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Students can download 10th Science Chapter 14 Transportation in Plants and Circulation in Animals 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 14 Transportation in Plants and Circulation in Animals

Samacheer Kalvi 10th Science Transportation in Plants and Circulation in Animals Text Book Back Questions and Answers

I. Choose the best answer:

Question 1.
Active transport involves:
(a) movement of molecules from lower to higher concentration
(b) expenditure of energy
(c) it is an uphill task
(d) all of the above
Answer:
(d) all of the above

Question 2.
Water which is absorbed by roots is transported to aerial parts of the plant through ______.
(a) cortex
(b) epidermis
(c) phloem
(d) xylem.
Answer:
(d) xylem.

Question 3.
During transpiration there is loss of:
(a) carbon dioxide
(b) oxygen
(c) water
(d) none of the above
Answer:
(c) water

Question 4.
Root hairs are ______.
(a) cortical cell
(b) projection of the epidermal cell
(c) unicellular
(d) both b and c.
Answer:
(b) projection of the epidermal cell

Question 5.
Which of the following process requires energy?
(a) active transport
(b) diffusion
(c) osmosis
(d) all of them
Answer:
(a) active transport

Question 6.
The wall of the human heart is made of ______.
(a) Endocardium
(b) Epicardium
(c) Myocardium
(d) All of the above.
Answer:
(d) All of the above.

Question 7.
Which is the sequence of correct blood flow?
(a) ventricle – atrium – vein – arteries
(b) atrium – ventricle – veins – arteries
(c) atrium – ventricle – arteries – vein
(d) ventricles – vein – atrium – arteries
Answer:
(c) atrium – ventricle – arteries – vein

Question 8.
A patient with blood group O was injured in an accident and has blood loss. Which blood group the doctor should effectively use for transfusion in this condition?
(a) O group
(b) AB group
(c) A or B group
(d) all blood group.
Answer:
(d) all blood group.

Question 9.
‘Heart of heart’ is called:
(a) SA node
(b) AV node
(c) Purkinje fibres
(d) Bundle of His
Answer:
(a) SA node

Question 10.
Which one of the following regarding blood composition is correct?
(a) Plasma – Blood + Lymphocyte
(b) Serum – Blood + Fibrinogen
(c) Lymph – Plasma + RBC + WBC
(d) Blood – Plasma + RBC + WBC + Platelets.
Answer:
(d) Blood – Plasma + RBC+ WBG + Platelets.

II. Fill in the blanks:

1. ……….. involves evaporative loss of water from aerial parts.
2. Water enters the root cell through a ………. plasma membrane.
3. Structures in roots that help to absorb water are …………
4. Normal blood pressure is …………
5. The normal human heartbeat rate is about ……….. time per minute.

Answer:

1. Transpiration
2. Semipermeable
3. Root hairs
4. 120 mm / 80 mm Hg
5. 72 – 75 times

III. Match the following:

Section – I

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

Section – II

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

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

1. The phloem is responsible for the translocation of food.
2. Plants lose water by the process of transpiration.
3. The form of sugar transported through the phloem is glucose.
4. In apoplastic movement the water travels through the cell membrane and enter the cell.
5. When the guard cells lose water the stoma opens.
6. Initiation and stimulation of heart beat take place by nerves.
7. All veins carry deoxygenated blood.
8. WBC defend the body from bacterial and viral infections.
9. The closure of the mitral and tricuspid valves at the start of the ventricular systole produces the first sound ‘LUBB’.

Answer:

1. True
2. True
3. False – The form of sugar transported through the phloem is sucrose.
4. False – Apoplastic movement does not involve crossing the cell membrane.
5. False – When the guard cells lose water the stoma closes.
6. False – Initiation and stimulation of heart beat take place by Sino-atrial node.
7. False – All veins carry deoxygenated blood except the pulmonary veins,
8. True
9. True.

V. Answer in a word or sentence:

Question 1.
Name two layered protective covering of human heart.
Answer:
Pericardium is the two layered protective covering of human heart.

Question 2.
What is the shape of RBC in human blood?
Answer:
The shape of RBC in human blood is biconcave and disc – shaped.

Question 3.
Why is the colour of the blood red?
Answer:
The blood is red in colour due to the presence of respiratory pigment – haemoglobin.

Question 4.
Which kind of cells are found in the lymph?
Answer:
Lymphocytes in the lymph, which defend the body from infection.

Question 5.
Name the heart valve associated with the major arteries leaving the ventricles.
Answer:
Aortic valve present at the base of aorta.

Question 6.
Mention the artery which supplies blood to the heart muscle.
Answer:
Coronary arteries.

VI. Short Answer Questions.

Question 1.
What causes the opening and closing of guard cells of stomata during transpiration?
Answer:
The general mechanism of stomatal movement is based on entry and exit of water molecules in guard cells. When the turgidity increases within the guard cells the stomata open. When the guard cells lose water it becomes flaccid and the stomata closes.

Question 2.
What is cohesion?
Answer:
The force of attraction between molecules of water is called Cohesion.

Question 3.
Trace the pathway followed by water molecules from the time it enters a plant root to the time it escapes into the atmosphere from a leaf.
Answer:
Soil → root hair → cortex → endodermis → Pericycle → xylem → stem → leaves.

Question 4.
What would happen to the leaves of a plant that transpires more water than its absorption in the roots?
Answer:
When transpiration exceeds than water absorption by roots, the plant dehydrates. It affects plant processes such as growth, Photosynthesis and transpiration.

Question 5.
Describe the structure and working of the human heart.
Answer:
The heart is thick muscular organ located in the thoracic cavity. It is covered by a two layered sac called pericardium. The heart is four chambered with two auricles and two ventricles. Right Auricle receives deoxygenated blood from superior and inferior vena cava and passes to right ventricles from right ventricles impure blood passes to pulmonary artery and reaches the lungs.

Question 6.
Why is the circulation in man referred to as double circulation?
Answer:
The blood circulates twice, through the heart in one complete cycle, in the circulation of blood in man. So it is called double circulation.

Question 7.
What are heart sounds? How are they produced?
Answer:
‘Lubb’ and ‘Dupp’ are the sounds of the heart. ‘Lubb’ sound is produced by the ventricle- contraction. It arises due to closing of mitral and tricuspid valve.
‘Dupp’ is produced by ventricular diastole. This arises due to the closing of the semi-lunar valves of two auricles.

Question 8.
What is the importance of valves in the heart?
Answer:
The valves are the muscular flaps, that regulate the flow of blood in a single direction and prevent backflow of blood.

Question 9.
Who discovered Rh factor? Why was it named so?
Answer:
Landsteiner and Wiener discovered Rh factor of blood in 1940. Rh factor is a protein CD antigen present on the surface of the red blood cells in majority of humans. This protein is similar to the protein present in Rhesus monkey, hence the term Rh.

Question 10.
How are arteries and veins structurally different from one another?
Answer:

Question 11.
Why is the Sinoatrial node called the pacemaker of the heart?
Answer:
Sinoatrial node is called the pacemaker of the heart because it is capable of initiating impulse, which can stimulate the heart muscles to contract.

Question 12.
Differentiate between systemic circulation and pulmonary circulation.
Answer:

Question 13.
The complete events of the cardiac cycle last for 0.8 sec. What is the timing for each event?
Answer:

1. Atrial Systole: Contraction of auricles (0.1 sec)
2. Ventricular Systole: Contraction of ventricles (0.3 sec)
3. Ventricular diastole: Relaxation of Ventricles (0.4 sec).

VII. Give reasons for the following statements:

Question 1.
Minerals cannot be passively absorbed by the roots.
Answer:
All minerals cannot be passively absorbed by the roots because, minerals are present in the soil as charged particles that cannot move across the cell membranes and the concentration of minerals in the soil is usually lower than the. concentration of minerals in the root. So, the minerals enter the root by active absorption through the cytoplasm of epidermal cells which needs energy.

Question 2.
Guard cells are responsible for opening and closing of stomata.
Answer:
Guard cells are responsible for the opening and closing of stomata because, during transpiration, the movement of ions (Potassium) in and out of the guard cells causes the opening and closing of Stomata. When the water moves inside the guard cells, causing them to swell up and become turgid making the stomata open. When guard cells cause water to move out of the cell and make guard cells shrunk and stomata pores close.

Question 3.
The movement of substances in the phloem can be in any direction.
Answer:
Phloem transports food from the source to sink. The source is part in which plant synthesize food, sink is the part that needs or stores food. Since the source-sink relationship is variable, the direction of movement in the Phloem can be upwards or downwards, i.e., bidirectional.

Question 4.
Minerals in the plants are not lost when the leaf falls.
Answer:
Minerals like Phosphorus, Sulphur, Nitrogen and Potassium are remobilised in the soil from older dying leaves to younger leaves. This phenomenon is seen in deciduous plants. So minerals in the plants are not lost, when the leaf falls.

Question 5.
The walls of the right ventricle are thicker than the right auricles.
Answer:
Because the right ventricle have to pump out the deoxygenated blood with force away from the heart through pulmonary artery to lungs.

Question 6.
Mature RBC in mammals does not have cell organelles.
Answer:
Mature RBC in mammals does not have cell organelles because

• The lack of Nucleus in RBC makes the cells, biconcave and disc – shaped. RBC involved in the transport of oxygen from lungs to tissues.
• They do not have cell organelles in order to accommodate maximum space for haemoglobin.
• The loss of endoplasmic reticulum allows more flexibility, for RBC to move through narrow capillaries.

VIII. Long Answer Questions:

Question 1.
How do plants absorb water? Explain.
Answer:
Water present in the soil must reach the xylem of roots. Root hair is in contact with soil water. Their cell wall is thin and water easily diffuses in the passage of water from the soil to leaf is
Soil water → Root hair → Epidermis → Cortex → Endodermis → Pericycle → Xylem → Stem and leaf.
Once water is absorbed by the root hairs, it can move deepers into root layers by two pathways. Apoplast and Symplast.

1. Apoplast : This is the non living path in plants. It occurs through the intercellular spaces and walls of the cells. This movement dependent on the gradient.
2. Symplast : This is the living passage. The movement of water from cell to cell through plasmodesmato and cytoplasm, Movement is again down a potential gradient.

Question 2.
What is transpiration? Give the importance of transpiration.
Answer:
Transpiration is the evaporation of water in plants through stomata in the leaves.
Importance of Transpiration

• Creates transpirational pull for the transport of water.
• Supplies water for photosynthesis.
• Transports minerals from the soil to all parts of the plant.
• Cools the surface of the leaves by evaporation.
• Keeps the cells turgid; hence, maintains their shape.

Question 3.
Why are leucocytes classified as granulocytes and agranulocytes? Name each cell and mention its functions.
Answer:
Depending on the presence and absence of granules, leucoytes are divided into two types Granulocytes and Agranulocytes.
Granulocytes : They are characterised by the presence of granules in cytoplasm. The granulocytes include neutrophils, eosinophils and basophils.

1. Neutrophils : They constitute about 60-65% of total WBC’s. They are large and have many lobed nucleus. They are phagocytic in nature and appear in large,number in and around the infected tissue.
2. Eosinophils : They have bilobed nucleus and constitute about 2-3% of total WBC’s. Eosinophil increase during certain types of parasitic infection and allergic reaction. It bring about detoxification of toxin.
3. Basophils: Basophil have lobed nucleus. They form 0.5 – 1.0% of the total leucocytes. They release chemicals during the process of inflammation.

Agranulocytes : They are characterised by the absence of granules in the cytoplasm. These are of two types Lymphocytes and Monocytes.

1. Lymphocytes: These are about 20-25% of the total leucocytes. They produce antibodies during bacterial and viral infections.
2. Monocytes: They are the largest leucocytes and amoeboid in shape. These cells form 5-6% of the total leucocytes. They are phagocytic and can engulf bacteria.

Question 4.
Differentiate between systole and diastole. Explain the conduction of the heartbeat.
Answer:
One complete contraction (Systole) and relaxation (diastole) of the atrium and ventricles of the heart constitutes Heartbeat. The human heart is myogenic. Contraction is initiated by the sino – atrial (SA) node, which is situated in the wall of the right atrium, near the opening of the superior vena cava. SA node is broader at the top and tapering below and made up of thin fibres.

SA node acts as the ‘pacemaker’ of the heart because it is capable of initiating impulse which can stimulate the heart muscles to contract. The impulse spreads like a wave of contraction over the right and left atrial wall, pushing the blood, through the atrioventricular valves into the ventricles and the same wave from SA node reaches the atrioventricular node (AV), to emit an impulse of contraction, spreading to the ventricular muscle, through the atrioventricular bundle and the Purkinje fibres. The expansion of the artery every time, the blood is forced into the arteries, is called pulse. Normal pulse rate ranges from 70 – 90 / min.

Question 5.
Enumerate the functions of blood.
Answer:

1. Transport of respiratory gases (Oxygen and CO₂).
2. Transport of digested food materials to the different body cells.
3. Transport of hormones.
4. Transport of nitrogenous excretory products like ammonia, urea and uric acid.
5. It is involved in protection of the body and defense against diseases.
6. It acts as buffer and also helps in regulation of pH and body temperature.
7. It maintains proper water balance in the body.

IX. Assertion and Reasoning:

Direction: In each of the following questions a statement of assertion (A) is given and a corresponding statement of reason (R) is given just below it. Mark the correct statement as.
(a) If both A and R are true and R is correct explanation of A.
(b) If both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) Both A and R are false.
1. Assertion: RBC plays an important role in the transport of respiratory gases.
Reason: RBC do not have cell organelles and nucleus.
Answer:
(a) If both A and R are true and R is correct explanation of A.

2. Assertion: Persons with AB blood group are called an universal recipients, because they can receive blood from all groups.
Reason: Antibodies are absent in persons with AB blood group.
Answer:
(a) If both A and R are true and R is correct explanation of A.

X. Higher Order Thinking Skills (HOTS):

Question 1.
When any dry plant material is kept in water, they swell up. Name and define the phenomenon involved in this change.
Answer:
The process is imbibition’. The type of diffusion in which a solid absorbs water and gets swelled up is Imbibition.

Question 2.
Why are the walls of the left ventricle thicker than the other chambers of the heart?
Answer:
The walls of the left ventricle are about three times thicker than the right ventricle. The left ventricle gives rise to the aorta, which carries oxygenated blood, to various organs of the body. The ventricle walls are thick because they have to pump out blood with force away from the heart.

Question 3.
Doctors use stethoscope to hear the sound of the heart. Why?
Answer:
A stethoscope is an instrument used to detect the sound produced by the internal organs of human body. It is an useful diagnostic tool to identify and localize health problems and diagnose disease.

Question 4.
How do the pulmonary artery and pulmonary vein differ in their function when compared to a normal artery and vein?
Answer:

• All arteries carry oxygenated blood, except pulmonary arteries, which carry deoxygenated blood from the right ventricle to the lungs.
• All veins carry deoxygenated blood, except pulmonary veins, which carry oxygenated blood from the lungs and to the left atrium of the Heart.

Question 5.
Transpiration is a necessary evil in plants. Explain.
Answer:
The water is lost from the leaves due to transpiration, pressure is created in, at the top to pull more water from the xylem to the leaves through the process of transpiration pull. This ensure the continuous flow of water from the roots of the leaves.

Samacheer Kalvi 10th Science Transportation in Plants and Circulation in Animals Additional Important Questions and Answers

I. Choose the correct answer:

Question 1.
Seed swells when placed in water due to:
(a) osmosis
(b) imbibition
(c) hydrolysis
(d) all of these
Answer:
(b) imbibition

Question 2.
Root hairs occurs in:
(a) Meristematic zone
(b) Cell elongation zone
(c) Cell maturation zone
(d) Old root
Answer:
(c) Cell maturation zone

The other name for Red blood corpuscles (RBC) is called ______.
Answer:
Erythrocytes.

Question 3.
Water in plants is transported by ascent of sap takes place through:
(a) cambium
(b) xylem
(c) phloem
(d) epidermis
Answer:
(b) xylem

Question 4.
Absorption of water is increased when:
(a) transpiration is increased
(b) photosynthesis is increased
(c) respiration is increased
(d) root pressure is increased
Answer:
(a) transpiration is increased

Question 5.
Opening of stomata is due to:
(a) Turgidity of guard cells
(b) Size of guard cells
(c) Number of guard cells
(d) Amount of CO₂ in the atmosphere.
Answer:
(a) Turgidity of guard cells

Question 6.
Guard cells help in:
(a) Fighting against infection
(b) Guttation
(c) Protecting against grazing
(d) Transpiration
Answer:
(d) Transpiration

Question 7.
Photosynthetic food material is transported in the form of:
(a) Glucose
(b) Sucrose
(c) Starch
(d) Fructose
Answer:
(b) Sucrose

Question 8.
Coronary artery supplies blood to:
(a) Mammary glands
(b) Rib muscles
(c) Skin
(d) Heart
Answer:
(d) Heart

Question 9.
All arteries carry oxygenated blood except:
(a) systematic
(b) hepatic
(c) pulmonary
(d) cardiac
Answer:
(c) pulmonary

Question 10.
The colour of lymph is:
(a) white
(b) pale yellow
(c) colourless
(d) milky
Answer:
(c) colourless

Question 11.
An artery can be distinguished from a vein in having:
(a) thicker wall
(b) elastic vessels
(c) no valves
(d) all of the above
Answer:
(d) all of the above

Question 12.
Purkinje fibres mainly help in contraction of:
(a) right auricle
(b) left auricle
(c) ventricles
(d) Aorta
Answer:
(c) ventricles

Question 13.
The ‘Lubb’ and ‘Dupp’ heart sound are due to:
(a) opening of heart valves
(b) action of papillary’ muscles
(c) closing of heart valves
(d) activity of pace maker
Answer:
(c) closing of heart valves

Question 14.
The closed circulatory system occurs in:
(a) cockroach
(b) fish
(c) mosquito
(d) house fly
Answer:
(b) fish

Question 15.
Normal pulse rate is:
(a) 80 mm Hg
(b) 120 mm Hg
(c) 40 mg Hg
(d) 90 mm Hg
Answer:
(d) 90 mm Hg

Question 16.
In the ABO system of blood groups, of both antigens are present but no antibody, the blood group of the individual would be?
(a) B
(b) O
(c) AB
(d) A
Answer:
(c) AB

Question 17.
Arteries are branches of:
(a) capillaries
(b) veins
(c) aorta
(d) none of these
Answer:
(c) aorta

Question 18.
Which type of WBCs are found in maximum number?
(a) Monocytes
(b) Basophils
(c) Eosinophils
(d) Neutrophils
Answer:
(d) Neutrophils

Question 19.
Which of the following are granular WBCs?
(a) Neutrophils, Basophils, Lymphocytes
(b) Eosinophil, Basophil, Monocytes
(c) Basophils, Monocytes, Lymphocytes
(d) Neutrophils, Eosinophil, Basophil
Answer:
(d) Neutrophils, Eosinophil, Basophil

Question 20.
RBCs are concerned with carriage of gases.
(a) CO₂
(b) O₂
(c) Respiratory
(d) CO₂ and SO₂
Answer:
(c) Respiratory

II. Fill in the blanks:

1. The contraction of heart is called ………..
2. Oxygenated blood is carried by ………..
3. Sphygmomanometer measures ………..
4. Heart of man is ………..
5. Pace maker of heart is ………..
6. The life span of RBC in human is ……… days.
7. Pulmonary artery carry ………….. blood.
8. ………. discovered the circulation of blood in man.
9. Semilunar valve is present at the base of ………….
10. Human heart beats ………. times in a minute at rest.
11. Red blood pigment is ………..
12. Transpiration helps in the absorption and ………… movement of water and minerals.
13. Water in plants is transported by ascent of sap takes place through ……….
14. The absorption water due to expenditure of energy is called ………….
15. In plants, the translocation of organic solutes takes place through ………….
Answer:
1. systole
2. pulmonary vein
3. blood pressure
4. myogenic
5. sino atrial node
6. 120
7. deoxygenated
8. William Harvey
9. Pulmonary artery
10. 72
11. Haemoglobin
12. upward
13. xylem
14. active absorption
15. phloem

III. Match the following:

Question 1.
Match the Column I with Column II.

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

Question 2.
Match the Column I with Column II.

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

IV. State whether true or false. If false write the correct statement:

1. The human heart beats 72 times per minute.
2. Right half of heart receives and pu mps off oxygenated blood.
3. Between right auricle and right ventricle seen Mitral valve.
4. The force of attraction between water molecule is cohesion.
5. By passive transport sucrose moves into the cells where it is utilised or stored.

Answer:

1. True
2. False – Right half of heart receives and pumps off deoxygenated blood.
3. False – Between right auricle and right ventricle seen tricuspid valve.
4. True
5. False – By active transport sucrose moves into the cells where it is utilised or stored.

