Plus One

Kerala Plus One Botany Previous Year Question Paper 2018

Time: 1 Hours
Cool off time : 10 Minutes

General Instructions to candidates :

• There is a ‘cool off time’ of 10 minutes each for Botany and Zoology in addition to the writing time ‘ of 1 hour each. Further there is a ‘ 5 minutes’ ‘preparatory time’ at the end of the Botany Examination and before the commencement of the Zoology Examination.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before you answering.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all questions from question numbers 1 to 4. Each carry one score. (4×1=4)

Question 1.
Fill in the blank.
In Anabaena specialized cells called…… help in nitrogen fixation.

Question 2.
There are different types of leucoplasts in plant cells. Name the leucoplast that store proteins.

Question 3.
Observe the relationship between the first two terms and fill in the blank.
Epipetalous Stamen – Brnjal
………… – Lily

Question 4.
Choose the correct answer from the bracket.
First stable product of carbon dioxide fixation in C₄ plant is ……….

(PGA, OAA, PEP, RUBP)

Answer any seven from question numbers 5 to 13. Each carries two scores. (7×2=14)

Question 5.
Match the following :

Question 6.
Complete the flowchart given below.

Question 7.
Ribosomes are organelles without a membrane found in all cells. Name another organelle devoid of membrane, seen in animal cells. Write its function

Question 8.
Plants are adversely affected by manganese toxicity. Mention the effects of manganese toxicity in plants.

Question 9.
Following figure shows the citric acid cycle

Identify the steps where FADH₂ and GTP are synthesized.

Question 10.
The given figure shows a plant belonging to liverworts. Identify the plant. Name the asexual buds seen on it and write their features.

Question 11.
State any four features of facilitated diffusion.

Question 12.
Write a note on the phenomenon plasticity exhibited by plants with an example.

Question 13.
The breakdown of glucose to pyruvic acid is called glycolysis. Where does it occur in a cell? How many ATP molecules are directly syn-thesized during this process?

Answer any four from question numbers 14 to 18. Each carries three scores.(4×3=12)

Question 14.
The tissue found between the upper and lower epidermis of a leaf is called mesophyll.
a. Write the type of cells found in this tissue in a dicot leaf.
b. Mention two differences between a dicot leaf and monocot leaf.

Question 15.
Photosynthesis is a process influenced by environmental factors as well as plant factors. Mention three factors under each category.

Question 16.
First phase of meiosis I is typically longer and complex one. Name it. Mention the five sub divisions of this phase.

Question 17.
Plant growth regulators can be employed for various agricultural and horizontal prac¬tices. Identify the growth regulators that can be used for the following purposes.
a. To have weedfree lawns
b. To increase sugarcane length
c. Fruitset in pineapples
d. Rooting in stem cuttings
e. Inhibiting seed germination
f. To promote female flowers in cucumber

Question 18.
Certain stages in cell cycle are given below. Karyokinesis, Prophase, Cytokinesis, Meta-phase, Anaphase, Telophase.
Choose the statements (from those given below) that match with these stages and prepare a table.
• Centromere split and chromosomes move to opposite poles.
• Chromosomes cluster at opposite poles and nuclear envelope assembles around.
• Chromosomes seems to be with two chromatids attached at centromere.
• Chromosomes arranged at spindle equator.
• Separation of daughter chromosomes.
• Division of cytoplasm.

Answers

Answer 1.
Heterocysts

Answer 2.
Aleuroplasts/ Proteinoplast

Answer 3.
Epiphyllous

Answer 4.
OAA

Answer 5.

Answer 6.
a. Secondary xylem
b. Metaxylem
c. Exarch
d. Stem

Answer 7.
Centrioles
Functions: Cell division, formation of basal body of flagella or cillia, spindle apparatus during cell division.

Answer 8.
• Chlorosis
• Necrosis
• Brown spots surrounded by chlorotic veins.
• Manganese competes with iron and magnesium for uptake.

Answer 9.
FADH₂ – Between the conversion of succinic acid to malic acid.
GTP -Between the conversion of α- ke- toglutaric acid to succinic acid.

Answer 10.
Plant : Marchantia ( Bryophyte)
Asexual buds : Gemma/ Gemmae
Features – Gemmae are green, multiel- lular, asexual/vegetative buds which develop in small receptacles. Gemmae become detached from the parent body and germinate to form new individuals.

Answer 11.
• Requires special membrane proteins.
• Highly selective.
• Not requires ATP energy.
• Transport saturates.
• Transport of molecules along the concentration gradient.

Answer 12.
Plants follow different pathways in response to environment or phases of life to form different kinds of structures. This ability is called plasticity.
E.g., Heterophylly in Cotton, Coriander and Larkspur etc.

Answer 13.
Glycolysis take place on cytoplasm or cytosol.
Glycolysis forms 2 ATP molecules per glucose molecule.

Answer 14.
a. Palisade parenchyma and spongy parenchyma

Answer 15.
Plant factor
Number, size, age, orientation of leaves, mesophyll cells, chloroplasts, internal C0₂ : concentration, amount of chlorophyll. External factor
Availability of sunlight, temperature, C0₂, concentration, water, availability of nutrients in the soil.

Answer 16.
Prophase I
Five sub divisions : Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

Answer 17.
a. 2,4-D (2,4- Dichlorophenoxy acetic acid)
b. Gibberellin
c. Ethylene
d. Auxin
e. ABA /Abscisic acid
f. Ethylene / Ethephon

Answer 18.

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Chemistry Model Question Paper 4

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to get familiar with questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before answering.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all questions from question numbers 1 to 7. Each carry one score.

Question 1:
Which of the following sets of quantum numbers are NOT possible?

Question 2:
‘A given compound always contains exactly the same proportion of elements by weight’. Name the above law.

Question 3:
Write the general outer electronic configuration of d-block elements.

Question 4:
What is dry ice?

Question 5:
The bond order value is an important property of a molecule. How is bond order related to bond length?

Question 6:
Name the gas law which gives the relation-ship between the pressure and temperature of a fixed amount of gas at a constant volume.

Question 7:
Sodium hexametaphosphate is commercially called

Answer any ten from question numbers 8 to 20. Each carries two scores.

Question 8:
Illustrate Markovnikov’s rule taking the example of propene.

Question 9:
Match the following by selecting the items from columns B and C which are most suitable to those in column A.

Question 10:
Write the IUPAC names of the following.

Question 11:
Consider the reaction given below:

i. Identify the major product obtained.
ii. Name the rule governing the formation of the major product.

Question 12:
The photon has a momentum as well as a wave length.
a. Which property of matter is revealed in the above statement?
b. A photon has a mass of 8.6 x 10^-30kg. Calculate its wavelength.
[Planck’s constant, h = 6.626 x 10^-34 Js]

Question 13:
Thermodynamics deals with energy changes of macroscopic systems.
a. Consider a chemical reaction taking place in a closed insulated vessel. To which type of thermodynamic system does it belong?
b. State the first law thermodynamics.

Question 14:
Mention any two similarities between Beryllium and Aluminium.

Question 15:
Analyze the following reaction:
CH₃ – CH = CH₂ + H – Br → ‘A’ + ‘B’
If ‘A’ is the major product and ‘B’ is the mi¬nor product, identify ‘A’ and ‘B’. Also name the related rule.

Question 16:
A graph of atomic radius versus atomic number is given below.

a. What do you understand from this graph?
b. Using the above graph, how will you account for the variation of ionization enthalpy in a period?

Question 17:
The threshold frequency v₀ for a metal is 6.2 x 10⁴S^-1. Calculate the K.E emitted of an electron when the radiation of frequency ν = 8.7 x 10¹⁴S^-1 strikes the metal.

Question 18:
Account for the following observations:
i. The density of ice is lower than that of water.
ii. Hard water doesn’t give ready lather with soap.

Question 19:
Represent Sawhorse and Newman projection formulae of staggered and eclipsed conformation of ethane.

Question 20:
Give the structures of the following compounds.

Answer any ten from question numbers 21 to 29. Each carries two scores.

Question 21:
A system in thermodynamics refers to that part of the universe in which observation are made.
a. What do you mean by an isolated system? Given an example.
b. Distinguish between intensive and extensive properties. Give two examples for each.

Question 22:
a. Name the species getting oxidized in the above reaction.
b. Balance the above equation by the oxidation number method.

Question 23:
The group 1 metals of the periodic table of elements are collectively called alkali metals.
a. Write the general electronic configuration of alkali metals.
b. Identify the alkali metal exhibiting anomalous properties. Explain.

Question 24:
Organic matter such as leaves, grass, trash etc. are major pollutants in water.
a. How do organic pollutants affect aquatic life?
b. What is biological oxygen demand (BOD)?

Question 25:
If the cocentration of the hydrogen ioni n a soft drink is 3 x 10^-3 M, calculate its pH.

Question 26:
Boron, Aluminium, Gallium, Indium and thallium belong to group 13 of the periodic table of elements.
a. How can you explain a higher stability of BCl₃ as compared to TlCl₃ ?
b. While Aluminium can be form the ion [AlF₆³, Boron is unable to form [BF₆ ]^3- ion. Explain.
c. State whether the compound BCl³ is acidic or basic.

Question 27:
The theory that attempts to explain the behaviour of gases is known as kinetic molecular theory.
a. On the basis of this theory, explain the compressible nature of gases and the temperature dependence on kinetic energy.
b. Liquid drops assume spherical shape. Why?

Question 28:
Neils Bohr was the first to explain quantitatively the general features of hydrogen atom structure and its spectrum.
a. Give the main postulates of Bohr’s model.
b. Find the maximum number of emission lines, when the excited electron of hydrogen atom in n = 6, drops to the ground state (n = 1).
c. Calculate the wave number of radiation due to transition of an electron from fourth orbit to second orbit. (R_H= 109677 cm^-1).

Question 29:
In the Celsius scale, melting point of ice is 0°C. Another scale of temperature is based on absolute zero.
a. Identify the scale.
b. What is the volume of a gas at absolute zero temperature?
c. Draw a graph showing the relationship between volume and temperature of an ideal gas at a constant pressure.

Answer any three from question numbers 30 to 33. Each carries four scores.

Question 30:
If the mass percent of the various elements of a compound is known, its empirical formula can be calculated.
a. What is mass percent? Give its mathematical expression.
b. A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molecular mass is 98.96. What are the empirical and molecular formulae?

Question 31:
The IUPAC names of alkanes are based on their chain structure.
a. Give the IUPAC name of

b. Represent 1-methyl-3- propyl cyclohexane using bond line notation.
c. What is the type of hybridization of C in CH₃⁺ Also predict its shape.
d. Name the type of bond fission resulting in the formation of free radicals.

Question 32:
The group 14 elements have four electrons in the outermost shell.
a. SiCl₄ can be easily hydrolyzed by water while CCl₄ cannot be hydrolyzed. Why?
b. How are fullerenes prepared?
c. Distinguish between silicones and silicates.

Question 33:
a. Draw the structures of the following compounds.
2,3 Dilbromo-1-phenylpentane
4-Ethyl-1-fluoro-2-nitrobenzene

b. Write all possible chain isomers of the compound with molecular formula C₅H₁₂

Answers

Answer 1:
n = 2, l = 2, ml= 0, m_s =+1/2
n = 3, l = 2, ml = -3, m_s = +1/2

Answer 2:
Law of definite proportions.

Answer 3:

Answer 4:
Solid CO₂

Answer 5:

Answer 6:
Gay Lussac’s law.

Answer 7:
Calgon.

Answer 8:
According to Markovnikov’s rule, when a hydrogen halide adds to an unsymmetrical alkene, halogen atom get attached to the doubly bonded C atom containing lesser number of hydrogen atom.

Answer 9:

Answer 10:
i. 2, 2, 4 – trimethyl pentane,
ii. Hex – 4 – ene -1 – oic acid.

Answer 11:

Answer 12:
a. Dual Behaviour of matter. This means that matter possesses dual character, i.e., particle nature and wave nature.

Answer 13:
a. Isolated system.
b. First law of thermodynamics. The energy of isolated system is constant.

Answer 14:
1. Both Be and Al have strong tendency to form complex.
2. Oxides and hydroxides of both Be and Al are amphoteric.

Answer 15:

Answer 16:
a. Atomic size decreases with increase in atomic radius.
b. Ionization enthalpy increases along a period.

Answer 17:

Answer 18:
i. Hydrogen bonding gives ice an open type structure. Hence ice has lower den sity than water.
ii. Hard water contains Mg²⁺ and Ca²⁺ions.
It is due to the presence of these ions that hard water does not lather with soap.

Answer 19:
Sawhorse method.

In staggered conformation, H atoms are far apart and the repulsion is minimum. Thus staggered conformation is more stable than eclipsed conformation.

Answer 20:

Answer 21:
a. A system which can neither exchange matter nor energy with the surrounding is called isolated system, eg., hot water in a thermoflask.
b. The properties of the system which depend upon the quantity of matter are called extensive properties, eg., mass, volume.
The properties of the system which are independent of quantity of matter are called intensive properties, eg., temperature, pressure.

Answer 22:

Answer 23:
a. ns¹
b. Lithium
Anamolous properties of lithium:

• Small size
• High ionisation enthalpy
• High polarising power
• Absence of d orbitals.

Answer 24:
a. Organic pollutants decreases the amount of dissolved oxygen in water, thereby caus ing harmful effects to aquatic life.
b. The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water is called Biochemical Oxygen Demand (BOD).

Answer 25:

Answer 26:
a. Because of inert pair effect, the +3 state of Tl is less stable. Thus BCla₃ is stable than TlCla₃. Since B is smaller than Tl it can undergo back bonding with Cl and hence BCla₃ is more stable than TlCla₃.
b. Boron does not have any d orbitals. While Al has vacant d orbitals. Thus, Al forms [AIFa₆]^3- ions while B cannot form [BF₆]^3- Hons.
c. BCla₃ is acidic because it is electron deficient and can accept electron pair.

Answer 27:
a. Gases consist of large number of identical particle. These are so small and so far a part. In kinetic theory it is considered that average K.E of gas molecules is directly proportional to the ab-solute temperature.ie as the temperature decreases, K.E of molecule decreases and intermolecular attraction between the molecule become significant. These attractive force make the molecule come closer and they become compressible.
b. Liquids assumes spherical shape due to surface tension. Liquids have tendency to minimise the surface area. Since for a sphere, surface area is minimum.

Answer 28:
a. 1. The electron in an atom revolve around the nucleus only in certain circular path called orbits.
2. The energy of the electron in a particular orbit remains constant.
3. Only those orbits are permitted in which the angular momentum of the electron is a whole number of multiple of \(\frac { h }{ 2\pi }\), where h is Plank’s constant.
4. Energy is emitted or absorbed by an atom only when an electron moves one from one orbit to another.
Maximum number of emission lines

Answer 29:

Answer 30:
a. Relative mass of each of the constituent element in 100 part of it by mass is called mass percent.

∴ Emperical formula of the compound

Answer 31:
a. 3 – ethyl-5-methylheptane

c. sp² hybridisation. Trigonal planer shape.
d. Homolytic fission.

Answer 32:
a. CCl₄ is not hydrolysed by water because carbon has no d-orbitals and hence can-not expand its co-ordination number beyond . However Si can expand its co-ordination number beyond 4 due to the availability of vacant d-orbitals. Therefore SiCl₄ readily undergoes hydrolysis.
b. Fullerenes are made by the heating of graphite in an electric arc in the presence of inert gases such as helium or argon.
c. Silicons are a group of inorganic polymers having general formula (R₂SiO)_n. The basic structural unit of silicates is SiO₄^4- units. When silicate units are linked together, they form chain, ring, sheet or three-dimensional structures.

Answer 33:

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Chemistry Model Question Paper 3

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to candidates

• There is a ‘cool off-time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to get familiar with questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before answering.
•  Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Answer all questions from question numbers 1 to 7. Each carry one score.

Question 1:
The number of protons, electrons and neutrons in a species are equal to 17, 18 and 18 respectively. Which of the following will be the proper symbol of this species?

