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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:

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:

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.

(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:

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 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.

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:

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 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:

Question 17.
Differentiate between Taxonomy & Systematics?
Answer:

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.

(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:

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:

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 Maths Chapter 8 Statistics and Probability Additional Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Additional Questions

I. Multiple Choice Questions

Question 1.
The range of the first 10 prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 is ______
(1) 28
(2) 26
(3) 29
(4) 27
Answer:
(4) 27
Hint:
Range = Largest value – Smallest value = 29 – 2 = 27

Question 2.
The least value in a collection of data is 14.1. If the range of the collection is 28.4, then the greatest value of the collection is _______
(1) 42.5
(2) 43.5
(3) 42.4
(4) 42.1
Answer:
(1) 42.5
Hint:
Given, S = 14.1; R = 28.4,
L = S + R = 28.4 + 14.1 = 42.5

Question 3.
The greatest value of a collection of data is 72 and the least value is 28. Then the coefficient of range is ______
(1) 44
(2) 0.72
(3) 0.44
(4) 0.28
Answer:
(3) 0.44
Hint:
Coefficient of range = \(\frac{L-S}{L+S}\)
= \(\frac{72-28}{72+28}=\frac{44}{100}\)
= 0.44

Question 4.
For a collection of 11 items, Σx = 132, then the arithmetic mean is _______
(1) 11
(2) 12
(3) 14
(4) 13
Answer:
(2) 12
Hint:
\(\bar{x}=\frac{\Sigma x}{n}=\frac{132}{11}=12\)

Question 5.
For any collection of n items, Σ(x – \(\bar{x}\)) = _____
(1) Σx
(2) \(\bar{x}\)
(3) n\(\bar{x}\)
(4) 0
Answer:
(4) 0
Hint:
We know that, For all collection of n items,
Σ(x – \(\bar{x}\)) = 0

Question 6.
For any collection of n items, (Σx) – \(\bar{x}\) = ________
(1) n\(\bar{x}\)
(2) (n – 2)\(\bar{x}\)
(3) (n – 1)\(\bar{x}\)
(4) 0
Answer:
(3) (n – 1)\(\bar{x}\)
Hint:

Question 7.
If t is the standard deviation of x, y, z, then the standard deviation of x + 5, y + 5, z + 5 is _______
(1) \(\frac{t}{3}\)
(2) t + 5
(3) t
(4) xyz
Answer:
(3) t
Hint:
The S.D. of distribution remains unchanged when each value is added or subtracted by the same quantity.

Question 8.
If the standard deviation of a set of data is 1.6, then the variance is _______
(1) 0.4
(2) 2.56
(3) 1.96
(4) 0.04
Answer:
(2) 2.56
Hint:
Variance = (S.D.)² = (1.6)² = 2.56

Question 9.
If the variance of a data is 12.25, then the S.D is _______
(1) 3.5
(2) 3
(3) 2.5
(4) 3.25
Answer:
(1) 3.5
Hint:
S.D = √Variance = √12.25 = 3.5

Question 10.
Variance of the first 11 natural numbers is ______
(1) √5
(2) √10
(3) 5√2
(4) 10
Answer:
(4) 10
Hint:
Variance = \(\frac{n^{2}-1}{12}=\frac{11^{2}-1}{12}=\frac{120}{12}=10\)

Question 11.
The variance of 10, 10, 10, 10, 10 is _______
(1) 10
(2) √10
(3) 5
(4) 0
Answer:
(4) 0
Hint:

Question 12.
If the variance of 14, 18, 22, 26, 30 is 32, then the variance of 28, 36, 44, 52, 60 is _______
(1) 64
(2) 128
(3) 32√2
(4) 32
Answer:
(2) 128
Hint:
Variance of the given numbers = 32;
S.D. = √32 = 4√2
Each data is multiplied by 2.
New S.D.= 4√2 × 2 = 8√2
Variance = (8√2)² = 128

Question 13.
The standard deviation of a collection of data is 2√2. If each value is multiplied by 3, then the standard deviation of the new data is ______
(1) √12
(2) 4√2
(3) 6√2
(4) 9√2
Answer:
(3) 6√2
Hint:
Given, S.D. = 2√2
Each value is multiplied by 3
New S.D. = 2√2 × 3 = 6√2

Question 14.
Given Σ(x – \(\bar{x}\))2 = 48, \(\bar{x}\) = 20 and n = 12. The coefficient of variation is ______
(1) 25
(2) 20
(3) 30
(4) 10
Answer:
(4) 10
Hint:

Question 15.
Mean and standard deviation of a data are 48 and 12 respectively. The coefficient of variation is _______
(1) 42
(2) 25
(3) 28
(4) 48
Answer:
(2) 25
Hint:
Coefficient of variation
C.V = \(\frac{\sigma}{\bar{x}} \times 100\) = \(\frac{12}{48} \times 100\) = 25

Question 16.
If Φ is an impossible event, then P(Φ) = ______
(1) 1
(2) \(\frac{1}{4}\)
(3) 0
(4) \(\frac{1}{2}\)
Answer:
(3) 0
Hint:
Probability of an impossible event is 0.

Question 17.
If S is the sample space of a random experiment, then P(S) = _______
(1) 0
(2) \(\frac{1}{8}\)
(3) \(\frac{1}{2}\)
(4) 1
Answer:
(4) 1
Hint:
Every event is a subset of S. Sample space contain all the possible element.
P(S) = 1

Question 18.
If p is the probability of an event A, then p satisfies _______
(1) 0 < p < 1
(2) 0 < p < 1
(3) 0 < p < 1
(4) 0 < p < 1
Answer:
(2) 0 < p < 1
Hint:
The Probability of an event is always greater than 0 and less than or equal to 1.

Question 19.
Let A and B be any two events and S be the corresponding sample space. Then P (\(\bar{A}\) ∩ B) = ______

(1) P(B) – P(A ∩ B)
(2) P(A ∩ B) – P(B)
(3) P(S)
(4) P[(A ∪ B)’]
Answer:
(1) P(B) – P(A ∩ B)
Hint:
P(\(\bar{A}\) ∩ B) means only B and not A

Question 20.
The probability that a student will score centum in mathematics is \(\frac{4}{5}\). The probability that he will not score centum is ______
(1) \(\frac{1}{5}\)
(2) \(\frac{2}{5}\)
(3) \(\frac{3}{5}\)
(4) \(\frac{4}{5}\)
Answer:
(1) \(\frac{1}{5}\)
Hint:
P(\(\bar{A}\)) = 1 – P(A) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)

Question 21.
If A and B are two events such that P(A) = 0.25, P(B) = 0.05 and P(A ∩ B) = 0.14, then P(A ∪ B) = _____
(1) 0.61
(2) 0.16
(3) 0.14
(4) 0.6
Answer:
(2) 0.16
Hint:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.25 + 0.05 – 0.14 = 0.16

Question 22.
There are 6 defective items in a sample of 20 items. One item is drawn at random. The that it is probability a non-defective item is ________
(1) \(\frac{1}{10}\)
(2) 0
(3) \(\frac{3}{10}\)
(4) \(\frac{2}{3}\)
Answer:
(1) \(\frac{1}{10}\)
Hint:
Non-defective item = 20 – 6 = 14
Probability of non-defective items = \(\frac{14}{20}\) = \(\frac{7}{10}\)