V. Creative Question:

Question 1.
Origin of heart beat and its conduction is represented by:
(a) AV node → Bundle of His → SA node → Purkinje fibres
(b) SA node → Purkinje fibres → AV node → Bundle of His
(c) Purkinje fibres → Purkinje node → AV fibres → Bundle of His
(d) SA node → AV node → Bundle of His → Purkinje fibres
Answer:
(d) SA node → AV node → Bundle of His → Purkinje fibres

Question 2.
The cardiac pacemaker in a patient fails to function normally. The doctors find that an artificial pacemaker is to be grafted in him. It is likely that it will be drafted at the site of:
(a) Purkinje system
(b) Sinu atrial node
(c) Atrio ventricular node
(d) Atrio ventricular bundle
Answer:
(b) Sinu atrial node

Question 3.
Doctor use stethoscope to hear the sounds produced during each cardiac cycle. The second sound is heard when:
(a) AV node receives signal from SA node
(b) AV valves open up
(c) Ventricular walls vibrate due to rushing in of blood from atria
(d) Semi lunar valves close down after the blood flows into vessels from ventricles
Answer:
(d) Semi lunar valves close down after the blood flows into vessels from ventricles

VI. Assertion and Reason:

(a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.
1. Assertion: RBC impart red colour to the blood due to the presence of respiratory pigment haemoglobin.
Reason: The young RBC contain nucleus in man.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

2. Assertion: Osmosis is the movement of solvent molecule from higher concentration to lower concentration.
Reason: Osmosis is the active movement of water.
Answer:
(c) Assertion is true, but the reason is false.

VII. Answer in a Sentence:

Question 1.
Name the components of circulatory system.
Answer:
The circulation system consists of the circulating fluids, the blood and lymph, the heart and the blood vessels namely arteries, veins and capillaries.

Question 2.
What is the Ascent of Sap?
Answer:
The upward movement of water and minerals from roots to different plant parts is called ascent of sap.

Question 3.
Mention the composition of plasma.
Answer:
Organic substances like protein, glucose, urea, enzyme, hormones, vitamins and minerals are present in plasma.

Question 4.
Define Adhesion?
Answer:
The force of attraction between molecules of different substances is called adhesion.

Question 5.
Why are auricles and ventricles separated by auricular and ventricular septum?
Answer:
The separation of chamber avoids mixing of oxygenated and deoxygenated blood.

Question 6.
What is Single Circulation?
Answer:
In fishes, amphibians and certain reptiles, the oxygenated and deoxygenated blood are mixed ‘ and pass through the heart only once, is called Single Circulation.

Question 7.
Name the instruments used to measure blood pressure.
Answer:
Sphygmomanometer is a clinical instruments used to measure blood pressure when a person is in a relaxed and resting condition. Normal blood pressure in human is 120 mm/ 80 mm Hg.

Question 8.
Which blood group is called as ‘Universal Donor’ and ‘Universal Recipient’.
Answer:
Person with ‘O’ blood group are called Universal Donor.
Person with ‘AB’ blood group are called Universal Recipient.

Question 9.
What is Pulse?
Answer:
During the expansion of the artery, every time, the blood is forced into the arteries is called pulse.

Question 10.
What is transpiration puil?
Answer:
Transpiration through stomata creates vacuum which creates a suction called transpiration pull.

Question 11.
What are Rh antibodies?
Answer:
Rh-negative persons do not have Rh antigen on the surface of RBC. Antibodies developed against this Rh antigen is called Rh antibodies.

Question 12.
What is Adhesion?
Answer:
The force of attraction between molecules of different substances is called . adhesion.

VIII. Short answer question:

Question 1.
What is blood pressure?
Answer:
Blood pressure is the force exerted during the flow of blood against the lateral walls of arteries. The blood pressure is high in the arteries gradually drop in the arterioles and capillaries and become very low in the veins. Blood pressure varies during conditions of physical exercise, anxiety, emotions, stress and sleep. Increase in blood pressure is known as hypertension and decrease in blood pressure is termed as hypotension.

Question 2.
Distinguish between open and closed circulation.
Answer:

Question 3.
Explain the phenomenon of Guttation.
Answer:
The Guttation is a phenomenon caused due to root pressure. It takes place through a specialized cells called hydathodes. Dews on the leaves of grass seen in the early morning, when the climate is humid and excess of water in the soil is an example of Guttation.

Question 4.
What is Coronary circulation?
Answer:
The supply of blood to the heart muscles is called as coronary circulation. Cardiac muscles receives oxygenated blood from coronary arteries that originate from the aortic arch. Deoxygenated blood from the cardiac muscles drains into the right atrium by the coronary sinuses.

IX. Long answer questions:

Question 1.
What is lymph? Write its functions.
Answer:
Lymph is a colourless fluid formed when plasma, proteins and blood cells escape into intercellular spaces in the tissues through the pores present in the walls of capillaries.
Functions of Lymph:

1. Supplies nutrition and oxygen to those parts where blood cannot reach.
2. It drains away excess tissue fluid and metabolites and returns proteins to the blood from tissue spaces.
3. The lymph also carries absorbed fats from small intestine to the blood. The lymphatic capillaries of intestinal villi (lacteals) absorb digested fats.
4. Lymphocytes in the lymph defend the body from infections.

Question 2.
(a) Name the main two components of blood.
(b) What does plasma contain?
Answer:
(a) 1. The fluid plasma
2. Formed elements (blood cells)

• RBC
• WBC
• Blood platelets

(b) Plasma is slightly alkaline, containing non – cellular substance, constitute 55 % of the blood. Organic substances like Proteins, Glucose, Urea, Enzymes, Hormones, Vitamins and Minerals are present in the Plasma.

Question 3.
Demonstrate the process of osmosis with thistle funnel.
Answer:
Aim : To demonstrate the process of osmosis, using Thistle funnel.
Materials Required : Beaker, Thistle funnel, semi permeable membrane, sucrose solution.
Procedure : A thistle funnel whose mouth is covered with a semipermeable membrane, is filled with sucrose solution. It is kept inverted in a beaker containing water.
Observation : The water will diffuse across the membrane due to osmosis and raise the level of the solution in the funnel.
Inference : Raise of liquid in the funnel indicates the process of osmosis.

Question 4.
Explain Double circulation.
Answer:
Diagrammatic representation of the double circulation.

Question 5.
Explain the steps involved in Ascent of sap.
Answer:
The upward movement of water, and minerals from roots to different plant parts is called ascent of sap. A number of factors play a role in ascent of sap and it takes places in following steps.
Root Pressure : Water from soil enters the root hairs due to osmosis. Root pressure is responsible for movement of water up to the base of the stem.

Capillary Action : Water or any liquid rises in a capillary tube because of physical forces, this phenomenon is called capillary action. In the same way, in stem water rises up to certain height because of capillary’ action.
Adhesion-cohesion of Water Molecules : Water molecules form a continuous column in the xylem because of forces of adhesion and cohesion among the molecules.
Cohesion : The force of attraction between molecules of water is called cohesion.
Adhesion : The force of attraction between molecules of different substances is called adhesion.Water molecules stick to a xylem because of force of adhesion.

Question 6.
Explain the causes of the sound of the Heart.
Answer:
The rhythmic closure and opening of the valves cause the sound of the heart.

• The first sound ‘LUBB’ is produced by the closure of the tricuspid and bicuspid valves, after the beginning of ventricular systole. It is a longer duration.
• The second sound ‘DUPP’ is produced by the closure of semilunar valves, at the end of ventricular systole. It is of short duration.

X. Higher Order Thinking Skills (HOTS)

Question 1.
An unconscious patient is rushed into the emergency room and need a fast blood transfusion. Because there is no time to check her medical history or determine her blood type, which type of blood should you as her doctor, give?
Answer:
O^+ve, it is universal donor. In O group individuals, Antigen A or B are absent on the surface of RBC. However the plasma contains both antibodies a and b.

Question 2.
What are the factors affecting the Ascent of Sap?
Answer:

• High Temperature
• Atmospheric Pressure
• Wind Velocity
• Low atmospheric humidity
• Soil water deficit.

Question 3.
Water vapour comes out from the plant leaf through the stomatal opening. Through the same stomatal opening carbon dioxide diffuses into the plant during photosynthesis. Reason out the above statements.
Answer:
Both processes can happen together because the diffusion co-efficient of water and CO₂ is different.

Question 4.
Write a few Strategies, to prevent heart disease.
Answer:

• Exercise for about 30 minutes every day.
• Smoking or use of Tobacco should be avoided.
• Eat a heart-healthy diet – Avoid too much of salt or sugar, use of a diet rich in Vegetables and Fruits.
• Maintain a healthy weight.
• Get enough quality sleep.
• Manage Stress.
• Get regular health screenings (B.P. monitoring, Cholesterol level, Diabetes Screening etc.).

XI. Give reason:

Question 1.
Valves are important in human heart. Give reason.
Answer:
The valves will prevent the back flow of blood.

Question 2.
During rainy season wooden door generally swells up – Give reason.
Answer:
It is due to imbibition. It is a type of diffusion in which a solid absorbs water and gets swelled up.

Question 3.
Para sympathetic neural signals affects the working of the heart. Give reason.
Answer:
It reduce both heart rate and cardiac output.

Question 4.
Grapes placed in salt solution shrink. Give reason.
Answer:
It is due to exosmosis.

Question 5.
Why WBC’s are known as phagocytes?
Answer:
WBC produce antibodies during bacterial and viral infection and engulf the germs.

Students can Download Tamil Nadu 12th Physics Model Question Paper 5 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 5 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.
If voltage applied on a capacitor is increased from V to 2V, Choose the correct conclusion,
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Question 2.
The electric field in the region between two concentric charged spherical shells
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Question 3.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be …………
(a) R
(b) 2R
(c) \(\frac{\mathbf{R}}{4}\)
(d) \(\frac{\mathbf{R}}{2}\)
Answer:
(c) \(\frac{\mathbf{R}}{4}\)

Question 4.
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(b) 45°

Question 5.
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is
(a) 1.2
(b) 0.83
(c) 0.12
(d) 0.9
Answer:
(b) 0.83

Question 6.
Faraday’s law of electromagnetic induction is related to the …………….
(a) Law of conservation of charge
(b) Law of conservation of energy
(c) Third law of motion
(d) Law of conservation of angular momentum
Answer:
(b) Law of conservation of energy

Question 7.
The dimension of \(\frac{1}{\mu_{0} \varepsilon_{0}}\) is ……….
(a) [L T^-1]
(b) [L² T^-2]
(c) [L^-1 T]
(d) [L^-2 T²]
Answer:
(b) [L² T^-2]

Question 8.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,…………..
(a) 2D
(b) \(\frac{\mathrm{D}}{2}\)
(c) \(\sqrt{2} \mathrm{D}\)
(d) \(\frac{\mathrm{D}}{\sqrt{2}}\)
Answer:
(a) 2D

Question 9.
The speed of light in an isotropic medium depends on,
(a) its intensity
(b) its wavelength
(c) the nature of propagation
(d) the motion of the source w.r.to medium
Answer:
(b) its wavelength

Question 10.
The mean wavelength of light from sun is taken to be 550 nm and its mean power is 3.8 x 10²⁶ W. The number of photons received by the human eye per second on the average from sunlight is of the order of ……………..
(a) 10⁴⁵
(b) 10⁴²
(c) 10⁵⁴
(d) 10⁵¹
Answer:
(a) 10⁴⁵
Hint:

Question 11.
As the intensity of incident light increases
(a) kinetic energy of emitted photoelectrons increases
(b) photoelectric current decreases
(c) photoelectric current increases
(d) kinetic energy of emitted photoelectrons decreases
Answer:
(c) photoelectric current increases
Hint: As the intensity of incident light increases, photoelectric current increases.

Question 12.
In J.J. Thomson e/m experiment, a beam of electron is replaced by that of muons (particle with same charge as that of electrons but mass 208 times that of electrons). No deflection condition is achieved only if
(a) B is increased by 208 times
(b) B is decreased by 208 times
(c) B is increased by 14.4 times
(d) B is decreased by 14.4 times
Answer:
(c) B is increased by 14.4 times
Hint: In the condition of no deflection \(\frac{e}{m}=\frac{\mathrm{E}^{2}}{2 v \mathrm{B}^{2}}\)
If m is increased by 208 times then B should be increased \(\sqrt{208}\) = 14.4 times

Question 13.
A forward biased diode is treated as
(a) An open switch with infinite resistance
(b) A closed switch with a voltage drop of 0V
(c) A closed switch in series with a battery voltage of 0.7V
(d) A closed switch in series with a small resistance and a battery.
Answer:
(d) A closed switch in series with a small resistance and a battery.

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.
Who is the father of the modem robotics industry formed the world’s first robotic company in 1956?
(a) Joliot
(b) Cormark
(c) Engelberger
(d) Edward purcell
Answer:
(c) Engelberger

Part – II

Answer any six questions in which Q. No 17 is compulsory.

Question 16.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from  – q to +q. The SI unit of dipole moment is coulomb meter (cm).
\(\vec{p}=q a \hat{i}-q a(-\hat{i})=2 q a \hat{i}\)

Question 17.
Calculate the electric flux through the rectangle of sides 5 cm and 10 cm kept in the region of a uniform electric field 100 NC^-1. The angle 0 is 60°. Suppose 0 becomes zero, what is the electric flux?
Answer:

Question 18.
What is meant by hysteresis?
Answer:
The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. Hysteresis means Tagging behind’.

Question 19.
What is meant by wattles current?
Answer:
The component of current (I_pMS sin φ), which has a phase angle of \(\frac{\pi}{2}\) with the voltage is called reactive component. The power consumed is zero. So that it is also known as ‘Wattless’ current.

Question 20.
What is Snell’s window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle i_c. The restricted illuminated circular area is called Snell’s window.

Question 21.
Define work function of a metal. Give its unit.
Answer:
The minimum energy needed for an electron to escape from the metal surface is called work function of that metal. It’s unit is electron volt (eV).

Question 22.
What is meant by radioactivity?
Answer:
The phenomenon of spontaneous emission of highly penetrating radiations such as α, β and γ rays by an element is called radioactivity.

Question 23.
What is the angular momentum of an electron in the third orbit of an atom?
Answer:
Here n = 3; h = 6.6 x 10^-34 Js
Angular momentum.

Question 24.
Explain the current flow in a NPN transistor
Answer:

• The conventional flow of current is based on the direction of the motion of holes
• In NPN transistor, current enters from the base into the emitter.

Part – III

Answer any six questions in which Q.No. 27 is compulsory. [6×3 = 18]

Question 25.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
\(C=\frac{Q}{V} \text { or } Q \propto V\)
The SI unit of capacitance is coulomb per volt or farad (F).

Question 26.
Distinguish between drift velocity and mobility
Answer:

SNo.	Drift Velocity	Mobility
1.	The drift velocity is the average velocity acquired by the electrons inside the conductor when it is subjected to an electric field.	Mobility of an electron is defined as the magnitude of the drift velocity per unit electric field.
2.	\(\overrightarrow{\mathrm{V}}_{d}=\vec{a} \tau\)	\(\mu=\frac{e \tau}{m} \text { or } \mu=\frac{\left|\overrightarrow{\mathrm{V}}_ {d}\right|}{\overrightarrow{\mathrm{E}}}\)
3.	It’s unit is ms-1.	It’s unit is m2 v_1 s_1

Question 27.
A coil of a tangent galvanometer of diametre 0.24 m has 100 turns. If the horizontal component of Earth’s magnetic field is 25 x 10^-6 T then, calculate the current which gives a deflection of 60°.
Answer:
The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T

Question 28.
Mention the ways of producing induced emf.
Answer:
Induced emf can be produced by changing magnetic flux in any of the following ways:

• By changing the magnetic field B
• By changing the area A of the coil and
• By changing the relative orientation 0 of the coil with magnetic field

Question 29.
A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.
Answer:
L = 400 x 10^-3 H; I_eff = 6 x 10^-3A; f= 1000 Hz
Inductive reactance, X_L = Lω = L x 2πf = 2 x 3.14 x 1000 x 0.4 = 2512Ω
Voltage across L, V – IX_L = 6 x 10^-3 x 2512 =15.072 V (RMS)

Question 30.
What are the sign conventions followed for lenses?
Answer:
The sign conventions for thin lenses differ only in the signs followed for focal lengths.
(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).

(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.

Question 31.
Write down the draw backs of Bohr atom’model.
Answer:
Limitations of Bohr atom model
The following are the drawbacks of Bohr atom model

• Bohr atom model is valid only for hydrogen atom or hydrogen like-atoms but not for
  complex atoms.
• When the spectral lines are closely examined, individual lines of hydrogen spectrum is
  accompanied by a number of faint lines. These are often called fine structure. This is not explained by Bohr atom model.
• Bohr atom model fails to explain the intensity variations in the spectral lines.
• The distribution of electrons in atoms is not completely explained by Bohr atom model.

Question 32.
What do you mean by leakage current in a diode?
Answer:
The leakage current in a diode is the current that the diode will leak when a reverse voltage is applied to it. Under the reverse bias, a very small current in μA, flows across the junction. This is due to the flow of the minority charge carriers called the leakage current or reverse saturation current.

Question 33.
Distinguish between wireline and wireless communication.
Answer:

Wireline communication	Wireless communication
It is a point-to-point communication.	It is a broadcast mode communication.
It uses mediums like wires, cable and optical fibres.	It uses free space as a communication medium.
These systems cannot be used for long distance transmission as they are connected.	These systems can be used for long distance transmission.
Ex. telephone, intercom and cable TV.	Ex. mobile, radio or TV broadcasting and satellite communication.

Part – IV

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

Question 34.
(a) Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field: Consider an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\) whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec{E}\)  in the direction of the field and charge -q will experience a force -q \(\vec{E}\) in a direction opposite to the field. Since the external field \(\vec{E}\) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O.

Using right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.
The magnitude of the total torque


where θ is the angle made by \(\vec{p} \text { with } \overrightarrow{\mathrm{E}}\) .
Since p = 2aq, the torque is written in terms of the vector product as \(\vec{\tau}=\vec{p} \times \vec{E}\)
The magnitude of this torque is τ = pE sin θ and is maximum
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field [/latex]\overrightarrow{\mathrm{E}}[/latex] . Once \(\overrightarrow{\mathrm{p}}\) is aligned with \(\overrightarrow{\mathrm{E}}\), the total torque on the dipole becomes zero.

[OR]

(b) Explain the equivalent resistance of a parallel resistor network.
Answer:
Resistors in parallel: Resistors are in parallel when they are connected across the same potential difference as shown in fig. (a).
In this case, the total current I that leaves the battery in split into three separate paths. Let I₁, I₂ and I₃ be the current through the resistors R_{1 ,}R₂ and R₃ respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I_{1 +} I₂ + I₃ ………………… (1)

Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have

Substituting these values in equation (1) we get.

Here R_p is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).
Note: The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Question 35.
(a) Obtain a relation for the magnetic induction at a point along the axis of a circular coil ‘ carrying current.
Answer:
Magnetic field produced along the axis of the current carrying circular coil: Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O. It is computed by taking two diametrically opposite line elements of the coil each of length \(d \bar{l} \) at C and D Let r be the vector joining the current element (I \(d \bar{l}\)) at C to the point P.
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
angle ∠CPO = ∠DPO = θ
According to Biot-Savart’s law, the magnetic field at P due to the current element \(d \bar{l} \) is


The magnitude of magnetic field due to current element I \(d \bar{l} \) at C and D are equal because of equal distance from the coil. The magnetic field dB due to each current element I dl is resolved into two components; dB sin θ along y-direction and dB cos θ along the z-direction. Horizontal components of each current element cancels out while the vertical components (dB cos θ \(\hat{k}\) ) alone contribute to total magnetic field at the point P.
If we integrate \(d \bar{l} \) around the loop, \(d \bar{B} \) sweeps out a cone, then the net magnetic field \(\overrightarrow{\mathrm{B}}\) at point P is

Using Pythagorous theorem r² = R² + Z² and integrating line element from 0 to 2πR, we get

Note that the magnetic field \(\bar{B} \) points along the direction from the point O to P. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point P to O.

[OR]

(b) Explain the working of a single-phase AC generator with necessary diagram.
Answer:
Single phase AC generator: In a single phase AC generator, the armature conductors are connected in series so as to form a single circuit which generates a single-phase alternating emf and hence it is called single-phase alternator.

The simplified version of a AC generator is discussed hire. Consider a stator core consisting of 2 slots in which 2 armature conductors PQ and RS are mounted to form single-turn rectangular loop PQRS. Rotor has 2 salient poles with field windings which can be magnetized by means of DC source.

Working: The loop PQRS is stationary and is perpendicular to the plane of the paper. When field windings are excited, magnetic field is produced around it. The direction of magnetic field passing through the armature core. Let the field magnet be rotated in clockwise direction by the prime mover. The axis of rotation is perpendicular to the plane of the paper.

Assume that initial position of the field magnet is horizontal. At that instant, the direction of magnetic field is perpendicular to the plane of the loop PQRS. The induced emf is zero. This is represented by origin O in the graph between induced emf and time angle.

When field magnet rotates through 90°, magnetic field becomes parallel to PQRS. The induced emfs across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Fleming’s right hand rule.