Question 2:
One-half of the difference between the number of electrons in the bonding and antibonding molecular orbitals is called …………

Question 3:
Name the phenomenon behind clean sing action of soap.

Question 4:
Name the process used for the industrial preparation of sodium carbonate,

Question 5:
In the Lassaigne’s test for halogens, they are precipitated as ……………….

Question 6:
Write the name of any one salt responsible for the permanent hardness of water.

Question 7:
Give the IUPAC name of the following compound: CH₃ CH₂ CO CH₂ CH₂ COOH

Answer any ten from question numbers 8 to 20. Each carries two scores.

Question 8:
Classify the following into intensive and extensive properties.
i. Internal energy
ii. Density
iii. Heat capacity
iv. Temperature.

Question 9:
Given the redox reaction:
CuO (s) + H₂ (g) → Cu (s) + H₂0 (g)
i. Identify the species which undergo reaction and which undergo oxidatlon.1
ii. Identify the reductant and oxidant in the above reaction.

Question 10:
Match the following :

Question 11:
Give the IUPAC name of the following compounds.

Question 12:
Give the name of the modified form of ideal gas equation and write down it.

Question 13:
Carbon has many allotropes.
a. Write the name of any two allotropic forms of carbon.
b. CCl₄ does not undergo hydrolysis. Give a reason.

Question 14:
i. Give the Arrhenius concept about acids and bases.
ii. Give one example each for Arrhenius acid and base.

Question 15:
When Sodium metal dissolves in liquid ammonia, it gives a deep blue coloured solution. Explain the reason.

Question 16:
A neon dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If the pressure of the mixture of gases in the cylinder is 25 bar, what are the partial pressures of O₂ and neon in the mixture?

Question 17:
Write the difference between extensive and intensive properties. Give one example of each.

Question 18:
Fluorine reacts with ice as given below.
H₂O(s) + F₂(g) → HF (g) + HOF (g)
Justify that this is a redox reaction.

Question 19:
a. H₂O₂ is a bleaching agent. Why?
b. Complete the following reaction?
Zn (s) + NaOH (aq) → …………………..

Question 20:
Complete the following chemical reactions:

Answer any seven from question numbers 21 to 29. Each carries three scores.

Question 21:
In redox reactions, oxidation and reduction occur simultaneously.
a. How are oxidation and reduction related
to the oxidation number?
b. During a group discussion, one of your friends argues that thermal decomposition of KC/O₃ is a redox reaction while that of CaCO₃ is not a redox reaction. Give your opinion and substantiate.

Question 22:
Give the IUPAC names of the following compounds:

Question 23:
When some sodium acetate is added to a solution of acetic acid, the concentration of unionzed acetic acid increases.
a. What is the phenomenon involved? Substantiate.
b. Consider the equilibrium,

The solubility of AgCl is 1.06 x 10^-5 mol/L at 298 K. Find out its K_sp at this temperature.
c. What happens to the value of solubility and solubility product when HCI is passed through an AgCl solution?

Question 24:
Atmospheric pollution increases the global average temperature and the phenomenon is called global warming.
a. What are the major gases which contribute to wards global warming?
b. What can we do to reduce global warming?

Question 25:
The solubility product of Al(OH)₃ is 1 x 10³⁶ Calculate the solubility of Al(OH)₃

Question 26:
Account for the following :
i. Ionization enthalpy of Nitrogen is greater than that of Oxygen.
ii. 2^nd period elements show anomalous behaviour.

Question 27:
a. Give any two factors influencing the formation of an ionic bond.
b. Give the shape of the following species.
i. NH₄⁺
ii. HgCl₂

Question 28:
Redox reactions can be considered as electron transfer reactions. In an experiment a copper rod is dipped in AgNO₃ solution.
a. What happens to the colour of the solution and why?
b. Identify the oxidizing and reducing agent in this reaction.
c. Calculate the oxidation number of Cr in
K₂Cr₂O₇ and P in H₂P₂O₅.

Question 29:
a. 1-alkynes are weakly acidic in nature.
Give any two reactions to showthe acidic character of ethyne.
b. From the following, select the one in which Markownikoffs rule is best applicable.

Answer any three from question numbers 30 to 33. Each carries four scores.

Question 30:
The enthalpy change in a process is the same, whether the process is carried out in a Single step or in several steps.
a. Identify the law stated here. 1
b. Calculate the enthalpy of formation of CH4 from the following data:

Question 31:
a. When nitrogen and hydrogen combines to form ammonia, the ratio between the volumes of gaseous reactants and products is 1:3:2. Name the law of chemical combination illustrated here.
A compound is made up of two elements A and B, has A= 70%, B= 30%. The relative number of moles of A andB in the compound are 1.25 and 1.88 respectively. If the molecular mass of the compound is 160, find the molecular of the compound.

Question 32:
Phenol exhibit resonance.
i. Draw the resonance structures of phenol.
ii. Predict the directive influence of -OH group in Benzene ring.

Question 33:
a. Hydrogen peroxide restore the colour of lead paintings. Give a reason.
b. How does the atomic hydrogen torch function for cutting and welding purposes?

Answers

Answer 1:
\(_{ 17 }^{ 35 }{ Cl }\)

Answer 2:
Bond order

Answer 3:
Surface tension.

Answer 4:
Solvay process.

Answer 5:
Silver halides.

Answer 6:
CaCl₂ or MgSO₄

Answer 7:

Answer 8:
Extensive properties: Internal energy, Heat capacity.
Intensive properties: Density, Temperature.

Answer 9:
i. Reduction species – CuO
Oxidation species – H₂
ii. Reductant- H₂, Oxidant – CuO

Answer 10:

Answer 11:
11. i. 3 – chloropropanal
ii. 3 – methylpentanenitrite.

Answer 12:
There is deviation from ideal behaviour at high pressure and low temperaure. This is due to two faulty assumptions in kinetic theory of gases. It is corrected by van der Waal’s equation.

where n is the number of moles of the gas a and b are van der Waal’s constants.

Answer 13:
a. Diamond, graphite, fullerene
b. This is due to the absence of vacant d- orbitals in carbon.

Answer 14:
i. Acid is a substance which dissociate to give H⁺ ion in aqueous solution. Base is a substance which dissociate to give OH^– ion in aqueous solution.
ii. HCl, H₂SO₄, CH₃COOH etc are acids which give free H⁺ ions in aqueous solution.

Answer 15:
Blue colour H₂ gas is evolved during the process. If M is alkali metal.

Blue colour due to the ammoniated electron which absorbs energy.

Answer 16:

Answer 17:
Extensive – Depends on the quantity of matter, eg., internal energy, heat capacity enthalpy volume, mass etc.
Intensive – Does not depend on the quantity of matter, eg., Density, S.T.P, T, etc.

Answer 18:
Oxidation number of fluorine decreases in HF while the oxidation no. of fluorine in-creases in HOF. So it is a redox reaction.

Answer 19:
a. H₂O₂ decomposes to release nascent oxygen that oxidises the colouring matter to colourless.
b. Zn(s)+2NaOH(aq) → Na₂ZnO₂(aq) +H₂(q)

Answer 20:

Answer 21:
a.Oxidation is a process which involves an increase in oxidation number of an element.
Reduction is a process which involves a decrease in oxidation number of an element.
IMG
The reaction (1) is a redox reaction because there is a change in oxidation state. But (2) is not a redox reaction because there is no change in oxidation state.

Answer 22:
a. 5, 5 -dimethyl heptan – 2 – ol
b. 2 – pentanone
c. 2, 4, 6 – tribromophenol.

Answer 23:
a. Common ion effect. It the suppression of dissociation of weak electrolyte
(CH₃COOH) by addition of a salt containing common ion (CH₃COONa).
IMG
c. When HCI gas is passed through AgCl solution. Solubility of AgCI decreases while the solubility product does not change.

Answer 24:
a. CO₂, methane, chlorofluoro carbons etc are the major gases which contribute to- wards global warming.
b. 1. Minimise the use of automobiles.
2. Plant more trees.
3. Avoid burning of dry leaves, wood.

Answer 25:

Answer 26:
i. because of the half filled electronic configuration of N. (N = 1s² 2s² 2p³)
ii. due to their small size, high electrone gativity and ionisation enthalpy,large charge/radius ratio and due to the absence of vacant d-orbitals.

Answer 27:
a. Low ionisation enthalpy of the electropositive atom (metal atom).
High negative electron gain enthalpy of the electronegative atom (non-metal atom)
High lattice enthalpy of the ionic compound formed.
i. NH⁴⁺ – Tetrahedral
ii. HgCl² – Linear

Answer 28:
a. The solution becomes blue colour due to the formation of Cu²⁺ ions.
b. Ag⁺ (AgNO₃) is the oxidising agent and Cu is the reducing agent.
c. Oxidation number of Cr in K₂Cr₂O₇ is +6 and P in H₂P₂O₅ is +4.

Answer 29:
a. Ethyne reacts with sodium ethynide and liberate H₂. It reacts with sodamide (NaNH₂) to form sodium ethynide and liberate ammonia.

Answer 30:

Answer 31:
a. Gay – Lussac’s law of Gaseous volumes
b. The relative number of moles means % / atomic mass

Element	%	Relative no.of moles	Simple ratio	Simplest whole no.ratio
A	70	1.25	1.25/1.25= 1	2
B	30	1.88	1.88/1.25 = 1.5	3

Empirical formula is A₂B₂
Atomic mass of A = % / no.of moles = 70 / 1.25 = 56
Atomic mass of B = % / no.of moles = 30/ 1.88 = 15.96
So, emp. Formula mass = 56 x 2 + 15.96 x 3 = 159.88 = 160
n = mol. mass / emp.Formula mass = 160/160 = 1
Molecular formula = (emp.Formula)n = (A₂B₃)X 1 = A₂B₃

Answer 32:

ii. -OH group is an ortho-para directing group.

Answer 33:
a. This is due to the oxidising action of H₂O₂.

b. Here H atoms are allowed to recombine on the surface to be welded to generate the temperature of 4000K.

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Chemistry Model Question Paper 2

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to candidates

•  There is a ‘cool off time’ of 15 minutes in addition to the writing time.
•  Use the ‘cool off time’ to get familiar with questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before answering.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all questions from question numbers 1 to 7. Each carry one score.

Question 1:
The ozone in the stratosphere is destroyed by

Question 2:
The geometry of SF₆ molecule is …………
i. tetrahedral
ii. planar
iii. octahedral
iv. triagonal bipyramidal

Question 3:
Syn gas is a mixture of …….
i. CO and H₂O
ii. CO and H₂
iii. CO₂ and H₂
iv. CH₄ and CO

Question 4:
Representation of orbital with quantum numbers n = 3, l = 1 is ………….

Question 5:
The metal present in the chlorophyll of plants ¡s …………

Question 6:
Thermodynamically, the most stable allot rope of carbon is ………..

Question 7:
The atomic number of the element with IUPAC nane Ununbium is ……….

Answer any ten from question numbers 8 to 20. Each carries two scores.

Question 8:
Explain sp³ hybridisation taking methane (CH₄) as an example.

Question 9:
Write the IUPAC name of the following compounds:

Question 10:
The Taj Mahal in India has been affected by ‘acid rain’. Explain the causes and harmful effects of acid rain.

Question 11:
Calculate the number of molecules in each of the following:
i. 1 g N₂
ii. 1 g CO₂
(Given that N_A is 6.02 x 10²³, molecular mass of N₂ is 28 and CO₂is 44).

Question 12:
a. How many moles of dioxygen are present in 64 g of dioxygen (Molecular mass of dioxygen is 32).

b. The following data were obtained when dinitrogen (N₂) and dioxygen (O₂) react together to form different compounds.

Question 13:
a. The number of electrons, protons and neutrons in a species are equal to 18,16 and 16 respectively. Assign the proper symbol to the species.

b. Among the following electronic configurations, which one is correct?

Question 14:
Give reasons:
i. KO₂ is paramagnetic.
ii. Solutions of alkali metals in liquid ammonia are blue in colour.

Question 15:
An alkene ‘A’ on ozonolysis gave two molecules of formaldehyde. Write the name of ‘A’ and the chemical equation of ozonolysis.

Question 16:
Photoelectric effect was first observed by Hertz.

1.  The number of electrons ejected in the photoelectric effect is proportional to ……….. of light used. [Frequency, intensity]
2. Select the correct statement related to the photoelectric effect:
   • Threshold frequency is the maximum frequency required to cause photoelectric emission from a particular metal.
   • The kinetic energy of the photoelectrons is directly proportional to the frequency of incident light.
   • Work function is the same for all metals.

Question 17:
Le Chatelier’s principle helps to explain the effect of change in conditions on equilibrium.
Discuss the effect of pressure in the following equilibrium on the basis of Le Chatelier’s principle:

Question 18:
Analyse the following graph between ionization enthalpy and atomic number.

What do you observe from the graph? Give justification for your observation.

Question 19:
Carbocations are formed by the heterolytic cleavage of a covalent bond.
a. What is heterolytic bond fission?
b. Arrange the following carbocations in the increasing order of stability:

Question 20:
Balance the following equation by the half-reaction method.

Answer any seven from question numbers 21 to 29. Each carries three scores.

Question 21:
Two important oxides of carbon are carbon monoxide and carbon dioxide.
a. Why is CO called a poisonous gas?
b. How is CO₂ responsible for global warming?
c. What are producer gas and water gas? Mention their uses.

Question 22:
Complete the following reactions:

Question 23:
Hybridization influences the bond length and bond enthalpy in organic compounds:
a. Compare the bond length and bond strength of C-H bonds formed by sp and sp₃ hybridized carbon atoms. Give reason.
b. How many s and p bonds are present in the following molecules?
i. CH₃ – CH₂ – CH₃
ii. CH₃ – C = CH

Question 24:
Particulate pollutants are the minute solid particles or liquid droplets in air.
a. Suggest two examples for non-viable particulate pollutants.
b. Write any two adverse effects of photochemical smog.

Question 25:
The behaviour of acids and bases can be explained using different concepts.
a. Select the Lewis acid from the following:
(NH₃, OH^–, BCl₃, Cl^–)
b. What are conjugate acid-base pairs? illustrate using a suitable example.

Question 26:
Valence Bond Theory(VBT) and Mole-cular Orbital Theory (MOT) are the two important theories of chemical bonding.
a. Out of the following, which is the hybridization of phosphorus in PCl₅?
(sp², sp³, dsp², sp²d)
b. Explain the geometry of the PCl₅ molecule and account for its high reactivity.

Question 27:
Free rotation is possible with respect to a C – C bond in the case of alkanes.
a. The repulsive interaction between the adjacent bonds in a conformation is called ………
b. Draw Newman’s projections of the two conformers of ethane. Which among these is more stable? Justify.

Question 28:
Equilibrium is possible only in a closed system at a given temperature.
a. Write the expression for equilibrium constant, K_c^– for the reaction.
4NH_3(g) +5O_2(g) ⇌ 4NO_(g) + 6H₂O_(g)
b. What happens to the value of the equilibrium constant (Kc) when the above reaction is reversed?

Question 29:
a. What is the number of hydrogen atoms in 1 mole of methane (CH₄)?
b. Calculate the amount of carbon dioxide formed by the complete combustion of 80g of methane as per the reactions
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂ O_(g)
(Atomic masses: C= 12.01u, H= 1.008u, O = 16u).

Answer any three from question numbers 30 to 33. Each carries four scores.

Question 30:
The Valence Shell Electron Pair Repulsion (VSEPR) theory helps in predicting the shapes of covalent molecules.
a. Arrange the bond pair electron and lone pair electron in the decreasing order of the repulsive interactions among them.
b. A molecule of the type. AB₃E₂ has three bond pairs and two lone pairs of electrons. Predict the most stable arrangement of electronpairs in this molecule.
c. The bond order value is an important property of a molecule. How is bond order related to bond length?
d. Write the electronic configuration of an oxygen molecule and justify its magnetic nature.

