind of herbs and roots. Nowadays forests are converted into National Parks attracting tourists from all over the world. That way also, trees bring us money. Thus forests help us in so many ways and so destroying them will be suicidal.

We all should emulate Wangari Maathai, the Kenyan lady and the Nobel Prize Winner, who led a crusade against deforestation. She wanted each person to plant some trees as his duty to help himself and also posterity. In India we had the Chipko Movement. When the contractors came to cut down the trees, the women from the locality rushed to the forest and stood near the trees embracing them. Chipko in Hindi means embrace. The men who came to cut down the trees had to go back because of the people embracing the trees.

John Keats, the famous English Romantic poet said, “A thing of beauty is a joy forever.” I believe a tree is a thing beauty and we should do our best to preserve our trees and plant new ones. Today let’s pledge that each one of us will plant at least ten trees. Remember it Is not enough to plant trees, but make sure that they grow by giving the trees adequate care.

Let’s make our campus green and contribute our share to make the earth green.
Thank you all!

Question 5.
Read the following line from the poem Rice and answer the given question.
Handloom dhoti stained with yellow mud’
What does this line imply?
“Handloom dhoti stained with yellow mud” – This line describes a farmer working in the field. He is wearing a handloom dhoti. The field is full of muddy water and since he is working in it, his dhoti is stained by the muddy water, ft implies the hard work of a farmer.

Question 6.
The Nature Club of your school wishes to visit the Botanical Garden at Thiruvananthapuram. Draft a letter to the Director of Botanical Garden, Thiruvananthapuram seeking permission to visit the garden and to enquire about their research projects on food crop cultivation.
The Secretary
Nature Club
NSSHSS, Pullur
7 June 2017

The Director
Botanical Garden
Thiruvananthapuram

Dear Sir,
Sub: Permission to Visit the Botanical Garden
The Nature Club Members of this school want to visit the Botanical Garden as part of their educational tour. We are 35 in number, including two teachers. We would lie to visit the place on Saturday the 25* of this month, from 10.00 a.m. We also want to make some enquiries regarding your research projects on food crop cultivation because we also want to do some agricultural production in our school. It has a lot of land which could be used for the cultivation of food crops.

Geeta Sankar
Secretary

Question 7.
Reads the lines from the poem ‘Rice’.
Son, we’ve stopped working on all the rice.
It was quite inconvenient.
a) ‘all the rice’ means ………..
b) What does it tell us about the farmers?
(Hints: change in attitude-lifestyle-profit motive-job preferences etc.)
a) Different type or rice/Everything in connection with rice cultivation.
b) It tells that there is a change in the attitude of the farmers. Their life style has changed. They are now driven by the profit motive. They are also reluctant to work in the fields as such work makes their bodies and clothes covered with mud.

Question 8.
The bar diagram shows the production (in tonnes) of wheat, rice, coarse grains and oilseeds of different countries. Analysefit and prepare a write-up.
Foodgrain And Oilseeds Production In The world
A study was conducted in April 2011 to see the production rate of wheat, rice, coarse grains and oil seeds in countries like India, the USA, China and the rest of the world. The study has shown the following results.

The highest rate of wheat production per hectare went to China with 4.7 tonnes per hectare. It was followed by the USA with 31.1 tonnes. India and the rest of the world were equal with 2.9 tonnes each. In the case of rice, the USA had the highest rate with 7.5 tonnes per hectare. China came second with 6.7 tonnes. India had only 2.2, whereas the rest of the world produced more than India with 4.3 tonnes per hectare.

Coarse grains had a different story to tell. USA and China were equal in this regard with 9.0 tonnes per hectare. The rest of the world had 3.5 whereas India had the lowest with just 2.5 tonnes.

In oil seeds, USA topped with 2.7 closely followed by China with 2.1. India came third with 1.2 . Quite strangely the rest of the world did not have any oil seeds production at all.

Question 9.
“Can we get some husk from the Centre, too, To make toys with it? I don’t know.”
The poem ‘Rice’ ends with these sarcastic lines. Do you think the poem is a satire? If so, what does the poet try to satirise? Consider the poem as a satire and prepare a paragraph on your views.
Chemmanam Chacko’s “Rice” is not just a satire but a biting satire. It shows the greed of some people and how they misguide farmers to change their crop from rice to cash crops like rubber. We can’t eat rubber! For our rice we have to depend on our neighbouring States and the rice we get is contaminated in so many ways. The height of the satire is that the boy went to North India to get his doctorate degree by researching the uses of husk. He spent 4 years there researching on the topic of making toys with husk.

Because his father is a farmer producing rice, there would be plenty of husk even in his own house to work with. But when he comes back with his doctorate degree what does he see? The rice fields have been converted to rubber plantations. Where will he get the husk now? Maybe he will ask the Central Government to give him some husk to make toys! His four years’ research is rendered useless now!

Activity – I: (Critical Appreciation)

Prepare a critical appreciation of the poem in the light of your responses to pie questions above.
The poem “Rice” written by Chemmanam Chacko’ and translated by Prof. Ayyappa Paniker is a superb satire hitting hard at the greed of some farmers here. The poem is laced with scorn and sarcasm. The son of a rice farmer goes to North India to do a research on making toys with husk. He works hard for four years, eating chapatti day after day, and finally succeeds in getting a doctorate. He must have thought of doing research on the possibilities of husk because as the son of a rice farmer he had plenty of husk at home and also in the homes nearby. By using husk for toy making, a lot of people could find employment in a state notorious for unemployment.

But then there is the anticlimax. When the son returns home with his well-earned doctorate on making toys with husk, there is no husk at all in his house or around. Farmers had shifted to cash crops, especially rubber, as they found rice cultivation is inconvenient and non- profitable. Moreover, a lot of incentives were given by promoters to cultivate cash crops. The son finds his father watching people setting up a machine for making rubber sheets. All the paddy fields are gone and in their place what he sees are rubber trees and dealwood trees.

There is biting sarcasm in the description of the Chief Minister flying to the Centre to request for more food grains to feed the people here. There is no more rice to eat. People have to eat wheat. The son comes home after four 4 years with a desire to eat his favourite ‘athirika’ rice. But his desire will remain an unfulfilled desire as he too has to eat the wheat that is given to the household as ration. And his doctoral degree? How can he find husk to make toys?

I would not call it an exquisite poem comparable to the poems of Wordsworth, Coleridge, Shelley or Keats. But it can stand comparison with the poetry of Alexander Pope who excelled in satire. The poem has excellent imagery. The poet has used a number of Malayalam words in the poem. The language is colloquial. The stanzas of the poem are irregular – ranging from 2 lines to 6. It is a prose-poem. Being a satirical poem, the style is simple, direct and lucid and it suits the theme. Chemmanam Chacko has done a good job in showing how the farmers have changed from simple people to shrewd commercially minded people.