Care has to be taken while applying this rule; the thumb indicates the direction of the motion of the conductor with respect to field. For clockwise rotating poles, the conductor appears to be rotating anti-clockwise. Flence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper. For PQ, it is downwards and for RS upwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.

For the rotation of 180° from the initial position, the field is again perpendicular to PQRS and the induced emf becomes zero. This is represented by point B. The field magnet becomes again parallel to PQRS for 270° rotation of field magnet. The induced emf is maximum but the direction is reversed. Thus the current flows along SRQP. This is represented by point C.

On completion of 360°, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature.

Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle. For this construction, the frequency of the induced emf depends on the speed at which the field magnet rotates.

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) Describe the Fizeau’s method to determine speed of light.
Answer:
Fizeau’s method to determine speed of light:
Apparatus: The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism.

The light passing through one cut in the wheel will, get reflected by a mirror M kept at a long distance d, about 8 km from the toothed wheel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.

Working: The angular speed c rotation of the toothed wheel was increased from zero to a value co until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.

Expression for speed of light: The speed of light in air v is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back 2d to the time taken t.
\(v=\frac{2 d}{t}\) ………………… (1)

The distance d is a known value from the arrangement. The time taken t for the light to travel the distance to and fro is calculated from the angular speed ω of the toothed wheel. The angular speed ω of the toothed wheel when the light disappeared for the first time is,
\(\omega=\frac{\theta}{t}\) ………………….. (2)

Here, θ is the angle between the tooth and the slot which is rotated by the toothed wheel within that time t

Rewriting the above equation for t
\(t=\frac{\pi}{\mathrm{N} \omega}\) ……………….. (3)
Substituting t from equation (3) in equation (1)
\(v=\frac{2 d}{\pi / \mathrm{N} \omega}\)
\(v=\frac{2 d \mathrm{N} \omega}{\pi}\) …………….(4)
Fizeau had some difficulty to visually estimate the minimum intensity blocked by the adjacent tooth, and his value for speed of light was very value.

Question 37.
(a) List out the laws of photoelectric effect.
Answer:
Laws of photoelectric effect:

• For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
• Maximum kinetic energy of the photo electrons is independent of intensity of the incident light.
• Maximum kinetic energy of the photo electrons from a given metal is directly proportional to the frequency of incident light.
• For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
• There is no time lag between incidence of light and ejection of photoelectrons.

[OR]

(b) Describe the working of nuclear reactor with a block diagram.
Answer:
Nuclear reactor:

• Nuclear reactor is a system in which the nuclear fission takes place in a self-sustained controlled manner and the energy produced is used either for research purpose or for power generation.
• The main parts of a nuclear reactor are fuel, moderator and control rods. In addition to this, there is a cooling system which is connected with power generation set up.

Fuel:

• The fuel is fissionable material, usually uranium or plutonium. Naturally occurring uranium contains only \(0.7 \% \text { of }_{92}^{235} \mathrm{U} \text { and } 99.3 \% \text { are only }^{238} \mathrm{g}_{2} \mathrm{U}\) . So the \(_{ 92 }^{ 235 }{ U }\) must be enriched such that it contains at least 2 to 4% of \(_{ 92 }^{ 235 }{ U }\) .
• In addition to this, a neutron source is required to initiate the chain reaction for the first time. A mixture of beryllium with plutonium or polonium is used as the neutron source. During fission of \(_{ 92 }^{ 235 }{ U }\), only fast neutrons are emitted but the probability of initiating fission by it in another nucleus is very low. Therefore, slow neutrons are preferred for sustained nuclear reactions.

Moderators:

• The moderator is a material used to convert fast neutrons into slow neutrons. Usually the moderators are chosen in such a way that it must be very light nucleus having mass comparable to that of neutrons. Hence, these light nuclei undergo collision with fast neutrons and the speed of the neutron is reduced
• Most of the reactors use water, heavy water (D₂0) and graphite as moderators. The blocks of uranium stacked together with blocks of graphite (the moderator) to form a large pile.

 

Control rods:

• The control rods are used to adjust the reaction rate. During each fission, on an average 2.5 neutrons are emitted and in order to have the controlled chain reactions, only one neutron is allowed to cause another fission and the remaining neutrons are absorbed by the control rods.
• Usually cadmium or boron acts as control rod material and these rods are inserted into the uranium blocks. Depending on the insertion depth of control rod into the uranium, the average number of neutrons produced per fission is set to be equal to one or greater than one.
• If the average number of neutrons produced per fission is equal to one, then reactor is said to be in critical state. In fact, all the nuclear reactors are maintained in critical state by suitable adjustment of control rods. If it is greater than one, then reactor is said to be in super-critical and it may explode sooner or may cause massive destruction.

Shielding:

• For a protection against harmful radiations, the nuclear reactor is surrounded by a concrete wall of thickness of about 2 to 2.5 m.

Cooling system:

• The cooling system removes the heat generated in the reactor core. Ordinary water, heavy water and liquid sodium are used as coolant since they have very high specific heat capacity and have large boiling point under high pressure.
• This coolant passes through the fuel block and carries away the heat to the steam generator through heat exchanger. The steam runs the turbines which produces electricity in power reactors.

Question 38.
(a) Explain the working principle of a solar cell. Mention its applications.
Answer:
Solar cell:
A solar cell, also known as photovoltaic cell, converts light energy directly into electricity or electric potential difference by photovoltaic effect. It is basically a p-n junction which generates emf when solar radiation falls on the p-n junction. A solar cell is of two types: p-type and n-type.

Both types use a combination of p-type and n-type Silicon which together forms the p-n junction of the solar cell. The difference is that p-type solar cells use p-type Silicon as the base with an ultra-thin layer of n-type Silicon as shown in Figure, while n-type solar cell uses the opposite combination. The other side of the p-Silicon is coated with metal which forms the back electrical contact. On top of the n-type Silicon, metal grid is deposited which acts as the front electrical contact. The top of the solar cell is coated with anti-reflection coating and toughened glass.

In a solar cell, electron-hole pairs are generated due to the absorption of light near the junction. Then the charge carriers are separated due to the electric field of the depletion region. Electrons move towards n-type Silicon and holes move towards p-type Silicon layer.

The electrons reaching the n-side are collected by the front contact and holes reaching p-side are collected by the back electrical contact. Thus a potential difference is developed across solar cell. When an external load is connected to the solar cell, photocurrent flows through the load.

Many solar cells are connected together either in series or in parallel combination to form solar panel or module. Many solar panels are connected with each other to form solar arrays. For  high power applications, solar panels and solar arrays are used.

Applications:

• Solar cells are widely used in calculators, watches, toys, portable power supplies, etc.
• Solar cells are used in satellites and space applications
• Solar panels are used to generate electricity.

(b) Elaborate on the basic elements of communication system with the necessary block diagram.
Answer:
Elements of an electronic communication system
Information (Baseband or input signal): Information can be in the form of a sound signal like speech, music, pictures, or computer data which is given as input to the input transducer.

Input transducer
A transducer is a device that converts variations in a physical quantity (pressure, temperature, sound) into an equivalent electrical signal or vice versa. In communication system, the transducer converts the information which is in the form of sound, music, pictures or computer data into corresponding electrical signals. The electrical equivalent of the original information is called the baseband signal. The best example for the transducer is the microphone that converts sound energy into electrical energy.

Transmitter:
It feeds the electrical signal from the transducer to the communication channel. It consists of circuits such as amplifier, oscillator, modulator, and power amplifier. The transmitter is located at the broadcasting station.

Amplifier:
The transducer output is very weak and is amplified by the amplifier. Oscillator: It generates high-frequency carrier wave (a sinusoidal wave) for long distance transmission into space. As the energy of a wave is proportional to its frequency, the carrier wave has very high energy.

Modulator:
It superimposes the baseband signal onto the carrier signal and generates the modulated signal.

Power amplifier:
It increases the power level of the electrical signal in order to cover a large distance.

Transmitting antenna:
It radiates the radio signal into space in all directions. It travels in the form of electromagnetic waves with the velocity of light (3 ^x 10⁸ ms^-1).

Communication channel:
Communication channel is used to carry the electrical signal from transmitter to receiver with less noise or distortion. The communication medium is basically of two types: wireline communication and wireless communication.

Noise:
It is the undesirable electrical signal that interfaces with the transmitted signal. Noise attenuates or reduces the quality of the transmitted signal. It may be man-made (automobiles, welding machines, electric motors etc .) or natural (lightening, radiation from sun and stars and environmental effects). Noise cannot be completely eliminated. However, it can be reduced using various techniques.

Receiver:
The signals that are transmitted through the communication medium are received with the help of a receiving antenna and are fed into the receiver. The receiver consists of electronic circuits like demodulator, amplifier, detector etc. The demodulator extracts the baseband signal from the carrier signal. Then the baseband signal is detected and amplified using amplifiers. Finally, it is fed to the output transducer.

Repeaters:
Repeaters are used to increase the range or distance through which the signals are sent. It is a combination of transmitter and receiver. The signals are received, amplified, and retransmitted with a carrier signal of different frequency to the destination. The best example is the communication satellite in space.

Output transducer:
It converts the electrical signal back to its original form such as sound, music, pictures or data. Examples of output transducers are loudspeakers, picture tubes, computer monitor, etc.

Attenuation:
The loss of strength of a signal while propagating through a medium is known as attenuation.

Range:
It is the maximum distance between the source and the destination up to which the signal is received with sufficient strength.

Students can download 10th Science Chapter 9 Solutions 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 9 Solutions

Samacheer Kalvi 10th Science Solutions Text Book Back Questions and Answers

I. Choose the best answer:

Question 1.
A solution is a mixture.
(a) homogeneous
(b) heterogeneous
(c) homogeneous and heterogeneous
(d) non-homogeneous
Answer:
(a) homogeneous

Question 2.
The number of components in a binary solution is ______.
(a) 2
(b) 3
(c) 4
(d) 5.
Answer:
(a) 2

Question 3.
Which of the following is the universal solvent?
(a) Acetone
(b) Benzene
(c) Water
(d) Alcohol
Answer:
(c) Water

Question 4.
A solution in which no more solute can be dissolved in a definite amount of solvent at a given temperature is called ______.
(a) Saturated solution
(b) Un saturated solution
(c) Supersaturated solution
(d) Dilute solution.
Answer:
(a) Saturated solution

Question 5.
Identify the non-aqueous solution.
(a) sodium chloride in water
(b) glucose in water
(c) copper sulphate in water
(d) sulphur in carbon-di-sulphide
Answer:
(d) sulphur in carbon-di-sulphide

Question 6.
When pressure is increased at a constant temperature the solubility of gases in liquid ______.
(a) No change
(b) increases
(c) decreases
(d) no reaction.
Answer:
(b) increases

Question 7.
Solubility of NaCl in 100 ml water is 36 g. If 25 g of salt is dissolved in 100 ml of water how much more salt is required for saturation:
(a) 12 g
(b) 11 g
(c) 16 g
(d) 20 g
Answer:
(b) 11 g

Question 8.
A 25% alcohol solution means ______.
(a) 25 ml of alcohol in. 100 ml of water
(b) 25 ml of alcohol in 25 ml of water
(c) 25 ml of alcohol in 75 ml of water
(d) 75 ml of alcohol in 25 ml of water.
Answer:
(c) 25 ml of alcohol in 75 ml of water

Question 9.
Deliquescence is due to:
(a) Strong affinity to water
(b) Less affinity to water
(c) Strong hatred to water
(d) Inertness to water
Answer:
(a) Strong affinity to water

Question 10.
Which of the following is hygroscopic in nature?
(a) ferric chloride
(b) copper sulphate pentahydrate
(c) silica gel
(d) none of the above.
Answer:
(c) silica gel

II. Fill in the blanks:

1. The component present in lesser amount, in a solution is called ……..
2. Example for liquid in solid type solution is ……….
3. Solubility is the amount of solute dissolved in ……… g of solvent.
4. Polar compounds are soluble in ……… solvents.
5. Volume percentage decreases with increases in temperature because ………

Answer:

1. solute
2. amalgam
3. 100
4. Polar
5. of expansion of liquid

III. Match the following:

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

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

1. Solutions which contain three components are called binary solution.
2. In a solution the component which is present in lesser amount is called solvent.
3. Sodium chloride dissolved in water forms a non-aqueous solution.
4. The molecular formula of green vitriol is MgSO₄. 7H₂O
5. When Silica gel is kept open, it absorbs moisture from the air, because it is hygroscopic in nature.

Answer:

1. False – Solutions which contain two components are called binary solution.
2. False – In a solution the component which is present in lesser amount is called solute.
3. False – Sodium chloride dissolved in water forms an aqueous solution.
4. False – The molecular formula of green vitriol is FeSO₄. 7H₂O
5. True

V. Short Answer Questions:

Question 1.
Define the term: Solution
Answer:
A solution is a homogeneous mixture of two or more substances.

Question 2.
What is mean by the binary solution?
Answer:
A solution must at least be consisting of two components. Such solutions which are made of one solute and one solvent are called binary solutions.
E.g., On adding CuSO₄ crystals to water.

Question 3.
Give an example each

1. gas in liquid;
2. solid in liquid;
3. solid in solid;
4. gas in gas.

Answer:

1. Gas in liquid – CO₂ in water
2. Solid in liquid – NaCl in water
3. Solid in solid – Alloys
4. Gas in gas – He – O₂ gas

Question 4.
What is the aqueous and non-aqueous solution? Give an example.
Answer:
Aqueous solution: The solution in which water act as a solvent is called aqueous solution. In general, ionic compounds are soluble in water and form aqueous solutions more readily than covalent compounds. E.g. Common salt in water.

Non – Aqueous solution: The solution in which any liquid, other than water act as a solvent is called non-aqueous solution. Alcohols, benzene, ethers, etc., are used as non – aqueous solvents. E.g. Sulphur dissolved in carbon disulphide.

Question 5.
Define Volume percentage.
Answer:
Volume percentage is defined as the percentage by volume of solute (in ml) present in the given volume of solution.

Question 6.
The aquatic animals live more in a cold region. Why?
Answer:
Aquatic animals live more in cold regions because the solubility of oxygen is more in cold water (at low temperature). Therefore, aquatic animals are more comfortable in cold water.

Question 7.
Define Hydrated salt.
Answer:
Ionic substances which crystallise out from their saturated aqueous solution with a definite number of molecules of water are called hydrated salts.

Question 8.
A hot saturated solution of copper sulphate forms crystals as it cools. Why?
Answer:
The capability of a solution to maintain a certain concentration of solute is temperature-dependent. When a saturated solution of copper sulphate at above room temperature is allowed to cool, the solution becomes supersaturated and in the absence of stirring or the return of the previous solution temperature, the solute starts to precipitate out. i.e., crystal formation occurs.

Question 9.
Classify the following substances into deliquescent, hygroscopic. Cone. Sulphuric acid, Copper sulphate penta hydrate, Silica gel, Calcium chloride and Gypsum salt.
Answer:

VI. Long answer questions:

Question 1.
Write notes on?

1. saturated solution
2. unsaturated solution

Answer:

1. Saturated solution: A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution, e.g. 36 g of sodium chloride in 100 g of water at 25°C forms a saturated solution.
2. Unsaturated solution: Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature, e.g. 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.

Question 2.
Write notes on various factors affecting solubility.
Answer:
There are three main factors which affects the solubility of a solute. They are

1. Nature of the solute and solvent
2. Temperature
3. Pressure

1. Nature of the solute and solvent : The nature of the solute and solvent plays an important role in solubility. Even though water is Universal solvent, all substances do not dissolve in water. Dissolution occurs when similarities exist between the solvent and the solute.

Ionic compounds are soluble in polar solvent like water and covalent compounds are soluble in non-polar solvents like ether, benzene, alcohol etc.

2. Effect of Temperature :
Solubility of solid in liquid : Generally solubility of a solid solute in a liquid increases with increase in temperature.
In Endothermic process : Solubility increases with increase in temperature.
In Exothermic process : Solubility decreases with increase in temperature.
Solubility of Gases in liquid : Solubility of gases in liquid decreases with increase in temperature.

3. Effect of Pressure : Effect of pressure is observed only in the case of solubility of a gas in a liquid. When the pressure is increased, the solubility of a gas in liquid increases.

Question 3.
(a) What happens when MgSO₄.7H₂O is heated? Write the appropriate equation
(b) Define solubility.
Answer:
(a) When Epsom salt MgSO₄.7H₂O crystals are gently heated, it loses seven water molecules and becomes anhydrous MgSO₄.

(b) Solubility is defined as the amount of solute in grams that can be dissolved in 100 g of the solvent to form its saturated solution at a given temperature and pressure.

Question 4.
In what way hygroscopic substances differ from deliquescent substances.
Answer:

Question 5.
A solution is prepared by dissolving 45 g of sugar in 180 g of water. Calculate the mass percentage of solute.
Answer:
Mass of the solute (sugar) = 45 g
Mass of the solvent (Water) = 180 g
Formula:
Mass percentage of solute (sugar)

The mass percentage of solute = 20%

Question 6.
3.5 litres of ethanol is present in 15 litres of aqueous solution of ethanol. Calculate volume percent of ethanol solution.
Answer:
Volume of ethanol = 3.5 lit = 3500 ml
Volume of water = 15 lit = 15000 ml
Formula:

The volume percentage of ethanol solution = 18.92

VII. HOT Questions

Question 1.
Vinu dissolves 50 g of sugar in 250 ml of hot water, Sarath dissolves 50 g of same sugar in 250 ml of cold water. Who will get a faster dissolution of sugar? and Why?
Answer:
Vinu will get a faster dissolution of sugar. Because generally solubility of a solid solute in a liquid solvent increases with increase in temperature. Therefore Vinu dissolves 50 g of sugar in 250 ml of hot water than Sarath dissolves 50 g of sugar in 250 ml of cold water.

Question 2.
‘A’ is a blue coloured crystalline salt. On heating it loses blue colour and to give ‘B’ When water is added, ‘B’ gives back to ‘A’. Identify A and B, write the equation.
Answer:
Since ‘A’ is a blue coloured crystalline salt, it is CuSO₄. 5H₂O (Blue vitriol). On heating it loses all five water molecules and becomes colourless anhydrous CuSO₄.

When water is added ‘B’ gives back A.

Question 3.
Will the cool drinks give more fizz at top of the hills or at the foot? Explain.
Answer:
At hilltops, the temperature will become less and pressure also decreases. Because temperature and pressure are directly proportional to each other. At low-pressure carbonate, cool drinks will give less fizz and give more fizz at the foot.

Samacheer Kalvi 10th Science Solutions Additional Important Questions and Answers

I. Choose the correct answer:

Question 1.
The dissolution of sugar and salt in water results in a solution.
(a) Binary
(b) Ternary
(c) Quaternary
(d) Saturated
Answer:
(b) Ternary

Question 2.
In a solution, the component which is present in a lesser amount is called ______.
(a) solvent
(b) dissolution
(c) solute
(d) mole.
Answer:
(c) solute

Question 3.
The supersaturated solution of NaCl in 100 g of water at 25°C contains:
(a) 40 g of NaCl
(b) 10 g of NaCl
(c) 20 g of NaCl
(d) 30 g of NaCl
Answer:
(a) 40 g of NaCl

Question 4.
How many component(s) present in binary solution?
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(b) 2

Question 5.
Formalin is an aqueous solution of:
(a) formic acid
(b) ammonia
(c) formaldehyde
(d) carbon tetrachloride
Answer:
(c) formaldehyde

Question 6.
The effect of pressure on the solubility of a gas in liquids is given by:
(a) Boyle’s Law
(b) Charle’s Law
(c) Henry’s Law
(d) Avogadro’s Law
Answer:
(c) Henry’s Law

Question 7.
Which one of the following is an example of an aqueous solution?
(a) Sugar in water
(b) Sulphur in carbon disulphide
(c) Iodine dissolved in carbon tetrachloride
(d) Benzoic acid in ethers.
Answer:
(a) Sugar in water

Question 8.
The type of solution when CO₂ is dissolved in water:
(a) solid/liquid
(b) liquid in liquid
(c) gas in liquid
(d) liquid in solid
Answer:
(c) gas in liquid

Question 9.
Tin amalgam is an example of ……… solution.
(a) solid in solid
(b) liquid in solid
(c) solid in liquid
(d) liquid in liquid
Answer:
(b) liquid in solid

Question 10.
In which case solubility decreases with increase in temperature?
(a) Endothermic process
(b) Exothermic process
(c) Both (a) and (b)
(d) None of these.
Answer:
(b) Exothermic process

Question 11.
Fat is soluble in:
(a) water
(b) alcohol
(c) CCl₄
(d) ether
Answer:
(d) ether

Question 12.
The deliquescent substance among the following is:
(a) con.H₂SO₄
(b) P₂O₅
(c) CaCl₂
(d) SiO₂
Answer:
(c) CaCl₂

Question 13.
Mass percentage is expressed as ______.
(a) v/v
(b) w/w
(c) v/w
(d) w/v.
Answer:
(b) w/w

Question 14.
Hygroscopic substances are used as ……… agents.
(a) foaming
(b) drying
(c) oxidising
(d) reducing
Answer:
(b) drying

Question 15.
The molecular formula of Epsom salt is ______.
(a) CuSO₄.5H₂O
(b) FeSO₄.7H₂O
(c) MgSO₄.7H₂O
(d) ZnSO₄.7H₂O.
Answer:
(c) MgSO₄.7H₂O

II. Fill in the blanks:

1. A true solution is a ……… mixture of solute and solvent.
2. Soil cannot store more nitrogen than it can hold because soil is said to be in a state of ………
3. In the dissolution of NaOH in water, the solubility …….. with increase in temperature.
4. Aquatic animals are more comfortable in cold water because as the temperature is less the solubility of dissolved oxygen ………
5. Hydrated salts contain ……… of crystallization.
6. He-O₂ mixture is a binary solution of …….. in ………. solution.
7. The solvent used for dissolving Sulphur is ……….
8. The solubility of NaOH at 25°C is ……….
9. According to Henry’s Law, the solubility of a gas in liquid is ………. proportional to the pressure of the gas over the solution at definite temperature.
10. Anhydrous Calcium chloride is a ………. substance.
11. ……… substances absorb enough water from the atmosphere and get completely dissolved.
12. When 90g of sodium bromide is dissolved in 100 g of water at 25°C it forms a ………. solution.
13. ………. is an example of a binary solution with liquid in Gas.
14. Air and sea water are important ……… solution.
15. A quaternary solution contains ……….. components.
16. The primary factor which determines the characteristic of a solution is ………..
Answer:
1. Homogeneous
2. saturation
3. decreases
4. increases
5. water
6. Gas, Gas
7. CS₂ (or) Carbon disulphide
8. 80 g
9. directly
10. Hygroscopic
11. deliquescent
12. Unsaturated
13. Cloud
14. Homogeneous
15. four
16. Physical state

III. Match the following

Question 1.
Match the column I with column II.

