Category: Class 11

Students can Download Tamil Nadu 11th Biology Model Question Paper 2 English Medium Pdf, Tamil Nadu 11th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Biology Model Question Paper 2 English Medium

General Instructions:

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

Time: 3:00 Hours
Maximum Marks: 70

Bio – Botany [Maximum Marks: 35]

PART – I

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
Red sea is red colour due to ……………………
(a) Dermacarpa sps
(b) Trichodesmium sps
(c) Scytonema sps
(d) Gloeocapsa sps.
Answer:
(b) Trichodesmium sps

Question 2.
Sago is obtained from …………………..
(a) Cycas revoluta
(b) Pinus roxburghil
(c) Pinus insularis
(d) Cedrus deodara
Answer:
(a) Cycas revoluta

Question 3.
Plants growing on the water surface are called as …………………… type of aquatic plants.
(a) Emergent
(b) Submerged
(c) Free floating
(d) Mangroves
Answer:
(c) Free floating

Question 4.
Synapsis occur between
(a) mRNA and ribosomes
(b) Spindle fibres and centromeres
(c) Two homologous chromosomes
(d) A male and a female gamete
Answer:
(c) Two homologous chromosomes

Question 5.
An example of feedback inhibition is ………………………
(a) Cyanide action on cytochrome
(b) Sulpha drug on folic acid synthesiser bacteria
(c) Allosteric inhibition of hexokinase by glucose-6-phosphate
(d) The inhibition of succinic dehydrogenase by malonate
Answer:
(c) Allosteric inhibition of hexokinase by glucose-6-phosphate

Question 6.
Rubber is a …………………..
(a) Latex
(b) Resin
(c) Alkaloid
(d) Drug
Answer:
(a) Latex

Question 7.
Identify the correct statement:
(I) Sulphur is essential for amino acids Cystine and Methionine
(II) Low level of N, K, S and Mo affect the cell division
(III) Non-leguminous plant Aims which contain bacterium Frankia
(IV) Denitrification carried out by Nitrosomonas and Nitrobacter
(a) (I), (II) are correct
(b) (I), (II), (III) are correct
(c) (I) only correct
(d) All are correct
Answer:
(b) (I), (II), (III) are correct

Question 8.
Which step is irrelevant with respect to aerobic respiration?
(a) Glycolysis
(b) Pyruvate oxidate
(c) Fermentation
(d) TCA cycle
Answer:
(c) Fermentation

PART – II

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

Question 9.
How milk is changed into curd, if a few drops of curd is added to it? What is the reason for its sourness?
Answer:
The change is brought by Lactobacillus lactis, a bacterium present in the curd. The sourness is due to the formation of lactic acid.

Question 10.
What are pyrenoids? Mention its role?
Answer:
Pyrenoids are proteinaceous bodies found in chromatophores of algae and assist in the synthesis and storage of starch.

Question 11.
From which type of flowers does the aggregate fruit develops?
Answer:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is develops into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit.

Question 12.
Define magnification. How will you calculate it?
Answer:
The optical increase in the size of an image is called magnification. It is calculated by the following formula;
Magnification =

Question 13.
Define Diffusion Pressure Deficit (DPD)?
Answer:
The difference between the diffusion pressure of the solution and its solvent at a particular . temperature and atmospheric pressure is called as Diffusion Pressure Deficit (DPD).

Question 14.
Name the two forms of phycobilins and also give an example?
Answer:
Phycobilins exists in two form. They are:

1. Phycocyanin found in Cyanobacteria.
2. Phycoerythrin found in red algae.

PART – III

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

Question 15.
Do you think shape of chloroplast is unique for algae? Justify your answer?
Answer:
Variation among the shape of the chloroplast is found in members of algae. It is Cup shaped (Chlartiydomonas), Discoid (Chara), Girdle shaped (Ulothrix), reticulate (Oedogonium), spiral (Spirogyra), stellate (Zygnema) and plate like (Mougeoutia).

Question 16.
Name few molecular markers used in molecular taxonomy?
Answer:
Allozymes, mitochondrial DNA, micro satellites, RFLP (Restriction Fragment Length Polymorphism), RAPD (Random amplified polymorphic DNA), AFLPs (Amplified Fragment Length Polymorphism), Single nucleotide Polymorphism – SNP, microchips or arrays.

Question 17.
List out the criteria for being as essential minerals?
Answer:
Amon and Stout (1939) gave criteria required for essential minerals:

1. Elements necessary for growth and development.
2. They should have direct role in the metabolism of the plant.
3. It cannot be replaced by other elements.
4. Deficiency makes the plants impossible to complete their vegetative and reproductive phase.

Question 18.
Draw a simplified diagram showing the various regions of root?
Answer:

Question 19.
What do you mean by the term – Basipetal transport and Acropetal transport?
Answer:
Basipetal means transport through phloem from shoot to root and acropetal means transport through xylem from root to shoot.

PART – IV

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

Question 20.
Describe in detail about the lytic cycle of phages with diagram?
Answer:
Lytic cycle:
During lytic cycle of phage, disintegration of host bacterial cell occurs and the progeny virions are released. The steps involved in the lytic cycle are as follows:

(I) Adsorption:
Phage (T₄) particles interact with cell wall of host (E coli). The phage tail makes contact between the two, and tail fibres recognize the specific receptor sites present on bacterial cell surface. The lipopolysaccharides of tail fibres act as receptor in phages.

The process involving the recognition of phage to bacterium is called landing. Once the contact is established between tail fibres of phage and bacterial cell, tail fibres bend to anchor the pins and base plate to the cell surface. This step is called pinning.

(II) Penetration:
The penetration process involves mechanical and enzymatic digestion of the cell wall of the host. At the recognition site phage digests certain cell wall structure by viral enzyme (lysozyme). After pinning the tail sheath contracts (using ATP) and appears shorter and thicker.

After contraction of the base plate enlarges through which DNA is injected into the cell wall without using metabolic energy. The step involving injection of DNA particle alone into the bacterial cell is called Transfection. The empty protein coat leaving outside the cell is known as ‘ghost’.

(III) Synthesis:
This step involves the degradation of bacterial chromosome, protein synthesis and DNA replication. The phage nucleic acid takes over the host biosynthetic machinery. Host DNA gets inactivated and breaks down.

Phage DNA suppresses the synthesis of bacterial protein and directs the metabolism of the cell to synthesis the proteins of the phage particles and simultaneously replication of phage DNA also takes place.

(IV) Assembly and Maturation:
The DNA of the phage and protein coat are synthesized separately and are assembled to form phage particles. The process of assembling the phage particles is known as maturation. After 20 minutes of infection about 300 new phages are assembled.

(V) Release:
The phage particle gets accumulated inside the host cell and are released by the lysis of the host cell wall.

[OR]

Draw a flow chart depicting the Bentham and Hooker Classification?
Answer:

Question 21.
With the help of diagram explain the possible route of water across root cells?
Answer:
There are three possible routes of water.
They are:-

1. Apoplast
2. Symplast and
3. Transmembrane route.

1. Apoplast:
The apoplast (Greek: apo = away; plast = cell) consists of everything external to the plasma membrane of the living cell. The apoplast includes cell walls, extra cellular spaces and the interior of dead cells such as vessel elements and tracheids.

In the apoplast pathway, water moves exclusively through the cell wall or the non-living part of the plant without crossing any membrane. The apoplast is a continuous system.

2. Symplast:
The symplast (Greek: sym = within; plast = cell) consists of the entire mass of cytosol of all the living cells in a plant, as well as the plasmodesmata, the cytoplasmic channel that interconnects them.

In the symplastic route, water has to cross plasma membrane to enter the cytoplasm of outer root cell; then it will move within adjoining cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane, till it reaches xylem.

3. Transmembrane route:
In transmembrane pathway water sequentially enters a cell on one side and exits from the cell on the other side. In this pathway, water crosses at least two membranes for each cell. Transport across the tonoplast is also involved.

Mechanism of Water Absorption Kramer (1949) recognized two distinct mechanisms which independently operate in the absorption of water in plants. They are,

1. Active absorption
2. Passive absorption.

[OR]

Explain the types of parasitic mode of nutrition in angiosperms?
Parasitic mode of nutrition in angiosperms:
Answer:
Organisms deriving their nutrient from another organism (host) and causing disease to the host are called parasites.

a. Obligate or Total parasite – Completely depends on host for their survival and produces haustoria.

(I) Total stem parasite:
The leafless stem twine around the host and produce haustoria. Example: Cuscuta (Dodder), a rootless plant growing on Zizyphns, Citrus and so on.

(II) Total root parasite:
They do not have stem axis and grow in the roots of host plants produce haustoria. Example: Rajflesia, Orobanche and Balanophora.

b. Partial parasite – Plants of this group contain chlorophyll and synthesize carbohydrates. Water and mineral requirements are dependent on host plant.

(I) Partial Stem Parasite:
Example: Loranthus and Viscum (Mistletoe) Loranthus grows on fig and mango trees and absorb water and minerals from xylem.

(II) Partial root parasite:
Example: Santalum album (Sandal wood tree) in its juvenile stage produces haustoria which grows on roots of many plants.

Bio – Zoology [Maximum marks: 35]

PART – I

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
Which is not a unit of taxonomic category?
(a) Series
(b) Glumaceae
(c) Class
(d) Phylum
Answer:
(b) Glumaceae

Question 2.
Study of ticks and mites is ……………………
(a) Acarology
(b) Entomology
(c) Malacology
(d) Carcinology
Answer:
(a) Acarology

Question 3.
Match the List – I and List – II.

List – I	List – II
1. Silver fish	(i) Book lung
2. Scorphion	(ii) Pharyngeal gills
3. Sea squirt	(iii) Lungs
4. Dolphin	(iv) Trachea

Answer:
(a) 1 – (iv), 2 – (i), 3 – (iii), 4 – (ii)
(b) 1 – (iv), 2 – (i), 3 – (ii), 4 – (iii)
(c) 1 – (i), 2 – (iii), 3 – (iv), 4 – (ii)
(d) 1 – (i), 2 – (iv), 3 – (iii), 4 – (ii)

Question 4.
MSH stands for ……………………
(a) Melanocyte stimulating hormone
(b) Malic stimulating hormone
(c) Myosin stimulating hormone
(d) Metabolic stimulating hormone
Answer:
(a) Melanocyte stimulating hormone

Question 5.
Which of the following is an correct statement?
(a) Ehler’s – Danlos syndrome – Affects collagen and results in facial abnormalities.
(b) Rhabdomyosarconca – Life threatening soft tissue tumour of head, neck and urinogenital tract.
(c) Rheumatoid arthritis – Progressive inability to secrete saliva and tears.
(d) Alzheiner’s disease – A degenerative disorder of the nervous system that affects movement often including tremors.
Answer:
(b) Rhabdomyosarconca – Life threatening soft tissue tumour of head, neck and urinogenital tract.

Question 6.
Which of the following is an incorrect statement?
(a) The functions of frontal region are Behaviour, intelligence, memory and movement.
(b) The functions of parietal region is intelligence and memory.
(c) The functions of temporal region are speech, hearing and memory.
(d) The functions of occipital region is visual processing.
Answer:
(b) The functions of parietal region is intelligence and memory.

Question 7.
Synovial fluid is present in …………………….
(a) Spinal cavity
(b) Cranial activity
(c) Freely movable joints
(d) Fixed joints
Answer:
(c) Freely movable joints

Question 8.
Glands responsible for secreting tears are ………………………….
(a) Glands of moll
(b) Lacrimal glands
(c) Meibomian glands
(d) Glands of Zeis
Answer:
(b) Lacrimal glands

PART – II

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

Question 9.
Why are spongin and spicules important to a sponge?
Answer:
Spongin and spicules provide support and supports the soft body parts of the sponges. The spicules give the sponges rigidity and form to the sponges.

Question 10.
What is water vascular system?
Answer:
The system which helps in nutrition and respiration in echinoderms is called water vascular system. Water enters into the body through special organs.

Question 11.
Difference between chordates and non-chordates?
Answer:

S.No	Chordates	Non-Chordates
1.	Notochord is present.	Absence of notochord.
2.	Pharynx perforated by gill slits.	Gill slits absent.
3.	A post anal tail is present.	Post anal tail is absent.

Question 12.
Draw the diagram of head region of Periplaneta americana?
Answer:

Question 13.
Name the different types of movement?
Answer:

1. Amoeboid movement.
2. Ciliary movement.
3. Flagellar movement.
4. Muscular movement.

Question 14.
What is endomysium?
Answer:
The connective tissue surrounding the muscle fibre is called the endomysium.

PART – III

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

Question 15.
What are the classical taxonomical tools?
Answer:
Taxonomical tools are the tools for the study of classification of organisms.
They include:-

Taxonomical keys:
Keys are based on comparative analysis of the similarities and dissimilarities of organisms. There are separate keys for different taxonomic categories.

Museum:
Biological Museums have collection of preserved plants and animals for study and ready reference. Specimens of both extinct and living organisms can be studied.

Zoological parks:
These are places where wild animals are kept in protected environments under human care. It enables us to study their food habits and behavior.

Marine parks:
Marine organisms are maintained in protected environments.

Printed taxonomical tools:
It consist of identification cards, description, field guides and manuals.

Question 16.
Differentiate Ehler’s – Danlos syndrome and stickler syndrome?
Answer:
Ehler’s – Danlos syndrome is the defect in the synthesis of collagen in the joints, heart valves, organ walls and arterial walls:
Stickler syndrome is a group of hereditary conditions affecting collagen and results in facial abnormalities.

Question 17.
Define Purkinje fibres?
Answer:
Two special cardiac muscle fibres originate from the auriculo ventricular node and are called the bundle of his which runs down into the interventricular septum and the fibres spread into the ventricles. These fibres are called the Purkinje fibres.

Question 18.
Draw the diagram of pelvic girdle with lower limb?
Answer:

Question 19.
What are the symptoms of acromegaly?
Answer:
Acromegaly is caused due to excessive secretion of growth hormone in adults. The symptoms of acromegaly are:

1. Overgrowth of hand bones, feet bones, jaw bones.
2. Malfunctioning of gonads.
3. Enlargement of viscera, tongue, lungs, heart, liver, splean and endocrine glands like thyroid, or adrenal glands.

PART – IV

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

Question 20.
Describe the digestive system of Lamipto mauritii?
Answer:
The digestive system of the earthworm consists of the alimentary canal and the digestive glands. The alimentary canal runs as a straight tube throughout the length of the body from the mouth to anus.

The mouth opens into the buccal cavity which occupies the 1s^stand 2^2nd segments. The buccal cavity leads into a thick muscular pharynx,which occupies the 3^rd and 4^th segments and is surrounded by the pharyngeal glands.

A small narrow tube, oesophagus lies in the 5^th segment and continues into a muscular gizzard in the 6^th segment. The gizzard helps in the grinding of soil particles and decaying leaves. Intestine starts from the 7^th segment and continues till the last segment.

The dorsal wall of the intestine is folded into the cavity as the typhlosole. This fold contains blood vessels and increases the absorptive area of the intestine. The imier epithelium consists of columnar cells and glandular cells. The alimentary canal opens to the exterior through the anus.

The ingested organic rich soil passes through the digestive tract where digestive enzymes breakdown complex food into smaller absorbable units. The simpler molecules are absorbed through the intestinal membrane and are utilized.

The undigested particles along with earth are passed out through the anus, as worm castings or vermicasts. The pharyngeal or salivary gland cells and the glandular cells of the intestine are supposed to be the digestive glands which secrete digestive enzymes for digestion of food.

[OR]

What are the effects of smoking?
Answer:
Today due to curiosity, excitement or adventure youngsters start to smoke and later get addicted to smoking. Research says about 80% of the lung cancer is due to cigarette smoking. Smoking is inhaling the smoke from burning tobacco. There are thousands of known chemicals which includes nicotine, tar, carbon monoxide, ammonia, sulphur-dioxide and even small quantities of arsenic.

Carbon monoxide and nicotine damage the cardiovascular system and tar damages the gaseous exchange system. Nicotine is the chemical that causes addiction and is a stimulant which makes the heart beat faster and the narrowing of blood vessels results in raised blood pressure and coronary heart diseases.

Presence of carbon monoxide reduces oxygen supply. Lung cancer, mouth cancer and larynx is more common in smokers than non – smokers. Smoking also causes cancer of the stomach, pancreas and bladder and lowers sperm count in men.

Smoking can cause lunb diseases by damaging the airways and alveoli and results in emphysema and chronic bronchitis. These two diseases along with asthma are often referred as Chronic Obstructive Pulmonary Disease (COPD).

When a person smokes, nearly 85% of the smoke released is inhaled by the smoker himself and others in the vicinity, called passive smokers, are also affected. Guidance or counselling should be done in such users to withdraw this habit.

Question 21.
Explain the structure of spinal cord?
Answer:
The spinal cord is a long, slender, cylindrical nervous tissue. It is protected by the vertebral column and surrounded by the three membranes as in the brain. The spinal cord that extends from the brain stem into the vertebral canal of the vertebral column up to the level of 1^st or 2^2nd lumbar vertebra.

So the nerve roots of the remaining nerves are greatly elongated to exit the vertebral column at their appropriate space. The thick bundle of elongated nerve . roots within the lower vertebral canal is called the cauda equina (horse’s tail) because of its appearance.

In the cross section of spinal cord, there are two indentations: the posterior median sulcus and the anterior median fissure. Although there might be slight variations, the cross section of spinal cord is generally the same throughout its length. In contrast to the brain, the grey matter in the spinal cord forms an inner butterfly shaped region surrounded by the outer white matter.

The grey matter consists of neuronal cell bodies and their dendrites, intemeurons and glial cells. White matter consists of bundles of nerve fibres. In the center of the grey matter there is a central canal which is filled with CSF. Each half of the grey matter is divided into a dorsal horn, a ventral hom and a lateral horn.

The dorsal horn contains cell bodies of interneurons on which afferent neurons terminate. The ventral hom contains cell bodies of the efferent motor neurons supplying the skeletal muscle. Autonomic nerve fibres, supplying cardiac and smooth muscles and exocrine glands, originate from the cell bodies found in the lateral horn.

In the white matter, the bundles of nerve fibres form two types of tracts namely ascending tracts which carry’ sensory impulses to the brain and descending tracts which carry motor impulses from the brain to the spinal nerves at various levels of the spinal cord. The spinal cord shows two enlargements,one in the cervical region and another one in the lumbosacral region. The cervical enlargement serves the upper limb and lumbar enlargement serves the lower limbs.

[OR]

What are the stages involved in rearing of chicken?
Answer:
Stages involved in rearing:
There are some steps involved in rearing of chicken.

1. Selection of the best layer:
An active intelligent looking bird, with a bright comb, not obese should be selected.

2. Selection of eggs for hatching:
Eggs should be selected very carefully. Eggs should be fertile, medium sized, dark brown shelled and freshly laid eggs are preferred for rearing. Eggs should be washed, cleaned and dried.

3. Incubation and hatching:
The maintenance of newly laid eggs in optimum condition till hatching is called incubation. The fully developed chick emerges out of egg after an incubation period of 21 – 22 days.

There are two types of incubation namely natural incubation and artificial incubation. In the natural incubation method, only a limited number of eggs can be incubated by a mother hen. In artificial incubation, more number of eggs can be incubated in a chamber (Incubator).

4. Brooding:
Caring and management of young chicks for 4 – 6 weeks immediately after hatching is called brooding. It can also be categorized into two types namely natural and artificial brooding.

5. Housing of Poultry:
To protect the poultry from sun, rain and predators it is necessary to provide housing to poultry. Poultry house should be moisture proof, rat proof and it should be easily cleanable and durable.

6. Poultry feeding:
The diet of chicks should contain adequate amount of water,carbohydrates, proteins, fats, vitamins and minerals.

Students can Download Tamil Nadu 11th Biology Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Biology Model Question Paper 3 English Medium

General Instructions:

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

Time: 3:00 Hours
Maximum Marks: 70

Bio – Botany [Maximum Marks: 35]

PART – 1

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
Identify the correct sequence regarding lytic cycle of viruses.
(i) Penetration
(ii) Adsorption
(iii) Assembly
(iv) Synthesis
(a) (ii), (i), (iv), (iii)
(b) (iii), (i), (ii), (iv)
(c) (ii), (iv), (i), (iii)
(d) (i), (iv), (ii), (iii)
Answer:
(a) (ii), (i), (iv), (iii)

Question 2.
Zygote meiosis is characterisitic of ………………………
(a) Marchantia
(b) Fucus
(c) Funaria
(d) Chlamydomonas
Answer:
(d) Chlamydomonas

Question 3.
Pinus roots are in symbiotic relationship with ……………………
(a) Blue green algae
(b) Mycorrhiza
(c) Euglena
(d) Rhizobium
Answer:
(b) Mycorrhiza

Question 4.
Climbers are also called as ……………………
(a) Herbs
(b) Trees
(c) Vines
(d) Shrubs
Answer:
(c) Vines

Question 5.
Arrangement of sepals and petals in flower bud is called ………………………
(a) Adhesion
(b) Aestivation
(c) Placentation
(d) Cohesion
Answer:
(b) Aestivation

Question 6.
Cell cycle was discovered by ………………………
(a) Singer & Nicholson
(b) Prevost & Dumans
(c) Schleider & Schwann
(d) Boveri
Answer:
(b) Prevost & Dumans

Question 7.
Parenchyma storing calcium carbonate crystals are called ……………………
(a) Leucoplasts
(b) Elaioplasts
(c) Idioblasts
(d) Chromoplasts
Answer:
(c) Idioblasts

Question 8.
Which of the following is a free – living bacterium?
(a) Rhizobium
(b) Clostridium
(c) Escherichia
(d) Cyanobacteria
Answer:
(b) Clostridium

PART – II

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

Question 9.
Distinguish between deoxyviruses and riboviruses?
Answer:

Deoxyviruses	Riboviruses
1. Viruses having DNA are called deoxyviruses. E.g. Animal viruses except HIV	1. Viruses having RNA are called riboviruses. E.g: Plant viruses except cauliflower mosaic virus (CMV)

Question 10.
What is plant morphology?
Answer:
Plant morphology also known as external morphology deals with the study of shape, size and structure of plants and their parts like (roots, stems, leaves, flowers, fruits and seeds).

Question 11.
Differentiate Regional Flora from continental flora?
Answer:

Regional Flora	Continental Flora
1. Flora covering a large geographical area or a botanical region Eg: flora of Madras Presidency.	1. Flora covering the entire continent. Eg: flora of Europaea.

Question 12.
Define C – Value?
Answer:
C-Value is the amount in picograms of DNA contained within a haploid nucleus.

Question 13.
Expand and Define TP?
Answer:
TP stands for Turgor pressure. When a plant cell is placed in pure water (hypotonie solution) the diffusion of water into the cell takes place by endosmosis. It creates a positive hydrostatic pressure on the rigid cell wall by the cell membrane.

Hence forth the pressure exerted by the ccli membrane towards the cell wall is Turgor Pressure (TP).

Question 14.
Mention the events of Photochernical phase of light reaction?
Answer:
Photocheinical Phase:

1. Photolysis of water and oxygen evolution
2. Electron transport and synthesis of assimilatory power.

PART – III

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

Question 15.
Name few fossil sites of India?
Answer:

Question 16.
What is a liana? Mention its importance?
Answer:
Liana is a vine that is perennial and woody. Lianas are major components in the tree canopy layer of some tropical forests. e.g., Ventilago.

Question 17.
Enumerate the steps ino1ed in herbarium preparation?
Answer:
Preparation of herbarium specimen includes the following steps.

1. Plant Collection
2. Documentation of field site data
3. Preparation of plant specimen
4. Mounting herbarium specimen
5. Herbarium labels
6. Protection of herbarium sheets against mold and insects

Question 18.
Draw and label the open vascular bundle?
Answer:

Question 19.
When does an essential mineral is considered as a “toxic”?
Answer:
Increase of mineral nutrients more than the normal concentration causes toxicity. A concentration, at which 10 % of the dry weight of tissue is reduced, is considered as toxic.

PART – IV

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

Question 20.
Tabulate the economic importance of algae?
Answer:

[OR]

Explain Datura metal in botanical terms. Draw floral diagram?
Answer:
1. Habit:
Large, erect and stout herb.

2. Root:
Branched tap root system.

3. Stem:
Stem is hollow, green and herbaceous with strong odour.

4. Leaf:
Simple, alternate, petiolate, entire or deeply lobed, glabrous exstipulate showing unicostate reticulate venation.

5. Inflorescence:
Solitary and axillary cyme.

6. Flower:
Flowers are large, greenish white, bracteate, ebracteolate, pedicellate, complete, heterochlamydeous, pentamerous, regular actinomorphic, bisexual and hypogyncas.

7. Calyx:
Sepals 5, green synsepalous showing valvate aestivation. Calyx is mostly persistant, odd sepal is posterior in position.

8. Corolla:
Petals 5, greenish white, sympetalous, plicate (folded like a fan) showing twisted aestivation, funnel shaped with wide mouth and 10 lobed.

9. Androecium:
Stamens 5, free from one another, epipetalous, altemipefalous and are inserted in the middle of the corolla tube. Anthers are basifixed, dithecous, with long filament, introse and longitudinally dehiscent.

10. Gynoecium:
Ovary bicarpellary, syncarpous superior ovary, basically biloculear but tetralocular due to the formation of false septum. Carpels are obliquely placed and ovules on swollen axile placentation. Style simple long and filiform, stigma two lobed.

11. Fruit:
Spinescent capsule opening by four special valves with persistent calyx.

12. Seed:
Endospermous

13. Floral Formula:

Question 21.
Describe the types of transpiration in plants?
Answer:
Types of Transpiration: Transpiration is of following three types:

1. Stomatal transpiration:
Stomata are microscopic structures present in high number on the lower epidermis of leaves. This is the most dominant form of transpiration and being responsible for most of the water loss (90 – 95%) in plants.

2. Lenticular transpiration:
In stems of woody plants and trees, the epidermis is replaced by periderm because of secondary growth. In order to provide gaseous exchange between the living cells and outer atmosphere, some pores which looks like lens – shaped raised spots are present on the surface of the stem called Lenticels. The loss of water from lenticels is very insignificant as it amounts to only 0.1% of the total.

3. Cuticular transpiration:
The cuticle is a waxy or resinous layer of cutin, a fatty substance covering the epidermis of leaves and other plant parts. Loss of water through cuticle is relatively small and it is only about 5 to 10 % of the total transpiration. The thickness of cuticle increases in xerophytes and transpiration is very much reduced or totally absent.

[OR]

Describe the concept of Phytochrome?
Answer:
Phytochrome is a bluish biliprotein pigment responsible for the perception of light in photo physiological process. But at (1959) named this pigment and it exists in two interconvertible forms:

(I) Red light absorbing pigment which is designated as P_r and

(II) Far red light absorbing pigment which is designated as P_fr. The P_fr form absorbs red light in 660 nm and changes to P_fr. The P_fr form absorbs far red light in 730 nm and changes to P_r.

The P_r form is biologically inactive and it is stable whereas Pfr form is biologically active and it is very unstable. In short day plants, P_r promotes flowering and P_fr inhibits the flowering whereas in long day plants flowering is promoted by P_fr and inhibited by P_r form.

P_fr is always associated with hydrophobic area of membrane systems while P_r is found in diffused state in the cytoplasm. The interconversion of the two forms of phytochrome is mainly involved in flower induction and also additionally plays a role in seed germination and changes in membrane conformation.

Bio – Zoology [Maximum Marks: 35]

PART – I

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
Entomology is concerned with the study of ……………………..
(a) Formation and properties of soil
(b) Agricultural practices
(c) Various aspects of human life
(d) Various aspects of insect
Answer:
(d) Various aspects of insect

Question 2.
Which is the site of production of blood cells?
(a) Cartilage
(b) Bone marrow
(c) Blood
(d) Plasma
Answer:
(b) Bone marrow

Question 3.
Match the List – I and List – II.