Activity – II (Write-up)

Chemmanam Chacko is a master satirist who has fought many a battle with the system through his writings, laced with scorn and sarcasm. He says, ‘Socio-political sphere is much meek and limited compared with earlier times. Society has changed, and, with it, have the mass sensibilities.’ Consider his poem ‘Rice’ as a satire on the farmers who are forced to switch to cash crops when the market for food crops fell. Prepare a write-up.
Chemmanam Chacko is very right in observing that society has changed and with it the mass sensibilities have also changed. Chacko was born in 1926, when Kerala was very much an agricultural land. People cultivated mainly food crops. Hardly anything was brought from outside. This situation continued until the 1960s and 70s, when there was an exodus of Kerala people, especially the youth, to the Gulf Countries, America and Europe.

In most families in Kerala there will be at least one member who will be working abroad. These people working abroad send money to their parents and relatives at home. Soon the Kerala farmers stopped working hard on their farms. Often they left them without cultivating because money was coming from outside and they could buy their food.

Many farmers started cultivating cash drops, especially rubber. As we saw in the poem, rice cultivation is inconvenient and non-profitable. We all have heard the saying that Indian agriculture is a gamble on the monsoons. Once we change into cash crops we are not so heavily dependent on the monsoons. But crops like rice heavily depend on the rains. If there is too much rain, there is a problem. If there is too little rain, there is a problem. If the rains don’t come in time, there is a problem. So rice farmers started shifting to cash crops. There were many promoters, including the government, who encouraged cash crops.

The glitter of money was very tempting to our rice farmers. Farmers can’t go into the fields well-dressed and they can’t even keep their body clean all the time as they have to work in the muddy fields. Their love of luxury also made the rice farmers either quit or shift to other cash crops. The results are obvious. Kerala State has become a Consumer State. Today we get our food grains, vegetables and fruits from neighbouring states. We can’t blame the farmers. We all want to catch the fish without wetting our fingers. Hard work is no more considered a virtue. One who makes quick money will be considered a successful and smart person. No wonder, the farmers also want to be smart!

Activity – III: (Paragraph writing)

Instead of a system of values, we have the market ruling us, making-decisions for us. Consider this statement in the light of the poem ‘Rice’ by Chemmanam Chacko.
These days we are not ruled by any system of value, but by the market. The market decides what we should eat, what we should drink and what we should wear. It even decides how we should worship, and how we should behave in our families. In the past people cultivated food crops and most of the things they consumed came from their farms. But today we survive on fast foods.

Fashion Designers decide what we should wear, Soft Drink companies decide what we should drink. Our worship is often controlled by Tour Operators who organize tours to places of Pilgrimage. The Gold and Diamond merchants decide how we should show our love to the family members by giving what gifts on which occasion. Marriages are controlled by videographers! Markets rule us and we do as they dictate. Even our religious festivals have become occasions for discount shopping.

You have now understood the importance of protecting the world around us. Will-this be enough? Can a society survive without ensuring that its members have a healthy life style? Won’t it be dangerous if people surrender themselves to different kinds of addiction? Let’s find out.

#### Rice (Poem) About the Poet

Prof. Chemmanam Chacko was born in 1926, at Mulakulam in the erstwhile Travancore. He is a popular poet in Malayalam. He is a master satirist and he is fond of poking fun at the trivialities of people and their customs.

#### Rice (Poem) About the translator

Prof. K. Ayyappa Paniker (1930-2006) is a poet, literary critic, an academic and a famous scholar. He is pioneer of modernism in Malayalam poetry and his book “Kurukshetram” is a turning point. He taught English in various colleges and universities. He retired as Director, Institute of English, University of Kerala.

#### Rice (Poem) Meaning Of The Poem in English

Stanza 1: After four years of research in North India, I return home. I have received a doctoral degree. I also received a lot of praise for my work on making toys with husk. I am bored with eating chapatti every day and now I want to eat a meal of athikira rice (a variety of rice popular in central Kerala).

Stanza 2: When I get back home, it will be the planting season. My father is in the fields below our house. He is wearing a handloom dhoti stained with yellow mud. He is excited about the water of the Varanganal canal. He will greet me amidst the shouts of ploughing with several oxen.

Stanza 3: The oxen will stop when they see me walking with my suitcase. There is a smile coming to my father’s lips and he does not show it. From the field itself he asks me when I started my journey from there.

Stanza 4: My little brother is carrying the tender saplings to be planted where the ploughing is done. When he sees me he will run and shout so that the people at home can hear him, saying “Mother, brother has arrived.”

Stanza 5: I walk carefully along the dyke so that I don’t upset the baskets kept there. They are full of seed. At last I reach home. Mother has drained the well-cooked rice.

I tell the train to run faster so that I can get home quickly and eat to my satisfaction.

II (Page 116)
Stanza 6: The bus stops on the road close to my house. When I left the place palm-thatched houses could be seen in the distance on the right side. But now there is nothing. There are only trees. The place has changed completely.

Stanza 7: All around me I see rows of rubber plants on the ridge. They have grown twice my height. I used to see modan and vellaran (varieties of paddy) here. I am confused even about the path leading to my home.

Stanza 8: There is no noise of people below. No shouts of ploughing. The whole field is planted with areca nut palms. In the corner, along the canal, there are the dealwood (wood that is soft and easy to saw, usually used for making packing cases and boxes) trees.

Stanza 9: I get into the house. On the southern side, my father is watching workers fixing up the machine for making rubber sheets. He looks happy and contented.

Stanza 10: My father tells with some pride that he has stopped producing rice. It was quite inconvenient. The farmer gained nothing. Only fools will try rice-farming for any gain. Rubber money is better. Good times have come. The government gives rice to those who don’t have paddy fields.

Stanza 11: My small brother runs to meet me. I am eager to eat a full meal of athikira rice. He is carrying the rations for the entire household. He trips over something and scatters the wheat all over the yard.

Stanza 12: A plane is flying above us. It is going north. The noise of the plane drowns my brother’s loud cries. The Chief Minister is flying to the Centre to get more grains. He is flying above the cash crops which are now growing like trees. No one promotes the farming of rice here.

Stanza 13: Can we get some husk from the Centre to make toys? I don’t know.