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

Question 2.
Match the column I with column II.

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

Question 3.
Match the column I with column II.

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

Question 4.
Match the column I with column II.

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

Question 5.
Match the column I with column II.

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

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

1. In an aqueous solution of copper sulphate, the solvent is copper sulphate.
2. A solution containing sugar and salt in water is a binary solution.
3. An example of a solid solution is alloy.
4. The difference between concentrated and dilute solution can be observed by means of colour (or) density.
5. A saturated solution contains 91 g of Glucose in 100 g of water at 25°C.
6. Fat is dissolved in the aqueous solvent ether.
7. The solubility of a gas in a liquid is inversely proportional to the pressure of the gas at a definite temperature.
8. Mass percentage of a solution is expressed as .
9. The white vitriol is represented by the formula ZnSO₄ . 7H₂O.
10. Ferric chloride is a Hygroscopic substance.
Answer:
1. False – In an aqueous solution of copper sulphate, the solvent is water.
2. False – A solution containing sugar and salt in water is a ternary solution.
3. True
4. True
5. True
6. False – Fat is dissolved in the non-aqueous solvent ether.
7. False – The solubility of a gas in a liquid is directly proportional to the pressure of the gas at a definite temperature.
8. True
9. True
10. False – Ferric chloride is a deliquescent substance.

V. Short answer questions:

Question 1.
(i) Which gas is dissolved in soft drinks?
(ii) What will you do to increase the solubility of this gas?
Answer:
(i) Carbon-di-oxide (CO₂) is dissolved in soft drinks.
(ii) An increase in pressure will increase the solubility of CO₂ gas.

Question 2.
Identify the type of binary solution given below.
Answer:

1. Alloys
2. Amalgam
3. Ethyl alcohol in water
4. Aerated drinks

Answer:

1. Solid in solid
2. liquid in solid
3. liquid in liquid
4. Gas in liquid

Question 3.
Explain why Nitrogen in soil is called a saturated solution in nature?
Answer:
Nitrogen in soil is an example of a saturated solution in nature. Soil cannot store more Nitrogen than it can hold.

Question 4.
Define Mass percentage.
Answer:
Mass percentage of a solution is defined as the percentage by mass of the solute present in the solution.

Question 5.
Define the term Molarity (M).
Answer:

Question 6.
Define the term Molality (m).
Answer:

Question 7.
Define the supersaturated solution.
Answer:
A supersaturated solution is one that contains more solute than the saturated solution at a given temperature.
E.g. 40 g of sodium chloride in 100 g of water at 25°C.

Question 8.
Justify the following statements with an explanation.
(i) Solubility of NH₄Cl increases with increase in temperature.
Answer:
Solubility of NH₄Cl increases with increase in temperature because it is an endothermic process.

(ii) Solubility of NaOH decreases with increase in temperature.
Answer:
Solubility of NaOH decreases with increase in temperature because it is an exothermic process.

Question 9.
Calculate the molarity of a solution containing 4 g of NaOH in 500 ml of water.
Answer:
Mass of NaOH = 4 g
Volume of solution = 500 ml

= 0.1 × 2
= 0.2 M

Question 10.
Calculate the molality of a solution containing 3 g of urea (molecular mass = 60) in 750 g of water.
Answer:
Mass of urea (solute) = 3 g
Mass of water (solvent) = 750 g
Formula:

Question 11.
Define dissolution.
Answer:
The process of uniform distribution of solute into solvent is called dissolution.

VI. Long answer questions:

Question 1.
Answer the blanks given in the table.
Answer:

Question 2.
Write a note on the type of solution based on the type of solvent.
Answer:
(i) Aqueous solution : The solution in which water acts as a solvent is called aqueous solution. In general, ionic compounds are soluble in water and form aqueous solutions more readily than covalent compounds.
Eg: Common salt in water, Sugar in water, Copper sulphate in water etc..

(ii) Non-Aqueous solution : The solution in which any liquid, otter than water, acts as a solvent is called non-aqueous solution. Solvent other than water is referred to as non-aqueous solvent. Generally, alcohols, benzene, ethers, carbon disulphide, acetone, etc., are used as non- aqueous solvents.
Eg: Sulphur dissolved in carbon disulphide, Iodine dissolved in carbon tetrachloride.

Question 3.
Justify the following statements with an explanation.

1. The solubility of calcium oxide decreases with increase in temperature,
2. What happens to the solubility in the exothermic process with regard to temperature?
3. In the endothermic process, solubility increases with increase in temperature.
4. At a given temperature, an increase in pressure increases the solubility of the gas

Answer:

1. In an exothermic process, the solubility decreases. When calcium oxide dissolves in water, an exothermic reaction takes place, and so the solubility of calcium oxide decreases
2. In an exothermic process, the solubility decreases with the increase in temperature, as there is already an evolution of heat and it is observed.
3. In an endothermic process, the solubility increases. The solubility of KNO₃ in water is an endothermic reaction and so solubility increases with the increase of temperature.
4. At a given temperature, an increase in pressure increases the solubility of gas according to Henry’s law. e.g. (CO₂ in soft drinks)

VII. Hot Questions.

Question 1.
50 ml tincture of benzoin, antiseptic solution contains 10 ml of benzoin. Calculate the volume of percentage of benzoin.
Answer:
Volume of the solute, Benzoin = 10 ml
Volume of the solution, tincture of benzoin = 50 ml

= 20% (v/v)

Question 2.
Neomycin, the antibiotic cream contains 300 mg of neomycin sulphate the active ingredient in 30 g of ointment base. Calculate the mass percentage of neomycin.
Answer:
Mass of neomycin sulphate(solute) = 300 mg
Mass of the ointment (solution) = 30 g
Formula:

VIII. Numerical problems:

Question 1.
Calculate the molality of the solution containing 18 g of Glucose (Molecular mass 180) in 2 kg of water.
Answer:
Mass of Glucose = 18 g
Molecular mass of Glucose = 180

= 0. 05 m

Question 2.
Calculate the molarity of a solution containing 5.85 g of sodium chloride in 500 ml of the water. (Molecular mass = 58.5)
Answer:
Mass of the solute = 5.85 g
Volume of the solution = 500
No. of moles of NaCl = \(\frac{5.85}{58.5}\) = 0.1

= \(\frac{0.1}{500}\) × 1000 = 0.2 M

Students can Download Tamil Nadu 12th Physics Model Question Paper 2 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 2 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.
Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between points A and C is……………………………..
(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) 1/4 μF

Answer:
(b) 2 μF

Question 2.
If the electric field in a region is given by \(\overrightarrow{\mathrm{E}}=5 \hat{i}+4 \hat{j}+9 \hat{k}\), then the electric flux through a surface of area 20 units lying in the
y – z plane will be …………………
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hint. The area vector
\(\overrightarrow{\mathrm{A}}=20 \hat{i} ; \overrightarrow{\mathrm{E}}=(5 \hat{i}+4 \hat{j}+9 \hat{k})\)
Flux ( φ)\(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\) = 5 x 20 = 100 units

Question 3.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is

(a) π Ω
(b) \(\frac{\pi}{2} \Omega\)
(c) 2π Ω
(d) \(\frac{\pi}{4} \Omega\)
Answer:
(b) \(\frac{\pi}{2} \Omega\)

Question 4.
A non-conducting charged ring of charge q, mass m and radius r is rotated with constant angular speed co. Find the ratio of its magnetic moment with angular momentum is …………..
(a) M
(b) \(\frac{3}{\pi} \mathrm{M}\)
(c) \(\frac{2}{\pi} \mathrm{M}\)
(d) \(\frac{1}{2} \mathrm{M}\)

Answer:
(b) \(\frac{3}{\pi} \mathrm{M}\)

Question 5.
A proton enters a magnetic field of flux density 1.5 Wb/m² with a speed of 2 x 10⁷ m/s at angle of 30° with the field. The force on the proton will be ……………….
(a) 0.24 x 10^-12 N
(b) 2.4 x 10 ^-12 N
(c) 24 x 10^-12 N
(d) 0.024 x 10^-12 N
Answer:
(b) 2.4 x 10 ^-12 N
Hint: F = Bqv sin θ = 1.5 x 1.6 x 10^-19 x 2 x 10⁷ x sin 30°= 2.4 x 10^-12 N

Question 6.
In an electrical circuit, R, L, C and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\frac{\pi}{3}\) Instead, if C is removed from the circuit, the phase difference is again \(\frac{\pi}{3}\) . The power factor of the of the circuit is ……………
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{\sqrt{2}}\)
(c) 1
(d) \(\frac{\sqrt{3}}{2}\)
Answer:
(c) 1

Question 7.
The inductance of a coil is proportional to…………………………………….
(a) its length
(b) the number of turns
(c) the resistance of the coil
(d) square of the number of turns
Answer:
(d) square of the number of turns

Question 8.
The electric and magnetic fields of an electromagnetic wave are……………….
(a) in phase and perpendicular to each other
(b) out of phase and not perpendicular to each other
(c) in phase and not perpendicular to each other
(d) out of phase and perpendicular to each other
Answer:
(a) in phase and perpendicular to each other

Question 9.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will ……………
(a) get shifted downwards
(b) get shifted upward
(c) will remain the same
(d) data insufficient to conclude

Answer:
(b) get shifted upward

Question 10.
The wavelength λ_e of an electron and λ_p of a photon of same energy E are related ………..

Answer:
(d) \(\lambda_{p} \propto \lambda_{e}^{2}\)

Question 11.
A system consists of N_o nucleus at t = 0. The number of nuclei remaining after half of a half-life (that  is, at time \(t=\frac{1}{2} \mathrm{T}_{\frac{1}{2}}\)
Answer:

Hint:

Question 12.
In a pure semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called……………………………….
(a) acceptor
(b) donor
(c) intrinsic semiconductor
(d) extrinsic semiconductor
Answer:
(c) intrinsic semiconductor
Hint: Pure semiconductors are called intrinsic semiconductors.

Question 13.
The light emitted in an LED is due to……………………………
(a) Recombination of charge carriers
(b) Reflection of light due to lens action
(c) Amplification of light falling at the junction
Answer:
(a) Recombination of charge carriers

Question 14.
The frequency range of 3 MHz to 30 MHz is used for………………………………..
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation

Question 15.
The materials used in Robotics are……………………..
(a) Aluminium and silver
(b) Silver and gold
(c) Copper and gold
(d) Steel and aluminium
Answer:
(d) Steel and aluminium

Part – III

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

Question 16.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Question 17.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross section of the conductor
\(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\)
The S.I. unit of current density. \(\frac{\mathrm{A}}{\mathrm{m}^{2}}(\text { or }) \mathrm{Am}^{-2}\)

Question 18.
What is magnetic susceptibility?
Answer:
It is defined as the ratio of the intensity of magnetisation \((\overrightarrow{\mathrm{M}})\) induced in the material due to magnetising field \((\overrightarrow{\mathrm{H}})\)
\(\chi_{m}=\left|\frac{\overrightarrow{\mathrm{M}}}{\overrightarrow{\mathrm{H}}}\right|\)

Question 19.
What is meant by electromagnetic induction?
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.

Question 20.
A coil of 200 turns carries a current of 0.4 A. If the magnetic flux of 4 mWb is linked with the coil, find the inductance of the coil.
Answer:
Number of turns, N = 200; Current, I = 0.4 A
Magnetic flux linked with coil, φ = 4 mWb = 4 x 10^-3 Wb
Induction of the coil , L
\(\mathrm{L}=\frac{\mathrm{N} \phi}{\mathrm{I}}=\frac{200 \times 4 \times 10^{-3}}{0.4}=\frac{800 \times 10^{-3}}{0.4}=2 \mathrm{H}\)

Question 21.
Why do stars twinkle?
Answer:
The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Question 22.
How many photons of frequency 10¹⁴ Hz will make up 19.86 J of energy?
Answer:
Total energy emitted per second = Power x time
19.863 = Power x is
∴ Power 19.86 W
Number of photons =

Question 23.
Define curie.
Answer:
One curie was defined as number of decays per second in 1 g of radium and it is equal to 3.7 x 10¹⁰ decays/s

Question 24.
A transistor having α =0.99 and V_BE = 0.7V, is given in the circuit. Find the value of the collector current.
Answer:

Part – III

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

Question 25.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 26.
A conductor of linear mass density 0.2 g m^-1 suspended ,by two flexible wire as shown in figure. Suppose the tension in the supporting wires is zero when it is kept inside the magnetic field of 1 T whose direction is into the x page. Compute the current inside the conductor and also the direction of the current. Assume g = 10 m s^-2 = 111.87.
Answer:

Question 27.
How is Eddy current produced? How do they flow in a conductor?
Answer:
Even for a conductor in the form of a sheet or plate, an emf is induced when magnetic flux linked with it changes. But the difference is that there is no definite loop or path for induced current to flow away. As a result, the induced currents flow in concentric circular paths. As these electric currents resemble eddies of water, these are known as Eddy currents. They are also called Foucault currents.

Question 28.
Explain the concept of intensity of electromagnetic waves.
Answer:
The energy crossing per unit area per unit time and perpendicular to the direction of propagation of electromagnetic wave is called the intensity.
Intensity, I = (u)c

Question 29.
If the focal length is 150 cm for a glass lens, what is the power of the lens?
Answer:
Given: focal length,f = 150 cm (or) f= 1.5 m
Equation for power of lens is, P = 1/f
Substituting the values,
\(P = \frac{1}{1.5}\)= 0.067 diopter
As the power is positive, it is a converging lens.

Question 30.
A proton and an electron have same kinetic energy. Which one has greater de Broglie wavelength. Justify.
Answer:
de-Broglic wavelength of the particle is \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m \mathrm{K}}}\) \(\text { i.e. } \lambda \propto \frac{1}{\sqrt{m}}\)
As m_e <<m_p ,so λ_e >>λ_P
Hence protons have greater de- broglic wavelength

Question 31.
Distinguish between avalanche and zener breakdown.
Answer:

Avalanche Breakdown	Zener Breakdown
It occurs injunctions which are lightly and have wide depletion widths.	It occurs in junctions which are heavily doped and have narrow depletion widths.
It occurs at higher reverse voltages when thermally generated electrons get enough kinetic energy to produce more electrons by collision.	It occurs due to rupture of covalent bonds by strong electric fields set up in depletion region by the reverse voltage.
At reverse voltage above 6V breakdown is due to avalanche effect.	At reverse voltage below 6V breakdown is due to zener effect.
Electric field produced is weak in nature.	A strong electric field is produced
Charge carriers obtain energy from the applied potential.	Zener current is independent of applied voltage.

Question 32.
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.

Question 33.
What are black holes?
Answer:
Black holes are end stage of stars Which are highly dense massive object. Its mass ranges from 20 times mass of the sun to 1 million times mass of the sun. It has very strong gravitational force such that no particle or even light can escape from it. The existence of black holes is studied when the stars orbiting the black hole behave differently from the other starts. Every galaxy has black hole at its center. Sagittarius A* is the black hole at the center of the Milky Way galaxy.

Part – IV

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

Question 34.
(a) How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies.

Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.

Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements
Δq₁, Δq₂, Δq₃ ….. Δq_n ……… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such change elements

Here Δ qi is the i^th charge element, r is the distance of the point P from the z^th charge element and \(\hat{r}_{i \mathrm{P}} \) is the unit vector from ith charge element to the point P.

However, the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit Δq → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form
\(\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \int \frac{d q}{r^{2}} \hat{r}\)

Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r}\) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\overrightarrow{\mathrm{F}}=q \overrightarrow{\mathrm{E}}\)

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = Q/L. Its unit is coulomb per meter (Cm^-1). The charge present in the infinitesimal length dl is dq = λ dl
The electric field due to the line of total charge Q is given by

(b) Surface charge distribution: If the charge Q is uniformly distributed on. a surface of area A, then surface charge density (charge per unit area) is \(\lambda=\frac{Q}{L}\). Its unit is coulomb per square meter (Cm^-2 ). The charge present in the infinitesimal area dA is dq = adA. The electric field due to a of total charge Q is given by

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by \(\rho=\frac{Q}{V}\) . Its unit is coulomb per cubic meter (Cm^-3 ). The charge present in the infinitesimal volume element dV is dq = ρdV.
The electric field due to a volume of total charge Q is given by

[OR]

(b) Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
Ohm’s law: The Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A.

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as
V = El
As we know, the magnitude of current density
\(\mathrm{J}=\sigma \mathrm{E}=\sigma \frac{\mathrm{V}}{l}\)
But \(\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}\),so we write the equation as
\(\frac{\mathrm{I}}{\mathrm{A}}=\sigma \frac{\mathrm{V}}{l}\)
By rearranging the above equations, we get
\(\mathrm{V}=\mathrm{I}\left(\frac{l}{\sigma \mathrm{A}}\right)\)

The quantity \(\frac{l}{\sigma \mathrm{A}}\) is called resistance of the conductor and it is denoted as R. Note that the resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Therefore, the macroscopic form of Ohm’s law can be stated as
V = IR …………….. (3)

Question 35.
(a) Calculate the magnetic held inside and outside of the long solenoid using Ampere’s circuital law.
Answer:
Magnetic field due to a long current carrying solenoid: Consider a solenoid of length L having N turns. The diametre of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

In order to calculate the magnetic field at any point inside the solenoid, we use Ampere’s circuital law. Consider a rectangular loop abed. Then from Ampere’s circuital law.
\(\oint_{C} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mu_{0} \mathrm{I}_{\text {enclosed }}\) = μ₀  x (total current enclosed by Amperian loop)

Since the elemental lengths along be and da are perpendicular to the magnetic field which is along the axis of the solenoid, the integrals.

Since the magnetic field outside the   solenoid is zero, the integral
\(\int_{c}^{d} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=0\)
For the path along ab, the integral is

where the length of the loop ab is h. But the choice of length of the loop ab is arbitrary. We can take very large loop such that it is equal to the length of the solenoid L. Therefore the integral is

Let NI be the current passing through the solenoid of N turns, then

The number of turns per unit length is given by \(\frac{\mathrm{NI}}{\mathrm{L}}=n\) then
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
Since n is a constant for a given solenoid and μ_o is also constant. For a fixed current I, the magnetic field inside the solenoid is also a constant.

[OR]

(b) Explain the construction and working of transformer.
Answer:
Construction and working of transformer:
Principle: The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.

Construction: In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core. The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.

The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.

Working: If the primary coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core. If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coil. This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils.

As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil ε_p is almost equal and opposite to the applied voltage υ_p and is given by
\(\mathrm{v}_{p}=\varepsilon_{p}=-\mathrm{N}_{p} \frac{d \Phi_{\mathrm{B}}}{d t}\)
The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil ε_s is given by
\(\varepsilon_{\mathrm{s}}=-N_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ………………… (1)
where N_p and N_s are the number of turns in the primary and secondary coil, respectively. If the secondary circuit is open, then  ε_s = υ_s where υ_s is the voltage across secondary coil.
\(v_{s}=\varepsilon_{s}=-\mathrm{N}_{s} \frac{d \Phi_{\mathrm{B}}}{d t}\) ……….. (2)

From equations (1) and (2),
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=K\) …………………… (3)

This constant K is known as voltage transformation ratio. For an ideal transformer,
Input power υ_p i_p= Output power υ_si_s
where i_pand i_s are the currents in the primary and secondary coil respectively. Therefore,
\(\frac{v_{s}}{v_{p}}=\frac{N_{s}}{N_{p}}=\frac{i_{p}}{i_{s}}\) ………………….. (4)

Equation (4) is written in terms of amplitude of corresponding quantities,
\(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}}=K\)

(i) If N_s> N_p ( or K > 1), ∴ V_s > V_p and I_s < I_p. This is the case of step-up transformer in which voltage is increased and the corresponding current is decreased.