List – I	List – II
1. Lipase	(i) Protein
2. Pepsin	(ii) Lipid
3. Renin	(iii) Starch
4. Ptyalin	(iv) Cassein

(a) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)
(b) 1 – (i), 2 – (ii), 3 – (iii), 4 – (iv)
(c) 1 – (iv), 2 – (iii), 3 – (i), 4 – (ii)
(d) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)
Answer:
(a) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)

Question 4.
ESV stands for ……………………..
(a) Endocard Systolic Volume
(b) End Systolic Volume
(c) Endocard Sound Volume
(d) Endocard Sinoatrial Volume
Answer:
(b) End Systolic Volume

Question 5.
Which of the following statement is correct?
(a) The descending limb of loop of Henle is impermeable to water
(b) The ascending limb of loop of Henle is permeable to water
(c) The ascending limb of loop of Henle is impermeable to water
(d) The descending limb of loop of Henle is permeable to electrolytes
Answer:
(c) The ascending limb of loop of Henle is impermeable to water

Question 6.
Which of the following statements are incorrect?
(a) Basal epithelial cells are sensitive portions of the taste
(b) Basal epithelial cells are stem cells which divide and differentiate into new gustatory cells
(c) Gustatory hairs project form the tip of the gustatory cells
(d) Gustatory cells are sensory portion of the taste
Answer:
(a) Basal epithelial cells are sensitive portions of the taste

Question 7.
Serum Calcium level is regulated by …………………….
(a) Thyroxine
(b) FSH
(c) Pancreas Assertion is true, but reason is false
(d) Thyroid and Parathyroid
Answer:
(d) Thyroid and Parathyroid

Question 8.
Ca⁺ metabolism is regulated by …………………..
(a) ACTH
(b) Thyroxin
(c) Parathormone
(d) Epinephrine
Answer:
(a) ACTH

PART – II

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

Question 9.
What is Phylogenetic tree?
Answer:
A phylogenetic tree or evolutionary tree is a branching diagram or “tree”, showing the inferred evolutionary relationships upon similarities and differences in their physical or genetic characteristics.

Question 10.
Write about Marbled cone snail?
Answer:
Marbled Cone Snail (Conus marmoreus):
This cone – shaped snail can deliver dangerous venom which may result in vision loss, respiratory failure, muscle paralysis and eventually death. There is no anti – venom available.

Question 11.
Differences between male and female cockroach?
Answer:

Question 12.
Define Floating ribs?
Answer:
The last 11^th and 12^th pairs of ribs are not connected ventrally. Therefore, they are called as ‘floating ribs’ or vertebral ribs.

Question 13.
Write a note on Tetany?
Answer:
Tetany is caused due to the hyposecretion of parathyroid hormone (PTH). Due to hyposecretion of PTH serum calcium level decreases (Hypocalcemia), as a result serum phosphate level increases.

Calcium and phosphate excretion level decreses. Generalized convulsion, locking of jaws increased heart beat rate, increases body temperature, muscular spasm are the major symptoms of tetany.

Question 14.
Draw the diagram of large intestine?
Answer:

PART – III

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

Question 15.
What is the need for classification?
Answer:
The basic need for classification are:

1. To identify and differentiate closely related species.
2. To know the variation among the species.
3. To understand the evolution of the species.
4. To create a phylogenetic tree among the different groups.

Question 16.
Distinguish between Diploblastic and Triploblastic animals?
Answer:

S.No	Diploblastic animals	Triploblastic animals
1.	The animals in which the cells are arranged in two embryonic layers, the ectoderm and endoderm are called diploblastic animals.	The animals in which the cells are arranged in three embryonic layers, the ectoderm, mesoderm and endoderm are called triploblastic animals.
2.	These are lower organisms, eg. Cnidaria, Ctenophora	These are higher organisms, eg. Platyhelminthes to mammalia.

Question 17.
Write a short note on connective tissue?
Answer:

1. Connective tissue develops from the mesoderm.
2. Connective tissue proper, Cartilage, bones and blood are the four main classes of connective tissues.
3. Binding, support, protection, insulation and transportation of substances are the major functions of connective tissue.

Question 18.
Draw the diagram of Lamnito mauritii?
Answer:

Question 19.
Define cross breedin?
Answer:
Breeding between a superior male of one breed with a superior female of another breed is known as cross breeding.

PART – IV

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

Question 20.
Who proposed the five kingdom classification? Represent the Five kingdom Classification in a tabular format?
Answer:
R.H. Whittaker (1969) proposed the Five kingdom Classification, the Kingdoms defined by him were Monera, Prgtista, Fungi, Plantae, and Animalia based on the cell structure, mode of nutrition, mode of reproduction and phylogenetic relationships. The table below gives a, comparative account of different characteristics of the five kingdoms.

[OR]

Explain the mechanism of breathing?
Answer:
The movement of air between the atmosphere and the lungs is known as ventilation or breathing. Inspiration and expiration are the two phases of breathing. Inspiration is the movement of atmospheric air into the lungs and expiration is the movement of alveolar air that diffuse out of the lungs.

Lungs do not contain muscle fibres but expands and contracts by the movement of the ribs and diaphragm. The diaphragm is a sheet of tissue which separates the thorax from the abdomen. In a relaxed state, the diaphragm is domed shaped.

Ribs are moved by the intercostal muscles. External and internal intercostal muscles found between the ribs and the diaphragm helps in creating pressure gradients.

Inspiration occurs if the pressure inside the lungs (intrapulmonary pressure) is less than the atmospheric pressure likewise expiration takes place when the pressure within the lungs is higher than the atmospheric pressure.

Inspiraton is initiated by the contraction of the diaphragm muscles and external intercostal muscles, which pulls the ribs and sternum upwards and outwards and increases the volume of the thoracic chamber in the dorso – ventral axis, forcing the lungs to expand the pulmonary volume.

The increase in pulmonary volume and decrease in the intrapulmonary pressure forces the fresh air from outside to enter the air passages into the lungs to equalize the pressure. This process is called inspiration.

Relaxation of the diaphragm allows the diaphragm and sternum to return to its dome shape and the internal intercostal muscles contract, pulling the ribs downward reducing the thoracic volume and pulmonary volume.

This results in an increase in the intrapulmonary pressure slightly above the atmospheric pressure causing the expulsion of air from the lungs. This process is called expiration.

Question 21.
What are the benefits of regular exercise?
Answer:
Regular exercises can produce the following beneficial physiological changes:

1. The muscles used in exercise grow larger and strongcr
2. More enzymes are synthesized in the muscle fibre
3. Ligaments and tendons become stronger
4. Influences hormonal &tivity
5. Prcvcnts Obesity
6. Aesthetically bettcr with good physique
7. Over all well-bcing with good quality of life
8. Prevents depression, stress and anxiety
9. The resting heart rate goes down
10. Joints become more flexible
11. Protection from hcart attack
12. Improves cognitive functions
13. Promotes confidence, esteem

During muscular exercise, there is an increase in metabolism. The O₂ need of the muscles is increased. This requirement is met with more oxygen rich RBCs available to the active sites. There is an increase in heart rate and cardiac output. Along with balanced diet, physical activity plays a significant role in strengthening the muscles and bones.

[OR]

What are the types of chicken breeds in Poultry Farming?
Answer:
There are more than 100 breeds. The commonly farmed chicken breeds are categorized into five based on the purpose for which it is farmed. They are egg layers, broiler type, dual type, games and ornamental types.

1. Egg layers: These are farmed mainly for the production of egg.

Leghorn:
This is the most popular commercial breed in India and originated from Italy. They are small, compact with a single comb and wattles with white, brown or black colour.

They mature early and begin to lay eggs at the age of 5 or 6 months. Hence these are preferred in commercial farms. They can also thrive well in dry areas.

Chittagong:
It is the breed chiefly found in West Bengal. They are golden or light yellow coloured. The beak is long and yellow in colour. Ear lobes and wattles are small and red in colour. They are good egg layers and are delicious.

2. Broiler type: These are well known for fast growth and soft quality meat.

White Plymouth rock:
They have white plumage throughout the body. It is commonly used in broiler production. This is an American breed. It is a fast growing breed and well suitable for growing intensively in confined farms.

3. Dual purpose breeds: These are for both meat and egg production purpose.

Brahma:
It is a breed popularly known for its massive body having heavy bones, well feathered and proportionate body. Pea comb is one of the important breed characters. It has two common varieties namely, Light Brahma and Dark Brahma.

4. Game breeds: Since ancient times, special breed of roosters have been used for the sport of cockfighting.

Aseel:
This breed is white or black in colour. The hens are not good egg layers but are good in incubation of eggs. It is found in all states of India. Aseel is noted for its pugnacity, high stamina, and majestic gait and dogged fighting qualities. Although poor in productivity, this breed is well-known for their meat qualities.

5. Ornamental breeds: Ornamental chicken are reared as pets in addition to their use for egg production and meat.

Silkie:
It is a breed of chicken has a typical fluffy plumage, which is said to feel like silk and satin. The breed has numerous additional special characters, such as black skin and bones, blue earlobes, and five toes on each foot, while the majority chickens only have four.

They are exhibited in poultry shows, and come out in various colours. Silkies are well recognized for their calm, friendly temperament. Silkie chicken is especially simple to maintain as pets.

Students can Download Tamil Nadu 11th Biology Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Biology Model Question Paper 1 English Medium

General Instructions:

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

Time: 3:00 Hours
Maximum Marks: 70

Bio – Botany [Maximum Marks: 35]

PART – 1

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
Syphilis is caused by …………………..
(a) Mycococcus candisans
(b) Treponema pallidum
(c) Yersinia pestis
(d) Mycobacterium leprae
Answer:
(b) Treponema pallidum

Question 2.
Who is called as the Father of Indian Phycology?
(a) M.O. Parthasarathy
(b) Y. Bharadwaja
(c) V.S. Sundaralingam
(d) V. Desikachary
Answer:
(a) M.O. Parthasarathy

Question 3.
Which of the following plant possess sessile leaves?
(a) Hibiscus
(b) Mangifera
(c) Psidium
(d) Gloriosa
Answer:
(d) Gloriosa

Question 4.
Histone proteins are seen in the DNA of …………………..
(a) Pseudokaryotes
(b) Prokaryotes
(c) Mesokaryotes
(d) Eukaryotes
Answer:
(d) Eukaryotes

Question 5.
Number of fatty acids in triglyceride is ……………………
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 6.
Principle behind the desalination of sea water is ……………………
(a) Endosmosis
(b) Diffusion
(c) Reverse osmosis
(d) Deplasmolysis
Answer:
(c) Reverse osmosis

Question 7.
In Emerson’s first effect, the photosynthetic yield was dropped in the region above ……………………
(a) 720 nm
(b) 620 nm
(c) 680 nm
(d) 600 nm
Answer:
(d) 600 nm

Question 8.
Which of the following plant hormone functions against auxin?
(a) Gibberellin
(b) Cytokinin
(c) Ethylene
(d) Abscissic acid
Answer:
(d) Abscissic acid

PART – II

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

Question 9.
What is Porin? How it helps the bacteria?
Answer:
Porin is an abundant polypeptide present in bacterial cell walls. It helps in the diffusion of solutes.

Question 10.
Mention various types of stem seen in angiosperms?
Answer:
Majority of angiosperm possess upright, vertically growing erect stem. They are

1. Excurrent
2. Decurrent
3. Caudex and
4. Culm.

Question 11.
How will you define inflorescence?
Answer:
An inflorescence is a group of flowers arising from a branched or unbranched axis with a definite pattern.

Question 12.
List out the disadvantages of Amitosis?
Answer:

1. Causes unequal distribution of chromosomes.
2. Can lead to abnormalities in metabolism and reproduction.

Question 13.
State Relay Pump theory?
Relay pump theory of Godlewski (1884)
Answer:
Periodic changes in osmotic pressure of living cells of the xylem parenchyma and medullary ray act as a pump for the movement of water.

Question 14.
Define anaerobic photosynthesis?
Answer:
In some bacteria, oxygen is not evolved and is called as non-oxygenic and anaerobic photosynthesis. Examples: Green sulphur, Purple sulphur and green filamentous bacteria.

PART – III

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

Question 15.
Who is called as founder of modern bacteriology? Mention his contribution?
Answer:
Robert Heinrich Hermann Koch is considered as the founder of modem bacteriology. He identified the causal organism for Anthrax, Cholera and Tuberculosis. He proved experimental evidence for the concept of infection (Koch’s postulates).

Question 16.
List down the key difference between roots and shoots?
Answer:

Roots	Shoots
1. Positively geotropic	1. Negatively geotropic
2. Negatively phototropic	2. Positively phototropic
3. Non-green in colour	3. Green in colour
4. Nodes, intemodes and buds are absent.	4. Nodes, intemodes and buds are present

Question 17.
Differentiate between Taxonomy & Systematics?
Answer:

Taxonomy	Systematics
1. Discipline of classifying organisms into taxa.	1. Broad field of biology that studies the diversification of species.
2. Governs the practices of naming, describing, identifying and specimen preservation.	2. Governs the evolutionary history and phylogenetic relationship in addition to taxonomy
3. Classification + Nomenclature = Taxonomy	3. Taxonomy + Phylogeny = Systematics

Question 18.
What do you mean by Phloem loading?
Answer:
The movement of photosynthates (products of photosynthesis) from mesophyll cells to phloem sieve elements of mature leaves is known as phloem loading.

Question 19.
Define Dark Reaction?
Answer:
Fixation and reduction of CO₂ into carbohydrates with the help of assimilatory power produced during light reaction. This reaction does not require light and is not directly light driven. Hence, it is called as Dark reaction or Calvin-Benson cycle.

PART – IV

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

Question 20.
List out the salient features of Basidiomycetes?
Answer:

• Basidiomycetes include puffballs, toad stools, Bird nest’s fungi, Bracket fungi, stink horns, rusts and smuts.
• The members are terrestrial and lead a saprophytic and parasitic mode of life.
• The mycelium is well developed, septate with dolipore septum (bracket like). Three types of mycelium namely primary (Monokaryotic), secondary (Dikaryotic) and tertiary are found.
• Clamp connections are formed to maintain dikaryotic condition.
• Asexual reproduction is by means of conidia, oidia or budding.
• Sexual reproduction is present but sex organs are absent. Somatogamy or spermatisation results in plasmogamy. Karyogamy is delayed and – dikaryotic phase is prolonged. Karyogamy takes place in basidium and it is immediately followed by meiotic division.
• The four nuclei thus formed are transformed into basidiospores which are borne on sterigmata outside the basidium (Exogenous). The basidium is club shaped with four basidiospores, thus this group of fungi is popularly called “Club fungi”. The fruit body formed is called Basidiocarp.

[OR]

List out the general characters of Pteridophytes?
Answer:
General characteristic features of Pteridophytes:

1. Plant body is sporophyte (2n) and it is the dominant phase. It is differentiated into root, stem and leaves.
2. Roots are adventitious.
3. Stem shows monopodial or dichotomous branching.
4. Leaves may be microphyllous or megaphyllous.
5. Stele is protostele but in some forms siphonostele is present (Marsilea)
6. Tracheids are the major water conducting elements but in Selaginella vessels are found.
7. Sporangia, spore bearing bag like structures are borne on special leaves called sporophyll. The sporophylls gets organized to form cone or strobilus. e.g., Selaginella and Equisetum.
8. They may be homosporous (produce one type of spores – Lycopodium) or Heterosporous (produce two types of spores -Selaginella). Heterospory is the origin for seed habit.
9. Development of sporangia may be eusporangiate (development of sporangium from group of initials) or leptosporangiate (development of sporangium from single initial).
10. Spore mother cells undergo meiosis and produce spores (n).
11. Spore germinates to produce haploid, multicellular green, cordate shaped independent gametophytes called prothallus.
12. Fragmentation, resting buds, root tubers and adventitious buds help in vegetative reproduction.
13. Sexual reproduction is Oogamous. Sex organs, namely antheridium and archegonium are produced on the prothallus.
14. Antheridium produces spirally coiled and multiflagellate antherozoids.
15. Archegonium is flask shaped with broad venter and elongated narrow neck. The venter possesses egg or ovum and neck contain neck canal cells.
16. Water is essential for fertilization. After fertilization a diploid zygote is formed and undergoes mitotic division to form embryo.
17. Pteridophytes show apogamy and apospory.

Question 21.
Draw a flow chart of Kreb’s cycle?
Answer:

[OR]

Give a detailed account on geometric growth rate?
Answer:
This growth occurs in many higher plants and plant organs and is measured in size or weight. In plant growth, geometric cell division results if all cells of an organism or tissue are active mitotically. Example: Round three in the given figure 15.5, produces 8 cells as 23 58 and after round 20 there are 220 5 1,048,576 cells.

The large plant or animal parts are produced this way. In fact, it is common in animals but rare in plants except when they are young and small. Exponential growth curve can be expressed as,

W₁ = W₀e^rt
W₁ = Final size at the beginning of the period
W₀ = Initial size at the beginning of the period
r = Growth rate
t = Time of growth
e = Base of the natural logarithms

Here ‘r’ is the relative rate and also a measure of the ability of the plant to produce new plant material, reffered to as efficiency index. Hence, the final size of W₁ depends on the initial size W₀.

Bio – Zoology [Maximum marks: 35]

PART – 1

Answer all the questions. Choose the correct answer. [8 × 1 = 8]

Question 1.
Which class of protozoa is totally parasitic …………………….
(a) Sporozoa
(b) Mastigophora
(c) Ciliate
(d) Sarcodina
Answer:
(a) Sporozoa

Question 2.
Medusa is the Reproductive organs of ……………………..
(a) Hydra
(b) Aurelia
(c) Obelia
(d) Sea anemone
Answer:
(b) Aurelia

Question 3.
Match the List I and List II.

List I	List II
1. Ball and socket	(i) Knee
2. Hinge	(ii) Humerous and pectoral of girdle
3. Pivot	(iii) Carpal and metacarpal of thumb
4. Saddle	(iv) Atlas and axis

(a) 1 – (i), 2 – (ii), 3 – (iii), 4 – (iv)
(b) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)
(c) 1 – (iv), 2 – (iii), 3 – (i), (4) – (ii)
(d) 1 – (iii), 2 – (iv), 3 – (ii), 4 – (i)
Answer:
(b) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)

Question 4.
LH – stands for ………………………
(a) Luteinising hormone
(b) Langerhans hormome
(c) Low secretion hormone
(d) Luteotrophic hormone
Answer:
(a) Luteinising hormone

Question 5.
Which of the following statement is correct?
(a) Calcitonin and thymosin are thyroid hormones.
(b) Pepsin and prolactin are selected in stomach.
(c) Secretin and rhodopsin are polypeptide hormones.
(d) Cortisol and aldosterone are steroid hormones.
Answer:
(d) Cortisol and aldosterone are steroid hormones.

Question 6.
Which of the following statement is incorrect regard to species …………………….
(a) They have similar morphological
(b) They are reproductively isolated
(c) They produce viable young ones
(d) They have similar anatomical features
Answer:
(b) They are reproductively isolated

Question 7.
The kidney of adult mammals is ……………………
(a) Opisthonephron
(b) Pronephros
(c) Mesonephros
(d) Metanephros
Answer:
(d) Metanephros

Question 8.
Deficiency of calciferol causes ………………………
(a) Scurvy
(b) Leucopenia
(c) Leukaemia
(d) Rickets
Answer:
(d) Rickets

PART – II

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

Question 9.
Why mule is sterile in nature?
Answer:
Mule gets one set of chromosomes (32) from male parent, horse and one set of chromosomes (31) from female parent, donkey. These two sets of chromosomes do not match with each other and cannot produce gametes by meiosis. Hence mule is sterile in nature.

Question 10.
Define macrophages?
Answer:
Macrophages are Immune cells derived from monocytes; engaged in phagocytosis of microbes and debris.

Question 11.
Differentiate between peristomium and prostomium in earthworm?
Answer:

Peristomium	Prostomium
1. The first segment of the body of earthworm is called peristomium.	1. A small flap overhanging the mouth is called prostomium or upper lip.

Question 12.
Draw the diagram of structure of alveoil?
Answer:

Question 13.
How is urea formed in the human body?
Answer:
More toxic ammonia produced as a result of breakdown of amino acids is converted into less toxic urea in the liver by a cyclic process called Ornithine cycle.

Question 14.
Write the symptoms of cretinism?
Answer:
Cretinism is caused due to hypothyroidism in infants. A cretin child shows the following symptoms

1. Retarded skeletal growth
2. Absence of sexual maturity
3. Retarded mental ability
4. Thick and short limbs
5. Thick wrinkled skin
6. Bloated face
7. Protruded enlarged tongue
8. Low BMR, slow pulse rate, subnormal body temperature and elevated blood cholesterol levels

PART – III

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

Question 15.
What is Canal system?
Answer:
The water transport system in sponges through which water enters through minute pores and goes out through the large opening called osculum. It helps is nutrition, circulation, respiration and excretion.

Question 16.
Draw the diagram of structure of the heart?
Answer:

Question 17.
What are spinal nerves?
Answer:
The 31 pairs of nerves which emerge out from the spinal cord through spaces called the intevertebral foramina found between the adjacent vertebrae are the spinal nerves.

Question 18.
Classify organisms on the basis of seven kingdom system?
Answer:

Question 19.
Distinguish between exocrine glands and endocrine glands?
Answer:

Exocrine glands	Endocrine glands
1. These glands release their products through ducts.	1. These are ductless gland and their secretions are released directly into the blood,
2. These secrete mucous, saliva, ear wax, oil, milk, digestive enzyme etc. e.g. Salivary glands.	2. These secrete hormones, e.g. Pituitary gland.

PART – IV

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

Question 20.
Explain the conditions which creates problems in oxygen transport?
Answer:
When a person travels quickly from sea level to elevations above 8000 ft, where the atmospheric pressure and partial pressure of oxygen are lowered, the individual responds with symptoms of acute mountain sickness (AMS)- headache, shortness of breath, nausea and dizziness due to poor binding of O₂ with haemoglobin.

When the person moves on a long – term basis to mountains from sea level is body begins to make respiratory and haematopoietic adjustments. To overcome this situation kidneys accelerate production of the hormone erythropoietin, which stimulates the bone marrow to produce more RBCs.

When a person descends deep into the sea, the pressure in the surrounding water increases which causes the lungs to decrease in volume.

This decrease in volume increases the partial pressure of the gases within the lungs. This effect can be beneficial, because it tends to drive additional oxygen into the circulation, but this benefit also has a risk, the increased pressure can also drive nitrogen gas into the circulation.

This increase in blood nitrogen content can lead to a condition called nitrogen narcosis. When the diver ascends to the surface too quickly a condition called ‘bends’ or decompression sickness occurs and nitrogen comes out of solution while still in the blood forming bubbles.

Small bubbles in the blood are not harmful, but large bubbles can lodge in small capillaries, blocking blood flow or can press on nerve endings. Decompression sickness is associated with pain in joints and muscles and neurological problems including stroke. The risk of nitrogen narcosis and bends is common in scuba divers.

During carbon-dioxide poisoning, the demand for oxygen increases. As the O₂ level in the blood decreases it leads to suffocation and the skin turns bluish black.

[OR]

What is an epithelium? Enumerate the characteristic features of different epithelia?
Answer:
Simple epithelium is a simple layered sheet of cells that covers the body surface or lines the body cavity.
Types:

1. Squamous epithelium: It is made of flattened cells with irregular boundaries. It is found in glomeruli, air sacs of lungs, lining of heart, blood vessels.
2. Cuboidal epithelium: It is made of cube like cells. It is found in kidney tubules, ducts and glands. It is important for secretion and absorption.
3. Columnar epithelium: It is made of column like cells. It lines the digestive tract. It is important for secretion and absorption.
4. Ciliated epithelium: It has cilia at the free end. It is found in bronchi, uterine tubes. It is helpful in propelling materials.
5. Glandular epithelium: Cuboidal or columnar epithelium specialized for secretion is called glandular epithelium. E.g., goblet cells and salivary gland.

Compound epithelium:

1. Compound epithelium is made up of multilayered cells.
2. These protect organs against chemical and mechanical stresses.
3. These cover the dry surface of the skin, moist surface of the buccal cavity, pharynx, inner lining of ducts of salivary glands and pancreatic ducts.

Classification of Compound epithelium:

Question 21.
Write a detailed account of gastro intestinal tract hormones?
Answer:
Group of specialized endocrine cells present in gastro – intestinal tract secretes hormones such as gastrin, cholecystokinin (CCK), secretin and gastric inhibitory peptides (GIP). Gastrin acts on the gastric glands and stimulates the secretion of HCl and pepsinogen. Cholecystokinin (CCK) is secreted by duodenum in response to the presence of fat and acid in the diet.

It acts on the gall bladder to release bile into duodenum and stimulates the secretion of pancreatic enzymes and its discharge. Secretin acts on acini cells of pancreas to secrete bicarbonate ions and water to neutralize the acidity. Gastric inhibitory peptide (GIP) inhibits gastric secretion and motility.

[OR]

Discuss the various techniques adopted in cattle breeding?
Answer:
Methods of Animal breeding:
There are two methods of animal breeding, namely inbreeding and outbreeding

1. Inbreeding:
Breeding between animals of the same breed for 4-6 generations is called inbreeding. Inbreeding increases homozygosity and exposes the harmful recessive genes. Continuous inbreeding reduces fertility and even productivity, resulting in “inbreeding depression”.

This can be avoided by breeding selected animals of the breeding population and they should be mated with superior animals of the same breed but unrelated to the breeding population. It helps to restore fertility and yield.

2. Outbreeding:
The breeding between unrelated animals is called outbreeding. Individuals produced do not have common ancestors for 4-6 generations.

Outbreeding helps to produce new and favourable traits, to produce hybrids with superior qualities and helps to create new breeds. New and favourable genes can be introduced into a population through outbreeding.

(I) Out crossing:
It is the breeding between unrelated animals of the same breed but having no common ancestry. The offspring of such a cross is called outcross. This method is suitable for breeding animals that are below average in productivity.

(II) Cross breeding:
Breeding between a superior male of one breed with a superior female of another breed. The cross bred progeny has superior traits (hybrid vigour or heterosis.)

(III) Interspecific hybridization:
In this method of breeding mating is between male and female of two different species. The progeny obtained from such crosses are different from their parents, and may possess the desirable traits of the parents. Mule was produced by the process of interspecific hybridization between a male donkey and a female horse.

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 5 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 5 English Medium

General Instructions:

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

Time: 2:30 Hours
Maximum Marks: 70

PART – 1

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
The equivalent mass of ferrous oxalate is …………………….
(a)
(b)
(c)
(d) None of these
Solution:

n = 1 + 2(1) = 3
Answer:
(c)

Question 2.
Time independent Schnodinger wave equation is ………………….
(a) \(\widehat{\mathrm{H}} \psi=\mathrm{E} \psi\)
(b) \(\nabla^{2} \psi+\frac{8 \pi^{2} m}{h^{2}}(E+V) \psi=0\)
(c) \(\frac{\partial^{2} \Psi}{\partial x^{2}}+\frac{\partial^{2} \Psi}{\partial y^{2}}+\frac{\partial^{2} \Psi}{\partial z^{2}}+\frac{2 m}{h^{2}}(E-V) \psi=0\)
(d) All of these
Answer:
(a) \(\widehat{\mathrm{H}} \psi=\mathrm{E} \psi\)

Question 3.
Choose the incorrect statement.
(а) The chemical symbol of nickel is Ni.
(b) An element is a material made up of different kind of atoms.
(c) The physical state of bromine is liquid.
(d) The physical and chemical properties of the elements are periodic functions of their atomic numbers.
Answer:
(b) An element is a material made up of different kind of atoms.

Question 4.
Assertion: Permanent hardness of water is removed by treatment with washing soda.
Reason: Washing soda reacts with soluble calcium and magnesium chlorides and sulphates in hard water to form insoluble carbonates.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false
(d) Both assertion and reason are false
Solution:
Ca²⁺ + Na₂CO₃ → CaCO₃↓+ 2Na⁺
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.