Question 23.
If A and B are mutually exclusive events and S is the sample space such that P(A) = \(\frac{1}{3}\) P(B) and S = A ∪ B, then P(A) = _____
(1) \(\frac{1}{4}\)
(2) \(\frac{1}{2}\)
(3) \(\frac{3}{4}\)
(4) \(\frac{3}{8}\)
Answer:
(1) \(\frac{1}{4}\)
Hint:
P(A) = \(\frac{1}{3}\) P(B)
P(B) = 3 P(A)
P(A ∪ B) = P(A) + P(B)
[A and B are mutually exclusive]
P(A ∪ B) = P(A) + 3 P(A)
1 = 4P(A) [But P(A ∪ B) = 1]
P(A) = \(\frac{1}{4}\)

Question 24.
The probabilities of three mutually exclusive events A, B and C are given by \(\frac{1}{3}\), \(\frac{1}{4}\) and \(\frac{5}{12}\). Then P(A ∪ B ∪ C) is _______
(1) \(\frac{19}{12}\)
(2) \(\frac{11}{12}\)
(3) \(\frac{7}{12}\)
(4) 1
Answer:
(4) 1
Hint:
P (A ∪ B ∪ C) = P(A) + P(B) + P(C)
\(\frac{1}{3}+\frac{1}{4}=\frac{5}{12}=\frac{4+3+5}{12}=\frac{12}{12}=1\)

Question 25.
If P(A) = 0.25, P(B) = 0.50, P(A ∩ B) = 0.14 then P(neither A nor B) = ______
(1) 0.39
(2) 0.25
(3) 0.11
(4) 0.24
Answer:
(1) 0.39
Hint:
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= 0.25 + 0.50 – 0.14
= 0.61
P (neither A nor B) = P(\(\bar{A}\) ∩ \(\bar{B}\))
= 1 – P(A ∪ B)
= 1 – 0.61
= 0.39

Question 26.
A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected at random, the probability that it is not red is _______
(1) \(\frac{5}{12}\)
(2) \(\frac{4}{12}\)
(3) \(\frac{3}{12}\)
(4) \(\frac{3}{4}\)
Answer:
(4) \(\frac{3}{4}\)
Hint:
P(\(\bar{R}\)) = 1 – P(R)
\(=1-\frac{3}{12}=\frac{9}{12}=\frac{3}{4}\)

Question 27.
Two dice are thrown simultaneously. The probability of getting a doublet is ________
(1) \(\frac{1}{36}\)
(2) \(\frac{1}{3}\)
(3) \(\frac{1}{6}\)
(4) \(\frac{2}{3}\)
Answer:
(3) \(\frac{1}{6}\)
Hint:
n(S) = 36
Let A be the event of getting a doublet
A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
n(A) = 6
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)

Question 28.
A fair die is thrown once. The probability of getting a prime or composite number is _______
(1) 1
(2) 0
(3) \(\frac{5}{6}\)
(4) \(\frac{1}{6}\)
Answer:
(3) \(\frac{5}{6}\)
Hint:
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
The required probability = \(\frac{5}{6}\)
[Science is neither prime not a composite number]

Question 29.
Probability of getting 3 heads or 3 tails in tossing a coin 3 times is _______
(1) \(\frac{1}{8}\)
(2) \(\frac{1}{4}\)
(3) \(\frac{3}{8}\)
(4) \(\frac{1}{2}\)
Answer:
(2) \(\frac{1}{4}\)
Hint:
S= {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(S) = 8
A = {HHH, TTT}; n(A) = 2
The required probability = \(\frac{2}{8}\) = \(\frac{1}{4}\)

Question 30.
A card is drawn from a pack of 52 cards at random. The probability of getting neither an ace nor a king card is _____
(1) \(\frac{2}{13}\)
(2) \(\frac{11}{13}\)
(3) \(\frac{4}{13}\)
(4) \(\frac{8}{13}\)
Answer:
(2) \(\frac{11}{13}\)
Hint:
n(S) = 52
Number of ace cards = 4
number of king cards = 4
n(non-ace and non-king cards) = 52 – 8 = 44
P(neither an ace nor a king) = \(\frac{44}{52}\) = \(\frac{11}{13}\)

Question 31.
The probability that a leap year will have 53 Fridays or 53 Saturdays is ______
(1) \(\frac{2}{7}\)
(2) \(\frac{1}{7}\)
(3) \(\frac{4}{7}\)
(4) \(\frac{3}{7}\)
Answer:
(4) \(\frac{3}{7}\)
Hint:
Leap year contains 52 weeks and 2 days
Sample space = {(sun, mon), (mon, tue), (tue, wed), (wed, thu), (thu, fri), (fri, sat), (sat, sun)}
n(S) = 7
The required probability = \(\frac{2}{7}+\frac{2}{7}-\frac{1}{7}=\frac{3}{7}\)

Question 32.
The probability that a non-leap year will have 53 Sundays and 53 Mondays is ________
(1) \(\frac{1}{7}\)
(2) \(\frac{2}{7}\)
(3) \(\frac{3}{7}\)
(4) 0
Answer:
(4) 0
Hint:
Non leap year contains 52 weeks and one day
Sample space (S) = {Sun, Mon, Tue, Wed, Thu, Fri, Sat}
n(S) = 7
The required probability = \(\frac{1}{7}+\frac{1}{7}-\frac{2}{7}\)
= \(\frac{2}{7}-\frac{2}{7}\)
= 0

Question 33.
The probability of selecting a queen of hearts when a card is drawn from a pack of 52 playing cards is ______
(1) \(\frac{1}{52}\)
(2) \(\frac{16}{52}\)
(3) \(\frac{1}{13}\)
(4) \(\frac{1}{26}\)
Answer:
(1) \(\frac{1}{52}\)
Hint:
n(S) = 52 [1 queen of hearts in 52 cards]
n(A) = 1
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{1}{52}\)

Question 34.
Probability of sure event is ______
(1) 1
(2) 0
(3) 100
(4) 0.1
Answer:
(1) 1

Question 35.
The outcome of a random experiment result in either success or failure. If the probability of success is twice the probability of failure, then the probability of success is ______
(1) \(\frac{1}{3}\)
(2) \(\frac{2}{3}\)
(3) 1
(4) 0
Answer:
(2) \(\frac{2}{3}\)
Hint:
n(A ∪ B) = 1; P(\(\bar{A}\)) = 1 – P(A)
Given P(A) = 2P(\(\bar{A}\))
P(A) = 2 [1 – P(A)] = 2 – 2 P(A)
3 P(A) = 2
P(A) = \(\frac{2}{3}\)

II. Answer the following questions.

Question 1.
Find the range and the coefficient of range of the following data.