(ii) If N_s < N_p (or K < 1) , ∴ V_s < V_p and I_s > I_p . This is step-down transformer where voltage is decreased and the current is increased.

Question 36.
(a) Discuss the source of electromagnetic waves.
Answer:
Sources of electromagnetic waves: Any stationary source charge produces only electric field. When the charge moves with uniform velocity, it produces steady current which gives rise to magnetic field (not time dependent, only space dependent) around the conductor in which charge flows. If the charged -particle accelerates, in addition to electric field it also produces magnetic field. Both electric and magnetic fields are time varying fields. Since the electromagnetic waves are transverse waves, the direction of propagation of electromagnetic waves is perpendicular to the plane containing electric and magnetic field vectors.

Any oscillatory motion is also an accelerating motion, so, when the charge oscillates (oscillating molecular dipole) about their mean position, it produces electromagnetic waves. Suppose the electromagnetic field in free space propagates along z direction, and if the electric field vector points along y axis then the magnetic field vector will be mutually perpendicular to both electric field and the propagation vector direction, which means
E_y =E₀ sin (kz-ωt)
B_r = B₀ sin(kz – ωt) where, E_o and B_o are amplitude of oscillating electric and magnetic field,\(\hat{k} \) is a wave number, ω is the angular frequency of the wave and k (unit vector, here it is called propagation vector) denotes the direction of propagation of electromagnetic wave.

Note that both electric field and magnetic field oscillate with a frequency (frequency of electromagnetic wave) which is equal to the frequency of the source (here, oscillating charge is the source for the production of electromagnetic waves). In free space or in vacuum, the ratio between E_o and B_o is equal to the speed of electromagnetic wave, which is equal to speed of light c.
\(c=\frac{E_{0}}{B_{0}}\)

In any medium, the ratio of E_o and B_o is equal to the speed of electromagnetic wave in that medium, mathematically, it can be written as
\(v=\frac{E_{0}}{B_{0}}<c\)
Further, the energy of electromagnetic waves comes from the energy of the oscillating charge.

[OR]

(b) Explain about compound microscope and obtain the equation for magnification.
Answer:
Compound microscope:
The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.

Magnification of compound microscope
From the ray diagram, the linear magnification due to the objective is,
\(m_{0}=\frac{h^{\prime}}{h}\)
from the figure ,\(\tan \beta=\frac{h}{f_{0}}=\frac{h^{\prime}}{L} \), then
\(\frac{h^{\prime}}{h}=\frac{L}{f_{0}} ; m_{o}=\frac{L}{f_{o}}\)

Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length L of the microscope as f_o and f_e are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification m_e of the eyepiece is,
\(m_{e}=1+\frac{D}{f_{e}}\)

The total magnification m in near point focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(1+\frac{D}{f_{e}}\right)\)

If the final image is formed at infinity (normal focusing), the magnification m_e of the eyepiece is
\(m_{e}=\frac{D}{f_{e}}\)

The total magnification m in normal focusing is,
\(m=m_{o} m_{e}=\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 37.
(a) Briefly explain the principle and working of electron microscope.
Answer:
Electron Microscope:
Principle:

• This is the direct application of wave nature of particles. The wave nature of the electron is used in the construction of microscope called electron microscope.
• The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under study. Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths.
• De Broglie wavelength of electron is very much less than (a few thousands less) that of the visible light being used in optical microscopes.
• As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope.
• Electron microscopes giving magnification more than 2,00,000 times are common in research laboratories.

Working:

• The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done.
• The electrons emitted from the source are accelerated by high potentials. The beam is made parallel by magnetic condenser lens. When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample.
• With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all branches of science.

[OR]

(b) Discuss the process of nuclear fission and its properties.
Answer:

• When uranium nucleus is bombarded with a neutron, it breaks up into two smaller nuclei of comparable masses with the release of energy.
• The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.
• The fission is accompanied by the release of neutrons. The energy that is released in the nuclear fission is of many orders of magnitude greater than the energy released in chemical reactions.
• Uranium undergoes fission reaction in 90 different ways. The most common fission reactions of
  
• Here Q is energy released during the decay of each uranium nuclei. When the slow neutron is absorbed by the uranium nuclei, the mass number increases by one and goes to an excited state. \(_{ 92 }^{ 236 }{ U }\) . But this excited state does not last longer than 10^-12s and decay into two daughter nuclei along with 2 or 3 neutrons. From each reaction, on an average, 2.5 neutrons are emitted.

Question 38.
(a) Transistor functions as a switch. Explain.
Answer:
The transistor in saturation and cut-off regions functions like an electronic switch that helps to. turn ON or OFF a given circuit by a small control signal.

Presence of dc source at the input (saturation region):
When a high input voltage (V. = +5V) is applied, the base current (I ) increases and in turn increases the collector current. The transistor will move into the saturation region (turned ON). The increase in collector current (I_c) increases the voltage drop across R_c .thereby lowering the output voltage, close to zero. The transistor acts like a closed switch and is equivalent to ON condition.

Absence of dc source at the input (cut-off region):
A low input voltage (V_in = OV), decreases the base current (I_B) and in turn decreases the collector current (I_c ). The transistor will move into the cut-off region (turned OFF). The decrease in collector current (I_c) decreases the drop across, thereby increasing the output voltage, dose to +5 V. The transistor acts as an open switch which is considered as the OFF condition.

It is manifested that, a high input gives a low output and a low input gives a high output. In addition, we can say that the output voltage is opposite to the applied input voltage. Therefore, a transistor can be used as an inverter in computer logic circuitry

[OR]

(b) What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation. There are 3 types of modulation based on which parameter is modified.
They are

1. Amplitude modulation,
2. Frequency modulation, and
3. Phase modulation.

1. Amplitude Modulation (AM): If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure
(a) is the message signal or baseband signal that carries information, figure
(b) shows the high-frequency carrier signal and figure
(c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal.

(ii) Frequency Modulation (FM):
The frequency of the carrier signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. An increase in the amplitude of the ‘ baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave.

Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction (A, C). The increase in amplitude in the negative half cycle (B, D) reduces the frequency of the modulated wave (Figure (c)).

(iii) Phase Modulation (PM)
The instantaneous amplitude of the baseband signal modifies the phase of the carrier signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency.

The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.

Students can download 10th Science Chapter 8 Periodic Classification of Elements 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 8 Periodic Classification of Elements

Samacheer Kalvi 10th Science Periodic Classification of Elements Text Book Back Questions and Answers

I. Choose the best answer:

Question 1.
The number of periods and groups in the periodic table are:
(a) 6, 16
(b) 7, 17
(c) 8, 18
(d) 7, 18
Answer:
(d) 7, 18

Question 2.
The basis of modem periodic law is ____.
(a) atomic number
(b) atomic mass
(c) isotopic mass
(d) number of neutrons.
Answer:
(a) atomic number

Question 3.
……….. group contains the member of the halogen family.
(a) 17th
(b) 15th
(c) 18th
(d) 16th
Answer:
(a) 17th

Question 4.
_______ is a relative periodic property.
(a) atomic radii
(b) ionic radii
(c) electron affinity
(d) electronegativity.
Answer:
(b) ionic radii

Question 5.
Chemical formula of rust is:
(a) Fe0.xH₂O
(b) Fe04.xH₂O
(c) Fe₂O₃. xH₂O
(d) FeO
Answer:
(c) Fe₂O₃. xH₂O

Question 6.
In the aluminothermic process the role of Al is:
(a) oxidizing agent
(b) reducing agent
(c) hydrogenating agent
(d) sulphurising agent
Answer:
(b) reducing agent

Question 7.
The process of coating the surface of the metal with a thin layer of zinc is called ____.
(a) painting
(b) thinning
(c) galvanization
(d) electroplating.
Answer:
(c) galvanization

Question 8.
Which of the following inert gas has electrons in the outermost shell?
(a) He
(b) Ne
(c) Ar
(d) Kr
Answer:
(a) He

Question 9.
Neon shows zero electron affinity due to ____.
(a) stable arrangement of neutrons
(b) stable configuration of electrons
(c) reduced size
(d) increased density.
Answer:
(b) stable configuration of electrons

Question 10.
……….. is an important metal to form amalgam.
(a) Ag
(b) Hg
(c) Mg
(d) Al
Answer:
(b) Hg

II. Fill in the blanks:

1. If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding is ………..
2. …………. is the longest period in the periodical table.
3. ………… forms the basis of modern periodic table.
4. If the distance between two Cl atoms in Cl₂ molecule is 1.98 Å, then the radius of Cl atom is ………..
5. Among the given species A^– A⁺, and A, the smallest one in size is ……….
6. The scientist who propounded the modern periodic law is …………
7. Across the period, ionic radii ………… (increases,decreases).
8. ……….. and ………… are called inner transition elements.
9. The chief ore of Aluminium is ………….
10. The chemical name of rust is ………….
Answer:
1. ionic
2. 6th (sixth) period
3. Atomic number
4. 0.99 Å
5. A⁺
6. Dimitri Mendeleev
7. decreases
8. Lanthanides, Actinides
9. bauxite
10. hydrated ferric oxide

III. Match the following:

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

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

1. Moseley’s periodic table is based on atomic mass.
2. Ionic radius increases across the period from left to right.
3. All ores are minerals; but all minerals cannot be called as ores.
4. Aluminium wires are used as electric cables due to their silvery white colour.
5. An alloy is a heterogenous mixture of metals.
Answer:
1. False – Moseley’s periodic table is based on atomic number.
2. True
3. True
4. False – Aluminium wires are used as electric cables because it is a good conductor of heat and electricity.
5. False – An alloy is an homogeneous mixture of metals.

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: The nature of bond in HF molecule is ionic.
Reason: The electronegativity difference between H and F is 1.9.
(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: Magnesium is used to protect steel from rusting.
Reason: Magnesium is more reactive than iron.
(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 3.
Assertion: An uncleaned copper vessel is covered with greenish layer. Reason: copper is not attacked by alkali.
(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:
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.

VI. Short answer questions:

Question 1.
A is a reddish brown metal, which combines with O₂ at < 1370 K gives B, a black coloured compound. At a temperature > 1370 K, A gives C which is red in colour. Find A, B and C with reaction.
Answer:
(A) is a reddish brown metal – Copper

A – Copper; B – Cupric oxide; C – Cuprous oxide.

Question 2.
A is a silvery white metal. A combines with O₂ to form B at 800°C, the alloy of A is used in making the aircraft. Find A and B.
Answer:
A – Silvery white metal – Aluminium

The alloys of aluminium, Duralumin and Magnalium are used in making the aircraft.
A – Aluminium; B – Aluminium oxide.

Question 3.
What is rust? Give the equation for the formation of rust.
Answer:
When iron is exposed to moist air, it forms a layer of brown hydrated ferric oxide on its surface. This compound is known as rust and the phenomenon of formation of rust is known as rusting.
4Fe + 3O₂ + xH₂O → 2Fe₂O₃. xH₂O (Rust).

Question 4.
State two conditions necessary for rusting of iron.
Answer:
(i) The presence of water and oxygen is essential for the rusting of iron.
(ii) Impurities in the iron, the presence of water vapour, acids, salts and CO₂ speeds up rusting.

VII. Long answer questions:

Question 1.
(a) State the reason for addition of caustic alkali to bauxite ore during purification of bauxite.
Answer:
Caustic alkali is added to bauxite, to dissolve bauxite ore and obtain a solution of sodium aluminate.

(b) Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.
Answer:
CaF₂ (Fluorspar) is added along with cryolite and alumina. It is added to reduce the high melting point of the electrolyte.

Question 2.
The electronic configuration of metal A is 2, 8, 18, 1.
The metal A when exposed to air and moisture forms B a green layered compound. A with con. H₂SO₄ forms C and D along with water. D is a gaseous compound. Find A, B, C and D.
Answer:
Metal (A) with electronic configuration- 2, 8, 18, 1 is copper.
2Cu + O₂ + CO₂ + H₂O → CuCO₃. Cu(OH)₂ (B)
Copper carbonate (Green layer)

(A) – Copper (Cu)
(B) – Copper Carbonate (CuCO₃. Cu(OH)₂)
(C) – Copper Sulphate (CuSO₄)
(D) – Sulphur dioxide (SO₂)

Question 3.
Explain the smelting process.
Answer:
The roasted ore of copper is mixed with powdered coke and sand and is heated in a blast furnace to obtain matte (Cu₂S + FeS) and slag. The slag is removed as waste.
2 FeS + 3O₂ → 2 FeO + 2 SO₂
2 Cu₂S + 3O₂ → 2 Cu₂O + 2SO₂
Cu₂O + FeS → Cu₂S + FeO
FeO + SiO₂ → FeSiO₂ (slag)

VIII. HOT questions:

Question 1.
Metal A belongs to period 3 and group 13. A in red hot condition reacts with steam to form B. A with strong alkali forms C. Find A, B and C with reactions.
Answer:
Metal A belongs to Period 3 and Group 13. So metal ‘A’ is aluminium.
(A) in red hot condition reacts with steam to form ‘B’.

‘A’ with strong alkali forms ‘C’

(A) – Aluminium
(B) – Aluminium oxide
(C) – Sodium meta aluminate

Question 2.
Name the acid that renders aluminium passive. Why?
Answer:
Dilute or concentrated nitric acid (HNO₃) renders aluminium passive. Because nitric acid does not attack aluminium but it renders aluminium passive due to the formation of an oxide film on its surface.

Question 3.
(a) Identify the bond between H and F in HF molecule.
Answer:
Ionic, because the electronegativity difference is more than 1.7.

(b) What property forms the basis of identification?
Answer:
Electronegativity.

(c) How does the property vary in periods and in groups?
Answer:
In a period, from left to right the electronegativity increases because of the increase in the nuclear charge.
In a Group, from top to bottom, the electronegativity decreases because of the increase in size of the elements.

Samacheer Kalvi 10th Science Periodic Classification of Elements Additional Important Questions and Answers

I. Choose the correct answer:

Question 1.
The shortest period in the periodic table contains elements.
(a) 18
(b) 8
(c) 2
(d) 32
Answer:
(c) 2

Question 2.
Group number of carbon family is _____.
(a) 13
(b) 15
(c) 17
(d) 14.
Answer:
(d) 14.

Question 3.
The ore forming elements, chalcogens are present in ……… group of the modern periodic table.
(a) 18th
(b) 1st
(c) 2nd
(d) 16th
Answer:
(d) 16th

Question 4.
Valency of all the alkali metals is _____.
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(a) 1

Question 5.
The largest atom in the 2nd period of the periodic table is:
(a) Li
(b) Be
(c) F
(d) Ne
Answer:
(a) Li

Question 6.
The covalent radii of Hydrogen, if the distance between the Hydrogen nuclei of the molecule is 0.74 Å is:
(a) 1.58 Å
(b) \(\frac{0.74}{4}\) Å
(c) 0.37 Å
(d) 0.74 Å
Answer:
(c) 0.37 Å

Question 7.
Pick out the correct ionic radii in increasing order for the following species – Na, Cl, Na⁺, Cl^– _____.
(a) Na > Cl > Na⁺ > Cl^–
(b) Cl > Na > Na⁺ > Cl^–
(c) Cl^– > Na > Na⁺ > Cl
(d) Cl < Na⁺ < Cl^– < Na.
Answer:
(d) Cl < Na⁺ < Cl^– < Na.
Hint:
Na = 186 pm,
Cl = 91 pm,
Na⁺ = 116 pm,
Cl^– = 167 pm.

Question 8.
In the third period, the first ionization potential is of the order:
(a) Na > Al > Mg > Si > P
(b) Mg > Na > Si > P > Al
(c) Na < Al < Mg < Si < P
(d) Na < Al < Mg < Si < P
Answer:
(c) Na < Al < Mg < Si < P

Question 9.
Which one of the following is the least electronegative element?
(a) Bromine
(b) Chlorine
(c) Iodine
(d) Hydrogen
Answer:
(d) Hydrogen

Question 10.
Which is a widely used a scale to determine the electronegativity?
(a) Pauling scale
(b) Moseley scale
(c) Mendeleev scale
(d) none of these.
Answer:
(a) Pauling scale

Question 11.
Which one of the following orders of ionic radii is correct?
(a) H^– > H⁺ > H
(b) Na⁺ > F^– > O^2-
(c) F > O^2- > Na⁺
(d) None of these
Answer:
(d) None of these

Question 12.
The percentage of carbon in Pig iron is:
(a) < 0.25%
(b) 0.25 – 2%
(c) 2 – 4.5%
(d) > 5%
Answer:
(c) 2 – 4.5%

Question 13.
The chemical formula of clay is _____.
(a) Al₂O₃
(b) Al₂O₃.2H₂O
(c) Al₂O₃. 2SiO₂.2H₂O
(d) Al₂O₃. 2SiO₂.H₂O.
Answer:
(c) Al₂O₃. 2SiO₂.2H₂O

Question 14.
The temperature in the combustion zone is maintained at:
(a) 1500°C
(b) 400°C
(c) 1000°C
(d) 1380°C
Answer:
(a) 1500°C

Question 15.
Oil used in Froth floatation method is _____.
(a) pine oil
(b) natural oil
(c) crude oil
(d) Synthetic oil.
Answer:
(a) pine oil

Question 16.
The first most abundant metal present in the Earth crust is:
(a) Iron
(b) Aluminium
(c) Zinc
(d) Copper
Answer:
(b) Aluminium

Question 17.
……….. metal is used for making calorimeters.
(a) Copper
(b) Tin
(c) Mercury
(d) Iron
Answer:
(a) Copper

Question 18.
More reactive metal is _____.
(a) Zn
(b) Fe
(c) Ag
(d) Na.
Answer:
(d) Na.

Question 19.
The chief ore of Iron is:
(a) Magnesite
(b) Galena
(c) Cinnabar
(d) Haematite
Answer:
(d) Haematite

Question 20.
The metal which melts at room temperature is:
(a) Zinc
(b) Lead
(c) Gallium
(d) Tin
Answer:
(c) Gallium

Question 21.
Conversion of bauxite into alumina is _____.
(a) Hall’s process
(b) Alumino thermic process
(c) Baeyer’s process
(d) Bessemerisation process.
Answer:
(c) Baeyer’s process

Question 22.
………. metal can be cut with knife.
(a) Potassium
(b) Gallium
(c) Mercury
(d) Gold
Answer:
(a) Potassium

Question 23.
………. is not a good conductor of heat and electricity.
(a) Silver
(b) Tungsten
(c) Copper
(d) Aluminium
Answer:
(b) Tungsten

Question 24.
Electrolyte used in Hall’s process ______.
(a) Pure alumina + molten cryolite + fluorspar
(b) Pure alumina + molten bauxite + fluorspar
(c) Pure bauxite + molten cryolite + fluorspar
(d) Pure bauxite + molten Haematite + fluorspar.
Answer:
(a) Pure alumina + molten cryolite + fluorspar

Question 25.
The foaming agent used for froth floatation process is:
(a) Coconut oil
(b) Pine oil
(c) Sodium chloride
(d) Groundnut oil
Answer:
(b) Pine oil

Question 26.
Three elements A, B and C are having the electronic configuration Is² 2s¹, Is² 2s² and Is² 2s² 2p¹ respectively. Which element will have the lowest ionization energy?
(a) A
(b) B
(c) C
(d) B and C
Answer:
(a) A

Question 27.
Metal used in household utensils is ______.
(a) Al
(b) Co
(c) Fe
(d) Na.
Answer:
(a) Al

Question 28.
Which one of the following pair is a metalloid?
(a) Na and K
(b) F and Cl
(c) Cu and Hg
(d) Si and Ge
Answer:
(d) Si and Ge

Question 29.
The highly metallic element will have the configuration of:
(a) 2, 8, 7
(b) 2, 8, 8, 5
(c) 2, 8, 8, 1
(d) 2, 8, 2
Answer:
(c) 2, 8, 8, 1

Question 30.
The metal used in electroplating is ______.
(a) Cu
(b) Al
(c) Fe
(d) Co.
Answer:
(a) Cu

Question 31.
The flux which is used when the gangue present in the ore is acidic:
(a) Silica
(b) Calcium oxide
(c) Calcium silicate
(d) Cuprous sulphide
Answer:
(b) Calcium oxide

Question 32.
Matte is a mixture of:
(a) Cu₂O + Cu₂S
(b) Cu₂O + FeS
(c) Cu₂S + FeS
(d) Cu₂O + PbS
Answer:
(c) Cu₂S + FeS

Question 33.
Fe reacts with dilute nitric acid in cold condition to give ______.
(a) Ferrous nitride
(b) Ferrous nitrate
(c) Ferric nitride
(d) Ferric nitrate.
Answer:
(b) Ferrous nitrate

Question 34.
In the brass alloy, which is solvent?
(a) Zn
(b) Co
(c) Ag
(d) Cu.
Answer:
(d) Cu.