Question 5.
Match the following:

Answer:

Question 6.
Compressibility factor for CO₂ at 400 K and 71.0 bar is 0.8697. The molar volume of CO₂under these conditions is …………………………
(a) 22.04 dm³
(b) 2.24 dm³
(c) 0.41 dm³
(d) 19.5 dm³
Solution:
Compressibility factor (z) = \(\frac{PV}{nRT}\)
V = \(\frac{z × nRT}{p}\) = \(\frac{0.8697 \times 1 \times 8.314 \times 10^{-2} \mathrm{bar} \mathrm{dm}^{3} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 400 \mathrm{K}}{71 \mathrm{bar}}\)
V = 0.41 dm³
Answer:
(c) 0.41 dm³

Question 7.
Thermodynamics is applicable to …………………….
(a) Macroscopic system only
(b) Microscopic system only
(c) Homogenous system only
(d) Heterogeneous system only
Answer:
(a) Macroscopic system only

Question 8.
In the reaction, Fe(OH)₃ (s) ⇄ Fe³⁺ (aq) + 3OH^–(aq), if the concentration of OH^– ions is decreased by \(\frac{1}{4}\) times, then the equilibrium concentration of Fe³⁺ will …………………….
(a) Not changed
(b) Also decreased by \(\frac{1}{4}\) times
(c) Increase by 4 times
(d) Increase by 64 times
Solution:

When concentration of OH^– ions declared by \(\frac{1}{4}\) times, then

To maintain K_C as constant, concentration of Fe³⁺ will increase by 64 times.
Answer:
(d) Increase by 64 times

Question 9.
The degree of dissociation a is equal to ……………………
(a) \(\frac{i-1}{n-1}\)
(b) \(\frac{(1-i)n}{n-1}\)
(c) \(\frac{i+1}{n+1}\)
(d) \(\frac{(1+i)n}{n+1}\)
Answer:
(a) \(\frac{i-1}{n-1}\)

Question 10.
The molecules having same hybridisation, shape and number of lone pair of electrons are ………………………..
(a) SeF₄, XeO₂F₂
(b) SF₄, XeF₂
(c) XeOF₄, TeF₄
(d) SeCl₄, XeF₄
Solution:
SeF₄, XeO₂F₂ – sp³d hybridisation
T-shaped, one lone pair on central atom.
Answer:
(a) SeF₄, XeO₂F₂

Question 11.
Connect pair of compounds which give blue colouration/precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
(a) NH₂NH₂HCl and ClCH₂ – CHO
(b) NH₂CS NH₂ and CH₃ – CH₂Cl
(c) NH₂CH₂ COOH and NH₂ CONH₂
(d) C₆H₅NH₂ and ClCH₂ – CHO
Answer:
(d) C₆H₅NH₂ and ClCH₂ – CHO

Question 12.
4-hydroxy phenol oxidised in the presence of K₂Cr₂O₇?H⁺ to give, …………………….
(a) Quinol
(b) Quinone
(c) Diol
(d) Alkane
Answer:
(b) Quinone

Question 13.
Methane gas is also called as ……………………
(a) Marsh gas
(b) Mass gas
(c) Molecular gas
(d) Model gas
Answer:
(a) Marsh gas

Question 14.
The catalyst used in Darzen halogenation of alcohol is ……………………..
(a) CCl₄
(b) Acetone
(c) Pyridine
(d) Ethene
Answer:
(c) Pyridine

Question 15.
Identify the wrong statement in the following.
(a) The clean water would have a BOD value of more than 5 ppm
(b) Greenhouse effect is also called as Global warming
(c) Minute solid particles in air is known as particulate pollutants
(d) Biosphere is the protective blanket of gases surrounding the earth
Answer:
(a) The clean water would have a BOD value of more than 5 ppm

PART – II

Answer any six questions in which question No. 20 is compulsory. [6 × 2 = 12]

Question 16.
First ionization potential of C-atom is greater than that of B-atom, where as the reverse is true for second ionization potential?
Answer:
G (Z = 6) 1s² 2s² 2p_x‘ 2p_y‘. The electron removal from p orbital is very difficult. So carbon has highest first ionization potential.
B (Z = 5) 1s² 2s² 2p¹. In boron nuclear charge is less than that of carbon, so boron has lowest first ionization potential.
I.E₁ C > I.E₁B

But it is reverse in the case of second ionization energy. Because in case of B⁺ the electronic configuration is 1s² 2s², which is completely filled and it has high ionization energy. But in C⁺ the electronic configuration is 1s² 2s² 2p¹, one electron removal is easy so it has low ionization energy.

Question 17.
Why H₂O₂ is used as mild antiseptic?
Answer:
The oxidising property of hydrogen peroxide and harmless nature of its products such as water and oxygen, leads to oxidation of pollutants in water and act as a mild antiseptic.

Question 18.
An alkali metal (x) forms a hydrated sulphate, X₂SO₄.10H₂O. Is the metal more likely to be sodium (or) potassium?
Answer:
X forms X₂SO₄. 10H₂O. The metal is more likely be sodium. So X is Na₂SO₄. 10H₂O. It is otherwise called as Glauber’s salt.

Question 19.
Define Zeroth law of thermodynamics (or) Law of thermal equilibrium?
Answer:
Zeroth law of thermodynamics states that ‘If two systems at different temperatures are separately in thermal equilibrium with a third one, then they tend to be in thermal equilibrium with themselves’.

Question 20.
2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of glucose and water?
Solution:
Mass of glucose = 2.82 g
No. of moles of glucose = \(\frac{2.82}{180}\) = 0.016
Mass of water = 30g = \(\frac{20}{18}\) = 1.67
x_H2O = \(\frac{1.67}{1.67+0.016}\) = \(\frac{1.67}{1.686}\) = 0.99
∴x_H2O + xx_glucose = 1
0.99 + x_glucose = 1
x_glucose = 1 – 0.99 = 0.01

Question 21.
Considering x-axis as molecular axis, which out of the following will form a sigma bond.

1. 1s and 2p_y
2. 2p_x and 2p_x
3. 2p_x and 2p_y
4. 1s and 2p_z

Answer:
Along X-axis as molecular axis, only 2p_x and 2p_y can form a sigma bond

2p_x + 2p_y and 1s and 2p₂ also cannot form σ bond.

Question 22.
What happen when nitrile undergoes acid hydrolysis?
Answer:
When alkyl nitrile undergoes acid hydrolysis to give amide, which on further hydrolysis to give carboxylic acid.

Question 23.
How ozone reacts with 2-methyl propene?
Answer:

Question 24.
What are Freons? Discuss their uses and environmental effects?
Answer:
Freons are the chlorofluoro derivatives of methane and ethane.
Freon is represented as Freon – cba
Where, c – number of carbon atoms, b = number of hydrogen atoms,
a = total number of fluorine atoms.

Uses of Freons:

1. Freons are used as refrigerants in refrigerators and air conditioners.
2. It is used as a propellant for aerosols and foams.
3. It is used as propellant for foams, to spray out deodorants, shaving creams and insecticides.

Environmental effects of Freons:

1. Freon gas is a very powerful greenhouse gas which means that it traps the heat normally radiated from the earth out into the space. This causes the earth’s temperature to increase, resulting in rising sea levels, droughts, stronger storms, flash floods and a host of other very unpleasant effect. ,

2. As freon moves throughout the air, its chemical ingredients causes depletion of ozone layer. Depletion of ozone increases the amount of ultraviolet radiations that reaches the earths surface, resulting in serious risk to human health. High levels of ozone, in turn, causes respiratory problems and can also kill plants.

PART – III

Answer any six questions in which question No. 26 is compulsory. [6 × 3 = 18]

Question 25.
The first ionisation energy (IE₁) and second ionisation energy (IE₂) of elements X, Y and Z are given below.

Which one of the above elements is the most reactive metal, the least reactive metal and a noble gas?
Answer:
Noble gases: Ioniation energy ranging from 2372 KJ mol^-1 to 1037 kJ mol^-1.
For element X, the IE₁, value is in the range of noble gas, moreover for this element both IE₁ and IE₂ are higher and hence X is the noble gas.

For Y, the first ionisation energy is low and second ionisation energy is very high and hence Y is most reactive metal.
For Z, both IE₁ and IE₂ are higher and hence it is least reactive.

Question 26.
An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a modulator in nuclear reaction. (A) adds on to a compound (C), which has the molecular formula C₃H₆ to give (D). Identify A, B, C and D?
Answer:
(I) An isotope of hydrogen Deuterium (A) reacts with diatomic molecule of element belongs to group number 16 and period number 2 oxygen O₂ to give a compound (B) which is heavy water D₂O. D₂O is used as a moderator in nuclear reaction:

(II) Deuterium reacts with C₃H₆ propane (C) to give Deutero propane C₂D₆ (D).

Question 27.
Give the uses of gypsum?
Answer:

1. The Alabaster variety qf gypsum was used by the sculptors.
2. Gypsum is used in making drywalls or plaster boards.
3. Gypsum is used in the production of Plaster of Paris, which is used as a sculpting material.
4. Gypsum is used in making surgical and orthopedic casts.
5. It plays an important role in agriculture as a soil additive, conditioner and fertilizer.
6. Gypsum is used in toothpaste, shampoo and hair products.
7. Calcium sulphate acts as a coagulator in making tofu.
8. It is also used in baking as a dough conditioner.
9. Gypsum is a component of Portland cement, where it acts as a hardening retarder to control the speed at which concrete sets.
10. Gypsum is used to give colour to cosmetics and drugs.
11. Gypsum plays a very important role in wine making.

Question 28.
Define inversion temperature?
Answer:
The temperature below which a gas obey Joule-Thomson effect is called inversion temperature (T_i).
T_i = \(\frac{2a}{2Rb}\)

Question 29.
The vapour pressure of pure benzene (C₆H₆) at a given temperature is 640 mm Hg. 2.2g , of non-volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Answer:

Question 30.
In CH₄, NH₃ and H₂O, the central atom undergoes sp³ hybridisation-yet their bond angles are different, why?
Answer:

1. In CH₄, NH₃ and H₂O the central atom undergoes sp³ hybridisation. But their bond angles are different due to .the presence of lone pair of electrons.
2. It can be explained by VSEPR theory. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs are not same.
3. Bond pair – Bond pair < Bond pair – Lone pair < Lone pair -Lone pair
   So due to the varying repulsive force the bond pairs and lone pairs are distorted from regular geometry and organise themselves in such a way that repulsion will be minimum and stability will be maximum.
4. In case of CH₄, there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry, i.e., tetrahedral with bond angle = 109° 28’.
5. H₂O has 2 bond pairs and 2 lone pairs. There is large repulsion between lp – lp. Again repulsion between lp – bp is more than that of 2 bond pairs. So 2 bonds are more restricted to form inverted V shape (eg;) bent shape molecule with a bond angle of 104° 35.
6. NH₃ has 3 bond pairs and 1 lone pair. There is repulsion between lp – bp. So 3 bonds are. more restricted to form pyramidal shape with bond angle equal to 107° 18’.

Question 31.
How does hyper conjugation effect explain the stability of alkenes?
Answer:

1. The relative stability of various classes of carbonium ions may be explained by the number of no bond resonance structures that can be written for them.
2. Such structures are arrived by shifting the bonding electrons from an adjacent C – H bond to the electron deficient carbon.
3. In this way, the positive charge originally on carbon is dispersed to the hydrogen. This manner of electron release by assuming no bond character in the adjacent C – H bond is called hyper conjugation or Baker – Nathan effect.
4. The greater the hyper conjugation, the greater will be the stability of the compound.
   The increasing order of stability can be shown as:

5. Alkyl group increases in the C = C double bond carbon, hyper conjugation increases and stability of that organic compound also increases.

Question 32.
What is BHC? How will you prepare BHC? Mention its uses?
Answer:

1. BHC is Benzene hexachloride.

2. Benzene reacts with three molecule of Cl₂ in the presence of sunlight or UV light to yield BHC. This is also called as gammaxane or Lindane.

3. BHC is a powerful insecticide.

Question 33.
Write a chemical reaction useful to prepare the following:

1. Freon-12 from carbon tetrachloride.
2. Carbon tetrachloride from carbon disulphide.

Answer:
1. Freon – 12 from carbon tetrachloride:
Freon – 12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride. This reaction is called “Swarts reaction.”

2. Carbon tetrachloride from carbon disulphide:
Carbon disulphide reacts with chlorine gas in the presence of anhydrous AlCl₃ as catalyst to give carbon tetrachloride.

PART – IV

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

Question 34 (a).
(I) What is the difference between molecular mass and molar mass? Calculate the molecular mass and molar mass for carbon monoxide.
(II) What are competitive electron transfer reaction? Give example.

[OR]

(b) (I) State the trends in the variation of electronegativity in period and group.
(II) The electron gain enthalpy of chlorine is 348 kJ mol^-1. How much energy in kJ is released when 17.5 g of chlorine is completely converted into Cl^– ions in the gaseous state?
Answer:
(a) (I) Molecular mass:

1. Relative molecular mass is defined as the ratio of the mass of the molecule to the unified atomic mass unit.
2. It can be calculated by adding the relative atomic masses of its constituent atoms.
3. For carbon monoxide (CO)

Molecular mass = Atomic mass of carbon + Atomic mass of oxygen 12 + 16 = 28 u.

Molar mass:

1. It is defined as the mass of one mole of a substance.
2. The molar mass of a compound is equal to the sum of the relative atomic masses of its constituent expressed in g mol^-1.
3. For carbon monoxide (CO)

12 + 16 = 28 g mol^-1.

Both molecular mass and molar mass are numerically same but the units are different.

(II) These are the reactions in which redox reactions take place in different vessels and it is an indirect redox reaction. There is a competition for the release of electrons among different metals.

Example:
Zn releases electrons to Cu and Cu releases electrons to Silver and so on.
Zn_(s) + Cu²⁺ → Zn²⁺_(aq) + Cu_(s) (Here Zn – oxidised; Cu²⁺ – reduced)
Cu_(s) + 2Ag⁺ → Cu²⁺_(aq) + 2Ag_(s) (Here Cu – oxidised; Ag⁺ – reduced)

[OR]

(b)
(I) Variation of electronegativity in a period:
The electronegativity increases across a period from left to right. Since the atomic radius decreases in a period, the attraction between the valence electron and the nucleus increases. Hence the tendency to attract shared pair of electrons increases, ‘therefore, electronegativity increases in a period.

(II) Water is amphoteric in nature and it behaves both as an acid as well as base. With acids stronger than itself (e.g., H₂S) it behaves as a base and with bases stronger than itself (e.g., NH, ) it acts as an acid.

Question 35 (a).
(I) Discuss the three types of Covalent hydrides?
(II) Write the chemical reactions to show the amphoteric nature of water?

[OR]

(b)
(I) Lithium forms monoxide with oxygen whereas sodium forms peroxide with oxygen. Why?
(II) Write about the uses of strontium?
Answer:
(a) (I)

1. They are the compounds in which hydrogen is attached to another element by sharing of electrons.
2. The most common examples of covalent hydrides are methane, ammonia, waterand hydrogen chloride.
3. Molecular hydrides of hydrogen are further classified into three categories as,
   • Electron precise (CH₄, C₂H₆, SiH₄, GeH₄)
   • Electron-deficient (B₂H₆) and
   • Electron-rich hydrides (NH₃, H₂O)
4. Since most of the covalent hydrides consist of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.

(II) Water is amphotenc in nature and it behaves both as an acid as well as base. With acids stronger than itself (e.g., H₂S) it behaves as a base and with bases stronger than itself
(e.g., NH₃) it acts as an acid.

1. As a base: H₂O_(l) + H₂S_(aq) → H₃O_(aq) – HS^–_(aq)
2. As a acid: H₂O_(l) + NH_3(aq) → OH^–_(aq) + NH⁺_(aq)

[OR]

(b) (I)

1. The fact that a small cation can stabilize a small anion and a large cation can stabilize a large anion explains the formation and stability of the oxides.
2. The size of Li⁺ ion is very small and it has a strong positive field around it. It can combine . with only small anion, O^2- ion, resulting in the formation of monoxide Li₂O.
3. The Na⁺ ion is a larger cation and has a weak positive field around it and can stabilize a bigger peroxide ion, \(\mathrm{O}_{2}^{2-} \text { or }[-\mathrm{O}-\mathrm{O}-]^{2-}\) resulting in the formation of peroxide Na₂O₂.

(II)

1. 90Sr is used in cancer therapy.
2. \(^{87} \mathrm{Sr} /^{86} \mathrm{Sr}\) ratio is used in marine investigators as well as in teeth, tracking animal migrations or in criminal forensics.
3. Dating of rocks.
4. Strontium is used as a radioactive tracer in determining the ‘source of archaeological materials ’ such as timbers and coins.

Question 36 (a).
(I) When the driver of an automobile applies brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward back of the car. Upon forward acceleration the passengers are pushed toward the front of the car. Why?
(II) Critical temperature of H₂O, NH₃ and CO₂ are 647.4, 405.5 and 304.2 K, respectively. When we start cooling from a temperature of 700 K which will liquefy first and which will liquefy finally?

[OR]

(b)
(I) The following water gas shift reaction is an important industrial process for the production of hydrogen gas.
C0(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)
At a given temperature K_p = 2.7. If 0.13 mol of CO, 0.56 mol of water, 0.78 mol of CO₂ and 0.28 mol of H₂ are introduced into a 2L flask, find out in which direction must the reaction proceed to reach equilibrium.

(II)
2H₂O(g) ⇄ 2H₂(g) + O₂(g) K_C = 4.1 × 10^-48 At 599 K
N₂(g) + O₂(g) ⇄ 2NO(g) K_C = 1 × 10^-30 at 1000 K
Predict the extent of the above two reactions.
Answer:
(a) (I)
1. When the driver of an automobile applies brake, the passengers are pushed toward the front of the car due to inertia of the body, but a helium balloon pushed toward back of the car. Helium balloon responds to the air around it. Helium molecules are lighter than air of our atmosphere, and so they move toward back by gravity as a result of the accelerating frame.

2. Upon forward acceleration, the passengers are pushed toward the front of the car,because the body in motion tends to stay in motion until acted upon by an outside force. Helium balloon is going to move opposite to this pseudo gravitational force.

(II) Critical temperature of a gas is defined as the temperature above which it cannot be liquefied even at high pressures.
∴ When cooling starts from 700 K, H₂O will liquefied first, then followed by ammonia and finally carbon dioxide will liquefied.

[OR]

(b) (I)
CO(g) + H₂O(g) ⇄ CO₂(g) + H₂(g)
Give K_P = 2.7
[CO] = 0.13 mol, [H₂O] = 0.56 mol
[CO₂] = 0.78 mol; [H₂] = 0.28 mol
V = 2L
K_P = K_C (RT)^∆ng
2.7 = K_C (RT)⁰
K_C = 2.7
_QC = \(\frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{H}_{2}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2} \mathrm{O}\right]}\) = \(\frac{\left(\frac{0.78}{2}\right)\left(\frac{0.28}{2}\right)}{\left(\frac{0.13}{2}\right)\left(\frac{0.56}{2}\right)}\)
Q = 3
Q > K_C, Hence the reaction proceed in the reverse direction.

(II) In the reactions, decomposition of water at 500 K and oxidation of nitrogen at 1000 K, the value of K_C is very less K_C < 10^-3. So reverse reaction is favoured.
∴ Products << reactants

Question 37 (a).
(I) CuCl is more covalent than NaCl. Give reason?
(II) Draw and explain the molecular orbital diagram of Boron molecule?

[OR]

(b)
(I) 0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water.
Calculate the percentage of carbon and hydrogen in it.

(II) Draw the fisher projection formula for tartaric acid.
Answer:
(a) (I)
1. Cations having ns² np⁶ nd¹⁰ configuration show greater polarising power than the cations with ns² np⁶ configuration. Hence they show greater covalent character.
2. CuCl is more covalent than NaCl. As compared to Na⁺ (1.13\(\overset { \circ }{ A } \)), Cu⁺ (0.6\(\overset { \circ }{ A } \)) is small and has 3s23p63d10 configuration.
3. Electronic configuration of Cu⁺: [Ar] 3s²3p⁶3d¹⁰
Electronic configuration of Na+: [He] 2s²2p⁶
So CuCl is more covalent than NaCl.

(II)

1. Electronic configuration of B = 1s² 2s² 2p³
2. Electronic configuration of B₂ = \(\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{1} \pi 2 p_{z}^{1}\)
3. Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \) = \(\frac{6-4}{2}\) = 1
4. B₂ molecule has two unpaired electrons hence it is paramagnetic.

[OR]

(b) (I) Weight of organic compound = 0.30 g
Weight of carbon-dioxide = 0.88 g
Weight of water = 0.54 g

Percentage of hydrogen:
18 g of water contains 2 g of hydrogen
∴ 0.54 g of water contain = \(\frac{2}{18}\) × 0.54g of hydrogen
∴% of hydrogen = \(\frac{2}{18}\) × \(\frac{0.54}{0.30}\) × 100 = \(\frac{2}{18}\) × \(\frac{54}{0.3}\)
% of H = 0.111 × 180 = 19.888 ~ 20%

Percentage of carbon:
44 g of water contains 12 g of hydrogen
0.88 g of water contain CO₂ contains = \(\frac{12}{44}\) × 0.88g of hydrogen
∴ % of carbon = \(\frac{12}{44}\) × \(\frac{0.88}{0.30}\) × 100 = \(\frac{12}{44}\) × \(\frac{88}{0.3}\) = \(\frac{24}{0.3}\)
% of carbon = 80%

(II)

Question 38 (a).
An organic compound (A) of a molecular formula C₆H₆ which is a simple aromatic hydrocarbon. A undergoes hydrogenation to give a cyclic compound (B). A reacts with chlorine in the presence of UV – light to give C which is used as insecticide. Identify A, B and C. Explain the reactions with equation.

[OR]

(b) An organic compound (A) of molecular formula CH₂ reacts with methyl magnesium iodide followed by acid hydrolysis to give (B) of molecular formula C₂H₆O. (B) on reaction with PCl gives (C).(C) on reaction with alcoholic KOH gives (D) an alkene as the product. Identify (A), (B), (C), (D) and explain the reactions involved.
Answer:

1. Simple aromatic hydrocarbon, C₆H₆ is benzene.
2. Benzene (A) reacts with H₂ in the presence of Pt to give cyclohexane (B).

3. Benzene (A) reacts with Cl₂ in presence of UV-light to give benzene hexachloride (C).

[OR]

(b) (I)
1. (A) Of molecular formula CH₂O is identified as HCHO, formaldehyde.
2. Formaldehyde reacts with CH₃MgI followed by hydrolysis to give ethanol, CH₃-CH₂OH B as the product.

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 4 English Medium

General Instructions:

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

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
When 6.3 g of sodium bicarbonate is added to 30 g of acetic acid solution, the residual solution is found to weigh 33 g. The number of moles of carbon dioxide released in the reaction is ……………………
(a) 3
(b) 0.75
(c) 0.075
(d) 0.3
Solution:

The amount of CO₂ released, x = 3.3 g
No. of moles of CO₂ released = 3.3/4.4 = 0.075 mol
Answer:
(c) 0.075

Question 2.
Two electrons occupying the same orbital are distinguished by …………………
(a) Azimuthal quantum number
(b) Spin quantum number
(c) Magnetic quantum number
(d) Orbital quantum number
Solution:
Spin quantum number For the first electron ms = + \(\frac{1}{2}\)
For the second electron ms = – \(\frac{1}{2}\)
Answer:
(b) Spin quantum number

Question 3.
Statement – 1: Ionization enthalpy of N is greater than that of O.
Statement – II: N has exactly half filled electronic configuration which is more stable than electronic configuration of O.
(a) Statement – I is wrong but statement – II is correct
(b) Statement – I is correct but statement – II is wrong.
(c) Statement – I and II are correct and statement – II is the correct explanation of statement – I.
(d) Statement – I and II are correct but statement – II is not the correct explanation of statement – I.
Answer:
(c) Statement – I and II are correct and statement – II is the correct explanation of statement – I.

Question 4.
Water gas is …………………….
(a) H₂O_(g)
(b) CO + H₂O
(c) CO + H₂
(d) CO + N₂
Answer:
(c) CO + H₂

Question 5.
Among the following the least thermally stable is ……………………
(a) K₂CO₃
(b) Na₂CO₃
(c) BaCO₃
(d) Li₂CO₃
Li₂CO₃ is least stable.
Answer:
(d) Li₂CO₃

Question 6.
Match the following.

Answer:

Question 7.
C(diamond) → C(graphite), ∆H = -ve, this indicates that …………………
(a) Graphite is more stable than diamond
(b) Graphite has more energy than diamond
(c) Both are equally stable
(d) Stability cannot be predicted
Answer:
(a) Graphite is more stable than diamond

Question 8.
In the equilibrium, 2A(g) ⇄ 2B(g) + C₂(g)
the equilibrium concentrations of A, B and C₂ at 400 K are 1 × 10^-4M, 2.0 × 10^-3M, 1.5 × 10⁴M respectively. The value of K_C for the equilibrium at 400 K is ……………………..
(a) 0.06
(b) 0.09
(c) 0.62
(d) 3 × 10^-2
Solution:
[A] = 1 × 10^-4M; [B] = 2 × 10^-3M; [C] = 1.5 × 10^-4M
2A(g) ⇄ 2B(g) + C₂(g)

= 6.0 × 10^-2 = 0.06
Answer:
(a) 0.06

Question 9.
Which of the following is a non-aqueous solution?
(a) Salt solution
(b) Sugar solution
(c) Br₂ in CCl₄
(d) Ethanol dissolved in water
Answer:
(c) Br₂ in CCl₄

Question 10.
Which of the following molecule does not exist due to its zero bond order?
(a) \(\mathrm{H}_{2}^{-}\)
(b) \(\mathrm{He}_{2}^{+}\)
(c) He₂
(d) \(\mathrm{H}_{2}^{+}\)
Answer:
(c) He₂

Question 11.
Which of the following is optically active?
(a) 3 – Chloropentane
(b) 2 – Chloropropane
(c) Meso – tartaric acid
(d) Glucose
Answer:
(d) Glucose

Question 12.
Which of the following represent a set of nucleophiles?
(a) BF₃, H₂O, NH^2-
(b) AlCl₃, BF₃, NH₃
(c) CN^–, RCH₂^–, ROH
(d) H⁺, RNH₃⁺, CCl₂
Answer:
(c) CN^–, RCH₂^–, ROH

Question 13.
Propyne on passing through red hot iron tube gives ……………………
(a)
(b)
(c)
(d) one of these
Answer:
(a)

Question 14.
Consider the following statements:

(I) E₂ reaction is a bimolecular elimination reaction of second order
(II) E₂ reaction takes place in two steps.
(III) E₂ reaction generally Jakes place in primary alkyl halides.

Which of the above statements is/are not correct?
(a) (I) only
(b) (II) only
(c) (III) only
(d) (I) & (III)
Answer:
(II) E₂ reaction takes place in two steps.

Question 15.
Photo chemical smog formed in congested metropolitan cities mainly consists of ………………….
(a) Ozone, SO₂ and hydrocarbons
(b) Ozone, PAN and NO₂
(c) PAN, smoke and SO₂
(d) Hydrocarbons, SO₂ and CO₂
Answer:
(b) Ozone, PAN and NO₂

PART – II

Answer any six questions in which question No. 23 is compulsory. [6 × 2 = 12]

Question 16.
Why interstitial hydrides have a lower density than the parent metal?
Answer:

1. d block elements form metallic or interstitial hydrides, on heating with dihydrogen under pressure.
2. Hydrogen atom being small in size occupy some in the metallic lattice producing distortion without any change in its type.
3. The densities of these hydrides are lower than those of metals from which they are formed since the crystal lattice expands due to the inclusion of dihydrogen.

Question 17.
Prove that calcium oxide is a basic oxide?
Answer:
Calcium oxide is a basic oxide. It combines with acidic oxides at high temperature.

1. 
2. 