Answer:
Largest value (L) = 690
Smallest value (S) = 610
Range R = L – S = 690 – 610 = 80
Coefficient of range = \(\frac{L-S}{L+S}=\frac{690-610}{690+610}=\frac{80}{1300}\) = 0. 06

Question 2.
Two dice are thrown simultaneously. What is the probability that
(i) 5 will not come up on either of them
(ii) 5 will come up at both dice
Answer:
S = {(1, 1)(1, 2)(1, 3)(1, 4)(1, 5)(1, 6), (2, 1)(2, 2)(2, 3)(2, 4)(2, 5)(2, 6), (3, 1)(3, 2)(3, 3)(3, 4)(3, 5)(3, 6), (4, 1)(4, 2)(4, 3)(4, 4)(4, 5)(4, 6), (5, 1)(5, 2)(5, 3)(5, 4)(5, 5)(5, 6), (6, 1)(6, 2)(6, 3)(6, 4)(6, 5)(6, 6)}
n(S) = 36
(i) Let A be the event of getting 5 on either of them.
A = {(1, 5) (2, 5) (3, 5) (4, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 5)}
n(A) = 11
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{11}{36}\)
Probability that 5 will not come up on either of them = 1 – P (A)
\(=1-\frac{11}{36}=\frac{36-11}{36}=\frac{25}{36}\)
(ii) Let B be the event of getting 5 will come up at both dice
B = {(5, 5)}
n(B) = 1
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{1}{36}\)

Question 3.
The king, Queen and Jack of clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards, find the probability of getting
(i) a card of clubs
(ii) a queen of diamond
Answer:
Sample space (S) = (52 – 3) = 49
n (S) = 49
(i) Let A be the event of getting a card of clubs.
n(A) = (13 – 3) = 10
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{10}{49}\)
(ii) Let B be the event of getting a queen of diamond
n(B) = 1
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{1}{49}\)

Question 4.
The standard deviation of 20 observations is √5. If each observation is multiplied by 2, find the standard deviation and variance of the resulting observations.
Answer:
Given a standard deviation of 20 observations = √5
Each observation is multiplied by 2 then,
New standard deviation = 2 × √5 = 2√5
New variance = (2√5)² = 20

Question 5.
Calculate the variance standard deviation of the following data 38, 40, 34, 31, 28, 26, 34.
Answer:
Arrange the given data in ascending order we get, 26, 28, 31, 34, 38, 40
Assumed mean = 34


Variance = 22
Standard deviation(σ) = √Variance = √22 = 4.69

Question 6.
Mean of 100 items is 48 and their standard deviation is 10. Find the sum of all the times and the sum of the squares of all items.
Answer:
The mean of 100 items = 48
The sum of 100 items (Σx) = 100 × 48 = 4800
standard deviation (σ²) = 10

Sum of the squares of all items (Σx²) = 240400

Question 7.
If n = 10, \(\bar{x}\) = 12 and Σx² = 1530, then calculate the coefficient of variation.
Answer:

Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100 \Rightarrow \frac{3}{12} \times 100=25\)

Question 8.
If the coefficient of variation of a collection of data is 57 and its standard deviation is 6, 84, then find the mean.
Answer:
Given the coefficient of variation = 57
Standard deviation (σ) = 6.84

Arithmetic mean = \(\bar{x}\) = 12

Question 9.
Find the standard deviation and the variance of first 23 natural numbers?
Answer:
Standard deviation of first “n” natural numbers = \(\sqrt{\frac{n^{2}-1}{12}}\)
Standard deviation of first “23” natural numbers = \(\sqrt{\frac{23^{2}-1}{12}}\)
\(=\sqrt{\frac{529-1}{12}}=\sqrt{\frac{528}{12}}=\sqrt{44}\)
= 6.63

Question 10.
Find the coefficient of variation of the following data: 18, 20, 15, 12, 25.
Answer:
Let us calculate the A.M of the given data.


The coefficient of variation is 24.6

Question 11.
Three rotten eggs are mixed with 12 good ones. One egg is chosen at random. What is the probability of choosing a rotten egg?
Answer:
Number of good eggs = 12
Number of rotton eggs = 3
Total number of eggs = 12 + 3 = 15
Sample space n(S) = 15
Let A be the event of choosing a rotten egg
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{15}=\frac{1}{5}\)
The Probability is \(\frac{1}{5}\)

Question 12.
Two coins are tossed together. What is the probability of getting at most one head?
Answer:
Sample space (S) = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
Let A be the event of getting atmost one head
A = {HT, TH, TT}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{4}\)
The probability is \(\frac{3}{4}\)

Question 13.
A number is selected at random from integers 1 to 100. Find the probability that it is
(i) a perfect square
(ii) not a perfect cube.
Answer:
Sample space (S) = {1, 2, 3, …,100}
n(S) = 100
(i) Let A be the event of getting a perfect square.
A = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
n(A) = 10
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{10}{100}=\frac{1}{10}\)
(ii) Let B be the event of getting a perfect cube.
B = {1, 8, 27, 64}
n(B) = 4
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{4}{100}=\frac{1}{25}\)
The Probability that the selected number is not a perfect cube is
P(\(\bar{B}\)) = 1 – P(B)
P(\(\bar{B}\)) = 1 – \(\frac{1}{25}\) = \(\frac{24}{25}\)

Question 14.
Three dice are thrown simultaneously. Find the probability of getting the same number on all the three dice.
Answer:
Sample space (S) = {(1, 1, 1) (1, 1, 2) (1, 1, 3) ….(6, 6, 6)}
n(S) = 63 = 216
Let A be the event of getting the same number on all the three dice.
A = {(1, 1, 1) (2, 2, 2) (3, 3, 3) (4, 4, 4) (5, 5, 5) (6, 6, 6)}
n(A) = 6
P(A) = \(\frac{n(A)}{n(S)}=\frac{6}{216}=\frac{1}{36}\)
The probability is \(\frac{1}{36}\)

Question 15.
If P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{10}\), P(A∪B) = 1, Find
(i) P(A ∩ B)
(ii) P(A’ ∪ B’)
Answer:

III. Answer the following questions.

Question 1.
The mean of the following frequency distribution is 53 and the sum of all frequencies is 100. compute the missing frequencies f₁ and f₂.

Answer:


f₂ = 29
Substitute the value of f₂ = 29 in (1)
f₁ + 29 = 47
⇒ f₁ = 47 – 29 = 18
The value of f₁ = 18 and f₂ = 29

Question 2.
Calculate the standard deviation of the following data.

Answer:
Assumed mean (A) = 13

Standard deviation = 6.32

Question 3.
The time (in seconds) taken by a group to walk across a pedestrian crossing is given in the table below.

Calculate the variance and standard deviation of the data.
Answer:
Assumed mean (A) = 17.5, c = 5


Standard deviation (σ) = √36.76 = 6.063
Variance 36. 76, Standard deviation = 6.06

Question 4.
The mean and standard deviation of 20 items are found to be 10 and 2 respectively. At the time of checking it was found that an item 12 was wrongly entered as 8. Calculate the correct mean and standard deviation.
Answer:
Given, mean of 20 items (\(\bar{x}\)) = 10
and standard deviation (σ) = 2



(i) Corrected mean = 10. 2
(ii) Corrected Standard deviaton = 1.99

Question 5.
Calculate the coefficient of variation of the following data: 20, 18, 32, 24, 26.
Answer:
Arrange in ascending order, we get 18, 20, 24, 26, 32
Assumed mean = 24

Question 6.
The marks scored by two students A, B in a class are given below.

Who is more consistent?
Answer:
Student = A
\(\bar{x}\) = 60




From student A and B the coefficient of variation for A is less than the coefficient of variation for B.
Student A is more Consistent.