Question 31:
Most of the naturally occuring processes are spontaneous.
a. Give the criteria for spontaneity of a process in terms of free energy change (∆G).
b. Exothermic reactions associated with a decrease in entropy are spontaneous at lower temperatures. Justify on the basis of Gibbs equation.
c. Find the temperature above which the reaction.
MgO_s + C_s → Mg_s + CO_(g)
becomes spontaneous. (Given ∆H° = 490 KJ mol^-1 and ∆S° = 198JK mol^-1)

Question 32:
What do you mean by the following terms?
a. Homolytic fission
b. Heterolytic fission
c. Nucleophiles
d. Electrophiles

Question 33:
The gases which obey Gas Laws at all temperatures and pressures are called ideal gases.
a. Give reasons for the deviation of real gases from the ideal gas behaviour.
b. Calculate the minimum pressure required to compress 500 ml of air at 1 atm to 300 ml at the same temperature.

Answers

Answer 1:
Cl
Answer 2:
Octahedral

Answer 3:
CO and H₂

Answer 4:
3p

Answer 5:
Magnesium

Answer 6:
graphite

Answer 7:
112

Answer 8:
i. It is defined as the phenomenon of inter-mixing of pure atomic orbitals of slightly different energies and shapes to form new hybrid orbitals of equivalent energies and identical shapes.
ii. Electronic configuration of C (6) is 1s² 2s² 2p_x¹ 2p_y¹ in ground state. It has electronic configuration 1s2 2s1 2px1 2py1 2pz1 in excited state (since C is tetravalent only in excited state).

Answer 9:
a. pent – 4 – ene – 2 – ol
b. 2, 5 – dimethylheptane

Answer 10:
Due to acid rain caused from exhaust of industries the marble layers of Taj Mahal get destroyed. Acid rain is caused by of acid ox-ides such as oxides of sulphur and nitrogen in the air.
Acid rain destroys crops, aquatic lives and buildings. The acid rain will carrode build ing material like marble, limestone etc. and dissolve heavy metals like Cu,Pb,Hg, etc. in water and make it unfit for drinking. It fades the colour of fabrics, leather and paper.

Answer 11:
i. Number of molecules present in 1g of
N² = 6.02 x 10²³/28 = 0.215 x 10²³
ii. Number of molecules present in 1 g of CO₂ = 6.02 x 10²³/44 = 0.136 x 10²³

Answer 12:
a. No. of moles = 64/32 = 2 mole
b. Law of multiple proportions.

Answer 13:
\(_{ 16 }^{ 32 }{ { S }^{ 2- } }\)
b. fig (iii) is the correct one. fig(i) and fig(ii) do not obey Hund’s rule.

Answer 14:
i. Presence of one unpaired e’.
ii. Due to ammoniated electrons.

Answer 15:

Answer 16:
a. Intensity.
b. Kinetic energy of the photoelectrons is directly proportional to the frequency of incident light.

Answer 17:

The forward reaction involves a decrease in number of molecules. Hence a high pressure favours forward reaction (formation of products).

Answer 18:
The graph shows variation of ionisation enthalpy along a group.
Moving down the group, atomic size increases and the outer electrons are far away from the nucleus. There is also an increase in the screening of nuclear charge by electron in inner shells. As a result the removal of electrons becomes easier down the group and ionization enthalpy decreases.

Answer 19:
a. When a bond between 2 atoms breaks in such a way that the shared pair of electrons stays on any one of the atoms. The process is called hetrolytic fission.

Answer 20:

Answer 21:
a. Carbon monoxide reacts with haemo globin to form carboxy haemoglobin which is highly stable complex. This prevent haemoglobin in red blood cells from carrying oxygen and ultimately resulting in death. Hence CO is considered as poisonous gas.
b. CO absorbs IR radiations. This lead to increase in green house effect and thereby raising the temperature of the atmosphere.
c. The mixture of CO and H₂ is called water gas. The mixture of CO and N₂ is called producer gas. Both water gas and pro ducer gas are very important industrial fuels.

Answer 22:

Answer 23:
a. Bond length of C-H bond for sp hybridised carbon is less and that of sp³ hybridised carbon is high. Bond strength is greater for C-H bond is sp hybridised carbon than that for sp³ hybridised carbon.
b. 1.10 σ bonds.
2. 6 σ bonds and 2 π bonds.

Answer 24:
a. Mist, smoke, dust
b. *Causes irritation to eyes.
*Causes health problems like head ache, chest pain, dryness of throat, cough and difficulty in breathing.
*PAN adversely affects green plants.

Answer 25:
a. BCl₃
b. The acid-base pair which is formed from each other by loss or gain of proton is called as conjugate acid-base pairs. Consider the reaction,

Water act as a base since it accepts proton. HCl act as an acid since it donate proton.
Cl^– is the conjugate base of acid HCl.
H₃O⁺ is the conjugate acid of base H₂O.

Answer 26:

In PCl₅, 5 sp³ d orbitals of phosphorus over¬lap with singly occupied P orbitals of chlorine atoms to form 5 P-Cl bonds. Three P- C/ bond lie in one plane and make bond angle of 120°. These bonds are equitorial bonds. The remaining 2 P-Cl bond lie above and below the equitorial plane making an angle of 90° with the plane. These bonds are called axial bonds. Axial bonds are longer and weaker than equitorial bonds since they suffer more repulsive interactions from equitorial bond pairs. As a result PCl₅ is highly reactive.

Answer 27:
a. Torsional strain.
b. Newman’s projection

Staggered conformation is more stable than eclipsed conformation. This is because in staggered conformation, the hydrogen atoms are at maximum distance apart and there is minimum repulsion and it is more stable. But in eclipsed conformation, H atoms of one carbon atom are directly behind those of other and thus repulsion is maximum and hence least stable.

Answer 28:

Answer 29:
a. 4 moles of hydrogen or 4 x 6.023 x 10²³ atoms of hydrogen.
b. According to the equation, 1 mole CH₄ gives 1 mole CO₂ number of moles of methane = \(\frac { 80 }{ 16 }\)
= 5 moles \ 5 moles of CO₂ is formed, ie., 5 x 44 = 220g CO₂

Answer 30:
a. Lone pair – lone pair > lone pair – bond pair > bond pair – bond pair.

Answer 31:
a. ∆G = ∆H – T∆S
When ∆G is negative (∆G < 0), the pro-cess is spontaneous, b. For exothermic reactions, ∆H = -ve and ∆S = -ve When ∆H = -ve, ∆S = -ve, ∆G become -ve only. When ∆H >T∆S. Such reactions are spontaneous only at low temperature.

The reaction is spontaneous above this temperature.

Answer 32:
a. When the bond between two atoms A-B breaks in such a way that each fragment carries one unpaired electron, the process is called homolytic fission,
b. When the bond between two atoms A-B breaks in such a way that the shared pair of electrons stays on any one of the atoms, the process is called heterolytic fission.
c. Electron rich species are called nucleo-philes, (nucleus loving). They has attraction for a positively charge centre.
eg:, OH^–, CN^–, I^–
d. These are the reagents which attacks electron rich centres in the substrate.

Answer 33:
a. Two postulates of kinetic theory of gases are defective:

i. Volume occupied by gas molecule is negligible as compared to the total volume of the gas.
ii. The force of attraction and repulsion among the gas molecule are negligible.

At low temperature and high pressure, gas is in compressed states. But molecules are incompressible. As a result the volume of the molecules in the gas cannot be ignored in comparison to the volume occupied by gas.
At low temperature and high pressure, the volume of the gas becomes small, and molecules come closer to each other. Under these conditions, the at¬tractive force between the molecules become operative. Hence under these conditions real gas deviate from ideal behaviour.

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Chemistry Model Question Paper 1

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to get familiar with questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before answering.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all questions from question numbers 1 to 7. Each carry one score.

Question 1:
Give the equation showing relation between empirical formula and molecular formula of a compound.

Question 2:
Which of the following transition of electron in the hydrogen atom will emit maximum energy ?
i. n₅ → n₄
ii. n₄ → n₃
iii. n₃ → n₂
iv. same in all transition

Question 3:
The geometry and type of hybridisation of central atom in BF₃ are
i. linear, SP
ii. tetrahedral, SP³
iii. trigonal planar,SP²
iv. Pyramidal, SP³

Question 4:
Arrange the following elements in the increasing order of first ionisation energy: Li, Be,B

Question 5:
The compressibility factor of an ideal gas is …………

Question 6:
What is the molecular formula of Buck ministers fullerene ?

Question 7:
Ozone layer depletion in Antarctica is due to the formation of the compound
i. acrolein
ii. formaldehyde
iii. SO₂ and SO₃
iv. chlorine nitrate

Answer any ten from question numbers 8 to 20. Each carries two scores.

Question 8:
In the two oxides of carbon, the mass ratio of oxygen is 1.33 : 2.66. Identify the law illustrated by the above ratio and state the law.

Question 9:
Using s, p, d notation describe the orbitals with following set of quantum number.

Question 10:
A set of molecules are given below:
NaCl, O₂, KCl, BF₃, PCl₅
a. Classify the molecules into ionic and covalent compounds.
b. Classify the above molecules into those obeying octet rule and those not obeying octet rule.

Question 11:
Identify the extensive and intensive properties from the following :
density, internal energy, heat capacity, molar heat capacity

Question 12:
Match the following:

Question 13:
Derive the relation between Kp and Kc of a gaseous reversible reaction.

Question 14:
a. What is Buffer solution ?
b. How a basic buffer can be prepared ?

Question 15:
Briefly explain the bonding in diborane.

Question 16:
Nitrogen, sulfur, halogen and phosphorous present in an organic compound are detected by Lassaigne’s test. Explain Lass aignes test for detecting nitrogen?

Question 17:
Write Van der Waals equation for one-mole gas and also write the physical significance of Van der Waals constants,

Question 18:
Complete the reaction.

Question 19:
Predict the products formed by addition reaction of HBr to but -1- ene in the presence of peroxide and in the absence of peroxide.

Question 20:
Give any two importance of green chemistry in day-to-day life.

Answer any seven from question numbers 21 to 29. Each carries three scores.

Question 21:
Calculate the number of atoms in each of the following.
a. 14 g of Nitrogen (N₂).
b. 14 mole atom Nitrogen (N₂)
c. 14 U of Nitrogen (N₂).

Question 22:
a. Identify the position of the elements in the periodic table having outer electronic configuration.
i. ns²np⁴ with n = 3
ii. (n-1)d² ns² with n=4.
b. “Removal of electron from an atom become easier on moving down the group.” Give reason.

Question 23:
a. Write the two postulates of Kinetic molecular theory of gases responsible for deviation of real gases from ideal behaviour.
b. What is Boyle point ?

Question 24:
a. Derive the equaion for Gibbs energy change (∆G ) in terms of ∆H and ∆S.
b. Write any two condition for which G becomes negative according to Gibbs equation.

Question 25:
Decomposition of PC15 can be represented as
PCl_5(g) → PCl_3(g)+ Cl_2(g)

∆H = 124.0KJmol^-1
Describe the effect of the following on the above equilibrium ?
a. addition of PCl₅
b. increase in pressure.
c. increase in temperature

Question 26:
Explain the following reactions.
a. CO is heated with ZnO
b. Boric acid is added to water
c. BF₃ with NaH.

Question 27:
Balance the following redox reaction using oxidation number or ion-electron method :

Question 28:
What causes the temporary and permanent hardness of water ? Give one method to remove permanent hardness ?

Question 29:
Analyze the given compound.
C₂H₅C(CH₃) = C(CH₃)C₂H₅
a. Write the IUPAC name of the compound.
b. Draw the structures of the geometrical isomers possible for the compound.

Answer any three from question numbers 30 to 33. Each carries four scores.

Question 30:
a. Write the Schrodinger wave equation and explain the terms.
b. Draw the boundary surface diagrams of any four 3d orbitals.

Question 31:
Compare the relative stability of the following species and indicate their magnetic properties based on molecular orbital theory.
O₂^– and O₂⁺

Question 32:
By passing Cl₂ gas through a solution, gets bleaching powder
a. Write the balanced chemical equation for above reaction.
b. What happens if CO₂ gas is passed instead of Cl₂ and write the balanced chemical equation.

Question 33:
Which of the following carbocation is most stable ?
(CH₃)₃C⁺, (CH₃)₂CH⁺, CH₃CH₂CH₂⁺
Justify your answer
b. Write the principle involved in the distillation of organic compound under reduced pressure ? Give an example.

Answers

Answer 1:
Relation between empirical and molecular formula,
MF = (EF) x n
Where n is a simple whole number

Answer 2:
n₃ → n₂

Answer 3:
Triagonal planar, SP²

Answer 4:
Be > B > Li

Answer 5:
The deviation from ideal behaviour is measured in terms of compressibility factor Z, which is the ratio of product PV and nRT.

Answer 6:
C₆₀

Answer 7:
Chlorine nitrate

Answer 8:
Law of Multiple Proortions
According to this law, if two elements can combine to form more than one compound the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Answer 9:
a. 4P
b. 3d

Answer 10:
a. lonic : NaCl, KCl
Covalent: BF₃, PCl₅₃, O₂
b. Obeying octet rule – NaCl, KCl, O₂
Not obeying Octet rule : BF₃, PCl₅

Answer 11:
Extensive : Heat capacity, Internal energy
Intensive : density, molar heat capacity

Answer 12:

Answer 13:
K_p = K_c (RT)^∆n

Answer 14:
a. The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions.
b. Basic buffer : Buffer mixture of a weak base and its salt with a strong acid,
e.g., NH₄OH and NH₄Cl (pH = 9.25)

Answer 15:

In boron atom lies one plane. Above and below this plane, there are 2 bridging H- atom. Each bridged H atom is bonded to 2 boron atom only by sharing of 2 electrons. Such a bond is called 3 centered 2 electron bond. Thus diborane have two (B-H-B) bonds, ie., two 3-centered 2 electron bond and four (B-H) bonds, ie., four 2 centered 2 electron bond.

Answer 16:
By fusing organic compound with sodium metal, the elements present in the compound forms corresponding water soluble salts.

Detection of Nitrogen:
N exist as NaCN in sodium fusion extract.

On heating with con.H₂SO₄, Fe²⁺ oxidised to Fe³⁺ to give prussion blue in colour.

Detection of Halogens:
Halogen exist as halides (max.) in fusion extract. Mien AgNO₃ and HNO₃ is added they form AgX.
X^– + Ag⁺ AgX; (X = Cl, Br, I)
AgCl – white ppt
Agl – yellow ppt
AgBr – pale yellow

Detection of Suphur:
Sulphur as Na2S is fusion extract when sodium nitroprusside added gives violet colour.
Na₂S + Na₂[Fe(CN)₅N0)] → Na₄[Fe(CN)₅ NOS] (violet colour)

Detection of Phosphorus:
The compound is heated with oxidising reagent (Sodium peroxide). The phosphorous present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.

Answer 17:
There is deviation from ideal behaviour at high pressure and low temperaure. This is due to two faulty assumptions in kinetic theory of gases. It is corrected by van der Waal’s equation.

where n is the number of moles of the gas a and b are van der Waal’s constants.

Answer 18:

Answer 19:

Answer 20:
Dry cleaning of clothes. Tetra chloro ethene (Toxic and contaminates ground water) which is used as solvent for dry cleaning is replaced by liquified CO₂.
• Bleaching of paper. Cl₂ gas is replaced H₂O₂.
• Synthesis of chemicals.

Answer 21:

Answer 22:
i. 1s² 2s² 2p⁶ 3s² 3p⁴
Block – p block
Atomic number – 16
Group – 16th group
Period – 3
Element – Sulphur

ii. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d² 4s²
Block – d block
Atomic number – 22
Group – 4 th group
Period – 4
Element – Ti

b. Down a group, the outermost electron being increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the electrons in the inner levels. In this case increase in shielding out weights the increasing nuclear charge and the removal of the outermost electrons requires less energy down a group.

Answer 23:
a. Maxwell, Boltzmann and others put forward a theoretical model of the gas. The theory is known as kinetic molecular theory of gases or microcopic model of gases.

Postulates of kinetic molecular theory are:
• All gases are made up of a large number of extremely small particles called molecules.
• The molecules are separated from one another by large distances so that the actual volume of the molecules is negligible as compared to the total volume of gas.
• The molecules are in a state of continuous rapid motion in all directions. During their motion, they keep on colliding with one another and also with the walls of the container.

b. The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point.

