Plus One

Kerala Plus One Botany Model Question Paper 5

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.

Qn.No.1 to 3. Answer all questions. Each question carries 1 score. (3×1=3)

Question 1.
Choose the correctly matched pair.
a. Telocentric chromosome
– Middle centromere
b. Metacentric chromosome
-Centromere slightly away from the middle
c. Acrocentric chromosome
-Centromere close to its end

Question 2.
Choose the correct answer.
Law of limiting factors is proposed by
a. Cornelius Van Niel
b. Blackman
c. Joseph Priestley
d. Engelmann

Question 3.
Observe the figure given below and identify the blue green alga.

Qn.No.4 to 11. Answer any 6 questions. Each question carries 2 scores.

Question 4.
Certain compounds formed during Krebs’ cycle are given below. Draw Krebs’cycle using the compounds.

Question 5.
Water from the root hairs move deeper into the root layers by two pathways. Name the two pathways. Which pathway is blocked in the endodermis? Give reason.

Question 6.
Analyse the table and fill in the blanks.

Question 7.
Artificial system and natural system are two systems of classification. Who are the proponents of these classifications.

Question 8.
Names of certain essential elements are given below:

Choose the macro nutrients and write one function of each of them.

Question 9.
Pili and fimbriae are surface structures seen in bacterial cells. Differentiate these structures.

Question 10.
Carbohydrates and fats are respiratory substrates. But their RQ is different. Define RQ. Write the RQ of these substrates.

Question 11.
Certain life cycle patterns of various plant groups are given below:

Choose the life cycle pattern shown by gymnosperms and angiosperms. Write the peculiarities of the identified life cycle pattern.

Qn. No 12 to 17. Answer any 5 questions. Each question carries 3 scores.

Question 12.
Observe the figure given below:

a. Source of which plant hormone is indicated in the figure.
b. Write four roles of the above identified hormone.

Question 13.
Observe the terms given below :
Vexillary, Axile, Marginal, Imbricate, Basal, Valvate.
Identify the three kinds of placentation from the above. Explain them.

Question 14.
Observe the figure.

a. Identify the above stage of mitosis.
b. Name the preceding stage of the above stage. Write its main events.

Question 15.
Observe the flow chart given below :

a. Identify A, B, C, Write the function of B.
b. Differentiate endarch and exarch xylem.

Question 16.
Observe the diagrammatic representation of Hatch and Slack pathway given below.

Identify the cells A and B. Explain the process of formation of C₄ acid specifying the enzyme involved.

Question 17.
Periderm is constituted by three kinds of tissues. Name them. Write one peculiarity of each of them.

Answers

Answer 1.
c. Acrocentric chromosome
-Centromere close to its end

Answer 2.
b. Blackman

Answer 3.
Nostoc

Answer 4.

Answer 5.
Apoplast and Symplast pathways are two pathways.
Apoplast pathway is blocked in the endodermis.
Due to the presence of casparian strips / suberin thickenings.

Answer 6.
a. golgi apparatus
b. basal body of cilia/ flagella/ spindle fibers
c. protein synthesis
d. mitochondria

Answer 7.
Artificial system- Carl Linnaeus
Natural system – George Bentham and Joseph Hooker
Artificial system – based on external characters
Natural system – based on both external and internal characters

Answer 8.
Potassium, Magnesium
Potassium – anoin cation balance/opening and closing of stomata/maintain turgidity of cells.
Magnesium – maintain ribosome structure/ constituent of chlorophyll/activates enzymes of photosynthesis.

Answer 9.
Pilli are elongated tubular structures. Fimbriae are small bristle like structures.

Answer 10.
Ratio of volume of Co₂ evolved to the ratio of oxygen absorbed is RQ.
RQ of carbohydrate is 1
RQ of fats are less than 1

Answer 11.
Diplotonic
Here diploid sporophytic phase is dominant.
In this type lifecycle the diploid sporophyte is the dominant, photosynthetic, independent phase of the plant. The gametophytic phase is represented by the single to fewcelled haploid gametophyte. This kind of lifecycle is termed as diplontic. All seed-bearing plants i.e., gymnosperms and angiosperms, follow this pattern.

Answer 12.
a. Auxin
b. Initiate rooting in stem cuttings
Induce parthenocarpy
Control xylem differenciation
Cause apical dominance

Answer 13.
Axial, Marginal and Basal
Axial : placenta is axial and ovules attached to it.
Marginal : Ovules are at margines of car-pels
Basal : Placenta at base of ovary

Answer 14.
a. Anaphase
b. Preceding stage – Metaphase .
Main events
Chromosomes are highly condensed they are arranged at equatorial plane.

Answer 15.
a. A – Parenchyma
B – Collenchyma
C – Sclerenchyma
Function of B – Give support
b.Endarch – protoxylem pointing towards centre
Exarch – Protoxylem pointing towards periphery

Answer 16.
A-mesophyll
B – Bundle sheath cell
The primary Co₂ acceptor is a 3-carbon molecule phosphoenol pyruvate (PEP) and is present in the mesophyll cells. The enzyme responsible for this fixation is PEP carboxylase or PEPcase. It is important to register that the mesophyll cells lack Ru- BisCO enzyme. The C₄ acid OAA is formed in the mesophyll cells. It then forms other 4-carbon compounds like malic acid or aspartic acid in the mesophyll cells itself, which are transported to the bundle sheath cells. In the bundle sheath cells these C₄ acids are broken down to release Co₂ and a 3-carbon molecule. The 3-carbon molecule is transported back to the mesophyll where it is converted to PEP again, thus, completing the cycle. The Co₂ released in the bundle sheath cells enters the C₃ or the Calvin pathway, a pathway common to all plants. The bundle sheath cells are rich in an enzyme Ribulose bisphosphate carboxylase-oxygenase (RuBisCO), but lack PEPcase. Thus, the basic pathway that results in the formation of the sugars, the Calvin pathway, is common to the C₃ and C₄ plants.

Answer 17.
Phellum, phellogen and phelloderm are the three kinds of tissue.

One type of meristematic tissue called cork cambium or phellogen develops, usually in the cortex region. Phellogen is a couple of layers thick. It is made of narrow, thin-walled and nearly rectangular Cells. Phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm. The cork is impervious to water due to suberin deposition in the cell wall. The cells of secondary cortex are parenchymatous. Phellogen, phellem, and phelloderm are collectively known as periderm.

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Botany Model Question Paper 4

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.