Question 7.
If for distribution Σ(x – 7) = 3, Σ(x – 7)² = 57 and total number of item is 20. find the mean and standard deviation.
Answer:


(i) Arithmetic mean = 7. 15
(ii) Standard deviation = 1.68

Question 8.
Two unbiased dice are rolled once. Find the probability of getting
(i) a sum 8
(ii) a doublet
(iii) a sum greater than 8
Answer:
When two dice are thrown, the sample space is
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 6 × 6 = 36
(i) Let A be the event of getting a sum 8
A= {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n(A) = 5
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{5}{36}\)
(ii) Let B be the event of getting a doublet.
B = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n(B) = 6
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting a sum greater than 8
C = {(3, 6) (4, 5) (4, 6) (5, 4) (5, 5) (5, 6) (6, 3) (6, 4) (6, 5) (6, 6)}
n(C) = 10
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}=\frac{10}{36}=\frac{5}{18}\)

Question 9.
A die is thrown twice. Find the probability that atleast one of the two throws conies up with the number 5 (use addition theorem).
Answer:
In rolling a die twice, the size of the sample space, n(S) = 36
Let A be the event of getting 5 in the first throw.
A = {(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)}
Thus, n(A) = 6, and P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{6}{36}\)
Let B be the event of getting 5 in the second throw.
B = {(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)}
Thus, n(B) = 6, and P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{6}{36}\)
A and B are not mutually exclusive events, since A ∩ B = {(5, 5)}
n(A ∩ B) = 1 and P(A ∩ B) = \(\frac{1}{36}\)
By addition theorem,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
\(=\frac{6}{36}+\frac{6}{36}-\frac{1}{36}=\frac{11}{36}\)

Question 10.
Let A, B, C be any three mutually exclusive and exhaustive events such that P(B) = \(\frac{3}{2}\) P(A) and P(C) = \(\frac{1}{2}\) P(B). Find P(A).
Answer:
Let P(A) = p
Now, P(B) = \(\frac{3}{2}\), P(A) = \(\frac{3}{2}\) p
Also, P(C) = \(\frac{1}{2}\) P(B)
= \(\frac{1}{2}\) (\(\frac{3}{2}\) p) = (\(\frac{3}{4}\) p)
Given that A, B and C are mutually exclusive and exhaustive events.
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) and S = A ∪ B ∪ C
Now, P(S) = 1
That is, P(A) + P(B) + P(C) = 1
p + \(\frac{3}{2}\) p + \(\frac{3}{4}\) p = 1
⇒ 4p + 6p + 3p = 4
Thus, p = \(\frac{4}{13}\)
Hence, P(A) = \(\frac{4}{13}\)

Question 11.
A bag contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If an item is chosen at random, find the probability that it is rusted or that it is a bolt.
Answer:
Sample space (S) = (50 + 150) = 200
n(S) = 200
Number of rusted bolts = \(\frac{1}{2}\) (50) = 25
Number of rusted nuts = \(\frac{1}{2}\) (150) = 75
Let A be the event of getting rusted bolts and nuts.
n (A) = 25 + 75 = 100
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{100}{200}\)
Let B be the event of getting a bolt.
n(B) = 50
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{50}{200}\)
Number of bolts which are rusted n(A ∩ B) = 25

Question 12.
Two dice are rolled simultaneously. Find the probability that that sum of the numbers on the faces is neither divisible by 3 nor by 4.
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Let A be the event of getting the sum is divisible by 3
A = {(1, 2) (2, 1) (1, 5) (5, 1) (2, 4) (4, 2) (3, 3) (3, 6) (6, 3) (4, 5) (5, 4) (6, 6)}
n(A) = 12
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{12}{36}\)
Let B be the event of getting a sum is divisible by 4.
B = {(1, 3) (2, 2) (2, 6) (3, 1) (3, 5) (4, 4) (5, 3) (6, 2) (6, 6)}
n(B) = 9
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{9}{36}\)

Question 13.
In a class, 40% of the students participated in Mathematics-quiz, 30% in Science -quiz and 10% in both the quiz programmes. If a students is selected at random from the class, find the probability that the students participated in Mathematics or science or both quiz programmes.
Answer:
Sample space (S) = 100
n(S) = 100
Let A be the event of getting a number of students participated in mathematics-quiz programme
n(A) = 40
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{40}{100}\)
Let B be the event getting a number of students participated in science-quiz programme
n(B) = 30

\(=\frac{40+30-10}{100}=\frac{60}{100}=\frac{3}{5}\)
The required probability = \(\frac{3}{5}\)

Question 14.
A two digit number is formed with the digits 2, 5, 9 (repetition is allowed). Find the probability that the number is divisible by 2 or 5.
Answer:
Sample space (S) = {22, 25, 29, 55, 52, 59, 99, 92, 95}
n(S) = 9
Let A be the event of getting number is divisible by 2.
A = {22, 52, 92}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{9}\)
Let B be the event of getting number is divisible by 5.
B = {25, 95, 55}
n(B) = 3
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{3}{9}\)
If A and B are mutually exclusive.
P(A ∪ B) = P(A) + P(B)
P(A ∪ B) = \(\frac{3}{9}+\frac{3}{9}=\frac{6}{9}=\frac{2}{3}\)
The required probability is \(\frac{2}{3}\)

Question 15.
The probability that A, B and C can solve a problem are \(\frac{4}{5}\), \(\frac{2}{3}\) and \(\frac{3}{7}\) respectively. The probability of the problem being solved by A and B is \(\frac{8}{15}\), B and C is \(\frac{2}{7}\), A and C is \(\frac{12}{35}\). The probability of the problem being solved by all the three is \(\frac{8}{35}\). Find the probability that the problem can be solved by atleast one of them.
Answer:
Given P(A) = \(\frac{4}{5}\)
P(B) = \(\frac{2}{3}\)
P(C) = \(\frac{3}{7}\)
P(A ∩ B) = \(\frac{8}{15}\)
P(B ∩ C) = \(\frac{2}{7}\)
P(A ∩ C) = \(\frac{12}{35}\)
P(A ∩ B ∩ c) = \(\frac{8}{35}\)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

The probability of the problem can be solved by at least one of them = \(\frac{101}{105}\)

Students can download Maths Chapter 8 Statistics and Probability Unit Exercise 8 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Unit Exercise 8

Question 1.
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f₁ and f₂.

Answer:
Arithmetic Mean (\(\bar{x}\)) = 62.8
Sum of all the frequencies (Σf_i) = 50
Let the missing frequencies be f₁ and f₂


Substitute of the value of f₂ in (1)
f₁ + 12 = 20
⇒ f₁ = 20 – 12 = 8
The Missing frequency is 8 and 12.

Question 2.
The diameter of circles (in mm) drawn in the design are given below.

Calculate the standard deviation.
Answer:
Assumed mean = 42. 5

Question 3.
The frequency distribution is given below

In table k is a positive integer, has a variance of 160. Determine the value of k.
Answer:
Assumed mean = 3k



The value of k = 7

Question 4.
The standard deviation of some temperature data in degree Celsius (°C) is 5. If the data were converted into degree Fahrenheit (°F) then what is the variance?
Solution:
F° = (C° × 1.8) + 32
\(\begin{array}{l}{\sigma_{c}=5^{\circ} \mathrm{C}} \\ {\sigma_{\mathrm{F}}=\left(1.8 \times 5^{\circ} \mathrm{C}\right) \cdot 9^{\circ} \mathrm{F}}\end{array}\)
Adding value to data doesn’t affect standard deviation.
New variance = \(\sigma_{\mathrm{F}}^{2}=81^{\circ} \mathrm{F}\)

Question 5.
If for a distribution, Σ (x – 5) = 3, Σ (x – 5)² = 43, and total number of observations is 18, find the mean and standard deviation.
Answer:


(i) Arithmetic mean (\(\bar{x}\)) = 5.17
(ii) Standard deviation (σ) = 1.53

Question 6.
Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?