Answer 24:
a. Neither decrease in enthalpy nor increase in entropy alone can determine the direction of spontaneous hange for these systems. For this purposes, we define a new thermodynamic function the Gibbs energy or Gibbs function, G as
G = H – TS
The change in Gibbs energy for the system
∆G_sys = ∆H_sys – T∆S_sys – S_sys ∆T
At constant temperature, ∆T = 0
∆G_sys = ∆H_sys

Usually the subscript ‘ system’ is dropped and we simply write this equation as
∆G = ∆H – T∆S
Gibbs energy change = enthalpy change – temperature x entropy change.
It is reffered to as the Gibbs equation.
b. 1. ∆G is always -ve for a spontaneous process.
2. ∆G became negative for a reactions which are spontaneous only at a low temperature.

Answer 25:
a. Shift the equilibrium in the forward reaction
b. Increase of pressure shift the equilibrium towards left (backward direction) where then no.of particle is less.
c. Increasing temperature shift the equilibrium in the forward direction as it is endothermic.

Answer 26:
a. ZnO(s) + CO (g) \(\underrightarrow { \Delta }\) Zn(s) + C0₂
Extraction of metals

b. When boric acid is added to water, it accepts electrons from -OH ion. Boric acid is sparingly soluble in cold water i however fairly soluble in hot water, boric acid behaves as a weak monobasic acid. It doesn’t act as proton donor, beh- 0 aves as a lewis acid, it accepts a pair of electrons.
B(OH)₃ + 2HOH → [B(OH)₄]^– + H₃O⁺
450 A:
c. 2BF₃ + 6 NaH \(\underrightarrow { 450K }\) B₂H₆ + 6 NaF
Diborane is produced on an industrial scale by the reaction of BF₃ with sodium hydride.

Answer 27:

Answer 28:
Temporary hardness is due to presence of bicarbonates of Ca and Mg in water. Permanent hardness is due to presence of chlorides and sulphates of Ca and Mg in water. Permanent hardness re¬moved by ion exchange method or permutit process (Zeolite).

Answer 29:

Answer 30:
a. Erwin Schrodinger developed an atomic model taking into account both the wave and particle nature of elecron. This equation is known as shrodinger wave equation. For a system the schrodinger equation is written as
H\(\Psi\) = E\(\Psi\), where h is a mathematical operatior called Hamiltonian. Schrodinger gave a recipe of constructing this operator from the expression for the total energy of the system. The total energy of the system takes into accountthe kinetic energiesofallthe subatomic particles attractive potential bet-ween the electrons and nuclei and repuisive potential among the electrons and nuclei individually. Solution of this equation gives E and \(\Psi\).

Answer 31:

Answer 32:
a. 2Ca(OH)₂ + 2Cl₂ → CaCl₂ + Ca(OCl)₂ + 2H₂O Bleaching powder
b. When CO₂ is passed through lime water Ca(OH)₂ it turns milky due to the formation of calcium carbonate.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Answer 33:
a. (CH₃)C⁺ >(CH₃)₂CH₊ > CH₃CH₂CH₂⁺

Carbocations are classified as primary, secondary or tertiary. Its stability follows the order 3° > 2° > 1° due to inductive effect and hyperconjugation.
b. This method is used to purify liquids having very high boiling points and those, which decompose at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. The pressure is reduced with the help of a water pump or vacuum pump.

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Chemistry Improvement Question Paper 2018

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to Candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to get familiar with questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before answering.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all questions from question numbers 1 to 7. Each carry one score.

Question 1:
Name the type of smog generally formed during cool and humid climate.

Question 2:
Predict the shape of XeF₄ molecule, according to VSEPR theory.

Question 3:
Which is the acidic oxide among the following?

Question 4:
How many angular nodes are present in a 5f-orbital?

Question 5:
The allotrope of carbon with the highest thermodynamic stability is …………

Question 6:
The critical temperatures of some gases are given in the following table:


Question 7:
The number of oxygen atoms present in 5 moles of glucose (C₆H₁₂O₆) is ……….

Answer any ten from question numbers 8 to 20. Each carries two scores.

Question 8:
By using the concept of hybridization, explain the structure of H₂O molecule.

Question 9:
Give the IUPAC names of the following compounds.

Question 10:
Write any two applications of green chemistry in day-to-day life.

Question 11:
Explain the effects of temperature and pressure on the following equilibrium.

Question 12:
Draw the structure of orthoboric acid. Why it is not a protonic acid?

Question 13:
Find the molecular formula of the compound with molar mass, 78g mol^-1, and empirical formula CH.

Question 14:
What is meant by entropy of a system? What happens to the entropy during the following changes?

Question 15:
Represent graphically, the variation of prob-ability density (\({ \Psi }^{ 2 }\) (r )) as a function of distance (r) of the electron from the nucleus for 1s and 2s orbitals.

Question 16:
Derive an equation relating molar mass of an ideal gas with its density.

Question 17:
What is Wurts reaction? Give an example.

Question 18:
Define buffer solutions and write one example for an acidic buffer.

Question 19:
Cycloheptatrienyl Cation is given below :

Is this ion aromatic or not? Justify the answer.

Question 20:
Write the thermochemical equation corresponding to the standard enthalpy of formation of benzene.
Hint: ∆_f H^\(\Theta\) of benzene = +49.0 Kjmol^-1

Answer any seven from question numbers 21 to 29. Each carries three scores.

Question 21:
Justify the following:
a. ‘Ne’ has positive value for electron gain enthalpy.
b. The electron gain enthalpy of ‘F’ is lower than that of ‘Cl’.
c. The size of ‘Al³⁺’ is lower than that of F.

Question 22:
Identify X, Y and Z in the following sequence of reactions :

Question 23:
Calculate the mass of oxalic acid dihydrate (H₇C₂O₄.2H₂O) required to prepare 0.1M, 250 ml of its aqueous solution.

Question 24:
Balance the following Redox reaction by the ion-electron method or oxidation number method (Acid medium)

Question 25:
Explain any one method of preparation and structure of diborane.

Question 26:
The value of the equilibrium constant is useful to predict the extent of reaction and the direction of the reaction at a given stage. Explain.

Question 27:
Give a reason for the following :
a. H₂O₂ is stored in wax-lined glass or plastic vessels in dark.
b. Hard water is not suitable for laundry.

Question 28:
Calculate the total pressure in a mixture of 3.5 g of dinitrogen and 16 g of dioxygen confined in a vessel of 2 dm³ at 27°C. (R=0.083 bar dm³K^-1 mol^-1).

Question 29:
The reaction of cyanamide (NH₂CN) with dioxygen was carried out in a bomb calorimeter and AM was found to be -742.7kJ mol^-1, at 298 K. Calculate enthalpy change for the reaction at 298 K.

Answer any three from question numbers 30 to 33. Each carries four scores.

Question 30:
Write the molecular orbital electronic configurations of N₂ and O₂ and calculate their bond orders. Give a comparison of their stability and magnetic behavior.

Question 31:
Briefly describe the principles of the following techniques, taking an example in each case.
a. Crystallization
b. Simple distillation
c. Distillation under reduced pressure
d. Paper chromatography

Question 32:
Give the postulates of Bohr model of hydrogenatom. Also write two merits and two limitations of this model.

Question 33:
Account for the following :
a. Blue coloured solutions are obtained when alkali metals are dissolved in liquid ammonia.
b. ‘Li’and ‘Mg’ show similar properties.
c. Aqueous solution of Na₂CO₃ is alkaline.
d. BeS0₄ and MgSO₄ are readily soluble in water.

Answers

Answer 1:
Classical smog

Answer 2:
Square planar

Answer 3:
a. Cl207

Answer 4:
3

Answer 5:
Graphite

Answer 6:
O₂

Answer 7:
5 mole glucose contains 30 moles oxygen.
So the number of Oxygen atoms present in 5 moles of glucose = 30 x 6.022 x 10²³

Answer 8:

The shape should have been tetrahedral if there were all bp but two Ip are present so the shape is distorted tetrahedral or angular. The reason is Ip-Ip repulsion is more than Ip- bp repulsion which is more than bp-bp repulsion. Thus, the angle is reduced to 104.5° from 109.5°

Answer 9:
a. 3 – Ethyl – 1,1 – dimethylcyclohexane
b. 3 – Bromo – 3 – chloroheptane

Answer 10:

• Dry cleaning of clothes. Tetrachloroethene
(Toxic and contaminates groundwater)
which is used as a solvent for dry cleaning is replaced by liquified CO₂.
• Bleaching of paper. Cl₂ gas is replaced H₂O₂.
• Synthesis of chemicals.

Answer 11:
Low temperature favours forward reaction.
High pressure favours forward reaction.

Answer 12:

Protonic acid are those acids which ionizes in solution and produce protons . But Boric Acid/orthoboric acid is monobasic and a weak acid. It doesn’t donating proton it accepts electrons of OH- which produces due to ionisation of water into H + and OH- hence we can say it act as a lewis acid and proton is remove from water instead of boric acid.

Answer 13:
The empirical formula of the compound = CH
Empirical formula mass =12 x 1 + 1 x 1 = 13
Molar mass = 78

Answer 14:
Measure of degree of disorders or randomness of a system.
a. Entropy decreases
b. Entropy increases

Answer 15:

Answer 16:
Ideal gas equation can be rearranged as follows:
\(\frac { n }{ V }\) = \(\frac { P }{ RT }\)
Replacing n by \(\frac { m }{ M }\)
\(\frac { m }{ MV }\) = \(\frac { P }{ RT }\)
We get \(\frac { d }{ M }\) = \(\frac { P }{ RT }\) ; where d is the density.
On rearranging equation we get the rela-tionship for calculating the molar mass of a gas.
M = \(\frac { dRT }{ P }\)

Answer 17:
Alkyl halides on treatment with sodium metal in dry ethereal (free from moisture) solution give higher alkanes. This reaction is known as Wurtz reaction and is used for the preparation of higher alkanes containing even number of carbon atoms.

Answer 18:
A buffer solution is one which can resist change in its pH value on the addition of small quantities of acid or base.
Example for acidic buffer is blood.

Answer 19:
Yes. It is aromatic. In the case of cycloheptatriene, there are seven carbon atoms. Except one carbon (one is in sp³ hybrid state), all are in sp² hybrid state and each has an unhybridized pure p orbital perpendicular to the molecular plane, in the case of cycloheptatrienyl cation contains 6 π electrons, fulfilling the Huckle rule and the carbocation also in sp² hybrid state leads continuous delocalization. Hence it is aromatic.

Answer 20:
6C_(s) + 3H_2(g) → C₆H₆ (I); ∆_f H^\(\Theta\) = + 49 kJ/mol

Answer 21:
a. Noble gases have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very un stable electronic configuration. It may be noted that electron gain enthalpies have large negative values toward the upper right of the periodic table preceding the noble gases.
b. Electron gain enthalpy of F is less negative than that of the succeeding element This is because when an electron is added to F, the added electron goes to the smaller n = 2 quantum level and j suffers significant repulsion from the i other electrons present in this level. For the n = 3 quantum level ( Cl), the ; added electron occupies a larger > region of space and the electron-electron repulsion is much less.
c. For iso electronic species, size of ion decreases with increase in nuclear charge atomic number or cation has a smaller size than anion.

Answer 22:

Answer 23:
Molarity = \(\frac { No. of moles in the solute }{ Volume of solution in ml }\) x 1000
Moles of oxalic acid dihydrate in the solution = 0.1 x 0.25 = 0. 025 moles.
Molecular mass of oxalic acid dihydrate is 2+12 x 2 + 16 x 4 + 2(2+16) = 126
Mass of the acid used = 126 x 0.025= 3.15g

Answer 24:

Answer 25:

In diborane 4 terminal H-atom and the 2 boron atom lies one plane. Above and below this plane, there are 2 bridging H-atom. Each bridged H atom is bonded to 2 boron atom only by sharing of 2 electrons. Such a bond is called 3 centered 2 elec¬tron bond. Thus diborane have two (B-H-B) bonds, ie., two 3-centered 2 electron bond and four (B-H) bonds, ie., four 2 centered 2 electron bond.

Answer 26:
Larger the value of K, greater is the extent to which the reactants are converted into the products.

• If K >10³, the product predominates over reactants at equilibrium. If K is very large the reaction proceeds nearly to completion.
• If K<10³, the product predominates over product. If K is very small the reaction proceeds to very small extend or rarely.
• If K is in the range of 10^-3 to 10³, appreciable concentrations of both reactants and products are present.

Prediction of direction of a reaction

The reaction quotient in terms of concentration or concentration Quotient (Q) with molar concentrations (Q_c) and with partial pressures (Q_p) is related to the equilibrium constant (K) for the reaction.
For a general reaction, aA+ bB ⇌ cC + dD

• If Q_c > K_c, the reaction will proceed in the direction of reactants (reverse reaction)
• If Q_c < K_c, the reaction will proceed in the direction of the products (forward reaction)
• If Q_c = K, the reaction is at equilibrium.

Answer 27:
a. H²O² decomposes in the presence of light and alkali content in glass. To prevent decomposition H²O² is stored in wax – lined glass or plasic vessels indark.
b. Hard water do not form lather easily with soap, due to the presence of dissolved salts in it.

Answer 28:

Answer 29:

Answer 30:

Since the bond order is 3, there is a triple bond in N² and the bond energy is very high. Since all the electrons are paired. It is diamagnetic.

Oxygen has unpaired electrons and hence I it is paramagnetic in nature.

Answer 31:
Crystallization:
This is one of the most commonly used techniques for the purification of solid organic compounds. It is based on the difference in the solubilities of the compound , and the impurities in a suitable solvent.
e.g., sugar and salt.

Simple distilation :
It is the process of eva-poration followed by condensation. It is used for separating a liquid which does not decompose on boiling from other liquids = having diferent boiling points or should contain a non-volatile component,
e.g., Benzene and Toluene. The vapours of a substance formed are condensed and the liquid is collected in the conical flask.

Distillation under reduced pressure: The method is used to separate those liquids which decompose at or below their normal ; boiling points. In this method, the distillation is carried at a lower pressure of 10 – i 20 mm Hg, so that the liquid can be boiled i at lower temperature without undergoing decomposition, e.g., Glycerol can be sep- arated from spentlye in the soap industry.

Paper chromatography: Partition chromaography is based on continuous differential partitioning of components of a mixture between stationary and mobile phases, Paper chromatography is a type of partition chromatography. In paper chromatograph, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase.

Answer 32:
1. The electron in an atom revolve around the nucleus only in certain circular path called orbits.
2. The energy of the electron in a particular orbit remains constant.
3. Only those orbits are permitted in which the angalar momentum of the electron is a whole number of multiple of h/27r, where h is plank’s constant.
4. Energy is emitted or absorbed by an atom only when an electron moves one from one orbit to another.

Merits

1.  It could account for the stability of hydrogen atom
2.  It could account for the line spectra of hydrogen atom and hydrogen like ions (for example, He⁺, Li²⁺, Be³⁺, and so on)

Demerits

1. Bohr’s theory was also unable to explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or an electric field (Stark effect).
2. It could not explain the ability of atoms to form molecules by chemical bonds

Answer 33:
a. Alkali metals dissolve in liquid ammonia giving deep blue solutions which are conducting in nature. The blue colour of the solution is due to ammoniated .electrons which absorb energy in the visible region of light.
b. Mg and Li shows diagonal relationships. Both have similar size and similar electronegativity.
c. Sodium carbonate acts as an alkali be-cause when dissolved in water, it dissociates into the weak acid, carbonic- acid and the strong alkali sodium hy droxide.
d. The hydration enthalpies of BeSO₄ and MgSO₄ are quite high because of the small size of Be²⁺ and Mg²⁺ ions. These hydration enthalpy values are higher than – their corresponding lattice enthalpies and therefore, BeSO₄ and MgSO₄ are highly soluble in water.