Question 1.
The class of fungi known as imperfect fungi is ……

Question 2.
Which one of the plant growth regulators induce immediate stomatal closure in leaves.

Question 3.
By observing the relationship of the first pair, fill in the blanks of the second pair
Auxins : Apical dominance
____: Overcome apical dominance

Question 4.
The reaction center of photosystems in green plants during light reaction is ……
a. Xanthophyll
b. Carotenoids
c. Chlorophyll b
d. Chlorophyll a

Answer any seven from question numbers 5 to 13. Each carries two scores.

Question 5.
Match the following:

Question 6.
Complete the diagram by filling the gaps.

Question 7.
Suggest a method for identification of stem and root in plants based on the arrangement of protoxylem.

Question 8.
A cell with a chromosome number 42 undergoes mitosis and another cell with 42 chromosomes undergoes meiosis. Give the chromosome number of daughter cells after mitosis and meiosis. Give reasons for your answer.

Question 9.
Observe the diagrammatic view of a cell and answer the questions.
a. Identify the phase in which DNA synthesis takes place.
b. Mention any one event that takes place in the stages marked as ‘X’ and ‘Y’.

Question 10.
Observe the given diagram. Analyze this diagram and explain the structure of plasma membrane.

Question 11.
Write any two distinguishing features of the algal class Rhodophyceae.

Question 12.
There is a clear division of labour within the chloroplast. Substantiate the given statement with an explanation stating two points.

Question 13.
Botany teacher brought a diseased plant in the classroom. Manoj identified the pathogen as a virus. Can you list any two symptoms of diseases, that helped him to identify the pathogen?

Answer any four from question numbers 14 to 18. Each carries 3 scores.

Question 14.
Three potato pieces of equal weight (2g) were left in three types of solutions (A, B and C) overnight. The weight change of potato pieces are as shown in the figure.

a. Identify the solutions A and C.
b. Discuss the reason for not having any change in the piece put in solution B.

Question 15.
C₄ plants are adapted to overcome a wasteful process found in C₃ plants and hence productivity and yields are better in these plants.
a. Name the wasteful process found in C₃ plants.
b. Identify the cells involved in the C₄ pathway.
c. Write any two differences between C₃ plants and C₄ plants.

Question 16.
Plants that are adapted to dry tropical regions have a special type of Co₂ fixation in addition to the C₃ cycle.
a. Name this pathway.
b. Can you identify any specialty in the leaf anatomy of such plants? If so, what is this anatomy called?
c. Which is the primary Co₂ acceptor in this pathway?
d. Write any one advantage of such plants over C₃ plants.

Question 17.
In most of the plants, roots are meant for absorption and fixation. But there are exceptions to these functions. Justify the statement citing at least three examples.

Question 18.
In a dicotyledonous stem, secondary growth takes place at two regions by the activity of two lateral meristems.
a. Identify the two lateral meristems.
b. List the new tissues formed from each of these meristems.

Answers

Answer 1.
Deuteromycetes

Answer 2.
Abscissic acid

Answer 3.
Cytokinin

Answer 4.
d. Chlorophyll a

Answer 5.
a. Synthesis and storage of energy – Mitochondria.
b. Packaging and delivery of materials – Golgi apparatus.
c. Digestion of intercellular materials – Lysosome.
d. Formation of basal body of cilia and flagella – Centriole.

Answer 6.
4C – Oxyloacetic acid
4C – Malic acid/ fumaric acid
5C – α-Ketoglutaric acid

Answer 7.
In root, the protoxylem are in exarch arrangement, while in stem, endarch.

Answer 8.
After mitosis chromosome doesn’t change. Because its division doesn’t cause any decrease in amount of chromosome. But meosis is known as reduction division. So it bring half of the chromosome number.

Answer 9.
a. DNA division occurs at S-phase.
b. X – Metaphase: During which chromosome align at metaphase plate.
Y – Anaphase: Centromere of the sister chromatids split and chromatids separate. Chromatids move to opposite poles.

Answer 10.
Fluid – mosaic model. According to this model quasi-fluid nature of lipid enables lateral movement of proteins within the overall bilayer. This ability to move within the membrane is measured as its fluidity. This fluid nature of the membrane is also important from the point of view of function like cell growth.

Answer 11.
• The members of Rhodophyceae are known as red algae.
• The food is stored as floridean starch.

Answer 12.
Within the chloroplast there is the membranous system consisting of grana, the stroma lamellae, and the fluid stroma. There is a clear division of labour within the chloroplast. The membrane system is respon¬sible for trapping the light energy and also for the synthesis of ATP and NADPH.

Answer 13.
In this plant, the symptoms are mosaic formation, leaf rolling and curling, yellowing and veinclearing, dwarfing and stunted growth.

Answer 14.
a. A – Hypertonic, C – Hypotonic
b. B – Solution is isotonic. The osmotic pressure of the external solution and cell sap are same. Hence no net flow of water towards inside or outside of potato pieces.

Answer 15.
a. Photo respiration
b. Bundle sheeth cells around the vascu lar bundles.
c. C₃ cycle C₄ cycle
1. Co₂ accepter is Co₂ accepter is PEP RuBp
2. The enzyme – The enzyme involved
RUBISCO – PEP case

Answer 16.
a. C₄ Pathway or Hatch and Slack pathway.
b. Yes. Kranz anatomy.
c. Phosphoenol pyruvate (PEP).
d. In C₄ plants photorespiration is absent so the productivity is high.

Answer 17.
Roots are modified for:
Storage of food: Tap roots of carrot, turnip and adventitious roots of sweet potato, get swollen and store food.
Respiration: Roots modified for physiological function like breathing are called breathing roots or pneumato phores. eg., Mangrove rees.
For mechanical support: Prop roots/ pillar roots of banyan trees are developing from the aerial branches and it grows vertically downwards.

Answer 18.
a. Vascular cambium and cork cambium.
b. From vascular cambium: Secondary xylem, secondary phloem, secondary medullary rays dulps.
From cork cambium: Phellem, phelloderm

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Botany Model Question Paper 3

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.

Questions 1-3. Answers all questions. Each question carries 1 score.

Question 1.
Choose the correctly matched pair
a. Golgi apparatus – Synthesis and storage of energy
b. Lysosome – Digestion of intercellular materials
c. Mitochondria – Packaging and delivery of materials

Question 2.
Identify the statement which is applicable to cyanobacteria.
a. Important decomposers
b. Form blooms in polluted water
c. Presence of prominent nucleus
d. Completely lack of a cell wall

Question 3.
The following diagram represents five petals of a flower in a plant belonging to a major family.

