Bhagya

Students can download 6th Science Term 2 Chapter 7 Parts of Computer Questions and Answers, Notes, Samacheer Kalvi 6th Science 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 6th Science Solutions Term 2 Chapter 7 Parts of Computer

Samacheer Kalvi 6th Science Parts of Computer Text Book Back Questions and Answers

I. Choose the correct answer:

Question 1.
Which one of the following is an output device?
(a) Mouse
(b) Keyboard
(c) Speaker
(d) Pen drive
Answer:
(c) Speaker

Question 2.
Name the cable that connects CPU to the Monitor
(a) Ethernet
(b) VGA cable
(c) HDMI
(d) USB
Answer:
(b) VGA cable

Question 3.
Which one of the following is an input device?
(a) Speaker
(b) Keyboard
(c) Monitor
(d) Printer
Answer:
(b) Keyboard

Question 4.
Which one of the following is an example for wireless connections?
(a) Wi-Fi
(b) Electric wires
(c) VGA
(d) USB
Answer:
(a) Wi-Fi

Question 5.
Pen drive is ……… device.
(a) Output
(b) Input
(c) Storage
(d) Connecting cable
Answer:
(c) Storage

II. Match the following

Answer:
1. – Connecting Cable
2. – wireless connection
3. – Output device
4. – Input device
5. – LDMI

III. Give Short Answer:

Question 1.
Name the parts of a computer.
Answer:
The computer has three parts like,
1. Input unit
2. Central Processing Unit
3. Output

Question 2.
Answer:

Activity

Look at the magic of connecting cables to desktop computer with 4,3,2,1 formula, start from 4 proceed till 1. Now your computer is ready to use.

By connecting the various parts of a computer we can assemble a computer. For the construction activity, students have to use 4-3-2-1 formula.

A system consist of mouse, key board, monitor, CPU, power cables, and connecting cables Students has to connect the four parts of a computer in row 4, using the cables in row 3, through the power cables in row 2 to construct a system.

Samacheer Kalvi 6th Science Parts of Computer Additional Important Questions and Answers

I. Choose the correct Answer:

Question 1.
Which one of the following is not important part of computer?
(a) Input device
(b) Output device
(c) Mouse
(d) Central Processing Unit
Answer:
(c) Mouse

Question 2.
The page on the monitor can be moved up and down using the _______
(a) Right button
(b) Scroll ball
(c) Left button
(d) Number key
Answer:
(b) Scroll ball

Question 3.
Which one of the following is an example for wireless connection.
(a) USB
(b) power cord
(c) HDMI
(d) wi-fi
Answer:
(d) wi-fi

Question 4.
The data is measured in units which is called as _______
(a) micron
(b) meter
(c) millimeter
(d) Bit
Answer:
(d) Bit

II. Fill in the Blanks

1. ………, ……….. plays a role of data in computer.
2. …….. controls the functions of all parts of the computer.
3. Memory unit divided into ………. types.
4. The data is measured in units which is called as ………..
5. Mouse is connected with the computer using ………..

Answer:

1. Numbers and alphabets
2. control unit
3. 2
4. Bit
5. USB cord or cable

III. Match the following

Answer:
1. – c
2. – d
3. – a
4. – e
5. – b

Students can download 10th Science Chapter 19 Origin and Evolution of Life Questions and Answers, Notes, Samacheer Kalvi 10th Science 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 Science Solutions Chapter 19 Origin and Evolution of Life

Samacheer Kalvi 10th Science Origin and Evolution of Life Text Book Back Questions and Answers

I. Choose the correct answer.

Question 1.
Biogenetic law states that:
(a) Ontogeny and phylogeny go together
(b) Ontogeny recapitulates phylogeny
(c) Phylogeny recapitulates ontogeny
(d) There is no relationship between phylogeny and ontogeny
Answer:
(b) Ontogeny recapitulates phylogeny

Question 2.
The ‘use and disuse theory’ was proposed by:
(a) Charles Darwin
(b) Ernst Haeckel
(c) Jean Baptiste Lamarck
(d) Gregor Mendel
Answer:
(c) Jean Baptiste Lamarck

Question 3.
Paleontologists deal with:
(a) Embryological evidences
(b) Fossil evidences
(c) Vestigial organ evidences
(d) All the above
Answer:
(a) Embryological evidences

Question 4.
The best way of direct dating fossils of recent origin is by:
(a) Radio-carbon method
(b) Uranium lead method
(c) Potassium-argon method
(d) Both (a) and (c)
Answer:
(a) Radio-carbon method

Question 5.
The term Ethnobotany was coined by:
(a) Khorana
(b) J.W. Harshberger
(c) Ronald Ross
(d) Hugo de Vries
Answer:
(b) J.W. Harshberger

II. Fill in the blanks.

1. The characters developed by the animals during their life time, in response to the environmental changes are called ………..
2. The degenerated and non-functional organs found in an organism are called ……….
3. The forelimb of bat and human are examples of ………. organs.
4. The theory of natural selection for evolution was proposed by ………..

Answer:

1. acquired characters
2. vestigial organs
3. homologous
4. Charles Darwin

III. State true or false. Correct the false statements.

1. The use and disuse theory of organs’ was postulated by Charles Darwin.
2. The homologous organs look similar and perform similar functions but they have different origin and developmental pattern.
3. Birds have evolved from reptiles.

Answer:

1. False – The use and disuse theory of organs’ was postulated by Jean Baptiste Lamarck.
2. False – The analogous organs look similar and perform similar functions but they have different origin and developmental pattern.
3. True

IV. Match the following.

Answer:
A. (iii)
B. (i)
C. (iv)
D. (ii)
E. (vi)
F. (v)

V. Answer in a word or sentence.

Question 1.
A human hand, a front leg of a cat, a front flipper of a whale and a bat’s wing look dissimilar and adapted for different functions. What is the name given to these organs?
Answer:
Homologous organ – as they have inherited from common ancestors with similar developmental pattern in embryos.

Question 2.
Which organism is considered to be the fossil bird?
Answer:
Archaeopteryx is the fossil bird, found in the Jurassic period.

Question 3.
Why is Archaeopteryx considered to be a connecting link?
Answer:
Archaeopteryx is considered to be a connecting link between reptiles and birds as it had wings with feathers like a bird and had a long tail, clawed digits and conical teeth like a reptile.

VI. Short Answers Questions

Question 1.
The degenerated wing of a kiwi is an acquired character. Why is it an acquired character?
Answer:
The characters developed by the animals during their life time in response to the environmental changes are called acquired character. The acquired characters are transmitted to the offspring by the process of inheritance.

Question 2.
Why is Archaeopteryx considered to be a connecting link?
Answer:
Archaeopteryx is the oldest known fossil bird. It is considered to be a connecting link between reptiles and birds. It had wings with feathers, like a bird. It had a long tail, clawed digits and conical teeth, like a reptile.

Question 3.
Define Ethnobotany and write its importance.
Answer:
Ethnobotany is the study of a region’s plants and their practical uses through the traditional knowledge of the local culture of people.

Question 4.
How can you determine the age of the fossils?
Answer:
The age of fossils is determined by radioactive elements present in it. The elements may be carbon, uranium, lead or potassium. Carbon consumption of animals and plants stops after death, and the decaying process of C14 occurs continuously. The time passed since the death of a plant or animal can be calculated by measuring the amount of C14 present in their body.

VII. Long Answer Questions

Question 1.
Natural selection is a driving force for evolution-How?
Answer:
Survival of the fittest or Natural selection : During the struggle for existence, the organisms which can overcome the challenging situation, survive and adapt to the surrounding environment. Organisms which are unable to face the challenges, are unfit to survive and disappear. The process of selection of organisms with favourable variation is called as natural selection.

Question 2.
How do you differentiate homologous organs from analogous organs?
Answer:

Question 3.
How does fossilization occur in plants?
Answer:
The process of formation of fossil in the rocks is called Fossilization. The common methods of fossilization include:
1. Petrifaction: Minerals like silica slowly penetrate in and replace the original organic tissue and forms a rock-like fossil. This method of fossilization can preserve hard and soft parts mostly bones and wood fossils are petrified.

2. Mould and cast: A replica of a plant or animal is preserved in sedimentary rocks. When the organism gets buried in sediment it is dissolved by underground water leaving a hollow depression called a mould. It shows the original shape but does not reveal the internal structure. Minerals or sediment fill the hollow depression and form a cast.

3. Preservation: Original remains can be preserved in ice or amber (tree sap). They protect the organisms from decay. The entire plant or animal is preserved.

4. Compression: When an organism dies, the hard parts of their bodies settle at the bottom of the sea bed and are covered by sediment. The process of sedimentation goes on continuously and fossils are formed.

5. Infiltration or Replacement: The precipitation of minerals takes place which later on infiltrates the cell wall. The process is brought about by several mineral elements such as silica, calcium carbonate and magnesium carbonate. Hard parts are dissolved and replaced by these minerals.

VIII. Higher Order Thinking Skills (HOTS).

Question 1.
Arun was playing in the garden. Suddenly he saw a dragon fly sitting on a plant. He observed the wings of it. He thought it looked similar to a wing of a crow. Is he correct? Give reason for your answer.
Answer:
No. Arun is not correct. It is called as Analogous organ. They look similar and perform similar functions, but they have different origin and developmental pattern.

Question 2.
Imprints of fossils tell us about evolution-How?
Answer:
Fossils provide solid evidence that organisms from the past are not the same as those found today. Fossils show a progression of evolution by comparing the anatomical record of fossils with modem and extinct species. Palaeontologist can infer the linkages of their species. For this the body parts such as shell bones or teeth are used. The resulting fossil record tells the story of the part and show the evolution of form over millions of years.

Question 3.
Octopus, cockroach and frog all have eyes. Can we group these animals together to establish a common evolutionary origin. Justify your answer.
Answer:
No, we cannot group these animals together because development of eye is not a point utilised in classification as well as in evolution.

Samacheer Kalvi 10th Science Origin and Evolution of Life Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
Life originated in:
(a) Air
(b) Earth
(c) Water
(d) Nitrogen
Answer:
(c) Water

Question 2.
Believers of spontaneous generation theory believed that:
(a) Life originated from a similar organism
(b) Life originated from the air
(c) Life originated only spontaneously
(d) Life originated from pre-existing life
Answer:
(c) Life originated only spontaneously

Question 3.
Which of the following organ in man is not vestigial?
(a) Vermiform appendix
(b) Nictitating membrane
(c) Ear muscles
(d) Epiglottis
Answer:
(a) Vermiform appendix

Question 4.
Analogous organs are:
(a) similar in origin
(b) similar in structure
(c) non-functional
(d) similar in function
Answer:
(b) similar in structure

Question 5.
Wings of locust, pigeon and bat are example of:
(a) Vestigial organs
(b) Analogous organs
(c) Homologous organs
(d) Evolution
Answer:
(a) Vestigial organs

Question 6.
The distance needs to be neither too hot nor too cold is called ______.
(a) habitable zone
(b) Goldilock zone
(c) Aquatic zone
(d) Terrestrial zone.
Answer:
(c) Aquatic zone

Question 7.
Fossils are dated by:
(a) Amount of calcium residue
(b) Amount of radio active carbon compound
(c) Association with other mammals
(d) Structure of bones
Answer:
(b) Amount of radio active carbon compound

Question 8.
A basket of vegetables contains carrot, potato and tomato. Which of them represent the correct homologous structures?
(a) Carrot and potato
(b) Carrot and tomato
(c) Radish and carrot
(d) Radish and potato
Answer:
(c) Radish and carrot

Question 9.
Most fossil have been found in:
(a) Black soil
(b) Lava flows
(c) Granite
(d) Sedimentary rocks
Answer:
(d) Sedimentary rocks

Question 10.
Birbal sahni was a:
(a) Zoologist
(b) Lounder of Central Drug Research Institute
(c) Ornithologist
(d) Palaeobotanist
Answer:
(d) Palaeobotanist

Question 11.
Darwin in his ‘Natural Selection Theory’ did not believe in any role of which one of the following in organic evolution.
(a) Parasites and Predators as natural enemies
(b) Survival of the fittest
(c) Struggle for existence
(d) Discontinuous Variation
Answer:
(d) Discontinuous Variation

Question 12.
Fossilizaticn can occur in:
(a) Animals are burned and preserved by natural process
(b) Animals are destroyed by scavangers
(c) Animals are eaten by predators
(d) Animals are destroyed by environmental conditions
Answer:
(a) Animals are burned and preserved by natural process

II. Fill in the blanks.

1. Branch of biology which deals with fossils is …………
2. Archaeopteryx is connecting link between reptiles and ………..
3. Darwin explained origin of species through …………
4. The ship on which Darwin worked as naturalist was …………
5. The book Origin of Species was published in the year …………
6. Mutation can be artificially induced by ………..
7. Extraterrestrial states that units of life called ………..
8. The idea of chemical evolution of life was developed by …………..
9. Living being able to reproduce more individuals and form the own progeny is called …………..
10. ……….. is the raw material which plays an important role in evolution.
Answer:
1. Palaeontology
2. Birds
3. Natural selection
4. H.M.S. Beagle
5.1859
6. Radiation
7. Panspermia (spores)
8. Oparin and Haldane
9. Over production
10. Variation

III. Match the following.

Answer:
A. (ii)
B. (iv)
C. (i)
D. (v)
E. (iii)

IV. State whether True or false. If false write the correct statement.

1. De Vries is called as the Father of Palaeontology.
2. NASA is developing the Mars 2020 astrobiology to investigate an astrobiologically details of Mars.
3. Competition among the individual of same species is Interspecific struggle.
4. Abiogenesis states that life originates from pre existing life.
5. The Big Bang theory explains the origin of Universe.

Answer:

1. False – Leonardo da vinci is called as the Father of Palaeontology.
2. True
3. False – Competition among the individual of same species is Intraspecific struggle.
4. False – Biogenesis states that life originates from pre existing life.
5. True

V. Answer in a word or sentence.

Question 1.
Who proposed the theory of Natural selection?
Answer:
Charles Darwin

Question 2.
What do you understand by evolution?
Answer:
Formation of new species due to changes in specific characters over several generations as response to natural selection is called evolution.

Question 3.
How do embryological studies provide evidences for evolution?
Answer:
The embryos from fish to mammals are similar in their early stages of development. The differentiation of their special characters appear in the later stages of development.

Question 4.
Mention three important features of fossils.
Answer:

1. Fossils are useful in classification of organisms.
2. Fossils are used in the field of descriptive and comparative analoms.
3. Fossils throw light on phylogeny and evolution of organisms.

Question 5.
Who is the Father of Indian Paleobotany? What is his contribution in this field.
Answer:
Birbal Sahani is the Father of Indian Paleobotany. He presented his research on two different areas of Paleobotany.

1. The anatomy and morphology of paleozoic ferns.
2. The fossil plants of the Indian Gondwana formation.

Question 6.
List out the methods of Fossilization.
Answer:
Common methods of fossilization includes petrifaction molds and cast, carbonization preservation, compression and infiltration.

Question 7.
What are extremophiles?
Answer:
The organisms which live in extreme environmental conditions on earth are called extremophiles.

Question 8.
What is Goldilock zone for life. Name the planet in the Goldilock zone.
Answer:
The Goldilock zone for life is the zone having a right mass to retain an atmosphere and have an orbit at the right distance from its star that allows liquid water to exist and the distance need to be neither too hot or cold. In our solar system – Earth is the only planet in the Goldilock zone.

Question 9.
What are the three types of struggle for existence?
Answer:

1. Intra specific struggle
2. Inter specific struggle
3. Environmental struggle

Question 10.
What is Atavism?
Answer:
The appearance of the ancestral character in some individuals is called Atavism. Eg: Presence of rudimentary tail in newborn babies.

VI. Short Answer Questions.

Question 1.
Distinguish between acquired and inherited trait with example.
Answer:

Question 2.
What are the aspects of ethnobotany?
Answer:
Ethnobotany has relevance with problems of nutrition, health care and life support system, faith in plants, cottage industries, economic upliftment, conservation of biodiversity and sustainable use of plant resources.

Question 3.
What is the Geological Time Scale?
Answer:
The geological time scale is a system of chronological dating that relates geological rock strata to time and is used by geologists, palaeontologists and other Earth scientists to describe the timing and relationships of events that have occurred during Earth’s history.

Question 4.
Write a note on types of Germinal Variation.
Answer:
Germinal variation are classified into two types:

1. Continuous variation,
2. Discontinuous variation

1. Continuous variation : These are small variations which occur among individuals of a population. They are also called as fluctuating variations. They occur by gradual accumulation in a population, e.g. skin colour, height and weight of an individual, colour of eye, etc.

2. Discontinuous variation : These are sudden changes which occur in an organism due to mutations. They do not have any intermediate forms. These large variations are not useful for evolution, e.g. short legged Ancon sheep, six or more digits (fingers) in human, etc.

Question 5.
Explain the existence for astro biology.
Answer:
The theory explains that any planets can support the existence of life, if it fulfills two important criteria.

1. It must have a right mass to retain an atmosphere.
2. It must have an orbit at just the right distance from its star (Sun) that it allows liquid water to exist. Thus, the distance need to be neither too hot or not too cold and is often referred as Goldilock Zone for life.

VII. Long Answer Questions.

Question 1.
Explain the various methods of Fossilization.
Answer:
Common methods of fossilization includes petrifaction, molds and cast, carbonization, preservation, compression and infiltration.
Petrifaction : Minerals like silica slowly penetrate in and replace the original organic tissue and forms a rock like fossil. This method of fossilization can preserve hard and soft parts. Most bones and wood fossils are petrified.

Mold and Cast : A replica of a plant or animal is preserved in sedimentary rocks. When the organism gets buried in sediment it is dissolved by underground water leaving a hollow depression called a mold. It shows the original shape but does not reveal the internal structure. Minerals or sediment fill the hollow depression and forms a cast.

Preservation : Original remains can be preserved in ice or amber (tree sap). They protect the organisms from decay. The entire plant or animal is preserved.

Compression : When an organism dies, the hard parts of their bodies settle at the bottom of the sea bed and are covered by sediment. The process of sedimentation goes on continuously and fossils are formed.

Infiltration or Replacement : The precipitation of minerals takes place which later on infiltrate the cell wall. The process is brought about by several mineral elements such as silica, calcium carbonate and magnesium carbonate. Hard parts are dissolved and replaced by these minerals.

Question 2.
Explain the principles of Lamarckism.
Answer:
Principles of Lamarckism:
(i) Internal vital force : Living organisms or their component parts tend to increase in size continuously. This increase in size is due to the inherent ability of the organisms.

(ii) Environment and new needs : A change in the environment brings about changes in the need of the organisms. In response to the changing environment, the organisms develop certain adaptive characters. The adaptations of the organisms may be in the form of development of new parts of the body.

(iii) Use and disuse theory : Lamarck’s use and disuse theory states that if an organ is used constantly, the organ develops well and gets strengthened. When an organ is not used for a long time, it gradually degenerates.

The ancestors of giraffe were provided with short neck and short – forelimbs. Due to shortage of grass, they were forced to feed on leaves from trees. The continuous stretching of their neck and forelimbs resulted in the development of long neck and long forelimbs which is an example for constant use of an organ. The degenerated wing of Kiwi is an example for organ of disuse.

(iv) Theory of Inheritance of acquired characters : When there is a change in the environment, the animals respond to the change. They develop adaptive structures. The characters developed by the animals during their life time, in response to the environmental changes are called acquired characters. According to Lamarck, the acquired characters are transmitted to the offspring by the process of inheritance.

VIII. Higher Order Thinking Skills (HOTS).

Question 1.
What is the relationship between mutation and variation?
Answer:
Mutation and variation are the two events involved in the process of evolution. Mutation arises due to errors occurring in DNA during replication or exposure to ultraviolet rays or chemicals. The mutation leads to variation. It brings about changes in a single individual.

Question 2.
Picture of the newborn baby to be given.

(i) Name the phenomenon responsible for the structure marked (a) in the Figure.
Answer:
Atavism

(ii) Give the definition of the phenomenon responsible for the development for the structure marked.
Answer:
The reappearance, ancestral character in some individuals.