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Chemistry Previous Year Question Paper 2017

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to Candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to get familiar with questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before answering.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Question 1:
a. Determine the number of moles present in 0.55 mg of electrons.
imgg
i. 1 mole
ii. 2 moles
iii. 1.5 moles
iv. 0.5 mole [1]

b. Give the empirical formula of the following. [2]
C₆H₁₂O₆, C₆H₆. CH₃COOH, C₆H₆C₁₆

c. Two elements, carbon and hydrogen combine to form C₂H₆, C₂H₄ and C₂H₂. Identify the law illustrated here. [1]

Question 2:
a. i. Write the electronic configuration of chromium (Z=24).
ii. Find the number of electrons in the subshells with azimuthal quantam number l=2.
iii. Represent the orbital with quantum numbers n = 1 and 1 = 0.
b. Give the mathematical representation of Heisenberg’s uncertainty principle and its one important significance.

Question 3:
Electron gain enthalpy is one of the important periodic property.
a. Define electron gain enthalpy. [1]
b. Explain any two factors affecting electron gain enthalpy. [2]
c. Write the oxidation state and covalency of Al in (A/ F₆)^3- [1]

Question 4:
The geometry of the molecule is decided by type of hybridization.
a. Discuss the shape of PCl₅ molecule using hybridization. [2]
b. Give the reason for the high reactivity of PCl₅ [2]
c. Isoelectronic species have the samebond order. Among the following, choose the pair having same bond order. Among the following, choose the pair having same bond order.
CN^–, O₂^– ,NO⁺, CN⁺

Question 5:
a. Give the reason behind the following.

i. The glass window pannels of old buildings are thicker at the bottom than at the top. [1]
ii. Sharp glass edges are heated for making them smooth. [1]

b. Maxwell and Boltzmann have shown that actual distribution of molecular speeds depends, on temperature and molecular mass.
i. What do you mean by most probable velocity? [1]
ii. At the same temperature which will move faster, N₂ or Cl₂ ? [1]

Question 6:
a. Some macroscopic properties are given below. Help Reena to classify them into two groups under suitable titles.
[Heat capacity, Entropy, Refractive in¬dex, Surface tension.] [2]
b. For the reaction
2A_(g) + 2B_(g) → 2D_(g)
∆U°= -10.5 kJ/mol
∆S°= – 44.1J/k/mol at 298 K.
Calculate ∆G° for the reaction. [2]

Question 7:
a. Classify the following solutions into acidic, basic and neutral
NaCl, NH₄NO₃, NaCN, NaNO₂ [2]
b. pH of blood remains constant in spite of the variety of goods and spices we eat. Give a reason. [1]
c. The solubility ofMg(OH)₂ at 298K is 1.5 x 10^-4.Calculate the solubility product.[2]

Question 8:
Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced equa¬tion for the reaction using oxidation number method.
Skeletal equation is
MnO₄^– + Br^– → MnO₂ + BrO₃^– [3]

Question 9:
Hydrogen is the most abundant element in the universe, But in free state it is almost not found in earth’s atmosphere.
a. Suggest any three methods for the preparation of H₂ gas by selecting suitable substance given below.

b. Do you expect carbohydrides of the type C_nH_2n+2to act as Lewis acid or base? Why?

Question 10:
The s-block of periodic table constitutes alkali metals and alkaline earth metals.
a. The hydroxides and carbonates of sodium and pottassium are more soluble than that of corresponding salts of Magnesium and Calcium. Explain. [2]
b. Write the chemical name of the following:
i. Caustic soda
ii. Baking soda
iii. Slaked lime
iv. Milk of lime [2]

Question 11:
Borax is an important compound of Boron.
a. The solution of borax is alkaline. Give a reason. [2]
b. Give any two uses of borax. [1]
c. Diamond has covalent bonding. Yet it has high melting point. Give a reason. [1 ]

Question 12:
a. Give the structural formula of the following compounds:
i. 2, 4, 7 – Trimethyloctane
ii. 2 – chloro – 4 – methyl pentane [2]
b. CH₃C₂ or (CH₃)₂CH -which is more stable? Explain. [2]
c. Explain the chemistry behind crystallization. [2]
Or
a. Give the IUPAC names of the following:

b. Which is more stable (CH₃)₃ C⁺ or CH₃C⁺H₂? Give a reason. [2]
c. Give the chemistry behind distillation under reduced pressure. [2]

Question 13:
Benzene and benzenoid compounds show aromatic character.
a. Select the aromatic compounds from the following: [1]

b. Suggest a method to convert Ethyne to benzene. [2]
c. Give the products formed when benzene reacts with
i. CH₃ Cl/AlCl₃
ii. Cl₂ /hυ [2]

Question 14:
Environmental pollution is the effect of undesirable change in surroundings that have harmful effect on plants, animals and human beings.
a. Explain the adverse effect of global warming. [2]
b. Choose the one which is not a component of photochemical smog.
i. NO₂
ii. O₃
iii. SO₂

Answers

Answer 1:
1. a.i. 1 mole
Mass of one electron = 9.1 x 10^-28 g
= 9.1 x 10^-25 mg
Mass of 1 mol of electron
= 9.1 x 10^-25 x 6.022 x 10^-23= 0.55 mg
So 0.55 mg of electron = 1 mol of electron

b. Empirical formula of C₆ H₁₂O₆ is CH₂O
Empirical formula of C₆H₆ is CH
Empirical formula of CH₃COOH is CH₂O
Enpiric formula of C₆H₆Cl₆ is CHCl
c. Law of multiple proportion.

Answer 2:
a. i. ₂₄Cr – 1s² , 2s², 2p⁶, 3s², 3p⁶, 4s¹, 3d⁵ or [Ar] 4s¹, 3d⁵
ii. 10 electrons in d subshell.
iii. 1s

b. Heisenberg’s uncertainty states that it is impossible to determine simultaneously and precisely both the exact position and exact momentum or velocity of a subatomic particle like an electron.
Mathematically, ∆x ∆p ≥ \(\frac { h }{ 4\pi }\)

where ∆p is the uncertainty in the momentum of the particle, ∆x is the uncertainty in position.
From this equation, it is seen that if ∆p increases, the ∆x decreases, and vice versa. It is also can be written as follows:
∆x ∆p ≥ \(\frac { h }{ 4\pi m }\)

Answer 3:
a. Electron gain enthalpy is the enthalpy change when an electron is added to the outer most shell of an isolated gaseous atom in the ground state.
b. The factors affecting electron gain enthalpy are atomic size, nuclear charge, shielding effect, electronic configuration etc. As the atomic size increases. ∆_eg H decreases. When nuclear charge increases, electron gain enthalpy increases.
c. Oxidation state is +3
Covalency is 6.

Answer 4:
a. In PCl₅, five sp³d orbitals of phosphorus overlap with singly occupied p-orbitals of chlorine atoms to form five P-Cl bonds. Three P-Cl bond lie in one plane and make bond angle of 120°. These bonds are equitorial bonds. The remaining two P-Cl bond lye above and below the equitorial plane making an angle of 90° with the plane. These bonds are called axial bonds.

b. Two P-Cl axial bonds are slightly elongaed to minimise repulsion from equitorial bond pairs. So axial bonds are weaker than equitorial bonds in PCl₅.
c. CN^– and NO⁺

Answer 5:
a.i. Glass is a very viscous liquid. So it has a tendency to flow very slightly. Hence the window panes of old buildings are thicker at the bottom than at the top.
ii. On heating, the glass melts and the surface of the liquid tend to take the rounded shape at the edges, which makes the edges smooth.
b. i. Most probable velocity is the velocity possessed by a maximum number of gas molecules.
ii. N₂.

b. Given All0 = -10.5 kJ/mol = -10.5 x 103 J/mol
AS0 = -44.1 J/K/ mol, T= 298 K We know that AH° = AU° + AnRT Here AH° = -10500 + (-1)x 8.314 x 298 = -12978 J/ mol
AG° = AH0 – T AS0 = -12978 – (298 X44.1) = -164 J/ mol = -0.164 kJ/ mol.

Answer 6:
a.

Extensive properties	Intensive properties
Heat capacity	Refractive Index
Entropy	Surface tension

b. Given ∆U° = -10.5 kJ/mol = -10.5 x 10³ J/mol
∆S° = -44.1 JIKI mol, T = 298 K
We know that ∆H° = ∆U° + ∆nRT
Here ∆H° = -10500 + (-1)x 8.314 x 298
= -12978 J/ mol
∆G° = ∆H° – T∆S° = -12978- (298 X 44.1)
= -164 J/ mol = -0.164 kJ/ mol.

Answer 7:
a. Neutral – NaCl
Acidic – NH₄NO₃
Basic – NaCN, NaNO₂
b. Because blood is a buffer solution.
c. Mg (OH)₂ → Mg²⁺ + 2 OH^–
Solubility: S S 2S
K_sp = [Mg²⁺][OH^–]² = S X (2S)² = 4S³
= 4 x (1.5 x 10^-4 )³ = 1.35 x 10^-11

Answer 8:
Step 1 : The skeletal equation is :
MnO₄^– +Br^– → MnO₂ + BrO₃^–

Step 2 : Assign oxidation number each element and identify the elements under-going change in oxidation number.
Mn⁺¹O₄^-2 + Br^-1 → Mn⁺⁴O^-2₂ + BC⁺⁵O₃^-2
Here the oxidation number of Mn and Br are changed.

Step 3 : Calculate the increase and decrease in oxidation number and make them equal by multiplying with suitable number.
2MnO₄^– + Br^– → 2 MnO₂+ BrO₃^–

Step 4 : Since the reaction takes place in the basic medium, 20H^– ions on the RHS to balance the ionic charges on both sides.
2MnO₊₄^– + Br^– → 2 MnO₂ + BrO₃^– + 2OH^–

Step 5 : Now balance hydrogen atom by adding a sufficient number of H₂O molecules. Here add one H₂0 molecule on LHS.
2MnO₊₄^–+ Br^– +H₂O → 2MnO₂+ BrO₃^–+ 2OH^–
This is the balanced equation.

Answer 9:
a. i. 2Na + 2H₂O → 2 NaOH + H₂
or
2Na + 2 HCl → 2NaCl + H₂
ii. Zn + 2 HCl → ZnCl₂ + H₂
OR
Zn + 2 NaOH → Na₂ZnO₂ + H₂
iii. CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
iv. 2Al + 6 HCl → 2AlCl₃ + 3H₂ (Any three)

b. C_n H_2n+1 is neither a Lewis acid nor a Lewis base due to the absence of vacant orbitals or lone pairs of electrons on carbon.

Answer 10:
a. This is due to greater atomic radii (size) or lesser lattice enthalpy of sodium and potassium ions.
b. i. Caustic soda : Sodium hydroxide (NaOH)
ii. Baking soda : Sodium bicarbonate (NaHCO₃)
iii. Slaked lime : Calcium hydroxide (Ca(OH)₂)
iv. Milk of lime : Magnesium hydroxide (Mg(0H)₂)

Answer 11:
a. Borax dissolves in water to give NaOH and orthoboric acid. Since NaOH is a strong alkali and orthoboric acid is a weak acid, the solution is basic in nature.
Na₂B₄O₇ + 7 H₂O → 2NaOH + 4H₃BO₃

b. Borax is used for the detection of transition elements by Borax bead test. It is used for the manufacture of heat-resistant glasses, glass-wool, and fiberglass. It is also used as a flux for soldering metals, for heat, scratch and stain resistant, glazed coating to earthenwares and as a  constituent of medicinal soaps. :
c. Diamond has a three-dimensional network structure with extended covalent bonding.
Since it is very difficult to break these covalent bonds, it has high m.p Or, due to the tetrahedral (sp³) hybridization of carbon atoms in diamond.

Answer 12:
a. i.

b. CH₃^– – CH₂^– is more stable than (CH₃)₂CH^– since for carbanions the stability order 1° > 2° > 3° due to inductive effect.

c. Crystallization is based on the difference in the solubilities of the compound which is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature. The solution is concentrated to get a nearly saturated solution at a higher temperature. On cooling the solution , pure compound crystallizes out and is removed by filtration.

OR

a. i. 3- Bromo-3-chloroheptane
ii. 4-Ethyl-2-methylaniline

b. (CH₃)₃C⁺ is more stable than CH₃ – CH₂⁺ since the stability of carbocations follow the order 3° > 2° > 1° due to inductive effect and hyperconjugation.

c. This method is used to purify liquids having very high boiling points and those, which decompose at or below their boiling points. Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface. The pressure is reduced with the
help of a water pump or vacuum pump.

Answer 13:
a. Aromatic compounds are :

b. Ethylene on passing through a red-hot iron tube at 873 K, we get benzene (C₆H₆).
c. i. Toluene (C₆H₅ – CH₃)

Answer 14:
a. Due to global warming, the average global temperature will increase. This will lead to the melting of polar ice caps and flooding of low lying areas all over the earth. Increase in global temperature results in infectious diseases like dengue, malaria, yellow fever, sleeping, sickness etc.
b. iii. SO₂

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Chemistry Previous Year Question Paper 2018

Time Allowed: 2 hours
Cool off time: 15 Minutes
Maximum Marks: 60

General Instructions to Candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to get familiar with questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before answering.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Answer all questions from question numbers 1 to 7. Each carry one score.

Question 1:
Name the quantum number which gives the spacial orientation of an orbital with respect to standard as et of coordinate axes.

Question 2:
Among N^3-, O², F^–, Na⁺ and Al³⁺, which one has the smallest size?

Question 3:
Which pollutant in water causes brown mottling of teeth ?

Question 4:
Examine the following graph and name the gas law corresponding to it.

Question 5:
If Z – axis is the internuclear axis, name the type of covalent bond formed by the overlapping of two Py – orbitals.

Question 6:
Which among the following measurements, contains the highest number of significant figures?

Question 7:
Write the formula of the basic structural unit of silicates.

Answer any ten from question numbers 8 to 20. Each carries two scores.

Question 8:
Write any two limitations of octet rule.

Question 9:
Draw the ‘sawhorse’ projections of the eclipsed and staggered conformations of ethane.

Question 10:
Give the relation between Kp and Kc for the reaction given below.

Question 11:
State and illustrate the law of multiple proportions.

Question 12:
a. Define ‘ normal boiling point’ of a liquid.
b. Give a reason for the use of pressure cooker to cook food, at high altitudes.

Question 13:
Write two important results observed during the photoelectric effect.

Question 14:
Differentiate state functions from path functions and give one example for each.

Question 15:
Define the terms, Biochemical Oxygen Demand (BOD) and Eutrophication.

Question 16:
Give the chemical equation for the conversion of hexane to benzene. Write the name of the process.

Question 17:
H₂0 and HSO₄, can act both as Bronsted acids and bases. For each case give he corresponding conjugate acid and conjugate base.

Question 18:
What is metamerism ? Write the metamers of C₄H₁₀O.

Question 19:
What are Zeolites ? Give any two uses of zeolites.

Question 20:
First law of thermodynamics can be stated as Δu = q + w. How can this equation be expressed for:
a. An isothermal reversible change ?
b. A process carried out at constant volume ?

Answer any seven from question numbers 21 to 29. Each carries three scores.

Question 21:
Predict the products :

Question 22:
Redox reactions are classified into four types. Describe any three of them with suitable examples.

Question 23:
Enthalpies of formation of some compounds are given below:

Using these data, calculate the enthalpy of reaction for N₂O_4(g) + 3CO_(g) → N₂ O_(g) + 3CO_2(g)

Question 24:
Sketch the structures of graphite and diamond. What is the impact of structure on physical properties of these allotropes?

Question 25:
Briefly explain the different types of hydrides.

Question 26:
Calculate the amount of CO₂(g) produced by the reaction of 32g of CH₄ and 32g of O₂.

Question 27:
Give reasons for the following :
a. ‘O’ has lower ionization enthalpy than N and F.
b. Cl has higher negative electron gain enthalpy than F.

Question 28:
Write the postulates of kinetic molecular theory of gases.

Question 29:
The ionization constant of nitrous acid is 4.5 i x 10^-4. Calculate the pH of 0.04M solution of j nitrous acid in water.