Identify the family.

Questions 4 -11. Answers any 6 questions. Each question carries 2 score.

Question 4.
Complete the given table of algal divisions and their main characteristics by filling ‘a’, ‘b’, ‘c’ and ‘d’.

Question 5.
Transpiration has more than one purpose in plants. Write any one purpose of transpiration in plants and list any two factors that affect this process.

Question 6.
Unlike a majority of the pteridophytes, genera like Selaginella and Salvinia show a unique feature in spore formation.
a. What is this feature?
b. Briefly comment on its significance.

Question 7.
Even though more than, sixty elements are found in different plants, all are not essential.
Write any two criteria for the essentially of an element.

Question 8.
C₄ plants have special features. List out any four specialities of C₄ plants compared to C₃ plants.

Question 9.
Glycolysis is a partial oxidation process. Justify.

Question 10.
Light reaction is otherwise called photo-phosphorylation

a. Locate the site of this reaction.
b. Write any two differences between cyclic photo phosphorylation and non-cyclic photo phosphorylation.

Question 11.
Analyse the given statements and correct them with respect to the underlined words.
a. Respiration is an anabolic pathway.
b. The site of perception of light by a plant for a photoperiodic response is a flower.

Questions 12-17. Answers any 5 questions. Each question carries 3 score.

Question 12.
The diagram given below shows parts of a true fruit.

a. Write the technical name of this fruit developed from a monocarpellary superior ovary.
b. Label the parts A and B.
c. Can you distinguish a parthenocarpic fruit from the given fruit.

Question 13.
Given below are the scientific terms related to photosynthesis. Classify them in appropriate manner, in the given table.

Question 14.
Observe the diagram and answer the following.

a. This is a peculiar type of plant cultivation. Name it.
b. Aeration is necessary for the conduct of this type of cultivation. Why?

Question 15.
Given below is the floral diagram of a family you have studied.

a. Identify the family and aestivation of corolla.
b. Describe three other floral characters from the floral diagram using technical terms.

Question 16.
In an anatomy lab, Ramu and Salim were taking transverse sections (T.S) of two specimens, A and B respectively. Their observations are given in the table. Complete the table.

Question 17.
Even though algae are primary producers on aquatic ecosystems, man is benefited by algae in a variety of ways. Write any four points in favour of this statement.

Answers

Answer 1.
b.

Answer 2.
Form blooms in polluted water.

Answer 3.
Fabaceae.

Answer 4.
a – Starch
b – Fucoxanthin
c – Rhodophyceae
d – Floridean starch

Answer 5.
Transpiration is a mechanism for water ex-change.
Transpiration is affected by:
External temperature, humidity, light.
Wind speed
Number and distribution of stomata.
Percentage of open stomata.

Answer 6.
a. Heterosporous.
b. It produces megaspores and micros-pores. Microspore develops into male gametophyte and megaspore develops into female gametophyte. The female gametophyte retained in the parent sporophyte for a short period and it is considered to be the seed habit.

Answer 7.
The mineral should support growth and re-production. The requirement of the element should be specific and not replaceable by another element. The element should directly involved in the metabolism of plant.

Answer 8.

Answer 9.
Glycolysis is a partial oxidation, glucose undergo oxidation to form two molecules of pyruvic acid. And this pyruvic acid undergo many type of reaction according to the organism and form the sub-stance which our body need. Pyruvic acid, in yeast like organism undergo fermentation and form Co₂ and ethanol. In animal cells it form lactic acid. Finally reach in Krebs cycle then undergo electron transport then it form ATP. Such a long cycle is needed to convert glucose to ATP. So we say that glycolysis is a partial oxidation.

Answer 10.
b. Grana.
c. Non-cyclic: When the two photosystems work in a series, first PS II and then the PS I, a process called non- cyclic photo-phosphorylation occurs. The two photo-systems are connected through an electron transport chain, as seen earlier -in the Z scheme. Both ATP and NADPH + H⁺ are synthesised by this kind of electron flow.

Cyclic: When only PS I is functional, the electron is circulated within the photosystem and the phosphorylation occurs due to cyclic flow of electrons. A possible location this could be happening is in the stroma lamellae. The excited electron does not pass on to NADP+ but is cycled back to the PS I complex through the electron transport chain. The cyclic flow hence, results only in the synthesis of ATP.

Answer 11.
a. The term anabolism means building up process. The term catabolism means break down process. Respiration involve the break down of organic molecules like glucose. But Intermediate compounds produced during the break down of glucose is used for synthesis of complex molecules like fats and proteins. Therefore respiratory pathway involve both anabolism and catabolism, hence can be considerate as amphibolic pathway.
b. The site of perception of light by a plant for a photo periodic responds is not a flower but leaves.

Answer 12.
a. Drupe.
b. A – Epicarp, B – Mesocarp
c. Parthenocarpic fruit is the fruit which formed without fertilization of the ovary. But in figure it is true fruit which formed by fertilization of monocarpellary superior ovary.

Answer 13.

Answer 14.
a. Hydroponics.
b. In this system plants are cultivated under greatful nutrient medium. A plant can live only in the presence of O₂. O₂ supply is necessary for the growth of plants/absorption of nutrients.

Answer 15.
a. Fabaceae. The aestivation of corella is vexillary.
b. Infloroscence – Racemose Flower – Bisexual, zygomorphic.
Calyx – Sepals five, gamosepalous.

Answer 16.

Answer 17.
Agar is obtained from gracillaria and geledium used for the preparation of icecream and jellies.
Chorella and spirullina are abundant in protein and used as food supplement by the space travellers.
70 species of marine algae are used as food. Half of the Co₂ fixation on earth is done by algae.

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Botany Model Question Paper 2

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.

Question 1.
Fill in the blank
Cork cambium is otherwise called …..

Question 2.
Glycolysis is present in all organisms and it is the only process of respiration in anaerobic organisms.
Where does glycolysis occur?

Question 3.
By observing the relationship of the first pair, fill in the blanks of the second pair
F. W. Went : Auxins
E. Kurosawa : …….

Question 4.
Out of the four statements given below related to respiration, the correct statements are:
i. Though respiration has traditionally been considered a catabolic process, it would be better to consider it as an amphibolic pathway.
ii. In muscles when oxygen is inadequate, lactic acid is reduced to pyruvic acid.
iii. When facts are used in respiration, the RQ is greater than one.
iv. In respiration, the energy of oxidation- reduction is utilized for phosphorylation,
a. i) and ii)
b. ii) and iii)
c. iii) and iv)
d. i) and iv)

Answer any seven from question numbers 5 to 13. Each carries two scores.