Students can download 6th Science Term 2 Chapter 6 Human Organ Systems Questions and Answers, Notes, Samacheer Kalvi 6th Science 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 6th Science Solutions Term 2 Chapter 6 Human Organ Systems

Samacheer Kalvi 6th Science Human Organ Systems Text Book Back Questions and Answers

I. Choose the correct answer:

Question 1.
The circulatory system transports these throughout the body
(a) Oxygen
(b) Nutrient
(c) Hormones
(d) All of these
Answer:
(d) All of these

Question 2.
Main organ of respiration in human body is
(a) Stomach
(b) Spleen
(c) Heart
(d) Lungs
Answer:
(d) Lungs

Question 3.
Breakdown of food into smaller molecules in our body is known as
(a) Muscle contraction
(b) Respiration
(c) Digestion
(d) Excretion
Answer:
(a) Digestion

II. Fill in the Blanks

1. A group of organs together make up an ………. system.
2. The part of the skeleton that protects the brain is …………
3. The process by which the body removes waste is ………..
4. The ………. is the largest sense organ in our body.
5. The endocrine glands produce chemical substances called …………

Answer:

1. Organ
2. Skull
3. Excretion
4. Skin
5. hormones

III. True or False. If False, give the correct answer.

1. Blood is produced in the bone marrow.
2. All the waste products of the body are excreted through the circulatory system.
3. The other name of the food pipe is an alimentary canal.
4. Thin tube-like structures which are the component of the circulatory system are called blood vessels.
5. The brain, the spinal cord, and nerves form the nervous system.

Answer:

1. False – RBC’s are produced in the bone marrow.
2. False – All the waste products are transported through the circulatory system.
3. False – The other name of the digestive tract is an alimentary canal.
4. True
5. True

IV. Match the following

Answer:
1. – c
2. – e
3. – b
4. – a
5. – d

V. Arrange in the correct sequence

Question 1.
Stomach → Large intestine → Oesophagus → Pharynx → Mouth → Small Intestine → Rectum → Anus
Answer:
Mouth → Pharynx → Oesophagus → Stomach → Small intestine → Large intestine → Rectum

Question 2.
Urethra → Ureter → Urinary Bladder → Kidney
Answer:
Kidney → Ureter → Urinary bladder → Urethra

VI. Analogy

1. Arteries : Carry blood from the heart:: ……….. : Carry blood to the heart.
2. Lungs : Respiratory system :: ……….. : Circulatory system.
3. Enzymes : Digestive glands :: ………… : Endocrine glands

Answer:

1. Veins
2. Heart
3. Hormones

VII. Give a Very short answer:

Question 1.
Describe the skeletal system.
Answer:

1. The skeletal system consists of bones, cartilages and joints.
2. Bones provide a framework for the body.
3. Bones along with muscles help in movements such as walking, running, chewing and dancing etc.

Question 2.
Write the functions of epiglottis?
Answer:

1. It prevents the entry of food into the windpipe.
2. It opens when the air enters the windpipe.

Question 3.
What are the three types of blood vessels?
Answer:

1. Arteries
2. Veins
3. Capillaries

Question 4.
Define the term “Trachea”
Answer:

1. The trachea commonly called “Windpipe” is a tube supported by cartilaginous rings.
2. It connects the pharynx and larynx to the lungs.
3. Allowing the passage of air.

Question 5.
Write any two functions of the digestive system.
Answer:

1. The digestive system is involved in the conversion of complex food substances into simple forms.
2. Absorption of digested food.

Question 6.
Name the important parts of the eye?
Three main parts.
Answer:

1. Cornea
2. Iris
3. Pupil

Question 7.
Name the five important sense organs.
Answer:

1. Eyes
2. Ears
3. Nose
4. Tongue
5. Skin

VIII. Give a short answer:

Question 1.
Write a short note on the rib cage.
Answer:

1. The rib cage is made up of 12 pairs of curved, flat rib bones.
2. It protects the delicate vital organs such as the heart and lungs.

Question 2.
List out the functions of the human skeleton.
Answer:

1. The skeletal system gives shape to the body.
2. Bones provide a framework for the body.
3. Bones along with muscles help in movements such as walking, running, chewing and dancing, etc.
4. It protects the soft internal organs.

Question 3.
Differentiate between the voluntary muscles and involuntary muscles.
Answer:

IX. Answer in detail:

Question 1.
List out the functions of the Endocrine system and Nervous system.
Answer:

1. Endocrine system regulates various functions of the body and maintains the internal environment.
2. Endocrine glands produce chemical substances called “Hormones’ which control various activities of the body.
   Eg. Growth hormone controls growth, the Adrenalin hormone acts at the time of fear stress, etc.

Functions of the nervous system:

1. Sensory input: The conduction of signals from sensory receptors.
2. Integration: The interpretation of the sensory signals and the formulation of responses.
3. Motor output: The conduction of signals from the brain and spinal card to effectors such as muscle and gland cells.

Question 2.
Label the diagram given below to show the four main parts of the urinary system and answer the following questions.

a. Which organ removes extra salts and water from the blood?
b. Where is the urine stored?
c. What is the tube through which urine is excreted out of the body?
d. What are the tubes that transfer urine from the kidneys to the urinary bladder called?
Answer:
a. The functional units of the kidney are called Nephrons which filter the blood and form the urine.
b. Urine is stored in the urinary bladder.
c. Urine is expelled out through the urethra.
d. The ureters that transfer urine from the kidneys to the urinary bladder.

X. Questions based on Higher Order Thinking Skills

Question 1.
What will happen if the diaphragm shows no movement?
Answer:

1. The diaphragm is the primary organ of breathing.
2. The movement of the diaphragm expands the lungs and creates a vacuum.
3. Due to this, the air is sucked in.
4. If the diaphragm does not move the lungs do not expand or contract and breathing stops.
5. The person will die.

Question 2.
Why is the heart divided into two halves by a thick muscular wall?
Answer:

1. The ventricles of the heart have thicker muscular walls than the atria.
2. The left ventricle also has a thicker muscular wall than the right ventricle.
3. This is due to the higher forces needed to pump blood through the systemic circuit compared to the pulmonary circuit.

Question 3.
Why do we sweat more in summer?
Answer:

1. Sweating plays an important health role as it helps to maintain constant body temperature by cooling us down.
2. When it is hot and we sweat that moisture evaporates and cools us immediately.
3. This is why we sweat more when the summer is very hot.

Question 4.
Why do we hiccup and cough sometimes when we swallow food?
Answer:
Reasons for hiccup and cough :

1. Eating too – quickly and swallowing air along with foods.
2. Eating too – many fatty or spicy foods in particular.
3. Drinking too much-carbonated beverages or alcohol can distend the stomach and irritate the diaphragm which can cause hiccups.

Samacheer Kalvi 6th Science Human Organ Systems Additional Important Questions and Answers

I. Choose the right answer:

Question 1.
A newborn baby has ……….. bones.
(a) 206
(b) More than 200
(c) More than 300
(d) 210
Answer:
(c) More than 300

Question 2.
_______ connect bone to muscle.
(a) Skeleton
(b) Tendons
(c) Cartilages
(d) Ligaments
Answer:
(b) Tendons

Question 3.
The walls of the heart is made up of
(a) Voluntary muscles
(b) Cardiae muscles
(c) Involuntary muscles
(d) Biceps muscle
Answer:
(b) Cardiac muscles

Question 4.
_______ muscles are found in the walls of the digestive tract, urinary bladder arteries, and other internal organs.
(a) Bone
(b) Smooth
(c) Cardiac
(d) triceps
Answer:
(b) Smooth

Question 5.
The bronchi divide further and end in small air sacs called
(a) Cerebrum
(b) Thymus
(c) Alveoli
(d) Pinna
Answer:
(c) Alveoli

II. Fill in the blanks:

1. The adult human skeletal system consists of 206 bones and a few ………., ……… and ………..
2. The brain is covered by a three-layered tissue covering called………..
3. When we are walking, running or climbing the balance of the body is maintained by ………….
4. The endocrine gland present in the chest is …………..
5. The ………. brings blood containing oxygen and urea from the aorta to the kidneys.
6. connects the brain to a different part of the body through nerves.

Answer:

1. cartilages, ligaments, and tendons
2. meninges
3. Ears
4. Thymus Gland
5. renal artery
6. spinal cord

III. Match the Following:

Answer
1. – iii.
2. – i.
3. – iv.
4. – v.
5. – ii.

IV. Arrange in the correct sequence

Question 1.
Trachea → Bronchi → Pharynx → Lungs → Larynx → Nasal cavity → Nostrils → Bronchiole → Alveolus
Answer:
Nostrils → Nasal cavity → Pharynx → Larynx → Trachea → Bronchi → Bronchiole → Alveolus

Question 2.
Middle ear → Pinna → Outer ear → Inner ear
Answer:
Pinna → Outer ear → Middle ear → Inner ear.

V. Analogy

1. Skull: Made up of cranial bones and facial bones :: ………… : Made up of 12 pairs of curved flat rib bones
2. Biceps : Bends the arm at the elbow :: ……….. : Straightens the elbow

Answer:

1. Ribcage
2. Triceps

VI. Very Short Answer:

Question 1.
What are the two major divisions of the skeletal system?
Answer:
The two major divisions of the skeletal system are.

1. Axial skeleton
2. Appendicular skeleton.

Question 2.
Name the auditory ossicles.
Answer:

1. Malleus
2. Incus
3. Stapes

Question 3.
Which is the longest bone in our body?
Answer:
The thigh bone (femur) is the longest bone.

Question 4.
What is the name of the muscles in the heart?
Answer:
Cardiac muscle.

Question 5.
What is the length of the Alimentary canal?
Answer:
9 metres.

VII. Answer in detail:

Question 1.
Tabulate the differences between arteries and veins.
Answer:
Arteries:

1. Carry blood from the heart to all parts of the body.
2. Carry oxygenated blood (Except pulmonary artery).
3. They have thick elastic muscular walls.
4. Valves are absent.
5. Blood flows under high pressure.

Veins:

1. Carry blood from all parts of the body back to the heart.
2. Carry deoxygenated blood (Except pulmonary vein).
3. They have thin non-elastic valves.
4. Valves are present to prevent the backward flow of blood.
5. Blood flows under low pressure.

Question 2.
Differentiate arteries from a vein and tabulate your answer.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.7 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.7

Choose the most suitable answer.

Question 1.
A zero of x³ + 64 is:
(a) 0
(b) 4
(c) 4i
(d) -4
Solution:
(d) -4
Hint:
x³ = -64
x³ = -4 × -4 × -4 = (-4)³
x = -4

Question 2.
If f and g are polynomials of degrees m and n respectively, and if h(x) = (f o g) (x), then the degree of h is:
(a) mn
(b) m + n
(c) mⁿ
(d) n^m
Solution:
(a) mn
Hint:
Let f(x) = x^m and g(x) = xⁿ
degree = m, degree = n
(f o g) (x) = f(g(x)) = f(xⁿ) = (xⁿ)^m =x^mn
degree = mn

Question 3.
A polynomial equation in x of degree n always has:
(a) n distinct roots
(b) n real roots
(c) n imaginary roots
(d) at most one root.
Solution:
(a) n distinct roots

Question 4.
If α, β and γ are the roots of x³ + px² + qx + r, then Σ \(\frac{1}{α}\) is:
(a) –\(\frac{q}{r}\)
(b) –\(\frac{p}{r}\)
(c) \(\frac{q}{r}\)
(d) –\(\frac{q}{p}\)
Solution:
(a) –\(\frac{q}{r}\)
Hint:
x³ + px² + qx + r = 0
α, β, γ are the roots
α + β + γ = -p
αβ + βγ + γα = q
αβγ = -r
Σ \(\frac{1}{α}\) = \(\frac{1}{α}\) + \(\frac{1}{β}\) + \(\frac{1}{γ}\) = \(\frac{βγ+αβ+αγ}{αβγ}\) = –\(\frac{q}{r}\)

Question 5.
According to the rational root theorem, which number is not possible rational root of 4x⁷ + 2x⁷ – 10x³ – 5?
(a) -1
(b) \(\frac{5}{4}\)
(c) \(\frac{4}{5}\)
(d) 5
Solution:
(c) \(\frac{4}{5}\)
Hint:
By rational root of theorem,
\(\frac{p}{q}\) is a root of a polynomial a₀ = -5, a_n = 4 and (4, -5) = 1, then p must divide 5 and q must divide 4.
Possible values of p are +1, -1, +5, -5
Possible values of q are +1, -1, 4, -4
∴ \(\frac{4}{5}\) is not possible rational roots.

Question 6.
The polynomial x³ – kx² + 9x has three real roots if and only if, k satisfies:
(a) |k | ≤ 6
(b) k = 0
(c) |k| > 6
(d) |k| ≥ 6
Solution:
(d) |k| ≥ 6
Hint:
Δ ≥ 0
k² – 4(1)(9) ≥ 0
k² ≥ 36
|k| ≥ 6

Question 7.
The number of real numbers in [0, 2π] satisfying sin⁴x – 2 sin²x + 1 is:
(a) 2
(b) 4
(c) 1
(d) ∞
Solution:
(a) 2
Hint:
sin⁴x – 2sin²x + 1 = 0
Put sin²x = t [t² – 2t + 1 = 0 ⇒ (t – 1)² = 0]
t = 1, t = 1
sin²x = 1
sin x = ± 1
∴ sin x = 1, sin x = -1
sin x = sin \(\frac{π}{2}\), sin x = sin (π + \(\frac{π}{2}\))
x = \(\frac{π}{2}\), x = \(\frac{3π}{2}\)
Number of real numbers in [0, 2, π] is 2.

Question 8.
If x³ + 12x² + 10ax + 1999 definitely has a positive root, if and only if:
(a) a ≥ 0
(b) a > 0
(c) a < 0
(d) a ≤ 0
Solution:
(c) a < 0
Hint:
∴ If a < 0, then only we can get one sign change.

Question 9.
The polynomial x³ + 2x + 3 has:
(a) one negative and two real roots
(b) one positive and two imaginary roots
(c) three real roots
(d) no solution
Solution:
(a) one negative and two real roots
Hint:
p(x) = x³ + 2x + 3, no sign changes ⇒ no positive real roots.
p(-x) = -x³ – 2x + 3, one sign changes ⇒ one negative real roots.
∴ It has 1 negative root and 2 imaginary roots.

Question 10.
The number of positive roots of the polynomials \(\sum_{r=0}^{n}\) ⁿC_r (-1)^r x^r is
(a) 0
(b) n
(c) <n
(d) r
Solution:
(b) n
Hint:
\(\sum_{r=0}^{n}\) ⁿC_r (-1)^r x^r = xⁿ – ⁿC₁ x^n-1 + ⁿC₂ x^n-2 + …. + (-1)ⁿ
P(x) has n changes
∴ It has n positive changes
P(-x) has no changes
∴ no negative roots.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.5 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5

Question 1.
Solve the following equations
(i) sin² x – 5 sin x + 4 = 0
Solution:
sin²x – 5sinx + 4 = 0
Let y = sin x
(y² – 5y + 4 = 0
(y – 1) (y – 4) = 0
(y – 1) = o or (y – 4) = 0
y = 1 or y = 4
sin x = 1 or sin x = 4 [not possible since sin x ≤ 1]
sin x = sin \(\frac{\pi}{2}\)
x = nπ + (-1)ⁿ α, n ∈ z.
x = nπ + (-1)ⁿ \(\frac{\pi}{2}\)

(ii) 12x³ + 8x = 29x² – 4 = 0
Solution:
12x³ – 29x² + 8x + 4 = 0

12x² – 5x – 2 = 0
12x² – 8x + 3x – 2 = 0
4x(3x – 2) + 1(3x – 2) = 0
(3x – 2)(4x + 1) = 0
3x = 2, 4x = -1
x = \(\frac{2}{3}\) or x = –\(\frac{1}{4}\)
The roots are 2, \(\frac{2}{3}\), –\(\frac{1}{4}\).

Question 2.
Examine for the rational roots of
(i) 2x³ – x² – 1 = 0
Solution:
Sum of co-efficients = 2 – 1 – 1 = 0
⇒ x = 1 is a factor.

Which is imaginary root
∴ x = 1 is rational root.

(ii) x⁸ – 3x + 1 = 0
Solution:
a_n = 1; a₀ = 1
If \(\frac{p}{q}\) is a root of the polynomial. (p, q) = 1
By rational root theorem, it has no rational roots.

Question 3.
Solve: 8x^{\(\frac{3}{2n}\)} – 8x^{\(\frac{-3}{2n}\)} = 63.
Solution:
Put k = \(\frac{3}{2n}\) ∴ 8x^k – 8x^-k = 63
8x^k – \(\frac{8}{x^k}\) = 63
8x^2k – 8 = 63x^k
8x^2k – 63x^k – 8 = 0
(x^k – 8) (8x^k + 1) = 0
x^k – 8 = 0
x^k = 8
x^{\(\frac{3}{2n}\)} = 8
x = 8^{\(\frac{3}{2n}\)} (2³)^{\(\frac{3}{2n}\)} = (2²)ⁿ = 4ⁿ
∴ x = 4ⁿ is a root of the equation.

Question 4.
Solve: 2\(\sqrt{\frac{x}{a}}\) + 3\(\sqrt{\frac{a}{x}}\) = \(\frac{b}{a}\) = \(\frac{6a}{b}\)
Solution:
put \(\sqrt{\frac{x}{a}}\) = y

Question 5.
Solve the equations
(i) 6x⁴ – 35x³ + 62x² – 35x + 6 = 0
Solution:
This is Type I even degree reciprocal equation. Hence it can be rewritten as

(1) ⇒ 6(y² – 2) – 35y + 62 = 0
6y² – 12 – 35y + 62 = 0
6y² – 35y + 50 = 0

2x² + 2 – 5x = 0 (or) 3x² + 3 = 10x
2x² – 5x + 2 = 0 (or) 3x² – 10x + 3 = 0
x = 2, \(\frac{1}{2}\) (or) x = 3, \(\frac{1}{3}\)
Roots are 2, \(\frac{1}{2}\), 3 and \(\frac{1}{3}\)

(ii) x⁴ + 3x³ – 3x – 1 = 0
Solution:
(x – 1) and (x + 1) is a factor.

(x – 1)(x + 1)(x² + 3x + 1) = 0
x – 1 = 0 (or) x + 1 = 0 (or) x² + 3x + 1 = 0
x = 1 (or) x = -1 (or) x² + 3x = -1

Question 6.
Find all real numbers satisfying
Solution:
4^x – 3 (2^x+2) + 2⁵ = 0
⇒ (2²)^x – 3(2^x . 2²) + 2⁵ = 0
(2²)^x – 12 . 2^x+ 32 = 0
Let y = 2^x
y² – 12y + 32 = 0
⇒ (y – 4) (y – 8) = 0
y – 4 = 0 or y – 8 = 0
Case (i): 2^x = 4
⇒ 2^x = (2)²
⇒ x = 2
Case (ii): 2x = 8
⇒ 2^x = (2)³
⇒ x = 3
∴ The roots are 2, 3

Question 7.
Solve the equation 6x⁴ – 5x³ – 38x² – 5x + 6 = 0 if it is known that \(\frac{1}{3}\) is a solution.
Solution:
6x⁴ – 5x³ – 38x² – 5x + 6 = 0

6(y² – 2) – 5y – 38 = 0
6y² – 12 – 5y – 38 = 0
6y² – 5y – 50 = 0
6y² – 20y + 15y  – 50 = 0
2y(3y – 10) + 5(3y – 10) = 0
(3y – 10)(2y + 5) = 0
3y = 10, 2y = -5

3x² + 3 = 10x, 2x² + 2 = -5x
3x² – 10x + 3 = 0, 2x² + 5x + 2 = 0
(x – 3)(3x – 1) = 0, (2x + 1)(x + 1) = 0
x = 3, x = \(\frac{1}{3}\) or x = \(\frac{-1}{2}\), x= -2
The roots are 3, \(\frac{1}{3}\), -2, \(\frac{-1}{2}\).