Answer:
Coefficients of variation of prices in city A.
Arrange in ascending order we get, 16, 19, 20, 22, 23
Assumed mean = 20

Coefficient of variation = \(\frac{\sigma}{\bar{x}} \times 100\)
= \(\frac{2.45}{20} \times 100\)
= 12. 25%
Coefficient of variation = 12.25%
Coefficients of variation of prices in city B.
Arrange in ascending order we get, 10, 12, 15, 18, 20
Assumed mean = 15


Prices in city A is more stable (since 12.25 < 24.6 %)

Question 7.
If the range and coefficient of range of the data are 20 and 0.2 respectively, then find the largest and smallest values of the data.
Answer:
Range of the data (R) = 20
L – S = 20 ……(1)
Coefficient of range = 0.2
Coefficient of range = \(\frac{L-S}{L+S}\)
0.2 = \(\frac{20}{L+S}\)

substituting the value of L = 60 in (2)
60 + S = 100
S = 100 – 60 = 40
Largest value = 60
Smallest value = 40

Question 8.
If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5.
Answer:
Sample space = {(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6),(2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6), (3, 1), (3, 2), (3, 3),(3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1),(6, 2), (6, 3), (6, 4),(6, 5), (6, 6)}
n(S) = 36
(i) Let A be the event of getting product of face value 6.
A = {(1, 6), (2, 3), (3, 2) (6, 1)}
n(A) = 4
P (A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{4}{36}\)
(ii) Let B be the event of getting difference of face value is 5.
B = {(6, 1)}
n(B) = 1

The probability is \(\frac{1}{9}\)

Question 9.
In a two children family, find the probability that there is at least one girl in a family.
Answer:
Sample space (S) = {(Boy, Boy) (Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(S) = 4
Let A be the event of getting atleast one Girl
A = {(Boy, Girl) (Girl, Boy) (Girl, Girl)}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{4}\)
Probability of atleast one girl in a family is \(\frac{3}{4}\)

Question 10.
A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.
Answer:
Let the number of black balls be “x”
Sample space (S) = x + 5
n(S) = x + 5
Let A be the event of drawing a black ball
n (A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{x}{x+5}\)
Let B be the event of getting a white ball
n(B) = 5
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{5}{x+5}\)
By the given condition,
\(\frac{x}{x+5}=2 \times\left(\frac{5}{x+5}\right)\)
⇒ \(\frac{x}{x+5}=\left(\frac{10}{x+5}\right)\)
⇒ 10x + 50 = x² + 5x
⇒ x² + 5x – 10x – 50 = 0
⇒ x² – 5x – 50 = 0
⇒ (x – 10)(x + 5) = 0
⇒ x = 10 or x = -5
Number of balls will not be negative.
Number of black balls = 10

Question 11.
The probability that a student will pass the final examination in both English and Tamil is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Tamil examination?
Answer:
Let A be the event of getting student pass in English
Let B be the event of getting student pass in Tamil


Probability of passing the tamil examination is \(\frac{13}{20}\)

Question 12.
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting
(i) a diamond
(ii) a queen
(iii) a spade
(iv) a heart card bearing the number 5.
Answer:
Total number of cards = 52
3 cards are removed
Remaining number of cards = 52 – 3 = 49
n(S) = 49
(i) Let A be the event of getting a diamond card.
n(A) = 13
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{13}{49}\)
(ii) Let B be the event of getting a queen card.
n(B) = (4 – 1) = 3
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{3}{49}\)
(iii) Let C be the event of getting a spade card.
n(C) = (13 – 3) = 10
P(C) = \(\frac{n(C)}{n(S)}=\frac{10}{49}\)
(iv) Let D be the event of getting a 5 of heart card.
n(D) = 1
P(D) = \(\frac{n(\mathrm{D})}{n(\mathrm{S})}=\frac{1}{49}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.5

Multiple Choice Questions.

Question 1.
Which of the following is not a measure of dispersion?
(1) Range
(2) Standard deviation
(3) Arithmetic mean
(4) Variance
Answer:
(3) Arithmetic mean
Hint:
Measures of dispersion are,
(i) Range
(ii) Mean deviation
(iii) Quartile deviation
(iv) Standard deviation
(v) Variance
(vi) coefficient of variation

Question 2.
The range of the data 8, 8, 8, 8, 8.. . 8 is
(1) 0
(2) 1
(3) 8
(4) 3
Solution:
(1) 0
Hint
R = L – S = 8 – 8 = 0

Question 3.
The sum of all deviations of the data from its mean is _______
(1) always positive
(2) always negative
(3) zero
(4) non-zero integer
Answer:
(3) zero

Question 4.
The mean of 100 observations is 40 and their standard deviation is 3. The sum of squares of all deviations is
(1) 40000
(2) 160900
(3) 160000
(4) 30000
Solution:
(2) 160900
Hint:
σ = 3

Question 5.
Variance of first 20 natural numbers is ______
(1) 32.25
(2) 44.25
(3) 33.25
(4) 30
Answer:
(3) 33.25
Hint:
Variance of 20 natural numbers is

Question 6.
The standard deviation of a data Is 3. If each value is multiplied by 5 then the new variance is
(1) 3
(2) 15
(3) 5
(4) 225
Solution:
σ = 3. 1f each is multiplied by 5. The new standard variation is also multiplied by 3.
∴ The new S.D = 5 × 3 = 15
Variance = 15² = 225

Question 7.
If the standard deviation of x, y, z is p then the standard deviation of 3x + 5, 3y + 5, 3z + 5 is ________
(1) 3p + 5
(2) 3p
(3) p + 5
(4) 9p + 15
Answer:
(2) 3p
Hint:
(i) Each value is added by any constant there is no change in standard deviation.
(ii) Each value is multiplied by 3 standard deviations also multiplied by 3.
The standard deviation is 3p.

Question 8.
If the mean and coefficient of variation of a data are 4 and 87.5% then the standard
deviation is
(1) 3.5
(2) 3
(3) 4.5
(4) 2.5
Solution:
(1) 3.5
Hint:
\(\bar{x}\) = 4, coefficient of variation is = 87. 5%

Question 9.
Which of the following is incorrect?
(1) P(A) > 1
(2) 0 ≤ P(A) ≤ 1
(3) P(Φ) = 0
(4) P(A) + P(\(\bar{A}\)) = 1
Answer:
(1) P(A) > 1
Hint:
Probability is always less than one or equal to one.

Question 10.
The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is _________
(1) \(\frac{q}{p+q+r}\)
(2) \(\frac{p}{p+q+r}\)
(3) \(\frac{p+q}{p+q+r}\)
(4) \(\frac{p+r}{p+q+r}\)
Answer:
(1) \(\frac{q}{p+q+r}\)
Hint:
Sample spaces = p + q + r
Let A be the event of getting red
n(A) = p
P(A) = \(\frac{q}{p+q+r}\)

Question 11.
A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is _______
(1) \(\frac{3}{10}\)
(2) \(\frac{7}{10}\)
(3) \(\frac{3}{9}\)
(4) \(\frac{7}{9}\)
Answer:
(2) \(\frac{7}{10}\)
Hint:
Here n(S)= 10 (given digit at imit place. It has two digit)
n(A) = 7
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{7}{10}\)

Question 12.
The probability of getting a job for a person is \(\frac{x}{3}\). If the probability of not getting the job is \(\frac{2}{3}\) then the value of x is.
(1) 2
(2) 1
(3) 3
(4) 1.5
Solution:
(2) 1
Hint:
Probability of getting a job = \(\frac{x}{3}\)
Probability of not getting a job = 1 – \(\frac{x}{3}\)