Question 5.
Match the following columns A,B.

Question 6.
Observe the incomplete schematic representation given below and answer the questions.
Identify this pathway common for both

a. aerobic and anaerobic respiration.
b. Complete the scheme by filling the boxes A, B, C and D.
c. Mention the three major ways in which different cells handle pyruvic acid produced by this pathway.

Question 7.
‘Diatoms, the chief producers in the oceans, are useful to us even after their death.’ Evaluate the statement.

Question 8.
In dicot stem, cambium occurs between xylem and phloem and the vascular bundles are arranged in a broken ring fashion. However, dicot stem grows in diameter in a circular manner. Explain briefly how this could happen.

Question 9.
Identify the cycle and name the scientist who traced the pathway.

Question 10.
Two crucial events of aerobic respiration takes place in two parts of mitochondria. Locate the two parts and mention the two events in one or two sentences each.

Question 11.
Observe the given floral diagram.

a. Identify the family.
b. Write any four floral characters of the identified family.

Question 12.
How can you differentiate active transport from facilitated diffusion?

Question 13.
Fermentation is the incomplete oxidation of pyruvic acid. Find the difference between two types of fermentations in micro-organisms.

Answer any four from question numbers 14 to 18. Each carries 3 scores.

Question 14.
Breakdown of glucose into two molecules of pyruvic acid is called glycolysis. Draw the scheme of glycolysis using the following, starting from glucose ending in pyruvic and 1,3-biphosphoglyceric acid, glyceraldehyde 3 phosphate—> dihydroxy acetone phosphate, fructose, 1, 6- biphosphate, 3-phosphoglycerate, fructose-6-phosphate phosphoenol pyruvate, glucose-6 phosphate, 3 phos-phoglyceric acid.

Question 15.
A diagram prophase I stage of a Meiosis is given below. Draw similar diagrams to il-lustrate the following stages.

Question 16.
Comment on the following statements.
a. Abscisic acid is also known as stress hormone.
b. There are hormones promoting apical dominance and those acting against apical dominance.

Question 17.
An anatomist observed peculiar type of large spherical bundle sheath cells in sugarcane leaf and a physiologist identified the presence of PEP carboxylase in that leaf mesophyll.
a. Name the peculiar leaf anatomy.
b. Explain the physiological advantage of such type of plants.

Question 18.
The use of radioactive C¹⁴ by Melvin Calvin in algal photosynthesis studies led to the discovery of Co₂ fixation in green plants.
a. Identify the first stable product in this Co₂ fixation cycle.
b. Which are the three main stages of this cycle?
c. Name the enzyme which catalyzes the first stage of this cycle.
d. Work out how many ATP and NADPH molecules will be required to make one molecule of glucose?

Answers

Answer 1.
Phellogen

Answer 2.
In cytoplasm.

Answer 3.
Gibberellin/GA

Answer 4.
d – i and iv

Answer 5.

Answer 6.
a. Glycolysis/EM P Pathway
b. A – Glucose 6- phosphate
B – Triose phosphate/glyceraldehyde- 3 phosphate
C – 2 -phosphoglyceric acid
D – 2 -phosphoenolpyruvic acid
c. Lactic acid fermentation in muscle cells. Alcoholic fermentation in Yeast. Aerobic respiration.

Answer 7.
In diatoms the cell wall form 2 thin over laping shells which fit together as in soap box. These walls are embeded with silica. So it is very tough and Indistricta Diatoms deposits their cell wall in their habitat. This accumulation of cell wall over a billion year is known diatomaceous earth used in pol-ishing, filtration of oils and syrups.

Answer 8.
The inter-fascicular cambium joins with in- tra-fascicular cambium and forms ring of cambium.

Answer 9.
Kreb’s cycle. Adolf Hanns Krebs

Answer 10.
For aerobic respiration to take place within the mitochondria, the final product of gly-colysis, pyruvate is transported from the cytoplasm into the mitochondria.
The crucial events in aerobic respiration are:
i. The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of C02.
ii. The passing on of the electrons removed as part of the hydrogen atoms to molecular O2 with simultaneous synthesis of ATP.
The first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria.

Answer 11.
a. Liliaceae / Lily family
b. a. Inflorescence: solitary/cymose; often umbellate clusters
Flower: bisexual; actinomorphic
Perianth tepal six (3+3), often united into tube; valvate aestivation
Androecium: stamen six, 3+3, epitepalous
Gynoecium: tricarpellary.syncarpous, ovary superior, trilocular with many ovules; axile placentation
Fruit : capsule, rarely berry
Seed: endospermous

Answer 12.
Active transport : Transport of substances from low concentration to a higher concentration (uphill transport). It requires ATP to carry substances across the cell mem-brane.
Facilitated diffusion: Transport of substances from higher concentration to lower concentration. Substances move across the cell membrane without the expenditure of energy.

Answer 13.
Alcoholic fermentation :- Pyruvic acid is converted to carbon dioxide and ethanol.
Lactic acid fermentation :- Pyruvic acid is converted to lactic acid.

Answer 14.
Glucose → glucose 6 – phosphate → Fructose 6 phosphate →fructose 1, 6- biphosphate (6c) → Dihydroxy acetone phosphate (3c)
↓
Glyceraldehydel, 3 phosphate (3c)
↓
1, 3 biphospho glyceric acid
↓
3 phospho glyceric acid
↓
2 phosphoglyceric acid
↓
2 phosphoenol pyruvate
↓
Pyruvic acid (3c)

Answer 15.

Answer 16.
a. ABA stimulates closure of stomata/ increases tolerance of plants to various kinds of stress
b. Apical dominance – Auxin/Ethylene. Cytokinin act against apical dominance.

Answer 17.
a. Kranz anatomy
b. 1. High temperature.
2. Photorespiration does not occur.
3. Several layers of bundle sheath cells.
4. Having large number of chloroplast.
5. No intercellular spaces.

Answer 18.
a. 3-Phosphoglyceric acid (3-PGA).
b. Carboxylation, Reduction and Regeneration.
c. RuBP carboxylase – oxygenase (RuBis CO)
d. 18 ATP molecules and 12 NADPH molecules.

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Botany Model Question Paper 1

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.