Students can download 10th Social Science Geography Chapter 4 Resources and Industries Questions and Answers, Notes, Samacheer Kalvi 10th Social Science 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 Social Science Solutions Geography Chapter 4 Resources and Industries

Samacheer Kalvi 10th Social Science Resources and Industries Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
Manganese is used in:
(a) Storage batteries
(b) Steel Making
(c) Copper smelting
(d) Petroleum Refining
Answer:
(b) Steel Making

Question 2.
The Anthracite coal has
(a) 80 to 95% Carbon
(b) Above 70% Carbon
(c) 60 to 7% Carbon
(d) Below 50% Carbon
Answer:
(a) 80 to 95% Carbon

Question 3.
The most important constituents of petroleum are hydrogen and:
(a) Oxygen
(b) Water
(c) Carbon
(d) Nitrogen
Answer:
(c) Carbon

Question 4.
The city which is called the Manchester of South India is
(a) Chennai
(b) Salem
(c) Madurai
(d) Coimbatore
Answer:
(d) Coimbatore

Question 5.
The first Jute mill of India was established at:
(a) Kolkata
(b) Mumbai
(c) Ahmedabad
(d) Baroda
Answer:
(a) Kolkata

Question 6.
The first Nuclear Power station was commissioned in
(a) Gujarat
(b) Rajasthan
(c) Maharashtra
(d) Tamil Nadu
Answer:
(c) Maharashtra

Question 7.
The most abundant source of energy is:
(a) Biomass
(b) Sun
(c) Coal
(d) Oil
Answer:
(b) Sun

Question 8.
The famous Sindri Fertilizer Plant is located in
(a) Jharkhand
(b) Bihar
(c) Rajasthan
(d) Assam
Answer:
(a) Jharkhand

Question 9.
The nucleus for the development of the chotanagpur plateau region is:
(a) Transport
(b) Mineral Deposits
(c) Large demand
(d) Power Availability
Answer:
(b) Mineral Deposits

Question 10.
One of the shore-based steel plants of India is located at
(a) Kolkata
(b) Tuticorin
(c) Goa
(d) Visakhapatnam
Answer:
(d) Visakhapatnam

II. Match the following

Question 1.
Match the Column I with Column II.

Answer:
A. (ii)
B. (i)
C. (iv)
D. (v)
E. (iii)

Question 2.
Match the Column I with Column II.

Answer:
A. (v)
B. (iv)
C. (i)
D. (ii)
E. (iii)

III. Answer the following questions briefly

Question 1.
Define the resource and state its types.
Answer:
Any matter or energy derived from the environment that is used by living things including humans is called a natural resource.
Types of Natural Resources are:

1. Renewable and
2. Non – renewable resources.

Question 2.
Name the states that lead in the production of Iron ore in India.
Answer:
Jharkhand is the leading producer of Iron ore (25%), other states are Odisha (21%), Chattisgarh (18%), Karnataka (20%), Andhra Pradesh (5%) and also the state of Tamil Nadu.

Question 3.
What are the minerals and its types?
Answer:
A mineral is a natural substance of organic or inorganic origin with definite chemical and physical properties.
Types of Minerals are:

1. Metallic minerals
2. Non – metallic minerals.

Question 4.
State the uses of Manganese.
Answer:

1. Manganese is a silvery grey element always available with iron, laterite and other minerals.
2. It is very hard and brittle. So it is used for making iron and steel to give strength.
3. It also serves as raw materials for alloying.
4. It is also used in the manufacture of bleaching powder, insecticides, paints and batteries.

Question 5.
What is natural gas?
Answer:
It usually accompanies petroleum accumulations.
[OR]
Natural gas is an important clean energy resources found in association with or without petroleum. It is used as a source of energy as well as an industrial raw material in the petrochemical industry. It is considered an eco-friendly fuel because of low carbondioxide emissions.

Question 6.
Name the different types of coal with their carbon content.
Answer:
Based on carbon content coal is classified into

1. Anthracite – contains 80% to 90% carbon.
2. Bituminous – contains 60% to 80% carbon.
3. Lignite – contains 40% to 60% carbon
4. Peat – less than 40% carbon

Question 7.
Mention the major areas of jute production in India.
Answer:
West Bengal, Andhra Pradesh, Bihar, Uttar Pradesh, Assam, Chattisgarh, Odisha are the major jute producing area.

Question 8.
Name the important oil producing regions of India.
Answer:
West coast: Mumbai high, Gujarat coast, Basseim, Aliabet (South of Bhavanagar), Ankleshwar, Cambay – Luni region, Ahmedabad -Kaloi region and Punjab – Haryana.

East Coast: Brahmaputra valley, Digboi, Nahoratitya, Moran-Hugrijan, Rudrasagar-Lawa (Assam region), Surrma Valley. Andaman and Nicobar, Gulf of Mannar, Punjab. Haryana, Baleshwar coast.

IV. Distinguish between

Question 1.
Renewable and non-renewable resources
Answer:
Renewable resources:

1. These resources can be replenished after utilization.
2. Abundantly available in nature.
3. Eg: Sun energy, Wind etc.

Non-Renewable resources:

1. These resources cannot be regained after utilization.
2. Limited stock take millions of years for formation.
3. Eg: Minerals

Question 2.
Metallic and non-metallic minerals
Answer:
Metallic minerals:

1. These minerals contain one or more metallic elements.
2. They can be further classified into ferrous and Non-ferrous minerals.
3. Eg: Iron-ore, Manganese

Non-Metallic minerals:

1. These minerals do not contain metal in them.
2. They can be classified into energy fuels and construction minerals.
3. Eg: Coal, Petroleum, lime stone, Gypsum.

Question 3.
Agra based industry and mineral based industry.
Answer:
Agro based industry:

1. These industries obtain raw materials from agricultural products.
2. Eg: Cotton textile industry, Jute mills, Silk industry, Sugar industry etc.
3. Labour intensive

Mineral based industry:

1. These industries obtain raw materials from metallic and non- metallic minerals.
2. Eg: Coal industry, Iron and Steel industry etc.
3. Capital intensive

Question 4.
Jute industry and sugar industry.
Answer:
Jute industry:

1. Second largest textile industry.
2. Jute fibre is the basic raw material.
3. Gunny bags, Canvas, pack sheets,Cordage are some of the Jute products.
4. Concentrated in and around West Bengal.

Sugar industry:

1. Second largest agro based industry.
2. Sugarcane is the basic raw material.
3. Sugar Jaggery, Khandsari are some of the products of sugar industry.
4. Uttar Pradesh have more than 50% of the sugar industries.

Question 5.
Conventional energy and non- conventional energy
Answer:
Conventional energy:

1. Non renewable energy.
2. Mainly energy is produced by burning fossil fuels.
3. Production of energy till the stock of mineral availability.
4. Thermal power – Coal, petroleum and Natural gas and Nuclear minerals.

Non-Conventional energy:

1. Renewable energy
2. Mainly energy is produced by harnessing power from nature.
3. Continuous flow of energy production is possible.
4. Solar energy, Wind energy. Bio mass energy, tidal and wave energy.

V. Answer the following in a paragraph

Question 1.
Write about the distribution of cotton textile industries in India.
Answer:
The cotton textile industries contribute about 7% of industrial output, 2% of India’s GDP and 15% of the country’s export earnings. It is one of the largest sources of employment generation in the country. At present there are 1,719 textile mills in the country. Out of which 188 mills are in public sector, 147 in cooperative sector and 1,284 in private sector. Currently, India is the third largest producer of cotton and has the largest loom arc and ring spindles in the world. At present, cotton textile industry is the largest organized modem industry of India. About 16% of the industrial capital, 14% of industrial production and over 20% of the industrial labour of the country are engaged in this industry.

The higher concentration of textile mills in and around Mumbai, makes it as “Manchester of India”. Presence of black cotton soil in Maharashtra, humid climate, presence of Mumbai port, availability of hydro power, good market and well developed transport facility favour the cotton textile industries in Mumbai. The major cotton textile industries are concentrated in the states of Maharashtra, Gujarat, West Bengal, Uttar Pradesh and Tamil nadu. Coimbatore is the most important centre in Tamil nadu with 200 mills out of its 435 and called as “Manchester ‘ of South India”. Erode, Tirupur, Karur, Chennai, Thirunelveli, Madurai, Thoothukudi, Salem and Virudhunagar are the other major cotton textiles centres in the state.

Question 2.
Explain the factors responsible for the concentration of jute industries in the Hoogly region.
Answer:
The following factors are responsible for the concentration of Jute industries in the Hoogly region in West Bengal.

1. Raw materials: West Bengal is the largest producer of Jute. Availability of raw Jute for production.
2. Processing: Jute require fresh water for processing. Abundant water supply is available by the riverines and continuous supply of fresh water is ensured due to Perennial nature.
3. Transport: Well connected by the network of water ways, road ways and railways.
4. Cheap labour: West Bengal is one of the densely populated area . So cheap labour is available.
5. Market: Kolkatta being one of the textile centre great demand for the product as well as port facilities available for export.

Question 3.
Write an account on the major iron and steel industries of India.
Answer:
Iron and Steel industry is the basic industry. The raw materials are obtained both from Metallic and Non-Metallic minerals.

Iron and Steel industries are located in close proximity to the coal fields or Iron ore mines.

VI. On the outline map of India mark the following.

Question 1.
iron ore production centres.
Answer:

Question 2.
Centres of Petroleum and Natural Gas production.
Answer:

Question 3.
Coal mining centres.
Answer:

Question 4.
Areas of cultivation of cotton.
Answer:

Question 5.
Iron and Steel industries.
Answer:

Samacheer Kalvi 10th Social Science Resources and Industries Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
The ………………. is called as mineral oil.
(a) Petroleum
(b) Coal
(c) Natural gas
(d) Mica
Answer:
(a) Petroleum

Question 2.
Manchester of India is ……..
(a) Delhi
(b) Mumbai
(c) Chennai
Answer:
(b) Mumbai

Question 3.
The ………………. is the largest oil field in India producing 65% of oil.
(a) Ankaleshwar
(b) Mumbai high
(c) Kalol
(d) Surma valley
Answer:
(b) Mumbai high

Question 4.
Chotta Nagpur plateau is noted for ……..
(a) Natural vegetation
(b) Mineral resource
(c) Cotton cultivation
Answer:
(b) Mineral resource

Question 5.
TamilNadu produces about ………………. % of the total thermal electricity produced in India.
(a) 5
(b) 20
(c) 18
(d) 90
Answer:
(a) 5

Question 6.
In India most of the Iron and Steel industries are located in the …… plateau.
(a) Chott Nagpur
(b) Deccan
(c) Malwa
Answer:
(a) Chott Nagpur

Question 7.
Areas near ………………. district has the largest concentrations of wind farm capacity at a single location in the world.
(a) Ramanathapuram
(b) Tuticorin
(c) Thiruvallur
(d) Kanyakumari
Answer:
(d) Kanyakumari

Question 8.
The resources that can be reproduced again and again is called ……
(a) Mineral resources
(b) Renewable resource
(c) Natural resource
Answer:
(b) Renewable resource

Question 9.
Bysinosis is an occupational lung disease caused by exposure to:
(a) natural gas
(b) cotton dust
(c) coal power
(d) automobile
Answer:
(b) cotton dust

Question 10.
Lignite is extracted in Tamil Nadu …….
(a) Kadaloor
(b) Neyveli
(c) Madurai
Answer:
(b) Neyveli

Question 11.
National News Print and Papermills (NEPA) is in ………………. state.
(a) Odisha
(b) West Bengal
(c) TamilNadu
(d) Madhya Pradesh
Answer:
(d) Madhya Pradesh

Question 12.
Sugar bowl of India is …….
(a) West Bengal
(b) Uttar pradesh and Bihar
(c) Mumbai
Answer:
(b) Uttar pradesh and Bihar

Question 13.
The ………………. is the largest producer of electronic goods in India.
(a) Bengaluru
(b) Mysuru
(c) Delhi
(d) Jaipur
Answer:
(a) Bengaluru

Question 14.
The leading producer of electronic goods is ……..
(a) Bangalore
(b) Coimbatore
(c) Hyderabad
Answer:
(a) Bangalore

Question 15.
………………. is an aluminium ore.
(a) Manganese
(b) Magnesium
(c) Bauxite
(d) Anthracite
Answer:
(c) Bauxite

II. Match the following

Question 1.
Match the Column I with Column II.

Answer:
A. (iv)
B. (iii)
C. (i)
D. (v)
E. (ii)

Question 2.
Match the Column I with Column II.

Answer:
A. (iii)
B. (v)
C. (ii)
D. (i)
E. (iv)

Question 3.
Match the Column I with Column II.

Answer:
A. (iii)
B. (iv)
C. (v)
D. (ii)
E. (i)

III. Answer the following questions briefly

Question 1.
What are the features of copper and aluminium?
Answer:

• Malleability
• Expandability
• Good conductor of heat and energy (electricity)

Question 2.
Where was the first hydro-electric power station in India established?
Answer:
The first hydro-electric power station in India was established at “Darjeeling” in 1897.

Question 3.
In which type of rocks is limestone found?
Answer:
Limestone is found in sedimentary rocks.

Question 4.
What is the origin of the word petroleum?
Answer:
The word petroleum is derived from two Latin words petro (Rock) and Oleum (oil) thus petroleum is oil obtained from rocks of the earth. It is also called as mineral oil.

Question 5.
Name any five software centres.
Answer:

1. Chennai
2. Bangalore
3. Mysore
4. Hyderabad
5. Coimbatore

Question 6.
When was the Nuclear power programme initiated in India and where was the first Atomic Power station was set up?
Answer:

1. Nuclear Power programme was initiated in the 1940s, when “ Tata Atomic Research Commission was incorporated in August 1948.
2. The 1st nuclear power station was set up at Tarapur near Mumbai in 1969.

Question 7.
What are the raw materials for the paper industry?
Answer:
Woodpulp, bamboo, salai, Sabai grasses, waste paper and bagasse

Question 8.
Write a note on NPCIL.
Answer:

1. The Nuclear Power Corporation of India Limited ( NPCIL) is wholly owned by the Government of India.
2. It is a Public sector undertaking.
3. Responsible for the generation of nuclear power for electricity.
4. Administered by the Department of Atomic Energy (DAE). It is responsible for designing, constructing and operating the Nuclear power stations in India.

Question 9.
What are the major centres of the automobile industry?
Answer:
Mumbai, Chennai, Kolkata, New Delhi, Pune, Ahmedabad, Lucknow, Satara and Mysore.

Question 10.
Where are the centres of IT parks located in India?
Answer:
Centres of IT parks in India are located in Chennai, Coimbatore, Thiruvananthapuram, Bengaluru, Mysuru, Hyderabad, Vishakapatnam, Mumbai, Pune, Indore, Gandhinagar, Jaipur, Noida, Mohali and Srinagar.

IV. Distinguish between

Question 1.
Paper industry and Iron and Steel industry.
Answer:
Paper industry

1. Forest-based industry.
2. Serve as an index for education and literacy.
3. Wood pulp, Bamboo, bagasse are some of the basic raw materials.

Iron and Steel industry:

1. Mineral-based industry.
2. Key industry for industrial development.
3. Iron ore and Manganese are the main raw materials.

Question 2.
Solar and Wind energy.
Answer:
Solar energy:

1. Conversion of sunlight into electricity.
2. Photo voltoic cells or use of lenses, mirror and tracking system is used.
3. Installation cost is more and different applications is required as per the need.
4. Occupy more space.

Wind energy:

1. Conversion of energy is from the flow of wind.
2. Wind turbines are used.
3. Installation cost is comparatively less.
4. Occupy less space.

Question 3.
Difference between Manganese and Mica.
Answer:
Manganese:

1. It is a Metallic mineral.
2. Very hard and brittle in nature.
3. Raw material for iron and steel basic raw material for alloying.

Mica:

1. It is a Non-Metallic mineral.
2. Translucent, splitable into thin sheets, elastic and incompressible.
3. Exclusively used in electrical goods and also for making lubricants, paints, varnishes etc.

Question 4.
Iron and Steel industry and Software industry.
Answer:
Iron and Steel industry:

1. Age old industry.
2. Depend upon minerals.
3. Large scale industry.
4. Began in 1907.

Software industry:

1. Recently developed industry.
2. Depend upon skill and knowledge (technical).
3. Small and medium scale industry.
4. Began in 1970.

V. Answer the following in a paragraph

Question 1.
Enumerate the automobile industry as the fast-growing industry of India.
Answer:
The first automobile industry was established at the Premier Automobiles Ltd at Kurla (Mumbai) in 1947 and the Hindustan Motor Ltd at Uttarpara (Kolkata) in 1948. At present India is the 7th largest producer of automobile manufactures which includes two wheelers, commercial vehicles, passenger cars, Jeep, scooty, scooters, motor cycles mopeds and three wheelers.

Among the production of two wheelers, motorcycles are manufactured at Faridabad, Haryana and Mysore. While scooters are manufactured at Lucknow, Satara, Akudi (Pune), Panki (Kanpur) and Odhai (Ahmedabad). The cars produced at Haryana, Kolkata, Mumbai and Chennai are Maruti, Ambassador, Fiat, Ford and Hyndai etc. Presenceof foreign car companies such as Mercedes Benz, Fiat, General Motors, Toyota and the recent entry of passenger cars manufacturers BMW, Audi, Volks Wagen and Volvo makes the Indian automobile sector a special one.

Several new joint venture agreements for the manufacture of cars have recently been signed by the Indian companies and renowned car manufacturers of the world. The Indian auto industry is said to take a big leep in the near fixture. This expected to provide much more competitive environment to the industry and a wide choice of ultra modem cars to the consumers. This is the fast growing industry in India.

Question 2.
What are the different forms of Iron ore and their nature?
Answer:
Iron ores are rocks and minerals from which metallic iron can be economically extracted.

The different forms of Iron ore are:

Nature: The ores are usually rich in iron oxides and vary in colour from dark grey, bright yellow or deep purple to rusty red.

Question 3.
Write about the uses of Coal and its areas of distribution.
Answer:

1. Coal is a non-renewable resource. It is available in the form of sedimentary rocks.
2. It has close association with the industrial development of any country.

Uses: Coal is an important source of energy in India with its varied and innumerable uses

1. It is converted into gas, oil and thermal power electricity.
2. Besides it forms a basic raw materials for the production of chemicals, dyes, fertilizers, paints, synthetic and explosives.
3. Distribution: Indian coal is mostly associated with Gondwana region and is primarily found in peninsular India.
4. The States of Jharkhand, Odisha, West Bengal and Madhya Pradesh alone account for nearly 90% of coal reserves of the country. About 2% of India’s coal is of tertiary type and is mostly in Assam and Jammu and Kashmir.
5. Jharkhand is the largest coal producing state in the country. Indian lignite (brown coal) deposits occur particularly in Tamil Nadu ( Neyveli), Puducherry and Kerala.

Question 4.
Give an account on Automobile industry in India.
Answer:
Automobile industry’ is one of the most dynamic industrial groups in India. India is the 7th largest producer of automobile manufacturers.

Distribution: The automobile industries are found in four clusters viz: Delhi, Gurgaon and Manesar in North India.

West India: Pune, Nasik, Halol and Aurangabad East India: Jamshedpur and Kolkatta South India: Chennai, Bengaluru and Hosur.

Major-Indian companies which manufacture commercial vehicles: Tata motors, Ashok Leyland, Mahindra and Mahindra, Eicher motors and Ford motors.

Foreign companies which manufacture commercial vehicles: Hyundai, Mercedes Benz , ITEC, MAN.

Automobile industries in India manufactures two wheelers passenger car, Jeep, Scooty, Scooter, Mopeds, Motorcycles and three wheelers.

Passenger car manufacturers: Tata motors, Maruti Suzuki , Mahindra and Mahindra and Hindustan motors are Indian companies.

Foreign car companies in India: Mercedez Benz, Fiat, General motors, Toyota, recent entry BMW, Audi, Volkswagon and Volvo.

Two wheeler manufacturing is dominated by Indian companies like Hero, Bajaj Auto and TVS.

Major centres of Automobile industries: Mumbai, Chennai, Jamshedpur, Jabalpur, Kolkata, Pune, New Delhi, Kanpur, Bengaluru, Satara, Lucknow • and Mysuru. Chennai is named as “Detroit of Asia” due to the presence of major automobile manufacturing units and allied industries around the city.

Question 5.
What are the major challenges faced by the Indian industries?
Answer:
Some major challenges faced by industries in India are:

1. Non-availability of large blocks of land.
2. Shortage and fluctuation in power supply.
3. Non-availability of cheap labourers.
4. Poor access to credit.
5. High rate of interest for borrowed loan.
6. Lack of technical and vocational training for employees.
7. In appropriate living conditions nearby industrial areas.