Question 13.
Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is \(\frac{1}{9}\), then the number of tickets bought by Kamalam is _______
(1) 5
(2) 10
(3) 15
(4) 20
Answer:
(3) 15
Hint:
n(S) = 135
P(A) = \(\frac{1}{9}\)

Question 14.
If a letter is chosen at random from the English alphabets {a, b, …, z}, then the probability that the letter chosen precedes x.
(1) \(\frac{12}{13}\)
(2) \(\frac{1}{13}\)
(3) \(\frac{23}{26}\)
(4) \(\frac{3}{26}\)
Solution:
(3) \(\frac{23}{26}\)
Hint:
n(S) = 26
Let A denote the letter chosen precedes x
A= {a, b, c, d, …, x}
n(A) = 23
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{23}{26}\)

Question 15.
A purse contains 10 notes of ₹ 2000, 15 notes of ₹ 500, and 25 notes of ₹ 200. One note is drawn at random. What is the probability that the note is either a ₹ 500 note or ₹ 200 note?
(1) \(\frac{1}{5}\)
(2) \(\frac{3}{10}\)
(3) \(\frac{2}{3}\)
(4) \(\frac{4}{5}\)
Answer:
(4) \(\frac{4}{5}\)
Hint:
Sample space (S) = 10 + 15 + 25 = 50
n(S) = 50
Let A be the event of getting ₹ 500 note
n (A) = 15
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{15}{50}\)
Let B be the event of getting ₹ 200 note
n (B) = 25
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}=\frac{25}{50}\)
Probability of the note is either a ₹ 500 note or ₹ 200 note
P(A) + P(B) = \(\frac{15}{50}+\frac{25}{50}\) = \(\frac{40}{50}\) = \(\frac{4}{5}\)

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.

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 Maths Chapter 8 Statistics and Probability Ex 8.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.4

Question 1.
If P (A) = \(\frac{2}{3}\), P(B) = \(\frac{2}{5}\), P(A ∪ B) = \(\frac{1}{3}\), then find P(A ∩ B).
Answer:
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)
\(\frac{1}{3}=\frac{2}{3}+\frac{2}{5}\) – P (A ∩ B)

Question 2.
A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A∩B)=016. Find (i) P(not A)
(ii) P(not B)
(iii) P(A or B)
Solution:
(a) P(A) = 0.42 ;
P(B) = 0.48
P(A∩B) = 0.16
(i) P(not A) = P(\(\overline{\mathbf{A}}\)) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = P(\(\overline{\mathbf{B}}\)) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.42 + 0.48 – 0.16
= 0.74

Question 3.
If A and B are two mutually exclusive events of a random experiment and P (not A) = 0.45, P (A ∪ B) = 0.65, then find P(B).
Answer:
P(not A) = 0.45
1 – P (A) = 0.45
P (A) = 1 – 0.45 = 0.55
P(A ∪ B) = P (A) + P (B)
0. 65 = 0.55 + P(B)
0. 65 – 0.55 = P(B)
0.10 = P (B)
P(B) = 0.1

Question 4.
The probability that atleast one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find P(\(\overline{\mathbf{A}}\)) + P(\(\overline{\mathbf{B}}\)).
Solution:
P(A∪B) = 0.6
P(A∩B) = 0.2
P(A) + P(B) = [1 – P(A∪B)] + [1 – P(A∩B)] = [1 – 0.6] + [1 – 0.2]
= 0.4 + 0.8 = 1.2

Question 5.
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability of neither A nor B happen.
Answer:
Here P(A) = 0.5, P (B) = 0.3
P(A ∪ B) = P (A) + P(B) [A and B are mutually exclusive]
= 0.5 + 0.3
= 0.8
Probability of neither A nor [P(A ∪ B)’] = 1 – P(A ∪ B) = 1 – 0.8 = 0.2

Question 6.
Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (S) = 36
Let A be the event of getting an even number on the first time
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (A) = 18
\(P(A)=\frac{n(A)}{n(S)}=\frac{18}{36}\)
(ii) Let B be the event of getting a total of face sum 8.
B = {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{36}\)
A ∩ B = {(2, 6) (4, 4) (6, 2)}
n(A ∩ B) = 3
P(A ∩ B) = \(\frac{3}{36}\)
P(A ∪ B) = P (A) + P (B) – P (A ∩ B)

The required probability = \(\frac{5}{9}\)

Question 7.
From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of its being either a red king or a black queen.
Answer:
n(S) = 52
Let A be the event of getting a red king
n(A) = 2
\(P(A)=\frac{n(A)}{n(S)}=\frac{2}{52}\)
Let B be the event of getting a black Queen king
n(B) = 2
\(P^{\prime}(B)=\frac{n(B)}{n(S)}=\frac{2}{52}\)
It A and B are mutually exclusive
P(A ∪ B) = P(A) + P(B)
\(=\frac{2}{52}+\frac{2}{52}=\frac{4}{52}=\frac{1}{13}\)
The required probability is \(\frac{1}{13}\)

Question 8.
A box contains cards numbered 3, 5, 7, 9,… 35, 37. A card is drawn at random from the box. Find the probability that the drawn card have either multiples of 7 or a prime number.
Answer:
Sample space = {3, 5, 7, 9,…….,35, 37}
n(S) = 18
Let A be the event of getting a multiple of 7
A = {7, 21, 35}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{18}\)
Let B be the event of getting a prime number
B = {3, 5, 7,11, 13, 17, 19, 23, 29, 31, 37}
n(B) = 11

Probability of getting a multiple of 7 or a prime number = \(\frac{13}{18}\)

Question 9.
Three unbiased coins are tossed once. Find the probability of getting atmost 2 tails or atleast 2 heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting atmost 2 tails.
A = {HTT, THT, TTH, HHT, HTH, THH, HHH}
n(A) = 7

Probability of getting atmost two tails or atleast 2 heads = \(\frac{7}{8}\)

Question 10.
The probability that a person will get an electrification contract is \(\frac{3}{5}\) and the probability that he will not get plumbing contract is \(\frac{5}{8}\). The probability of getting atleast one contract is \(\frac{5}{7}\). What is the probability that he will get both?
Answer:
Let A and B represent the event of getting electrification control and plumbing contract.


Probability of getting both the job is \(\frac{73}{280}\)

Question 11.
In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years?
Answer:
Total number of people in a town is 8000.
n(S) = 8000
Total number of females = 3000
Let A be the event of getting number of females
n(A) = 3000
\(P(A)=\frac{n(A)}{n(S)}=\frac{3000}{8000}\)
Number of people over 50 years = 1300
Let B be the event of getting number of people over 50 years.
n(B) = 1300
\(P(B)=\frac{n(B)}{n(S)}=\frac{1300}{8000}\)
Given 30% of the females are over 50 years.


Proability of getting either a female or over 50 years = \(\frac{17}{40}\)

Question 12.
A coin is tossed thrice. Find the probability of getting exactly two heads or atleast one tail or two consecutive heads.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting exactly two heads.
A = {HHT, HTH, THH}
n(A) = 3
\(P(A)=\frac{n(A)}{n(S)}=\frac{3}{8}\)
Let B be the event of getting atleast one tail
B = {HHT, HTH, HTT, THH, THT, TTH, TTT}
n(B) = 7
\(P(B)=\frac{n(B)}{n(S)}=\frac{7}{8}\)
Let C be the event of getting consecutively
C = {HHH, HHT, THH}
n(C) = 3



The probability is 1.