Questions 1-3. Answers all questions. Each question carries 1 score.

Question 1.
Choose the correctly matched pair
a. Banyan tree – Assimillatorry roots
b. Ribizophora – Prop roots
c. Radish – Fusiform root

Question 2.
The name ‘Virus’ that means venom or poisonous fluid was given by …..
a. Pasteur
b. Ivanowsky
c. Beijerinck
d. Stanley

Question 3.
Observe the figure given below and identify the cell organalle.

Questions 4-11. Answers any 6 questions. Each question carries 2 score.

Question 4.
Calvin and his co-workers worked out the whole pathway of Co₂ fixation and showed that the pathway is operated in a cyclic manner.
Draw this cycle using the given terms in the correct order. 3 phosphoglycerate, Regeneration, Carboxylation, Triose phosphate, Reduction, Ribulose, 1,5 biphosphate, Co₂, Sucrose.

Question 5.
Most pteriodophytes are homosporous but there are exceptions.
‘Heterospory is a precursor to seed habit’. Point out any one common character found in heterospory and seed habit.

Question 6.
Analyse the table and fill in the blanks,

Question 7.
Proteins in the membrane are responsible for facilitated diffusion and active transport and hence both show common characteristics. List any two such characteristics.

Question 8.
‘Double fertilization is an event unique to angiosperms’.
a. Mention the two fusions in this event.
b. Name the products obtained as a result of these fusions.

Question 9.
In a plants, deficiency of N₂ is visible inorder parts and that of Ca is visible in younger parts’. Critically evaluate the statement.

Question 10.
Photosynthesis can be considered as the most significant physicochemical process on earth. Evaluate this statement citing any two significances.

Question 11.
A diagrammatic view of a cell cycle is given below.

a. Identify the phase in which
i. DNA synthesis takes place
ii. Chromosomes are arranged at the equator of the spindle,
b. Mention two significant points of mitosis in the life of an organism.

Questions 12 – 17. Answers any 5 questions.Each question carries 3 scores.

Question 12.
a. Identify the cell organelle found both in eukaryotic and prokaryotic cells.
b. Justify its presence in both types of cells.

Question 13.
Observe the given diagram of cell division.
a. Identify the state.

b. Label the parts A and B.
c. Mention any one peculiarity for the pachytene stage of meiosis.

Question 14.
In the anatomy lab, Eugin observed the following features in the T.S of a plant part.
a. adial and polyarch xylem bundles.
b. Parenchymatous (homogenous) cortex
c. Large Path
d. Epidermis with unicellular hairs.
e. Pericycle.
ii. Re-arrange the given regions from the periphery to the center in their correct sequence.
iii. Give an account of casparian strips.

Question 15.
Observe the flow chart given below :

a. Identify A, B, C. Write the function of B
b. Differentiate radial & conjoint arrangements.

Question 16.
Pigments in plants are organized into two photochemical light harvesting com-plexes.
a. Name the reaction center of PS-1 and write its absorption peak.
b. Name the reaction center of PS-II and write its absorption peak.
c. Explain non-cyclic photo phosphorylation giving reasons for the naming of the phenomenon, with the help of a diagram.

Question 17.
Prophase of the first meiotic division is typically longer and complex. It has been further subdivided into five phases. The major events of these substages are given below:
a. Pairing of homologous chromosomes.
b. Terminalization of chiasmata.
c. Occurs crossing over.
d. Chromosomes-are visible under a light microscope.
e. Formation of chiasmata.
i. Identify the stages a,b,c,d and e.
ii. Arrange these stages in correct order of occurrence.

Answers

Answer 1.
c.

Answer 2.
a. Pasteur

Answer 3.
Mitochondria

Answer 4.

Answer 5.
Heterospory means production of two kinds of spores ie, megaspore and microspore. Megaspore germinate and give rise to fe-male gametophyte. Micro- spore germinate and give rise to male gametophyte. So the development of zygote into embryo takes place in the female gametophyte.

Answer 6.
a. Auxin
b. Bolting
c. Overcome apical dominance

Answer 7.
Active transport is carried out by some membrane-protein like facilitated transport. Rate of both of them reach maximum when all membrane proteins are used up. Both are more specific.

Answer 8.
a. Syngamy and triple fusion.
b. By sygamy zygote is formed, by triple fusion primary endosperm nucleus (PEN) is formed.

Answer 9.
The elements that are actively mobilised within the plants and exported to young developing tissue the deficiency symp-toms appears first at the older tissues, i.e., the deficiency of N₂, K⁺ Mg⁺ are visible at the older parts.

The elements that are relatively in mobile and are not transported out the mature organs show deficiency sympto-ms of younger parts, i.e., the deficiency of S, Ca²⁺ are visible at younger parts.

Answer 10.
It is the only natural process by which light energy is converted to chemical energy. It provide food either directly or indirectly. It maintains the equilibrium of oxygen in at-mosphere. It provide vast reserves of en-ergy in the form of petro-lium products like diesel, natural gase etc.

Answer 11.
a. i. Synthesis phase
ii. Metaphase
b. Growth of multicellular organisms is due to mitosis.
Restore the nucleo cytoplasmic ratio.

Answer 12.
a. Ribosome.
b. Ribosome is a site of protein synthesis in both prokaryotes and eukaryotes.

Answer 13.
a. Metaphase.
b. A – Spindle fibers, B – Chromatids.
c. Crossing over occurs between the chromatids of homologous chromosomes. It gives the appearance of recombination nodule.

Answer 14.
i. Monoroot root
ii. Epidermis with unicellular hairs.
Parenchymatous cortex.
Endodermis with casparian strips.
Pericycle.
Radial and polyarch xylem bundles.
Large pith.
iii. The tangential, as well as radial walls of the endodermal cells, have a deposition i of water-impermeable, waxy material suberin in the form of casparian strips.

Answer 15.
a. A- Epidermis, B – Stomata
C – Epidermal appendages
b. Radial: When xylem and phloem within a vascular bundle are arranged in alternate manner on different radii, e.g., roots.
Conjoint : Xylem and phloem are situated at the same radius of vascular bundles. Such vascular bundles are common in stems and leaves. The conjoint vascular bundles usually have the phloem located only on the outer side of xylem.

Answer 16.
a. P₇₀₀ – Peak is 700 nm
b. P₆₈₀ – Peak is 680 nm

The schematic representation showing Z-scheme.

Answer 17.
i. a. Zygotene
b. Diplotene
c. Pachytene
d. Leptotene
e. Pachytene
ii. Leptotene, Zygotene, Pachytene, Diplotene.