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## Kerala Plus Two Botany Chapter Wise Questions and Answers Chapter 1 Reproduction in Organisms

### Plus Two Botany Reproduction in Organisms One Mark Questions and Answers

Plus Two Botany Chapter Wise Questions And Answers Pdf  Question 1.
Syngamy means:
(a) Fusion of similar spores
(b) Fusion of dissimilar spores
(c) Fusion of cytoplasm
(d) Fusion of gametes
(d) Fusion of gametes

Plus Two Botany Chapter Wise Questions And Answers Question 2.
Fill in the blanks by observing the relationship of the first.

1. Amoeba: Binary fission
Yeast: …………..
2. Reptiles: Oviparous
Mammals: ………….

1. Yeast: Budding
2. Mammals: Viviparous

Plus Two Botany Questions And Answers Chapter 1 Question 3.
Internal buds of Sponges are:,
(a) Spores
(b) Gemmules
(c) Planula
(d) Blastos
(b) Gemmules

12th Botany 1st Lesson Important Questions  Question 4.
In honeybees and lizards, the female gamete undergoes development to form new organism without fertilization. Name this phenomenon.
Parthenogenesis

Plus Two Botany Previous Question Papers Chapter Wise Question 5.
The life span of man is_____years.
(a) 60
(b) 70
(c) 85
(d) 100
(d) 100

Plus Two Botany Chapter Wise Questions And Answers Hsslive Question 6.
What is meiocyte?
(a) The cell undergoes meiosis
(b) The cell undergoes mitosis
(c) Both a and b
(d) None of the above.
(a) The cell undergoes meiosis

Plus Two Botany Chapter Wise Previous Questions And Answers Pdf Question 7
All individuals produced are genetically identical
(a) clone
(b) offspring
(c) fission
(d) monoecious
(a) clone

Plus Two Botany Previous Year Questions And Answers Chapter 1 Question 8.
Hydra reproduces asexually by
(a) budding
(b) binary fission
(c) Multiple fission
(d) Vegetative propagation
(a) budding

Plus Two Botany Focus Area Questions And Answers Chapter 1 Question 9.
Find the propagules of ginger
(a) Runner
(b) Bulbil
(c) Rhizome
(d) Bulb
(c) Rhizome

Hsslive Botany Question Bank Plus Two Chapter 1 Question 10.
Bulb acts as a propagule in
(a) Onion
(b) Garlic
(c) Agave
(d) Both and b
(d) Both and b

Plus Two Biology Chapter Wise Questions And Answers Pdf Question 11.
In some algae_____are motile
(a) male gametes
(b) female gametes
(c) both male and female gametes
(d) none of the above.
(c) both male and female gametes

Plus Two Botany Chapter Wise Previous Questions And Answers Question 12.
A population of genetically identical plants derived from a single parent is called………..
Clone

Plus Two Botany Previous Questions Chapter Wise Chapter 1 Question 13.
Embryo sac is found in:
(a) Endosperm
(b) Embryo
(c) Ovule
(d) Seed
(c) Ovule

Plus Two Botany Previous Questions Chapter Wise Pdf Chapter 1 Question 14.
Types of asexual reproduction found in Hydra is………
(a) Gemmule formation
(b) Budding
(c) Sporulation
(d) Multiple fission
(b) Budding

Reproduction In Organisms Class 12 Objective Questions Question 15.
Observe the relation and fill in the blanks. Syncarpous: the pistils remain united: the pistils remain free.
Apocarpous

Reproduction In Organisms Class 12 Questions And Answers Pdf Question 16.
In honeybees and turkey new organisms are formed through a peculiar phenomenon. Name that phenomenon.
Parthenogenesis

Botany Chapter Wise Questions And Answers Chapter 1 Question 17.
Choose the correct answer: Chlamydomonas reproduces asexually through
(a) Gemmules
(b) Conidia
(c) Bud
(d) Zoospores
(d) Zoospores

Question 18.
In Marchantia male and female thallus are separate. This condition is called
(a) Dioecious
(b) Hermaphrodite
(c) Monoecious
(d) Bisexual
(a) Dioecious

Question 19.
Find the odd one. Bulbil, Bulb, Rhizome, Stem tuber.
Bulbil

Question 20.
Identify the correctly matched pair.
(a) Chlamydomonas – conidia
(b) Sponge – Zoospore
(c) Hydra – bud
(c) Hydra – bud

### Plus Two Botany Reproduction in Organisms Two Mark Questions and Answers

Question 1.
Sugarcane, Banana, Ginger, and Onion are flowering plants. But these are mostly cultivated by vegetative propagation.

1. What is meant by vegetative propagation.
2. Can you identify which part of the above plants are used for vegetative propagation.

1. The regeneration of new plants from the portions of vegetative organs like stem, root, and leaves.
2. plants are used for vegetative propagation.
• Sugar cane – stem
• Banana – Rhizome
• Ginger – Rhizome
• Onion – Bulb

Question 2.
Leaves not only produce foods, but also produce young ones. Comment on this statement.
In some plants, leaves are organ of vegetative propagation i.e new plants are formed from leaf margins eg-Bryophyllum.

Question 3.
How is pistia and zingiber officinale different in their mode of Asexual reproduction?
The above two plants shows vegetative propagation. The vegetative propagation is a form of asexual reproduction The vegetative propagule of pistia is offset (short and thick internode) and zingiber is rhizome (underground stem).

Question 4.
Fill in the blanks by observing the relationship of the first.

1. Male and female flowers on the same plant: monoecious
Male and female flowers on different plants: ………….
2. Fusion of gametes: Syngamy
Offspring from unfertilized female gamete: ……………

1. Dioecious
2. Parthenogenesis

Question 5.
Plant can reproduce both vegetatively and sexually. Name any 4 vegetative propagule.

1. Eye of potato,
2. rhizome of ginger,
3. bulbil of agave,
4. offset of water hyacinth.

Question 6.
Given below is a diagram showing particular types of reproduction.

1. Name the process of reproduction.
2. Briefly explain the process

1. Budding
2. It is the mode of asexual reproduction in which the unicellular organism divides unequally and the buds remain attached to the parent cell. Eg. yeast.

Question 7.
Given below the figures of different types of reproduction present in organisms. Identify and name the type of reproduction.

1. Zoospores in chlamydomanas
2. Conidia of penicillium
3. Eyes of potato
4. Leaf buds of bryophyllum.

Question 8.
Match the following: (Hint: Name of organisms in column A and asexual reproductive structure in Column B)

 A B 1. Chlamydomonas 2.  Penicillium 3.  Hydra 4.  Sponge 1. Gemmules 2. Buds 3. Zoospores 4. Conidia

 A B 1.  Chamydomonas 2.  Pencillium 3.  Hydra 4.  Sponge Zoospores Conidia Buds Gemmules

Question 9.
In diploid organisms, meiosis and gametogenesis are always interlinked. Justify.
In diploid organism, reproductive cells undergoes reduction division – meiosis and produces haploid gametes. Gametogenesis means the process of gamete formation. So meiosis and gametogenesis are always interlinked.