VI. On the outline map of India mark the following

Question 1.
Any two places producing limestone
Answer:

Question 2.
Any two Nuclear power stations
Answer:

Question 3.
Any two Thermal power stations in Tamil Nadu
Answer:

Question 4.
Wind farm places (any 2)
Answer:

Students can Download Tamil Nadu 12th Physics Model Question Paper 1 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 1 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• 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
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is …………….
(a) 80 x 10^-17 J
(b) -8.80 x 10^-17 J
(c) 4.40 x 10^-17 J
(d) 5.80 x 10^-17 J
Answer:
(a) 80 x 10^-17 J

Question 2.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

(a) 2 ohm
(b) 4 ohm
(c) 8 ohm
(d) 1 ohm
Answer:
(a) 2 ohm

Question 3.
Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P which is at exactly at  R/2 distance between two coils is …………..

Answer:

Question 4.
An electron moves on a straight line path XY as shown in the figure. The coil abed is adjacent to the path of the electron. What will be the direction of current, if any, induced in the coil?

(a) The current will reverse its direction as the electron goes past the coil
(b) No current will be induced
(c) abcd
(d) adcb
Answer:
(a) The current will reverse its direction as the electron goes past the coil

Question 5.
The resistance of an ideal ammeter is ……………
(a) zero
(b) small
(c) high
(d) infinite
Answer:
(a) zero

Question 6.
In a step-down transformer the input voltage is 22 kV and the output voltage is 550 V. The ratio of the number of turns in the secondary to that in the primary is ……………
(a) 1 : 20
(b) 20 : 1
(c) 1 : 40
(d) 40 : 1
Answer:
(c) 1 : 40

Question 7.
If the magnetic monopole exists, then which of the Maxwell’s equation to be modified?

Answer
\(\oint \overrightarrow { { E } } \cdot d\overrightarrow { { A } } =0\)

Question 8.
An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is,
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm
Answer:
(c) 12 cm
Hint. Let d₁ = 5 cm and d₂ = 3 cm ; n = 1.5
Actual width is the sum of real depth from 2 sides
Thickness of slab = d₁n + d₂n
= (5 x 1.5) +(3 x 1.5)= 12 cm

Question 9.
When a ray of light enters a glass slab from air
(a) its wavelength decreases
(b) its wavelength increases
(c) its frequency increases
(d) neither its wavelength nor its frequency changes
Answer:
(a) its wavelength decreases
Hint: \(Wavelength,\lambda =\frac { { Velocity } }{ { Frequency } } =\frac { u }{ v } \)
When light travels from air to glass, frequency ν remains unchanged, velocity u decreases and hence wave length λ also decreses

Question 10.
A particle of mass 3 x 10^-6 g has the same wavelength as an electron moving with a velocity
6 x 10⁶ m s^-1. The velocity of the particle is…………………………………………
(a) 1.82 x 10^-18 ms^-1
(b) 9 x 10^-2 ms^-1
(c) 3 x 10^-31 ms^-1
(d) 1.82 x 10¹⁵ ms^-1
Answer:
(d) 1.82 x 10¹⁵ ms^-1
Hint: de – Broglie wavelength of electron

Question 11.
If the nuclear radius of ²⁷A1 is 3.6 fermi, the approximate unclear radius of ⁶⁴Cu is
(a) 2.4
(b) 1.2
(c) 4.8
(d) 3.6
Answer:
(c) 4.8
Hint :

Question 12.
Energy of characteristic X-ray is a consequence of ………….
(a) energy of projectile electron
(b) thermal energy of target
(c) transition in target atoms
(d) none of the above
Answer:
(c) transition in target atoms

Question 13.
The specific characteristic of a common emitter amplifier is ……………
(a) High input resistance
(b) Low power gain
(c) Signal phase reversal
(d) Low current gain
Answer:
(c) Signal phase reversal

Question 14.
The variation of frequency of carrier wave with respect to the amplitude of the modulating signal is called ………………
(a) Amplitude modulation
(b) Frequency modulation
(c) Phase modulation
(d) Pulse width modulation
Answer:
(b) Frequency modulation

Question 15.
The particle size of ZnO material is 30 nm. Based on the dimension it is classified as ……………..
(a) Bulk material
(b) Nanomaterial
(c) Soft material
(d) Magnetic material
Answer:
(b) Nanomaterial

Part – II

Answer any six questions in which Q. No 17 is compulsory. [6 x 2 = 12]

Question 16.
Write a short note on the superposition principle.
Answer:
According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
\(\overrightarrow{\mathrm{F}}_{1}^{\mathrm{tot}}=\overrightarrow{\mathrm{F}}_{12}+\overrightarrow{\mathrm{F}}_{13}+\overrightarrow{\mathrm{F}}_{14}+\cdots+\overrightarrow{\mathrm{F}}_{1 n}\)

Question 17.
If an electric field of magnitude 570 N C^-1, is applied in the copper wire, find the acceleration experienced by the electron.
Answer:

Question 18.
State Ampere’s circuital law.
Answer:
The line integral of magnetic field over a closed loop is p0 times net current enclosed by the loop.
\(\oint_{C} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0} \mathrm{I}_{\text {enclosed }}\)

Question 19.
Write down the equation for a sinusoidal voltage of §0 Hz and its peak value is 20 V. Draw the corresponding voltage versus time graph.
Answer:

Question 20.
What is displacement current?
Answer:
The displacement current can be defined as the current which comes into play in the region in which the electric field and the electric flux are changing with time.

Question 21.
Why do clouds appear white?
Answer:
Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 22.
How many photons per second emanate from a 50 mW laser of 640 nm?
Answer:

Question 23.
An ideal diode and a 5 Ω resistor are connected in series with a 15 V power supply as shown in figure below. Calculate the current that flows through the diode.
Answer:
The diode is forward biased and it is an ideal one. Hence, it acts like a closed switch with no barrier voltage.Therefore, current that flows through the diode can be calculated using Ohm’s law.

V = IR
\(I=\frac{V}{R}=\frac{15}{5}=3 \mathrm{A}\)

Question 24.
What do you mean by Internet of Things?
Answer:
Internet of Things (IoT), it is made possible to control various devices from a single device. Example: home automation using a mobile phone.

Part – III

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

Question 25.
Define electrostatic potential energy?
Answer:
The potential energy of a system of point charges may be defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 26.
If the resistance of coil is 3 Ω at 20°C and a = 0.004/°C then determine its resistance at 100°C.
Answer:
R₀ =3Ω = 100°C, T₀ = 20°C
α = 0.004/°C, R_T = ?
Rr= R₀(1+∝ (T-T₀))
R₁₀₀ = 3(1 + 0.004 x 80)
⇒ R₁₀₀ = 3(1 + 0.32)
R₁₀₀= 3(1.32)
⇒ R₁₀₀ = 3.96 Ω

Question 27.
State ‘Tangent Law’.
Answer:
When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.

Question 28.
What are step-up and step-down transformers?
Answer:
If the transformer converts an alternating current with low voltage into an alternating current with high voltage, it is called step-up transformer. On the contrary, if the transformer converts alternating current with high voltage into an alternating current with low voltage, then it is called step-down transformer.

Question 29.
One type of transparent glass has refractive index 1.5. What is the speed of light through this glass?
Answer:

Question 30.
What is a photo cell? Mention the different types of photocells.
Answer:
Photocells: Photo electric cell or photo cell is a device which converts light energy into electrical energy. It works on the principle of photo electric effect.
Types:

• Photo emissive cell
• Photo voltaic cell
• Photo conductive cell

Question 31.
Half lives of two radioactive elements A and B are 20 minutes and 40 minutes respectively. Initially, the samples have equal number of nuclei. Calculate the ratio of decayed numbers of A and B nuclei after 80 minutes.
Answer:
80 minutes = 4 half lives of A = 2 half live of B
Let the initial number of nuclei in each sample be N.

Question 32.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:

Question 33.
Distinguish between Nanoscience and Nanotechnology.
Answer:

Part – IV

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

Question 34.
(a) Obtain the expression for electric field due to an uniformly charged spherical shell. Electric field due to a uniformly charged
Answer:
spherical shell: Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.

Case (a) At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially
outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,

The electric field \(\overrightarrow { { E } } { and }\quad d\overrightarrow { { A } } \) point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of \(\overrightarrow { { E } } \) is also the same at all points due to the spherical symmetry of the charge distribution.

But = total area of Gaussian surface  = 4πr² Substituting this value in equation (2)

The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b): At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
\(\overrightarrow{\mathrm{E}}=\frac{Q}{4 \pi \varepsilon_{0} R^{2}} \hat{r}\) ………. (4)

Case (c) At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law.

Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes
E = 0    (r < R)  ………….. (6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

[OR]

(b) Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:

An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The bridge consists of four resistances P, Q, R and S connected. A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.
Applying Kirchhoff’s current rule to junction B.
I₁ – I_G – I₃ = 0   ………….. (1)
Applying Kirchhoff’s current rule to junction D.
I₂ + I_G – I₄ = 0   ………….. (2)
Applying Kirchhoff’s voltage rule to loop ABDA,
I₁P + I_GG – I₂R = 0  ………….. (3)
Applying Kirchhoff’s voltage rule to loop ABCDA,
I₁P + I₃Q – I₄S – I₂R = 0  ………….. (4)
When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B through galvanometer (I_Q = 0). Substituting I_G = 0 in equation, (1), (2) and (3), we get
I₁ = I₃  ………………….. (5)
I₁ = I₄ ……………………. (6)
I₁P =I₂R ……………….(7)
Substituting the equation (5) and (6) in equation (4)
I₁P + I₁Q – I₁S – I₁R = 0
I₁( P + Q) = I₂( P + S) …………………(8)
Dividing equation (7)

This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance  (fourth one) can be determined.

Question 35.
(a) Obtain the magnetic induction at a point on the equatorial line of a bar magnet. Magnetic field at a point along the equatorial line due to a magnetic dipole (bar magnet)
Answer:
Consider a bar magnet NS. Let N be the north pole and S be the south pole of the bar magnet, each with pole strength q_m and separated by a distance of 21. The magnetic field at a point C (lies along the equatorial line) at a distance r from the geometrical center O of the bar magnet can be computed by keeping unit north pole (q_mC = 1 A m) at C. The force experienced by the unit north pole at C due to pole strength N-S can be computed using Coulomb’s law of magnetism as follows:

The force of repulsion between North Pole of the bar magnet and unit north pole at point C (in free space) is
\(\overrightarrow{\mathrm{F}}_{\mathrm{N}}=-\mathrm{F}_{\mathrm{N}} \cos \theta \hat{i}+\mathrm{F}_{\mathrm{N}} \sin \theta \hat{j}\) ……………. (1)
Where \(\overrightarrow{\mathrm{F}}_{\mathrm{S}}=\frac{\mu_{0}}{4 \pi} \frac{q_{m}}{r^{\prime 2}}\) . The force of attraction (in free space) between south pole of the bar magnet and unit north pole at point C is

From equation (1) and equation (2), the net force at point C is \(\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{N}}+\mathrm{F}_{\mathrm{S}}\). This net force is equal to the magnetic field at the point C.

If the distance between two poles in a bar magnet are small (looks like short magnet) when compared to the distance between geometrical center O of bar magnet and the location of point C i.e., r>> l, then,

Therefore , using equation (7) in equation (6) ,we get
\(\overrightarrow{\mathrm{B}}_{\text {equatorial }}=-\frac{\mu_{0}}{4 \pi} \frac{p_{m}}{r^{3}} \hat{i}\)
Since in genaral , the magnetic filed at equatorial points is given by

Note that magnitude of B_axial is twice that of magnitude of B_equatoral and the direction of B_axial and B_equatoral are opposite.

[OR]

Question 35.
(b) Give the advantage of AC in long distance power transmission with an example.
Answer:
Advantages of AC in long distance power transmission: Electric power is produced in a large scale at electric power stations with the help of AC generators. These power stations are classified based on the type of fuel used as thermal, hydro electric and nuclear power stations. Most of these stations are located at remote places. Hence the electric power generated is transmitted over long distances through transmission lines to reach towns or cities where it is actually consumed. This process is called power transmission.

But there is a difficulty during power transmission. A sizable fraction of electric power is lost due to Joule heating (i²R) in the transmission lines which are hundreds of kilometer long. This power loss can be tackled either by reducing current I or by reducing resistance R of the transmission lines. The resistance R can be reduced with thick wires of copper or aluminium. But this increases die cost of production of transmission lines and other related expenses. So this way of reducing power loss is not economically viable.

Since power produced is alternating in nature, there is a way out. The most important property of alternating voltage is that it can be stepped up and stepped down by using transformers could be exploited in reducing current and thereby reducing power losses to a greater extent. At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer.

Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss. At the receiving point, the voltage is decreased and the current is increased to appropriate values by using step-down transformer and then it is given to consumers. Thus power transmission is done efficiently and economically.

Illustration: An electric power of 2 MW is transmitted to a place through transmission lines of total resistance, say R = 40 Ω, at two different voltages. One is lower voltage (10 kV) and the other is higher (100 kV). Let us now calculate and compare power losses in these two cases.

Case (i) : P = 2 MW; R = 40 Ω; V = 10 kV
Power, P= VI
Power loss = Heat produced = I²R = (200)² x 40 = 1.6 x 10⁶ W
% of power loss = \(\frac{1.6 \times 10^{6}}{2 \times 10^{6}} \times 100 \%\) = 0.8 x 100% = 80%

Case (ii): P = 2 MW; R = 40 Ω; V = 100 kV
∴ Current, \(I=\frac{P}{V}=\frac{2 \times 10^{6}}{100 \times 10^{3}}=20 \mathrm{A}\)
Power loss = Heat produced = I²R = (20)² x 40 = 0.016 x 10⁶ W
% of power loss = \(\frac{0.016 \times 10^{6}}{2 \times 10^{6}} \times 100 \%=\) = 0.008 x 100% = 0.8%

Question 36.
(a) What is emission spectra? Give their types.
Answer:
Emission spectra: When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum can be divided into three types:

(i) Continuous emission spectra (or continuous spectra): If the light from incandescent lamp (filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours. Thus, it consists of wavelengths containing all the visible colours ranging from violet to red. Examples: spectrum obtained from carbon arc, incandescent solids, liquids gives continuous spectra.

(ii) Line emission spectrum (or line spectrum): Suppose light from hot gas is allowed to pass through prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra are sharp lines of definite wavelengths or frequencies. Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element which means it is different for different elements. Examples: spectra of atomic hydrogen, helium, etc

(iii) Band emission spectrum (or band spectrum): Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces, known as band spectra. This spectrum has a sharp edge at one end and fades out at the other end. Such spectra arise when the molecules are excited. Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube etc.

[OR]

(b) Obtain the equation for radius of illumination (or) Snell’s window.
Answer:
The radius of Snell’s window can be deduced with the illustration as shown in figure. Light is seen from a point A at a depth d. The Snell’s law in product form, equation n₂ sin i = n₂ sin r for the refraction happening at the point B on the boundary between the two media is,
n₁ sin i_c = n₂ sin 90° ………………. (1)
n₁ sin i_c = n₂
∵ sin 90° = 1
sin i_c = \(\frac{n_{2}}{n_{1}}\) ………………….(2)
From the right angle triangle ΔABC,

Question 37.
(a) Explain why photoelectric effect cannot be explained on the basis of wave nature of light.
Failures of classical wave theory:
Answer:
From Maxwell’s theory, light is an electromagnetic wave consisting of coupled electric and magnetic oscillations that move with the speed of light and exhibit typical wave behaviour. Let us try to explain the experimental observations of photoelectric effect using wave picture of light.

When light is incident on the target, there is a continuous supply of energy to the electrons. According to wave theory, light of greater intensity should impart greater kinetic energy to the liberated electrons (Here, Intensity of light is the energy delivered per unit area per unit time). But this does not happen. The experiments show that maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light.

According to wave theory, if a sufficiently intense beam of light is incident on the surface, electrons will be liberated from the surface of the target, however low the frequency of the radiation is. From the experiments, we know that photoelectric emission is not possible below a certain minimum frequency. Therefore, the wave theory fails to explain the existence of threshold frequency.

Since the energy of light is spread across the wavefront, the electrons which receive energy from it are large in number. Each electron needs considerable amount of time (a few hours) to get energy sufficient to overcome the work function and to get liberated from the surface. But experiments show that photoelectric emission is almost instantaneous process (the time lag is less than 10^-9 s after the surface is illuminated) which could not be explained by wave theory.

[or]

(b) Obtain the law of radioactivity.
Answer:
Law of radioactive decay:
At any instant t, the number of decays per unit time, called rate of decay \(\left(\frac{d \mathrm{N}}{d t}\right) \) to the number of nuclei at the same instant.
\(\frac{d \mathrm{N}}{d t} \propto \mathrm{N}\)

By introducing a proportionality constant, the relation can be written as
\(\frac{d \mathrm{N}}{d t}=-\lambda \mathrm{N}\) ……………………. (1)

Here proportionality constant X is called decay constant which is different for different radioactive sample and the negative sign in the equation implies that the N is decreasing with time. By rewriting the equation (1), we get
dN = -XNdt …………… (2)

Here dN represents the number of nuclei decaying in the time interval dt. Let us assume that at time t = 0 s, the number of nuclei present in the radioactive sample is N_o. By integrating the equation (2), we can calculate the number of undecayed nuclei N at any time t.
From equation (2), we get

N = N_o e^-λt …………… (4)
[Note: e^lnx = e^y ⇒ x = e^y ]
Equation (4) is called the law of radioactive decay. Here N denotes the number of undecayed nuelei present at any time t and N₀ denotes the number of nuclei at initial time t = 0. Note that the number of atoms is decreasing exponentially over the time. This implies that the time taken for all the radioactive nuclei to decay will be infinite. Equation (4) is plotted.

We can also define another useful quantity called activity (R) or decay rate which is the number of nuclei decayed per second and it is denoted as R = \(R=\left| \frac { dN }{ dt } \right| \)

Note : That activity R is a positive quantity. From equation (4), we get.

The equation (6) is also equivalent to radioactive law of decay. Here R₀ is the activity of the sample at t = 0 and R is the activity of the sample at any time t. From equation (6), activity also shows exponential decay behavior. The activity R also can be expressed in terms of number of undecayed atoms present at any time t. From equation (6), since N = N₀ e ^-λt we write
R = λN    ………………… (7)
Equation (4) implies that the activity at any time t is equal to the product of decay constant and number of undecayed nuclei at the same time t. Since N decreases over time, R also decreases.

Question 38.
(a) Explain the construction and working of a full wave rectifier.
Answer:
Full wave rectifier:
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier. It consists of two p-n junction diodes, a center tapped transformer, and a load resistor (R_L). The centre is usually taken as the ground or zero voltage reference point. Due to the centre tap transformer, the output voltage rectified by each diode is only one half of the total secondary voltage.

During positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal M is positive, G is at zero potential and N is at negative potential. This forward biases diode D₁ and reverse biases diode D₂. Hence, being forward biased, diode D₁ conducts and current flows along the path MD₁ AGC As a result, positive half cycle of the voltage appears across R_L in the direction G to C.

During negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal N is positive, G is at zero potential and M is at negative potential. This forward biases diode D, and reverse biases diode D_r Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂ BGC. As a result, negative half cycle of the voltage appears across R_L in the same direction from G to C

Hence in a full wave rectifier both positive and negative half cycles of the input signal pass through the circuit in the same direction as shown in figure Though both positive and negative half cycles of ac input are rectified, the output is still pulsating in nature. The efficiency (r|) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %. It is because both the positive and negative half cycles of the ac input source are rectified.

[OR]

(b) Explain the three modes of propagation of electromagnetic waves through space. Propagation of electromagnetic waves:
Answer:
The electromagnetic wave transmitted by the transmitter travels in three different modes to reach the receiver-according to its frequency range:
(i) Ground wave propagation (or) surface wave propagation (nearly 2 kHz to 2 MHz)
(ii) Sky wave propagation (or) ionospheric propagation (nearly 3 MHz to 30 MHz)
(iii) Space wave propagation (nearly 30 MHz to 400 GHz)

(i) Ground wave propagation
If the electromagnetic waves transmitted by the transmitter glide over the surface of the earth to reach the receiver, then the propagation is called ground wave propagation. The corresponding waves are called ground waves or surface waves.

Increasing distance: The attenuation of the signal depends on

• power of the transmitter
• frequency of the transmitter, and
• condition of the earth surface.