Answer any three from question numbers 30 to 33. Each carries four scores.

Question 30:
Name the commercial process used to prepare sodium carbonate and write the chemical equations of the steps involved in it.

Question 31:
What is ‘sodium fusion extract’? How the presence of N, S and halogens inorganic compounds are detected?

Question 32:
The diatomic species Ne₂ does not exist, but Ne₂ can exist. Explain on the basis of molecular orbital theory.

Question 33:
Explain how the different series of lines are formed in the hydrogen spectrum. Derive an equation to find the waW number of a line in the hydrogen spectrum.

Answers

Answer 1:
Magnetic quantum number

Answer 2:
Al³⁺

Answer 3:
Fluoride (P)

Answer 4:
Gay-Lussac’s law

Answer 5:
π bond

Answer 6:
a

Answer 7:
SiO₄^4-

Answer 8:
The incomplete octet of the central atom
In some compounds, the number of electrons surrounding the central atom is less than eight. This is especially the case with elements having less than four valence electrons.

Odd-electron molecules
In molecules with an odd number of electrons like nitric oxide, NO and nitrogen dioxide, N02, the octet rule is not satisfied for all the atoms

Answer 9:

Answer 10:
K_p = K_c (RT)^∆n
∆n = 3 – 2 = 1
K_p = K_c (RT)¹= K_c(RT)

Answer 11:
Law of Multiple Proportions
According to this law, if two elements can combine to form more than one compound the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers

Here the masses of oxygen which com¬bine with a fixed mass of hydrogen bear a simple ratio, i.e., 16: 32 or 1: 2.

Answer 12:
a. The temperature at which a real gas obeys ideal gas law over an appreciable range of pressure is called Boyle temperature or Boyle point.
b. At high altitudes atmospheric pressure is low. Therefore liquids at high altitudes boil at lower temperatures in comparison to that at sea level. Since water boils at low temperature on hills, the pressure cooker is used for cooking food.

Answer 13:
i. There is no time lag between this striking of light beam and ejection of an electron from the metal surface.
ii. The no.of electron ejected is proportional to the intensity or brightness of light

Answer 14:
A state function is a property of the system whose values depends upon the initial and final states of the system and independent of the path by which the state is reached.
Path function is a property of the system whose values depend on the path followed,
eg. Heat (q), Work (w).

Answer 15:
BOD is Biochemical Oxygen Demand. It is defined as the amount of oxygen in milligrams dissolved in water needed to break down the organic matter present in one litre of water for five days at 20 °C.Pure water contains BOD of 0-3 ppm. 5 ppm means water is contaminated.
The process in which the nutrient enriched water bodies support a dense plant population, which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as eutrophication

Answer 16:
Alkanes having six or more carbon atoms on heating to 773 K at 10^-20 atmospheric pressure in presence of Cr₂O₃ or V₂O₅ or MO₂O₃ catalyst give aromatic compounds. This process is known as aromatization.

Answer 17:
H₂O : Conjugate acid – H₃O⁺ Conjugate base – OH^–
HS0₄^– : Conjugate acid – HSO₄^–
Conjugate base – SO₄^2-

Answer 18:
Metamerism: It arises due to different alkyl chains on either side of the functional group in the molecule. For example, C₄H₁₀O represents methoxypropane (CH₃OC₃H₇) and ethoxyethane (C₂H₅OC₂H₅).

Answer 19:
Zeolites : If aluminum atoms replace few silicon atoms in three-dimensional network of silicon dioxide, overall structure known as aluminosilicate, acquires a negative charge. Cations such as Na⁺, K⁺ or Ca²⁺ balance the negative charge. Examples are feldspar and zeolites. Zeolites are idely used as a catalyst in petrochemical industries for cracking of hydrocarbons and isomerisation, e.g., ZSM- 5 (A type of zeolite) used to convert alcohols directly into gasoline. Hydrated zeolites are used as ion exchangers in softening of “hard” water.

Answer 20:
a. q = -w = 2.303 RT log \(\frac { { V }_{ f } }{ { V }_{ i } }\)
q = -w
b. ∆u = ∆w

Answer 21:
a. CH₃– CH₂– CH₂– Br 1 – bromopropane

Answer 22:
1. Combination reactions. The reaction involves combination of a compound from two or more components atleast one of which must be in the elemental form or both the components are in the elemental state.
eg., CO + O₂ \(\underrightarrow { \Delta }\) C0₂

2. Decomposition reactions. The reaction involves decomposition of a compound into two or more components atleast one of which must be in the elemental state.
eg., 2NaH \(\underrightarrow { \Delta }\) 2Na + H₂

3. Displacement reactions. In a displace-ment reaction, an ion or an atom in a com-pound is replaced by an ion or an atom of another element,
eg., A + BC → AB + C

a. Metal displacement reaction.
A metal in a compound can be displaced by another metal in the uncombined state.

b. Non-metal displacement reaction.
This include hydrogen displacement and a rarely occuring reaction involving oxygen displacement.

Answer 23:
Enthalpy of reaction
(∆_r H)= [81 +3(-393)] -[9.7 +3(-110)]
= [81 – 1179] – [9.7 – 330] = -778kJmol^-1

Answer 24:
Graphite. Graphite has layered structure. Distance between 2 layers is 340 pm. C-C bond length within the layer is 141.5 pm. Each carbon atom in hexagonal ring undergoes sp² hybridisation with 3 neigh-bouring carbon atoms. The 4^th electron present in the unhybridised p- orbital is used for forming a π bond.

Diamond: Purest form of carbon. In dia-mond, each carbon atom is in sp3 hybridised state and is tetrahedrally bonded to 4 other carbon atom. The C-C bond length is 154 pm and C-C-C bond angle is 109°28. This 3 dimensional net work structure makes diamond harder and a bad conducor of electricity.

Answer 25:
Binary compounds of hydrogen with other elements are called hydrides. It can be ex-pressed as EH_x(MgH₂) or E_mH_n (B₂H₆).

ionic or Saline Hydrides
There are stoichiometric compounds of H₂ formed with most of the s-block elements j which are highly electropositive in character. The ionic hydrides are crystalline, nonvolatile and non-conducting in solid state.
In molten state, ionic hydrides conduct I electricity. Salts like hydrides are strong bases such as LiH, BeH₂, MgH₂.

Covalent or Molecular Hydrides
Covalent hydrides are formed by most of the p-block elements. It have very low con¬ductivity and are acidic in nature. Covalent hydrides are soft and have very low melting and boiling points,
eg., CH₄, NH₃

Molecular hydrides are further classified j according to the relative numbers of electrons and bonds in their lewis structure,

a. Electron-deficient hydrides. These hydrides are deficient in electrons to write their conventional lewis structure.
eg: group 13 elements (B₂H₆).

b. Electron-precise hydrides. These hydrides have the required no: of electro¬ns to write their conventional lewis structure,
eg., group 14 elements (CH₄).

c. Electron-rich hydrides. These hydrides have excess which are present as lone pairs, eg. group 15-17 elements (NH₃ has 1 lone pair, H₂O has 2 lone pairs).

Metallic or Interstitial Hydrides (Non- stoichiometric)
These hydrides are formed by many d- block and f-block elements. Metallic hydrides are hard, possess metaliclusture, conduct electricity and show magnetic properties. eg., LaH_2.87, Y₆H_2.55. These hydrides are non-stoichiometric being deficient in hydrogen.

Answer 26:
To solve such questions you must write a balanced reaction first.
CH₄ (g)+2O₂ (g) → CO₂ (g) +2H₂O (g)
The molecular mass of methane is 16
32 grams of Methane mean 2 mole
32 grams of Oxygen mean 1 mole
2 moles of Oxygen for every mole of methane. So that makes oxygen the limiting reagent. So 1 moles of 02 will form 0.5 moles of C0₂.
0.5 mole of CO₂ = 0.5*44 = 22 grams
So 22 grams of CO₂will be produced

Answer 27:
a. Oxygen has outer electronic configuration 2s² 2p⁴ removal of one electron from the 2p subshell will make its configuration2s² 2p³ thus it attains a half-filled configuration thus lesser energy is required to remove an electron from oxygen where-as in nitrogen the electronic configuration 1s² 2s² 2p³ it has a stable half filled configuration and thus more energy is required to remove an electron salts.

from nitrogen, in case of flourine, the electronic configuration is 2s²2p⁵ it has to gain one more electron to attain noble gas configuration and become stable thus more energy is required to remove an I electron from flourine thus ionisation enthalpy of oxygen is lesser than nitrogen or fluorine.

b. Cl has more negative electron gain enthalpy than F. F has less negative electron gain enthalpy because in it the added electron goes to the smaller { energy level (n=2) and hence suffers ‘ significant repulsion from the electrons already present in this shell also flurine possesses high charge density.

Answer 28:
All gases are made up of a large number of extremely small particles called molecules.

• The molecules are separated from one another by large distances so that the actual volume of the molecules is negligible as compared to the total volume of gas.
•  The molecules are in a state of continuous rapid motion in all directions. During their motion, they keep on colliding with one another and also with the walls of the: container.
• Molecular collisions are perfectly elastic, there is no net loss or gain of energy in their collisions. However, there may be a redistribution of energy during such collisions,
• There are no attractive forces between the molecules. The move completely independent of each other.

Answer 29:
NO_-2 + H₂O ⇌ HNO₂ + OH^–
K_h= [ HNO₂ ] [ OH^– ] /NO^– ₂]
Kw / ka = 10^-14 / 4.5 x 10^-4 = 0.22 x 10^-10
Now, If x moles of the salt undergo hydrolysis,  then the concentration of various species present i in the solution will be:
[NO^-2] = 0.04 -x;0.04
[ HN0₂ ] = x [ OH-] = x
K_h = x² / 0.04 = 0.22 x 10^-10 x2
x² = 0.0088 x 10^-10
x = 0.093 x 10^-5
[ OH^– ] = 0.093 x 10^-5M
[H₃0⁺] = 10^-14] / 0.093 x 10^-5] = 10.75 x 10^-9] M
pH = -log( 10.75 x 10^-9] ) = 7.96

Answer 30:
Sodium carbonate is generally prepared by Solvay Process. In this process, advantage is taken of the low solubility of sodium hydrogencarbonate whereby it gets precipitated in the reaction of sodium chloride with ammonium hydrogencarbonate. The latter is prepared by passing C02 to a concentrated solution of sodium chloride saturated with ammonia, where ammonium carbonate followed by ammonium hydro-gencarbonate are formed. The equations for the complete process may be written as :

2NH₃ + H₂0 + C0₂ → (NH₄)₂CO₃
(NH₄)₂CO₃+ H₂O + CO₂ → 2NH₄HCO₃
NH₄HCO₃+ NaCl → NH₄Cl + NaHCO₃

Sodium hydrogencarbonate crystal sepa¬rates. These are heated to give sodium carbonate.

2NaHCO₃ →Na₂CO₃ + C0₂ + H₂0

In this process NH₃ is recovered when the solution containing NH₄CI is treated with Ca(OH)₂. Calcium chloride is obtained as a by product.

Answer 31:
Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “Lassaigne’s test”. The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.

By fusing organic compound with sodium metal, the elements present in the compound forms corresponding water soluble salts.

Detection of Nitrogen:
N exist as NaCN in sodium fusion extract.

CN^– + Fe²⁺ \(\underrightarrow { FeS{ O }_{ 4 } }\) [Fe(CN)₆]^4-

On heating with con.H² SO⁴ , Fe²⁺ oxidised to Fe³⁺ to give prussion blue in colour.

3[Fe(CN)₆]^4- + 4Fe³⁺ \(\underrightarrow { x{ H }_{ 2 }O }\) Fe₄[Fe (CN)₆]₃.xH₂O Prussian blue

Detection of Halogens:
Halogen exist as halides (max.) in fusion extract. When AgNO₃ and HNO₃ is added they form AgX.
X^– + Ag⁺ → AgX ; (X = Cl, Br, I)

AgCI – white ppt
Agl – yellow ppt
AgBr – pale yellow

Detection of Sulphur:
Sulfur as Na2S is fusion extract when sodium nitroprusside added gives violet color.
Na²S + Na²[Fe(CN)⁵NO)] → Na⁴[Fe(CN)⁵ NOS] (violet colour)

Answer 32:
Electronic configuration of Ne2 is

Answer 33:
When electric discharge is passed through hydrogen gas enclosed in a discharge tube under low pressure and the emit¬ted light is analysed by a spectroscope, the spectrum consists of a large number of lines which are grtouped into different series. These series appear in different regions of light and are named after their discoveries. The complete spectrum is known as hydrogen spectrum.
According to Bohr’s model for Hydrogen
atom, E_n = -R_H\((\frac { 1 }{ { n }^{ 2 } } )\)
and ∆E = E_f -E_i
where E_fis the energy associated with final state while E_i is the energy of the initial state.

Here n_f and n_i represent final and initial electron orbits.

The frequency (υ) for the absorption and emission of the photon can be calculated as follows:

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Physics Model Question Paper 4

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

General Instructions to candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read the instructions carefully.
• Read questions carefully before you answering.
• Calculations, figures, and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provided.
• Give equations wherever necessary.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall.

Answer any four questions from question numbers 1 to 5. Each carries one score.

Question 1.
Name the weakest force among the fundamental forces.

Question 2.
The work done during an isochoric process is …………….

Question 3.
Highway police detect over speeding vehicles by using ……………….
a. Magnus effect
b. Pascals law
c. Doppler effect
d. Bernoulli’s theorem

Question 4.
Two forces 3N and 4N are acting perpendicular to each other. The magnitude of the resultant force is

Question 5.
Say true/false: “Trade winds are produced due to conduction.”

Answer any five questions from question numbers 6 to 11. Each carries two scores.

Question 6.
The displacement (S) of a body in time ‘t’ is given by S = at² + bt. Find the dimensions of a and b.

Question 7.
Give the magnitude and direction of the net force on a stone of mass 0.1 kg.
a. Just after it is dropped from the window of a train accelerating 1 ms².
b. Lying on the floor of a train which is accelerating with 1 ms^-2, the stone being at rest relative to the train.

Question 8.
A body is rolling on a horizontal surface. Derive an equation for its kinetic energy,

Question 9.
The stress-strain graphs for two materials A and B are shown below (the graphs are drawn using the same scale) Which one is more elastic? Why?

Question 10.
“A heavy and a light body have the same kinetic energy.” Which one has greater momentum? Why?

Question 11.
The following figures refer to the steady flow of a nonviscous liquid. Which of the two figures is correct? Why?

Answer any five questions from question num 1.5 m numbers 12 to 17. Each carries three scores.

Question 12.
The side of a cube is measured as 3.405 cm.
a. How many significant figures are there in the measurement?
b. If the percentage error in the measurement of the side of the cube is 3%, find; the percentage error in its volume.

Question 13.
According to the conservation of energy “energy can neither be created nor be destroyed”
a. Prove law of conservation of mechanical energy in the case of a freely falling body.
b. The bob of a pendulum of length 1.5 m is released from the position A shown in the figure. What is the speed with which the bob arrives at the lowermost point B, given that 5% of its initial energy is dissipated against air resistance?

Question 14.
Acceleration due to gravity on earth changes with depth and height.
a. What is the weight of a body placed at the center of the earth? Why?
b. Find the height at which the acceleration due to gravity is 1/4^th that at the surface of the earth.

Question 15.
A metal sphere of density ‘p’ and radius ‘a is falling through an infinite column of liquid of density ‘o’ and coefficient of viscosity Ty
a. Name any two forces acting on the; sphere.
b. With the help of Stokes theorem, derive an equation for the terminal velocity of I the sphere.

Question 16.
Conduction is the mode of transfer of heat in solids. of Write the unit of thermal conductivity.
b. “Burns produced by steam is severe than that produced by boiling water”Why?

Question 17.
A gas has ‘f’ degrees of freedom.
a. Calculate its Cp, Cv, and γ.
b. Define the mean free path.

Answer any five questions from question numbers 18 to 22. Each carries two scores.