Question 13.
If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P (A ∩ B) = \(\frac{1}{6}\), P(B ∩ C) = \(\frac{1}{4}\), P(A ∩ C) = \(\frac{1}{8}\), P(A ∪ B ∪ C) = \(\frac{9}{10}\) and P (A ∩ B ∩ C) = \(\frac{1}{15}\), then find P(A), P(B) and P(C)?
Answer:
By the given condition,
P(B) = 2 P(A), P(C) = 3 P(A)
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) – P(A ∩ B) – P(B ∩ C) – P(A ∩ C) + P(A ∩ B ∩ C)

Question 14.
In a class of 35, students are numbered from 1 to 35. The ratio of boys and girls is 4 : 3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with prime roll number or a girl with composite roll number or an even roll number.
Answer:
Sample space (S) = {1, 2, 3,… ,35}
n(S) = 35
Total number of students = 35
Number of boys = \(\frac{4}{7}\) × 35 = 20 [Boys Numbers = {1, 2, 3,…, 20}]
Number of girls = \(\frac{3}{7}\) × 35 = 15 [Girls Numbers = { 21, 22,…, 35}]
Let A be the event of getting a boy role number with prime number
A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\frac{8}{35}\)
Let B be the event of getting girls roll number with composite number.
B = {21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35}
n(B) = 12
P(B) = \(\frac{n(\mathrm{B})}{n(\mathrm{S})}\) = \(\frac{12}{35}\)
Let C be the event of getting an even roll number.
C = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34}
n(C) = 17


Probability of getting roll number is \(\frac{29}{35}\)

Students can download Maths Chapter 8 Statistics and Probability Ex 8.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.3

Question 1.
Write the sample space for tossing three coins using tree diagram.
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Question 2.
Write the sample space for selecting two balls from a bag containing 6 balls numbered 1 to 6 (using tree diagram).
Answer:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Question 3.
If A is an event of a random experiment such that P(A) : P(\(\bar{A}\)) = 17 : 15 and n(s) = 640 then find (i) P(\(\bar{A}\))
(ii) n(A)
Answer:

Question 4.
A coin is tossed thrice. What is the probability of getting two consecutive tails?
Answer:
Sample space = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting consecutive tails
A = {HTT, TTH, TTT}
n(A) = 3
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}=\frac{3}{8}\)

Question 5.
At a fete, cards bearing numbers 1 to 1000, one number on one card are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square number greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize
(ii) the second player wins a prize if the first has won?
Solution:
22² = 484
31² = 961
23² = 529
32² = 1024
23, 24, 25, 26, 27, 28, 29, 30, 31 has squares below 500 × 1000.
(i) P(first player wins a prize) = \(\frac{9}{1000}\)
(ii) P(second player ins if first has won) = \(\frac{8}{999}\)

Question 6.
A bag contains 12 blue balls and x red balls. If one ball is drawn at random (i) what is the probability that it will be a red ball? (ii) If 8 more red balls are put in the bag, and if the probability of drawing a red ball will be twice that of the probability in (i), then find x.
Answer:
Sample space = 12 + x
n(S) = x + 12
(i) Let A be the event of getting a red ball
n(A) = x
P(A) = \(\frac{n(\mathrm{A})}{n(\mathrm{S})}\) = \(\left(\frac{x}{x+12}\right)\)
(ii) 8 more red balls are added
Sample space = x + 12 + 8 = x + 20
Number of red balls = x + 8
Probability of drawing red ball = \(\frac{x+8}{x+20}\)
By the given condition
\(\frac{x+8}{x+20}=2\left(\frac{x}{x+12}\right)\)
(x + 8)(x + 12) = 2x(x + 20)
x² + 20x + 96 = 2x² + 40x
x² + 20x – 96 = 0
(x + 24)(x – 4) = 0
x = -24 (or) x = 4
The value of x = 4 (Number of balls will not be negative)
The probability of getting red balls = \(\left(\frac{x}{x+12}\right)=\frac{4}{16}=\frac{1}{4}\)

Question 7.
Two unbiased dice are rolled once. Find the probability of getting
(i) a doublet (equal numbers on both dice)
(ii) the product as a prime number
(iii) the sum as a prime number
(iv) the sum as 1
Answer:
(i) Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
Let A be the event of getting doublet
A = {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of getting a product is a prime number.
B = {(1, 2) (1, 3) (1, 5) (2, 1) (3, 1) (5, 1)}
n(B) = 6
\(P(B)=\frac{n(B)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)
(iii) Let C be the event of getting a sum is a prime number
C = {(1, 1) (1, 2) (1, 4) (1, 6) (2, 1) (2, 3) (2, 5) (3, 2), (3, 4) (4, 1) (4, 3) (5, 2) (5, 6) (6, 1) (6, 5)}
n(C) = 15
\(P(C)=\frac{n(C)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)
(iv) Let D be the event of getting a sum is 1
n(D) = 0
\(P(D)=\frac{n(D)}{n(S)}=\frac{0}{36}=0\)
Probability of getting a sum is 1 is 0

Question 8.
Three fair coins are tossed together. Find the probability of getting
(i) all heads
(ii) atleast one tail
(iii) atmost one head
(iv) atmost two tails
Solution:
Possible outcomes = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
No. of possible outcomes = 2 × 2 × 2 = 8
(i) Prob(all heads) = \(\frac{1}{8}\)
(ii) Atleast one tail = {(THH), (HTH), (HHT), (TTH), (THT), (HTT), (TTT)}
Prob(atleast one tail) = \(\frac{7}{8}\)
(iii) Atmost one head = {(HTT), (THT), (TTH), (TTT)}
∴ Prob(atmost one head) = \(\frac{4}{8}=\frac{1}{2}\)
(iv) Atmost two tail = {(HHH), (THH), (HTH), (HHT), (TTH), (THT), (HTT)}
∴ Prob(atmost two tail) = \(\frac{7}{8}\)

Question 9.
Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Answer:
1st dice A = {1, 2, 3, 4, 5, 6}
2nd dice B = {1, 1, 2, 2, 3, 3}
Sample Space (S) = {(1, 1), (1, 1), (1, 2), (1, 2), (1, 3), (1, 3), (2, 1), (2, 1), (2, 2), (2, 2), (2, 3), (2, 3), (3, 1), (3, 1), (3, 2), (3, 2), (3, 3), (3, 3), (4, 1), (4, 1), (4, 2), (4, 2), (4, 3), (4, 3), (5, 1), (5, 1), (5, 2), (5, 2), (5, 3), (5, 3),(6, 1), (6, 1), (6, 2), (6, 2), (6, 3), (6, 3)}
n(S) = 36
(i) Let A₁ be the event of getting sum is 2
A1 = {(1, 1) (1, 1)}
n(A₁) = 2
\(P\left(A_{1}\right)=\frac{n\left(A_{1}\right)}{n(S)}=\frac{2}{36}=\frac{1}{18}\)
(ii) Let A₂ be the event of getting a sum is 3.
A₂ = {(1, 2) (1, 2) (2, 1) (2, 1)}
n(A₂) = 4
\(P\left(A_{2}\right)=\frac{4}{36}=\frac{1}{9}\)
(iii) Let A₃ be the event of getting a sum is 4.
A₃ = {(1, 3) (1, 3) (2, 2) (2, 2) (3, 1) (3, 1)}
n(A₃) = 6
\(P\left(A_{3}\right)=\frac{6}{36}=\frac{1}{6}\)
(iv) Let A₄ be the event of getting a sum is 5.
A₄ = {(2, 3) (2, 3) (3, 2) (3, 2) (4, 1) (4, 1)}
n(A₄) = 6
\(P\left(A_{4}\right)=\frac{6}{36}=\frac{1}{6}\)
(v) Let A₅ be the event of getting a sum is 6.
A₅ = {(3, 3) (3, 3) (4, 2) (4, 2) (5, 1) (5, 1)}
n(A₅) = 6
\(P\left(A_{5}\right)=\frac{6}{36}=\frac{1}{6}\)
(vi) Let A₆ be the event of getting a sum is 7.
A₆ = {(4, 3) (4, 3) (5, 2) (5, 2) (6, 1) (6, 1)}
n(A₆) = 6
\(P\left(A_{6}\right)=\frac{6}{36}=\frac{1}{6}\)
(vii) Let A₇ be the event of getting a sum is 8.
A₇ = {(5, 3) (5, 3) (6, 2) (6, 2)}
n(A₇) = 4
\(P\left(A_{7}\right)=\frac{4}{36}=\frac{1}{9}\)
(viii) Let A₈ be the event of getting a sum is 9.
A₈ = {(6, 3) (6, 3)}
n(A₈) = 2
\(P\left(A_{8}\right)=\frac{2}{36}=\frac{1}{18}\)