Absorption of energy by the Earth: When the transmitted signal in the form of EM wave is in contact with the Earth, it induces charges in the Earth and constitutes a current. Due to this, the earth behaves like a leaky capacitor which leads to the attenuation of the wave.

Tilting of the wave: As the wave progresses, the wavefront starts gradually tilting according to the curvature of the Earth. This increase in the tilt decreases the electric field strength of the wave. Finally, at some distance, the surface wave dies out due to energy loss.

The frequency of the ground waves is mostly less than 2 MHz as high frequency waves undergo more absorption of energy at the earth’s atmosphere. The medium wave signals received during the day time use surface wave propagation.

It is mainly used in local broadcasting, radio navigation, for ship-to-ship, ship-to-shore communication and mobile communication.

(ii) Sky Wave Propagation:
The mode of propagation in which the electromagnetic waves radiated from an antenna, directed upwards at large angles gets reflected by the ionosphere back to earth is called sky wave propagation or ionospheric propagation. The corresponding waves are called sky waves.

The frequency range of EM waves in this mode of propagation is 3 to 30 MHz. EM waves of frequency more than 30 MHz can easily penetrate through the ionosphere and does not undergo reflection. It is used for short wave broadcast services. Medium and high frequencies are for long-distance radio communication. Extremely long distance communication, is also possible as the radio waves can undergo multiple reflections between the earth and the ionosphere. A single reflection helps the radio waves to travel a distance of approximately 4000 km.

Ionosphere acts as a reflecting surface. It is at a distance of approximately 50 km and spreads up to 400 km above the Earth surface. Due to the absorption of ultraviolet rays, cosmic ray, and other high energy radiations like a, (3 rays from sun, the air molecules in the ionosphere get ionized. This produces charged ions and these ions provide a reflecting medium for the reflection of radio waves or communication waves back to earth within the permitted frequency range. The phenomenon of bending the radio waves back to earth is nothing but the total internal reflection.

(iii) Space wave propagation:
The process of seeding and receiving information signal through space is called space wave communication. The electromagnetic waves of very high frequencies above 30 MHz are called as space waves. These waves travel in a straight line from the transmitter to the receiver. Hence, it is used for a line of sight communication (LOS).

For high frequencies, the transmission towers must be high enough so that the transmitted and received signals (direct waves) will not encounter the curvature of the earth and hence travel with less attenuation and loss of signal strength. Certain waves reach the receiver after getting reflected from the ground.

Students can download 10th Science Chapter 7 Atoms and Molecules Questions and Answers, Notes, Samacheer Kalvi 10th Science 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 Science Solutions Chapter 7 Atoms and Molecules

Samacheer Kalvi 10th Science Atoms and Molecules Text Book Back Questions and Answers

I. Choose the best answer:

Question 1.
Which of the following has the smallest mass?
(a) 6.023 × 10²³ atoms of He
(b) 1 atom of He
(c) 2 g of He
(d) 1-mole atoms of He
Answer:
(b) 1 atom of He

Question 2.
Which of the following is a triatomic molecule?
(a) Glucose
(b) Helium
(c) Carbon dioxide
(d) Hydrogen.
Answer:
(c) Carbon dioxide
Hint:
(a) Glucose = C₆H₁₂O₆ (Polyatomic molecule)
(b) Helium = He (Monoatomic molecule)
(c) Carbon dioxide = CO₂ (Triatomic molecule)
(d) Hydrogen = H₂ (Diatomic molecule)
So, (c) is the correct answer.

Question 3.
The volume occupied by 4.4 g of CO₂ at S.T.P:
(a) 22.4 litre
(b) 2.24 litre
(c) 0.24 litre
(d) 0.1 litre
Answer:
(b) 2.24 litre

Question 4.
Mass of 1 mole of Nitrogen atom is _____.
(a) 28 amu
(b) 14 amu
(c) 28 g
(d) 14 g.
Answer:
(b) 14 amu
Hint: Atomic mass of Nitrogen is 14.00674 grams. It is equal to 1 mole of Nitrogen atoms.
So, answer (b) is correct.

Question 5.
Which of the following represents 1 amu?
(a) Mass of a C – 12 atom
(b) Mass of a hydrogen atom
(c) 1/12 th of the mass of a C – 12 atom
(d) Mass of O – 16 atom
Answer:
(c) 1/12 th of the mass of a C – 12 atom

Question 6.
Which of the following statement is incorrect?
(a) One gram of C – 12 contains Avogadro’s number of atoms.
(b) One mole of oxygen gas contains Avogadro’s number of molecules.
(c) One mole of hydrogen gas contains Avogadro’s number of atoms.
(d) One mole of electrons stands for 6.023 × 10²³ electrons.
Answer:
(a) One gram of C – 12 contains Avogadro’s number of atoms.
Hint: 12 g of Carbon contains 6.023 × 10²³ atoms,
1 g of Carbon contain \(\frac{6.023 \times 10^{23}}{12}\) = 5.018 × 10²² atoms and its is not Avogadro’s number of atoms.
So (a) is the incorrect statement.

Question 7.
The volume occupied by 1 mole of a diatomic gas at S.T.P is:
(a) 11.2 litre
(b) 5.6 litre
(c) 22.4 litre
(d) 44.8 litre
Answer:
(c) 22.4 litre

Question 8.
In the nucleus of ₂₀Ca⁴⁰, there are
(a) 20 protons and 40 neutrons
(b) 20 protons and 20 neutrons
(c) 20 protons and 40 electrons
(d) 40 protons and 20 electrons
Answer:
(b) 20 protons and 20 neutrons

Question 9.
The gram molecular mass of oxygen molecule is_____.
(a) 16 g
(b) 18 g
(c) 32 g
(d) 17 g.
Answer:
(c) 32 g
Hint: By definition, the gram molecular mass of oxygen molecule O₂ is 32 g.
So the answer (c) is correct.

Question 10.
1 mole of any substance contains molecules.
(a) 6.023 × 10²³
(b) 6.023 × 10^-23
(c) 3.0115 × 10²³
(d) 12.046 × 10²³
Answer:
(a) 6.023 × 10²³

II. Fill in the blanks:

1. Atoms of different elements having ……… mass number, but ………. atomic numbers are called isobars.
2. Atoms of different elements having same number of ………. are called isotones.
3. Atoms of one element can be transmuted into atoms of other element by ………….
4. The sum of the numbers of protons and neutrons of an atom is called its …………
5. Relative atomic mass is otherwise known as …………
6. The average atomic mass of hydrogen is ……….. amu.
7. If a molecule is made of similar kind of atoms, then it is called ……….. atomic molecule.
8. The number of atoms present in a molecule is called its ………….
9. One mole of any gas occupies ………… ml at S.T.P
10. Atomicity of phosphorous is …………
Answer:
1. same, different
2. neutrons
3. artificial transmutation
4. mass number
5. standard atomic weight
6. 1.008
7. homo
8. atomicity
9. 22, 400
10. four

III. Match the following:

Answer:
A. (ii)
B. (iii)
C. (v)
D. (i)
E. (iv)

IV. True or False: (If false give the correct statement)

1. Two elements sometimes can form more than one compound.
2. Nobel gases are diatomic.
3. The gram atomic mass of an element has no unit.
4. 1 mole of Gold and Silver contain same number of atoms.
5. Molar mass of CO₂ is 42 g.

Answer:

1. True
2. False – Noble gases are Monoatomic.
3. False – The unit of gram atomic mass of an element is gram.
4. True
5. False – Molar mass of CO₂ is 44 g.

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: Atomic mass of aluminium is 27
Reason: An atom of aluminium is 27 times heavier than 1/12 th of the mass of the C-12 atom.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 2.
Assertion: The Relative Molecular Mass of Chlorine is 35.5 a.m.u.
Reason: The natural abundance of Chlorine isotopes are not equal.
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(c) Assertion is wrong, Reason is correct.

VI. Short answer questions:

Question 1.
Define: Relative atomic mass.
Answer:
Relative atomic mass of an element is the ratio between the average mass of its isotopes to \(\frac{1}{12^{th}}\)part of the mass of a carbon-12 atom. It is denoted as A_r.
[OR]

Question 2.
Write the different types of isotopes of oxygen and its percentage abundance.
Answer:
Oxygen has three stable isotopes. They are

Question 3.
Define Atomicity.
Answer:
The number of atoms present in the molecule is called its ‘Atomicity’.

Question 4.
Give any two examples for heteroatomic molecules.
Answer:
HI, HCl, CO, HBr, HF.

Question 5.
What is Molar volume of a gas?
Answer:
One mole of any gas occupies 22.4 litres.
(or)
22400 ml at S.T.R This volume is called as molar volume.

Question 6.
Find the percentage of nitrogen in ammonia.
Answer:
Molar mass of NH₃ = 1(14) + 3(1) = 17 g

VII. Long answer questions:

Question 1.
Calculate the number of water molecule present in one drop of water which weighs 0.18 g.
Answer:
The molecular mass of water (H₂O) is 18.
18 g of water molecule = 1 mole.
0. 18 g of water = \(\frac{1}{18} \times 0.18\) = 0.01 mole.
1 mole of water (Avogadro’s number) contains 6.023 × 10²³ water molecules.
0. 01 mole of water contain \(\frac{6.023 \times 10^{23}}{1} \times 0.01\) = 6.023 × 10²¹ molecules.

Question 2.
N₂ + 3 H₂ → 2 NH₃
(The atomic mass of nitrogen is 14, and that of hydrogen is 1)
1 mole of nitrogen (……..g) +
3 moles of hydrogen (………g) →
2 moles of ammonia (………g)
Answer:
1 mole of nitrogen (28 g) +
3 moles of hydrogen (6 g) →
2 moles of ammonia (34 g)

Question 3.
Calculate the number of moles in
(i) 27 g of Al;
(ii) 1.51 × 10²³ molecules of NH₄Cl.
Answer:
(i) 27 g of Al
Given mass atomic mass = \(\frac{Given Mass}{Atomic Mass}\) = \(\frac{27}{27}\)
= 1 mole

(ii) 1.51 x 10²³ molecules of NH₄Cl
Number of moles

Question 4.
Give the salient features of “Modern atomic theory”.
Answer:
The salient features of “Modem atomic theory” are,

1. An atom is no longer indivisible.
2. Atoms of the same element may have different atomic mass.
3. Atoms of different elements may have the same atomic masses.
4. Atoms of one element can be transmuted into atoms of other elements. In other words, an atom is no longer indestructible.
5. Atoms may not always combine in a simple whole-number ratio.
6. Atom is the smallest particle that takes part in a chemical reaction.
7. The mass of an atom can be converted into energy [E = mc²].

Question 5.
Derive the relationship between Relative molecular mass and Vapour density.
Answer:
Relative molecular mass : The relative molecular mass of a gas or vapour is the ratio between the mass of one molecule of the gas or vapour to mass of one atom of hydrogen.
Vapour density : Vapour density is the ratio of the mass of certain volume of a gas or vapour, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.

According to Avogadro’s law equal volumes of all gases contain equal number of molecules.
Let the number of molecules in one volume = n, then

When cancelling ‘n’ which is common at STP, we get

Since hydrogen is diatomic,

2 × Vapour density = Relative Molecular mass of a gas
[OR]
Relative Molecular Mass = 2 × Vapour density

VIII. HOT Question:

Question 1.
Calcium carbonate is decomposed on heating in the following reaction
CaCO₃ → CaO + CO₂

1. How many moles of Calcium carbonate is involved in this reaction?
2. Calculate the gram molecular mass of calcium carbonate involved in this reaction.
3. How many moles of CO₂ are there in this equation?

Answer:
CaCO₃ → CaO + CO₂

1. 1 mole of CaCO₃ is involved in this reaction.
2. Gram molecular mass of calcium carbonate
   CaCO₃ = (40 + 12 + 3 × 16) = 52 + 48 = 100 g
3. 1 mole of CO₂ is in this equation.

IX. Solve the following problems:

Question 1.
How many grams are there in the following?
(i) 2 moles of a hydrogen molecule, H₂
(ii) 3 moles of chlorine molecule, Cl₂
(iii) 5 moles of sulphur molecule, S₈
(iv) 4 moles of a phosphorous molecule, P₄
Solution:
(i) 2 moles of a hydrogen molecule, H₂
Mass of 1 mole of hydrogen molecule = 2 g
Mass of 2 moles of hydrogen molecule = 2 × 2 = 4 g.

(ii) 3 moles of chlorine molecule, Cl₂
Mass of 1 mole of chlorine molecule = 71 g
Mass of 3 moles of chlorine molecules = 71 × 3 = 213 g.

(iii) 5 moles of sulphur molecule, S₈
Mass of 1 mole of sulphur molecule = 32 g
Mass of 5 moles of sulphur molecules = 32 × 5 = 160 g.

(iv) 4 moles of the phosphorous molecule, P₄
Mass of 1 mole of phosphorous molecule = 30.97 g
Mass of 4 moles of phosphorous molecules = 30.97 × 4 = 123.88 g.

Question 2.
Calculate the % of each element in calcium carbonate. (Atomic mass: C – 12, O – 16, Ca – 40)
Answer:
Formula to find % of each element

Question 3.
Calculate the % of oxygen in Al₂(SO₄)₃.
(Atomic mass: Al – 27, O – 16, S – 32)
Answer:
Formula:

Molar mass of Al₂(SO₄)₃ = [2(Atomic mass of Al) + 3(Atomic mass of S) + 12(Atomic mass of O)]
= 2(27) + 3(32) + 12(16) = 342 g
% of Oxygen = \(\frac{12(16)}{342}\) × 100 = 56.14%.

Question 4.
Calculate the % relative abundance of B – 10 and B – 11, if its average atomic mass is 10.804 amu.
Answer:
% of relative abundance can be calculated by the formula.
Average atomic mass of the element
= Atomic mass of 1st isotope × abundance of 1st isotope + Atomic mass of 2nd isotope × abundance of 2nd isotope
∴ Average atomic mass of Boron
= Atomic mass of B – 0 × abundance of B -10 + Atomic mass of B – 11 × abundance of B – 11
Let the abundance of B – 10 be ‘x’ and B – 11 be (1 – x)
So, 10.804 = 10 × x + 11 (1 – x)
10.804 = 10x + 11 – 11x
x = 11 – 10.804
x = 0.196
1 -x = 1 – 0.196 = 0.804
Therefore % abundance of B – 10 is 19.6% and B – 11 is 80.4%
[OR]
Let the % of the isotope B – 10 = x
Then the % of the isotope B – 11 = 100 – x

1100 – x = 1080.4
x = 19.6
% abundance of B – 10 = 19.6%
% abundance of B – 11 = 80.4%

Samacheer Kalvi 10th Science Atoms and Molecules Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
The first scientific theory of an atom was proposed by:
(a) Ruther Ford
(b) Newland
(c) John Dalton
(d) Neils Bohr
Answer:
(c) John Dalton

Question 2.
Identify the pair that indicates isobars among the following _____.
(a) \(_{1} \mathrm{H}^{2},_{1} \mathrm{H}^{3}\)
(b) \(_{17} \mathrm{Cl}^{35},_{17} \mathrm{Cl}^{37}\)
(c) \(\text { (c) }_{18} \mathrm{Ar}^{40},_{18} \mathrm{Ca}^{40}\)
(d) \(_{6} \mathrm{C}^{13},_{7} \mathrm{N}^{14}\).
Answer:
(c) \(\text { (c) }_{18} \mathrm{Ar}^{40},_{18} \mathrm{Ca}^{40}\)

Question 3.
Which one of the following represents 180 g of water?
(a) 5 moles of water
(b) 90 moles of water
(c) 6.023 × 10²⁴ molecules of water
(d) 6.023 × 10²² molecules of water
Answer:
(c) 6.023 × 10²⁴ molecules of water

Question 4.
The isotope of Carbon-12 contains _____.
(a) 6 protons and 7 electrons
(b) 6 protons and 6 neutrons
(c) 12 protons and no neutrons
(d) 12 neutrons and no protons.
Answer:
(b) 6 protons and 6 neutrons

Question 5.
Which contains the greatest number of moles of oxygen atoms?
(a) 1 mol of water
(b) 1 mole of NaOH
(c) 1 mole of Na₂CO₃
(d) 1 mole of CO
Answer:
(c) 1 mole of Na₂CO₃

Question 6.
The mass of proton or neutron is approximately _____.
(a) 1 amu
(b) 1.609 × 10^-19 g
(c) 1 g
(d) 6.023 × 10^-23 g.
Answer:
(a) 1 amu

Question 7.
The natural abundance of C-12 and C-13 are 98.90% and 1.10% respectively. The average atomic mass of carbon is:
(a) 12 amu
(b) 12.011 amu
(c) 14 amu
(d) 12.90 amu
Answer:
(b) 12.011 amu

Question 8.
The relative atomic mass of magnesium-based on C – 12 scale is _____.
(a) 24 g
(b) 24
(c) 24 amu
(d) 24 kg
Answer:
(b) 24

Question 9.
If 1.5 moles of oxygen combine with Al to form Al₂O₃, the mass of Al in g (atomic mass of Al = 27) used in the reaction is:
(a) 2.7
(b) 54
(c) 40.5
(d) 81
Answer:
(b) 54

Question 10.
The atomicity of methane is:
(a) 5
(b) 4
(c) 3
(d) 6
Answer:
(a) 5

Question 11.
Find the odd one out _____.
(a) \(_{8} \mathrm{O}^{16}\)
(b) \(_{8} \mathrm{O}^{17}\)
(c) \(_{6} \mathrm{O}^{12}\)
(d) \(_{8} \mathrm{O}^{18}\).
Answer:
(c) \(_{6} \mathrm{O}^{12}\)

Question 12.
The volume occupied by 3 moles of HCl gas at STP is:
(a) 22.4 L
(b) 44.8 L
(c) 2.24 L
(d) 67.2 L
Answer:
(d) 67.2 L

Question 13.
The mass percentage of hydrogen in ethane (C₂H₆) is:
(a) 25%
(b) 75%
(c) 80%
(d) 20%
Answer:
(d) 20%

Question 14.
Which one of the following is a homo diatomic molecule?
(a) H₂
(6) CO
(c) NO
(d) O₃.
Answer:
(a) H₂

Question 15.
The percentage of nitrogen in urea is about:
(a) 38.4
(b) 46.6
(c) 59.1
(d) 61.3
Answer:
(b) 46.6

Question 16.
Out of the following the largest number of atoms are contained in:
(a) 11 g of CO₂
(b) 4 g of H₂
(C) 5 g of NH₃
(d) 8 g of SO₂
Answer:
(b) 4 g of H₂

Question 17.
Which of the following is an example of a homo triatomic molecule?
(a) Phosphorous
(b) Sulphur
(c) Bromine
(d) Ozone.
Answer:
(d) Ozone.

Question 18.
For the reaction A + 2B → C, 5 moles of A and 8 moles of B will produce:
(a) 5 moles of C
(b) 4 moles of C
(c) 8 moles of C
(d) 13 moles of C
Answer:
(b) 4 moles of C

Question 19.
The vapour density of a gas is 32. Its relative molecular mass will be:
(a) 32
(b) 16
(c) 64
(d) 96
Answer:
(c) 64

Question 20.
Find the odd one out _____.
(a) Silver
(b) Potassium
(c) Iron
(d) Phosphorous.
Answer:
(d) Phosphorous.

II. Fill in the blanks.

1. The volume occupied by 16 g of oxygen is ………..
2. One mole of a triatomic gas contains ………… atoms.
3. Equal volume of all gases under the same conditions of temperature and pressure contain equal number of …………
4. The mass of an atom can be converted into energy by using the formula …………
5. The percentage composition is useful to determine the ………… formula and ………… formula.

Answer:

1. 11.2 L
2. 3 × 6.023 × 10²³
3. molecules
4. E = me²
5. empirical, molecular

III. Match the following:

Question 1.
Match the Column I with Column II.

Answer:
A. (ii)
B. (iii)
C. (i)
D. (v)
E. (iv)

Question 2.
Match the Column I with Column II.

Answer:
A. (iii)
B. (i)
C. (iv)
D. (v)
E. (ii)

Question 3.
Match the Column I with Column II.

Answer:
A. (iv)
B. (v)
C. (ii)
D. (iii)
E. (i)

Question 4.
Match the Column I with Column II.