Question 18.
A satellite moves in a circular orbit of radius ‘r’ with an orbital velocity.
a. Derive an equation for the orbital velocity of a satellite.
b. The time taken by Saturn to complete one orbit around the Sun is 29.5 times the earth year. If the distance of the earth from the Sun is 1.5 × 108km, then what will be the distance of the Saturn from the Sun?

Question 19.
In the simple harmonic motion, force is directly proportional to the displacement from the mean position.
a. Give an example of a harmonic oscillator.
b. Derive equations for the kinetic and potential energies of a harmonic oscillator.
c. Show graphically the variation of kinetic energy’ and potential energy of a harmonic oscillator.

Question 20.
A stretched string can be used as a musical instrument.
a. What is the fundamental frequency of a stretched string?
b. With neat diagrams, derive equations for the second and third harmonics of a stretched string.

Question 21.
A body having an initial velocity ‘v₀’ has an acceleration ‘a’.
a. Using the velocity-time graph, derive an equation for displacement of the above body.
b. Draw the velocity Time graph and speed Time graph of a body thrown vertically in the air.

Question 22.
A javelin is thrown with an initial velocity ‘ V₀‘ at an angle ” with the horizontal.
a. What are the horizontal and vertical velocities of the body
i. At the point of projection
ii. At maximum height
b. Find the angle of projection at which the maximum height attained by the javelin is equal to the horizontal range.

Answer any three questions from question numbers 23 to 26. Each carries five scores.

Question 23.
a. What is meant by ‘banking of roads’?
b. With a neat diagram, derive an equation for the maximum velocity of a car on a banked road.
c. What is the optimum speed of the car along the banked road?

Question 24.
The moment of inertia of a thin rod of mass M and length 1 about an axis perpendicular to the rod at its midpoint is \(\frac { { Ml }^{ 2 } }{ 12 }\).
a. What is the radius of gyration in the above case?
b. A student has to find the moment of inertia of the above rod about an axis (AB) perpendicular to the rod and passing through one end of the rod. Name and state the law used for this case.
c. Using the theorem, find the moment of inertia of the rod about AB.

Question 25.
Small drops of water assume spherical shape due to surface tension.
a. Define surface tension.
b. Derive an equation for the excess pressure inside a liquid drop of radius ‘R’ having surface tension σ.
c. Why do farmers plow the fields before summer?

Question 26,
Carnot engine is considered as an ideal heat engine.
a. Draw the PV graph of Carnot’s cycle.
b. Derive an equation to find the work done during an adiabatic process.
c. Calculate the efficiency of a heat engine working between ice point and steam point.

Answers

Answer 1.
Doppler effect

Answer 2.
Zero

Answer 3.
Doppler effect

Answer 4.
7N

Answer 5.
False

Answer 6.
[S] = [L]
[at²] = [L]
a = [LT^-2]
[bt] = [L]
[b] = [LT^-1]

Answer 7
a. Only force is gravitational. F = mg = 0.1 × 9.8 = 9.8 N downward j
b. Gravitational force is cancelled by normal I reaction.
∴ F₂ = ma = 0.1 × 1 = 0.1 N, direction of motion of train.

Answer 8.

Answer 9.
In the two graphs, the slope of a graph of material A is greater than the slope of a graph of material B. So material A is more elastic than B. For material A the break-even point (D) is higher.

Answer 10.

Momentum is greater for a heavy body.

Answer 11.
Figure b is correct. According to an equation of continuity, the speed of liquid is larger at a smaller area. From Bernoulli’s theorem due to larger speed, the pressure will be lower at a smaller area and therefore the height of liquid column will also be at lesser height, while in Fig(a) height of liquid column at the narrow area is higher.

Answer 12.

Answer 13.
a. Law of conservation of energy. Energy can neither be created nor be destroyed, but it can be transformed from one form into another. Consider a body of mass’s’ placed at

b. Changing in PE after dissipation.

Answer 14.

Answer 15.
a. i. Weight, F, = mg acting downward
ii. Viscous force, F2 acting upward,
b. By strokes, formula F = 6πrηV Viscous force = Apparent weight of sphere in the solid

Answer 16.
a. W m^-1K^-1
b. Boiling water contains only a specific amount of heat energy required for it to boil. However, as steam is formed from boiling water, it contains the heat energy of boiling water, along with the latent heat of vaporization.i.e., 1kg of steam at 100°C contains 22.6 × 10⁵ J more heat than 1 kg of water at 100°C. Hence, as steam has more heat energy, it can cause more severe burns than boiling water.

Answer 17.

b. Mean free path is an average distance between two successive collisions.

Answer 18.
a. It is the velocity required to put the satellite into its orbit around the earth.

The gravitational force on the satellite

The centripetal force required by the satellite to stay in this orbit is

in this orbit is In equilibrium the centripetal force is given by the gravitational force

Answer 19.
a. Oscillation of simple pendulum Oscillation of loaded spring
b. Let m be the mass of the particle executing SHM. Let v be the velocity at any instant,

Potential energy is the work required to take a particle against the restoring., force. Let a particle be displaced through a distance x from the mean position. Then restoring force, F = – kx, where k is the force constant. Now if we displace the particle further through a distance dx, Small work done, dw = – Fdx = kx dx Total work done from 0 to x

Answer 20.
a. Fundamental mode (or) First harmonic: If the string is plucked in the middle and released, then it vibrates in one segment with nodes at its ends and an antinode in the middle.

This is the lowest frequency with which string vibrates.
b. Second harmonic If the string is pressed in the middle and plucked at one-fourth of its length, then the string vibrates in two segments.


Third harmonic If the striping is pressed at one-third of its length from one end and plucked at one-sixth its length, it will vibrate in three segments.

Thus a collection of all possible mode is called harmonic series and n is called harmonic number.

Answer 21.

The area under the velocity-time graph gives the displacement of the body. Displacement, x = area OABD x = area of triangle ABC+ area of rectangle OACD.

Answer 22.
a.
i. Horizontal V_x = V₀ cosθ Vertical V_y = V₀sinθ
ii. Horizontal V’_x = VO cosθ

Answer 23.
a. To avoid skidding and damage to tires of vehicles, the outer part of a road is slightly raised than the inner part. This is known as banking of roads.

The forces on the car are:
1. The weight of the car vertically downwards.
2. Normal reaction Racing normal to the road.
3. Frictional force acting parallel to the road.
Since there is no vertical acceleration,
R cosθ = mg + F sinθ
or R cosθ – F sinθ = mg …(1)
Now for maximum speed, F = μ, R
The centripetal force is provided by horizontal components of Rand Fas shown in the figure.

Answer 24.
a. The radius of gyration (k). It is the defined as the distance from an axis of rotation at which, if the whole mass of the body was concentrated, then its moment of inertia about that point would be the same as the moment of inertia of actual distribution of mass. l = Mk²
The radius of gyration (k) of a body is the square root of a ratio of the moment of inertia and a total mass of the body.
ie., a radius of gyration, k= \(k=\sqrt { \frac { l }{ M } }\)

b. Theorem of parallel axes: This theorem is good for any shape. The moment of inertia of the body about any axis is equal to the sum of a moment of inertia of a.parallel axis passing through the center of mass and product of its mass of the body and square of the distance between the two parallel axes.

where I am the moment
c. Using parallel axes theorem, the moment of inertia about AB,

Answer 25.
a. Surface tension (a) is the property due to which the free surface of a liquid at rest behaves like an elastic stretched membrane tending to contract so as to occupy a minimum surface area.

Thus it is measured as the force acting per unit length of an imaginary line drawn on the liquid surface, the direction of force being perpendicular to this line and tangential to the liquid surface.

b. Consider a liquid drop of radius R and surface tension o. Let P be the excess pressure inside the drop. The work done by the force due to excess pressure is

c. On plowing, the gap between sand particles act as a capillary tube, so that groundwater reaches the surface easily due to capillary rise.

Answer 26.

b. Work was done in the adiabatic process: We have a small amount of work done when volume changes through at pressure P.

(volume changes from v₁ to v₂ diabolically)

Plus One Physics Previous Year Question Papers and Answers

Kerala Plus One Hindi Previous Year Question Paper 2015

Time : 2 1/2 Hours
Cool-oft time: 15 MInutes
Maximum : 80 Score

General Instructions Candidates

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2% hrs.
• Your are not allowed to write your answers nor to discuss anything with others during the ‘cool off time’.
• Read questions carefully before answering.
• All questions are compulsory and only internal choice is allowed.
• When you select a question, all the sub-questions must be answered from the same question itself.
• Electronic devices except non-programmable calculators are not allowed in the Examination Hall

सूचना : निम्नमिलिखित कवितांश पढ़िये और 1 से 4 तक के प्रश्नों का उत्तर लिखिए।

मनु नहीं मनु-पुत्र है यह सामने, जिसकी,
कल्पना की जीभ में भी धार होती है,
बाण ही होते विचारों के नहीं केवल,
स्वप्न के भी हाथ में तलवार होती है।
स्वर्ग के सम्राट को जाकर खबर कर दे
रोज़ ही आकाश चढ़ते जा रहे है ये;
रोकिए, जसे बने, इन स्वप्नवालों का;
स्वर्ग की हो और बढ़ते आ रहे हैं ये।

प्रश्न 1.
यह कवितांश किस कविता से लिया गया है?
उत्तर :
चाँद और कवि

प्रश्न 2.
कविता के ‘धार’ शब् का समानार्थी शब्द कोष्ठक से चुनकर लिखिए
(तलवार, तेज़, बाण, रोज़)
उत्तर :
तेज़

प्रश्न 3.
मनु-पुत्र की कल्पना कैसी होती है?
उत्तर :
मनु-पुत्र की कल्पना में तलवार की जैसी तीक्ष्णता है। मनुष्य के विचारों में ही नहीं, सपने में भी तलवार की शक्ति होती है।

प्रश्न 4.
कवितांश की आस्वादन टिप्पणी लिखिए।

सूचना : निम्नलिखित खंड पढ़िये और 5 से 10 तक के प्रश्नों का उत्तर लिखिए।
दिवाली की रात को मैं अपने गाँव के युवा मंडल के कार्यालय में अपने साथियों के साथ बैठा था। तभी हमें पता चला की गाँव में आग लग गई है। मैंने सभी युवाओं को आग बुझाने के लिए जल्दी जाने को कहा। सभी युवा लोग आग लगे स्थान की तरफ भागे। किंतु एक युवक बड़ी मस्त चाल से चल रहा था।

मैंने उसे भी जल्दी चलने को कहा तो वह बोला, “भागना किसलिए हैं? अपने को तो तमाशा देखना है। कुछ क्षण बाद देख लेंगे।” तभी उसके भाई की लड़की दौड़ती हुई आयी और युवक को संबोधित करते हुए बोली, “चाचा अपने घर में आग लग गई है।” लड़की की बात सुनते ही युवक दौड़ने लगा। क्योंकि अब उसका अपना घर जल रहा था।
(कार्यालय – office, मस्त चाल – खुशी से)
उत्तर :
प्रगतिशील कवि रामधारी सिंह दिनकर द्वारा लिखित कविता है ‘चाँद और कवि’। कविता में चाँद और कवि के बीच बातचीत चलती है। कवि की कल्पना को पानी को बुलबुले कहकर उपहास करनेवाले चाँद को ललकारते हुए कवि पूछते हैं – हे चाँद, अब तुम्हारे सामने मनु नहीं है, मनु-पुत्र है। उनकी कल्पना में अद्भुत शक्ति है। कल्पना में तलवार की तीक्ष्णता है, समाज में क्रांति लाने की क्षमता है।

उनके विचारों में ही नहीं, सपनों में भी तलवार की जैसी शक्ति है। फिर कवि चाँद से कहते हैं – ‘हे चाँद, तुम स्वर्ग के सम्राट को जाकर यह खबर दें कि उन्हें जीतने केलिए ये स्वप्नवाले आ रहे हैं। रूढ़ियों को दूर करके समाज में प्रगति लाने केलिए वे आ रहे हैं। चाहे तो तुम उन्हें रोक लो। कवि के अनुसार व्यक्ति, समाज और राष्ट्र की प्रगति में बाधा डालनेवाली शक्तियाँ स्वर्ग के सम्राट है। कवि अपनी कविताओं के ज़रिए इन शक्तियों पर विजय पाने की तैयारी में हैं।

प्रश्न 5.
दिवाली की रात को लेखक कहाँ बैठा था?
उत्तर :
दीवाली की रात को लेखक गाँव के युवामंडल के कार्यालय में बैठा था।

प्रश्न 6.
लेखक ने सभी युवाओं को क्यों जल्दी जाने को कहा?
उत्तर :
लेखक ने सभी युवाओं से आग बुझाने केलिए दौड़कर जाने को कहा।

प्रश्न 7.
लड़की की बात सुनते ही युवक दौड़ने लगा। क्यों?
उत्तर :
युवक पहले धीमी गति से चल रहा था। भागकर वहाँ पहुँचकर क्या तमाशा देखना है, यही उनका विचार था। लेकिन अंत में लड़की ने उन्हें अपने घर पर आग लगने की सूचना दी तब युवक बेहताश दौडने लगा।

प्रश्न 8.
उस दिन के लेखक की डायरी कल्पना करके लिखिए।
उत्तर :

रविवार
6.5.2015

कल दिवाली की रात थी, मेरे अविस्मरणीय दिन भी। आज के मनुष्य जितना स्वार्थी है, वह आज ही मुझे पता चला। गाँव के युवा मंडल कार्यालय में बैठ रहे थे। साथ अनेक युवा साथी भी थे। तभी हमें पता चला गाँव में आग लग गई है। मैंने सभी युवाओं को आग बुझाने केलिए भेजा। सब लोग आग लगे स्थान की तरफ़ तेज़ी से दौड़े। लेकिन उनमें एक युवक बड़ी मस्त चाल चल रहा था। मेरे पूछने पर उसने कहा कि भागना किसलिए हैं। तमाशा देखना है तो बाद में देख लेंगे। तभी उसके भाई की लड़की उसके घर में आग लगने की सूचना देती है। यह सुनते ही युवक घर की तरफ़ दोडने लगा। क्योंकि अब उसका घर जल रह था। अपना हमेशा अपना ही होता है। मनुष्य की स्वार्थता पर में शर्मिन्दा हुआ।

प्रश्न 9.
खंड़ का संक्षेपण कीजिए।
उत्तर :
दिवाली की रात में युवा मंडल के कार्यालय में बैठते वक्त हमें पता चला कि गाँव में आग लग गई है। तुरंत ही युवा साथ गाँव की तरफ़ भागे। एक युवक बड़ी मस्त चल रहा था कि उसके लिए यह सिर्फ तमाशा थी। लेकिन जब भाई की लड़की आकर उसे अपने घर में आग लगने की सूचना देती है तो युवक बेहताश दौडने लगा।

प्रश्न 10.
संक्षेपण के लिए शीर्षक लिखिए। सूचना : निम्नलिखित 11 से 16 तक के प्रश्नों का उचित हिंदी शब्द चुनकर मिलान कीजिए। (6)
(अगला, चरण, सार्वजनिक, साज्ञा करें, बातचीत, प्रारूप, कूड़ेदान, संकेत)
उत्तर :
दिवाली की रात

प्रश्न 11.
Chats :
उत्तर :
बातचीत

प्रश्न 12.
Public :
उत्तर :
सार्वजनिक

प्रश्न 13.
Trash :
उत्तर :
कूड़ेदान

प्रश्न 14.
Next step:
उत्तर :
अगला चरण

प्रश्न 15.
Share:
उत्तर :
साझा करें

प्रश्न 16.
Format :

सूचना : निम्नलिखित दोहा पढ़िये और 17 एवं 18 के उत्तर लिखिए।

दुख में सुमिरन सब करे, सुख में करे न कोय।
जो सुख में सुमिरन करे, तो दुख काहे होय।।

उत्तर :
प्रारूप

प्रश्न 17.
‘स्मरण’ शब्द का समानार्थी शब्द दोहे से ढूँढकर लिखिए।
उत्तर :
सुमिरन

प्रश्न 18.
इस दोहे का भावार्थ लिखिए।

सूचना : जुलूस नाट्यरूपांतर के बीरबलसिंह का निम्नलिखित कथन पढ़िये और प्रश्न 19 का उत्तर लिखिए।
बीरबलसिंह

हमारा हुक्म क्या आपको सुनाई नहीं पड़ा?
फिर से सोच लें ! बहुत नुकसान उठाना पड़ेगा!
सिपाहियो, लाठी चार्ज करो!