Question 18.
Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means?
Answer:

1. The mathematical relationship between the volume of a gas and the number of moles is V ∝ n
2. \(\frac { V_{ 1 } }{ n_{ 1 } } \) = \(\frac { V_{ 2 } }{ n_{ 2 } } \) Constant, where V₁ and n₁ are the volume and number of moles of a gas and V₂ and n₂
   and n₂ are the values of volume and number of moles of same gas at a different set of conditions.
3. If the volume of the gas increase then the number of moles of the gas also increases.
4. At a certain temperature and pressure, the volume of a gas is directly proportional to the number of the moles of the gas.

Question 19.
Why pressure has no effect on the synthesis of HI?
Answer:
When the total number of moles of gaseous reactants and gaseous products are equal, the change in pressure has no effect on system at equilibrium.
H₂(g) + I₂(g) ⇄ 2HI(g)
Here the number of moles of reactants and products are equal. So the pressure has no effect on such equilibrium with ∆n_g = 0.

Question 20.
Draw the lewis structure of PCl₅ and SF₆
Answer:

Question 21.
How are naphthalene and camphor purified?
Answer:

1. Naphthalene, camphor and benzoic acid when heated, pass directly from solid to vapour without melting. On cooling the vapours will give back solid. This phenomenon is known as sublimation. This technique is used to purify naphthalene, camphor from non volatile impurities.

2. Substances to be purified is taken in a beaker. It is covered with a watch glass. The beaker is heated for a while and the resulting vapours condense on the bottom of the watch glass. Then the watch glass is removed and the crystals are collected.

Question 22.
How will you convert ethyl chloride into
(I) Ethane
(II) n – butane
Answer:
(I) Conversion of ethyl chloride into ethane:

(II) Conversion of ethyl chloride into n – butane:
Wurtz reaction:

Question 23.
Chloroform is kept with a little ethyl alcohol in a dark coloured bottle, why?
Answer:
(I) Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carboxyl chloride (phosgene), it is therefore stored in closed dark coloured bottles completely filled so that air is kept out.

(II) With the use of 1 % ethanol we can stabilise chloroform, because ethanol can convert the poisonous COCl₂ gas into non poisonous diethyl carbonate.
COCl₂ + 2C₂H₅OH → CO(OC₂H₅)₂ + 2HCl.

Question 24.
How does classical smog differ from photochemical smog?
Answer:
Classical smog:

1. Classical smog is caused by coal-smoke and fog.
2. It occurs in cold humid climate.
3. The chemical composition is the mixture of SO₂, SO₃ gases and humidity.
4. Chemically it is reducing in nature because of high concentration of SO₂ and so it is also called reducing smog.
5. It is primarily responsible for acid rain.
6. It also causes bronchial irritation.

Photochemical smog:

1. Photochemical smog is cause by photochemical oxidants.
2. It occurs in warm, dry and sunny climate.
3. The chemical composition is the mixture of NO₂ and O₃ gases.
4. Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃ and so it is also called oxidising smog.
5. It causes irritation to eyes, skin and lungs and increase the chances of asthma.
6. It causes corrosion of metals, stones and painted surfaces.

PART – III

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

Question 25.
An ice cube at 0°C is placed in some liquid water at 0°C, the ice cube sinks – Why?
Answer:

1. In an ice cube, each atom is surrounded tetrahedrally by four water molecules through hydrogen bond and its density is low.
2. Liquid water at 0°C has the density as 999.82 kg/cm³. Maximum density is attained by water only at 4°C as 1000 kg/cm³.
3. When the temperature changed from 4°C to 0°C, the density of water decreases rather than increases. This is called anomalous expansion of water.
4. The reason for this phenomenon lies in the structure of ice lattice and hydrogen bonding in water,
5. At 0°C, ice cube sinks in liquid water at 0°C because of the lesser density and greater volume of water.

Question 26.
Write the chemical equations for the reactions involved in Solvay process of preparation of sodium carbonate?
Answer:
Solvay process:
The Solvay process is represented by the below chemical equations:

Question 27.
Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if
(a) It Is compressed to a smaller volume at constant temperature
(b) The temperature is raised while keeping the volume constant
(c) More gas is introduced into the same volume and at the same temperature
Answer:
(a) If a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.

(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.

(c) If more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. If the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.

Question 28.
Calculate \(\Delta \mathrm{H}_{\mathrm{r}}^{0}\) for the reaction
CO₂(g) + H₂(g) → CO(g) + H₂O (g)
given that \(\Delta \mathrm{H}_{\mathrm{f}}^{0}\) for CO₂(g), CO(g) and H₂O(g) are – 393.5, – 111.31 and – 242 kJ mol^-1 respectively.
Answer:
Given:
\(\Delta \mathrm{H}_{\mathrm{f}}^{0}\) CO₂ = -393.5 KJ mol^-1

Question 29.
Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O₂ is paramagnetic?
Answer:
(I) Electronic configuration of O atom is 1s² 2s² 2p⁴

(II) Electronic configuration of O₂ molecule is

(III) Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \) = \(\frac{10-6}{2}\) = 2

(IV) Molecule has two unpaired electrons, hence it is paramagnetic.

Question 30.
Give the principle involved in the estimation of halogen in an organic compound by Carius method?
Answer:
Estimation of halogens: Carius method
(I) A known mass of the organic compound is heated with fuming HNO₃ and AgNO₃.

(II) C, H and S gets oxidised to CO₂, H₂O and SO₂ and halogen combines with AgNO₃ to form a precipitate of silver halide.

(III) The precipate AgX is filtered, washed, dried and weighted.

(IV) From the mass of AgX and the mass of organic compound taken, the percentage of halogen are caluculated.
(V)

Question 31.
What polymerisation? Explain with suitable example?
Answer:
A polymer is a larga molecule formed by the combination of large number of small molecules (monomers). This process is known as polymerisation, a few examples are:

Question 32.
Compare \(\mathbf{S}_{\mathrm{N}^{1}}\) and \(\mathbf{S}_{\mathrm{N}^{2}}\) reaction mechanisms?
Answer:

Question 33.
From where does ozone come in the photochemical smog?
Answer:
(I) Photochemical smog is formed by the combination of smoke, dust and fog with air pollutants in the presence of sunlight.

(II) Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃. So it is also called oxidising smog.

(III) Photochemical smog is formed by following reactions:
N₂ + O₂ 2NO
2NO + O₂ 2NO₂

(O) + O₂ O₃
O₃ + NO NO₂ + (O)

(IV) NO and O₃ are strong oxidising agents and they can react with unbumt hydrocarbons in polluted air to form formaldehyde, acrolein and PAN.

PART – IV

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

Question 34 (a).
(I) An atom of an element contains 35 electrons and 45 neutrons. Deduce

1. The number of protons
2. The electronic configuration for the element
3. All the four quantum numbers for the last electron

(II) How many unpaired electrons are present in the ground state of Fe²⁺ (z = 26), Mn²⁺ (z = 25) and argon (z=18)?

[OR]

(b)
(I) Explain why hydrogen is not placed with the halogen in the periodic table.
(II) Complete the following reactions.
Al₄C₃ + D₂O → ?
CaC₂ + D₂O → ?
Mg₃N₂, + D₂O → ?
Ca₃P₂ + D₂O → ?
Answer:
(a) (I) An element X contains 35 electrons and 45 neutrons

1. The number of protons must be equal to the number of electrons. So the number of protons = 35.
2. Number of electrons = 35. So the electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵.
3. The last electron i.e. 5^th electron in 4p orbital has the following quantum numbers. n = 4, l = 1, m = +1, s = –\(\frac{1}{2}\)

(II) Fe → Fe²⁺ + 3e^–
Fe (Z = 26) Fe³⁺ = number of electrons = 23
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25). Electronic configuration is
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

[OR]

(b) (I)

1. Hydrogen resembles alkali metals as well as halogens.
2. Hydrogen resembles more alkali metals than halogens.
3. Electron affinity of hydrogen is much less than that of halogen atom. Hence the tendency to form hydride ion is low compared to that of halogens.
4. In most of its compounds hydrogen exists in +1 oxidation state. Therefore it is reasonable to place the hydrogen in group 1 along with alkali metals as shown in the latest periodic table published by IUPAC.

(II)

Question 35 (a).
(I) Why alkafr metals have high chemical reactivity? How this changes along the group?
(II) Distinguish between alkali metals and alkaline earth metals?

[OR]

(b)
(I) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude?
(II) Explain the graphical representation of Charles’ law?
Answer:
(a) (I) Alkali metals exhibit high chemical reactivity because of their low ionization enthalpy and their larger size.
The reactivity of alkali metals increases from Li to Cs, since the value of ionization energy decreases down the group (Li to Cs). All the alkali metals are highly reactive towards the more electronegative elements such as oxygen and halogens.

(II)
Alkali Metals:

1. Alkali metals are soft.
2. They have a single electron in the valence shell and their electronic configuration is [noble gas] ns¹.
3. They have low melting points.
4. Hydroxides are strongly basic.
5. Carbonates do not decompose.
6. Nitrates give corresponding nitrites and oxygen as products.
7. They show +1 oxidation states.
8. Their carbonates are soluble in water except Li₂CO₃.
9. Except Li, alkali metals do not form complex compounds.

Alkaline earth metals:

1. Alkaline earth metals are hard.
2. They have two electrons in the valence shell and their electronic configuration is [noble gas] ns².
3. They have relatively high melting points.
4. Hydroxides are less basic.
5. Carbonates decompose to form oxide, when heated to high temperatures.
6. Nitrates give corresponding oxides, nitrogen dioxide and oxygen as products.
7. They show +2 oxidation states.
8. Their carbonates are insoluble in water.
9. They can form complex compounds.

[OR]

(b)
(I) The volume of the gas is inversely proportional to pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease.
As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

(II)

1. Variation of volume of the gas sample with temperature at constant pressure.
2. Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P₁ to P₅.
3. i.e. P₁ < P₂ < P₃ < P₄ < P₅. When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.
4. All gases are becoming liquids, if they are cooled to sufficiently low temperatures.
5. In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.

Question 36 (a).
(I) Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?
(II) Derive the relationship between standard free energy (∆G°) and equilibrium constant (K_eq)

[OR]

(b)
(I) 2.56g of Sulphur is dissolved in 100 g of carbon disulphide. The solution boils at 319. 692 K. What is the molecular formula of Sulphur in solution. The boiling point of CS₂ is 319. 450K. Given that K_b for CS₂ = 2.42 K kg mol^-1
(II) Show that the sum of mole fraction of a solution is equal to one?
Answer:
(a) (I) A substance has a perfectly ordered arrangement only at absolute zero. Hence, entropy is zero only at absolute zero. Enthalpy of formation is the heat change involved in the formation of one mole of the substance from its elements. An element formed from its constituents means no heat change.

(II)

1. In a reversible process, system is at all times in perfect equilibrium with its surroundings.

2. A reversible chemical reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up.

3. This means that the reactions in both the directions should proceed with decrease in free energy, which is impossible.

4. It is possible only if at equilibrium, the free energy of a systepi is minimum.

5. Lets consider a general equilibrium reaction,
A + B ⇄ C + D
The free energy change of the above reaction in any state (∆G) is related to the standard free energy change of the reaction (∆G°) according to the following equation.
∆G = ∆G° +RTIn Q ………………. (1)
where Q is reaction quotient and is defined as the ratio of concentrajion of the products to the concentration of the reactants under non-equilibrium condition.

6. When equilibrium is attained, there is no further free energy change i.e. ∆G = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes, ∆G° =-RTln K_eq ……………… (2)
This equation is known as Van’t Hoff equation.
∆G° = -2.303 RTlogKeq ………………. (3)
We also know that,
∆G° = ∆H° -T∆S° = – RT In K …………….. (4)

[OR]

W₂ = 2.56 g; W₁ = 100 g
T = 319.692; K_b = 2.42 K kg mol^-1
∆T_b = (319.692 – 319.450) K = 0.242 K
M₂ = image 28
M₂ = 256 g mol^-1
Molecular mass of sulphur in solution = 256 g mol^-1
Atomic mass of one mole of sulphur atom = 32
No. of atoms in a molecule of sulphur = \(\frac{256}{2}\) = 8
Hence, molecular formula of sulphur is S₈.

(II) Consider a solution containing two components A and B whose mole fractions are x_A and x_B respectively. Let the number of moles of two components A and B are n_A
and n_B respectively.

Question 37 (a).
(I) Explain about the procedure and calculation behind the carius method of estimation of sulphur?
(II) What is the difference between distillation, distillation under reduced pressure and steam distillation?

[OR]

(b)
(I) An organic compound (A) of a molecular formula C₂H₄ which is a simple alkene. A reacts with dil H₂SO₄ to give B. A again reacts with Cl₂ to give C. Identify A, B and C and write the equations.
(II) Why chloro acetic acid is stronger acid than acetic acid?
Answer:
(a) (I) Carius method

• Procedure: A known mass of the organic compound is taken in a clean carius tube and few mL of fuming HNO₃ is added and then the tube is sealed. It is then placed in an iron tube and heated for 5 hours. The tube is allowed to cool and a hole is made to allow gases to escape.
• The carius tube is broken and the content collected in a beaker. Excess of BaCl₂ is added to the beaker. H₂SO₄ formed is converted to BaSO₄ (white ppt.) The precipitate is filtered, washed, dried and weight. From the mass of BaSO₄, percentage of S is calculated.

(II) Calculation:
Mass of organic compound = Wg
233 g of BaSO₄ contains 32 g of sulphur
Percentage of sulphur = (\(\frac{32}{233}\) × \(\frac{x}{w}\) × 100)%

(II) Distillation is used in case of volatile liquid mixed with a non-volatile impurities.
Distillation under reduced pressure:
This method is used to purify such liquids which have very high boiling points and which decompose at or below their boiling points.
Steam distillation is used to purify steam volatile liquids associated with water immiscible impurities.

[OR]

(b) (I)

1. C₂H₄ is CH₂ = CH₂ is a simple alkene. A is ethylene.
2. Ethylene (A) reacts with dil H₂SO₄ to give ethanol (B)

3. Ethylene (A) reacts with Cl₂ to give 1, 2 dichloro ethane (C)

(II) Chloro acetic acid: image 31
Chloro acetic acid has Cl – group and it has high electronegativity and shows -I effect. Therefore Cl – atom to facilitate the dissociation of O – H bond very fastly. Whereas in the case of acetic acid, has CH₃ group and it shows +1 effect, therefore dissociation of O – H bond will be more difficult. Thus chloro acetic acid is stronger acid than acetic acid.

Question 38 (a).
(I) Write a chemical reaction useful to prepare the following:

1. Freon – 12 from carbon tetrachloride.
2. Carbon tetrachloride from carbon disulphide.

(II) What are ambident nucleophiles? Explain with an example.

[OR]

(I) Write about hydrosphere (or) Why Earth is called as Blue planet?
(II) Even though the use of pesticides increases the crop production, they adversely affect the living organisms. Explain the function and the adverse effects of the pesticides.
Answer:
1. Freon-12 from carbon tetrachloride:
Freon-12 is prepared by the action of hydrogen fluoride on carbon tetrachloride in the presence of catalytic amount of antimony pentachloride. This reaction is called “Swarts reaction.”

2. Carbon tetrachloride from carbon disulphide:
Carbon disulphide reacts with chlorine gas in the presence of anhydrous AlCl₃ as catalyst to give carbon tetrachloride.

(II) Nucleophiles which can attack through two different sites are called ambident nucleophiles. For example, cyanide group is a resonance hybrid of two contributing structures and therefore it can act as a nucleophile in two different ways:

It can attack through carbon to form cyanides and through nitrogen to form isocyanides or carbylamines.

[OR]

(b)

1. Hydrosphere include all types of water sources like oceans, seas, rivers, lakes, streams, underground water, polar ice – caps, clouds etc.
2. It covers about 75% of the earth’s surface. Hence earth is called as Blue planet:

(II) Pesticides are the chemicals that are used to kill or stop the growth of unwanted organims. But these pesticides can affect the health of human beings. Pesticides are classified as

1. Insecticides
2. Fungicides and
3. Herbicides.

1. Insecticides:
Insecticides like DDT, BHC, Aldrin can stay in soil for a long period of time and are absorbed by soil. They contaminate root crops like carrot, radish..

2. Fungicides:
Organomercury compounds dissociate in soil to produce mercury which is highly toxic.

3. Herbicides:
They are used to control unwanted plants and are also known as weed killers. Eg, Sodium chlorate, sodium nitrate. They are toxic to mammals.

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 3 English Medium

General Instructions:

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

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
40 ml of methane is completely burnt using 80ml of oxygen at room temperature. The volume of gas left after cooling to room temperature. The volume of gas left after cooling to room temperature is …………………….
(a) 40 ml of CO₂ gas
(b) 40 ml of CO₂ gas and 80 ml of H₂O gas
(c) 60 ml of CO₂ gas and 60 ml H₂O gas
(d) 120 ml of CO₂ gas
Solution:
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(l)

Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.
Answer:
(a) 40 ml of CO₂ gas

Question 2.
What are the values of n, l, m and s for 3p_x electron?
(a) 3, 2, 1,0
(b) 3, 1,-1, +\(\frac{1}{2}\)
(c) 3, 2, +1, –\(\frac{1}{2}\)
(d) 3, 0, 0, +\(\frac{1}{2}\)
Solution:
3p_x electron; n = 3 (main shell)
for p_x orbital, l = 1, m = -1, s = \(\frac{1}{2}\)
Answer:
(b) 3, 1,-1, +\(\frac{1}{2}\)

Question 3.
Which of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electrons from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.
Answer:
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having high n value.

Question 4.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 5.
Lithium shows diagonal relationship with ………………………
(a) Sodium
(b) Magnesium
(c) Calcium
(d) Ahuninium
Answer:
(b) Magnesium

Question 6.
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\) \((\frac { \partial V }{ \partial T } )\)
(a) T
(b) 1/T
(c) P
(d) None of these
Solution:

Answer:
(b) 1/T

Question 7.
Heat of combustion is always …………………….
(a) Positive
(b) Negative
(c) Zero
(d) Either positive or negative
Answer:
(b) Negative

Question 8.
If in a mixture where Q = K, then what happens?
(a) The reaction shift towards products
(b) The reaction shift towards reactants
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate
(d) Nothing happens
Answer:
(c) Nothing appears to happen, but forward and reverse reactions are continuing at the same rate

Question 9.
Which one of the following gases has the lowest value of Henry’s law constant?
(a) N₂
(b) He
(c) CO₂
(d) H₂
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.
Answer:
(c) CO₂

Question 10.
In the molecule O_A = C = O_B, the formal charge on O_A, C and O_B are respectively.
(a) – 1, 0, + 1
(b) + 1, 0, -1
(c) – 2, 0, + 2
(d) 0, 0, 0
Solution:

Formal charge of O_A/O_B = N_V – (N_e + \(\frac { N_{ b } }{ 2 } \)) = 6 – (4 + \(\frac{4}{2}\)) = 6 – 6 = 0
Formal charge of C = 4 – (0 + \(\frac{8}{2}\)) = 4 – 4 = 0
Answer:
(d) 0, 0, 0

Question 11.
In the hydrocarbon CH₃ – CH₃ – CH = CH – CH₂ – C = CH the state of hybridisation of carbon 1, 2, 3, 4 and 7 are in the following sequence.
(a) sp, sp, sp³, sp², sp³
(b) sp², sp, sp³, sp², sp³
(c) sp, sp, sp², sp, sp³
(d) None of these
Answer:
(a) sp, sp, sp³, sp², sp³

Question 12.
Enzyme present in apple is ………………………..
(a) Polyphenol oxidase
(b) Polyphenol reductase
(c) Polyphenol
(d) Polyphenol hydrolase
Answer:
(a) Polyphenol oxidase

Question 13.
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is ………………………….
(a) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
(c) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
(d) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer:
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

Question 14.
In Finkelstein reaction, the mechanism followed is ……………………….
(a) S_N1
(b) E₁
(c) E₂
(d) S_N2
Answer:
(d) S_N2

Question 15.
Which sequence for greenhouse gases is based on GWP?
(a) CFC > N₂O > CO₂ > CH₄
(b) CFC > CO₂ > N₂O > CH₄
(c) CFC > N₂O > CH₄ > CO₂
(d) CFC > CH₄ > N₂O > CO₂
Answer:
(c) CFC > N₂O > CH₄ > CO₂

PART – II

Answer any six questions in which question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
Define Avogadro Number?
Answer:
Avogadro number is the number of atoms present in one mole of an element or number of molecules present in one mole of a compound. The value of Avogadro number (N) = 6.023 × 10²³.

Question 17.
Explain the meaning of the symbol 4P. Write all the four quantum numbers for these electrons?
Answer:
4f²: It means that the element has 2 electrons in outermost 4f shell.
Quantum number values are,
n = principal quantum number = 4
l = azimuthal quantum number = 3
m = magnetic quantum number = – 3, -2
s = spin quantum number = +\(\frac{1}{2}\), –\(\frac{1}{2}\)

Question 18.
Is the definition given below for ionization enthalpy is correct?
“Ionization enthalpy is defined as the energy required to remove the most loosely bound electron from the valence shell of an atom”?
Answer:
No, It is not correct. Thq accurate and absolute definition is as follows:
Ionization energy is defined as the minimum amount of energy required to remove the most loosely bound electron from the valence shell of the isolated neutral gaseous atom in its ground state.

Question 19.
What are the uses of calcium hydroxide?
Calcium hydroxide is used
Answer:

1. In the preparation of mortar, a building material.
2. In white wash due to its disinfectant nature.
3. In glass making and tanning industry.
4. For the preparation of bleaching powder and for the purification of sugar.

Question 20.
30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK^-1 mol^-1. Calculate the melting point of sodium chloride?
Answer:
Given:
∆H_f = 30.4 kJ = 30400 J mol^-1
∆S_f (NaCl) = 28.4 kJ^-1 mol^-1
T_f = ?
∆S_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}}\); T_f = \(\frac{\Delta \mathrm{H}_{\mathrm{f}}}{\Delta \mathrm{S}_{\mathrm{f}}}\)
T_f = \(\frac{30400 \mathrm{J} \mathrm{mol}^{-1}}{28.4 \mathrm{JK}^{-1} \mathrm{mol}^{-1}}\) = 1070.4 K

Question 21.
How is a gas-solution equilibrium exist?
Answer:
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecules in the gaseous state and those dissolved in the liquid.
Example: In carbonate beverages the following equilibrium exists.
CO₂(g) ⇄ CO₂ (solution).

Question 22.
What type of hybridisations are possible in the following geometeries?

1. Octahedral
2. Tetrahedral
3. Square planar

Answer:

1. Octahedral geometry is possible by sp³d² (or) d²sp³ hybridisation.
2. Tetrahedral geometry is possible by sp³ hybridisation.
3. Square planar geometry is possible by dsp² hybridisation.

Question 23.
How will you prepare Lassaigne’s extract?
Lassagine’s extract preparation:
Answer:

1. A small piece of Na dried by pressing between the folds of filter paper is taken in a fusion tube and it is gently heated. When it melts to a shining globule, a pinch of organic compound is added.
2. The tube is heated till reaction ceases and become red hot. Then it is plunged in 50 ml of distilled water taken in a china dish and the bottom of the tube is broken by striking it against the dish.
3. The contents of the dist is boiled for 10 minutes and then it is filtered. The filtrate is known as Lassaigne’s extract.

Question 24.
Complete the reactions and identify the products?

Answer:

PART – III

Answer any six questions in which question No. 32 is compulsory. [6 × 3 = 18]

Question 25.
The balanced equation for a reaction is given below 2x + 3y → 41 + m
When 8 moles of x react with 15 moles of y, then

1. Which is the limiting reagent?
2. Calculate the amount of products formed.
3. Calculate the amount of excess reactant left at the end of the reaction.

Answer:
2x + 3y → 41 + m
1. 2x reacts with 3y to give products.
8x reacts with 15y means, y is the excess because 8 moles of x should react with 4 × 3y = 12y moles of y to give products.
In this reaction 15y moles are used.
Therefore, 3 moles of y is excess and x is the limiting agent.

2. When 8 moles of x react with 12 moles of y, the product formed will be 4 × 41 i.e. 161 and 4m as product.
8x+ 12y → 161 + 4m

3. At the end of the reaction, the excess reactant left is 3 moles of y.

Question 26.
Which would you expect to have a higher melting point, magnesium oxide or magnesium fluoride? Explain your reasoning.
Answer:

1. Magnesium oxide has very strong ionic bonds as compared to magnesium fluoride.
2. Mg²⁺ and O^2- have charges of +2 and -2, respectively.
3. Oxygen ion is smaller than fluoride ion.
4. The smaller the ionic radii, the smaller the bond length in MgO and the bond is stronger than MgF₂.
5. Due to more strong bond nature in MgO, it has high melting point than MgF₂.

Question 27.
A sample of solid KClO₃ (potassium chlorate) was heated in a test tube to obtain O₂ according to the reaction
2KClO_3(s) → 2KCl + 3O₂
The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mm of Hg at 300K. What is the partial pressure of the oxygen gas?
Answer:

Question 28.
List the characteristics of entropy?
Characteristics of entropy:
Answer:

• Entropy is a thermodynamic state function that is a measure of the randomness or disorderliness of the system.
• In general, the entropy of gaseous system is greater than liquids and greater than solids. The symbol of entropy is S.
• Entropy is defined as “for a reversible change taking place at a constant temperature (T), the change in entropy (∆S) of the system is equal to heat energy absorbed or evolved (q) by the system divided by the constant temperature (T).
  \(\Delta S_{\mathrm{sys}}=\frac{q_{\mathrm{rev}}}{T}\)
• If heat is absorbed, then ∆S is positive and there will be increase in entropy. If heat is evolved, ∆S is negative and there is a decrease in entropy,
• The change in entropy of a process represented by ∆S and is given by the equation,
  ∆S_sys = S_f – S_i
• If S_f > S_i, ∆S is positive, the reaction is spontaneous and reversible.
  If S_f < S_i, ∆S is negative, the reaction is non-spontaneous and irreversible.
• Unit of entropy: SI unit of entropy is J K^-1.

Question 29.
Derive the values of K_C and K_P for the synthesis of HI.
Answer:
H₂(g) + I₂(g) ⇄ 2HI(g)
Let us consider the formation of HI in which V moles of hydrogen, ‘b’ moles of iodine gas are allowed to react in an container of volume ‘V’.
Let ‘x’ moles of each of H₂ and I₂ react together to form 2x moles of HI.

Applying mass of action

Caluculation of K_P: K_P = K_C.RT^∆ng
Here ∆n_g = n_p – n_r = 2 -2 = 0
Hence, K_P = K_C
K_P = \(\frac{4 x^{2}}{(a-x)(b-x)}\)

Question 30.
Describe the classification of organic compounds based on their structure?
Answer:
Classification of organic compounds based on the structure

Question 31.
For the following bond cleavages use curved-arrows to show the electron flow and classify each as homolytic or heterolytic fission. Identify reactive intermediate produced as free radical, carbocation and carbanion?

Answer:

Question 32.
An alkyl halide with molecular formula C₆H₁₂Br on dehydrohalogenation gave two isomeric alkenes X and,Y with molecular formula C₆H₁₂. On reductive ozonolysis, X and Y gave four compounds CH₃COCH₃, CH₃CHO, CH₃CH₂CHO and (CH₃)₂ CHCHO. Find the alkyl halide?
Answer:

1. C₆ H₁₃ Br is 3 – Bromo – 4 – methylpentane.
2. 3 – Bromo – 4 – methylpentane on dehydration give two isomers X and Y as follows:

Therefore C₆H₁₃ Br is 3 – Bromo – 4 – methylpentane

Question 33.
What do you mean by ozone hole? What are its consequences?
Answer:
Depletion of ozone layer creates some sort of holes in the blanket of ozone which surrounds us in the atmosphere and this is known as ozone hole.