Plus One Chemistry Previous Year Question Papers and Answers

Kerala Plus One Botany Previous Year Question Paper 2017

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.

Question 1.
Who among the following scientists is related with the identification of cytokinins?
a. E. Kurosawa
b. F. Skoog
c. C.Darwin
d. F. W. Went

Question 2.
Read the following statements and select the correct statements from the given choices.
i. Methanogens are present in the guts of ruminant animals
ii. Mycoplasma has distinct cell wall.
iii. Viroids are infectious naked DNA molecule.
iv. Algal component of lichen is phycobiont.
a. ii and iii
b. i and iv
c. i and iii
d. ii and iv.

Question 3.
Algae are useful to man in a variety of ways.
Suggest any four uses of algae.

Question 4.
The root nodules of leguminous plants contain two necessary biochemical compounds for N₂ fixation. Identify these compounds.

Question 5.
An important difference between C₃ and C₄ plants is photorespiration. Explain how photorespiration occurs in C₃ plants.

Question 6.
Write any four agricultural applications of ethylene.

Question 7.
Define Blackman’s law of limiting factors and identify any two important factors which influence the rate of photosynthesis in plants.

Question 8.
The movement of water through the root layers is ultimately symplastic in the endodermis. Give a reason.

Question 9.
a. The arrangement of flowers on the floral axis is called ……
i. Aestivation
ii. Phyllotaxy
iii.Placentation
iv.Inflorescence
b. How can you differentiate an actinomorphic flower from a zygomorphic flower ?

Question 10.
a. The tissues involves in secondary growth of dicot plants are vascular cambium and ….
b. Compare the formation of vascular cambia in dicot stem and dicot root.

Question 11.
a. Identify a cell organelle that contains hydrolytic enzymes.
b. Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER) are morphologically and functionally different. Justify this statement.

Question 12.
Given below is the metaphase of mitosis. Analyze the diagram and draw a sketch of anaphase. Write any two events of anaphase.

Question 13.
Glycolysis is the partial oxidation of glucose to produce two molecules of pyruvic acid.
a. Where does glycolysis occur ?
b. Steps of glycolysis are given below. Fill up the blank boxes.


The complete oxidation of pyruvic acid yields three molecules of C0₂ by a cyclic process in the matrix of mitochondria.
a. Who first develop this cycle ?
b. Draw a diagramatic sketch of the identified cycle.

Answers

Answer 1.
b. F. Skoog

Answer 2.
b. i and iv.

Answer 3.
Helps in Co₂ fixation by photosynthesis, in-crease level of oxygen, primary producers on ocean, used as food.

Answer 4.
Nitrogeneous enzyme / Mo – Fe protein leg hemoglobin

Answer 5.
In C₃ plants O₂ binds to RuBisCO. The RuBP instead of being converted to 2 molecules of PGA, binds with O₂ to form one molecule of phosphoglycerate and phosphogly collate in a pathway called photorespiration. Condition for photorespiration are dual nature of RuBisCO, increases O₂ concentration, moderate temperature etc.

Answer 6.
1. Ethylene breaks seed and bud dormancy, initiates germination in peanut seeds, sproutind of potatao tubers.
2. Ethylene promotes rapid internode/petiole elongation in deep water rice plants, It helps leaves/ upper parts of the shoot to remain above water.
3. Ethylene also promotes root growth and root hair formation.
4. Ethylene is used to initiate flowering and synchronising fruit – set in pineapples.

Answer 7.
The law states that, if a chemical process is affected by more than one factor, then its rate will be determined by the factor which directly affects the process if its quantity is changed.
For example, despite the presence of a green leaf and optimal light and Co₂ conditions, the plant may not photosynthesis, if the temperature is very low. This leaf, if given the optimal temperature will start photosynthesizing.

Answer 8.
Simplistic system is the movement of water occurs from cell to cell through plasmodesmata. Most of the water flow in the roots occur via apoplast since the cortical cells are loosly packed, hence offer no resistance to water movement. The continuation of cell walls breaks in the endodermis of the root due to presence of waxy casperian strips which are impermeable. So the movement of substances in endodermis occurs any simplest.

Answer 9.
a. iv. Inflorescence
b.The flower can be divided into two equal halves along more than one longitudinal plane, i.e., they are radially symmetrical. Such flowers are actinomorphic. Zygomorphic flower cannot be divided into two similar halves in any plane.

Answer 10.
a. Cork cambium
b.

Answer 11.
a. Lysosomes
b. In some ER, spherical granular structures called ribosomes are attracted on the surface. This type of ER is called RER. It occurs in almost all cells which are actively engaged in protein synthesis.
In SER, ribosomes are not attached with ER and the surface is smooth. It is well developed in cells that synthesize steroid hormones.

Answer 12.
Anaphase

Centromeres split and chromatids separate.Chromatids move to opposite poles.
The centromere of each chromosome is towards the pole arms towards the equator.

Answer 13.
a. Cytoplasm (Cytosol)
b. A. Glucose – 6 – phosphate
B. Fructose – 6 – phosphate
C. Fructose – 1, 6 – biphosphate
D. 1, 3 biphosphoglyceric acid / Triose biphosphate
E. 2 – phosphoglycerate / 2 – phosphoglyceric acid
F. Phosphoenol pyruvate / Phospho enolpyruvic acid
OR
a. Hans Kreb
b.

Plus One Chemistry Previous Year Question Papers and Answers

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

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.

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 3

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.

Questions 15 carry 1 score each. Answer any I four questions.

Question 1.
The branch of Physics which deals with the study of phenomena related to light.

a. Optics
b. Thermodynamics
c. Mechanics
d. Electronics

Question 2.
State the theorem of perpendicular axes on a moment of inertia.

Question 3.
Select a true statement from the following.
a. The distance traveled by light in one year through vacuum or air is an astronomical unit.
b. The range of time in Physics varies from 10^-14 sec to 10^-25 sec.
c. Lightyear is the largest practical unit of length.
d. Chandra Shekhar Limit (C.S.L) is the largest practical unit of mass.

Question 4.
When two objects collide, after collision they could move together, the collision is ……………………
(elastic, completely elastic, inelastic, completely inelastic)

Question 5.
List any two conditions for a motion of a body to be simple harmonic.

Questions 6 to 11 carry 2 scores each. Answer any 5 questions.

Question 6.
The possibility of falling backward with the ladder is more when you are high up on the ladder than when you just begin to climb. Explain why.