उत्तर :
कबीरदास कहते हैं – जब मनुष्य दुःख का अनुभव करते हैं तब ईश्वर का स्मरण करते हैं। सुख और आनंद के समय के ईश्वर का स्मरण नहीं करते। ईश्वर का नाम तक भूल जाते हैं। लेकिन जब कोई दुख आता है तब वे तुरंत ईश्वर की ओर मुड़ जाते हैं, मंदिर मस्जिदों में जाकर प्रार्थना करने लगते हैं। विपत्ति से बचने केलिए उन्हें ईश्वर की कृपा-दृष्टि की जरूरत पड़ती है। कबीरदास कहते हैं कि यदि सुख के समय भी ईश्वर का स्मरण करें तो जीवन में दुख कभी नहीं होता।

प्रश्न 19.
इस कथन के आधार पर बीरबलसिंह के चरित्र पर टिप्पणी लिखिए।
उत्तर :
बीरबलसिंह भारतीय है लेकिन अंग्रेजों के पिठू बन गए हैं। भारतीय होकर भी वह अंग्रेज़ पुलीस के साथ मिलकर स्वराजियों के जुलूस को रोकते हैं। वे देशप्रेमी नहीं है। उनके मन में अपने भाइओं के प्रति सहानुभूति नहीं है। वह बड़ा जल्लाद है।

प्रश्न 20.
निम्नलिखित अंग्रेजी बातचीत का हिंदी में अनुवाद कीजिए। (6)

Teacher : Why are you late Puneeth?
Puneeth : Sorry teacher, I saw a tragic incident on the road.
Teacher : Oh! What happened?
Puneeth : Aperson was lying on the road. He was severely injured.
Teacher : What did you do then?
Puneeth : It took him to the hospital.
(Tragic incident – दुखात्मक घटना, severely – बुरी तरह, injured – घायल हुआ)

उत्तर :
टीचर : पुनीत, इतना देर क्यों हुआ?
पुनीत : क्षमा कीजिए टीचर, मैंने सड़क पर दुखात्मक घटना देखी।
टीचर : क्या हुआ?
पुनीत : एक आदमी सड़क पर पड़ा हुआ था। वह बुरी तरह से घायल था।
टीचर : फिर तुमने क्या किया?
पुनीत : मैंने उसे अस्पताल पहुंचाया।

प्रश्न 21.
निम्नलिखित सहायक बिंदु के आधार पर विज्ञापन तैयार कीजिए। बाज़ार में रिंग-टोन कंपनी की ओर से नया मोबाईल फोन आया है। उसकी बिक्री बढ़ाने के लिए एक विज्ञापन तैयार कीजिए।
सहायक बिंदु:

1. बहुरंगी
2. डेढ़ साल की वारंटी
3. 2 जीबी मेमरी मुफ्त
4. अदला-बदला (Exchange) की सुविधा (7)

सूचना: निम्नलिखित 22 से 26 तक के प्रश्नों में से किन्हीं तीन प्रश्नों का उत्तर लिखिए।
उत्तर :

प्रश्न 22.
निम्नलिखित सहायक बिंदु के आधार पर किसी मनपसंद फिल्म की समीक्षा लिखिए।
सहायक बिंदु:

1. फिल्म का कथासार
2. पात्रों का अभिनय
3. निदेशक की भूमिका
4. पटकथा, संवाद, छायांकन, संपादन, गीत आदि
5.  फिल्म की समग्रता पर अपना दृष्टिकोण (8)

उत्तर :
तारे ज़मीन पर (फिल्मी समीक्षा)
बच्चे ओस की बूंदों की तरह एकदम् शुद्ध और पवित्र होते हैं। वे कल के नागरिक हैं, लेकिन दुःख की बात है कि बच्चों को अनुशासन के नाम पर तमाम बंदिशों में रहना पड़ता है। लेकिन, आजकल हर घर से ‘टॉपर्स’ और ‘रैकर्स’ तैयार करने की कोशिश की जा रही है। कोई यह नहीं सोचता कि उनके मन में क्या है? वे क्या सोचते हैं? उनके क्या विचार है?

इन्हीं सवालों और बातों को आमिरखान ने अपनी फ़िल्म ‘तारे ज़मीं पर’ में उठाया है। फ़िल्म की कहानी इस प्रकार है : आठ वर्षीय ईशान अवस्थी (दर्शील सफारी) का मन पढ़ाई के बजाय कुत्तों, मछलियों और पेंटिंग में लगता है। उसके माता-पिता चाहते हैं कि वह अपनी पढ़ाई पर ध्यान दें, लेकिन कोई नतीजा नहीं निकलता। ईशान घर पर माता-पिता की डॉट खाता है और स्कूल में शिक्षकों की। कोई भी यह जानने की कोशिश नहीं करता कि ईशान पढ़ाई पर ध्यान क्यों न दे रहा है। इसके बजाय वे ईशान को बोर्डिंग स्कूल भेज देते हैं।

खिलखिलाता ईशान वहाँ जाकर मुरझा जाता है। वह हमेशा सहमा और उदास रहने लगता है। उस पर निगाह जाती है आर्ट टीचर रामशंकर निकुंभ (आमिर खान) की। निकुंभ उसकी उदासी का पता लगाते हैं और उन्हें पता चलता है कि ईशान बहुत प्रतिभाशाली है, लेकिन डिसलेक्सिया की बीमारी से पीडित है। उसे अक्षरों को पढ़ने में तकलीफ़ होती है। अपने प्यार और दुलार से निकुंभ ईशान के अंदर छिपी प्रतीभा को सबके सामने लाते हैं।

कहानी सरल है, जिसे आमिरखान ने बेहद प्रतिभाशाली तरीके से परदे पर उतारा है। पटकथा की बुनावट एकदम चुस्त है। छोटे-छोटे भावनात्मक दृश्य रखे गए हैं, जो सीधे दिल को छू जाते हैं। ईशान की भूमिका में दर्शील सफारी इस फ़िल्म की जान है।

आमिर खान मध्यांतर में आते हैं और छा जाते हैं। टिस्का चोपडा (ईशान की मम्मी) ने एक माँ की बेचैनी को उम्दा तरीके से पेश किया है। विपिन शर्मा (ईशान के पापा) सचेत इंजीनीयर और सारे अध्यापकों का अभिनय भी अच्छा है। निर्देशन के मैदान में पहली बार उतरे आमिर खान ने दिखा दिया है कि फ़िल्म माध्यम पर उनकी समझ और पकड अद्भुत हैं। प्रसून जोशी द्वारा लिखे गीत बहुत कुछ कहते हैं और परदे पर उनको देखते समय उनका प्रभाव और बढ़ जाता है। शंकर-अहसान-लॉय का संगीत भी अच्छा है। फिल्म की फोटोग्राफी बहुत ही प्रभावशाली है।

प्रश्न 23.
निम्नलिखित सहायक बिंदु के आधार पर संपादकीय तैयार कीजिए।
आजकल संसार भर निरपराध बच्चों के प्रति अत्याचार और आक्रमण हो रहा है। इस प्रसंग में बच्चों के प्रति अत्याचार रोकें’ इस विषय पर संपादकीय तैयार कीजिए।
सहायक बिंदु:

1. बच्चों के विरुद्ध अत्याचार
2. बच्चे कल के नागरिक
3. मानविकता का अभाव
4. सुरक्षाहीन वातावरण
5. मख्त नियमों का पालन
6. मनोवृत्ति में बदलाव
7. अपराधियों को कठिन दंड (8)

उत्तर :

दैनिक हिंदुस्तान
बुधवार 20 अप्रैल 2014
बच्चों के प्रति अत्याचार रोकें

देश का भविष्य बच्चों पर निर्भर रहता है। जो आज बच्चे हैं वे कल देश के नेता होंगे, नागरिक होंगे। बच्चों की कोमल निर्मल मुस्कान में ही देश का उजला भविष्य नज़र आता है। इसलिए ही नेहरूजी ने कहा है – बालकों की आत्मा पवत्र होती है। लेकिन अफसोस की बात यह है कि हमारे देश में इन मासूम बच्चों पर कई प्रकार के अत्याचार हो रहे हैं। उन्हें जीविका चलाने के लिए मजदूरी करना पड़ता है।

माता-पिता की प्रताड़ना सहना पडता है। स्कूलों में कॉपी किताबों और दोस्तों के बीच बितानेवाले जिंदगी उन्हें घरों में झाड पोंछे और कारखानों में औजारों के बीच बिताना पडते हैं। आज सरकार ने आठवीं तक की शिक्षा को अनिवार्य और निशुल्क कर दिया है, फिर भी स्कूली शिक्षा से वंचित कई बच्चे अब भी हमारे देश में हैं। बच्चों के माता-पिता सिर्फ इस वजह से उन्हें स्कूल नहीं भेजते क्योंकि उनके स्कूल जाने से परिवार की आमदनी कम हो जाएगी। दुनिया में सबसे ज्यादा बालमजदूर भारत में ही है।।

मजदूरी से परिवार की आमदनी ज़रूर बढ़ती है, लेकिन खेल-फूद, पढ़ाई और मस्ती बिताने योग्य उनके ये बचपन का ही हम बर्बाद कर रहे हैं। मज़दूर के जगहों पर ये निरीह बच्चे शारीरिक प्रताडन के शिकार हो रहे हैं। गरीबी और सुरक्षाहीन वातावरण इस बालमजदूरी के मुख्य कारण हैं। ‘

सरकार ने बालमजदूरी एवं बालकों पर होनेवाले अत्याचार को रोकने केलिए कानून तो बना दिए हैं और बालमजदूरी को अपराध भी घोषित किया है। लेकिन अब भी हमारे देश से यह भयावह समस्या पूरी तरह खतम नहीं हुआ है। इसे जड़ से खतम करना है तो पहले गरीबी को खतम करना पडेगा। सिर्फ सरकार ही नहीं आम जनता की भी इसमें सहमागिता ज़रूरी है। लोगों की मनोवृत्ति में परिवर्तन लाना भी ज़रूरी है। हर एक व्यक्ति जो आर्थिक रूप से सक्षम हो अगर ऐसे एक बच्चे को भी ज़िम्मेदारी लेने लगे तो सारा परिदृष्य ही बदल जाएगा।

प्रश्न 24.
निम्न सुखियाँ पढ़िए।

स्कूल और आसपास को स्वच्छ रखने के लिए सरकार द्वारा किए जानेवाले कारवाइयों की जानकारी पाने के लिए सुमन, सदस्य, अध्यापक अभिभावक संघ, सरकारी उच्च माध्यमिक स्कूल, पालक्काड की ओर से सार्वजनिक सूचना अधिकारी, शिक्षा विभाग, केरल सरकार, तिरुवनंतपुरम् के नाम सूचना
अधिकार पत्र तैयार कीजिए।
उत्तर :

प्रश्न 25.
निम्नलिखित सहायक बिंदु के आधार पर वार्तालाप लिखिए।
यात्री शाम को घर पहुंचा। उसकी परेशानी देखकर पत्नी ने कारण पूछा। यात्री ने असलम की मृत्यु की खबर सुनाई। यात्री और पत्नी के बीच का वार्तालाप लिखिए।
सहायक बिंदु:

1. असलम की मृत्यु की खबर
2. असलम के प्रति अपना व्यवहार
3. हमदर्दी का अभाव
4. पश्चात्ताप से उत्पन्न अनुताप

उत्तर :

सेवा : अरे, जल्दी एक गिलास पानी दो
पत्नी : क्या हुआ यार, बहुत परेशान दिख रहे हैं?
यात्री : पहले तुम पानी दे दो, में सब कुछ बताऊँगा। (पत्नी उन्हें पानी देती है)
पत्नी : अब बताओ, क्या हुआ आपको?
यात्री : तुम्हें असलम की याद है न?
पत्नी : क्या आप उस रिक्शेवाले की बात कर रहे हैं?
यात्री : हाँ।
पत्नी : उसे क्या हुआ?
यात्री : कल उसकी मृत्यु हुई।
पत्नी : अरे ! मुझे भरोसा नहीं आती। कल भी वह आप को लेने आया था न?
यात्री : हाँ, वही मुझे दफ्तर छोड़ दिया था। लौटते वक्त रास्ते में ही उसकी मृत्यु हुई।
पत्नी : उन्हें कोई बीमारी है क्या?
यात्री : उनकी दोनों गुर्दे खराब थी। कल गाड़ी चलाते समय उन्होने कुछ परेशानियाँ दिखाई थी।
पत्नी : तो आप ने क्या किया था?
यात्री : मैंने सोचा यह उनका नाटक होगा, इसलिए रिक्शे पर बैठा रहा। अब मुझे लगता है में ही उनकी मृत्यु का ज़िम्मेदार हूँ।
पत्नी : ऐसा मत सोच लो यार, शायद भगवान ने उसे छोटी ज़िदगी ही दी गई होगी।
यात्री : फिर भी मुझे उस समय गाड़ी से उतरना चाहिए था।
पत्नी : फिक्र मत करो यार, अभी आराम कर लें। कल हम अस्लम के घर जाकर उनके बीवी और बच्चों से मिलेंगे।
यात्री : हमें जरूर जाना है, नहीं तो मुझे तसल्ली नहीं मिलेगी।

प्रश्न 26.
निम्नलिखित सहायक बिंदु के आधार पर निबंध लिखिए।
हिंदी भारत की राजभाषा एवं राष्ट्रभाषा है – ‘सीखें हिंदी, सिखाएँ हिंदी’ इस विषय पर निबंध लिखिए।
सहायक बिंदु:

1. भारत की संपर्क भाषा
2. अधिकांश जनता की बोलचाल की भाषा
3. एकता की कड़ी
4. हिंदी हमारी संस्कृति
5. हिंदी का प्रचार – नागरिकों का कर्तव्य

उत्तर :

सीखे हिंदी, सिखाएँ हिंदी

भाषा हमारे लिए बोलचाल का माध्यम होती है। संप्रेषण का माध्यम होती है। इसके द्वारा ही हम अपने विचार व्यक्त कर सकते हैं, बातचीत कर सकते हैं। किसी भी देश की राष्ट्रभाषा उसे ही बनाया जाता है जो उस देश में व्यापक रूप में फैली जाती है।

इस दृष्टि से देखें तो हिंदी ही भारत की राष्ट्रभाषा बनने योग्य है। हिंदी हमारे देश के प्रमुख राज्य जैसे मध्यप्रदेश, उत्तरप्रदेश, बिहार, राजस्थान, हरियाणा हिमाचल प्रदेश और दिल्ली में प्रमुख रूप से बोली जाती है। राष्ट्रभाषा संपूर्ण देश में भावात्मक एवं सांस्कृतिक एकता स्थापित करने का प्रधान साधन होती है हिंदी भाषा को इसलिए राष्ट्रभाषा का पद दिया कि हिंदी भाषा बोलने में, लिखने में, पढ़ने में सरल है।

अहिंदी प्रदेश के लोग भी आसानी से हिंदी सीख सकते हैं साथ ही हिंदी भाषा में कहानी, कविता, उपन्यास, नाटक जैसे प्रचुर साहित्य उपलब्ध है। हिंदी भारत की स्वयं सिद्ध राष्ट्रभाषा है। इसे बोलने वालों का प्रतिशत 65 से भी अधिक है। हमारी हिंदी भाषा की एक विशेषता यह भी है कि हमारे देश वें रहने वाले भारतीय लोग किसी भी प्रांत के रहनेवाले हो, उनकी कोई भी मातृभाषा हो, वे हिंदी समझते हैं और किसी न किसी रूप में व्यवहार में लाते हैं। राष्ट्रभाषा होने के नाते हिंदी सीखना हमारा कर्तव्य है और इसका प्रचार करना भी।

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