1. With the depletion of the ozone layer, UV radiations filters into the troposphere which leads to ageing of skin, cataract, sunburn, skin cancer etc.
2. By killing many of the phytoplanktons, it can damage the fish productivity.
3. Evaporation rate increases through the surface and stomata of leaves which can decrease the moisture content of the soil.

PART – IV

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

Question 34 (a).
(I) Balance the following equations by ion electron method?
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
(II) Boric acid, H₃BO₃ is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H₃BO₃. What is the mass of boric acid in the sample?

[OR]

(b)
(I) How many unpaired electrons are present in the ground state of Fe³⁺ (z = 26), Mn²⁺ (z = 25) and argon (z = 18)?
(II) Explain about the significance of de Broglie equation?
Answer:
KMnO₄ + SnCl₂ + HCl → MnCl₂ + SnCl₄ + H₂O + KCl
Oxidation half reaction: (loss of electron)

Reduction of halfa reaction: (gain of electron)

Add H₂O to balance oxygen atoms

Add Hcl to balance hydrogen atoms
KMnO₄ + 5e^– + 8 HCl → MnCl₂ + 4H₂O …………………. (4)

To equalize the number of electrons equation (1) × 5 and equation (2) × 2

(II) Molecular mass of H₃BO₃ = (1 × 3) + (11 × 1) + (16 × 3) = 62
Boric acid sample contains 0.543 mole.
Mass of 0.543 mole of Boric acid = Molecular mass x mole
= 62 × 0.543
= 33.66 g

[OR]

(b) (I) Fe → Fe³⁺ + 3e^–
Fe (Z = 26)
Fe³⁺ = number of electrons = 23
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² for Fe atom.
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ for Fe³⁺ ion.
So, it contain 5 unpaired electrons.
Mn (Z = 25) Electronic configuration is
1s² 2s² 2p⁵ 3s² 3p⁶ 4s² 3d⁵
Mn → Mn²⁺ + 2e^–
Number of unpaired electrons in Mn²⁺ = 5
Ar (Z = 18). Electronic configuration is
1s² 2s² 2p⁶ 3s² 3p⁶.
All orbitals are completely filled. So, no unpaired electrons in it.

(II) Significance of de Broglie equation:

1. λ = \(\frac{h}{mv}\) This equation implies that a moving particle can be considered as a wave and a wave can exhibit the properties of a particle.
2. For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.
3. For a microscopic particle such as an electron, the mass is of the order of 10^-31 kg, hence the wavelength is much larger than the size of atom and it becomes significant.
4. For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 35 (a).
(I) Mention any two anomalous properties of second period elements?
(II) Arrange Na⁺, Mg²⁺ and Al³⁺ in the increasing order of ionic radii. Give reason?

[OR]

(b)
(I) How do you expect the metallic hydrides to be useful for hydrogen storage?
(II) Write a note about ortho water and para water?
Answer:
(a) (I) Anomalous properties of second period elements:

1. In the 1^st group, lithium forms compounds with more covalent character while the other elements of this group form only ionic compounds.
2. In the 2^nd group, beryllium forms compounds with more covalent character while the other elements of this family form only ionic compounds.

(II) Na⁺, Mg²⁺ and Al³⁺ are isoelectronic cations.

The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Hence the increasing order of ionic radii is, \(\mathbf{r}_{\mathrm{Na}^{+}}\), > \(\mathbf{r}_{\mathrm{Mg}^{2+}}\), > \(\mathbf{r}_{\mathrm{Al}^{3+}}\).

[OR]

(b)
(I) In metallic hydrides, hydrogen is adsorbed as H-atoms. Due to the adsorption of H atoms the metal lattice expands and become unstable. Thus, when metallic hydride is heated, it decomposes to form hydrogen and finely divided metal. The hydrogen evolved can be used as fuel.

(II)
1. Water exists in space in the interstellar clouds, in proto-planetary disks, in the comets and icy satellites on the solar system, and on the Earth.
2. In particular, the ortho-to-para ratio (OPR) of water in space has recently received attention. Like hydrogen, water can also be classified into ortho – H₂O and para – H₂O, in which the directions are antiparallel.

3. At the temperature conditions of the Earth (300 K), the OPR of H₂O is 3.
4. At low temperatures below (< 50 K) the amount of para – H₂O increases. It is known that the OPR of water in interstellar clouds and comets has more para – H₂O (OPR = 2.5) than on Earth.

Question 36 (a).
Derive the values of critical constants from the Van der Waals constants?

[OR]

(b) Derive the values of K_p and K_C for dissociation of PCl₅?
Answer:
(a) Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
\(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT for 1 mole.
From this equation, the values of critical constant P_C, V_C and T_C are derived in terms of a and b the Vander Waals constants.

The above equation (4) is an cubic equation of V, which can have three roots. At the critical point, all the three values of V are equal to the critical volume Vc. i.e. V = V_C„
i.e; V = V_C
V – V_C = 0 ………………… (5)
(V – V_C)³ ………………….. (6)
(V³ – 3V_CV²) + 3V_C²V – V_C³ …………………. (7)
As the equation (4) is identical with equation (7), comparing the ‘V’ terms in (4) and (7),

Divide equation (11) by (10)

When equation (12) is substituted in (10)

Substituting the values of V_C and P_C in equation (9)

Critical constants a and b can be calculated rising Vander Waals constants as follows:

(b) Consider that V moles of PCl₅is taken in container of volume‘V’
Let x moles of PCl₅ be dissociated into x moles of PCl₃ and x moles of Cl₂.

Applying law of mass action

K_p caluculation: K_p = K_C. \(\mathrm{RT}^{\mathrm{An}}\); ∆n_g = 2 – 1 = 1
We know that PV = nRT
RT = \(\frac{PV}{n}\)
Where ‘n’ is the total number of moles at equilibrium
n = a – x + x + x = a + x

Question 37 (a).
(I) Solubility of a solid solute in a liquid solvent increases with increase in temperature. Justify this statement?
(II) Explain how non-ideal solutions shows positive deviation from Raoult’s law?

[OR]

(b)
(I) How will you distinguish between electrophiles and nucleophiles?
(II) Complete the following reactions and identify the products?
(a)
(b)
Answer:
(a) (I) When the temperature is increased,the average kinetic energy of the molecules of the solute and the solvent increases. The increase in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.
(II)

1. Let us consider the positive deviation shown by a solution of ethyl alcohol and water
2. In this solution, the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alco¬hol and water-water interaction).
3. This results in the increased evaporation of both components from the aqueous solution of ethanol.
4. Consequently, the vapour pressure of the solution is greater than the vapour pressure
   predicted by Raoult’s law.
5. Here, the mixing process is endothermic i.e., DHmixing > 0 and there will be a slight increase in volume (DVmixing > 0)
6. 

(b)
(I)
Electrophiles:

1. They are electron deficient.
2. They are cations.
3. They are lewis acids.
4. Accept an electron pair.
5. Attack on electron rich sites.

Nucleophiles:

1. They are electron rich.
2. They are anions.
3. They are lewis bases.
4. Donate an electron pair.
5. Attack on electron deficient sites.

(II)
(a)
(b)

Question 38 (a).
(I) Is it possible to prepare methane by Kolbe’s electrolytic method?
(II) Explain how 2-butyne reacts with
(a) Lindlar’s catalyst
(b) Sodium in liquid ammonia.

[OR]

(b) (I) Discuss the aromatic nucleophilic substitution reactions of chlorobenzene?
(II) CCl₄ > CHCl₃ > CH₂C₁₂ > CH₃Cl is the decreasing order of boiling point in haloalkanes. Give reason?
Answer:
(a) (I) Kolbe’s electrolytic method is suitable for the preparation of symmetrical alkanes, that is alkanes containing even number of carbon atoms. Methane has only one carbon, hence it cannot be prepared by Kolbe’s electrolytic method.
(II) (a) 2-butyne reacts with Lindlar’s catalyst: 2-butyne can be reduced to cis – 2 butene using CaCO₃ supported in Pd -metal partially deactivated with sulphur.
This reaction is stereo specific giving only the cis – 2 – butene.

(b) 2-butyne reacts with sodium in liquid ammonia:
2-butyne can also be reduced to trans – 2 – butene using sodium in liquid ammonia. This reaction is stereospecific giving only the trans – 2 – butene.

[OR]

(b) (I) Aromatic nucleophilic substitution reactions:
Dow’s process:

1.

2.

3.

(II) The boiling point of chloro, bromo and iodoalkanes increases with increase in the number of halogen atoms. So the correct decreasing order of boiling point of haloalkanes is:
CCl₄ > CHCl₃ > CH₂Cl₂ > CH₃Cl.

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 2 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 2 English Medium

General Instructions:

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

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
An element X has the following isotopic composition ²⁰⁰X = 90 %, ¹⁰⁰X = 8 % and ²⁰²X = 2 %.
The weighted average atomic mass of the element X is closest to ………………………….
(a) 201 u
(b) 202 u
(c) 199 u
(d) 200 u
\(\frac { (200\times 90)+(199\times 18)+(202\times 2) }{ 100 } \)
= 199.96 = 200 u
Answer:
(d) 200 u

Question 2.
Which of the following is not used in writing electronic configuration of an atom?
(a) Aufbau principle
(b) Hund’s rule
(c) Pauli’s exclusion principle
(d) Heisenberg’s uncertainty principle
Answer:
(d) Heisenberg’s uncertainty principle

Question 3.
Assertion (A): Cr with electronic configuration [Ar]3d⁵ 4s¹ is more stable than [Ar] 3d⁴ 4s².
Reason (R): Half filled orbitals have been found to have extra stability than partially filled orbitals.
(a) A and R are correct and R is the correct explanation of A.
(b) A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.
Answer:
(a) A and R are correct and R is the correct explanation of A.

Question 4.
The type of H-bonding present in ortho nitro phenol and p-nitro phenol are respectively ……………………..
(a) Inter molecular H-bonding and intra molecular H – bonding
(b) Intra molecular H-bonding and inter molecular H – bonding
(c) Intra molecular H – bonding and no H – bonding
(d) Intra molecular H – bonding and intra molecular H – bonding

Answer:
(b) Intra molecular H-bonding and inter molecular H – bonding

Question 5.
Match the following:

Answer:

Question 6.
Which of the following is the correct expression for the equation of state of van der Waals gas?
(a) \(\left(P+\frac{a}{n^{2} V^{2}}\right)\) (V – nb) = nRT
(b) \(\left(P+\frac{n a}{n^{2} V^{2}}\right)\) (V – nb) = nRT
(c) \(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT
(d) \(\left(P+\frac{n^{2} a^{2}}{V^{2}}\right)\) (V – nb) = nRT
Answer:
(c) \(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT

Question 7.
In a reversible process, the change in entropy of the universe is ………………………
(a) > 0
(b) >0
(c) <0
(d) = 0
Answer:
(d) = 0

Question 8.
Which of the following is not a general characteristic of equilibrium involving physical process?
(a) Equilibrium is possible only in a closed system at a given temperature
(b) The opposing processes occur at the same rate and there is a dynamic but stable condition
(c) All the physical processes stop at equilibrium
(d) All measurable properties of the system remains constant
Solution:
Correct statement: Physical processes occurs at the same rate at equilibrium.
Answer:
(c) All the physical processes stop at equilibrium

Question 9.
Stomach acid, a dilute solution of HCl can be neutralised by reaction with Aluminium hydroxide
Al(OH)₃ + 3HCl(aq) → AlCl₃ + 3H₂O
How many millilitres of 0.1 M Al(OH)₃ solution are needed to neutralise 21 mL of 0.1 M HCl?
(a) 14 mL
(b) 7 mL
(c) 21 mL
(d) None of these
Solution:
M₁ × V = M₂ × V₂
∵ 0.1 M Al(OH)₃ gives 3 × 0.1 = 0.3 M OH^– ions
0.3 × V₁ = 0.1 × 21
V₁ = \(\frac{0.1×21}{0.3}\) = 7 ml
Answer:
(b) 7 mL

Question 10.
Shape and hybridisation of IF₅ are ……………………..
(a) Trigonal bipyramidal, sp³d²
(b) Trigonal bipyramidal, sp³d
(c) Square pyramidal, sp³d²
(d) Octahedral, sp³d²
Solution:
IF₅ – 5 bond pair + 1 lone pair
∴ hybridisation sp³d²

Answer:
(c) Square pyramidal, sp³d²

Question 11.
Consider the following statements:

1. It is not possible for the carbon to form either C⁴⁺ (or) C^4- ions.
2. Carbon can form ionic bonds.
3. In compounds of carbon, it form covalent bonds.

Which of the above statement is/are not correct?
(a) (I) and (II)
(b) (III) only
(c) (I) only
(d) (II) only
Answer:
(d) (II) only

Question 12.
For the following reactions
(A) CH₃CH₂CH₂Br + KOH → CH₃-CH + KBr + H₂O
(B) (CH₃)₃CBr + KOH → (CH₃)₃ COH + KBr
(C)
Which of the following statement is correct?
(a) (A) is elimination, (B) and (G) are substitution
(b) (A) is substitution, (B) and (C) are elimination
(c) (A) and (B) are elimination and (C) is addition reaction
(d) (A) is elimination, B is substitution and (C) is addition reaction.
Answer:
(d) (A) is elimination, B is substitution and (C) is addition reaction.

Question 13.
Which of the following compound used for metal cleaning solvent?
(a) Methylene chloride
(b) Methyl chloride
(c) Chloroform
(d) Ethane
Answer:
(a) Methylene chloride

Question 14.
The number of possible isomers of C₆H₁₂ is ……………………..
(a) 2
(b) 3
(c) 5
(d) 6
Answer:
(c) 5

Question 15.
Ozone layer is depleted by the reactive ……………………..
(a) Hydrogen atom
(b) Oxygen atom
(c) Fluorine atom
(d) Chlorine atom
Answer:
(d) Chlorine atom

PART – II

Answer any six questions in which question No. 19 is compulsory. [6 × 2 = 12]

Question 16.
Calculate the average atomic mass of naturally occurring magnesium using the following data?

Answer:
Solution: Isotopes of Mg.
Atomic mass = Mg²⁴ = 23.99 × 78.99/100 = 18.95
Atomic mass = Mg²⁶ = 24.99 × 10/100 = 2.499
Atomic mass = Mg²⁵ = 25.98 × 11.01/100 = 2,860
Average Atomic mass = 24.309
Average atomic mass of Mg = 24.309

Question 17.
What are quantum numbers?
Answer:

1. The electron in an atom can be characterized by a set of four quantum numbers, namely – principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m) and spin quantum number (s).
2. When Schrodinger equation is solved for a wave function φ, the solution contains the first three quantum numbers n, 1 and m.
3. The fourth quantum number arises due to the spinning of the electron about its own axis.

Question 18.
How 2-ethylanthraquinone helps to prepare hydrogen peroxide?
Answer:
On an industrial scale, hydrogen peroxide is prepared by auto oxidation of 2-alkyl anthraquinol.

Question 19.
Calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume 6 dm³ at 70°C assuming it is an ideal gas?
Answer:
We will use the ideal gas equation for this calculation as below:
P = \(\frac{nRT}{V}\) = image 7 = 9.39 atm.

Question 20.
What are the important features of lattice enthalpy?
Answer:

1. Higher lattice energy shows greater electrostatic attraction and therefore a stronger bond in the solid.
2. The lattice enthalpy is greater for ions of higher charge and smaller radii.

Question 21.
What are aqueous and non-aqueous solution? Give example?
Answer:

1. If the solute is dissolved in the solvent water, the resultant solution is called as an aqueous solution, e.g., salt in water.
2. If the solute is dissolved in the solvent other than water such as benzene, ether, CCl₄ etc, the resultant solution is called a non aqueous solution, e.g., Br₂ in CCl₄.

Question 22.
What is bond enthalpy? How they relate with bond strength?
Answer:
The bond enthalpy is defined as the minimum amount of energy required to break one mole of a particular bond in molecules in their gaseous state. Larger the bond enthalpy stronger will be the bond.

Question 23.
What is triad system? Give example?
Answer:
(I) In this system hydrogen atom oscillates between three polyvalent atoms. It involves 1, 3 – migration of hydrogen atom from one polyvalent atom to other with in the molecule

(II) The most important type of triad system is keto-enol tautomerism and the two groups of tautomers are keto form and enol form.

(III)

Question 24.
What is stone leprosy? How is it formed?
Answer:

1. The attack on, the marble of buildings by acid rain is called stone leprosy.
2. Acid rain causes extensive damage to buildings made up of marble.

CaCO + H₂SO₄ → CaSO₄ + H₂O + CO₂↑

PART – III

Answer any six questions in which question No. 26 is compulsory. [6 × 3 = 18]

Question 25.
Calculate the equivalent mass of hydrated ferrous sulphate?
Answer:
Hydrated ferrous sulphate = FeSO₄.7H₂O
Ferrous sulphate – Reducing agent
Ferrous sulphate reacts with an oxidising agent in acid medium according to the equation.

16 parts by mass of oxygen oxidised 304 g of FeSO₄.
8 parts by mass of oxygen will oxidise \(\frac{304}{16}\) × 8 parts by mass of FeSO₄ = 152
Equivalent mass of Ferrous sulphate (Anhydrous) = 152
Equivalent mass of crystalline Ferrous sulphate FeSO₄.7H₂O = 152 + 126 = 278

Question 26.
Calculate the uncertainty in position of an electron, if ∆v = 0.1% and v = 2.2 × 10⁶ ms^-1.
Answer:
Mass of an electron = m = 9.1 × 10^-31 kg.
∆v = Uncertainty in velocity = \(\frac{0.1}{100}\) × 2.2 × 10⁶ ms^-1
∆v = 0.22 × 10⁴ = 2.2 × 10³ ms^-1
∆x. ∆v.m = \(\frac{h}{4π}\)

∆x = 2.635 × 10^-8
Uncertainty in position = 2.635 × 10^-8

Question 27.
Distinguish between diffusion and effusion?
Answer:
Diffusion:

1. Diffusion is the spreading of molecules of a substance throughout a space or a second subsance.
2. Diffusion refers to the ability of the gases to mix with each other.
3. E.g; Spreading of something such as brown tea liquid spreading through the water in a tea cup.

Effusion:

1. Effusion is the escape of gas molecules through a very small hole in a membrane into an evacuated area.
2. Effusion is a ability of a gas to travel through a small pin-hole.
3. E.g; Pouring out something like the soap studs bubbling out from a bucket of water.

Question 28.
For a chemical reaction the values of ∆H and ∆S at 300 K are – 10 kJ mol^-1 and – 20 JK^-1 mol^-1 respectively. What is the value of ∆G of the reaction? Calculate the ∆G of a reaction at 600K assuming ∆H and ∆S values are constant. Predict the nature of the reaction?
Answer:
Given:
∆H = -10 kJ mol^-1 = -10000 J mol^-1
∆S = – 20 JK^-1 mol^-1
T = 300 K

∆G?
∆G = ∆H – T∆S
∆G = -10 kJ mol^-1 – 300 K × (-20 × 10^-3) kJ K^-1 mol^-1
∆G = (-10 + 6) kJ mol^-1
∆G = – 4 kJ mol^-1

At 600 K,
∆G = – 10 kJ mol^-1 – 600 K × (-20 × 10^-3) k^-1 mol^-1
∆G = (-10 + 12) kJ mol^-1
∆G + 2 kJ mol^-1
The value of ∆G is negative at 300K and the reaction is spontaneous, but at 600K the value ∆G becomes positive and the reaction is non-spontaneous.

Question 29.
For the reaction: A₂(g) + B₂(g) ⇄ 2AB(g); H is -∆ve.
The following molecular scenes represent different reaction mixture (A – light grey, B-dark grey)

1. Calculate the equilibrium constant K_p and K_c.
2. For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
3. What is the effect of increase in pressure for the mixture at equilibrium?

Answer:

K_c > Q i.e; forward reaction is favoured.

(III) Since ∆n_g = 2 -2 = 0, thus, pressure has no effect. So by increasing the pressure, equilibrium will not be affected.

Question 30.
Derive the relationship between the relative lowering of vapour pressure and mole fraction of the solute?
Answer:

Question 31.
Differentiate between the principle of estimation of nitrogen in an organic compound by

1. Dumas method
2. Kjeldahl’s method

Answer:

1. Dumas method:
The organic compound is heated strongly with excess of CuO (Cupic Oxide) in an atmosphere of CO₂ where free nitrogen, CO₂ and H₂O are obtained.

2. Kjeldahl’s method:
A known mass of the organic compound is heated strongly with cone. H₂SO₄, a little amount of potassium sulphate arid a little amount of mercury (as catalyst). As a result of reaction, the nitrogen present in the organic compound is converted to ammonium sulphate.

Question 32.
In what way free radical affect the human body?
Answer:

1. Free radicals can disrupt cell membranes.
2. Increase the risk of many forms of cancer.
3. Damage the interior lining of blood vessels.
4. Eads to a high risk of heart disease and stroke.

Question 33.
Dissolved oxygen in water is responsible for aquatic life. What processes are responsible for the reduction in dissolved oxygen in water?
Answer:

1. Organic matter such as leaves, grass, trash can pollute water. Microorganisms present in water can decompose these organic matter and consume dissolved oxygen in water.
2. Eutrophication is a process by which water bodies receive excess nutrients that stimulates excessive plant growth.
3. This enhanced plant growth in water bodies is called algal bloom.
4. The growth of algae in extreme abundance covers the water surface and reduces the oxygen concentration in water. Thus, bloom-infeded water inhibits the growth of other living organisms in the water body.
5. This process in which the nutrient rich water support a dense plant population, kills animal life by depriving it of oxygen and results in loss of biodiversity is known as eutrophication.

PART – IV

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

Question 34 (a).
(I) Calculate the equivalent mass of sulphuric acid?
(II) The reaction between aluminium and ferric oxide can generate temperatures up to 3273 K and is used in welding metals.
(Atomic mass of Al = 27 u Atomic mass of O = 16 u)
2Al + Fe₂O₃ → Al₂O₃ + 2Fe; If, in this process, 324 g of aluminium is allowed to react with 1.12 kg of ferric oxide.
(a) Calculate the mass of Al₂O₃ formed.
(b) How much of the excess reagent is left at the end of the reaction?

[OR]

(b) (I) Consider the following electronic arrangements for the d⁵ configuration?

(a)

(b)

(c)

1. Which of these represents the ground state?
2. Which configuration has the maximum exchange energy?

(II) An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion?
Answer:
(a) (I) Equivalent mass of sulphuric acid:
Sulphuric acid = H₂SO₄
Molar mass of Sulphuric acid = 2 + 32 + 64 = 96
Basicity of Sulphuric acid = 2

= \(\frac{96}{2}\) = 49 g eq^-1

(II) (a)

As per balanced equation 54 g Al is required for 112 g of iron and 102 g of Al₂O₃.
54 g of Al gives 102 g of Al₂O₃.
∴ 324 g of Al will give \(\frac{102}{54}\) × 324 = 612 g of AlcO₃

(b) 54 g of Al requires 160 g of Fe₂O₃ for welding reaction.
∴ 324 g of Al will require \(\frac{160}{54}\) × 324 = 960 g of Fe\(\frac{102}{54}\)O₃
∴ Excess Fe₂O₃ -Unreacted Fe₂O₃ = 1120 – 960 = 160 g 160 g of exces reagent is left at the end of the reaction.

[0R]

(b)
(I) 1.
2.

(II) Let the no. of electrons in the ion = x
∴ The no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac{30.4x}{100}\) = x + 0.304 x
Now, mass number of ion = Number of protons + Number of neutrons
= (x + 3) + (x + 0.304 x)
∴ 56 = (x + 3) + (x + 0.304 x) or 2.304 x = 56 – 3 = 53
x = \(\frac{53}{2.304}\) = 23
Atomic number of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Question 35 (a).
(I) State the Newland’s law of octaves?
(II) What are the two exceptions of block division in the periodic table?

[OR]

(b) (I) Complete the following reactions.
(a)

(b) 2 BeCl₂ + LiaH₄ →?

(II) What happens when quick lime reacts with
(a) H₂O and
(b) CO₂?

(I) The Law of octaves states that, “when elements are arranged in the order of increasing atomic weights, the properties of the eighth element are a repetition of the properties of the first element”.

(II)

1. Helium has two electrons. Its electronic configuration is 1s². As per the configuration, it is supposed to be placed in ‘s’ block, but actually placed in 18th group which belongs to ‘p’ block. Because it has a completely filled valence shell as the other elements present in 18th group. It also resembles with 18th group elements in other properties. Hence helium is placed with other noble gases.

2. The other exception is hydrogen. It has only one s-electron and hence can be placed in group 1. It can also gain an electron to achieve a noble gas arrangement and hence it can behave as halogens (17^th group elements). Because of these assumptions, position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table.

[OR]

(b) (I) (a) Beryllium oxide is heated with carbon and chloride to get BeCl₂

(b) Beryllium chloride is treated with LiAlH₄ to get beryllium hydride.
2BeCl₂ + LiAlH₄ → 2BeH₂ + Licl + Alcl₃

(II) (a) CaO + H₂O → Ca(OH)₂ (calcium hydroxide)
(b) CaO + CO₂ → CaO₃ (calcium carbonate)

Question 36 (a).
(I) State the first law of thermodynamics?
(II) Calculate the enthalpy of combustion of ethylene at 300 Kc at constant pressure, if its heat of combustion at constant volume (∆U) is -1406 kJ?

[OR]

(b)
(I) Explain how the equilibrium constant Kc predict the extent of a reaction? (3)
(II) Explain about the effect of catalyst in an equilibrium reaction? (2)
Answer:
(a) (I) The first law of thermodynamics states that “the total energy of an isolated system . remains constant though it may change from one form to another” (or) Energy can neither be created nor destroyed, but may be converted from one form to another.
(II) The complete ethylene combustion reaction can be written as,

C₂H_4(g) + 3O_2(g) → 2CO_2(g) + 2H₂O_(l)
ΔU = -1406 kJ
Δn = n_p(g) – n_r(g)
Δn = 2 – 4 – 2
ΔH = ΔU + RTΔn_g
ΔH = -1406 + (8.314 × 10^-3 × 300 × (-2)) ΔH = -1410.9 kJ

[OR]

(b) (I)

1. The value of equilibrium constant K_C tells us the extent of the reaction i.e., it indicatehow far the reaction has proceeded towards product formation at a given temperature.
2. A large value of K_C indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of K_C indicates that the reaction reaches equilibrium with low product yield.
3. If K_C > 10³, the reaction proceeds nearly to completion.
4. If K_C < 10^-3 the reaction rarely proceeds.
5. It the K_C is in the range 10^-3 to 10³, significant amount of both reactants and products are present at equilibrium.

(II) Addition of catalyst does not affect the state of equilibrium. The catalyst increases the rate of both the forward and reverse reactions to the same extent. Hence it does not change the equilibrium composition of the reaction mixture.

Question 37 (a).
(I) Briefly explain geometrical isomerism in alkenes by considering 2- butene as an example.
2 – butene: Geometrical isomerism: CH₃ – CH = CH – CH₃ 

(II) What is meant by condensed structure? Explain with an example.

[OR]

(b)
(I) Why cut apple turns a brown colour?
(II) Predict the product for the following reaction,

1.

2.

3.

Answer:
(a) (I)

• Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.
• In 2-butene, the carbon-carbon double bond is sp² hybridised. The carbon-carbon double bond consists of a s bond and a p bond. The presence of p bond lock the molecule in one position. Hence, rotation around C = C bond is not possible.
• 
• These two compounds are termed as geometrical isomers and are termed as cis and transform.
• The cis isomer is the one in which two similar groups are on the same side of the double bond. The trans isomer is that in which two similar groups are on the opposite side of the double bond. Hence, this type of isomerism is called cis-trans isomerism.

(II) The bond line structure can be further abbreviated by omittiilg all the these dashes representing covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. The resulting expression of the compound is called a condensed structural formula.
e.g., 1,3 – butadiene.