Question 7.
The displacement of y (in cm) of an oscillating particle varies with time t (in a sec) according to the equation. y = 2 cos (0.5 πt + π/3) Find the amplitude and period of the particle

Question 8.
Draw the stress-strain graph of a loading wire. Mark the following points:

1. Elastic limit
2. Fracture point
3. Plastic region
4. Elastic region.

Question 9.
Calculate the work done in lifting a body of mass 10 kg to a height of 10 m above the ground. 10 kg

Question 10.
a. Which among the following possess the highest specific heat capacity?

1. Water
2. Silver
3. Copper
4. Steel

b. You are in a restaurant waiting for your friend and ordered coffee. It has arrived. Do you add sugar to your friend’s coffee and then wait for him or do you add sugar after he arrives? Explain with respect to the concept of cooling.

Question 11.
A collision between two particles need not be the physical contact of two particles as in the case of scattering of the particle by a nucleus.
a. What is the quantity that remains conserved in all types of collisions?
b. Suppose an electron and a proton are projected with equal kinetic energy, What will be the ratio of their linear momentums if the proton is 1830 times heavier than an electron?

Questions 12 to 17 carry 3 scores each. Answer any five questions.

Question 12.
Motion along a plane is called two-dimensional motion. A body moving in two dimensions is found to have an acceleration in one dimension.

a. Identify the motion.

b. A ball thrown by a player reaches another player in 2s. What is the maximum height attained by the ball above the point of projection? (Take g= 10ms^-2)

c. In the figure, the point P on a wheel of radius R is in contact with the ground. What is the displacement of the point, when the wheel rolls a half revolution?

Question 13.
All physical quantities can be expressed in terms of dimension.
a. Write the physical quantities of the following dimensions:
i. [M¹L¹T¹ ]
ii. [M²L ²T²]
b. Check whether the equation T = \(T=2\pi \sqrt { \frac { m }{ g } }\) is dimensionally correct
T → Time period of a simple pendulum
m → mass of the bob
g → acceleration due to gravity

Question 14.
Kinetic theory of gases is based on the molecular picture of matter.
a. Write any two postulates of kinetic theory of gases.
b. Write short notes on:

1. Equipartition of energy
2. Mean free path

Question 15.
The stress-strain graph for wires of two materials A and B are given below.

a. Which material is more ductile?
b. When spring balances are continuously used for a long time, they show a wrong reading. Explain why?

Question 16.
Force is required to lift a body from the ground to a height h and work is measured as the product of force and magnitude of displacement.
a. Name the energy possessed by the body at maximum height. Write an equation for it.
b. A man of mass 60 kg carries a stone of mass 20 kg to the top of a multi-story building of height 50m. Calculate the total energy spent by him? (g = 9.8 m/s²)

Question 17.
The acceleration due to gravity o,n the surface of the moon is 1.7 m/s². what is the time period of a simple pendulum on the moon, if its time period on Me earth is 3.5 second?

Questions 18 to 22 carry 4 scores each. Answer any 4 questions.

Question 18.
A stone is thrown with the help of a sling with initial velocity V_Q at an angle θ from the horizontal.

a. Working of a sling is based on a law of vector addition.
b. With the help of a vector diagram, state this law.
c. Derive the expression for the maximum height reached by the

Question 19.
The velocity of sound depends on density (p) and modulus of elasticity (E). (The dimensional formula of E is ML^-1 T^-2)
a. State the principle of homogeneity.
b. Using the above principle, arrive at an expression for the velocity of sound. (Take K = 1).

Question 20.
Hooke’s law states that stress a strain.
a. What is the necessary condition for the above law to be valid?
b. Explain with the help of a graph, the relation between stress and strain for a given solid material under increasing tensile stress.

Question 21.
a. The figure shows the position-time graph of a body moving along a straight line.

1. Draw the velocity-time graph of the body.
2. From the graph, find the displacement in 20 seconds.

b. From the velocity-time graph of a body moving with uniform acceleration, deduce the velocity-time relation and the velocity displacement relation.

Question 22.
The motion represented by the equation, y(t) = A cos (ωt + Φ) is called simple harmonic motion (SHM).
a. Which one of the following examples ‘ closely represents SHM? Substantiate your answer.

1. The rotation of the earth about its axis.
2. Oscillations of a swing.

b. A vibrating simple pendulum of period T is placed in a lift which is accelerating downwards. What is the effect of this on the time period of the pendulum?

Questions from 23 to 26 carry 4 scores. Answer any 3 questions.

Question 23.
According to Newton’s law of motion, the force depends on the rate of change of momentum.
a. Name the law that helps to measure force.
b. Using the above law, deduce an expression for a force.
c. A man jumping out of a moving bus falls with his head forward. What should he do in order to land safely?

Question 24.
The flow of an ideal fluid in a pipe of a varying cross-section is shown.

a. Differentiate between streamline flow and turbulent flow.
b. State and prove Bernoulli’s principle.

Question 25.
The figure given below depicts the schematic representation of an engine.

a. Which type of engine is this, a heat engine or a refrigerator?
b. Write the 4 steps of operation in the Carnot cycle.
c. A Carnot engine is working with a temperature of 27°C and 327°C. Find its efficiency (η).

Question 26.
In a hammer throw event, a solid sphere of mass 16 kg is tied to a light 50 cm long chain. A sportsman gives it a constant moment of 30 Nm for 10 seconds and then throws the sphere. Consider the sphere as a point mass.
a. Find the moment of inertia about the axis of rotation.
b. If ‘L’is the angular momentum and τ is the torque. Show that τ = \(\frac { dL }{ dt }\)
c. Write an example for the motion in which an angular momentum remains constant.

Answers

Answer 1.
Optics

Answer 2.
This theorem is good for flat bodies which means, thickness much less than length and breadth. The theorem states that “the moment of inertia of a plana body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two mutually perpendicular axes passing through the same point, lying on the plane”.

Answer 3.
d

Answer 4.
Completely inelastic.

Answer 5.
1. acceleration is directly proportional to displacement. i.e., a α y
2. Periodic motion.

Answer 6.

As the distance from the axis of rotation (r) increases, the torque also increases. When we are high on the ladder the, r is greater and so a small force in the perpendicular direction can cause a large torque. So there is a greater chance of falling.