Question 10.
Categorize the following into pre and post-fertilization changes in plants. Gamete transfer, pericarp formation, ovule development, embryogenesis.
Pre fertilization-ovule development and gamete transfer. Post fertilization -embryogenesis and pericarp formation.

Question 11.
Events of sexual reproduction is given below.

1. Write the events in a sequential order. Fertilization, gamete transfer, zygote, gameto – genesis, embryogenesis.
2. Point out the differences between gametogenesis and embryogenesis

1. Gametogenesis, gamete transfer, fertilization, zygote, embryogenesis.

Question 12.
‘Water-hyacinth’ is commonly called ‘Terror of Bengal’. Give reason.
‘Water-hyacinth’ is a highly spreading aquatic weeds, which drains oxygen from the water and leads to the death of fishes.

Question 13.
Bamboo plant and strobilanthus kunthianus (Neela Kurunji) have similarity and dissimilarity in their reproductive and vegetative character. Justify.

1. Similarity – Both Bamboo and neela kurinji are perennial plants.
2. Dissimilarity – Bamboo flowers only once in their life time (in the period of 50-100 years) and die whereas neelakurinji flowers once in 12 years.

Question 14.
Zygote is called as the vital link. Why?
Zygote is the product of fertilization, that maintains the continuity of species between organisms of one generation and the next.

Question 15.
Post fertilization changes in flowering plants are given, by observing the given pair complete the others.
(a) Zygote: Embryo
(b) Ovule: ……….
(c) Ovary: ………
(d) Nucellus: Perlsperm
(e) Integument: ……….
(f) PEN: ………..
(b) Ovule: Seed
(c) Ovary: Fruit
(e) Integument: seed coat
(d) PEN: Endosperm

Question 16.
Leaves are used not only for photosynthesis but also for reproduction.

1. Name the type of reproduction
2. Give an example.

1. Vegetative propagation
2. Eg-Broyophyllum

Question 17.
Observe the relationship between first two terms and fill up the blanks.
1. Asexual reproduction – Unparental
Sexual reproduction……………….

2. Zygote: Diploid nucleus: Endosperm: ………….

3. Pea: non albuminous: wheat ………….

1. biparental
2. Triploid nucleus
3. albuminous

Question 18.
In yeast and Amoeba the parent cell divides to give rise to two new individual cells. How does the cell division differ in these two organisms?

• In yeast, the cell division is unequal and small buds are produced.
• In Amoeba, the cell division is equal and identical daughter cells are produced.

Question 19.
Offsprings formed due to sexual reproduction have better chances of survival. Why?
Sexual reproduction combines the characters of two parents and introduces variations which make the offsprings better in environmental adaptation.

Question 20.
Some organism are capable of producing fertilized egg but others are not possible.

1. Name the organism producing fertilized egg.
2. Why are offspring of oviparous-animals subjected to greater risk as compared to offsprings of viviparous animals.

1. oviparous animals
2. In oviparous orgaisms development of zygote take place outside the body of the female parent while in viviparous organisms development of zygote take place inside the body of female organisms. Hence viviparous type get proper embryonic care and protection.

Question 21.
In one type plant Adventitious buds develops into new leafy shoot by leaf as means of vegetative propagation but in others by rhizome. Give examples for leaf and rhizome used as unit of vegetative propagation.

• Leave – bryophyllum
• Rhizome – banana, and Ginger

Question 22.
In certain lower plants, equal and un equal daughter cells are produced by asexual reproduction. Name these Asexual Reproduction with examples
In binary fission equal daughter cells are produced. Example bacteria and Paramecium. In budding two unequal daughter cells are produced. Example Yeast and hydra.

Question 23.
In lower plants movement of gametes takes place by water but in seed plants male gametes are non motile. Which part of plant helps the movement of gametes?
Pollen tube

Question 24.
Haploid gametes are produced in bryophytes and angiosperms by two types of cell division. Name it

• bryophytes-Mitosis
• angiosperms-meiosis

Question 25.
Sexual reproduction is very common in higher plants to produce young ones that shows variation.

1. Name the product formed in sexual reproduction is considered as vital link
2. Give its significance

1. product-zygote
2. It maintains continuity of species for many generation.

Question 26.
Nature prefer internal fertilization than external fertilisaiton. Do you agree with this statement.
Yes, In nature the advanced plants and animals show internal fertilisation.

Question 27.
In fishes gamete fusion takes place in external medium. Give the disadvantage of this method.
The offsprings formed through external fertilisation are subjected to the attack of predators.

Question 28.
Animals are categorized into viviparous and oviparous based on the development of the Zygote takes place outside the body or inside. In which of these two types the chances of survival is greater. Why?
In oviparous animals like reptiles and birds, the fertilised eggs after the period of incubation young ones hatch out. In viviparous animals, the zygote develops into a young one inside the body of the female organism. After the period of growth, they are delivered out. The chances of survival is greater in viviparous animals.

Question 29.
Mention the characteristic feature and a function of zoospore in some algae.

• Zoospores are flagellated, motile aquatic bodies
• On germination give rise to new plants

Question 30.
Name the site of occurrence of syngamy in amphibians and reptiles.

• In amphibians syngamy occurs outside the body of organism i.e external.
• In reptile syngamy occur inside the body of the organism i.e internal

Question 31.
What are the three major phases in the life cycle of an organism? Define each phase.
Juvenile phase, Reproductive phase, senescent phase

• Juvenile phase-The phase of growth in the organisms before reproductive maturity
• Reproductive phase- In this phase organism attains reproductive maturity
• Senescent phase- The phase between reproductive maturity and death

Question 32.
How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?
The progeny formed from asexual reproduction is the product of single parent and does not have genetic variations.

Question 33.
In yeast and amoeba the parent cell divides and give rise to two new individual cells. How does the cell division differ in these two organisms?
In yeast, cell division is unequal and small buds are produced that remain attached to parent cell. In amoeba the cell division is equal and produces identical daughter cells.

Question 34.
Mention the site where syngamy occurs in amphibians and reptiles respectively.
In amphibians syngamy occurs outside the body of organisms in the external medium (water). In reptiles syngamy occurs inside the body of an organism.

Question 35.
The turkey usually produces females for several. generations. How is this possible?
In a turkey, female gametes undergo development without fertilization. This phenomenon is called parthenogenesis.

Question 36.
Leaves not only produce food but also produce young ones. Substantiate.
In few plants the leaves are the means of vegetative propagation in addition to photosynthesis. In such plants, plantlets are formed on the leaves eg- Bryophyllum.