[OR]

(b) (I)

1. Apples contains an enzyme called polyphenol oxidase (PPO) also known as tyrosinase.
2. Cutting an apple exposes its cells to the atmospheric oxygen and oxidizes the phenolic compounds present in apples. This is called the “enzymatic browning” that turns a cut apple brown.
3. In addition to apples, enzymatic browning is also evident in bananas, pears, avocados and even potatoes.

Question 38 (a).
Suggest the route for the preparation of the following from benzene?

1. 3 – chloro-nitrobenzene
2. 4 – chlorotoluene
3. Bromobenzene
4. m – dinitrobenzene

[OR]

(b) A hydrocarbon C₃H₆ (A) reacts with HBr to form compound (B). Compound (B) reacts with aqueous potassium hydroxide to give (C) of molecular formula C₃H₆O. What are (A) (B) and (C). Explain the reactions?
Answer:
(a)
1. Preparation of 3 – chloronitro – benzene from benzene: Benzene undergoes nitration and followed by chlorination and it leads to the formation of 3 – chloronitrobenzene.

2. Preparation 4-chlorotoulene from benzene: Benzene undergoes Friedal craft’s alkylation followed by chlorination and it leads to the formation of 4 – chlorotoulene.

3. Preparation of Bromobenzene from benzene: Benzene undergo bromination to give bromobenzene.

4. Preparation of m-dinitrobenzene from benzene: Benzene undergo twice the time nitration to give m-dinitrobenzene.

[OR]

(b)
1. The hydrocarbon with molecular formula C₃H₆ (A) is identified as propene,
CH – CH = CH₂
2. Propene reacts with HBr to form bromopropane CH₃ – CH₂ – CH₂Br as (B).

3. 1 – bromopropane react with aqueous potassium hydroxide to give 1 – propanol CH₃ – CH₂ – CH₂OH as (C).
4. 2 – bromo propane reacts with aqueous KOH to give 2-propanol as (C)

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 1 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Chemistry Model Question Paper 1 English Medium

General Instructions:

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

Time: 2:30 Hours
Maximum Marks: 70

PART – I

Answer all the Questions: [15 × 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is …………………….
(a) 0%
(b) 4.4%
(c) 16%
(d) 8.4%
MgCO₃ → MgO + CO₂↑
MgCO₃: (1 × 24) + (1 × 12) + (3 × 16) = 84g
CO₂: (1 × 12) + (2 × 16) = 44g
100% pure 84 g MgCO₃ on heating gives 44 g CO₂
Given that 1 g of MgCO₃ on heating gives 0.44 g CO₂
Therefore, 84 g MgCO₃ sample on heating gives 36.96 g CO₂
Percentage of purity of the sample = \(\frac { 100% }{ 44gCO_{ 2 } } \) × 36.96 g CO₂ = 84%
Percentage of impurity = 16%
Answer:
(c) 16%

Question 2.
The electronic configuration of Eu (atomic no. 63) Gd (atomic no. 64) and Tb (atomic no. 65) are …………………….. [NEET – Phase II]
(a) [Xe] 4f⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁸ 5d¹ 6s²
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²
(c) [Xe] 4f⁷ 6s² [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
(d) [Xe] 4f ⁶ 5d¹ 6s², [Xe] 4f⁷ 5d¹ 6s² and [Xe] 4f⁹ 6s²
Solution:
Eu : [Xe] 4f⁷, 5d⁰, 6s²
Gd : [Xe] 4f⁷, 5d¹, 6s²
Tb : [Xe] 4f⁹, 5d⁰, 6s²
Answer:
(b) [Xe] 4f⁷ 6s² [Xe] 4f⁸ 6s² and [Xe] 4f⁹ 6s²

Question 3.
Match the following:

Answer:

Question 4.
Various successive ionization enthalpies (in kJ mol^-1) of an element are given below.

The element is ……………………….
(a) Phosphorus
(b) Sodium
(c) Aluminium
(d) Silicon
Answer:
(c) Aluminium

Question 5.
Non-stoichiometric hydrides are formed by ……………………..
(a) Palladium, vanadium
(b) Carbon, nickel
(c) Manganese, lithium
(d) Nitrogen, chlorine
Answer:
(a) Palladium, vanadium

Question 6.
Which is the function of sodium – potassium pump?
(a) Maintenance of ion balance
(b) Used in nerve impulse conduction
(c) Transmitting nerve signals
(d) Regulates the blood level
Answer:
(c) Transmitting nerve signals

Question 7.
∆S is expected to be maximum for the reaction ………………………
(a) Ca_(s) + 1/2O_2(g) → CaO_(s)
(b) C_(s) + O_2(g) → CO_2(g)
(c) N_2(g) + O_2(g) → 2NO_(g)
(d) CaCO_3(s) → CaO_(s) + CO_2(g)
Solution:
In CaCO_3(s) → CaO_(s) + CO_2(g) entropy change is positive. In (a) and (b) entropy change is negative; in (c) entropy change is zero.
Answer:
(d) CaCO_3(s) → CaO_(s) + CO_2(g)

Question 8.
Which one of the following is a reversible reaction?
(a) Ripening of a banana
(b) Rusting of iron
(c) Tarnishing of silver
(d) Transport of oxygen by Hemoglobin in our body
Solution:
All the other three reactions are irreversible reactions. But the hemoglobin combines with O₂ in lungs to form oxyhemoglobin. The oxyhemoglobin has a tendency to form hemoglobin by releasing O₂. So it is a reversible reaction.
Answer:
(d) Transport of oxygen by Hemoglobin in our body

Question 9.
P₁ and P₂ are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x₁ represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ………………………….
Solution:
(a) P₁ + x₁ (P₂ – P₁)
(b) P₂ – x₁ (P₂ + P₁)
(c) P₁ – x₂ (P₁ – P₂)
(d) P₁ + x₂ (P₁ – P₂)
P_total = P₁ + P₂
= P₁x₁ + P₂ x₂
= P₁(1 – x₂) + P₂x₂ [∵x₁ + x₂ = 1, x₁ = 1 – x₂]
= P₁ – P₁x₂ + P₂x₂ = P₁ – x₂ = P₁ – x₂ (P₁ – P₂)
Answer:
(c) P₁ – x₂ (P₁ – P₂)

Question 10.
Which of the following molecule contain no n bond?
(a) SO₂

(b) NO₂

(c) CO₂

(d) H₂O

Solution:
Water (H₂O) contains only σ bonds and no π bonds.
Answer:
(d) H₂O

Question 11.
IUPAC name of  is …………………….
(a) Trimethylheptane
(b) 2 -Ethyl -3, 3- dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 – Butyl -2 -methyl – 3 – ethyl-butane.
Answer:
(c) 3, 4, 4 – Trimethyloctane

Question 12.
Which one of the following is an example for free radical initiators?
(a) Benzoyl peroxide
(b) Benzyl alcohol
(c) Benzyl acetate
(d) Benzaldehyde.
Answer:
(a) Benzoyl peroxide

Question 13.
Some meta-directing substituents in aromatic substitution are given. Which one is most – deactivating?
(a) – COOH
(b) – NO₂
(C) – C ≡ N
(d) – SO₃H
Answer:
(b) – NO₂

Question 14.
Consider the following statements.
(I) S_N² reaction is a bimolecular nucleophilic first order reaction.
(II) S_N² reaction take place in one step.
(III) S_N² reaction involves the formation of a carbocation. Which of the above statements is/are not correct?
(a) (II)
(b) (I) only
(c) (I) & (III)
(d) (III)
Answer:
(c) (I) & (III)

Question 15.
The questions given below consists of an assertion and the reason. Choose the correct option out of the choices given below each question.
Assertion (A): If BOD level of water in a reservoir is more than 5 ppm it is highly polluted.
Reason(R): High biological oxygen demand means high activity of bacteria in water.
(a) Both (A) and R are correct and (R) is the correct explanation of (A)
(b) Both (A) and R are correct and (R) is not the correct explanation of (A)
(c) Both (A) and R are not correct
(d) (A) is correct but( R) is not correct
Answer:
(d) (A) is correct but( R) is not correct

PART – II

Answer any six questions in which question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
What is the actual configuration of copper (Z = 29)? Explain about its stability?
Answer:
Copper (Z = 29)
Expected configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹ 4s²
Actual configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹
The reason is that fully filled orbitals have been found to have extra stability.

Copper has the electronic configuration [Ar] 3d¹⁰ 4s¹ and not [Ar] 3d⁹ 4s² due the symmetrical distribution and exchange energies of d electrons.

Symmetry leads to stability. The full filled configuration have symmetrical distribution of electrons and hence they are more stable than unsymmetrical configuration.

Question 17.
What is screening effect?
Answer:
Screening effect:
The repulsive force between inner shell electrons and the valence electrons leads to a decrease in the electrostatic attractive forces acting on the valence electrons by the nucleus. Thus the inner shell elections act as a shield between the nucleus and the valence electrons. This effect is called shielding effect (or) screening effect.

Question 18.
Ice is less dense than water at 0°C. Justify this statement?
Answer:
In ice, each oxygen atom is surrounded tetrahedrally by through hydrogen bonds to four water molecules. That is, the presence of two hydrogen atoms and two lone electron pairs (on oxygen) in each water molecule results in a three-dimensional structure. The arrangements creates an open structure, which in turn accounts for the fact that ice is less dense than water at 0°C.

Question 19.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions.

So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire, or leave it in the direct sunlight, even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

1. The gas pressure increases.
2. More of the liquefied propellant turns into a gas.

Question 20.
One mole of PCl₅ is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant?
Answer:
Given that [PCl₅]_initial = \(\frac { 1mol }{ dm^{ 3 } } \)
[Cl₂]_eq = 0.6 mol dm^-3
PCl₅⇄ PCl₃ + Cl₂
[PCl₅]_eq = 0.6 mole dm^-3
[PCl5]_eq = 0.4 mole dm^-3
∴ K_C = \(\frac{0.6×0.6}{0.4}\)
K_C = 0.9

Question 21.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Mostly dissolution of gases in liquid is an exothermic process. It is because the fact that this process involves decrease of entropy.

Thus, increase of temperature tends to push the equilibrium towards backward direction as a result of which solubility of the gas decrease with rise in temperature.
(Gas + Solvent ⇄ Solution + Heat)

Question 22.
Give the general formula for the following classes of organic compounds?
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines.
Answer:
(a) Aliphatic monohydric alcohol

(b) Aliphatic ketones

(c) Aliphatic amines

Question 23.
Identify which of the following shows +1 and -I effect?

(I) -NO₂
(II) -SO₃H
(III) -I
(IV) -OH
(V) CH₃O^–
(VI) CH_3-

Answer:

Question 24.
Write the products A & B for the following reaction?

Answer:

PART – III

Answer any six questions in which question No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Balance by oxidation number method: Mg + HNO₃ → Mg(N0₃)₂ + NO₂ + H₂O.
Answer:
Step 1:

Step 2:
Mg + 2HNO₃ → Mg(NO₃)₂ + NG₂ + H₂O

Step 3:
To balance the number of oxygen atoms and hydrogen atoms 2HNO₃ is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + H₂O

Step 4:
To balance the number of hydrogen atoms, the H₂O molecule is multiplied by 2.
Mg + 4HNO₃ → Mg(NO₃)₂ + 2NO₂ + 2H₂O

Question 26.
For each of the following, give the sub level designation, the allowable m values and the number of orbitals?
Answer:

1. n = 4, l = 2
2. n = 5, l = 3
3. n = 7, l = 0

1. n = 4, l = 2
If l = 2, ‘m’values are -2,-1, 0, +1, +2. So, 5 orbitals such as d_xy, d_yz ,d_xz, \(d_{ x^{ 2 }-y^{ 2 } }\) and \(d_{ z^{ 2 } }\)

2. n = 5, l = 3
If l = 3, ‘m’ values are -3, -2, -1, 0, +1, +2, +3
So, 7 orbitals such as \(\mathrm{f}_{\mathrm{z}}^{3}, \mathrm{f}_{\mathrm{xz}}^{2}, \mathrm{f}_{\mathrm{yz}}^{2}, \mathrm{f}_{\mathrm{xyz}}, \mathrm{f}_{\mathrm{z}\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{x}\left(\mathrm{x}^{2}-3 \mathrm{y}^{2}\right)}, \mathrm{f}_{\mathrm{y}\left(3 \mathrm{x}^{2}-\mathrm{z}^{2}\right)}\)

3. n = 7, l = 0
If l = 0, ‘m’ values are 0. Only one value. So, 1 orbital such as 7s orbital.

Question 27.
Prove that ionization energy is a periodic property?
Answer:
Variation in a period:
On moving across a period from left to right, the ionization enthalpy value increases.
This is due to the following reasons:

1. Increase of nuclear charge in a period
2. Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus, thus ionization enthalpy increases. Hence, ionization energy is a periodic property.

Variation in a group:
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons:

1. A gradual increase in atomic size
2. Increase of screening effect on the outermost electrons due. to the increase of number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 28.
Explain how heat absorbed at constant pressure is measured using coffee cup calorimeter with neat diagram?
Answer:

1. Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
2. We know that ∆H = qp (at constant P) and therefore, heat absorbed or evolved, q_p at constant pressure is also called the heat of reaction or enthalpy of reaction, ∆H_r
3. In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and ∆H_r will Reaction also be negative. Mixture
4. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ∆H_r will also be positive.

Question 29.
Consider the following reactions,
(a) H₂(g) + I₂(g) ⇄ 2 HI(g)
(b) CaCO₂(s) ⇄ CaO(s) + CO₂(g)
(c) S(s) + 3F₂(g) ⇄ SF₆(g)
In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product?
Answer:
(a) H₂(g) + I₂2(g) ⇄ 2HI(g)
In the above equilibrium reaction, volume of gaseous mdlecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of product.

(b)
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product CO₂, the pressure should be decreased or volume should be increased.

(c)
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.

Question 30.
You are provided with a solid ‘A’ and three solutions of A dissolved in water one saturated, one unsaturated, and one super saturated. How would you determine each solution?
Answer:
(I) Saturated solution:
When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.

(II) Unsaturated solution:
When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.

(III) Super saturated solution:
It is a solution that holds more solute than it normally could in its saturated form.

Example:

1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCl in 1 litre of water at 25°C.
2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCl in 1 litre of water at 25°C.
3. A super saturated solution is the solution in which crystals can start growing. 500 g of NaCl in 1 litre of water at 25°C.

Question 31.
Explain about the salient features of molecular orbital theory?
Answer:

• When atoms combine to fonn molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.
• The shape of molecular orbitals depend upon the shapes of combining atomic orbitals.
• The number of molecular orbitals formed is the same as the number of combining atomic orbitals. Half the number of molecular orbitals formed will have lower energy and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.
• The bonding molecular orbitals are represented as σ (sigma), π (pi), δ (delta) and the corresponding anti-bonding orbitals are called C*, 7t* and 5*.
• The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Aufbau’s Principle, Pauli’s exclusion principle and Hund’s rule as in the case of filling of electrons in the atomic orbitals.
• Bond order gives the number of covalent bonds between the two combining atoms.

The bond order of a molecule can be calculated using the following equation:
Bond order = \(\frac { N_{ b }-N_{ a } }{ 2 } \)
N_b = Number of electrons in bonding molecular orbitals.
N_a = Number of electrons in anti-bonding molecular orbitals.
(vii) A bond order of zero value indicates that the molecule does not exist.

Question 32.
Explain the types of addition reactions?
Answer:
Addition reactions are classified into three types they are,

1. Electrophilic addition reaction
2. Nucleophilic addition reaction
3. Free radical addition reaction

1. Electrophilic addition reaction:
An electrophilic addition reaction can be described as an addition reaction in whifth a reactant with multiple bonds as in a double or triple bond undergoes has its n bond broken and two new a bond are formed.

(ethane) Br Bf

2. Nucleophilic addition reaction:
A nucleophilic addition reaction is an addition reaction where a chemical compound with an electron deficient or electrophilic double or triple bond, a n bond, reacts with a nucleophilic which is an electron rich reactant with the disappearance of the double bond and creation of two new single or a bonds.

3. Free radical addition reaction:
It is an addition reaction in organic chemistry involving free radicals. The addition may occur between a radical and a non radical or between two radicals.

Question 33.
Complete the following reaction and identify A, B and C?

Answer:

PART – IV

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

Question 34 (a).
(I) How many radial nodes for 2s, 4p, 5d and 4f orbitals exhibit? How many angular nodes?
(II) How many unpaired electrons are present in the ground state of

(a) Cr³⁺ (Z = 24)
(b) Ne (Z = 10)

[OR]

(b) (I) The electronic configuration of an atom is one of the important factor which affects the value of ionization potential and electron gain enthalpy. Explain?
(II) Explain why cation are smaller and anions are larger in radii than their parent atoms?
Answer:
(a) (I) Formula for total number of nodes = n – 1
For 2s orbital: Number of radial nodes = 1.
For 4p orbital: Number of radial nodes = n – l – 1.
= 4 – 1 – 1 = 2
Number of angular nodes = l
∴Number of angular nodes = l.
So, 4p orbital has 2 radial nodes and 1 angular node.

For 5d orbital:
Total number of nodes = n – 1
= 5 – 1 =4 nodes
Number of radial nodes = n – l – 1
= 5 – 2 – 1 = 2 radial nodes.
Number of angular nodes = l = 2
∴5d orbital have 2 radial nodes and 2 angular nodes.

For 4f orbital:
Total number of nodes = n – 1
= 4 – 1 = 3 nodes
Number of radial nodes = n – l – 1
= 4 – 3 – 1 = 0 node.
Number of angular nodes = l
= 3 nodes
∴ 5d orbital have 0 radial node and 3 angular nodes.

(II) (a) Cr (Z = 24) 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹
Cr³⁺ – 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴.
It contains 4 unpaired electrons.
(b) Ne(Z=10) 1s² 2s² 2p⁶. No unpaired electrons in it.

[OR]

(b) (I)

• Electronic configuration of an atom affects the value of ionization potential and electron gain enthalpy.
• Half filled valence shell electronic configuration and completely filled valence shell electronic configuration are more stable than partially filled electronic configuration.
• For e.g. Beryllium (Z = 4) 1s² 2s² (completely filled electronic configuration)
  Nitrogen (Z = 7) 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹ (half filled electronic configuration)
  Both beryllium and nitrogen have high ionization energy due to more stable nature.
• In the case of beryllium (1s² 2s²), nitrogen (1s² 2s² 2p³) the addition of extra electron will disturb their stable electronic configuration and they have almost zero electron affinity.
• Noble gases have stable ns² np⁶ configuration and the addition of further electron is unfavorable and they have zero electron affinity.

(II) A cation is smaller than the parent atom because it has fewer electrons while its nuclear charge remains the same. The size of anion will be larger than that of parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

Question 35 (a).
(I) Arrange NH₃, H₂O and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement?
(II) Can we use concentrated sulphuric acid and pure zinc in the preparation of dihydrogen?

[OR]

(b) (I) Beryllium halides are covalent whereas magnesium halides are ionic. Why?
(II) What happens when
Sodium metal is dropped in water?

1. Sodium metal is heated in free supply of air?
2. Sodium peroxide dissolves in water?

Answer:
(a) 1. Increasing magnitude of hydrogen bonding among NH₃, H₂O and HF is
HF > H₂O > NH₃

2. The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding.

3. Among N, F and O the increasing order of their electronegativities are
N < O < F 4. Hence the expected order of the extent of hydrogen bonding is HF > H₂O > NH₃

(II) Conc. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO₂.
Zn + 2H₂SO₄ (Conc) → ZnSO₄ + 2H₂O + SO₂ Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.

[OR]

(b)
(I) Beryllium ion (Be²⁺) is smaller in size and it is involved in equal sharing of electrons with halogens to form covalent bond, whereas magnesium ion (Mg²⁺) is bigger and it is involved in transfer of electrons to form ionic bond.

(II)

1. 2Na + 2H₂O → 2NaOH + H₂
2. 2Na + O₂ → Na₂O₂
3. Na₂O₂ + 2H₂O → 2NaOH + H₂O₂

Question 36 (a).
(I) Define Gibb’s free energy?
(II) You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below?
Answer:

[OR]

(b) (I) Define mole fraction.
(II) Differentiate between ideal solution and non-ideal solution.
Answer:
(a) (I) Gibbs free energy is defined as the part of total energy of a system that can be converted (or) available for conversion into work.
G = H – TS, where G = Gibb’s free energy
H – enthalpy;
T = temperature;
S = entropy

(II) For ethanol:
Given: T_b = 78.4°C = (78.4 + 273) = 351.4 K

[OR]

(b) (I) Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all components present in the solution.

(II)
Ideal Solution:

1. An ideal solution is a solution in which each component obeys the Raoult’s law over entire range of concentration.
2. For an ideal solution,
   ΔH_missing = 0, ΔV_mixing = 0
3. Example: Benzene and toulene n – Hexane and n – Heptane.

Non – Ideal Solution:

1. The solution which do not obey Raults’s law over entire range of concentrations are called non-ideal solution.
2. For an ideal solution,
   ΔH_mixing ≠ 0, ΔV_mixing ≠ 0
3. Example: Ethyl alcohol and Cyclo hexane, Benzene and aceton.

Question 37 (a).
(I) What is dipole moment?
(II) Describe Fajan’s rule?

[OR]

(b) (I) How does Huckel rule help to decide the aromatic character of a compound?
(II) Draw cis-trans isomers for the following compounds
(a) 2- chloro-2-butene
(b) CH₃-CCl=CH-CH₂CH₃
Answer:
(a)
(I)

1. The polarity of a covalent bond can be measured in terms of dipole moment which is defined as: m = q × 2d, where m is the dipole moment, q is the charge, 2d is the distance between the two charges.
2. The dipole moment is a vector quantity and the direction of the dipole moment points from the negative charge to positive charge.
3. The unit of dipole moment is Coulomb metre (C m). It is usually expressed in Debye unit (D).
4. 1 Debye = 3.336 × 10^-3o C m

(II)

1. The ability of a cation to polarise an anion is called its polarising ability and the tendency of ah anion to get polarised is called its polarisibility. The extent of polarisation in an ionic compound is given by the Fajans rule.

2. To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the cation greater will be the attraction on the electron cloud of the anion. Similarly higher the magnitude of negative charge on anion, greater is its polarisability. For example, Na⁺ < Mg²⁺ < Al³⁺, the covalent character also follows the order:
NaCl < MgCl₂ < AlCl₃.

3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation, e.g., LiCl is more covalent than NaCl.

4. Cation having ns² np⁶ nd⁰ configuration shows greater polarising power than the cations with ns² np⁶ configuration, e.g., CuCl is more covalent than NaCl.

[OR]

(b) (I) A compound is said to be aromatic, if it obeys the following rules:

1. The molecule must be cyclic.
2. The molecule must be co-planar.
3. Complete delocalisation of rc-electrons in the ring.
4. Presence of (4n + 2)π electrons in the ring where n is an integer (n = 0, 1, 2 …)

This is known as Huckel’s rule.
Example –  – Benzene

1. It is cyclic one.
2. It is a co-planar molecule.
3. It has six delocalised n electrons.
4. 4n + 2 = 6

4n = 6 – 2
4n = 4
⇒ n = 1
It obey Huckel’s rule, with n = 1, hence benzene is aromatic in nature.

(II) (a) 2-Chloro-2-butene:

(b)

Question 38 (a).
(I) Reagents and the conditions used in the reactions are given below. Complete the table by writing down the product and the name of the reaction?
Answer:

(II) What is the IUPAC name of the insecticide DDT? Why is their use banned in most of the countries?

[OR]

(b) (I) Explain about green chemistry in day-to-day life?
(II) How acetaldehyde is commercially prepared by green chemistry?
Answer:
(a)
(I)

(II)

•  The IUPAC name of the insecticide DDT is p, p’-dichloro-diphenyl trichloroethane.
• Even DDT is an effective insecticide. Now-adays it is banned because of its long term toxic effects.
• DDT is very persistent in the environment and it has a high affinity for fatty tissues. As a result, DDT gets accumulated in animal tissue fat, in particular that of birds of prey with subsequent thinning of their eggs shells and impacting their rate of reproduction. That is why DDT is banned in most of the countries.

[OR]

(b)
(I)
1. Dry cleaning of clothes:
Solvents like tetrachloroethylene used in dry cleaning of clothes, pollute the ground water and are carcinogenic. In place of tetrachloro ethylene, liquefied CO₂ with suitable detergent is an alternate solvent used. Liquefied CO₂ is not harmful to the ground water. Nowadays H₂O₂ is used for bleaching clothes in laundry, gives better result and utilises less water.

2. Bleaching of paper:
Conventional method of bleaching was done with chlorine. Nowadays H₂O₂ can be used for bleaching paper in the presence of catalyst.

1. Instead of petrol, methaftol is used as a fuel in automobiles.
2. Neem based pesticides have been synthesised, which are more safer than the chlorinated hydrocarbons.

(II) Acetaldehyde is commericially prepared by one step oxidation of ethene in the presence of ionic catalyst in aqueous medium with 90% yield.

Students can Download Tamil Nadu 11th Physics Model Question Paper 5 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 5 English Medium

General Instructions:

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

Time: 3 Hours
Maximum Marks: 70

PART – I

Answer all the questions: [15 × 1 = 15]

Question 1.
Two protons are travelling. along the same straight path but in opposite directions. The relative velocity between the two is ……………………
(a) c
(b) \(\frac{c}{2}\)
(c) 2c
(d) 0
Hint:
One of the velocity V₁ = V; Other velocity V₂ = -V
Relative velocity (V_rel) = \(\frac{V_{1}-V_{2}}{\left[1+\frac{V^{2}}{C^{2}}\right]}\) = \(\frac{V-(-V)}{\left[1+\frac{V^{2}}{C^{2}}\right]}\) = \(\frac{2v}{2}\)
V = C
V_relative = C
Answer:
(a) c

Question 2.
If the Earth stops rotating about its own axis, g remains unchanged at …………………..
(a) Equator
(b) Poles
(c) Latitude of 45°
(d) No where
Answer:
(b) Poles

Question 3.
When train stops, the passenger moves forward. It is due to ……………………
(a) Inertia of passenger
(b) Inertia of train
(c) Gravitational pull by Earth
(d) None of the above
Answer:
(a) Inertia of passenger

Question 4.
A particle of mass m moves in the xy plane with a velocity v along the straight line AB. If the angular momention of the particle with respect to origin O is L_A when it is at A and L_B when it is at B, then …………………….
(a) L_A = L_B
(b) L_A < L_B
(c) L_A > L_B
(d) The relationship between L_A and L_B depends uopn the slope of the line AB

Hint:
Magnitude of L is, L = mvr sin ϕ = mvd
d = r sin ϕ is the distance of closest approach of the particle so the origin, as ‘d’ is same for both particles.
So, L _A = L_B
Answer:
(a) L_A = L_B

Question 5.
A couple produces ……………………….
(a) Pure rotation
(b) Pure translation
(c) Rotation and translation
(d) No motion
Answer:
(a) Pure rotation

Question 6.
A body starting from rest has an acceleration of 20 ms~2 the distance travelled by it in the sixth second is ……………………..
(a) 110 m
(b) 130m
(c) 90m
(d) 50 m
Hint:
Distance travelled in n^th second, u = 0
S_n = u + \(\frac{1}{2}\)a(2n – 1)
S₆ = 0 + \(\frac{1}{2}\) × 20 × (2 × 6 – 1); S₆ = 110 m
Answer:
(a) 110 m

Question 7.
A lift of mass 1000 kg, which is moving with an acceleration of 1m/s² in upward direction has tension has developed in its string is ………………………
(a) 9800 N
(b) 10800 N
(c) 11000 N
(d) 10000 N
Hint:
Tension, T = mg + ma = m(g + a) = 1000 (10 + 1)
T = 11000 N
Answer:
(c) 11000 N

Question 8.
The relation between acceleration and displacement of four particles are given below ………………………..
(a) a_x = 2x
(b) a_x = + 2x²
(c) a_x = -2x²
(d) a_x = -2x
Answer:
(d) a_x = -2x

Question 9.
A sonometer wire is vibrating in the second overtone. In the wire there are, ……………………..
(a) Two nodes and two antinodes
(b) One node and two antinodes
(c) Four nodes and three antinodes
(d) Three nodes and three antinodes
Answer:
(d) Three nodes and three antinodes

Question 10.
Which of the following is the graph between the light (h) of a projectile and time (t), when it is projected from the ground ………………………..
(a)
(b)
(c)
(d)
Answer:
(c)

Question 11.
According to kinetic theory of gases, the rms velocity of the gas molecules is directly proportional to ………………………
(a) \(\sqrt{T}\)
(b) T³
(c) T
(d) T⁴
Hint:
The rms velocity, V_rms = \(\sqrt{3KT/m}\) ⇒ V_rms ∝ \(\sqrt{T}\)
Answer:
(a) \(\sqrt{T}\)

Question 12.
A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. The ratio of K.E of colliding body before and after collision will be ……………………..
(a) 1 : 1
(b) 2 : 1
(c) 4 : 1
(d) 9 : 1
Hint:
KE of colliding bodies before collision = \(\frac{1}{2}\) mv²
After collision the mass = m + 2m = 3m
velocity becomes V’ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)v = \(\frac{mv}{3m}\) = \(\frac{v}{3}\)
KE after collision = \(\frac{1}{2}\)m (\(\frac{V}{3}\)² = \(\frac{1}{9}\) (\(\frac{1}{2}\) mv²)
\(\frac{\mathrm{KE}_{\text {before }}}{\mathrm{KE}_{\text {after }}}=\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{9}\left(\frac{1}{2} \mathrm{mv}^{2}\right)}=9: 1\)
Answer:
(d) 9 : 1

Question 13.
Four particles have velocity 1, 0, 2 and 3ms^-1. The root mean square velocity of the particles is ……………………..
(a) 3.5 ms^-1
(b) \(\sqrt{3.5}\) ms^-1
(c) 1.5ms^-1
(d) Zero
Hint:

Answer:
(b) \(\sqrt{3.5}\) ms^-1

Question 14.
Two vibrating tuning forks produce progressive waves given by y₁ = 4 sin 500 πt and y₂ = 2 sin 506 πt where t is in seconds Number of beat produced per minute is ………………………
(a) 360
(b) 180
(c) 3
(d) 60
Hint:

f₂ – f₁ = 3 = beats per sec and 3 × 60 = 180 beats per min

Answer:
(b) 180

Question 15.
Workdone by a simple pendulum in one complete oscillation is …………………………..
(a) Zero
(b) Jmg
(c) mg cos θ
(d) mg sin θ
Answer:
(a) Zero

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
A girl is swinging in the sitting position. How will the period of the swing be changed if she stands up?
Answer:
This can be explained using the concept of a simple pendulum. We know that the time period of a simple pendulum is given by
T = 2π\(\sqrt{l/g}\) i.e; T ∝\(\sqrt{l}\)
When the girl stands up, the distance between the point of suspension and the centre of mass of the swinging body decreases i.e., l decreases, so T will also decrease.