Answer 7.
A pendulum whose period is 2 seconds is called seconds pendulum. y = A cos (ωt + ∅)

1. x = 20 cos (62.Bt + π/14) Amplitude = 20 icm
2. Frequency a> = 62.8
3. Initial phase = 14 Strain

Answer 8.

Answer 9.
Work was done = Force × displacement
W = mg × h = 10 × 9.8 × 10 = 980 J

Answer 10.
a. Water.
b. Add sugar as soon as the coffee arrives and then wait. This is Newton’s law of cooling. Thus when sugar is added to coffee, the temperature of coffee decreases. Hence the rate of cooling also decreases.

Answer 11.
a. Momentum is conserved
b. The kinetic energy of an electron

Answer 12.
a. Projectile motion

Answer 13.
a.

1. Momentum or impulse
2. Work, energy, torque

So the equation is wrong

Answer 14.
a. AverageThe average energy of a molecule of an ideal gas is proportional to the absolute temperature. PV = nRT at all pressure and temperature.
b.

1. According to equipartition theorem, total energy of a system is equally distributed among various degree of freedom.
2. Mean free. a path is the mean distance traveled by the molecules between two successive collisions.

Answer 15.
a. Material A is more ductile.
b.This is due to a phenomenon known as fatique. Due to continuous use for a long time, the elasticity of materials decreases. So they show a wrong reading.

Answer 16.
a. Potential energy U = mgh
b. m = 60 + 20 = 80 kg;
h = 50m; g= 9.8 m/s²
U= mgh = 80 × 9.8 × 50
= 39200J

Answer 17.
g_m = 1.7 m/s², T_m = ? T= 3.5 s

Answer 18.
a. Parallelogram law of vector addition.
b. It states that “if two vectors are represented both in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, then the resultant vector is represented both in magnitude and direction by the diagonal of the parallelogram passing through the same point”.
c. Horizontal range (R). It is the horizontal distance covered by a projectile from the point of projection on the ground to the point on the ground where the projectile returns again. R = horizontal velocity x time of flight (The horizontal velocity v cosQ of the projectile will be a constant)

The above equation shows that R is maximum when sin 20 is maximum, ie., when 0 = 45°
The maximum horizontal range Rmax = \({ R }_{ max }=\frac { { u }^{ 2 } }{ g }\)

Answer 19.
a. Acca ordering to a principle of homogeneity, the dimensions of the same fundamental quantity must be the same on both sides of the equation.

Answer 20.
a. Hooke’s law states that within elastic limit stress is directly proportional to strain.

Consider a uniform wire suspended freely from a fixed point. The face end is subjected to a stress and strain is noted. In the above graph, stress is on Y-axis, and strain on X-axis.

In the initial stages as stress increases strain also increases linearly from 0 to A. Till A stress a brain. For the all-region AB the proportionality is lost but even in this region, the wire will partially recover its initial state.

Beyond B (B is called the elastic limit). The strain increases very rapidly and it is seen that the wire retains the law dimension’s on the removal of the force. Now its behavior is plastic.

Beyond this point, even without increasing the stress any further the wire extends and ultimately at D, it breaks. This point D is called Breakpoint or yield point.

Answer 21.
a. A slope of a displacement-time graph (x-t) gives velocity. Velocity in the interval 0 – 5 sec.

When we draw a velocity-time graph, we get a graph as shown in the figure

ii. An area under the velocity-time graph gives displacement. Area A = 2 × 5 + 2 × 5 = 0 displacement = 0
b. Consider a body moving with uniform acceleration ‘a’. Let u be the initial velocity at a time t = 0 and v be the final velocity at time t.

But from equation (1), we get t = \(\frac { v-u }{ a }\)

Answer 22.
a. A simple harmonic motion is special type of motion which is periodic and takes place continuously at regular intervals of time in which a body moves to and fro on a straight line about a fixed point under a restoring force which is directed always towards the fixed point and directly proportional to displacement at that instant from the fixed point.

An ideal simple pendulum is made of a heavy point mass body suspended from rigid support by an inextensible, weightless and perfectly flexible string. The body is free to oscillate in a vertical plane. When an undisturbed pendulum is in the position ‘SO’. ‘S’ is the point of suspension and ‘O’ is the center of gravity of the bob.

The position SO is the equilibrium position. If the bob is slightly pulled aside and released, the pendulum oscillates between Q and P. Q & P represents the position of a center of gravity of the bob in the displaced position during the oscillation, then the force at P is shown in the figure.

1. Weight, mg of the bob acting vertically downwards.
2. The tension along the string T.

Resolving mg we get mg sinθ and mg cosθ as shown in the figure. Here mg cosθ balances the tension and mg sinθ acts tangentially at P hence it acts as the restoring force.
Let SO = SQ = SP = L
Let OP = x
Restoring force F = mg sin θ (Restoring force tends to reduce angular displacement θ hence the ve sign).
If the displacement ‘x’ is small compared to 1 θ is small hence sinθ = θ Restoring force F = – mg θ

Answer 23.
a. Newton’s 2^nd law of motion.
b. The law state that the rate of change of linear momentum is directly proportional to the applied force and it takes place in the direction of applied force. According to the second law,


c. A man jumping out of a moving bus falls with his head forward because his body gets a forward momentum and has a tendency to continue in the state of motion (inertia of motion) so in order to land safely he should not stand stationary but continue in the state of motion by running for a while.

Answer 24.
a. Streamline flow is defined as a flow in which all the liquid particles that pass any given point follow the same path at the same speed.

When the velocity of flow exceeds the critical velocity, the flow becomes Turbulent. The flow of a liquid is said to be turbulent flow if the speed and direction of the liquid particles passing any point change with time.

b. According to Bernoulli’s theorem, the steady flow of an ideal liquid, the total energy per unit volume remains constant throughout the flow. France by Bernoulli’s theorem, kinetic energy + potential energy + pressure energy = a constant throughout the flow

Consider the steady flow of a liquid through a nonuniform tube as shown in the figure. The pressure area of cross-section and speed at A are P₁, a₁, v₁ and P₂, a₂, v₂, at B. A and B are at height h₁ and h₂ from the horizontal. Let in time at liquid near

Total energy at A = Total energy at B Thus Bernoulli’s Theorem is proved.

Answer 25.
Heat engine.

Adiabatic expansion → BC
Isothermal compression → CD
Adiabatic compression → DA
c. The efficiency of a Carnot engine working between two temperature T₁ and T₂ of hot and cold reservoir is given by

Answer 26.

c. A motion of the earth around the sun.

Plus One Physics Previous Year Question Papers and Answers