II. Fill in the blanks:

1. The major component of the matte is ………….
2. The modern periodic table is based on ………..
3. The valency of alkali metals is …………
4. The unreactive elements are present in group ………..
5. In the 2nd period, the smallest atom is ……….
6. The size of a cation is ………… than the neutral atom.
7. ……….. is the unit of ionization energy.
8. The ionization energy ……… down the group in the periodic table.
9. The electron affinities of noble gases are …………
10. ………. is the process of extracting the ore from the Earth’s crust.
11. Galena is the ore of …………..
12. The silvery white metal which is a good conductor of heat and electricity is …………
13. The slag formed during Bessemerisation process is ………….
14. Blister copper contains ………. percentage of copper.
15. Haematite ore is concentrated by ……….. washing.
Answer:
1. Cu₂S
2. atomic number
3. one
4. 18
5. Neon
6. smaller
7. KJ/mol
8. decreases
9. zero
10. Mining
11. lead
12. aluminium
13. Iron silicate or FeSiO₃
14. 98%
15. hydraulic

III. Match the following:

Question 1.
Match the column I with column II.

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

Question 2.
Match the column I with column II.

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

Question 3.
Match the column I with column II.

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

Question 4.
Match the column I with column II.

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

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

1. Alkali metals are generally extracted by the electrolysis of their ores in fused state.
2. Every mineral is an ore but every ore is not a mineral.
3. Slag is a product formed during smelting by combination of flux and impurities.
4. Reactive metals occur in native state.
5. Malachite is a sulphide ore of copper.
6. Lanthanides are present in the 6th group of the periodic table.
7. Atomic radius increases as we go across the period due to increase in size.
8. As the positive charge increases, the size of the cation decreases.
9. If the difference in electronegativity is greater than 1.7, the bond is considered to be covalent.
10. Siderite is the carbonate ore of calcium.
Answer:
1.True
2. True
3. True
4. False – Reactive metals always occur in the combined state.
5. False – Malachite is the carbonate ore of copper.
6. False – Lanthanides are present in the 6th period of the periodic table.
7. False – Atomic radius increases as we go across the period due to decrease in size.
8. True
9. False – If the difference in electronegativity is greater than 1.7, the bond is considered to be ionic.
10. False – Siderite is the carbonate ore of Iron.

V. Assertion and Reason:

Answer the following questions using the data given below:
Question 1.
Assertion: Zinc blende is concentrated by Froth floatation process.
Reason: Zinc blende is a sulphide ore.
(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: In thermite welding, aluminium powder and Fe₂O₃ are used.
Reason: Aluminium powder is a strong reducing agent.
(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 3.
Assertion: To design the body of an aircraft, aluminium alloys are used.
Reason: Aluminium becomes passive when it is treated with dil or con.HNO₃
(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:
(d) Assertion and Reason are correct, Reason doesn’t explains Assertion.

Question 4.
Assertion: Tinstone and the impurity wolframite are seperated by magnetic separation.
Reason: Tinstone is magnetic and wolframite is non-magnetic in nature.
(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 5.
Assertion: Bauxite is purified by leaching.
Reason: Bauxite undergoes thermal decomposition.
(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:
(b) Assertion is correct, Reason is wrong.

VI. Short answer questions:

Question 1.
State the modern periodic law.
Answer:
The physical and chemical properties of the elements are the periodic function of their atomic number.

Question 2.
‘X’ is a silvery white metal. X reacts with O₂ to form Y. The same compound is obtained from the metal on reaction with steam with the liberation of hydrogen gas. Identify X and Y.
Answer:
(i) Silvery white metal ‘X’ is Aluminium.
(ii) It reacts with O₂ to form ‘Y’

(iii) Y can also be obtained on reaction with steam with the liberation of H₂.

Question 3.
Write any four characteristics of periods.
Answer:

• In a period, the electrons are filled in the same valence shell of all elements.
• As the electronic configuration changes along the period, the chemical properties of the elements also change.
• The atomic size of the elements in a period decreases from left to right
• In a period, die metallic character of the element decreases, while their non-metallic character increases.

Question 4.
Write the Principle of Hydraulic washing.
Answer:
The difference in the densities or specific gravities of the ore and the gangue is the main principle behind this method.

Question 5.
What are coinage metals?
Answer:
Copper, silver and gold are called coinage metals, as they are used in making coins and jewellery.

Question 6.
How will you separate tinstone from wolframite?
Answer:
Magnetic separation method. Tinstone is magnetic in nature.
Method: The crushed ore is placed over a conveyer belt which rotates around two metal wheels, one of which is magnetic. The magnetic particles are attracted to the magnetic wheel and fall separately apart from the non¬magnetic particles.

Question 7.
What are ores?
Answer:
The mineral from which a metal can be readily and economically extracted on a large scale is said to be ore.
eg. Bauxite Al₂O₃.2H₂O is the ore of Aluminium

Question 8.
Define electronegativity.
Answer:
It is the tendency of an element in a covalent bond to attract the shared pair of electrons towards itself. It is a relative property.

Question 9.
In what period and group will an element with z = 118 will be present?
Answer:
Elements with z = 118 will be present in Period number ‘7’ and Group number 18.

Question 10.
Why flux is added during metallurgy?
Answer:
Flux is the substance added to the ore to reduce the fusion temperature and to remove impurities.
e.g. CaO, SiO₂

Question 11.
State the trends in the electronegativity in a Group and period.
Answer:
In a Group: Electronegativity decreases in a group because of the increased number of energy levels.
In a Period: The electronegativity increases because the increase in the nuclear charge.

Question 12.
Write a note about smelting.
Answer:
Smelting is a process of reducing the roasted metallic oxide to metal in a molten condition. In this process, impurities are removed by the addition of flux as slag.

Question 13.
Write the formula of the ores of Aluminium.
Answer:

Question 14.
Explain the action of Aluminium with air.
Answer:
On heating at 800°C, aluminium bums in the air very brightly forming its oxide and nitride.
4Al + 3O₂ → 2Al₂O₃ (Aluminium oxide)
2Al + N₂ → 2AlN (Aluminium nitride).

Question 15.
What happen when Aluminium reacts with steam?
Answer:
When steam is passed over red hot aluminium, H₂ gas is evolved.

Question 16.
Write the reaction of Aluminium with Sodium hydroxide?
Answer:

Question 17.
Explain the electrolytic refining of copper.
Answer:
Cathode: A thin plate of pure copper metal.
Anode: A block of impure copper metal.
Electrolyte: Copper sulphate solution + dilute H₂SO₄
When an electric current is passed through the electrolytic solution, pure copper gets deposited at the cathode and the impurities are settled at the bottom of the anode as anode mud.

Question 18.
Mention the uses of Aluminium.
Answer:
Aluminium is used in

1. household utensils
2. electrical cable industry
3. making aeroplanes and other industrial machine parts.

Question 19.
What are the methods employed to make an alloy?
Answer:

1. By fusing the metals together. Eg: Brass is made by melting zinc and copper.
2. By compressing finely divided metals. Eg: wood mexai.

Question 20.
Write the components of wood metal.
Answer:
Wood metal is an alloy of Lead, Tin, Bismuth and Cadmium.

Question 21.
What are the uses of copper?
Answer:

• Copper is used in manufacturing electric cables and other electric appliances.
• Copper is used for making utensils, containers, calorimeters and coins.
• Copper is used in electroplating.
• Copper is alloyed with gold and silver for making coins and jewels.

Question 22.
Give example for non-ferrous copper and aluminium alloys. Non-ferrous copper alloys: Brass (Cu, Zn), Bronze (Cu, Sn)
Answer:
Non-ferrous aluminium alloys: Duralumin (Al, Mg, Cu, Mn), Magnalium (Al, Mg)

Question 23.
How is rust formed?
Answer:
When iron is exposed to moist air, it forms a layer of brown hydrated Ferric oxide on its surface. This compound is known as rust.
4Fe + 3O₂ + xH₂O → 2Fe₂O₃. xH₂O (Rust).

Question 24.
Why are the alloys prepared?
Answer:

1. To modify appearance and colour.
2. To modify chemical activity.
3. To lower the melting point.
4. To increase hardness and tensile strength.
5. To increase resistance to electricity.

Question 25.
Define corrosion.
Answer:
Corrosion is the gradual destruction of metals by chemical or electrochemical reaction with the environment.

Question 26.
What are alloys? How are they prepared?
Answer:

• An alloy is a homogeneous mixture of a metal with other metals or with non-metals that are fused together. e.g. Brass is an alloy of zinc (solute) in copper (solvent)
• Alloys are prepared by fusing the metals together.
• Alloys are prepared by compressing finely divided metals one over the other.

Question 27.
Which is known as Wet corrosion or Electrochemical corrosion?
Answer:
The corrosive action in the presence of moisture is called wet corrosion. It occurs as a result of electrochemical reaction of metal with water or aqueous solution of salt or acids or bases.

Question 28.
Write a note on Cathodic protection.
Answer:
It is the method of controlling corrosion of a metal surface protected is coated with the metal which is easily corrodible. The easily corrodible metal is called Sacrificial metal to act as anode ensuring cathodic protection.

Question 29.
What are the methods used to prevent corrosion?
Answer:
Corrosion of metals is prevented

• by coating with paints
• by coating with oil and grease
• by alloying with other metals
• by the process of galvanization
• by electroplating
• by sacrificial protection

Question 30.
A reddish brown metal ‘A’ reacts with dil.HCl in the presence of O₂ and forms the compound ‘B’. ‘B’ can also be prepared by heating the metal A with Cl₂. Identify A and B.
Answer:
Reddish brown metal ‘A’ is copper.
(A) reacts with dil.HCl in the presence of O₂ and forms CuCl₂ which is ‘B’.

(B) can also prepared by the action of Cl₂.

Question 31.
Write the uses of copper.
Answer:

1. It is extensively used in manufacturing electric cables and other electric appliances.
2. It is used for making utensils, containers, calorimeters and coins.
3. It is used in electroplating.
4. It is alloyed with gold and silver for making coins and jewels.

Question 32.
Write the name and formula of the ores of iron.
Answer:

Question 33.
Define periodicity.
Answer:
The electronic configurations of elements help us to explain the periodic recurrence of physical and chemical properties. Anything which repeats itself after a regular interval is called periodic and this behaviour is called periodicity.

Question 34.
What happens in the combustion zone during the extraction of iron.
Answer:
The temperature in the combustion zone is 150°C. In this region coke bums 02 to form CO₂, when the charge comes in contact with a hot blast of air.

Question 35.
Explain the reactions taking place in the reduction zone.
Answer:
In the upper region of reduction zone, the temperature is at 400°C. In this region CO reduces ferric oxide to form spongy iron.

Question 36.
Define Metallic radius.
Answer:
It is defined as half the distance between the nuclei of adjacent metal atoms.

Question 37.
Complete the following reactions.

1. 4Fe + 10HNO₃ → 4Fe(NO₃)₂ + ………. + 3H₂O
2. 2Fe + 6H₂SO₄ → Fe₂(SO₄)₃ + ………. + 6H₂O

Answer:

1. NH₄NO₃
2. 3SO₂

Question 38.
What happens when steam is passed over red hot iron?
Answer:
When steam is passed over red hot iron magnetic oxide is formed.
3Fe + 4H₂O (steam) → Fe₃O₄ + 4H₂↑

Question 39.
Define Electron affinity.
Answer:
Electron affinity is the amount of energy released when a gaseous atom gains an electron to form its anion. It is also measured in kJ / mol.

Question 40.
Complete the table.
Answer:

Question 41.
Define Metallurgy.
Answer:
Metallurgy is a science of extracting metals from their ores and modifying the metals into alloys for various uses, based on their physical and chemical properties and their structural arrangement of atoms.

Question 42.
Write a short note on leaching or chemical process.
Answer:
This method is employed when the ore is in a very pure form. The ore is treated with a suitable reagent such that the ore is soluble in it but the impurities are not. The impurities are removed by filtration. The solution of the ore, ie., the filtrate is treated with a suitable reagent which precipitates the ore.
E.g. Bauxite Al₂O₃.2H₂O, the ore of aluminium.

Question 43.
Relate all the four columns of the table with their unique properties.
Answer:

Question 44.
Guess Who I am?
(i) I am preserved in Kerosene.
Answer:
Sodium

(ii) My ore is leached with NaOH.
Answer:
Aluminium

(iii) I sacrifice myself to protect my friend Iron.
Answer:
Magnesium

(iv) I am being used in propellers
Answer:
Nickel steel

Question 45.
Explain the method of making alloys.
Answer:

• By fusing the metals together. E.g. Brass is made by melting zinc and copper.
• By compressing finely divided metals. E.g. Wood metal: an alloy of lead, tin, bismuth and cadmium powder is a fusible alloy.

Question 46.
Write the differences between a mineral and a ore.
Answer:

VII. Long answer questions:

Question 1.
Write the reactions taking place during Bessemerisation of copper.
Answer:
2 FeS + 3O₂ → 2FeO + 2SO₂↑

2Cu₂S + 3O₂ → 2 Cu₂O + 2 SO₂↑
2Cu₂O + Cu₂S → 6 Cu + SO₂↑

Question 2.
How do electronegativity values help to find out the nature of bonding between atoms?
Answer:

• If the difference in electronegativity between two elements is 1.7, the bond has 50% ionic character and 50 % covalent character.
• If the difference is less than 1.7, the bond is considered to be covalent.
• If the difference is greater than 1.7, the bond is considered to be ionic.

Question 3.
Explain Froth floatation with diagram.
Answer:
Principle: This process depends on the preferential wettability of the ore with oil (pine oil) and the gangue particles by water. Lighter ores, such as sulphide ores, are concentrated by this method. Eg: Zinc blende (ZnS).

Question 4.
Explain the Baeyer’s process of conversion of Bauxite into alumina.
Answer:
(i) Bauxite ore is finely ground and heated under pressure with a solution of concentrated caustic soda solution at 150°C to obtain sodium meta aluminate.
(ii) On diluting sodium meta aluminate with water, a precipitate of aluminium hydroxide is formed.
(iii) The precipitate is filtered, washed, dried and ignited at 1000°C to get alumina.

Question 5.
Explain the Hall’s Process of electrolytic reduction of alumina with diagram.
Answer:
Hall’s Process:
Aluminium is produced by the electrolytic reduction of fused alumina (Al₂O₃) in the electrolytic cell.
Cathode : Iron tank linked with graphite
Anode : A bunch of graphite rods suspended in molten electrolyte.
Electrolyte : Pure alumina + molten cryolite + fluorspar (fluorspar lowers the fusion temperature of electrolyte)

Temperature: 900 – 950 °C
Voltage used: 5 – 6 V
Overall reaction: 2 Al₂O₃ → 4 Al + 3O₂↑

Question 6.
Write the reaction involved in the middle region of blast furnace during the extraction of iron.
Answer:
The Middle Region (Fusion Zone): The temperature prevails at 1000°C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂.

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

Question 7.
What are the three different types of iron? Write their uses.
Answer:
(i) Pig iron (Iron with 2-4.5% of carbon): It is used in making pipes, stoves, radiators, railings, manhole covers and drain pipes.
(ii) Steel (Iron with < 0.25% of carbon): It is used in the construction of buildings, machinery, transmission cables and T. V towers and in making alloys.
(iii) Wrought iron (Iron with 0.25-2% of wraught carbon): It is used in making springs, anchors and electromagnets.

Question 8.
What is corrosion? Write the chemistry behind the formation of rust.
Answer:
(i) The slow and steady destruction of a metal by chemical or electro chemical reaction with the environment.
(ii) When the surface of iron is exposed to moisture and other gases present in the atmosphere, the following chemical reaction takes place.
Fe → Fe²⁺ + 2e^–
O₂ + 2H₂O + 4e^– → 4OH^–
O₂ + 4H⁺ + 4e^– → 2H₂O
The Fe²⁺ ions are oxidised to Fe³⁺ ions.
The Fe³⁺ ions combine OH^– ions to form Fe(OH)₃. This becomes rust which is hydrated ferric oxide with the formula Fe₂O₃.xH₂O. It is a reddish brown substance.

Question 9.
Explain the methods of preventing corrosion.
Answer:
(i) Alloying : The metals can be alloyed to prevent the process of corrosion. Eg: Stainless Steel

(ii) Surface Coating : It involves application of a protective coating over the metal. It is of the following types:
(a) Galvanization: It is the process of coating zinc on iron sheets by using electric current.
(b) Electroplating: It is a method of coating one metal over another metal by passing electric current.
(c) Anodizing: It is an electrochemical process that converts the metal surface into a decorative, durable and corrosion resistant. Aluminium is widely used for anodizing process.
(d) Cathodic Protection: It is the method of controlling corrosion of a metal surface protected is coated with the metal which is easily corrodible. The easily corrodible metal is called sacrificial metal to act as anode ensuring cathodic protection.

Question 10.
Discuss the main featured of Periods in the modern periodic table (or) long form of periodic table.
Answer:
The horizontal rows are called periods.There are seven periods in the periodic table.

1. First period (Atomic number 1 and 2): This is the shortest period. It contains only two elements (Hydrogen and Helium).
2. Second period (Atomic number 3 to 10): This is a short period. It contains eight elements (Lithium to Neon).
3. Third period (Atomic number 11 to 18): This is also a short period. It contains eight elements (Sodium to Argon).
4. Fourth period (Atomic number 19 to 36): This is a long period. It contains eighteen elements (Potassium to Krypton). This includes 8 normal elements and 10 transition elements.
5. Fifth period (Atomic number 37 to 54): This is also a long period. It contains 18 elements (Rubidium to Xenon). This includes 8 normal elements and 10 transition elements.
6. Sixth period (Atomic number 55 to 86): This is the longest period. It contains 32 elements (Caesium to Radon). This includes 8 normal elements, 10 transition elements and 14 inner transition elements (Lanthanides).
7. Seventh period (Atomic number 87 to 118): Like the sixth period, this period also accommodates 32 elements. Recently 4 elements have been included by IUPAC.

Question 11.
Discuss the main feature of Groups in the long form of periodic table.
Answer:
(i) The vertical columns in the periodic table starting from top to bottom are called groups. There are 18 groups in the periodic table.

(ii) Based on the common characteristics of elements in each group, they can be grouped as various families.

(iii) The Lanthanides and Actinides, which form part of Group 3 are called inner transition elements.

(iv) Except ‘group O’, all the elements present in each group have the same number of electrons in their valence shell and thus have the same valency. Eg: all the elements of group 1 have one electron in their valence shells (Is¹). So, the valency of all the alkali metals is ‘ 1’.

(v) As the elements present in a group have identical valence shell electronic configurations, they possess similar chemical properties.

(vi) The physical properties of the elements in a group such as melting point, boiling point and density vary gradually.

(vii) The atoms of the ‘group 0’ elements have stable electronic configuration in their valence shells and hence they are unreactive.

VIII. Hot Questions:

Question 1.
Why noble gases have zero electron affinity value?
Answer:
Noble gases show no tendency to accept electrons because the outers and p orbitals of noble gases are completely filled. No more electrons can be added to them and hence their electron affinities are zero.

Question 2.
Arrange the following ions in order of their increasing ionic radii.
Answer:
Li⁺, Mg²⁺, K⁺ Al³⁺
Al³⁺ < Li⁺ < Mg²⁺ < K⁺

Question 3.
Cationic radius is smaller than its corresponding neutral atom. Why?
Answer:
When a neutral atom lose one or more electrons it forms a cation.
Na → Na⁺ + e^–
The radius of this cation (r_Na+)is decreased than its parent atom (r_Na).
When an atom is charged to cation, the number of nuclear charges becomes greater than the number of orbital electrons. Florence the remaining electrons are more strongly attracted by the nucleus. Hence the cationic radius is smaller than its corresponding neutral atom.