Question 17.
A car starts to move from rest with uniform acceleration 10 ms^-2 then after 2 sec, what is its velocity?
Answer:
a =10 ms^-2;
t = 2s;
w = 0;
v = ?
v = u + at
v = 0 + 10 × 2
= 20 ms^-1

Question 18.
State Lami’s theorem?
Answer:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces. The constant of proportionality is same for all three forces.

Question 19.
Due to the action of constant torque, a wheel from rest makes n rotations in t seconds? Show that the angular acceleration of a wheel as \(\frac{4 \pi n}{t^{2}}\) rad s^-2
Answer:
Initial angluar velocity ω₀ = 0
Number of rotations in t seconds = n
angular displacement θ = 2πn
but, θ = ω₀t + \(\frac{1}{2}\)αt²
2πn = \(\frac{1}{2}\)αt²
α = \(\frac{4 \pi n}{t^{2}}\)

Question 20.
Why a given sound is louder in a hall than in the open?
Answer:
In a hall, repeated reflections of sound take place from the walls and the ceiling. These reflected sounds mix with original sound which results in increase the intensity of sound. But in open, no such a repeated reflection is possible. .’. sound will not be louder as in hall.

Question 21.
What are the differences between connection and conduction?
Answer:
Conduction:
Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.

Convection:
Convection is the process in which heat transfer is by actual movement of molecules in fluids such as liquids and gases. In convection, molecules move freely from one place to another.

Question 22.
Why two holes are made to empty an oil tin?
Answer:
When oil comes out through a tin with one hole, the pressure inside the tin becomes less than the atmospheric pressure, soon the oil stops flowing out. When two holes are made in the tin, air keeps on entering the tin through the other hole and maintains pressure inside.

Question 23.
If the length of the simple pendulum is increased by 44% from its original length, calculate the percentage increase in time period of the pendulum?
Answer:
Since T ∝ \(\sqrt{l}\) = Constant \(\sqrt{l}\)

∴ T_f = 1.2 T_i = T_i + 20% T_i

Question 24.
When do the real gases obey more correctly the gas equation PV = nRT?
Answer:
An ideal gas is one whose molecules have zero volume and no mutual force between them. At low pressure, the volume of a gas is large and so the volume occupied by the molecules is negligible in comparison to the volume of the gas.

At high temperature, the molecules have large velocities and so the intermolecular force has no influence on their motion. Hence at low pressure and high temperature, the behaviour of real gases approach the ideal gas behaviour.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
A stone is thrown upwards with a speed v from the top of a tower. It reaches the ground with a velocity 3v. What is the height of the tower?
Answer:
From equation of motion,
v’ = u + at ……………….. (1)
h = ut + \(\frac{1}{2}\) at²
here, v’ = av; u = v; a = +g
Using equ. (1)
3v = v + gt ⇒ 3v – v = gt
t = \(\frac{2v}{g}\)

Substitute ‘t’ value in equ. (2)
h = v(\(\frac{2v}{g}\)) + \(\frac{1}{2}\)g(\(\frac{2v}{g}\))² = \(\frac { 2v^{ 2 } }{ g } \) + \(\frac{1}{2}\)g (\(\frac{2v}{g}\))²
h = \(\frac { 2v^{ 2 } }{ g } \) + \(\frac { 2v^{ 2 } }{ g } \)
= \(\frac { 4v^{ 2 } }{ g } \) g = 10 ms^-2
= \(\frac { 4v^{ 2 } }{ 10 } \); h = \(\frac { 2v^{ 2 } }{ 5 } \)

Question 26.
An object is projected at an angle such that the horizontal range is 4 times of the maximum height. What is the angle of projection of the object?
Answer:
Given:
Horizontal range = 4 H_max
Horizontal range = \(\frac{u^{2} \sin 2 \theta}{g}\) = \(\frac{2 u^{2} \sin \theta \cos \theta}{g}\)
Maximum height = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
as given, \(\frac{2 u^{2} \sin \theta \cos \theta}{g}\) = \(\frac{4 u^{2} \sin ^{2} \theta}{2 g}\)
2 cos θ = 2 sin θ
tan θ = 1
∴ θ = 45°

Question 27.
A room contains oxygen and hydrogen molecules in the ratio 3 : 1. The temperature of the room is 27°C. The molar mass of 0₂ is 32 g mol^-1 and for H₂, 2 g mol^-1. The value of gas constant R is 8.32 J mol^-11 k^-1. Calculate rms speed of oxygen and hydrogen molecule?
Answer:
(a) Absolute Temperature T = 27°C = 27 + 273 = 300 K.
Gas constant R = 8.32 J mol^-1 K^-1
For Oxygen molecule: Molar mass

M = 32 gm/mol = 32 × 10^-3 kg mol^-1
rms speed v_rms = \(\sqrt { \frac { 3RT }{ M } } \) = \(\sqrt{\frac{3 \times 8.32 \times 300}{32 \times 10^{-3}}}\) = 483.73 ms^-1 ~ 484ms^-1

For Hydrogen molecule: Molar mass M = 2 × 10^-3 kg mol^-1
rms speed v_rms = \(\sqrt { \frac { 3RT }{ M } } \) = \(\sqrt{\frac{3 \times 8.32 \times 300}{2 \times 10^{-3}}}\) = 1934 ms^-1 = 1.93 K ms^-1

Note that the rms speed is inversely proportional to \(\sqrt{M}\) and the molar mass of oxygen is 16 times higher than molar mass of hydrogen. It implies that the rms speed of hydrogen is 4 times greater than rms speed of oxygen at the same temperature. \(\frac{1934}{484}\) ~ 4.

Question 28.
Explain about an angle of friction?
Answer:
The angle of friction is defined as the angle between the normal force (N) and the resultant force (R) of normal force and maximum friction force (\(f_{s}^{\max }\))

In the figure the resultant force is R = \(\sqrt{\left(f_{s}^{\max }\right)^{2}+\mathrm{N}^{2}}\)
tan θ = \(\frac{f_{s}^{\max }}{\mathrm{N}}\) ………………….. (1)

But from the frictional relation, the object begins to slide when \(f_{s}^{\max }=\mu_{\mathrm{s}} \mathrm{N}\)
or when \(\frac{f_{s}^{\max }}{\mathrm{N}}\) = µ_s ………………….. (2)

From equations (1) and (2) the coefficient of static friction is
µ_s = tan θ ……………………. (3)
The coefficient of static friction is equal to tangent of the angle of friction.

Question 29.
How does resolve a vector into its component? Explain?
Answer:
component of a resolve:
In the Cartesian coordinate system any vector \(\vec { A } \) can be resolved into three components along x, y and z directions. This is shown in figure. Consider a 3-dimensional coordinate system. With respect to this a vector can be written in component form as
\(\vec { A } \) = A_x \(\hat { i } \) + A_y\(\hat { j } \) + A_z\(\hat { k } \)
Components of a vector in 2 dimensions and 3 dimensions

Here Ax is the x-component of \(\vec { A } \), Ay is the y-component of \(\vec { A } \) and A_z is the z component of \(\vec { A } \).
In a 2-dimensional Cartesian coordinate system (which is shown in the figure) the vector \(\vec { A } \) is given by
\(\vec { A } \) = A_x \(\hat { i } \) + A_y \(\hat { j } \)

If \(\vec { A } \) makes an angle θ with x axis, and A_x and A_y are the components of A along x-axis and y-axis respectively, then as shown in figure,
A_x θ = A cos θ, A = A sin θ
where ‘A’ is the magnitude (length) of the vector \(\vec { A } \), A = \(\sqrt{\mathrm{A}_{x}^{2}+\mathrm{A}_{y}^{2}}\)

Question 30.
Derive an expression for energy of an orbiting satellite?
Answer:
The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,
U = \(-\frac{\mathrm{GM}_{s} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\) ………………….. (1)

Here M_s -mass of the satellite, M_E -mass of the Earth, R_E – radius of the Earth.
The Kinetic energy of the satellite is
K.E = \(\frac{1}{2}\) M_sv² ………………….. (2)

Here v is the orbital speed of the satellite and is equal to
v = \(\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}}\)

Substituting the value of v in (2) the kinetic energy of the satellite becomes,
K.E = \(\frac{1}{2} \frac{\mathrm{GM}_{\mathrm{E}} \mathrm{M}_{s}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\)

Therefore the total energy of the satellite is
E = \(\frac{1}{2}\) \(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{M}_{s}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}-\frac{\mathrm{GM}_{s} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\)
E = \(-\frac{\mathrm{GM}_{s} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\)
The total energy implies that the satellite is bound to the Earth by means of the attractive gravitational force.

Note:
As h approaches ∞, the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth’s gravity and is not bound to Earth at large distance.

Question 31.
Explain in detail Newton’s law of cooling?
Answer:
Newton’s law of cooling:
Newton’s law of cooling body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{dQ}{dT}\) ∝(T – T_s) …………………. (1)
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,

T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.

Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,
dQ = msdT …………………….. (2)

Dividing both sides of equation (2) by dt
\(\frac{dQ}{dT}\) = \(\frac{msdT}{dt}\) ………………….. (3)

From Newton’s law of cooling
\(\frac{dQ}{dT}\) ∝(T – T_s)
\(\frac{dQ}{dT}\) = -a(T – T_s) ………………………… (4)

Where a is some positive constant.
From equation (3) and (4)
-a (T – T_s) = ms\(\frac{dT}{dt}\)
\(\frac{d \mathrm{T}}{\mathrm{T}-\mathrm{T}_{s}}\) = -a\(\frac{a}{ms}\) dt …………………… (5)

Integrating equation (5) on both sides,

Where b₁ is the constant of integration. Taking exponential both sides we get,
T = T_s + \(b_{2} e^{\frac{-a}{m s} t}\) …………………….. (6)
Here b₂ = e^b1 Constant

Question 32.
Explain Laplace’s correction?
Answer:
Laplace’s correction: In 1816, Laplace satisfactorily corrected this discrepancy by assuming that when the sound propagates through a medium, the particles oscillate very rapidly such that the compression and rarefaction occur very fast.

Hence the exchange of heat produced due to compression and cooling effect due to rarefaction do not take place, because, air (medium) is a bad conductor of heat. Since, temperature is no longer considered as a constant here, sound propagation is an adiabatic process. By adiabatic considerations, the gas obeys Poisson’s law (not Boyle’s law as Newton assumed), which is
PV^γ = Constant …………………. (1)
where, γ = \(\frac { C_{ p } }{ C_{ v } } \) which is the ratio between specific heat at constant pressure and specific heat at constant volume. Differentiating equation (1) on both the sides, we get
\(V^{ \gamma }dP+P(\gamma ^{ V\gamma -1 }dV)=0\)

or

\(\gamma \mathrm{P}=-\mathrm{V} \frac{d p}{d \mathrm{V}}=\mathrm{B}_{\mathrm{A}}\) ………………….. (2)
where, B_A is the adiabatic bulk modulus of air. Now, substituting equation (2) in equation
V = \(\sqrt{\frac{B}{\rho}}\), the speed of sound in air is
v_A = \(\sqrt{\frac{\mathrm{B}_{\mathrm{A}}}{\rho}}=\sqrt{\frac{\gamma \mathrm{P}}{\rho}}=\sqrt{\gamma v_{\mathrm{T}}}\)
Since air contains mainly, nitrogen, oxygen, hydrogen etc, (diatomic gas), we take γ = 1.47. Hence, speed of sound in air is v_A = (\(\sqrt{1.4}\)) (280 m s^-1) = 331.30 ms^-1, which is very much closer to experimental data.

Question 33.
Explain types of equilibrium?
Answer:

PART – IV

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

Question 34 (a).
What are the applications of dimensional analysis?
Verify s = ut + \(\frac{1}{2}\)at² by dimensional analysis?
Answer:

Given equation is dimensionally correct as the dimensions on the both side are same. Applications of dimensional analysis

1. Convert a physical quantity from one system of units to another.
2. Check the dimensional correctness of a given physical equation.
3. Establish relations among various physical quantities.

[OR]

(b) Explain the types of equilibrium with suitable examples?
Answer:
Translational motion – A book resting on a table.
Rotational equilibrium – A body moves in a circular path with constant velocity.
Static equilibrium – A wall-hanging, hanging on the wall.
Dynamic equilibrium – A ball decends down in a fluid with its terminal velocity.
Stable equilibrium – A table on the floor A pencil
Unstable equilibrium – standing on its tip.
Neutral equilibrium – A dice rolling on a game board.

Question 35 (a)
Explain the motion of block connected by a string in vertical motion?
Answer:
When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁ > m₂) connected by a light and inextensible string that passes over a pulley as shown in Figure 1.

Let the tension in the string be T and acceleration a.

When the system is released, both the blocks start, Two blocks connected by a string moving, m₂ vertically upward and m₁ downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂.

The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure 2.

Applying Newton’s second law for mass m₂
T\(\hat { j } \) – m₂\(\hat { j } \)g = m₂ a\(\hat { j } \)

The left hand side of the above equation is the total force that acts on m₂ and the right hand side is the product of mass and acceleration of m₂ in direction.

By comparing the components on both sides, we get
T -m₂g = m₂a ……………………. (1)

Similarly, applying Newton’s law second law of for mass m₁
T\(\hat { j } \) – m₁g\(\hat { j } \) = -m₁a\(\hat { j } \)

As mass m₁ moves downward (-\(\hat { j } \)), its accleration is along (-\(\hat { j } \))
By comparing the components on both sides, we get
T – m₁g = -m₁a
m₁g – T = m₁a …………………….. (2)

Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁g – m₂)g = (m₁ + m₂)a …………………….. (3)

From equation (3), the acceleration of both the masses is
a = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)g ……………………… (4)

If both the masses are equal (m₁ = m₂), from equation (4)
a = 0

This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest. To find the tension acting on the string, substitute the acceleration from the equation (4)
T – m₂g = m₂ \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)g
T = m₂g + m₂\(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\)g ……………………… (5)
By taking m₂g common in the RHS of equation (5)

Equation (4) gives only magnitude of accleration.

For mass m₁ g the accleration vector is given by \(\vec{a}=-\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g \hat{j}\)
For mass m₂ g the accleration vector is given by \(\vec{a}=-\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) g \hat{j}\)

[OR]

(b) Derive the kinematic equation of motion for constant acceleration?
Answer:
Consider an object moving in a straight line with uniform or constant acceleration ‘a’. Let u be the velocity of the object at time t = 0, and v be velocity of the body at a later time t.

Velocity – time relation:

(I) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time,
a = \(\frac{dv}{dt}\) or dv = a dt
Integrating both sides with the condition that as time changes from 0 to t, the velocity changes from u to v. For the constant acceleration,

Displacement – time relation:

(II) The velocity of the body is given by the first derivative of the displacement with respect to time.
v = \(\frac{ds}{dt}\) or ds = vdt
and since v = u + at,
We get ds = (u + at)dt
Assume that initially at time t = 0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time-independent, we have
\(\int_{0}^{s} d s=\int_{0}^{t} u d t+\int_{0}^{t} a t d t \text { or } s=u t+\frac{1}{2} a t^{2}\) ……………………. (2)
Velocity – displacement relation

(III) The acceleration is given by the first derivative of velocity with respect to time.
a = \(\frac{dv}{dt}\) = \(\frac{dv}{ds}\) \(\frac{ds}{dt}\) = \(\frac{dv}{ds}\) v [since ds/dt = v] where s is displacement traversed.
This is rewritten as a = \(\frac{1}{2}\) \(\frac { dv^{ 2 } }{ s } \) or ds = \(\frac{1}{2a}\) d(v²)
Integrating the above equation, using the fact when the velocity changes from u² to v², displacement changes from u² to v², we get

We can also derive the displacement s in terms of initial velocity u and final velocity v.
From equation we can write,
at = v – u
Substitute this in equation, we get
s = ut + \(\frac{1}{2}\) (v -u)t
s = \(\frac{(u+v)t}{2}\)

Question 36 (a).
State and prove perpendicular axis theorem?
Answer:
Perpendicular axis theorem: This perpendicular axis theorem holds good only for plane laminar objects.

The theorem states that the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of moments of inertia about two perpendicular axes lying in the plane of the body such that all the three axes are mutually perpendicular and have a common point.

Let the X and Y-axes lie in the plane and Z-axis perpendicular to the plane of the laminar object. If the moments of inertia of the body about X and Y-axes are Ix and IY respectively and Iz is the moment of inertia about Z-axis, then the perpendicular axis theorem could be expressed as,
I_z = I_x + I_y

To prove this theorem, let us consider a plane laminar object of negligible thickness on which lies the origin (O). The X and Y-axes lie on the plane and Z-axis is perpendicular to it as shown in figure. The lamina is considered to be made up of a large number of particles of mass m. Let us choose one such particle at a point P which has coordinates (x, y) at a distance r from O.

The moment of inertia of the particle about Z-axis is, mr².
The summation of the above expression gives the moment of inertia of the entire lamina about Z-axis as,
I_z = Σ mr²
Here, r² = x² + y²
Then, I_z = Σm(x² + y²)
I_z = Σmx² + Σmy²
In the above expression, the term Σmx² is the moment of inertia of the body about the Y-axis and similarly the term Σmy² is the moment of inertia about X-axis. Thus,
I_X = Σmy² and I_Y = Σmx²
Substituting in the equation for I_Z gives, I_Z = I_X + I_Y
Thus, the perpendicular axis theorem is proved.

[OR]

(b) Explain in detail the triangle law of addition?
Answer:
Let us consider two vectors \(\vec { A } \) and \(\vec { B } \) as shown in figure.

To find the resultant of the two vectors we apply the triangular law of addition as follows:

Represent the vectors \(\vec { A } \) and \(\vec { B } \) by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector \(\vec { A } \) is connected to the tail of the second vector \(\vec { B } \). Let θ be the angle between A and B. Then R is the resultant vector connecting the tail of the first vector \(\vec { A } \) to the head of the second vector \(\vec { B } \).

The magnitude of \(\vec { R } \) (resultant) is given geometrically by the length of \(\vec { R } \)(OQ) and the direction of the resultant vector is the angle between \(\vec { R } \) and \(\vec { A } \). Thus we write \(\vec { R } \) = \(\vec { A } \) + \(\vec { B } \).
\(\overline { OQ } \) = \(\overline { OP } \) + \(\overline { PQ } \)

1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows.

From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\vec { A } \) and \(\vec { B } \).

For ∆OBN, we have OB² = ON² + BN²
⇒ R² = (A + B cos θ)² + (B sin θ)²
⇒ R² = A² + B² cos² θ + 2AB cos θ + B²sin² θ
⇒ R² = A² + B²(cos² θ + sin² θ) + 2AB cos θ
⇒R² = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)

2. Direction of resultant vectors:
If θ is the angle between \(\vec { A } \) and \(\vec { B } \), then
|\(\vec { A } \) + \(\vec { B } \)| = \(\sqrt{A^{2}+B^{2}+2 A B \cos \theta}\)
If \(\vec { R } \) makes an angle α with \(\vec { A } \), then in ∆OBN,
tan α = \(\frac{BN}{ON}\) = \(\frac{BN}{OA+AN}\)
tan α = \(\frac{B \sin \theta}{A+B \cos \theta}\) ⇒ α = tan^-1\(\left(\frac{B \sin \theta}{A+B \cos \theta}\right)\)

Question 37 (a).
Explain in detail the various types of errors?
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.

(I) Systematic errors:
Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the .experiment. Systematic errors can be classified as follows.

(1) Instrumental errors:
When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.

(2) Imperfections in experimental technique or procedure:
These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied

(3) Personal errors:
These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.

(4) Errors due to external causes:
The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.

(5) Least count error:
Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.

(II) Random errors:
Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called “chance error”.

When different readings are obtained by a person every time he repeats the experiment, personal error occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.
If n number of trial readings are taken in an experiment, and the readings are
a₁, a₂, a₃, ……………………….. a_n. The arithmetic mean is

[OR]

(b) To move an object, which one is easier, push or pull? Explain?
Answer:
When a body is pushed at an arbitrary angle θ (0 to \(\frac{π}{2}\)), the applied force F can be resolved into two components as F sin θ parallel to the surface and F cos θ perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ …………………. (1)
As a result the maximal static friction also increases and is equal to
\(f_{s}^{\max }=\mu_{r} N_{\text {push }}=\mu_{s}(m g+F \cos \theta)\) …………………. (2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.

When an object is pulled at an angle 0, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is
N_pull = mg – F cos θ …………………….. (3)

Question 38 (a).
Describe the method of measuring angle of repose?
Answer:
Angle of Repose Consider an inclined plane on which an object is placed, as shown in figure. Let the angle which this plane makes with the horizontal be θ. For small angles of θ, the object may not slide down. As θ is increased, for a particular value of θ, the object begins to slide down. This value is called angle of repose. Hence, the angle of repose is the angle of inclined plane with the horizontal such that an object placed on it begins to slide.

Let us consider the various forces in action here. The gravitational force mg is resolved into components parallel (mg sin θ) and perpendicular (mg cos θ) to the inclined plane. The component of force parallel to the inclined plane (mg sin θ) tries to move the object down. The component of force perpendicular to the inclined plane (mg cos θ) is balanced by the Normal force (N).
N = mg cos θ ……………… (1)

When the object just begins to move, the static friction attains its maximum value
f_s = \(f_{s}^{\max }\)

This friction also satisfies the relation
\(f_{s}^{\max }\) = µ_s = mg sin θ …………………… (2)

Equating the right hand side of equations (1) and (2),
\(\left(f_{s}^{\max }\right) / \mathrm{N}=\sin \theta / \cos \theta\)

From the definition of angle of friction, we also know that
tan θ = µ_s ……………………… (3) in which θ is the angle of friction.

Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.

[OR]

(b) A block of mass m slides down the plane inclined at an angle 60° with an acceleration g/2. Find the co-efficient of kinetic friction?
Answer:
Kinetic friction comes to play as the block is moving on the surface.
The forces acting on the mass are the normal force perpendicular to surface, downward gravitational force and kinetic friction along the surface.
Along the x-direction

There is no motion along the y-direction as normal force is exactly balanced by the mg cos θ.
mg cos θ = N = mg/2

[OR]

(c) Write a note on triangulation method and radar method to measure larger distances?
Answer:
Triangulation method for the height of an accessible object:
Let AB = h be the height of the tree or tower to 6e measured. Let C be the point of observation at distance x from B. Place a range finder at C and measure the angle of elevation, ACB = θ as shown in figure.

From right angled triangle ABC,
tan θ = \(\frac{AB}{BC}\) = \(\frac{h}{x}\)
(or) height h = x tan θ
Knowing the distance x. the height h can be detennined.

RADAR method:
The word RADAR stands for radio detection and ranging. A radar can be used to measure accurately the distance of a nearby planet such as Mars. In this method, radio waves are sent from transmitters which, after reflection from the planet, are detected by the receiver.

By measuring, the time interval (t) between the instants the radio waves are sent and received, the distance of the planet can be determined as where v is the speed of the radio wave. As the time taken (t) is for the distance covered during the forward and backward path of the radio waves, it is divided by 2 to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.