Question 37.
Observe the figures given below, identify and name the organisms and their asexual reproductive structures.

Organisms – Asexual reproductive Structures

1. Chlamydomonas – Zoospore
2. Penicilium – Conidia

Question 38.
The chromosome number in the meiocyte of housefly is 12 and that of its gamete is 6. Write the reason for the change in chromosome number of meiocyte and gamete.
Meiocyte is a diploid cell undergo reduction division to form haploid gametes.

Question 39.
In coconut, male and female flowers are separate. Write the technical term for the male and female flowers. Write the condition of the flower.

1. Female flower – pistillate
2. Male flower – staminate
3. Condition of the flower – unisexual

Question 40.
Observe the figures given below. Identify the gametes in A and B. Justify your answer.

1. Homogametes
2. Heterogametes

In first figure gamates come to fusion have similar structure but in second figure gametes have dissimilar structure.

Question 41.
Match the items of column A with B

 A B a. Offset i. Ginger b. Bulb ii. Agave c. Rhizome iii. Onion d. Adventitious leaf buds iv. Water hyacinth v. Bryophyllum

Question 42.
Analyse the table given below and fill in the blanks.

1. Hermaphrodite or bisexual
2. Female gamete develops into new organism without fertilisation
3. majority of algae or fishes or amphibians
4. Fertilisation occurs inside the body

Question 43.
The chances of survival of young ones is greater in viviparous organisms than oviparous organisms. Justify this statement.
In viviparous organisms the zygote develops into young one inside the body of the female organism. So during the course of development proper embryonic care and protection is provided inside the female organism.

Question 44.
In some organisms male and female reproductive organs are seen in the same individual. Name that condition. Give two examples of such organisms.
Hermaphrodite Eg.Earthworm, Leech.

Question 45.
Name of certain plants and their vegetative propagules are given below. Make correct pairs using them. Agave, Offset, Ginger, Bulb, Bulbil, Water hyacinth, Potato, Bryophyllum, Rhizome

1. Agave – Bulbil
2. Ginger-Rhizome
3. Water hyacinth – offset
4. Onion-Bulb

Question 46.
Gametogenesis and gamete transfer are the two prefertilisation events. Write the differences between the two.
Gametogenesis – It is the formation of gametes.
Gamete transfer – It is the transfer of male gamete to the female gamete.

Question 47.
Based on the nature of reproduction, organisms are classified as continuous breeders and seasonal breeders. Write the difference between the two.

1. Continuous breeders – They are reproductively active throughout their reproductive phase.
2. Seasonal breeders – They reproduce only during favourable seasons in their reproductive phase.

Question 48.
Observe the relationship between the first two terms and fill in the blanks.

1. Hydra: Bud; …….: Gemmule
2. Birds: Internal fertilization; …….External fertilization.

1. Sponge: gemmule
2. Algae or fishes or amphibian

Question 49.
Fertilisation in some organisms occur outside the body. Name that kind of fertilization. Write its disadvantage.
External fertilization
In this type young ones are subjected to threat of predators. So, it is the disadvantage of external fertilisation.

Question 50.
The offspring formed by asexual reproduction is referred to as clone. Justify this statement.
Offsprings formed are morphologically and genetically similar among themselves and to their parents. So such individuals are called clone.

### Plus Two Botany Reproduction in Organisms Three Mark Questions and Answers

Question 1.
During sexual reproduction-fertilization is an important process. But in most aquatic organisms like fishes, amphibians, etc. external fertilization occur. In most terrestrial organisms internal fertilization is common.

1. Point out the differences between external and internal fertilization.
2. Write the disadvantages of external fertilization.

(a) 1. External Fertilization:
Syngamy occurs in the external medium – water i.e. outside the body of organisms. Eggs and sperms are released outside the female and male parents respectively and move through water before fusion of gametes.

2. Internal Fertilization:
Syngamy occurs inside the body of organisms. Egg is formed inside the female body where they fuse with male gamete that is released by the male parent.

(b) Offsprings formed by external fertilization, are extremely vulnerable to predators threatening their survival up to adulthood.

Question 2.
Write technical terms for the following.

1. Morphologically different types of gametes.
2. Process of formation of male and female gametes.
3. Formation of new organisms without fertilization.

1. Heterogametes
2. Gametogenesis
3. Parthenogenesis

### Plus Two Botany Reproduction in Organisms NCERT Questions and Answers

Question 1.
In bacteria and lower organisms offspring formed are close similar among themselves and to their parents.

1. Name offsprings derived through such process
2. Write down the 2 characterstics of such offsprings

1. Clone
2. Morphologically and genetically similar

Question 2.
The offsprings formed by a sexual reproduction have better chances of survival. Why?
Sexual reproduction combines the characters of two parents and introduces variations which make the offsprings better in environmental adaption.

Question 3.
How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?
The progeny formed from asexual reproduction is the product of single parent and does not have genetic variations.

Question 4.
In some lower and higher plants offsprings are produced shows greater similarity and it occurs mainly through asexual reproduction, in higher plants the process occurs through vegetative propagation.

1. What is vegetative propagation?
2. Give two suitable examples.

1. The process of multiplication, in which parts or fragments of the plant act as reproductive unit or propagule to form new individuals is called vegetative propagation.

2. Examples,

• Buds (eyes) of Potato.
• Rhizome of Ginger.

Question 5.
Distinguish between asexual and sexual reproduction. Why is vegetative reproduction also considered as a type of asexual reproduction?

 Asexual reproduction Sexual reproduction 1. It is always uniparental 2. There is no formation and fusion of gametes. 3. Only mitotic cell division takes place. 4. Offsprings are genetically identical to the parent. 1. It is usually biparental 2. Formation and fusion of gametes take place. 3. It involves both meiosis and mitosis. 4. Offsprings genetically differ from the parent.

Vegetative reproduction involves single parent and its offsprings are genetically identical hence it is considered as a type of asexual reproduction.

Question 6.
Higher organisms have resorted to sexual reproduction in spite of its complexity. Why?
Because sexual reproduction induces genetic variability(adaptations) which make the offsprings better equipped for the struggle of existence.

Question 7.
Some organism are capable of producing fertilized egg but others are not possible.

1. Name the organism producing fertilized egg.
2. Why are offspring of oviparous animals at a greater risk as compared to offsprings of viviparous animals.

1. oviparous animals
2. In oviparous orgaisms development of zygote take place outside the body of the female parent, (they lay fertilized /unfertilized egg). In viviparous organisms development of zygote develops into a young one inside the body of the female organism.