Question 10.
A bag contains 5 red balls, 6 white balls, 7 green balls, 8 black balls. One ball is drawn at random from the bag. Find the probability that the ball drawn is
(i) white
(ii) black or red
(iii) not white
(iv) neither white nor black
Answer:
Sample space (S) = 5 + 6 + 7 + 8
n(S) = 26
(i) Let A be the event of getting a white ball
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(P(A)=\frac{6}{26}=\frac{3}{13}\)
(ii) Let A be the event of getting a black ball
n(A) = 8
\(P(A)=\frac{n(A)}{n(S)}=\frac{8}{26}\)
Let B be the event of getting a red ball
n(B) = 5
\(P(B)=\frac{n(B)}{n(S)}=\frac{5}{26}\)
Probability of getting black or red ball
P(A ∪ B) = P (A) + P (B)
= \(\frac{8}{26}+\frac{5}{26}=\frac{13}{26}=\frac{1}{2}\)
(iii) Not white probability of getting white ball
P(A) = \(\frac{3}{13}\) from (i)
Probability of not getting white ball P(\(\bar{A}\)) = 1 – P(A)
\(1-\frac{3}{13}=\frac{13-3}{13}=\frac{10}{13}\)
(iv) Probability of getting a white ball.
P(A) = \(\frac{6}{26}\) (from 1)
Let B be the event of getting a black ball
n(B) = 8
\(P(B)=\frac{n(B)}{n(S)}=\frac{8}{26}\)
P(A ∪ B) = P(A) + P(B) = \(\frac{6}{26}+\frac{8}{26}=\frac{14}{26}\)
Probability of neither white nor black P(A ∪ B)’ = 1 – P(A ∪ B)
= \(1-\frac{14}{26}\)
= \(\frac{26-14}{26}=\frac{12}{26}=\frac{6}{13}\)

Question 11.
In a box there are 20 non-defective and some defective bulbs. If the probability that a bulb selected at random from the box found to be defective is \(\frac{3}{8}\) then, find the number of defective bulbs.
Answer:
Let the number of defective bulbs be “x”
Sample space (S) = 20 + x
n(S) = 20 + x
Let A be the event of getting to be defective
n(A) = x
\(P(A)=\frac{n(A)}{n(S)}\)
⇒ \(\frac{3}{8}=\frac{x}{20+x}\)
⇒ 8x = 3(20 + x) = (60 + 3x)
⇒ 8x – 3x = 60
⇒ 5x = 60
⇒ x = \(\frac{60}{5}\)
⇒ x = 12
Number of defective bulbs = 12

Question 12.
The king and queen of diamonds, queen and jack of hearts, jack and king of spades are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a clavor
(ii) a queen of red card
(iii) a king of black card
Answer:
King diamond + Queen diamonds = 1 + 1 = 2 …….(1)
Queen hearts + Jack of hearts = 1 + 1 = 2 …….(2)
Jack spade + King of spades =1 + 1 = 2 …….(3)
Remaining number of cards = 52 – (6)
n(S) = 46
(i) Let A be the event of getting a clavor
n (A) = 13
\(P(A)=\frac{n(A)}{n(S)}=\frac{13}{46}\)
(ii) Let B be the event of getting a queen of red card
n(B) = 2
But the above two cards are removed from (1) and (2)
n(B) = 0
\(P(B)=\frac{n(B)}{n(S)}=\frac{0}{46}=0\)
(iii) Let B be the event of getting a king of black card
n(B) = (2 – 1) [from (3) one black card is removed]
n (B) = 1
\(P(B)=\frac{n(B)}{n(S)}=\frac{1}{46}\)

Question 13.
Some boys are playing a game, in which the stone thrown by them landing in a circular region given in the figure is considered as win and landing other than the circular region is considered as a loss. What is the probability to win the game?
Answer:

Area of a rectangle = l × b sq. feet = 3 × 4 sq. feet = 12 sq. feet
sample space (S) = 12
n(S) = 12
Let A be the event of getting the stone landing in a circular region
n(A) = Area of a circle
= πr²
= π × 1 × 1 (radius of a circle = 1 feet)
= π

Probability to win the game = \(\frac{11}{42}\) (or) \(\frac{157}{600}\)

Question 14.
Two customers Priya and Amuthan are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another day. What is the probability that both will visit the shop on
(i) the same day
(ii) different days
(iii) consecutive days?
Answer:
Sample space (S) = 6 × 6 = 36
n(S) = 36
[priya and Amuthan are visiting a particular shop in any one of 6 days is 6 × 6 = 36]
(i) Let A be the event of getting both are shopping on the same day
A = {(Mon, Mon) (Tue, Tue) (Wed, Wed) (Thu, Thu) (Fri, Fri) (Sat, Sat)}
n(A) = 6
\(P(A)=\frac{n(A)}{n(S)}\)
\(=\frac{6}{36}=\frac{1}{6}\)
(ii) Let B be the event of shopping in different days.
n(B) = 36 – 6 = 30
\(P(B)=\frac{n(B)}{n(S)}\)
\(=\frac{30}{36}=\frac{5}{6}\)
(iii) Let C be the event of shopping consecutive days
C = {(Mon, Tue) (Tue, Wed) (Wed, Thu) (Thu, Fri) (Fri, Sat)}
n(C) = 5
P(C) = \(\frac{n(\mathrm{C})}{n(\mathrm{S})}\) = \(\frac{5}{36}\)

Question 15.
In a game, the entry fee is ₹ 150. The game consists of tossing a coin 3 times. Dhana bought a ticket for entry. If one or two heads show, she gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise, she will lose. Find the probability that she
(i) gets double entry fee
(ii) just gets her entry fee
(iii) loses the entry fee.
Answer:
Sample space (S) = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
(i) Let A be the event of getting double entry fee (only getting 3 heads)
n(A) = 1
\(P(A)=\frac{n(A)}{n(S)}=\frac{1}{8}\)
(ii) Let B be the event of getting her entry fee (one or two heads to show)
n(B) = Probability of one head + Probability of 2 head
= \(\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\)
(iii) To loss the entry means not getting the head (only tail)
n(C) = 1
\(P(C)=\frac{n(C)}{n(S)}=\frac{1}{8}\)