[OR]

(d) Jupiter is at a distance of 824.7 million km from the earth. Its angular diameter is measured to be 35.75”. Calculate the diameter of jupiter?
Answer:
Given,
Distance of Jupiter = 824.7 × 10⁶ km = 8.247 × 10¹¹ m
angular diameter = 35.72 × 4.85 × 10^-6 rad = 173.242 × 10^-6 rad
= 1.73 × 10^-4rad
∴ Diameter of Jupiter D = θ × d = 1.73 × 10^-4 rad × 8.247 × 10¹¹ m
14.267 × 10⁷ m = 1.427 × 10⁸ m (or) 1.427 × 10⁵ km

Students can Download Tamil Nadu 11th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 11th Physics Model Question Paper 4 English Medium

General Instructions:

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

Time: 3 Hours
Maximum Marks: 70

PART – 1

Answer all the questions: [15 × 1 = 15]

Question 1.
In some region, the gravitational field is zero. The gravitational potential in this region is …………………….
(a) A variable
(b) A constant
(c) Zero
(d) Can’t be zero
Answer:
(b) A constant

Question 2.
The angle between two vectors 2\(\hat { i } \) + 3\(\hat { j } \) + \(\hat { k } \) and -3\(\hat { j } \) + 6k is …………………..
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Hint:
The angle between the two vector is 90°.
[cos θ = \(\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\mathrm{AB}}\)]
Answer:
(d) 90°

Question 3.
A stretched rubber has
(a) Increased kinetic energy
(b) Increased potential energy
(c) Decreased kinetic energy
(d) The axis of rotation
Answer:
(b) Increased potential energy

Question 4.
The direction of the angular velocity vector is along ……………………..
(a) The tangent to the circular path
(b) The inward radius
(c) The outward radius
(d) The axis of rotation
Answer:
(d) The axis of rotation

Question 5.
For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is …………………….
(a) \(\frac{1}{2}\)
(b) \(\frac { 1 }{ \sqrt { 2 } } \)
(c) 2
(d) \(\sqrt{2}\)
Hint:

Answer:
(a) \(\frac{1}{2}\)

Question 6.
If the linear momentum of the object is increased by 0.1 %, then the kinetic energy is increased by ………………………..
(a) 0.1%
(b) 0.2%
(c) 0.4%
(d) 0.01%
Hint:
The relation b/w linear momentum and kinetic energy
P = \(\sqrt{2 \mathrm{mE}_{\mathrm{K}}} \Rightarrow \mathrm{E}_{\mathrm{K}}=\frac{\mathrm{P}^{2}}{2 \mathrm{m}}\)
If L is increased by 0.1% 1′ = P + \(\frac{0.1}{100}\)P

Answer:
(c) 0.4%

Question 7.
A block of mass 4 kg is suspended through two light spring balances A and B. Then A and B will read respectively ………………………
(a) 4 kg and 0 kg
(b) 0 kg and 4 kg
(c) 4 kg and 4 kg
(d) 2kg and 2 kg
Hint:
Tension is uniformly transmitted if the springs are massless.
Answer:
(c) 4 kg and 4 kg

Question 8.
At the same temperature, the mean kinetic energies of molecules of hydrogen and oxygen are in the ratio of ………………………
(a) 1 : 1
(b) 1 : 16
(c) 8 : 1
(d) 16 : 1
Hint:
Average kinetic energy of a molecule is proportional to the absolute temperature.
Answer:
(a) 1 : 1

Question 9.
A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. Which of the following plots shows the correct variation of speed v with height h from the lower end?
(a)
(b)
(c)
(d)
Answer:

(d)

Question 10.
In a simple hormonic oscillation, the acceleration against displacement for one complete oscillation will be ………………………
(a) An ellipse
(b) A circle
(c) A parabola
(d) A straight line
Answer:
(d) A straight line

Question 11.
A shell is fired from a canon with velocity v m/s at an angle 0 with the horizontal direction. At the highest point in the path it explodes into two pieces of equal mass. One of the pieces retraces it path of the cannon and the speed in m/s of the other piece immediately after the. explosion is ………………………..
(a) 3v cos θ
(b) 2v cos θ
(c) \(\frac{3}{2}\)v cos θ
(d) \(\frac { \sqrt { 3 } }{ 2 } \) cos θ
Hint:
Velocity at the highest point = horizontal component of velocity = V cos θ
Momentum of shell before explosion = mv cos θ
Momentum of two pieces after explosion = \(\frac{m}{2}\) (- V cos θ) + \(\frac{m}{2}\) v
Law of conservation of momentum mv cos θ = –\(\frac{mv}{2}\)cos θ + \(\frac{m}{2}\) v
∴ V = 3V cos θ
Answer:
(a) 3v cos θ

Question 12.
In which process, the p – v indicator diagram is a straight line parallel to volume axis?
(a) Isothermal
(b) Adiabatic
(c) Isobaric
(d) Irreversible
Answer:
(c) Isobaric

Question 13.
For a liquid to rise in capillary tube, the angle of contact should be ………………………….
(a) Acute
(b) Obtuse
(c) Right
(d) None of these
Answer:
(a) Acute

Question 14.
The increase in internal energy of a system is equal to the work done on the system which process does the system undergo?
(a) Isochoric
(b) Adiabatic
(c) Isobaric
(d) Isothermal
Answer:
(b) Adiabatic

Question 15.
The change in frequency due to Doppler effect does not depend on …………………………
(a) The speed of the source
(b) The speed of the observer
(c) The frequency of the source
(d) Separation between the source and the observer
Answer:
(d) Separation between the source and the observer

PART – II

Answer any six questions in which Q. No 23 is compulsory. [6 × 2 = 12]

Question 16.
What are the advantages of SI system?
Answer:

1. This system makes use of only one unit for one physical quantity, which means a rational system of units
2. In this system, all the derived units can be easily obtained from basic and supplementary units, which means it is a coherent system of units.
3. It is a metric system which means that multiples and submultiples can be expressed as powers of 10.

Question 17.
Using the principle of homogunity of dimensions, check dimensionally the given equations are correct,
(a) \(\mathbf{T}^{2}=\frac{4 \pi^{2} r^{3}}{\mathbf{G}}\)
(b) \(T^{2}=\frac{4 \pi^{2} r^{3}}{G M}\)?
Answer:
Here G-gravitational constant
r – radius of orbit M – mass
Dimensional formula for T = T
Dimensional formula for r = L
Dimensional formula for G = M^-1L³T^-2
Dimensional formula for M = M
(a) T² M° L° = [L³] [M L^-3T^-2]
T² M° L° = L° MT² – Not correct

(b) T² = [L³][ML^-3T²][M^-1]
T² = T² – Dimensionally correct

Question 18.
Find out the workdone required to extract water from the well of depth 20 m. Weight of water and backet is 2.8 kg wt?
Answer:
Workdone, W = mgh
W = (Weight) × depth
= 2.8 × 20
W = 56 J

Question 19.
Is a single isolated force possible in nature?
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 20.
State the factors on which the moment of inertia of a body depends?
Answer:

1. Mass of body
2. Size and shape of body
3. Mass distribution w.r.t. axis of rotation
4. Position and orientation of rotational axis

Question 21.
If a drop of water falls on a very hot iron, it takes long time to evaporate. Explain why?
Answer:
When a drop of water falls on a very hot iron it gets insulated from the hot iron due to a thin layer of water vapour which is a bad conductor of heat. It takes quite long to evaporate as heat is conducted from hot iron to the drop through the insulating layer of water vapour very slowly.

Question 22.
What is meant by gravitational field. Give its unit?
Answer:
The gravitational field intensity \(\vec { E } \)₁, at a point is defined as the gravitational force experienced by unit mass at that point. It’s unit N kg^-1.

Question 23.
Why dimensional methods are applicable only up to three quantities?
Answer:
Understanding dimensions is of utmost importance as it helps us in studying the nature of physical quantities mathematically. The basic concept of dimensions is that we can add or subtract only those quantities which have same dimensions. Also, two physical quantities are equal if they have same dimensions, these basic ideas help us in deriving the new relation between physical quantities, it is just like units.

Question 24.
Why does a porter bend forward while carrying a sack of rice on his back?
Answer:
When a porter carries a sack of rice, the line of action of his centre of gravity will go away from the body. It affects the balance, to avoid this he bends. By which centre of gravity will realign within the body again. So balance is maintained.

PART – III

Answer any six questions in which Q.No. 29 is compulsory. [6 × 3 = 18]

Question 25.
Explain variation of ‘g’ with latitude?
Answer:
When an object is on the surface for the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
OP_z, cos λ = \(\frac{PZ}{OP}\) = \(\frac{R’}{R}\)
R’ = R cos λ

where λ is is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
a_PQ = ω²R cos λ = ω²R cos² λ
Since R’ = R cos λ
Therefore, g’ = g – ω’²R cos²λ
From the above expression, we can infer that at equator, λ = 0, g’ = g – ω²R. The acceleration due to gravity is minimum. At poles λ = 90; g1 = g. it is maximum. At the equator, g’ is minimum.

Question 26.
Cap you associate a vector with
(a) the length of a wire bent into a loop
(b) a plane area
(c) a sphere.
Answer:
(a) We cannot associate a vector with the length of a wire bent into a loop, this is cause the length of the loop does not have a definite direction.

(b) We can associate a vector with a plane area. Such a vector is called area vector and its direction is represented by a normal drawn outward to the area.

(c) The area of a sphere does not point in any difinite direction. However, we can associate a null vector with the area of the sphere. We cannot associate a vector with the volume of a sphere.

Question 27.
What is mean by inertia? Explain its types with example?
Answer:
The inability of objects to move on its own or change its state of motion is called inertia.
Inertia means resistance to change its state. There are three types of inertia:

1. Inertia of rest: The inability of an object to change its state of rest is called inertia of rest.
Example:

• When a stationary bus starts to move, the passengers experience a sudden backward push.
• A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion: The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

• When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
• An athlete running in a race will continue to run even after reaching the finishing point.

3. Inertia of direction: The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

• When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
• When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Question 28.
A light body and body with greater mass both are having equal kinetic energy. Among these two which one will have greater linear momentum?
Answer:
Given Data:
E₁ – E₂

then P₂ > P₁
i.e; a heavier body has greater linear momemtum.

Question 29.
Three particles of masses m₁ = 1 kg, m₂ = 2 kg and m₃ = 3 kg are placed at the corners of an equilateral triangle of side lm as shown in Figure. Find the position of center of mass?
Answer:
The center of mass of an equilateral triangle lies at its geometrical center G.
The positions of the mass m₁, m₂ and m₃ are at positions A, B and C as shown in the Figure. From the given position of the masses, the coordinates of the masses m₁ and m₂ are easily marked as (0,0) and (1,0) respectively.

To find the position of m3 the Pythagoras theorem is applied. As the ∆DBC is a right angle triangle,
BC² = CD² + DB²
CD² = BC² – DB²
CD² = 1² – (\(\frac{1}{2}\))² = 1 – (\(\frac{1}{4}\)) = \(\frac{3}{4}\)
CD = \(\frac { \sqrt { 3 } }{ 2 } \)

The position of mass m₃ is or (0.5, 0.5\(\sqrt{3}\))
X coordìnate of center of mass,
y_CM = \(\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}\)
y_CM = \(\frac { \sqrt { 3 } }{ 4 } \) m
∴ The coordinates of center of mass G (x_CM, y_CM) is (\(\frac{7}{12}\), \(\frac { \sqrt { 3 } }{ 4 } \))

Question 30.
Define precision and accuracy. Explain with one example?
Answer:
The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision of a measurement is a closeness of two or more measured values to each other.

The true value of a certain length is near 5.678 cm. In one experiment, using a measuring instrument of resolution 0.1 cm, the measured value is found to be 5.5 cm. In another experiment using a measuring instrument of greater resolution, say 0.01 cm, the length is found to be 5.38 cm. We find that the first measurement is more accurate as it is closer to the true value, but it has lesser precision. On the contrary, the second measurement is less accurate, but it is more precise.

Question 31.
Derive an expression for total acceleration in the non uniform circular motion?
Answer:
If the speed of the object in circular motion is not constant, then we have non-uniform circular motion. For example, when the bob attached to a string moves in vertical circle, the speed of the bob is not the same at all time. Whenever the speed is not same in circular motion, the particle will have both centripetal and tangential acceleration as shown in the figure.

The resultant acceleration is obtained by vector sum of centripetal and tangential acceleration. Since centripetal acceleration is \(\frac { v^{ 2 } }{ r } \), the magnitude of this resultant acceleration is given by
a_R = \(\sqrt{a_{t}^{2}+\left(\frac{v^{2}}{r}\right)^{2}}\)

This resultant acceleration makes an angle θ with the radius vector as shown in figure.
This angle is given by tan θ = \(\frac{a_{t}}{\left(v^{2} / r\right)}\)

Question 32.
Calculate the value of adiabatic exponent for monoatomic molecule?
Answer:
Monoatomic molecule:
Average kinetic energy of a molecule = [\(\frac{3}{2}\)kT]
Total energy of a mole of gas = \(\frac{3}{2}\) kT × N_A = \(\frac{3}{2}\)RT
For one mole, the molar specific heat at constant volume

Question 33.
At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth’s surface, (mass of oxygen molecule : 2.76 × 10^-26kg Boltzmann’s constant (k_B) = 1.38 × 10^-23 J mol^-1 k^-1
Answer:

PART – IV

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

Question 34 (a).
State kegler’s laws of planetary motion?
Answer:
1. Law of Orbits:
Each planet moves around the Sun in an elliptical orbit with the Sun at one of the foci.

2. Law of area:
The radial vector (line joining the Sun to a planet) sweeps equal areas in equal intervals of time.

3. Law of period:
The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi-major axis of the ellipse.
T² ∝ a³
\(\frac { T^{ 2 } }{ a^{ 3 } } \) = Constant

(b) The distance of planet Jupiter from the sun is 5.2 times that of the earth. Find the period of resolution of Jupiter around the sun?
Answer:
Here r₁ = 5.2 r_e; T_J = ?; T_e = 1 year

= 11.86 years

[OR]

(c) Explain the propagation of errors in multiplication?
Answer:
Error in the product of two quantities:
Let ∆A and ∆B be the absolute errors in the two quantities A, and B, respectively. Consider the product Z = AB,
The error AZ in Z is given by Z ± AZ = (A ± ∆A) (B ± ∆B)
= (AB) ± (A ∆ B) ± (B ∆ A) ± (∆A • ∆B)
Dividing L.H.S by Z and R.H.S by AB, we get,
1 ± \(\frac{∆Z}{Z}\) = 1 ± \(\frac{∆B}{B}\) ± \(\frac{∆A}{A}\) ± \(\frac{∆A}{A}\). \(\frac{∆B}{B}\)
As ∆A/A, ∆B/B are both small quantities, their product term \(\frac{∆A}{A}\).\(\frac{∆B}{B}\) can be neglected.
The maximum fractional error in Z is
\(\frac{∆Z}{Z}\) = ± (\(\frac{∆A}{A}\) + \(\frac{∆B}{B}\))

(d) The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42s, 271s and 2.80s respectively. Calculate the average absolute error?
Answer:
Mean absolute error = \(\frac{\Sigma\left|\Delta \mathrm{T}_{i}\right|}{n}\)
∆T_m = \(\frac{0.01+0.06+0.20+0.09+0.18}{5}\)
∆T_in = \(\frac{0.54}{5}\) = 0.108s = 0.1 1s (Rounded of 2^nd decimal place).

Question 35 (a).
Explain in detail the triangle law of vector addition?
Answer:
Let us consider two vectors \(\vec { A } \) and \(\vec { B } \) as shown in figure.

To find the resultant of the two vectors we apply the triangular law of addition as follows:

Represent the vectors \(\vec { A } \) and \(\vec { B } \) by the two adjacent sides of a triangle taken in the same order. Then the resultant is given by the third side of the triangle as shown in figure.

To explain further, the head of the first vector \(\vec { A } \) is connected to the tail of the second vector \(\vec { B } \). Let θ be the angle between \(\vec { A } \) and \(\vec { B } \). Then \(\vec { R } \) is the resultant vector connecting the tail of the first vector A to the head of the second vector B.

The magnitude of R (resultant) is given geometrically by the length of \(\vec { R } \) (OQ) and the direction of the resultant vector is the angle between \(\vec { R } \) and \(\vec { A } \). Thus we write \(\vec { R } \) = \(\vec { A } \) + \(\vec { B } \).

1. Magnitude of resultant vector:
The magnitude and angle of the resultant vector are determined by using triangle law of vectors as follows. From figure, consider the triangle ABN, which is obtained by extending the side OA to ON. ABN is a right angled triangle.

From figure, let R is the magnitude of the resultant of \(\vec { A } \) and \(\vec { B } \).
cos θ = \(\frac{AN}{B}\) ∴AN = B cos θ ans sin θ = \(\frac{BN}{B}\) ∴ BN = B sin θ
For ∆OBN, we have OB² = ON² + BN²

2. Direction of resultant vectors: If θ is the angle between \(\vec { A } \) and \(\vec { B } \), then

(b) State and prove Bernoulli’s theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid?
Answer:
Bernoulli’s theorem:
According to Bernoulli’s theorem, the sum of pressure energy, kinetic energy, and potential energy per unit mass of an incompressible, non-viscous fluid in a streamlined flow remains a constant. Mathematically,
\(\frac { P }{ \rho } \) + \(\frac{1}{2}\)v² + gh = Constant
This is known as Bernoulli’s equation.

Proof: Let us consider a flow of liquid through a pipe AB. Let V be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time. Let a_A, v_A and P_A be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at A respectively.

Let the force exerted by the liquid at A is
F_A = P_Aa_A

Distance travelled by the liquid in time t is d = v_at

Therefore, the work done is W = F_Ad = P_Aa_Av_At
But a_Av_At = a_Ad = V, volume of the liquid entering at A.

Thus, the work done is the pressure energy (at A), W = P_Ad = P_AV
Pressure energy per unit volume at
A = \($\frac{\text { Pressure energy }}{\text { Volume }}$\) = \(\frac{P_{A} V}{V}\) = P_A

Pressure energy per unit mass
A = \($\frac{\text { Pressure energy }}{\text { Mass }}$\) = \(\frac{P_{A} V}{m}\) = \(\frac{P_{A}}{\frac{m}{V}}=\frac{P_{A}}{\rho}\)

Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
EP_A = P_AV = P_AV × (\(\frac{m}{m}\)) = m\(\frac { P_{ A } }{ \rho } \)

Potential energy of the liquid at A,
PE_A = mg h_A

Due to the flow of liquid, the kinetic energy of the liquid at A,
KE_A = \(\frac{1}{2} m v_{\mathrm{A}^{2}}\)

Therefore, the total energy due to the flow of liquid at A,
E_A = EP_A + KE_A + PE_A
E_A = m \(m \frac{P_{A}}{\rho}+\frac{1}{2} m v_{A}^{2}+m g h_{A}\)

Similarly, let a_B, v_B and P_B be the area of cross section of the tube, velocity of the liquid and pressure exerted by the liquid at B. Calculating the total energy at F_B, we get
E_B = m \(m \frac{P_{B}}{\rho}+\frac{1}{2} m v_{B}^{2}+m g h_{B}\)
From the law of conservation of energy,
E_A = E_B

Thus, the above equation can be written as
\(\frac{P}{\rho g}+\frac{1}{2} \frac{v^{2}}{g}\) + h = Constant

The above equation is the consequence of the conservation of energy which is true until there is no loss of energy due to friction. But in practice, some energy is lost due to friction.

This arises due to the fact that in a fluid flow, the layers flowing with different velocities exert frictional forces on each other. This loss of energy is generally converted into heat energy. Therefore, Bernoulli’s relation is strictly valid for fluids with zero viscosity or non-viscous liquids. Notice that when the liquid flows through a horizontal pipe, then
h = 0 ⇒ \(\frac{P}{\rho g}+\frac{1}{2} \frac{v^{2}}{g}\) = Constant

Question 36 (a).
What are the limitations of dimensional analysis?
Answer:
Limitations of Dimensional analysis

1. This method gives no information about the dimensionless constants in the formula like 1, 2, …………….. π, e, etc.
2. This method cannot decide whether the given quantity is a vector or a scalar.
3. This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
4. It cannot be applied to an equation involving more than three physical quantities.
5. It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation.

For example, using dimensional analysis, s = ut + \(\frac{1}{3}\)at² is dimensionally correct whereas the correct relation is s = ut + \(\frac{1}{2}\)at²

(b) The escape velocity v of a body depends on

1. The acceleration due to gravity ‘g’ of the planet
2. The radius R of the planet. Establish dimensionally the relation for the escape velocity?

Answer:
\(v \propto g^{a} \mathrm{R}^{b} \Rightarrow v=k g^{a} \mathrm{R}^{b}\), K → dimensionally proportionality constant.
[v] = [g]^a [R]^b
[M⁰L¹T^-1] = [M⁰L¹T^-2]^a [M⁰L¹T¹⁰]^b
equating powers
1 = a + b
-1 = -2a ⇒ a = \(\frac{1}{2}\)
b = 1 – a = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ v = k\(\sqrt{gR}\)

[OR]

(c) Discuss the law of transverse vibrations In stretched string?
Answer:
Laws of transverse vibrations in stretched strings:
There are three laws of transverse vibrations of stretched strings which are given as follows:

(I) The law of length:
For a given wire with tension T (which is fixed) and mass per unit length p (fixed) the frequency varies inversely with the vibrating length. Therefore,
f ∝\(\frac{1}{l}\) ⇒ f = \(\frac{C}{2}\)
⇒1 × f = C, where C is constant

(II) The law of tension:
For a given vibrating length I (fixed) and mass per unit length p (fixed) the frequency varies directly with the square root of the tension T,
f ∝\(\sqrt{T}\)
⇒f = A\(\sqrt{T}\), where A is constant

(III) The law of mass:
For a given vibrating length l (fixed) and tension T (fixed) the frequency varies inverely with the square root of the mass per unit length µ,
f = \(\frac{1}{\sqrt{\mu}}\)
⇒f = \(\frac{B}{\sqrt{\mu}}\), where B is constant

(d) Explain how to determine the frequency of tuning for k using sonometer?
Answer:
Working: A transverse stationary or standing wave is produced and hence, at the knife edges P and Q, nodes are formed. In between the knife edges, anti-nodes are formed. If the length of the vibrating element is l then
l = \(\frac{\lambda}{2}\) ⇒ λ = 2l
Let f be the frequency of the vibrating element, T the tension of in the string and µ the mass per unit length of the string. Then using equation ,we get
f = \(\frac{v}{\lambda}=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}\) in Hertz …………………. (1)

Let ρ be the density of the material of the string and d be the diameter of the string. Then the mass per unit length µ,

Question 37.
(a) To move an object, which one is easier, push or pull? Explain?
Answer:
When a body is pushed at an arbitrary angle θ (o to \(\frac{π}{2}\)), the applied force F can be resolved into two components as F sin e parallel to the surface and F cos θ perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is
no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ ………………………. (1)
As a result the maximal static friction also increases and is equal to
\(f_{s}^{\max }=\mu_{r} N_{p u s h}=\mu_{s}(m g+F \cos \theta)\) ………………… (2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.
When an object is pulled at an angle θ, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is
N_pull = mg – F cos θ ……………….. (3)

Equation (3) shows that the normal force needs is less than N_push. From equations (1) and (3), it is easier to pull an object than to push to make it move.

[OR]

(b) Derive an expression for the velocities of two objects colliding elastically in one dimension?
Answer:
Consider two elastic bodies of masses m₁ and m₂ moving in a straight line (along positive x direction) on a frictionless horizontal surface as shown in figure.

In order to have collision, we assume that the mass m₁ moves, faster than mass m₂ i.e., u₁ > u₂. For elastic collision, the total linear momentum and kinetic energies of the two bodies before and after collision must remain the same.

From the law of conservation of linear momentum,
Total momentum before collision (ρ_i) = Total momentum after collision (ρ_f)
m₁u₁ + m₂u₂ = m₁v₁ + m₂ v₁ …………………. (1)
or m₁ (u₁ – v₁) = m₂(v₂ – u₂) …………………… (2)
Furthur,

For elastic collision,
Total kinetic energy before collision KE_i = Total kinetic energy after collision KE_f
\(\frac{1}{2} m_{1} u_{1}^{2}+\frac{1}{2} m_{2} u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}\) ………………….. (3)
\(m_{1}\left(u_{1}^{2}-v_{1}^{2}\right)=m_{2}\left(v_{2}^{2}-u_{2}^{2}\right)\)
After simplifying and rearranging the terms,
Using the formula a² – b² = (a + b) (a – b), we can rewrite the above equation as
\(m_{1}\left(u_{1}+v_{1}\right)\left(u_{1}-v_{1}\right)=m_{2}\left(v_{2}+u_{2}\right)\left(v_{2}-u_{2}\right)\) ……………….. (4)
Dividing the equation (4) by (2) we get

Equation (5) can be written as
u₁ – u₂ = -(v₁ – v₂)
This means that for any elastic head on collision, the relative speed of the two elastic bodies after the collision has the same magnitude as before collision but in opposite direction. Further note that this result is independent of mass.
Rewriting the above equation for v₁ and v₂,
v₁ = v₂ + u₂ – u₁ …………………….. (6)
or v₂ = u₁ + v₁ – u₁ …………………….. (7)
To find the final velocities V₁ and v₂:
Substituting equation (5) in equation (2) gives the velocity of m₁ as
m₁ (u₁ – v₁) = m₂ (u₁ + v₁ – u₂ – u₂)
m (u₁ – v₁) = m₂ (u₁ + v₁ – 2u₂)
m₁u₁ – m₁v₁ = m₂u₁ + m₂v₁ + 2m₂u₂
m₁u₁ – m₂u₁ + 2m₂u₂ = m₁v₁ + m₂v₁
(m₁ – m₂)u₁ + 2m₂u₂ = (m₁ + m₂) v₁
or v₁ = \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)\) u₁  + \(\left(\frac{2 m_{2}}{m_{1}+m_{2}}\right)\) u₂
Similarly, by substituting (6) in equation (2) or substituting equation (8) in equation (7), we get the final m₂ as
v₂ = \(\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right)\) u₁ + \(\left(\frac{m_{2}-m_{1}}{m_{1}+m_{2}}\right)\) u₂ …………………. (9)

Question 38 (a).
Prove that at points near the surface of the Earth the gravitational potential energy of the object is v = mgh?
Answer:
When an object of mass m is raised to a height h, the potential energy stored in the object is mgh. This can be derived using the general expression for gravitational potential energy. Consider the Earth and mass system, with r, the distance between the mass m and the Earth’s centre. Then the gravitational potential energy.
U = –\(\frac{\mathrm{GM}_{e} m}{r}\) …………………… (1)

Here r = R_e + h, where Re is the radius of the Earth, h is the height above the Earth’s surface

By using Binomial expansion and neglecting the higher order

Replace this value and we get,
U = \(-\frac{\mathrm{GM}_{e} m}{\mathrm{R}_{e}}\left(1-\frac{h}{\mathrm{R}_{e}}\right)\) ………………… (4)

We know that, for a mass m on the Earth’s surface,
\(\mathrm{G} \frac{\mathrm{M}_{e} m}{\mathrm{R}_{e}}=m g \mathrm{R}_{e}\) …………………… (5)

Substituting equation (4) in (5) we get,
U = -mgR_e + mgh ………………….. (6)

It is clear that the first term in the above expression is independent of the height h. For example, if the object is taken from height h₁ to h₂, then the potential energy at h₁ is
U(h₁) = -mgR_e + mgh₁ …………………… (7)
and the potential energy at h₂ is
U(h₂) = -mgR_e + mgh₂ …………………… (8)

The potential energy differenqe between h₁ and h₂ is
U(h₂) – U(h₁) = mg(h₁-h₂) …………………. (9)

The term mgR_e in equation (7) and (8) plays no role in the result. Hence in the equation (6) the first term can be omitted or taken to zero. Thus it can be stated that the gravitational potential energy stored in the particle of mass m at a height h from the surface of the Earth is U = mgh. On the surface of the Earth, U = 0, since h is zero.

[OR]

(b) Derive an expression for Radius of gyration?
Answer:
For bulk objects of regular shape with uniform mass distribution, the expression for moment of inertia about an axis involves their total mass and geometrical features like radius, length, breadth, which take care of the shape and the size of the objects.

But, we need an expression for the moment of inertia which could take care of not only the mass, shape and size of objects, but also its orientation to the axis of rotation. Such an expression should be general so that it is applicable even for objects of irregular shape and non-uniform distribution of mass. The general expression for moment of inertia is given as,
I = MK²

where, M is the total mass of the object and K is called the radius of gyration. The radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object.

As the radius of gyration is distance, its unit is m. Its dimension is L. A rotating rigid body with respect to any axis, is considered to be made up of point masses m₁, m₂, m₃, . . . mn at perpendicular distances (or positions) r₁, r₂, r₃ . . . r_n respectively as shown in figure. The moment of inertia of that object can be written as,

I = \(\sum m_{1} r_{1}^{2}=m_{1} r_{1}^{2}+m_{2} r_{2}^{2}+m_{3} r_{3}^{2}+\ldots .+m_{n} r_{n}^{2}\)
If we take all the n number of individual masses to be equal
m = m₁ = m, = m₂ = m₃ = ……………… = m_n
then

I = MK²

where, nm is the total mass M of the body and K is the radius of gyration.

The expression for radius of gyration indicates that it is the root mean square (rms) distance of the particles of the body from the axis of rotation. In fact, the moment of inertia of any object could be expressed in the form, I = MK²

For example, let us take the moment of inertia of a uniform rod of mass M and length l. Its moment of inertia with respect to a perpendicular axis passing through the center of mass is,
I = \(\frac{1}{12}\)Ml²
In terms of radius of gyration, I = MK²
Hence, MK² = \(\frac{1}{12}\)Ml²
K² = \(\frac{1}{12}\)l²
K = \(\frac{1}{12}\) or K = \(\frac{1}{2 \sqrt{3}} l\) or K = (0.289)l