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TN State Board 12th Commerce Model Question Paper 5 English Medium

Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II. III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part -I

Choose the correct answer. Answer all the questions: [20 x 1 = 20]

Question 1.
…………… is a global and universal concept.
(a) Management
(b) Process
(c) Art
(d) Science
Answer:
(a) Management

Question 2.
………… is the subsidiary function of management.
(a) Planning
(b) Co-ordination
(c) Innovation
(d) Controlling
Answer:
(c) Innovation

Question 3.
The Indian financial system can be broadly classified into………….. sector.
(a) Two
(b) Three
(c) One
(d) Four
Answer:
(a) Two

Question 4.
Money market is a market for purely …………..
(a) Short term funds
(b) Long term funds
(c) Medium term funds
(d) None of these
Answer:
(a) Short term funds

Question 5.
Which is not a foreign stock exchange?
(a) London stock exchange
(b) Bombay stock exchange
(c) Tokyo stock exchange
(d) New york stock exchange
Answer:
(b) Bombay stock exchange

Question 6.
SEBI got the statutory powers in the year …………..
(a) 1988
(b) 1992
(c) 1969
(a) 1980
Answer:
(b) 1992

Question 7.
The internal source of recruitment are …………..
(i) promotion
(ii) e-recruitment
(iii) retention
(iv) advertisements
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (iii) and (iv)
Answer:
(b) (i) and (iii)

Question 8.
The types of Aptitude Test are …………..
(i) Numerical Reasoning Test
(ii) Attitude Test
(iii) Vocabulary Test
(iv) Interest Test
(a) (i) and (ii)
(b) (i) and (iii)
(c) (I) and (iv)
(d) (ii) and (iii)
Answer:
(b) (i) and (iii)

Question 9.
On the job training is given to the employee at the ……………
(a) College
(b) Home
(c) Work place
(d) None of the above
Answer:
(c) Work place

Question 10.
Find Out which is not suitable? The products marketed in commodity exchange are…………..
(a) Crude oil
(b) Rice
(c) Copper
(d) Gold
Answer:
(b) Rice

Question 11.
The term ‘Caveat emptor’ is a Latin term, which means ………..
(a) Let the seller beware
(b) Let the buyer beware
(c) Consumer
(d) Marketer
Answer:
(b) Let the buyer beware

Question 12.
The consumer is the of the modern marketing.
(a) Manager
(b) Director
(c) King
(d) None of these
Answer:
(c) King

Question 13.
The national commission should have ……….. members.
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
(c) 5

Question 14.
It is a situation in which a country loses its most educated and talented workers to other countries is known as…………
(a) Liberalisation
(b) Foreign trade
(c) Brain Drain
(d) Nationalisation
Answer:
(c) Brain Drain

Question 15.
A bill of exchange drawn on a specified banker is ………..
(a) promissory note
(b) cheque
(c) hundi
(d) share
Answer:
(b) cheque

Question 16.
The entrepreneurs classified on the basis of type of business are …………..
(i) Industrial entrepreneur
(ii) Technical entrepreneur
(iii) Professional entrepreneur
(iv) Business entrepreneur
(a) (z) and (ii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
Answer:
(b) (i) and (iv)

Question 17.
The first two stages of formation of a company are ………..
(i) Issue of prospectus
(ii) Promotion
(iii) Issue of share certificate
(iv) Registration
(a) (i) and (ii)
(b) (ii) and (iv)
(c) (ii) and (iii)
(d) (i) and (iii)
Answer:
(b) (ii) and (iv)

Question 18.
A company may appoint more than 15 directors after passing a ……….. resolution.
(a) Special
(b) Ordinary
(c) Usual
(d) Commanding
Answer:
(a) Special

Question 19.
A person can hold the position of Directorship in different companies upto the maximum of …………
(a) 15
(b) 10
(c) 18
(d) 20
Answer:
(d) 20

Question 20.
A statutory meeting can be held within months.
(a) 10
(b) 5
(c) 6
(d) 3
Answer:
(c) 6

Part – II

Answer any seven in which Question No.30 is compulsory. [7 x 2 = 14]

Question 21.
State the meaning of Authority.
Answer:
Authority means the right of a superior to give the order to his subordinates. This is the issue of commands followed responsibility for their consequences.

Question 22.
What is debt market?
Answer:
Debt Market is the financial market for trading in Debt Instrument (i.e. Government Bonds or Securities, Corporate Debentures or Bonds).

Question 23.
What is Government Securities Market?
Answer:
A market whereby the Government or gilt-edged securities can be bought and sold is called ‘Government Securities Market’.

Question 24.
Write any 5 Stock Exchanges in India.
Answer:

• The Bombay Stock Exchange
• Bangalore Stock Exchange Ltd.
• The Madras Stock Exchange Ltd.
• The Hydrabad Stock Exchange Ltd.
• The Cochin Stock Exchange Ltd.

Question 25.
What is intelligence test?
Answer:
Intelligence tests are one of the psychological tests, that is designed to measure a variety of mental ability, individual capacity of a candidate.

Question 26.
Define Product.
Answer:
“A product is anything that can be offered to a market for attention, acquisition, use or consumption that might satisfy a want or a need”, says Philip Kotler.

Question 27.
Give two examples of adulteration.
Answer:

• Powdered rice/wheat is adulterated with starch.
• Coffee powder is adulterated with tamarind seed.

Question 28.
List down the essential elements of a contract of sale.
Answer:
Following essential elements are necessary for a contract of sale:

• Two Parties
• Transfer of Property
• Goods
• Price
• Includes both ‘Sale’ and ‘Agreement to Sell’.

Question 29.
Mention the other name for corporate entrepreneur.
Answer:
Corporate entrepreneur is called promoter. He/she takes initiative necessary to start an entity under corporate format.

Question 30.
What is Resolution?
Answer:
As per the Companies Act 2013, for taking any decision or executing any transaction, the consent of the shareholders, the Board of Directors and other specified is required. The decisions taken at a meeting are called resolutions.

Part – III

Answer any seven in which question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
What are the Process of MBE?
Answer:

• Primarily, it is necessary to set objectives or norms with predictable or estimated results.
• These performances are assessed and get equated to the actual performance.
• Next, the deviation gets analysed. With an insignificant or no deviation, no action is required and senior managers can concentrate on other matters.

Question 32.
Explain any two functions of Capital Market.
Answer:
(i) Savings and Capital Formation: In capital market, various types of securities help to, mobilize savings from various sectors of population (Individuals, Corporate, Govt., etc.). The twin features of reasonable return and liquidity in stock exchange are definite incentives to the people to invest in securities. This accelerates the capital formation in the country.

(ii) Permanent Capital: The existence of a capital market/stock exchange enables companies to raise permanent capital. The investors cannot commit their funds for a permanent period but companies require funds permanently.

Question 33.
What are the features of Human resources?
Answer:

• Human resource exhibits innovation and creativity.
• Human resource alone can think, act, analyse and interpret.
• Human resources are emotional beings.
• Human resources can be motivated either financially or non-financially.
• The behaviour of human resources are unpredictable.
• Over years human resources gains value and appreciates.
• Human resources are movable.
• Human resource can work as a team.

Question 34.
What can be marketed in the Market?
Answer:
The dynamic items that can be marketed are listed below:

• Goods
• Services
• Experiences
• Events
• Persons
• Places
• Properties
• Organisations
• Information
• Ideas

Question 35.
Who are the members of the National Commission?
Answer:
Members: The National Consumer Disputes Redressal Commission has been constituted by a Notification.

• The National Commission should have five members.
• One should be from judiciary.
• Four other members of ability, knowledge and experience from any other fields.
• It should include a woman.

Question 36.
Write about any three internal environmental factors of business.
Answer:
The major internal factors affecting business decisions are:

• Values system: The values of the founder/owner of the business , percolates down to the entire organisation and has a profound effect on the organisation.
• Vision and objectives: The vision and objectives of a business guides its operations and strategic decisions.
• Management structure: The structure of management/board and their style of functioning, the level of professionalism of management, the composition of the board are the various factors which affect the decision making.

Question 37.
Distinguish between Bill of Exchange and Promissory Note.
Answer:

S. No.	Basis of Difference	Bill of Exchange	Promissory Note
1.	Nature of Undertaking	A bill of exchange contains an unconditional order to pay money.	A promissory note contains an unconditional undertaking to pay money.
2.	No. of Parties	There are three parties in a bill of exchange – drawer, drawee and payee.	In a promissory note there are only two parties the maker and the payee.
3.	Drawer of the instrument	A creditor draws a bill on a debtor.	A debtor executes a promissory note in favour of a creditor.
4.	Identity of the parties	In a bill of exchange, both the drawer and the payee can be one and the same person.	In a promissory note, the maker himself cannot be the payee because the same person cannot be both the promisor and the promisee.

Question 38.
List down few examples of pure entrepreneurship.
Answer:
Pure entrepreneurs are individuals who are propelled to enter into venture by psychological and economic motives. They nurture desire of starting a particular venture and earning high profit there from and thus attaining a social status. They apply their knowledge, skill and insight in making the venture a great success in order to earn maximum profit out of the venture. Dhirubai Ambani, Jamshedji Tata, T.V. Sundaram Iyengar, Seshadriji, Birla, Narayanamurthi, and Azim Premji are few examples of pure entrepreneurship.

Question 39.
Expand the following: STEP, JAM, TREAD, M-S1PS, SEED and New Gen IEDC.
Answer:

• STEP: Support to Training and Employment Programme for women
• JAM: Jan-Dhan-Aadhaar-Mobile
• M-SIPS: Modified Special Incentive Package Scheme
• SEED: Science for Equity Empowerment and Development
• New Gen IEDC: New Gen Innovation and Entrepreneurship Development Centre
• TREAD: Trade Related Entrepreneurship Assistance and Development.

Question 40.
Who is a shadow director?
Answer:
A shadow director is a person who is not the member of Board but has some power to run it and can be appointed as the director but according to his/her wish.

Part – IV

Answer all the following questions. [7 x 5 = 35]

Question 41.
(a) Explain the principles of modern management.
Answer:
The Father of Modem Management is Mr.Henry Fayol, and according to him there are 14 major principles of management which every manager has to practice for the success of the organisation.
(i) Division of Work: According to this principle the whole work is divided into small tasks.
This leads to specialization which increases the efficiency of labour.
(ii) Authority and Responsibility: This is the issue of commands followed by responsibility for their consequences.
(iii) Discipline: It is obedience, proper conduct in relation to others, respect of authority, etc.
It is essential for the smooth functioning of all organisations.
(iv) Unity of Command: This principle states that each subordinate should receive orders and be accountable to one and only one superior.
(v) Unity of Direction: All related activities should be put under one group, there should be one plan of action for them, and they should be under the control of one manager.
(vi) Subordination of Individual Interest to Mutual Interest: The management must put aside personal considerations and put company objectives firstly.
(vii) Remuneration: Workers must be paid sufficiently as this is a chief motivation of employees and therefore greatly influences productivity.

(viii) The Degree of Centralization: The amount of power wielded with the central management depends on company size.
(ix) Line of Authority/Scalar Chain: This refers to the chain of superiors ranging from top management to the lowest rank.
(x) Order: Social order ensures the fluid operation of a company through authoritative procedure.
(xi) Equity: Employees must be treated kindly, and justice must be enacted to ensure a just workplace.
(xii) Stability of Tenure of Personnel: Stability of tenure of personnel is a principle stating that in order for an organisation to run smoothly, personnel (especially managerial personnel) must not frequently enter and exit the organisation.
(xiii) Initiative: Using the initiative of employees can add strength and new ideas to an organisation.
(x/v) Esprit de Corps/Team Spirit: This refers to the need of managers to ensure and develop morale in the workplace; individually and communally.

[OR]

(b) Discuss the disadvantages of MBE.
Answer:

• The main disadvantage of MBE is, only managers have the power over really important decisions, which can be demotivating for employees at a lower level.
• Furthermore, it takes time to pass the issues to managers. Managing employees who deviate from the normal procedures. Because of compliance failures are considered difficult to manage and typically find themselves with limited job duties and ultimately dismissed/ terminated.

Question 42.
(a) Discuss the role of financial market.
Answer:

• Savings Mobilization: Obtaining funds from the savers or ‘surplus’ units such as household individuals, business firms, public sector units, Government is an important role played by financial markets.
• Investment: Financial market plays a key role in arranging the investment of funds thus
  collected, in those units which are in need of the same.
• National Growth: Financial markets contribute to a nation’s growth by ensuring an unfettered flow of surplus funds to deficit units. Flow of funds for productive purposes is also made possible, ft leads to overall economic growth.
• Entrepreneurship Growth: Financial markets contribute to the development of the entrepreneurial class by making available the necessary financial resources.
• Industrial Development: The different components of financial markets help an accelerated growth of industrial and economic development of a country and thus contributing to raising the standard of living and the society’s well-being.

[OR]

(b) Explain the features and types of Commercial Bills.
Answer:

• Drawer
• Acceptor
• Payee
• Discounter
• Endorser
• Assessment
• Maturity
• Credit Rating

Types:

• Demand and Usance Bills: A demand bill is one wherein no specific time of payment is mentioned. So, demand bills are payable immediately when they are presented to the drawee.
• Clean Bills and Documentary Bills: Bills that are accompanied by documents of title
  to goods are called documentary bills. Clean bills are drawn without accompanying any document.
• Inland Bills and Foreign Bills: Bills that are drawn and payable in India on a person who is resident in India are called inland bills.
• Indigeneous Bills: The drawing and acceptance of indigenous bills are governed by native custom or usage of trade.
• Accommodation and Supply Bills: Accommodation bills are those which do not arise out of genuine trade of transactions.

Question 43.
(a) What are the benefits of Dematerialisation?
Answer:

• The risks relating to physical certificates like loss, theft, forgery are eliminated completely with a Demat Account.
• The risk of paper work enables quicker transactions and higher efficiency in trading.
• The shares which are created through mergers and consolidation of companies are credited automatically in the Demat account.
• There is no stamp duty for transfer of securities.
• Certain banks also permit holding of both equity and debt securities in a single account.
• A Demat account holder can buy or sell any amount of shares.

[OR]

(b) Elaborate on the factors affecting recruitment.
Answer:
Some of the factors that affect recruitment are:

• Size of the Organisation: Bigger organisations find recruitment less problematic than organisations that are smaller in size.
• Recruiting Policy: Most organisations prefer recruiting through internal sources, because own employees know the organisation and they can well fit into the organisation’s culture.
• Nature of Post: The recruitment process varies according to type of personnel to be recruited. For example, recruitment process of a bank officer will differ from that of an IPS officer.
• Demographic factors: Like the details of possible employees- age, religion, literacy level, gender, occupation, economic status, etc.
• Unemployment rate: If the unemployment rate is high, the number of applicants will be more, making it easy for the recruiters to select from a wide range.
• Labour Laws: These cover working conditions, compensation, retirement benefits, and safety and health of employees in industrial undertakings.
• Competitors: Time to time the organisations have to change their recruitment policies and manuals according to the policies being followed by the competitors.

Question 44.
(a) Explain the benefits of training.
Answer:
(i) Benefits to the Organisation:

• It improves the skill of employees and increase the productivity.
• It reduces wastages of materials and idle time.
• It minimizes the time for supervision.
• It reduces the frequent accidents at workplace and consequent payment of compensation.

(ii) Benefits to the Employees:

• It increases the knowledge, skill of the employees.
• It enables him to gain promotion in shorter time.
• It improves the employees productivity.
• Employees get higher earnings through incentives and rewards.

(iii) Benefits of Customer:

• Customers get better quality of product and service.
• Customers get innovative products or value added or feature rich products.

[OR]

(b) Narrate the Elements of Marketing mix.
Answer:
Marketing mix means a marketing programme that is offered by a firm for the satisfaction of human wants. There are four Elements of Marketing mix:

• Product: A Product is the main element of marketing. Without a product, there can be no marketing.
• Price: Price is the value of a product expressed in monetary terms. It is the amount Charged for the product.
• Place (Physical Distribution): An excellent quality product, with a good price, will be waste, if it is not transferred from the production place to consumption place.
• Promotion: An excellent product with competitive price cannot achieve a desired success and acceptance in market, with special features are conveyed to the consumers.

Question 45.
(a) What are the objectives of Consumer Protection Act, 1986?
Answer:
The Central Government enacted a comprehensive law called the Consumer Protection Act in 1986. This Act came into force with effect from 15.04.1987. It is in short, called as ‘COPRA’.
Objectives:

• Consumer protection Act protects the interests of the consumers.
• This Act provides safeguard against defective goods and deficient services, unfair trade practices.
• It also gives settlement of consumer disputes.
• It is applicable to public sector, financial and co-operative enterprises.

[OR]

(b) Classify goods under the Sale of Goods Act.
Answer:
The term goods mean every kind of movable property other than actionable claim and money.
The goods are classified as follows:
(1) Existing Goods- These goods are owned or possessed by the seller at the time of contract of sale. Existing goods may again be divided as:

• Specific Goods- It denotes goods identified and agreed upon at the time of contract of sale.
• Ascertained Goods- The term ‘ascertained goods’ is also used as similar in meaning to specific goods.
• Unascertained Goods- These are goods which are not identified and agreed upon at the time of contract of sale.

(2) Future Goods- These are goods which a seller does not possess at the time of contract of
sale, but which will be manufactured or produced or acquired by him after entering into the contract.
(3) Contingent Goods- These are the goods, the acquisition of which by the seller depends upon a contingency (an event which may or may not happen).

Question 46.
(a) What are the requisites for a valid endorsement?
Answer:
If an endorsement is to be valid, it must possess the following requisites:

• Endorsement is to be made on the face of the instrument or on its back.
• When there is no space for making further endorsements a piece of paper can be attached
• Endorsement for only a part of the amount of the instrument is invalid.
• Endorsement is complete only when delivery of the instrument is made.
• Signing in block letters does not constitute regular endorsement.
• If the payee is an illiterate person, he can endorse it by affixing his thumb impression on the instrument.

[OR]

(b) Distinguish between the rural and urban entrepreneur.
Answer:

S. No.	Rural Entrepreneur	Urban Entrepreneur
1.	It refers to the person who starts business in rural areas.	It refers to the person who commences business in urban areas.
2.	These entrepreneurs start doing business in the villages and small towns.	They will do their business in state capital, towns, district headquarters, municipalities, etc.
3.	They may be agricultural and trading entrepreneurs.	They may be industrial or corporate entrepreneur.
4:	The availability of material and labour is easy. So the cost of operation tends to be low.	The availability of material and labour may be difficult. So the cost of operation may be high.

Question 47.
(a) What formalities need to be fulfilled for a companies having share capital to commence business?
Answer:
A public limited company having its share capital has to pass through two more stages. One of them is capital subscription, steps to be taken at this stage are listed below:

• The fulfilling formalities to raise necessary capital.
• Following the SEBI guidelines in this regard.
• Issuing prospectus.
• Appointing official banker of the company for receiving application from the investors,
• Passing resolution for making allotments by director.
• Despatch allotment letters to allottees.
• Filing allotment returns with the Registrar.
• Issuing share certificates in exchange for their allotment letter.
• Ensuring collection of minimum subscription.

[OR]

(b) List the disqualification of a directors.
Answer:
Disqualifications of a Director: Section 164 of Companies Act, 2013, has mentioned the disqualification as follows: A person shall not be capable of being appointed director of a company, if the director is

• Of unsound mind
• An undercharged insolvent
• Has been convicted by a court for any offence
• Has not paid any call in respect of shares of the company held by him
• An order disqualifying him for appointment as director has been passed by a court
• He has not got the Director Identification Number.

Students can download 10th Science Chapter 6 Nuclear Physics 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 6 Nuclear Physics

Samacheer Kalvi 10th Science Nuclear Physics Text Book Back Questions and Answers

I. Choose the correct answer.

Question 1.
Man-made radioactivity is also known as;
(a) Induced radioactivity
(b) Spontaneous radioactivity
(c) Artificial radioactivity
(d) (a) and (c)
Answer:
(a) Induced radioactivity

Question 2.
Unit of radioactivity is _____.
(a) roentgen
(b) curie
(c) becquerel
(d) all the above.
Answer:
(d) all the above.