Answer:
A. (v)
B. (iii)
C. (iv)
D. (i)
E. (ii)

IV. True or False: (If false give the correct statement)

1. Atoms always combine in a simple whole number ratio.
2. 2 × RMM = VD
3. The average atomic mass of Beryllium is 9.012 because of the presence of isotopes.
4. The noble gases are diatomic.
5. The number of atoms present in one mole of phosphorus(P₄) is 4 × 6.023 × 10²³

Answer:

1. False -Atoms may not combine always in a simple whole number ratio.
2. False – 2 × VD = RMM
3. True
4. False – The noble gases are mono atomic.
5. True

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: The standard unit for expressing mass of atom is amu.
Reason: Atomic mass unit is one-twelth of the mass of a C-12 atom
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

Question 2.
Assertion: The volume occupied by 44 g of CO₂ is 22.4 L
Reason: The volume occupied by one mole of any gas is 22.4 L
(a) Assertion and Reason are correct, Reason explains the Assertion.
(b) Assertion is correct, Reason is wrong.
(c) Assertion is wrong, Reason is correct.
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.
Answer:
(a) Assertion and Reason are correct, Reason explains the Assertion.

VI. Short answer questions:

Question 1.
Define vapour density.
Answer:
Vapour density is the ratio of the mass of certain volume of a gas or vapour, to the mass of an equal volume of hydrogen measured under the same condition of temperature and pressure.
[OR]

Question 2.
What are isobars? Give an example.
Answer:
Atoms of different elements that have same atomic masses but different atomic numbers are called isobars.
e.g., \(_{18} \mathrm{Ar}^{40}\), \(_{20} \mathrm{Ca}^{40}\).

Question 3.
Write the differences between an atom and a molecule.
Answer:

Question 4.
What is artificial transmutation?
Answer:
Atoms of one element can be transmuted into atoms of other elements. In other words, the atom is no longer indestructible. It is known as artificial transmutation.

Question 5.
Classify the following based on atomicity.
Answer:
(i) Bromine
2 – Diatomic

(ii) Argon
1 – Monoatomic

(iii) Ozone
3 – Triatomic

(iv) Sulphur
8 – Polyatomic

Question 6.
Define atomic mass unit.
Answer:
Atomic mass unit is one-twelfth of the mass of carbon – 12 atom, as an isotope of carbon which contains 6 protons and 6 neutrons. It is amu.

VII. Long answer questions:

Question 1.
Explain how Avogadro hypothesis is used to derive the value of atomicity.
Answer:
(i) The Avogadro’s law states that “equal volumes of all gases under similar conditions of temperature and pressure contain the equal number of molecules”.

(ii) Let us consider the reaction between hydrogen and chlorine to form hydrogen chloride gas.
H_{2 (g)} + Cl_{2 (g)} → 2HCl _(g)
⇒ 1 volume + 1 volume → 2 volumes.

(iii) According to Avogadro’s law, 1 volume of any gas is occupied by “n” number of molecules,
“n” molecules + “n” molecules → “2n” molecules
If “n” = 1, then
1 molecule + 1 molecule → 2 molecules.
\(\frac { 1 }{ 2 }\) molecule + \(\frac { 1 }{ 2 }\) molecule → 1 molecule.

(iv) 1 molecule of hydrogen chloride gas is made up of \(\frac { 1 }{ 2 }\) molecule of hydrogen and \(\frac { 1 }{ 2 }\) molecule of chlorine.

(v) \(\frac { 1 }{ 2 }\) molecule of hydrogen contains 1 atom.
So, 1 molecule of hydrogen contains 2 atoms.
So, hydrogen atomicity is 2. Similarly, chlorine atomicity is also 2.
So, H₂ and Cl₂ are diatomic molecules.

Question 2.
Write a note on the following,
(i) Isotopeos
(ii) Isobars
(iii) Relative atomic mass.
Answer:
(i) Isotopes : Atoms of same element with different mass number. Eg: ₁₇Cl³⁵, ₁₇Cl³⁷.
(ii) Isobars : Atoms of different elements with same mass number. Eg: ₁₈Ar⁴⁰, ₂₀Ca⁴⁰
(iii) Relative Atomic Mass (RAM) :

Question 3.
Sodim bicarbonate breaks down on heating as follows:
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
(Atomic mass of Na = 23, H = 1, C = 12, O = 16)
(i) How many moles of NaHCO₃ are there in the equation?
Answer:
2 moles.

(ii) What is the mass of CO₂ produced in the equation?
Answer:
44 g

(iii) How many moles of water molecules are produced in the equation?
Answer:
1 mole.

(iv) What is the mass of NaHCO₃ used in this equation?
Answer:
= 2[23 + 1 + 12 + 3(16)]
= 2[84]
= 168 g

(v) What is the volume occupied by CO₂ in this equation?
Answer:
22.4 lit.

VIII. Hot Questions

Question 1.
Why do we take an atomic mass of Carbon – 12 as standard?
Answer:
Carbon – 12 is the standard while measuring the atomic masses. Because no other nuclides other than C – 12 have exactly whole-number masses in this scale. This is due to the different mass of neutrons and protons acting to change the total mass in nuclides with proton/neutron ratio other than 1 : 1 ratio of carbon – 12.

Question 2.
The cost of common salt (NaCl) is Rs 18 per kg. Calculate the cost of the salt per mole.
Answer:
Gram molar mass of NaCl = 23 + 35.5
= 58.5 g
1000 g of NaCl costs = Rs 18
∴ 58.5 g of NaCl costs = \(\frac{18}{1000}\) × 58.5
= Rs 1.053
The cost of one mole of NaCl = Rs 1.053

Question 3.
What will be the mass of one \(^{12} \mathbf{C}\) atom in g?
Answer:
1 mol of \(^{12} \mathbf{C}\) atoms = 6.022 × 10²³ atoms = 12 g
Thus, 6.022 × 10²³ atoms of \(^{12} \mathbf{C}\) have mass = 12 g
∴ 1 atom of \(^{12} \mathbf{C}\) will have mass = \(\frac{12}{6.022 \times 10^{23}}\) g = 1.9927 × 10^-23 g.

IX. Solve the following problems:

Question 1.
Calculate the average atomic mass of carbon, if the natural abundance of C – 12 and C – 13 are 98.90% and 1.10% respectively.
Solution:
Average atomic mass of carbon
\(=\left(12 \times \frac{98.9}{100}\right)+\left(13 \times \frac{1.1}{100}\right)\)
= (12 × 0.989) +(13 × 0.011)
= 11.868 + 0.143
= 12.011 amu.

Question 2.
Find how many moles are there in
(a) 98 g of H₂SO₄
Answer:
(a) 98 g of H₂SO₄
GMM of H₂SO₄ = 2(1) + 32 + 4(16)
= 98 g
Number of moles = \(\frac{Given Mass}{Mol. Mass}\)
= \(\frac{98}{98}\)
= 1 mole

(b) 18.069 × 10²³ atoms of calcium
Answer:
Number of moles

= 3 moles

(c) 4.48 L of CO₂
Answer:
Number of moles

Question 3.
Calculate the number of moles in
(i) 12.046 × 10²³ atoms of copper
(ii) 27.95 g of iron
(iii) 1.51 × 10²³ molecules of CO₂
Answer:
(i) 12.046 × 10²³ atoms of copper
6.023 × 10²³ atoms of copper = 1 mole
12.046 × 10²³ atoms of copper = \(\frac{1 \times 12.046 \times 10^{23}}{6.023 \times 10^{23}}\) = 2 moles of copper

(ii) 27.95 g of iron
55.9 g of iron = 1 mole
27.95 g of iron = \(\frac{1}{55.9}\) × 27.95 = 0.5 mole of iron.

(iii) 1.51 × 10²³ molecules of CO₂
No of moles = \(\frac{\text { No. of molecules }}{\text { Avogadro number }}\)
= \(\frac{1.51 \times 10^{23}}{6.023 \times 10^{23}}\)
= 0.25 mole of CO₂

Question 4.
Calculate the number of atoms of mercury present in 1 kg of Mercury. [Atomic mass of Hg = 200.6]
Answer:
200.6 g of Hg contains 6.023 × 10²³ Hg atoms
∴ 1 Kg of Hg will contain = \(\frac{6.023×10^{23}}{200.6}\) × 1000
= 30.02 × 10²³ Hg atoms

Question 5.
How many molecules are present in 7 × 10^-3 m³ of NH₃ at STP?
Answer:
Molar volume = 22.4 dm³ = 2.24 × 10^-2 m³
2.24 × 10^-2 m³ of NH₃ at STP contains 6.023 × 10²³ NH₃ molecules.
:. 7 × 10^-3 m³ of NH₃ will contain
= \(\frac{6.023×10^{23}}{2.24×10^{-2}}\) × 7 × 10^-3
18.82 × 10²² NH₃ molecule

Question 6.
What is the mass in grams of the following?
(a) 3 moles of NaOH
(b) 6.023 × 10²² atoms of Ca
(c) 224 L of CO₂
Answer:
Formula:

(a) 3 moles of NaOH
Mass of 3 moles of NaOH = 3 × mol. mass of NaOH
GMM of NaOH = 23 + 16 + 1 = 40 g
Mass of 3 moles of NaOH = 3 × 40 = 120 g

(b) 6.023 × 10²² atoms of Ca = n × atomic mass of ca

= 4 g

(c) 224 L of CO₂
Mass of 224 L of CO₂

= 10 × 44
= 440 g

Question 7.
How many grams are therein:
(i) 5 moles of water
(ii) 2 moles of Ammonia
(iii) 2 moles of Glucose
Solution:
(i) 5 moles of water.
Mass of 1 mole of water (H₂O) = 18 g (2 + 16)
Mass of 5 moles of H₂O = 18 × 5 = 90 g.

(ii) 2 moles of ammonia.
Mass of 1 mole of ammonia (NH₃) = 17 g (14 + 3)
Mass of 2 moles of ammonia = 17 × 2 = 34 g.

(iii) 2 moles of glucose.
Mass of 1 mole of glucose (C₆H₁₂O₆) = 180 g (72 + 12 + 96)
Mass of 2 moles of glucose = 180 × 2 = 360 g.

Question 8.
Calculate tbe molar mass of the following compounds.
(a) Urea (NH₂CONH₂)
(b) Ethanol(C₂H₅OH);
(c) Boric acid (H₃BO₃)
[Atomic mass of N – 14, H – 1, C – 12, B – 11, O – 16]
Answer:
(a) Urea (NH₂CONH₂) = 2(14) + 4(1) + 1(16) + 1(12)
= 28 + 4 + 16 + 12
= 60 g

(b) Ethanol(C₂H₅OH) = 2(12) + 6(1) + 1(16)
= 24 + 6 + 16 = 46 g

(c) Boric acid (H₂BO₃) = 3(1) + 1(11) + 3(16)
= 3 + 11 + 48
= 62 g

Question 9.
Mass of one atom of an element is 6.645 × 10^-23 g. How many moles of element are there in 0.320 kg.
Answer:
Mass of one atom of an element = 6.645 × 10^-23 g
∴ Mass of 1 mol of atom = 6.645 × 10^-23 × 6.023 × 10²³ = 40 g
Number of moles =

Students can download 10th Social Science Geography Chapter 2 Climate and Natural Vegetation of India Questions and Answers, Notes, Samacheer Kalvi 10th Social Science 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 Social Science Solutions Geography Chapter 2 Climate and Natural Vegetation of India

Samacheer Kalvi 10th Social Science Climate and Natural Vegetation of India Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
Meteorology is the science of:
(a) Weather
(b) Social
(c) Political
(d) Human
Answer:
(a) Weather

Question 2.
We wear cotton during ……….
(a) Summer
(b) Winter
(c) Rainy
(d) Northeast monsoon
Answer:
(a) Summer

Question 3.
Western disturbances cause rainfall in:
(a) Tamilnadu
(b) Kerala
(c) Punjab
(d) Madhya Pradesh
Answer:
(c) Punjab

Question 4.
………. helps in quick ripening of mangoes along the coast of Kerala and Karnataka.
(a) Loo
(b) Norwester
(c) Mango showers
(d) Jet stream
Answer:
(c) Mango showers

Question 5.
……………. is a line joining the places of equal rainfall.
(a) Isohyets
(b) Isobar
(c) Isotherm
(d) Latitudes
Answer:
(a) Isohyets

Question 6.
The climate of India is labelled as ………
(a) Tropical humid
(b) Equatorial Climate
(c) Tropical Monsoon Climate
(d) Temperate Climate
Answer:
(c) Tropical Monsoon Climate

Question 7.
The monsoon forests are otherwise called as
(a) Tropical evergreen forest
(b) Deciduous forest
(c) Mangrove forest
(d) Mountain forest
Answer:
(b) Deciduous forest

Question 8.
……. forests are found above 2400 m Himalayas.
(a) Deciduous forests
(b) Alpine forests
(c) Mangrove forests
(d) Tidal forests
Answer:
(b) Alpine forests

Question 9.
Sesahachalam hills, a Biosphere reserve is situated in
(a) Tamil Nadu
(b) Andhra Pradesh
(c) Madhya Pradesh
(d) Karnataka
Answer:
(b) Andhra Pradesh

Question 10.
………… is a part of the world network biosphere reserves of UNESCO .
(a) Nilgiri
(b) Agasthiyamalai
(c) Great Nicobar
(d) Kachch
Answer:
(a) Nilgiri

II. Match the following

Answer:
A. (iv)
B. (v)
C. (ii)
D. (i)
E. (iii)

III. Consider the given statements and choose the correct option from the given below ones.

Question 1.
Assertion(A): Monsoons are a complex meteorological phenomenon. Reason(R): Meteorologists have developed a number of concepts about the origin of monsoons.
(a) Both (A) and (R) are true: R explains A
(b) Both (A) and (R) are true: R does not explain A.
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(a) Both (A) and (R) are true: R explains A

Question 2.
Assertion(A): The Himalayas acts as a climatic barrier.
Reason(R): The Himalayas prevents cold winds from central Asia and keep the Indian Sub-continent warm.(Give option for this questions)
(a) Both (A) and (R) are true: R explains A
(b) Both (A) and (R) are true: R does not explain A.
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(a) Both (A) and (R) are true: R explains A

IV. Choose the inappropriate answer

Question 1.
Tidal forests are found in and around ……..
(a) Desert
(b) The deltas of Ganga and Brahmaputra
(c) The delta of Godavari
(d) The delta of Mahanadhi
Answer:
(a) Desert

Question 2.
Climate of India is affected by:
(a) Latitudinal extent
(b) Altitude
(c) Distance from the sea
(d) Soil
Answer:
(d) Soil

V. Answer briefly

Question 1.
Define ‘Meteorology’.
Answer:
The branch of science concerned with the processes and phenomena of the atmosphere, especially as a means of forecasting the weather.

Question 2.
What is meant by ‘normal lapse rate’?
Answer:
Temperature decreases when the altitude increases .Normal Lapse rate is a phenomenon in which temperature decreases at the rate of 6.5°C for every 1000mts of ascent. Hence places in the mountains are cooler than the places on the plain.

Question 3.
What are ‘jet streams’?
Answer:
In the upper layers of the atmosphere, there are strong westerly winds concentrated in a relatively narrow and shallow stream known as “Jet streams” They cause heavy rainfall in North-west India.

Question 4.
Write a short note on ‘Monsoon wind’.
Answer:

1. The Word ‘Monsoon’ is derived from the Arabic word “Mausim” which means season.
2. Monsoons are a complex meteorological phenomenon .
3. These winds originates due to the seasonal migration of planetary winds and pressure belts following the position of the sun.
4. During summer they blow from South west to North east ( South West monsoon) June-September and from the North east to South west (North East monsoon) October and November.

Question 5.
Name the four distinct seasons of India.
Answer:

1. Winter or cold weather season (Jan-Feb)
2. Pre Monsoon or Summer (March-May)
3. Southwest Monsoon or rainy season (June-September)
4. Northeast Monsoon season (October-December)

Question 6.
What is ‘burst of monsoon’?
Answer:

1. Prior to the onset of the Southwest monsoon the temperature in North India reaches up to 46°C.
2. The sudden approach of monsoon wind over South India with lightning and thunder is termed as the “break or burst of monsoon”.

Question 7.
Name the areas which receive heavy rainfall.
Answer:

1. Middle Ganga Valley
2. Western Ghats
3. Eastern Maharashtra
4. Madhya Pradesh and Odisha

Question 8.
State places of mangrove forest in India.
Answer:

1. Mangrove forests occur in and around the deltas, estuaries and creeks prone to tidal influences.
2. In India they are found in deltas of Ganga, Brahmaputra (largest), the deltas of Mahanadi, Godavari and Krishna rivers.

Question 9.
Name the trees of tropical evergreen forest.
Answer:
The most important trees are rubber, mahogany, ebony, rosewood, coconut, bamboo, cinchona, candes, palm, iron wood and cedar.

Question 10.
Write any five Biosphere Reserves in India.
Answer:

1. Biosphere Reserves are the protected areas of land coastal environments where in people are an integral component of the system.
2. The Indian Government has established 18 Biosphere Reserves. Gulf of Mannar Nilgiris, Agasthiamalai, Kanjanjunga, Great Nicobar etc.

Question 11.
What is ‘Project Tiger’?
Answer:
Project Tiger was launched in April 1973 with the aim to conserve tiger population in specifically constituted “Tiger Reserves” in India. This project is benefited tremendously, with an increase of over 60% – the 1979 consensus put the population at 3,015.

VI. Distinguish between

Question 1.
Weather and Climate.
Answer:
Weather:

1. State of the atmosphere of a place at a given point of time.
2. Weather changes occur daily.
3. Temperature, wind pressure, humidity duration of sunlight and rainfall decides the weather.

Climate:

1. Accumulation of daily and seasonal weather events of a given location over a period of time.
2. It is an average of weather condition over a period of 30-35 years.
3. Climate of a place is determined by Latitude, altitude, distance from the seas, relief features etc.

Question 2.
Tropical Evergreen Forest and Deciduous Forest.
Answer:
Tropical Evergreen forest:

1. These forests are found in the areas where the annual rainfall is 200 cm and more.
2. The tress of these forests do not shed their leaves.
3. Rubber, iron wood, Mahogony are some of the main trees.

Deciduous Forest:

1. These forests are found in the areas with 100 to 200cm annual rainfall.
2. The trees of these forests shed their leaves in spring and early summer.
3. Sandalwood, teak, sal, padak are some of the main trees.

Question 3.
North East Monsoon and South West Monsoon.
Answer:
North East Monsoon:

1. These monsoon winds blow from North-east to South west.
2. They blow during the month of October-November.
3. TamilNadu and Andhra Pradesh gets rainfall from this monsoon.

South West Monsoon:

1. These monsoon winds blow from South west towards North east.
2. They blow during the month of June-September.
3. 75% of India’s rainfall is from this monsoon winds.

VII. Give reasons for the following topics

Question 1.
Western Coastal plain is narrow.
Answer:
It lies between the Western Ghats and Arabian Sea.
It is a narrow plain, which stretches from Gujarat to Kerala with an average width of 50 – 80 km. It is mainly characterised by sandy beaches, coastal sand, dunes, mud flats, lagoons, estuary, laterite platforms and residual hills.

Question 2.
India has a tropical monsoon climate.
Answer:
Latitudinally India lies between 8°4′ N and 37°6’N latitudes. Tropic of Cancer (23° 30’N) divides the country into two equal halves . Most parts of the country’ lie in the Tropical Zone and receives rainfall from monsoon winds which is the dominating climatic factor. Hence India has Tropical monsoon climate.

Question 3.
Mountains are cooler than the plains.
Answer:
When the altitude increases, the temperature decreases. Temperature decreases at the rate of 6.5°C for every 1000 metres of ascent. Hence, places in the mountains are cooler than the places on the plains. That is why the places located at higher altitudes even in South India have cool climate. Ooty and several other hill stations of South India and of the Himalayan ranges like Mussoorie, Shimla, etc., are much cooler than the places located on the Great Plains.