This offers proper embryonic care and protection and have better chances of survival than young ones of viviparous organisms.

Question 8.
Prefertilisation events events of sexual reproduction in all organisms are gametogenesis and gamete transfer.

1. What are the post fertilization events?
2. Name the parts of plants such as ovary, ovule and ovary wall develop in post fertilization process.

1. Zygote formation Embrogenesis
2. Ovary-fruit, ovule-seed, ovary wall -pericarp

### Plus Two Botany Reproduction in Organisms Multiple Choice Questions and Answers

Question 1.
In papaya, the flowers are:
(a) Unisexual
(b) Bisexual
(c) neuter
(d) Flowers are not formed
(a) Unisexual

Question 2.
In vegetative propagation, characters of parent plants are:
(a) Changed
(b) Not preserved
(c) preserved
(d) Exchanged
(c) preserved

Question 3.
Rhizopus reproduces asexually by:
(a) Conidia
(b) Spores
(c) Gemma
(d) Bulbil
(b) Spores

Question 4.
In potato, vegetative propagation takes place by:
(a) Root
(b) Leaf
(c) Grafting
(d) Stem tuber
(d) Stem tuber

Question 5.
Vegetatively propagated plants are:
(a) Genetically similar
(b) Genetically dissimilar
(c) Do not bear roots
(d) Do not form buds
(a) Genetically similar

Question 6.
Syngamy means:
(a) Fusion of similar spores
(b) Fusion of dissimilar spores
(c) Fusion of cytoplasm
(d) Fusion of gametes
(d) Fusion of gametes

Question 7.
In which pair both the plants can be vegetatively propagated by leaf pieces?
(a) Bryophyllum and Kalanchoe
(b) Chrysanthemum and Agave
(c) Agave and Kalanchoe
(d) Asparagus and Bryophyllum
(a) Bryophyllum and Kalanchoe

Question 8.
Regeneration of a plant, cell to give rise to new plant is called :
(a) Reproduction
(b) Budding,
(c) Totipotency
(d) Pleuripotency
(c) Totipotency

Question 9.
Hydra reproduces asexually through:
(a) iragmettatwn
(b) Budding
(c) binary fission
(d) Sporulation
(d) Sporulation

Question 10.
Animals giving birth to young ones are
(a) Oviparous
(b) Ovoviviparous
(c) viviparous
(d) both b and c
(c) viviparous

Question 11.
The offspring that are exactly identical to one another as well as identical to their parents are called as
(a) clone
(b) twins
(c) replicates
(d) drones
(a) clone

Question 12.
Buds and conidia are asexual reproductive structures of
(a) hydra&rhizopus
(b) rhizopus & penciriium
(c) hydra & pencillium
(d) both a&b
(c) hydra & pencillium

Question 13.
Banana is multiplied through
(a) suckers
(b) seeds
(c) rhizome
(d) stolen
(c) rhizome

Question 14.
In rotifers, the type of reproduction responsible for the organisms to multiply is
(a) apomixis
(b) parthenogenesis
(c) asexual reproduction
(d) sexual reproduction
(b) parthenogenesis

Question 15.
Bulbils are employed for multiplication of
(a) bryophyllum
(b) crocus
(c) Agave
(d) strawberry
(c) Agave

Question 16.
During favourable condition Amoeba reproduces by
(a) binary fission
(b) buds
(c) multiple fission
(d) both a and c
(a) binary fission

Question 17.
After post-fertilization, the ovary and ovule develops into
(a) pericarp and fruit
(b) fruit and seed
(c) fruit and seed coat
(d) none of the above
(b) fruit and seed

Question 18.
The plants which flower every year after attaining certain maturity are called
(a) monocarpic
(b) polycarpic
(c) perennials
(d) annuals
(b) polycarpic

Question 19.
The type of reproduction occurs in ciliated protozoans is
(a) syngamy
(b) conjugation
(c) budding
(d) cross fertilization
(b) conjugation

## Plus Two Chemistry Notes Chapter Wise HSSLive Kerala

HSE Kerala Board Syllabus HSSLive Plus Two Chemistry Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium are part of SCERT Kerala HSSLive Plus Two Notes. Here HSSLive.Guru has given Higher Secondary Kerala Plus Two Chemistry Chapter Wise Quick Revision Notes based on CBSE NCERT syllabus.

 Board SCERT, Kerala Text Book NCERT Based Class Plus Two Subject Chemistry Chapter All Chapters Category Kerala Plus Two

## Kerala Plus Two Chemistry Notes Chapter Wise

We hope the given HSE Kerala Board Syllabus HSSLive Plus Two Chemistry Notes Chapter Wise Pdf Free Download in both English Medium and Malayalam Medium will help you. If you have any query regarding Higher Secondary Kerala Plus Two Chemistry Chapter Wise Quick Revision Notes based on CBSE NCERT syllabus, drop a comment below and we will get back to you at the earliest.

## Kerala Plus Two Sociology Chapter Wise Questions and Answers Chapter 3 Social Institutions: Continuity and Change

### Social Institutions: Continuity and Change Questions and Answers

Social Institutions Continuity And Change Important Questions Chapter 3  Question 1.
In the Census taken by the British Government in the year ………… information about the caste was included.
1860

Social Institutions Continuity And Change Questions And Answers Chapter 3 Question 2.
From the names given below, who is not considered to be a leader of the lower castes of South India?
a) Sri Narayana Guru
b) Ayyankali
c) Chattambi Swamikal
d) Gandhiji
Gandhiji

Social Institutions: Continuity And Change Questions And Answers Chapter 3 Question 3.
From… the British Government began to take regular census every 10 years.
a) 1781
b) 1881
c) 1891
d) 2001
1881

Social Institutions Questions And Answers Chapter 3 Question 4.
From among the given things what is not an important feature of the Upper Classes?
a) right to education
b) right to enter temples
c) Not drinking alcohol
d) political power
Not drinking alcohol

Sociology Class 12 Chapter 3 Questions And Answers Chapter 3 Question 5.
From the following chose the family in which members of only two generations live: Joint family, Nuclear family, Extended Family.
Nuclear family

In India The Institution Of Family Has A Trend Towards Chapter 3 Question 6.
The family in which newly married couples stay with the family of the father of the groom is called …….
Patriarchal family

Social Institutions Continuity And Change Notes Chapter 3 Question 7.
A family of a mother with her children and their children is called ………
Matriarchal family

Social Institutions Continuity And Change Chapter 3 Question 8.
What are the 3 most important social institutions of a society?
Jati, Tribe (Gotram), Family

Caste System Multiple Choice Questions Chapter 3 Question 9.
The 4-phase division of the society is called ……….
Caste System

Social Institutions Continuity And Change Pdf Chapter 3 Question 10.
Who is the father of Indian Sociology?
a) VenierElvin
b) G. S. Ghurye
c) M.N. Srinivas
d) Periyar
G.S. Ghurye

Sociology Class 12 Important Questions Chapter 3 Question 11.
Match the following

 A B C Caste Matrififcal Fought against caste system Herbert Risley Social Revolution Sanskritization M.N. Srinivas Matriarchal family Patrilocal Sri Narayana Guru 1901 Patriarchal family Authority Dominant Caste Information about Caste Dwelling place Regional classification More than 1000 sub-castes

Multiple Choice Questions On Social Institutions Chapter 3 Question 12.
Explain the different levels of the meaning of ‘Jati’.
Jati is translated as ‘caste’ in English. Caste actually comes from Portuguese. It means ‘pure breed’. Jati shows a social arrangement of people based on birth. Varna is another word that means Jati. Both these words ‘Jati’ and ‘Varna’ are often used as synonyms. But actually there are some differences between them. Varna is a Sanskrit word. It means color. The ‘Varna’ system originated during the Veda period. At that time the people were divided into four groups – Brahmin, Kshatriya, Vaisya, and Sudra. This 4-class division is called Varna, in these 4 groups, a good number of people were not included.

They were excluded from the 4-Varna system. They included those ostracised from the Jati, foreigners, slaves and the people of the defeated nations. These people who were excluded from the 4-Varna system were called ‘Panchamar’ or ‘the fifth group’. Jati is a common name to indicate a group. In this, even inanimate things, plants, animals and persons were included. Thus Jati is simply an institution and S does not mean Varna.

Question 13.
Explain the relation between Jati and Varna.
In fact, the actual relation between Jati and Varna is often debated by scholars. There are many imaginary concepts about this. Varna system is an all-India system. The 4-Varna system can be seen everywhere in India. But Jati is a regional or local institution. A Jati found in one area of India may not be found in another area. The Jati chain is different in each area. The main difference between Jati and Varna is in the number. Vamas are 4. But there are hundreds of Jatis and Updates. The Jati system is highly complicated.

Question 14.
What are the main features of Jati?

• Jati is decided by birth.
• Marriage in the same group (Swagana Vivaham).
• Restrictions in food items and sharing of food.
• Hierarchical system by giving people high, low and middle status.
• Upajatis and upa-upajatis
• Kulathozhil (specified family jobs)

Question 15.
What re the contributions of Jyoti Rao Phule?
Jyoti Rao Phule was later known as Mahatma Phule. He started a Social Reformation Movement in Maharashtra.

• He gave leadership to a fight against Brahmin domination and Jati system.
• He worked for the education of the lower castes and women.
• He rejected the idea of ‘purity-pollution’ (suddha- asuddha) concepts.
• He worked for righteousness and human rights of the lower castes.
• • He found the Satyajodhak Samaj.

Question 16.
Jati system is a combination of two kinds of principles. Explain.
Theoretically, Jati system is a combination of two kinds of principles. One kind is based on differences and distance. The other kind is based on universality and power hierarchy (chain).
Each Jati is supposed to be different from others. Therefore each Jati was expected to keep away from other Jatis. Most of the rules and regulations of Jati are formulated on the principle of preventing the mingling of Jatis. Thus there were restrictions regarding marriage and ‘panthibhojanam’ (eating together).

At the same time a Jati can’t survive by itself. It can survive only as part of larger community. And this is its universality. The community (society) as a whole is not based on any principle of equality. There is a hierarchical order there. It is like a ladder. In this ladder each Jati is given its place.
Jati is a combination of two kinds of concepts:
a) Differences and distance from other Jatis.
b) Part of a larger community and part of a power hierarchy.

Question 17.
Explain the purity-pollution (Suddha-Asuddha) concept of Jati system.
Purity and Pollution are like antonyms, opposite to each other. Purity is associated with the higher classes whereas impurity or pollution is associated with the lower classes. Purity is related to those things which are considered sacred. Thus some rituals in temples can be performed by only the Upper Castes like Brahmins. But those who do lowertypes of jobs, like cleaning, sweeping, etc. are considered impure and even their touch was avoided polluting (untouchability) by the Upper Classes.

Question 18.
What changes did Colonialism bring in the Jati system?
Scholars say1 that colonialism brought drastic changes in the Jati system. There are some scholars who even argue the Jati-system as we see it today is not part of the Indian system, but is a creation of colonialism. However, these systems were not deliberately introduced but happened over centuries, it is true that the colonialists did make some changes in the existing system.

Question 19.
Who is Periyar? What were his contributions?
E . V. Ramaswami Naicker is called Periyar. He was a rationalist and a leader of the lower caste people of South India. He taught that all people are equal and freedom and equality were the birthrights of everyone. He formed the ‘Swabhiman Movement.’ He took part in the Vaikom Satyagraha, in Kerala.

Question 20.
What were the reforms made for the prosperity of the lower castes by the Colonial rulers?
Towards the end of their rule, the Colonial Masters took some interest in bringing prosperity to the so called lower classes of people. As part of the reform activities, in 1935, the Government prepared a schedule of the lower classes. This schedule was officially approved by a law. Those that were approved in the schedule were known as Scheduled Castes and Scheduled Tribes and they got special considerations from the Government.

All the lower class people who even suffered from the inhuman, horrible ‘untouchability’ were in the schedule. This way the Colonial rulers brought some good changes in the rigid caste system in India.

Question 21.
What were the factors that brought changes in the Indian Jati System after independence?
Changes in the Jati System were brought by different factors working together. They included industrialization, urbanization and spread of modem education.
Developmental activities by the nation and the growth of private industries brought economic changes. This indirectly affected Jati system. Modern industries created a lot of new employment opportunities. The Jati laws did not apply to those jobs. When people of different castes began go work together in factories and business enterprises, many of Jati restrictions became impractical and irrelevant.

Urbanization made life according to Jati laws impossible. People of different Jatis had to live together in cities. It was not practical to live in the city following Jati system. There people had to live together.

Question 22.
Codify the ideas of Sree Narayana Guru.

• He propagated the principle of universal brotherhood.
• He fought against the Varna system and dominance by the Upper Class.
• He tried to wipe away the evil practices among the Ezhavas.
• He formed the SNDPYogam.
• He brought a silent revolution in the Kerala Society. He spread the message of One Jati, One Religion and One God for Man.

Question 23.
What is Sanskritization?
Sanskritization is a process by which members of the lower classes adapt the lifestyles and social conventions of the Higher classes with the intention of raising their own status in society. It is a way through which the lower class people try to enter the higher classes. They do it through imitation.