VIII. Write in detail

Question 1.
Write about South West Monsoon.
Answer:

1. Southwest monsoon season is also known as the rainy season in India.
2. The onset of Southwest monsoon season takes place over the southern tip of the country’ by the first week of June, advances along the Konkon coast in early June and covers the whole country by 15th July.
3. Jet stream and ELNino are the two factors that determine the occurrence of Southwest monsoon.
4. Due to the high temperature over north India creates a low-pressure trough which draws the moisture-laden winds from the Indian Ocean towards the Indian landmass.
5. The sudden approach of monsoon wind over south India with ‘ lightning and the thunder indicates the onset of south-west monsoon.
   This is also known as “break or burst of monsoon”.
6. The monsoon winds strike against the southern tip of the Indian landmass and get divided into two branches.
   • (a) Arabian sea branch (of Southwest monsoon)
   • (b) The Bay of Bengal branch (of Southwest monsoon)

(a) Arabian sea branch of South west monsoon:

• The Arabian sea branch of south west monsoon gives heavy rainfall to the west coast of India as it is located in the wind wards side of the Western Ghats.
• The other part which advances towards north strikes against the Himalayan mountains results in heavy rainfall in north.

As Aravalli mountains lie parallel to the wind direction. Rajasthan and the western parts do not get much rainfall.

(b) Bay of Bengal branch of south west monsoon:

• The Wind from Bay of Bengal moves towards north east India and Myanmar.
• This wind is trapped by Garo, Khasi and Jaintia hills and gives the heaviest rainfall to Meghalaya (at Mawsynram).
• This wind gets deflected towards west.
• When the wind moves from east to west rainfall decreases as it looses its moisture.
• Tamil Nadu receives only a meager rainfall as the state is located on the leeward side.

About 75% of India’s rainfall is from the south west monsoon.

Question 2.
Describe the forests of India.
Answer:
Natural Vegetation refers to a plant community which has grown naturally without human aid
and has been left undisturbed by human for a long time.

Tropical Evergreen Forest:
These forests are found in areas with 200 cm or more annual rainfall.The annual temperature is about more than 22°C and the average annual humidity exceeds 70 percent in this region. Western Ghats in Maharashtra, Karnataka, Kerala, Andaman-Nicobar Islands, Assam, West Bengal, Nagaland, Tripura, Mizoram, Manipur and Meghalaya states have this type of forests. The most important trees are rubber, mahogany, ebony, rosewood, coconut, bamboo, cinchona, candes, palm, iron wood and cedar.

Tropical Deciduous Forest:
It covers a large area of the Peninsula and northern India, where the rainfall is from 100 to 200 m. These trees of the forests shed their leaves for a few weeks in early summer. The main trees are teak, Sandalwood, deodars, sisam, sal and redwood.

Tropical Dry Forest:
These are found in the areas with 50 to 100 cm. annual rainfall.They represent a transitional type of forests. These are found in east Rajasthan, Haryana, Punjab, Western Uttar Pradesh, Madhya Pradesh, Eastern Maharashtra, Telangana, West Karnataka and East Tamilnadu. The important species are mahua, banyan, amaltas, palas, haldu, kikar, bamboo, babool, khair etc.,

Mountain or Montane Forest:
These forests are classified on the basis of altitude and amount of rainfall. Accordingly two different types of forests namely Eastern Himalayas Forests and Western Himalayas Forests. Eastern Himalayan Forest: These are found on the slopes of the mountains in north-east states. They receive rainfall of more than 200 cm. The vegetation is of evergreen type. Sal, oak, laurel, amura, chestnut, cinnamon are the main trees from 1200 to 2400 m altitude oak, birch, silver, fir, fine, spruce and juniper are the major trees from 2400 to 3600 m height. Western Himalayan Forest: The rainfall of this region is moderate. They are found n the states of Jammu and Kashmir, Himachal Pradesh and Uttarakhand. In altitude from 900 to 1800 m, chir tree is the most common tree. The other important trees of this region are sal, semal, dhak, jamun and jujube.

Alpine Forest:
It occurs all along the Himalayas with above 2400 m altitude. These are purely having coniferous trees. Oak, silver fir, pine and juniper are the main trees of these forests. The eastern parts of Himalayas has large extent of these forests.

Tidel Forest:
These forests occur in and around the deltas, estuaries and creeks prone to tidal influences and as such are also known as delta or swamp forests. The delta of the Ganga-Brahmaputra has the largest tidal forest. The deltas of Mahanadi, Godavari and Krishna rivers are also known for tidal forests. These are also known as mangrove forest.

Coastal Forest:
These are littoral forests. Generally, coastal areas have these types of forests. Casurina, palm and coconut are the dominant trees. Both the eastern and western coasts have this type of forests.

Riverine Forest:
These forests are found along the rivers on Khadar areas. These are known for tamarisk and tamarind trees. The rivers of Great Plains are more prominent for this type of natural vegetation.

Question 3.
Write the names of biosphere reserves and their location in India.
Answer:
The Indian Government has established 18 Biosphere Reserves which protect larger areas of natural habitat and often include one or more national parks along with the buffer zones that are open to some economic uses.

IX. Map

Mark the following on the outline map of India.

Question 1.
Direction of South West Monsoon wind.
Answer:

Question 2.
Direction of North East Monsoon wind.
Answer:

Question 3.
Areas of heavy rainfall.
Answer:

Question 4.
Mountain forests.
Answer:

Question 5.
Panna biosphere reserve
Answer:

Question 6.
Agasthiyamalai biosphere reserve
Answer:

TB. PNo: 99
Find Out

Question 1.
Find out the temperature of Ooty (2240m). If it is 35°C in Chennai (6.7 m).
Answer:
For every 1000 mt altitude temperature decreases by 6.5°C . So the temperature of Ooty is nearly 15°C.

TB. PNo: 102
HOTS

Question 1.
Why is Mawsynram, the wettest place in the world?
Answer:
Mawsynram in Meghalaya receives the highest rainfall (1141 cm) in the world. Almost daily it rains. Thus make this place a swampy and the wettest due to very dense vegetation.

Samacheer Kalvi 10th Social Science Climate and Natural Vegetation of India Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
The most dominant factor which affects the climate of India is:
(a) Relief
(b) Monsoon winds
(c) Temperature
(d) Soil
Answer:
(b) Monsoon winds

Question 2.
The Local storms in the north eastern part of India during the weather seasons are called
(a) Loo
(b) Norwesters
(c) Mangoshowers
Answer:
(a) Loo

Question 3.
Seventy-five percentage of Indian rainfall is from this wind:
(a) Western Disturbance
(b) North East monsoon
(c) Norwesters
(d) South West monsoon
Answer:
(d) South West monsoon

Question 4.
The mountain which stands parallel to the direction of the south-west monsoon wind ….
(a) Vindhya
(b) Aravalli
(c) Satpura
Answer:
(b) Aravalli

Question 5.
The ……………. forests are found in the region where the rainfall is 200 cm and above.
(a) Tropical evergreen
(b) Tidal
(c) Tropical deciduous
(d) Alpine
Answer:
(a) Tropical evergreen

Question 6.
……… experience continental climate.
(a) Chennai
(b) Mumbai
(c) Delhi
Answer:
(c) Delhi

Question 7.
The term ……………. includes animals of any habitat in nature.
(a) Mammals
(b) Forests
(c) Insects
(d) Wild life
Answer:
(d) Wild life

Question 8.
In India, 85% of the rain is received from the ……….
(a) North-east
(b) South-west
(c) South-east
Answer:
(b) South-west

Question 9.
In the year ……………. United nations Convention on Biological Diversity (CBD) recognizes the sovereign rights of the states to use their own Biological Resources.
(a) 1992
(b) 1982
(c) 1979
(d) 1929
Answer:
(a) 1992

Question 10.
The ……… prevents the cold polar winds blowing from Central Asia.
(a) Aravalli Hills
(b) Bay of Bengal
(c) Himalayas
Answer:
(c) Himalayas

Question 11.
The highest rainfall region in India is located in this state:
(a) Assam
(b) Bihar
(c) Meghalaya
(d) Manipur
Answer:
(c) Meghalaya

Question 12.
There is a peculiar uniformity in the climate of India due to its unique ……..
(a) Geography
(b) Physiography
(c) Demography
Answer:
(b) Physiography

II. Match the following

Question 1.
Match the Column I with Column II.

Answer:
A. (iii)
B. (i)
C. (v)
D. (ii)
E. (iv)

Question 2.
Match the Column I with Column II.

Answer:
A. (vi)
B. (iii)
C. (v)
D. (ii)
E. (iv)
F. (i)

Question 3.
Match the Column I with Column II.

Answer:
A. (v)
B. (i)
C. (ii)
D. (vi)
E. (iv)
F. (iii)

III. Consider the given statements and choose the correct option from the given below ones

Question 1.
Assertion(A): Peninsular India enjoys equable climate.
Reason(R): The peninsular region is surrounded by the seas on three sides, not very far from the sea.
(a) Both (A) and (R) are true: R explains A
(b) Both (A) and (R) are true: R does not explain A
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(c) (A) is correct (R) is false

Question 2.
Assertion(A): Two different types-of mountain forests are in India namely Eastern Himalayan forests and Western Himalayan forests.
Reason(R): Mountain forests are classified on the basis of altitude and amount of Rainfall.
(a) Both (A) and (R) are true: R explains A
(b) Both (A) and (R) are true: R does not explain A
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(b) Both (A) and (R) are true: R does not explain A

Question 3.
Assertion(A): Systematic change in the direction of planetary winds is known as monsoons.
Reason(R): Monsoon winds originates due to the seasonal migration of planetary winds and pressure belts following the position of the Sun.
(a) Both (A) and (R) are true: R explains A
(b) Both (A) and (R) are true: R does not explain A
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(c) (A) is correct (R) is false

Question 4.
Assertion(A): Rajasthan remains as desert.
Reason(R): As Aravalli mountain is located parallel to the Arabian Sea branch South west Monsoon winds the western part (Rajasthan) do not receive much rainfall. To the Bay of Bengal Branch of South west Monsoon wind it is located on the leeward side.
(a) Both (A) and (R) are true: R explains A
(b) Both (A) and (R) are true: R does not explain A
(c) (A) is correct (R) is false
(d) (A) is false (R) is true
Answer:
(a) Both (A) and (R) are true: R explains A

IV. Choose the inappropriate answer

Question 1.
The pre-monsoon showers are known as mango showers in:
(a) Kerala and Karnataka
(b) Punjab and Haryana
(c) Hot weather season
(d) Helps quick ripening of mangoes
Answer:
(b) Punjab and Haryana

Question 2.
Tropical deciduous forests trees are of economic importance:
(a) Have valuable trees like teak, sal.
(b) To make fragnant oil, varnish, perfumes.
(c) Do not shed their leaves.
(d) Found in rainfall areas of 100 to 200 cm.
Answer:
(c) Do not shed their leaves.

Question 3.
Tidal forests are also known as swamp or delta forests.
(a) These forests are not prone to tidal waves.
(b) Known as mangrove forests.
(c) Occur in and around deltas, estuaries.
(d) Prone to tidal influences.
Answer:
(a) These forests are not prone to tidal waves.

Question 4.
Biosphere Reserves are protected areas of coastal and land environment.
(a) Indian Government has established 18 Biosphere Reserves.
(b) Do not protect areas of natural habitat.
(c) People are an integral component of the system.
(d) One or more National parks preserves along with buffer zone.
Answer:
(b) Do not protect areas of natural habitat.

V. Answer briefly

Question 1.
Name the classification of Natural vegetation.
Answer:
Natural Vegetation can be classified as:

1. Tropical Evergreen forests
2. Tropical deciduous forests
3. Tropical dry forests
4. Desert and semi-desert
5. Mountain forests
6. Alpine forests and
7. Tidal forests

Question 2.
Name the factors affecting the climate of India.
Answer:
Climate of India is affected by the factors of latitude, distance from the seas, Monsoon wind, relief features and jet stream.

Question 3.
The climate and weather conditions in India are governed by the three-atmosphere conditions. Name them.
Answer:

1. The Pressure and Surface winds.
2. Upper air Circulation
3. Western cyclonic disturbances and tropical conditions

Question 4.
What is ElNino? How does it affect the climate of India?
Answer:
ElNino is a complex global phenomena of weather that appears once eveiy five to ten years. It is a cause for the delay of onset of the southwest monsoon in India.

Question 5.
What is the Characteristic features of monsoon rains in India?
Answer:

1. The Monsoon rains are pulsating in nature
2. They can cause heavy rainfall in one part and drought in the other.
3. They are known for their uncertainties.

Question 6.
What is the role of Jet stream regarding the climate of India?
Answer:

1. The arrival and departure of monsoon winds is determined by Jet stream.
2. Cause tropical depressions both during South west monsoon and North east monsoon.
3. Plays a vital role in bringing western disturbances to India there by helping winter wheat cultivation.

Question 7.
What values are associated with the monsoon in India?
Answer:

• Monsoon acts as unifying bond for India.
• The unifying influence of the monsoon on the Indian Subcontinent is quite perceptible.
• The Seasonal alteration of the wind systems and the associated weather conditions provide a rhythmic cycle of seasons.
• The Indian landscape, its animal and plant life, its entire agricultural calender and the life of the people, including their festivities, revolve around this phenomenon.
• Year after year, people of India from North to South and from east to west, eagerly await the arrival of the monsoon. These monsoon winds binds the whole country by providing water to set the agricultural activities in motion.
• The rivers valleys which carry this water also unite as a single river valley unit.

Question 8.
Give a brief note on the Indian Board for Wild life:
Answer:

1. The Indian Board for wild life was constituted in 1952 to suggest means of protection, conservation and management of wild life to the Government of India.
2. The Government of India enacted wild life (protection) Act in 1972 with the objective of effectively protecting the wild life.

Question 9.
Explain the steps taken to conserve wildlife.
Answer:
The Wild Life Act Of India protects and conserves this grest heritage of nation. The first weak of October is observed as Wild life Week Of India.

Question 10.
Give the Full form of these abbreviations.

1. IBWL
2. CBD
3. ITCZ

Answer:

1. IBWL : Indian Board for Wild Life.
2. CBD: Convention on Biological Diversity.
3. ITCZ : Inter Tropical Convergence Zone.

Question 11.
Why does the coromandal coast face frequent cyclones?
Answer:
The north east monsoon winds by crossing the Bay of Bengal absorbs moisture and gives heavy rain to the coromandal coast. These are frequent cyclones formed in the Bay of Bengal and they cause heavy damage to life and property along the coromandal coast.

Question 12.
Give a brief description of the Mangrove forests.
Answer:

• The mangrove tidal forests are found in the areas of coasts influenced by tides. Hence, mangroves are the common varities with roots of the plants submerged under water. The deltas of the Ganga, the Mahanadi, the Krishna, the Godavari and the Kaveri are covered by such vegetation. In the Ganga-Brahmaputra delta sundari trees are found which provides durable hard timber.
• Palm, coconut, Keora agar also grow in some parts of the delta.
• Royal Bengal Tiger is the famous animal in these forests. Turtles, crocodiles, gharials and snakes are also found in these forests.

VI. Distinguish between

Question 1.
Winter/Cold weather and Summer/Hot weather season.
Solution:
Winter or Cold weather season:

1. In India,winter weather period is January-February.
2. The mean temperature increases from North to South.
3. Western Disturbances occurs in this season.
4. Generally fine weather and low tempertaure.

Summer or Hot weather season:

1. The summer season is from March- May.
2. The mean tempertaure increases from South to North.
3. Mango showers, Norwesters or Kalbaisakhi occurs in this season.
4. Generally hot and dry weather.

Question 2.
Heavy rainfall and less/low rainfall region.
Answer:
Heavy rainfall region:

1. These areas get annual rainfall of 200 cms and above.
2. Western coast, Assam, South Meghalaya are heavy rainfall areas.

Less or low rainfall region:

1. These areas get annual rainfall less than 100 cms.
2. Rajasthan, Punjab, Haryana, Western and South western parts of Uttar Pradesh, plateau region gets less rainfall.

Question 3.
Windward side and Leeward side.
Answer:
Windward side:

1. The slope of the mountain that lie on the path of rain bearing winds.
2. The windward side gets more rainfall.
3. Western slopes of Western Ghats – West coastal plains.

Leeward side:

1. The slope of the mountain that do not face the rain bearing winds.
2. The Leeward side gets less or no rainfall.
3. Eastern slopes of Western Ghats – Deccan Plateau.

Question 4.
Eastern and Western Himalayan forests.
Answer:
Eastern Himalayan forests:

1. The vegetation of this forest is’ evergreen as these forests receive more than 200cm rainfall.
2. Pine, Sal Fir, Oak, Laurel are some of the main trees of these forests.
3. These forests are found in the North Eastern states.

Western Himlayan forests:

1. The rainfall of this region is moderate and have varied vegetation regards to altitude.
2. Upto 900m altitude semi desert vegetation from 900m – 1800m chir tree and sal, at 1800m conifer trees.
3. These forests are found in the states of Jammu and Kashmir, Himachal Pradesh and UttaraKhand.

VII. Give reasons for the following topics

Question 1.
Kolkatta receives more rainfall than Bikaner in Rajasthan.
Answer:
Air near the coast has more moisture and greater potential to produce precipitation. Due to this fact Kolkatta near the coast receives more rainfall (119cm) than Bikaner (24cm) which is located in the interior part.

Question 2.
In India Conservation and management of Biodiversity is necessary why?
Answer:
Hunting, Poaching, deforestation and encroachments in the natural habitats have caused extinction of some species and many are in the danger of extinction. The role of wild life in maintaining the ecological balance made it necessary for conservation and management of Biodiversity.

Question 3.
Relief of India has a great bearing on climate.
Answer:
The elements of climate such as temperature, atmospheric pressure, direction of winds and the amount of rainfall are affected by the presence of relief features as they act as a climatic barrier like the Himalayas and Western Ghats.

VIII. Write in detail

Question 1.
What are the major determinants of climate of a place? Explain them.
Answer:
The major determinants of climate of a place are,

1. Latitude
2. Altitude
3. Distance from the seas
4. Monsoon winds
5. Relief features (Mountains)
6. Jet Streams

1. Latitude:
Latitudinally, India lies between 8°4’N and 37°6’N latitudes. The Tropic of cancer divides the country into two equal halves. The area located to the south of Tropic of cancer experiences high temperature and no severe cold season throughout the year whereas, the areas to the north of this parallel enjoys subtropical climate. Here, summer temperature may rise above 40°C and it is close to freezing point during winter.

2. Altitude:
As one goes from the surface of the earth to higher altitudes, the atmosphere becomes less dense and temperature decreases. The hills are therefore cooler during summer.

3. Distance from the seas:
Distance from the sea does not cause only temperature and pressure variations but also affects the amount of rainfall. As the distance from the sea increased its moderating influence decreases and the people experience extreme weather conditions. This condition is known as continentality.(i.e.) Very hot during summers and very cold during winter.

4. Monsoon winds:
The most dominant factor which affects the climate of India is the monsoon winds. These are seasonal reversal winds and India remains in the influence of these winds for a considerable part of a year. Though, the sun’s rays are vertical over the central part of India during the mid-June, the summer season ends in India by the end of May. It is because the onset of southwest monsoon brings down the temperature of the entire India and causes moderate to heavy rainfall in many parts of the country. Similarly, the climate of southeast India is also influenced by northeast monsoon.

5. Relief features (Mountains):
Relief of India has a great bearing on major elements of climate such as temperature, atmospheric pressure, direction of winds and the amount of rainfall. High mountains act as barriers for cold or hot winds; they may also cause precipitation if they are high enough and lie in the path of rain-bearing winds. The leeward side of mountains remains dry.

6. Jet Streams:
Jet streams are the fast moving winds blowing in a narrow zone in the upper atmosphere. According to the Jet stream theory, the onset of southwest monsoon is driven by the shift of the sub tropical westerly jet from the plains of India towards the Tibetan plateau. The easterly jet streams cause tropical depressions both during southwest monsoon and retreating monsoon.

Question 2.
Write about the factors that affect the climate of India.
Answer:
Climate of India is affected by the factors of latitude, altitude, distance from the seas, monsoon winds and jet streams.

Latitude:

1. Latitudinally India is located in the Tropical belt between 8°4’N and 37°6’N latitudes.
2. The tropic of cancer divides the country into two equal halves.
3. This enables the places south of Tropic of cancer (23°30’N) to experience high temperature and no severe cold.
4. The places north of Tropic of cancer enjoy sub tropical climate.

Altitude:

1. When the altitude increases, temperature decreases as per the phenomena. “Normal Lapse rate”. For every 1000 mts of ascent the temperature decreases at the rate of 6.5°C.
2. Hence the places in the mountains are cooler than the places on the plains.
3. Even in South India places at higher altitude have cool climate.