Question 24.
What is meant by dominant Caste (Prabala Jati)? What are they?
Through the land reform measures after independence, many people got ownership of their lands. Among these new owners of land, the castes that had big numbers were called dominant castes by M.N. Srinivas. The land reforms took away the rights of many high caste estate owners to their estates. They were not making any contributions to the economic system except that they collected axes from the farmers. The lands that were confiscated from these landlords were given to people at the next lower level. The people who newly got the land rights were not farmers. They were simply managers of agriculture. They were middle-class Dalits who had a lot of members. Here are some dominant castes:
b) Vokkaligas of Karnataka.
c) Reddys and Khammans in Andhra Pradesh.
d) Marathis of Maharashtra
e) Jats of Punjab, Haryana and Western UP.
f) Patidars of Gujarat

Question 25.
What were the special features of the dominant castes?
The special features are:

• Large number
• Land and economic power
• Political power

Question 26.
What are the permanent features of ‘Gotras’ (Tribes)?
The permanent features of Gotras include their region, language, special physical features, and environmental habitat. The Tribals in India are scattered in many regions of the country. But in some places, there are some concentrations. Among the Tribals, some 85% live in Central India. Central India is a wide area extending to Western Gujarat and Rajasthan, Eastern West Bengal and Orissa. Madhya Pradesh, Jharkhand, Chhatisgarh, Parts of Maharashtra and Andhra Pradesh come in this vast area. Of the remaining 15%, more than 11% live in North-East States, and the remaining in different other States of India.

If we look State-wise, most of them are found in the North-East States. Except in Assam, all the North East States have Tribals which come to more than 30% of the total population. In Arunachal Pradesh, Mizoram and Nagaland 60 to 95% of the people belong to Tribal groups. IntherestoflndiatheTribalpopulation is negligible. Except in Orissa and Madhya Pradesh, the Tribals are less than 12% of the population. Hills, forests, village plains and even some industrial ‘ belts of cities are the habitats of these Tribals.

Question 27.
Based on language, how are the Tribal groups categorized?
Based on language, Tribal groups are categorized into 4:
a) Indo-Aryan
b) Dravidian
c) Austric
d) Tibeto-Burman
The first two, Indo-Aryan and Dravidian, are also spoken by other people of India. Among the Tribals, 1% speak Indo-Aryan and 3% speak Dravidian languages. Austric and Tibeto-Burman are the most popular languages of the Tribals.
Based on physical and racial features, the Tribals of India are divided into 5:
a) Negrito
b) Australoid
c) Mongoloid
d) Dravidian
e) Aryan.
Of these the last two – Dravidian and Aryan, are shared by other Indians also.

Question 28.
Tribal Groups are categorized based on their acquired features. Explain.
Acquired features include the ways they earn their livelihood and their mixing with Hindu communities. Sometimes both these happen simultaneously. Based on the means of their livelihood, Tribais are grouped as fishermen, collectors of food, hunters, shifting cultivators, peasants, estate workers, and industrial workers.
Adapting Hindu ways of life is another criterion for the categorization of Tribal Groups. In Sociology, political science and public matters this criterion is more widely used.

Question 29.
In 1940 there was a controversy regarding separation/ integration of the Tribal groups in India. This was the result of a feeling that Tribal Groups should themselves separately from the mainstream society. The spokesman for the Separation theory was Verrier Elwin and the spokesman for Integration was G.S. Ghurye.

The proponents of the Separation theory argued that the Tribals should be kept separate from the mainstream society. They said that these Tribals should be protected from traders, financiers, Hindu and Christian missionaries. All these people are trying to make the Tribals laborers without land and thus make them non- Tribals. The Separation Theory supporters argued that the close relation between Tribals and mainstream society would result in their ruin.

On the other hand, the Integration theory supporters argued that Tribals are part of the Hindu community. GS. Ghurye called them backward Hindus. He argued that the Tribals should be integrated into the mainstream Hindu community. They also said that since the Tribals are backward Hindus, they also should be treated like the other Scheduled Castes and given all considerations.

This argument created a lot of noise in the Indian Constitutional Assembly. People supported both sides. Finally, it was agreed that the Tribals should be integrated into the mainstream gradually. It was this approach -gradual or controlled integration – that resulted in many welfare schemes for the Tribals. There were many welfare schemes for them, provisions for them in the Five Year Plans, specific Tribal schemes, Tribal welfare blocks, multiple-aim schemes and so on. But the integration of the Tribals created a basic problem. The Five Year Plans during the Nehru Era gave prominence to Industrial and agricultural development. Later government followed the same policy The stress was given to the construction of huge dams and factories and exploiting mineral wealth for the development of country.

Question 30.
What are the problems faced by the Tribals?
Tribals depended on the forest for their livelihood. The loss of forests was a big blow to them. During the British rule forests were exploited. This tendency continued even after independence. The coming of the land under private ownership also adversely affected the Tribals. When private landowners had their own private lands, the Tribals held their land collectively. This collective ownership proved harmful to them. For example, when a series of dams were constructed on Narmada River, ail the communities did not equally share their advantages and disadvantages. It helped the private landowners but it was harmful to the Tribals, who owned the land collectively.

Many of the regions where* Tribals are concentrated are becoming the targets, and thus victims, of national development schemes. Non-Tribals migrate into their areas in large numbers. It proves a great threat to ihe Tribals and their ways of life. It also reduces their population. For example, in Jharkhand, because of the migration by non-tribals into the new industrial areas, the number of Tribals has been reduced drastically there. But the most dramatic development was in North-Easter States. In States like Tripura, the Tribal population has come down to half in just one decade. The same thing was seen in Arunachal Pradesh. ,

Question 31.
What are the reasons for the rise of Tribal Movements?
There are mainly two reasons. One of them is the problem related to the control of land, forests and such other important economic resources. The second problem in connected with the racial-cultural being of the Tribals.

Question 32.
What are the special features of the internal structure of the family?
The following are the features:

• A family can be nuclear or joint.
• It can be patriarchal or matriarchal.
• The hereditary rights can be paternal or maternal.

Question 33.
What are the differences between a nuclear family and an extended family?
Nuclear family is the smallest. It is also called the primary family. In a nuclear family, there are the parents and their children. It consists of members belonging to two generations.
An Extended family is quite different. It is commonly known as joint family. There are different types of extended families. In an extended family, more than one couple and their children live. This can be a group of brothers and their families. It could be the family of an old couple whose children and grandchildren stay with them. An extended family is often seen as a sign of India. But that was never a strong-knit family. It was limited to some regions and some communities. An extended family is not a strong form even now.

Question 34.
What are the other forms of families?