Distance from the sea:

1. Distance from the sea not only causes temperature and pressure variations but also affects the amount of rainfall.
2. The entire area of peninsular region as it is surrounded by seas on three sides having the temperature equable throughout the year.
3. Areas of central and north India experiences much seasonal variation in temperature due to the absence of seas.

Monsoon winds:

1. The most dominant factor which affects the climate of India is the. monsoon winds.
2. India is influenced by the South west monsoon and North east monsoon from June to September and October, November respectively.
3. Indian agriculture is at the mercy of monsoons.
4. India experiences Tropical monsoon climate.

Relief:

1. Relief of India has a great bearing on major elements of climate such as temperature, atmospheric pressure direction of wind and amount of rainfall.
2. The mighty Himalayas in the north act as a climatic barrier preventing the freezing cold winds from central Asia away from India.
3. The Western ghats act as the barrier for south west monsoon winds causing rainfall to the western slopes, and the eastern slopes remain as rain shadow region.

Jet Stream:

1. Jet streams are the fast moving winds blowing in the narrow zone of upper layers of atmosphere.
2. These air currents play a dominant role in bringing the onset of South west monsoon and withdrawal of monsoon winds.
3. More over they are the main reason in bringing the Western Disturbances to the North Western part of Punjab and Haryana causing rainfall and enable the cultivation of winter wheat.
4. It causes snow fall in Jammu and Kashmir region.

Thus the Tropical monsoon = climate of India revolve around these factors

IX. Map Work

Mark the following on the outline map of India.

Question 1.
Areas of low rainfall
Answer:

Question 2.
Any 4 Biosphere Reserve
Answer:

Question 3.
Any 5 National parks
Answer:

Question 4.
Any 3 Wild animal sanctuary
Answer:

Question 5.
Any 2 Bird Sanctuary
Answer:

Question 6.
Top 5 states having maximum forest cover
Answer:

Question 7.
Top 5 states where forest cover is decreased
Answer:

Students can Download Tamil Nadu 12th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Physics Model Question Paper 4 English Medium

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• 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
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 x 1 = 15]

Question 1.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Question 2.
The work done in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is  ………….
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) Zero
(c)  \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}} \)
(d) infinite
Answer:
(b) Zero
Hint: The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in a electric field.

Question 3.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is……………………..
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Question 4.
An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor when the plates of the capacitor are kept under constant magnetic field of induction \(\overrightarrow{\mathrm{B}}\) is ……….


Answer:
(d)

Question 5.
In a series resonant RLC circuit, the voltage across 100Ω resistor is 40 V. The resonant frequency co is 250 rad/s. If the value of C is 4 pF, then the voltage across L is……………….
(a) 600 V
(b) 4000 V
(c) 400 V
(d) IV
Answer:
(c) 400 V

Question 6.
During the propagation of electromagnetic waves in a medium:
(a) electric energy density is double of the magnetic energy density
(b) electric energy density is half of the magnetic energy density
(c) electric energy density is equal to the magnetic energy density
(d) both electric and magnetic energy densities are zero
Answer:
(c) electric energy density is equal to the magnetic energy density

Question 7.
First diffraction minimum due to a single slit of width 1.0 x 10^-5 cm is at 30°. Then wavelength of light used is, …………………
(a) 400 Å
(b) 500 Å
(c) 600 Å
(d) 700 Å
Answer:
(b) 500 Å
Hint. For diffraction minima, d sin θ = nλ
\(\lambda=\frac{d \sin \theta}{n}=\frac{1 \times 10^{-5} \times 10^{-2} \times \sin 30^{\circ}}{1}=0.5 \times 10^{-7}\)
λ = 500 Å

Question 8.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light

Question 9.
Kinetic energy of emitted electron depends upon
(a) frequency
(b) intensity
(c) nature of atmosphere surrounding the electron
(d) none of these
Answer:
(a) frequency
Hint: Kinetic energy of emitted electron depends on the frequency of incident radiation.

Question 10.
The ratio between the first three orbits of hydrogen atom is…………….
(a) 1:2:3
(b) 2:4:6
(c) 1:4:9
(d) 1:3:5
Answer:
(c) 1:4:9
Hint :
E_n = \(\frac{-13.6 \times z^{2}}{n^{2}}\)
n = 1; E₁ =- 13.6 eV/ atom
n = 2; E₂ = – 3.4 eV/ atom
n = 3; E₃ = -1.51 eV/atom                             ’
The ratio of three orbits E₁ : E₂ : E₃ = 13.6 : 3.4 : 1.51 = 1 : 4 : 9

Question 11.
Bohr’s theory of hydrogen atom did not explain fully
(a) diameter of H-atom
(b) emission spectra
(c) ionisation energy
(d) the fine structure of even hydrogen spectrum
Answer:
(d) the fine structure of even hydrogen spectrum
Hint: Bohr theory could not explain the five structure of hydrogen spectrum.

Question 12.
If a half-wave rectified voltage is fed to a load resistor, which part of a cycle the load current will flow?
(a) 0° – 90°
(b) 90° – 180°
(c) 0° – 180°
(d) 0° – 360°
Answer:
(c) 0° – 180°

Question 13.
Diamond is very hard because ……………..
(a) it is covalent solid
(b) it has large cohesive energy
(c) high melting point
(d) insoluble in all solvents
Answer:
(b) it has large cohesive energy

Question 14.
The internationally accepted frequency deviation for the purpose of FM broadcasts.
(a) 75 kHz
(b) 68 kHz
(c) 80 kHz
(d) 70 kHz
Answer:
(a) 75 kHz

Question 15.
The blue print for making ultra durable synthetic material is mimicked from
(a) Lotus leaf
(b) Morpho butterfly
(c) Parrot fish
(d) Peacock feather
Answer:
(c) Parrot fish

Part – II

Answer any six questions in which Q. No 17 is compulsory.   [6 x 2 = 12]

Question 16.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines  is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 17.
Determine the number of electrons flowing per second through a conductor, when a current of 32 A flows through it.
Answer:
I = 32A, t= 1 s
Charge of an electron, e = 1.6 x 10^-19 C
The number of electrons flowing per second, n =?

Question 18.
Define magnetic flux.
Answer:
The number of magnetic field lines crossing per unit area is called magnetic flux φ_B
\(\phi_{\mathrm{B}}=\overrightarrow{\mathrm{B}} \cdot\overrightarrow{\mathrm{A}}=\mathrm{B} \mathrm{A} \cos \theta=\mathrm{B} \perp \mathrm{A}\)

Question 19.
Give any one definition of power factor.
Answer:
The power factor is defined as the ratio of true power to the apparent power of an a.c. circuit.
It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit.

Question 20.
State the laws of reflection.
Answer:

• The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
• The angle of incidence i is equal to the angle of reflection r. i = r

Question 21.
Why do metals have a large number of free electrons?
Answer:
In metals, the electrons in the outer most shells are loosely bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving inside the metal in a random manner.

Question 22.
The radius of the 5th orbit of hydrogen atom is 13.25 A. Calculate the wavelength of the electron in the 5th orbit.
Answer:
2πr = nλ
2 x 3.14 x 13.25 Å = 5 x λ
.’. λ = 16.64 Å

Question 23.
Draw the output waveform of a full wave rectifier.
Answer:

Question 24.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Part – III

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

Question 25.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge.
This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Question 26.
What is electric power and electric energy?
Answer:
Electric power: It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
\({ P }=\frac { { W } }{ t } ={ VI }={ I }^{ 2 }{ R }=\frac { { V }^{ 2 } }{ { R } } \)
Electric energy: It is the total workdone in maintaining an electric current in an electric circuit for a given time.
W = Vt = VIr joule = I²R? joule.

Question 27.
A bar magnet having a magnetic moment \(\overrightarrow{\mathrm{M}}\) is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Answer:
Consider a bar magnet of magnetic moment \(\overrightarrow{\mathrm{M}}\). When a bar magnet first cut in two pieces along the axis, their magnetic moment is \(\frac{\overrightarrow{\mathrm{M}}}{2}\)

Question 28.
The current in an inductive circuit is given by 0.3 sin (200t – 40°) A. Write the equation for the voltage across it if the inductance is 40 mH.
Answer:
L = 40 x 10^-3H; i = 0.1 sin (200 t – 40°); X_L = ωL = 200 x 40 x 10^-3 = 8 Ω
V_m = I_m X_L = 0.3 x 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90° Therefore,
υ = V_m sin (ωt + 90°)
υ = 2.4 sin (200f – 40°+ 90°)
υ = 2.4 sin (200f +50°)volt

Question 29.
Writedown the integral form of modified Ampere’s circuital law.
Answer:
This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.

Question 30.
Two light sources have intensity of light as I_o. What is the resultant intensity at a point where the two light waves have a phase difference of π/3?
Answer:
Let the intensities be I_o
The resultant intensity is, I = 4 I_o cos² (φ/2)
Resultant intensity when, ϕ = π/ 3, is I = 4I₀ cos² (π / 6) I
= 4I₀(√3 / 2)² =3I₀

Question 31.
Write the properties of cathode rays.
Answer:

• Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 10⁷m s^-1
• It can be deflected by application of electric and magnetic fields.
• When the cathode rays are allowed to fall on matter, they produce heat.
• They affect the photographic plates and also produce fluorescence when they fall on   certain crystals and minerals.
• When the cathode rays fall on a material of high atomic weight, x-rays are produced.
• Cathode rays ionize the gas through which they pass.
• The speed of cathode rays is up \(\left( \frac { 1 }{ 10 } \right) ^{ th }\) of the speed of light.

Question 32.
Distinguish between wireline and wireless communication.
Answer:

Question 33.
What is the difference between Nano materials and Bulk materials?
Answer:

• The solids are made up of particles. Each of the particle has a definite number of atoms, which might differ from material to material. If the particle of a solid is of size less than 100 nm, it is said to be a ‘nano solid’.
• When the particle size exceeds 100 nm, it is a ‘bulk solid’. It is to be noted that nano and bulk solids may be of the same chemical composition.
• For example, ZnO can be both in bulk and nano form.
• Though chemical composition is the; same, nano form of the material shows strikingly different properties when compared to its bulk counterpart.

Part – IV

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

Question 34.
(a) Calculate the electric field due to a dipole on its equatorial plane.
Answer:
Electric field due to an electric dipole at a point on the equatorial plane. Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure.

Since the point C is equi­distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of \(\overrightarrow{\mathrm{E}}_{+}\) is along BC \(\overrightarrow{\mathrm{E}}_{-} \) and the direction of E is along CA. \(\overrightarrow{\mathrm{E}}_{+} \) and \(\overrightarrow{\mathrm{E}}_{-} \) and E are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(\left|\overrightarrow{\mathrm{E}}_{+}\right| \sin \theta \) and \(\left|\overrightarrow{\mathrm{E}}_{-}\right| \sin \theta \) are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of E₊ and E and its direction is \(\overrightarrow { { E } } _{ + }\quad and\quad \overrightarrow { { E } } _{ – }\) and its direction is along \(-\hat{p}\)

(b) How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer: To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bf and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ₁ and ξ₂ and to be compared are connected to the terminals M₁, N₁ and M₂, N₂, of the DPDT switch. The positive terminals of Bf ξ₁ and ξ₂ should be connected to the same end C.

The DPDT switch is pressed towards M₁, N₁ so that cell ξ₁, is included in the secondary circuit and the balancing length l₁, is found by adjusting the jockey for zero deflection. Then the second cells ξ₂, is included in the circuit and the balancing length l₂, is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have
ξ₁ = Irl₁ ……………….. (1)
ξ₂ = Irl₂ ……………….. (2)
By dividing equation (1) by (2)
\(\frac{\xi_{1}}{\xi_{2}}=\frac{l_{1}}{l_{2}}\)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Question 35.
(a) Discuss the working of cyclotron in detail.
Answer:
Cyclotron : Cyclotron is a device used accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle : When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.

Construction : The particles are allowed to move in between two semicircular metal containers called Dees (hollow D – shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working: Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.

\(\frac { mv^{ 2 } }{ r } =qv{ B }\Rightarrow r=\frac { m }{ q{ B } } v\Rightarrow r\propto v\)
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the Very important condition in cyclotron operation is the resonance condition. It happens when the frequency { at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc From equation
\(f_{\text {osc }}=\frac{q \mathrm{B}}{2 \pi m} \Rightarrow \mathrm{T}=\frac{1}{f_{\text {osc }}}\)
The time Period of oscillation is \(\mathrm{T}=\frac{2 \pi m}{q \mathrm{B}}\)
The kinetic energy of the charged particle is \(\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{q^{2} \mathrm{B}^{2} r^{2}}{2 m}\)

Limitations of cyclotron

• the speed of the ion is limited
• electron cannot be accelerated
• uncharged particles cannot be accelerated

[OR]

(b) Give the uses of Foucault current.
Answer:
Though the production of eddy current is undesirable in some cases, it is useful in some other cases. A few of them are

(1) Induction stove : Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone. When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan. The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.

(2) Eddy current brake : This eddy current braking system is generally used in high speed trains and roller coasters. Strong electromagnets are fixed just above the rails. To stop the train, electromagnets are switched on. The magnetic field of these magnets induces eddy currents in the rails which oppose or resist the movement of the train. This is Eddy current linear brake.

In some cases, the circular disc, connected to the wheel of the train through a common shaft, is made to rotate in between the poles of an electromagnet. When there is a relative motion between the disc and the magnet, eddy currents are induced in the disc which stop the train. This is Eddy current circular brake.

(3) Eddy current testing : It is one of the simple non-destructive testing methods to find defects like surface cracks and air bubbles present in a specimen. A coil of insulated wire is given an alternating electric current so that it produces an alternating magnetic field. When this coil is brought near the test surface, eddy current is induced in the test surface. The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.

(4) Electro magnetic damping : The armature of the galvanometer coil is wound on a soft iron cylinder. Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder. The damping force due to the flow of eddy current brings the armature to rest immediately and then galvanometer shows a steady deflection. This is called electromagnetic damping.

Question 36.
(a) Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:

• Electromagnetic waves are produced by any accelerated charge.
• Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.
• Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.
• Electromagnetic waves travel with speed which is equal to the speed of light in vacuum
  or free space, \(c=\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } } =3\times 10^{ 8 }{ ms }^{ -1 }\)
• The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

[OR]

(b) Explain the Young’s double slit experimental setup and obtain the equation for path difference.
Answer:
Experimental setup
Wavefronts from S₁ and S₂ spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.

S₁ and S₂ travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.

The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.

Equation for path difference
Let d be the distance between the double slits S₁ and S₂ which act as coherent sources of wavelength λ. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S₁ and S₂ is C and the midpoint of the screen O is equidistant from S₁ and S₂. P is any point at a distance y  from O.

The waves from S₁ and S₂ meet at P either inphase or out-of-phase depending upon the path difference between the two waves. The path difference S between the light waves from S₁ and S₂ to the point P is, δ = S₂P and S₁P
A perpendicular is dropped from the point S₁, to the line S₂P at M to find the path difference more precisely.
δ = S₂P – MP = S₂M
The angular position of the point P from C is θ. ∠OCP = θ.
From the geometry, the angles ∠OCP and ∠S₂S₁ M are equal.
∠OCP = ∠S₂S₁ M = θ
In right angle triangle ΔS₁S₂M, the path difference, S₂M = d sin θ
δ = d sin θ
If the angle θ is small, sin θ ≈ tan θ ≈ θ From the right angle triangle ΔOCP, tan θ = y/D. The path difference, δ =dy/D

Question  37.
(a) Give the construction and working of photo emissive cell.
Answer:
Photo emissive cell:
Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.

• It consists of an evacuated glass or quartz bulb in which two metallic electrodes – that is, a cathode and an anode are fixed.
• The cathode C is semi-cylindrical in shape and is coated with a photo sensitive material. The anode A is a thin rod or wire kept along the axis of the semi-cylindrical cathode.
• A potential difference is applied between the anode and the cathode through a galvanometer G.

Working:

• When cathode is illuminated, electrons are emitted from it. These electrons are attracted by anode and hence a current is produced which is measured by the galvanometer
• For a given cathode, the magnitude of
  the current depends on the intensity to incident radiation and
  the potential difference between anode and cathode.

[OR]

Question 37.
(b) Explain the J.J. Thomson experiment to determine the specific charge of electron.
Answer:
In 1887, J. J. Thomson made remarkable improvement in the scope of study of gases in discharge tubes. In the presence of electric and magnetic fields, the cathode rays are deflected. By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.

A highly evacuated discharge tube is used and cathode rays (electron beam) produced at cathode are attracted towards anode disc A. Anode disc is made with pin hole in order to allow only a narrow beam of cathode rays. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage. Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed. This is achieved by coating the screen with zinc sulphide.

(i) Determination of velocity of cathode rays
For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O. This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field. Let e be the charge of the cathode rays, then eE = eBv
\(v=\frac{E}{B}\) …………….. (1)

(ii) Determination of specific charge
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
\(e \mathrm{V}=\frac{1}{2} m v^{2} \Rightarrow \frac{e}{m}=\frac{v^{2}}{2 \mathrm{V}}\)
Substituting the value of velocity from equation (1) , we get
\(\frac{e}{m}=\frac{1}{2 \mathrm{V}} \frac{\mathrm{E}^{2}}{\mathrm{B}^{2}}\) ………..(2)
Substituting the value of E ,B and V, the specific charge vam be determined as
\(\frac{e}{m}=1.7 \times 10^{11} \mathrm{Ckg}^{-1}\)

(iii) Deflection of charge only due to uniform electric field
When the magnetic field is turned off, the deflection is only due to electric field. The
deflection in vertical direction is due to the electric force.
F_e = eE ………………….. (3)
Let m be the mass of the electron and by applying Newton’s second law of motion,
acceleration of the electron is
\(a_{e}=\frac{1}{m} \mathrm{F}_{e}\) ………………… (4)
Substituting equation (4) in equation (3),
\(a_{e}=\frac{1}{m} e E=\frac{e}{m} E\)
Let y be the deviation produced from original position on the screen. Let the initial upward velocity of cathode ray be μ = 0 before entering the parallel electric plates. Let l be the time taken by the cathode rays to travel in electric field. Let l be the length of one of the plates, then the time taken is
\(t=\frac{l}{v}\) …………(5)

Hence, the deflection y’ of cathod rays is( note : u = 0 and \(a_{e}=\frac{e}{m} \mathrm{E}\))

Therefore, the deflection y on the screen is
y α y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get

Substituting the values on RHS, the value of specific charge is calculated as
e/m = 1.7 x 10¹¹Ckg^-1

Question 38.
(a) What is an LED? Give the principle of operation with a diagram.
Answer:
Light Emitting Diode (LED): LED is a p-n junction diode which emits visible or invisible light when it is forward biased. Since, electrical energy is converted into light energy, this process is also called electroluminescence. The cross-sectional view of a commercial LED is shown in figure (b). It consists of a p-layer, n-layer and a substrate. A transparent window is used to allow light to travel in the desired direction. An external resistance in series with the biasing source is required to limit the forward current through the LED. In addition, it has two leads; anode and cathode.

When the p-n junction is forward biased, the conduction band electrons on n-side and valence band holes on p-side diffuse across the junction. When they cross the junction, they become excess minority carriers (electrons in p-side and holes in n-side). These excess minority carriers recombine with oppositely charged majority carriers in the respective regions, i.e. the. electrons in the conduction band recombine with holes in the valence band as shown in the figure (c). During recombination process, energy is released in the form of light (radiative) or heat (non-radiative). For radiative recombination, a photon of energy hv is emitted. For non- radiative recombination, energy is liberated in the form of heat.

The colour of the light is determined by the energy band gap of the material. Therefore, LEDs are available in a wide range of colours such as blue (SiC), green (AlGaP) and red (GaAsP). Now a days, LED which emits white light (GalnN) is also available.

[OR]

(b) Give the applications of ICT in mining and agriculture sectors.
Answer:
(i) Agriculture
The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the Challenges and risk factors.

• ICT is widely used in increasing food productivity and farm management.
• It helps to optimize the use of water, seeds and fertilizers etc.
• Sophisticated technologies that include robots, temperature and moisture sensors, aerial images, and GPS technology can be used.
• Geographic information systems are extensively used in farming to decide the suitable place for the species to be planted.

(ii) Mining

• ICT in mining improves operational efficiency, remote monitoring and disaster locating system.
• Information and communication technology provides audio-visual warning to the trapped underground miners.
• It helps to connect remote sites.