Question 3.
Artificial radioactivity was discovered by:
(a) Becquerel
(b) Irene Curie
(c) Roentgen
(d) Neils Bohr
Answer:
(b) Irene Curie

Question 4.
In which of the following, no change in mass number of the daughter nuclei takes place _____.
(i) α decay
(ii) β decay
(iii) γ decay
(iv) neutron decay
(a) (i) is correct
(b) (ii) and (iii) are correct
(c) (i) & (iv) are correct
(d) (ii) & (iv) are correct.
Answer:
(b) (ii) and (iii) are correct

Question 5.
……. isotope is used for the treatment of cancer.
(a) Radio Iodine
(b) Radio Cobalt
(c) Radio Carbon
(d) Radio Nickel
Answer:
(b) Radio Cobalt

Question 6.
Gaming radiations are dangerous because of _____.
(a) it affects eyes & bones
(b) it affects tissues
(c) it produces genetic disorder
(d) it produces an enormous amount of heat.
Answer:
(c) it produces genetic disorder

Question 7.
……… aprons are used to protect us from gamma radiations.
(a) Lead oxide
(b) Iron
(c) Lead
(d) Aluminium
Answer:
(c) Lead

Question 8.
Which of the following statements is/are correct?
(i) a particles are photons.
(ii) Penetrating power of γ radiation is very low.
(iii) Ionization power is maximum for a rays.
(iv) Penetrating power of γ radiation is very high.
(a) (i) & (ii) are correct
(b) (ii) & (iii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct
Answer:
(d) (iii) & (iv) are correct

Question 9.
Proton-Proton chain reaction is an example of _____.
(a) Nuclear fission
(b) α – decay
(c) Nuclear fusion
(d) β – decay.
Answer:
(c) Nuclear fusion

Question 10.
In the nuclear reaction the value of A and Z.
(a) 8, 6
(b) 8, 4
(c) 4, 8
(d) cannot be determined with the given data
Answer:
(b) 8, 4

Question 11.
Kamini reactor is located at:
(a) Kalpakkam
(b) Koodankulam
(c) Mumbai
(d) Rajasthan
Answer:
(a) Kalpakkam

Question 12.
Which of the following is/are correct?
(i) Chain reaction takes place in a nuclear reactor and an atomic bomb.
(ii) The chain reaction in a nuclear reactor is controlled
(iii) The chain reaction in a nuclear reactor is not controlled
(iv) No chain reaction takes place in an atom bomb
(a) (i) only correct
(b) (i) & (ii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct.
Answer:
(b) (i) & (ii) are correct

II. Fill in the blanks.

1. One roentgen is equal to ………… disintegrations per second.
2. Positron is an …………
3. Aneamia can be cured by ………… isotope.
4. Abbreviation of ICRP …………..
5. ………… is used to measure exposure rate of radiation in humans.
6. ………. has the greatest penetration power.
7. _ZY^A → _Z+1Y^A + X; Then, X is …………..
8. _ZX^A → _ZY^A This reaction is possible in …………. decay.
9. The average energy released in each fusion reaction is about …………. J.
10. Nuclear fusion is possible only at an extremely high temperature of the order of ……….. K.
11. The radio isotope of …………. helps to increase the productivity of crops.
12. If the radiation exposure is 100 R, it may cause ………….
Answer:
1. 1.6 × 10¹⁵ disintegrations / second
2. antiparticle [₁e⁰]
3. Radio iron Fe⁵⁹
4. International Commission on Radiological Protection
5. Dosimeter
6. Gama rays
7. X is _-1e⁰
8. gamma
9. 3.84 × 10¹² J
10. 10⁷ to 10⁹ K
11. Radio phosphorus P – 32
12. fatal diseases like leukemia.

III. State whether the following statements are true or false: If false, correct the statement.

1. Plutonium -239 is a fissionable material.
2. Elements having atomic number greater than 83 can undergo nuclear fusion.
3. Nuclear fusion is more dangerous than nuclear fission.
4. Natural uranium U-238 is the core fuel used in a nuclear reactor.
5. If a moderator is not present, then a nuclear reactor will behave as an atom bomb.
6. During one nuclear fission on an average, 2 to 3 neutrons are produced.
7. Einstein’s theory of mass energy equivalence is used in nuclear fission and fusion.
Answer:
1. True
2. False – Elements having atomic number greater than 83 can undergo nuclear fission.
3. True
4. False – Natural U – 238 not used as fuel in a nuclear reactor.
5. True
6. True
7. True

IV. Match the following.

Question 1.
Match the column I with column II.

Answer:
(a) – (iii)
(b) – (i)
(c) – (iv)
(d) – (ii)

Question 2.
Match the column I with column II.

Answer:
(a) – (iv)
(b) – (ii)
(c) – (iii)
(d) – (i)

Question 3.
Match the column I with column II.

Answer:
(a) – (ii)
(b) – (iv)
(c) – (i)
(d) – (iii)

Question 4.
Match the column I with column II.

Answer:
(a) – (iv)
(b) – (iii)
(c) – (ii)
(d) – (i)

Question 5.
Match the column I with column II.

Answer:
(a) – (iii)
(b) – (iv)
(c) – (ii)
(d) – (i)

V. Arrange the following in the correct sequence.

Question 1.
Arrange in descending order, on the basis of their penetration power. Alpha rays, beta rays, gamma rays, cosmic rays
Answer:
gamma rays < beta rays < alpha rays < cosmic rays.

Question 2.
Arrange the following in the chronological order of discovery.
Nuclear reactor, radioactivity, artificial radioactivity, discovery of radium.
Answer:
Discovery of radium, Radioactivity, Artifical radioactivity, Nuclear reactor.

VI. Use the analogy to fill in the blank.

1. Spontaneous process : Natural Radioactivity, Induced process : ………..
2. Nuclear Fusion : Extreme temperature, Nuclear Fission : ………..
3. Increasing crops : Radio phosphorous, Effective functioning of heart : …………
4. Deflected by electric field : a ray, No Deflection : ………….

Answer:

1. artifical radioactivity
2. higher temperature
3. Radio sodium
4. γ-ray

VII. Numerical problems.

Question 1.
₈₈Ra²²⁶ experiences three a – decay. Find the number of neutrons in the daughter element.
Answer:
Mass number of ₈₈Ra²²⁶ is 266
Atomic number of ₈₈Ra²²⁶ is 88
Mass number of α = 4
Atomic number of α = 2
After 3 a decay
Mass number of the daughter element is
= 266 – (3 × 4)
= 266 – 12
= 254
Atomic number of the daughter element is
= 88 – (3 × 2)
= 88 – 6
= 82
Number of neutrons is
= 254 – 82
= 172

Question 2.
A cobalt specimen emits induced radiation of 75.6 millicurie per second. Convert this disintegration in to becquerel (one curie = 3.7 × 10¹⁰ Bq).
Answer:
1 curie = 3.7 × 10¹⁰ Bq
∴ 75.6 × 10^-3 curie = 75.6 × 10^-3 × 3.7 × 10¹⁰
= 279.72 × 10⁷ Bq
= 279.72 × 10⁷ Becquerel

VIII. Assertion and reason type questions.

Question 1.
Assertion: A neutron impinging on U²³⁵, splits it to produce Barium and Krypton.
Reason: U – 235 is a fissile material.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 2.
Assertion: In a 3 – decay, the neutron number decreases by one.
Reason: In 3 – decay atomic number increases by one.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 3.
Assertion: Extreme temperature is necessary to execute nuclear fusion.
Reason: In a nuclear fusion, the nuclei of the reactants combine releasing high energy.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 4.
Assertion: Control rods are known as ‘neutron seeking rods’.
Reason: Control rods are used to perform sustained nuclear fission reaction.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

IX. Answer in one or two words (VSA).

Question 1.
Who discovered natural radioactivity?
Answer:
Henri Becquerel

Question 2.
Which radioactive material is present in the ore of pitchblende?
Answer:
Uranium is the radioactive material is present in the ore of pitchblende.

Question 3.
Write any two elements which are used for inducing radioactivity?
Answer:
Boron and aluminium

Question 4.
Write the name of the electromagnetic radiation which is emitted during natural radioactivity.
Answer:
Gamma rays are the electromagnetic radiation which is emitted during natural radioactivity.

Question 5.
If A is a radioactive element which emits an α – particle and produces ₁₀₄Rf²⁵⁹. Write the atomic number and mass number of the element A.
Answer:

Mass number A = 259 + 4 = 263
Atomic number Z = 104 + 2 = 106

Question 6.
What is the average energy released from a single fission process?
Answer:
200 MeV

Question 7.
Which hazardous radiation is the cause for the genetic disease?
Answer:
γ – ray (Gamma-ray) or any high energy nuclear particle passes through a human being, it disrupts the entire normal functioning of the biological system and the effect may be an either pathological or genetic disease.

Question 8.
What is the amount of radiation that may cause death of a person when exposed to it?
Answer:
600 R.

Question 9.
When and where was the first nuclear reactor built?
Answer:
The first nuclear reactor was built in 1942 at Chicago, USA.

Question 10.
Give the SI unit of radioactivity.
Answer:
The SI unit of radioactivity is becquerel.

Question 11.
Which material protects us from radiation?
Answer:
Lead

X. Answer the following questions in a few sentences.

Question 1.
Write any three features of natural and artificial radioactivity.
Answer:

Question 2.
Define critical mass.
Answer:
The minimum mass of fissile material necessary to sustain the chain reaction is called ‘critical mass (m_c)’. It depends on the nature, density and the size of the fissile material.

Question 3.
Define one roentgen.
Answer:
One roentgen is defined as the quantity of radioactive substance which produces a charge of 2.58 × 10^-4 coulomb in 1 kg of air under standard conditions of pressure, temperature and humidity.

Question 4.
State Soddy and Fajan’s displacement law.
Answer:
(i) When a radioactive element emits an alpha particle, a daughter nucleus is formed whose mass number is less by 4 units and the atomic number is less by 2 units, than the mass number and atomic number of the parent nucleus.

(ii) When a radioactive element emits a beta particle, a daughter nucleus is formed whose mass number is the same and the atomic number is more by 1 unit, than the atomic number of the parent nucleus.

Question 5.
Give the function of control rods in a nuclear reactor.
Answer:
Control rods are used to control the number of neutrons in order to have a sustained chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.

Question 6.
In Japan, some of the newborn children are having congenital diseases. Why?
Answer:
In Japan, there was a leakage of nuclear radiations in some areas. In those areas, if the new born children handled with careless they are exposed to harmful radiations then they have congenital diseases.

Question 7.
Mr. Ramu is working as an X – ray technician in a hospital. But, he does not wear the lead aprons. What suggestion will you give to Mr. Ramu?
Answer:
The intensity of X-rays is very low. So, X-rays do not produce any severe effects. There are certains clothes that can be used while operating X-ray machines. Ramu can use them and so he may not be severely affected by X-rays, he should avoid eating and drinking when he is working with X-rays.

Question 8.
What is stellar energy?
Answer:
Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called stellar energy.

Question 9.
Give any two uses of radio isotopes in the field of agriculture?
Answer:
The radio isotope of phosphorous (P-32) helps to increase the productivity of crops. The radiations from the radio isotopes can be used to kill the insects and parasites and prevent the wastage of agricultural products.

Question 10.
What is stellar energy?
Answer:
Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called as stellar energy.

XI. Answer the following questions in detail.

Question 1.
Explain the process of controlled and uncontrolled chain reactions.
Answer:
(i) Controlled chain reaction: In the controlled chain reaction the number of neutrons released is maintained to be one. This is achieved by absorbing the extra neutrons with a neutron absorber leaving only one neutron to produce further fission. Thus, the reaction is sustained in a controlled manner. The energy released due to a controlled chain reaction can be utilized for constructive purposes. Controlled chain reaction is used in a nuclear reactor to produce energy in a sustained and controlled manner.

(ii) Uncontrolled chain reaction: In the uncontrolled chain reaction the number of neutrons multiplies indefinitely and causes fission in a large amount of the fissile material. This results in the release of a huge amount of energy within a fraction of a second. This kind of chain reaction is used in the atom bomb to produce an explosion.

Question 2.
Compare the properties of alpha, beta and gamma radiations.
Answer:

Question 3.
What is a nuclear reactor? Explain its essential parts with their functions.
Answer:
A Nuclear reactor is a device in which the nuclear fission reaction takes place in a self-sustained and controlled manner to produce electricity.
Components of a nuclear reactors:
The essential components of a nuclear reactor are

1. fuel,
2. moderator,
3. control rod,
4. coolant and
5. protection wall.

1. Fuel : A fissile material is used as the fuel. The commonly used fuel material is uranium.
2. Moderator : A moderator is used to slow down the high energy neutrons to provide slow neutrons. Graphite and heavy water are the commonly used moderators
3. Control rod : Control rods are used to control the number of neutrons in order to have sustained chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.
4. Coolant : A coolant is used to remove the heat produced in die reactor core, to produce steam. This steam is used to run a turbine in order to produce electricity. Water, air and helium are some of the coolants.
5. Protection wall : A thick concrete lead wall is built around the nuclear reactor in order to prevent the harmful radiations from escaping into the environment.

XII. HOT Questions.

Question 1.
The mass number of a radioactive element is 232 and its atomic number is 90. When this element undergoes certain nuclear reactions, it transforms into an isotope of lead with a mass number 208 and an atomic number 82. Determine the number of alpha and beta decay that can occur.
Answer:
\(_{90} \mathrm{x}^{232} \stackrel{4 \beta}{\longrightarrow} 94^{\mathrm{Y}^{232}+4_{-4} \mathrm{e}^{\circ}}\) (4β decay)
From the result of β decay,
\(94^{232} \stackrel{6 \alpha}{\longrightarrow} 82^{\mathrm{Pb}^{208}+6}_{} \mathrm{He}^{4}\) (6α decay)
The number of α decay = 6
The number of β decay = 4.

Question 2.
‘X – rays should not be taken often’. Give the reason.
Answer:
X-rays have some low intensity. Even then if a person undergoes the exposure of X-ray, his skin may be affected. The tissues near by the bones may be damaged.

Question 3.
Cell phone towers should be placed far away from the residential area – why?
Answer:
From the cell phone towers micro-waves are scattered for a particular area. This may affect small children and aged people. If people are exposed with those rays very often then they may be severly affected.

Samacheer Kalvi 10th Science Nuclear Physics Additional Important Questions and Answers

I. Choose the correct answer.

Question 1.
The elements that undergo spontaneous radioactivity:
(a) Uranium
(b) Radium
(c) Uranium and Radium
(d) None of the above
Answer:
(c) Uranium and Radium

Question 2.
The artificial (or) induced radioactive element is _____.
(a) Uranium
(b) Boron
(c) Radium
(d) none of these.
Answer:
(b) boron

Question 3.
The nucleons in a nucleus are attracted by:
(a) electrostatic force
(b) gravitational force
(c) nuclear force
(d) magnetic force
Answer:
(c) nuclear force

Question 4.
In p decay by one.
(a) mass number decreases
(b) atomic number decreases
(c) neutron number decreases
(d) proton number decreases
Answer:
(c) neutron number decreases

Question 5.
The charge of the α rays _____.
(a) 2e
(b) -e
(c) 0
(d) none of these.
Answer:
(a) 2e

Question 6.
In which process positron is emitted?
(a) natural radioactivity
(b) artifical radioactivity
(c) nuclear fission
(d) nuclear fussion
Answer:
(b) artifical radioactivity

Question 7.
The atomic number of the elements that exhibit natural radioactivity is:
(a) less than 83
(b) more than 83
(c) more than 42
(d) more than 82
Answer:
(b) more than 83

Question 8.
The ionising power of the α rays _____.
(a) 100 times greater than the β rays
(b) 1000 times greater than the γ rays
(c) very less ionization
(d) both (a) and (b).
Answer:
(d) both (a) and (b).

Question 9.
In the reaction  is:
(a) ₆C¹³
(b) ₆C¹²
(c) ₇N¹³
(d) ₈O¹⁶
Answer:
(b) ₆C¹²

Question 10.
Ionising power of β rays _____.
(а) 100 times greater than α rays
(b) 100 times less than α rays
(c) comparatively very less ionization power
(d) 1000 times less than γ rays.
Answer:
(b) 100 times less than α rays

Question 11.
During α – decay the mass number of the daughter element is:
(a) more by 2 units
(b) less by 2 units
(c) more by 4 units
(d) less by 4 units
Answer:
(d) less by 4 units

Question 12.
When α – particle is emitted the atomic number of daughter element is:
(a) more by 4 units
(b) less by 4 units
(c) less by 2 units
(d) more by 2 units
Answer:
(c) less by 2 units

Question 13.
Which power has more penetrating power?
(a) α rays
(b) γ rays
(c) β rays
(d) X rays.
Answer:
(b) γ rays

Question 14.
What is the charge of alpha particle?
(a) -e
(b) +e
(c) zero
(d) +2e^–
Answer:
(d) +2e^–

Question 15.
Which of the following has speed from 1/10 to 1/20 times the speed of light?
(a) β
(b) α
(c) γ
(d) all the above
Answer:
(b) α

Question 16.
In α – decay _____.
(a) atomic number decreases by 2 and the mass number decreases by 4
(b) atomic number increases by 2 and the mass number decrease by 4
(c) there is no change in atomic and mass number
(d) neutron number decreases by one.
Answer:
(a) atomic number decreases by 2 and the mass number decreases by 4

Question 17.
Which of the following is a fertile material?
(a) Thorium – 232
(b) Uranium – 238
(c) Plutonium – 240
(d) All the above
Answer:
(d) All the above

Question 18.
If the mass of the fissile material is less than the critical mass, then it is termed as:
(a) supercritical
(b) critical
(c) subcritical
(d) none of the above
Answer:
(b) critical

Question 19.
1 curie is equal to _____.
(a) the radioactivity of 1 gram of uranium
(b) 1 decay/second
(c) 3.7 × 10¹⁰ becquerel
(d) 1.6 × 10¹² decay / second.
Answer:
(c) 3.7 × 10¹⁰ becquerel

Question 20.
During nuclear fusion which of the following are emitted:
(a) alpha rays
(b) positrons
(c) neutrons
(d) all the above
Answer:
(d) all the above

Question 21.
Anaemia can be diagonised by:
(a) ₁₅P³²
(b) ₂₆Fe⁵⁹
(c) ₁₁Na²⁴
(d) ₂₈Fe⁵⁹
Answer:
(b) ₂₆Fe⁵⁹

Question 22.
I – 131 is used in the treatment of _____.
(a) cancer
(b) skin disease
(c) disease in the thyroid gland
(d) tumour.
Answer:
(c) disease in the thyroid gland

Question 23.
In which decay atomic number of daughter element is increased by 1 unit?
(a) α – decay
(b) β – decay
(c) γ – decay
(d) neutron decay
Answer:
(b) β – decay

Question 24.
in nuclear fusion ………… is satisfied.
(a) law of conservation of energy
(b) law of conservation of momentum
(c) law of conservation angular momentum
(d) mass energy relation
Answer:
(d) mass energy relation

Question 25.
The safe limit of receiving the radiation is about _____.
(a) 100 R
(b) 600 R
(c) 250 R
(d) 250 mR per week.
Answer:
(d) 250 mR per week.

Question 26.
Hydrogen bomb is based on the principle of:
(a) natural radioactivity
(b) induced radioactivity
(c) nuclear fusion
(d) nuclear fission
Answer:
(c) nuclear fusion

Question 27.
Uncontrolled chain reaction yields _____.
(a) nuclear reactor
(b) atom bomb
(c) hydrogen bomb
(d) breeder reactor.
Answer:
(b) atom bomb

Question 28.
In the following reaction.
₁H³ + ₁H² → X + Y + energy X and Y are:
(a) ₂He³, ₀n¹
(b) ₂He⁴, ₀n¹
(c) ₁H², ₀n¹
(d) ₁H³, ₀n¹
Answer:
(b) ₂He⁴, ₀n¹

Question 29.
The number of α and ß particles emitted when ₉₂U²³⁸ decays to ₈₂Pb²⁰⁶ are respectively:
(a) 6, 8
(b) 8, 6
(c) 4, 3
(d) 3, 4
Answer:
(b) 8, 6

Question 30.
The fusion process can be carried out only at a temperature of _____.
(a) 1000°C
(b) 100°C
(c) 104°C
(d) 107 K.
Answer:
(d) 107 K.

Question 31.
Which of the following is used to cure goiter?
(a) Na²⁴
(b) I¹³¹
(c) P³²
(d) Fe⁵⁹
Answer:
(b) I¹³¹

Question 32.
Which radio isotope is used to check the function of heart?
(a) Na²⁴
(b) Fe⁵⁹
(c) Co⁶⁰
(d) I¹³¹
Answer:
(a) Na²⁴

Question 33.
The fuel of the prototype fast breeder reactor is _____.
(a) low enriched uranium
(b) carbides of uranium
(c) high enriched uranium
(d) oxides of plutonium and uranium.
Answer:
(d) oxides of plutonium and uranium.

Question 34.
Which radio isotope is used as a smoke detector?
(a) Cf²³²
(b) I¹³¹
(c) Co⁶⁰
(d) Am²⁴¹
Answer:
(d) Am²⁴¹

Question 35.
The control rod used in a nuclear reactor is _____.
(a) barium
(b) carbon
(c) cadmium
(d) Na.
Answer:
(c) cadmium

II. Fill in the blanks.

1. Radium was discovered by ………..
2. The photoelectric effect is an instantaneous process Natural radioactivity is a ………..
3. The elements with atomic number ……… undergo spontaneous radioactivity.
4. In natural radioactivity emission of radiation is due to ………. of a nucleus.
5. One disintegration per second is equal to ………..
6. S.I unit of radioactivity is ………….
7. The speed of ………. can go upto 9/10 times the velocity of light.
8. α ray is …………
9. ……….ray consists of photons.
10. ………… and ………… are deflected by electric and magnetic fields.
11. When uranium decays into thorium ……….. particle is emitted.
12. In ………… decay there is no change in the mass number of the product nuclear.
13. The minimum mass of a fissile material necessary to sustain the chain reaction is called …………
14. The value of 1MeV in joules is ………..
15. In the transformation ₁H² + ₁H² → X + energy. X is …………
16. Einstein’s mass energy relation is ………….
17. ……….causes the mutation in the human gene.
18. ……….. is also used in treatment of skin cancer.
19. The safe limit for receiving the radiation is ………….
20. The substances that are used as coolants are ………….. and
21. ………… reactors convert non-fissionable material units.
22. The expansion of BARC is ………….
23. The number of nuclear reactor that are operating in India are ……………
24. Identify X in the following transformation ₁₂Mg²⁴ + ₀n¹ → X + ₁n¹. X is ………….
25. In the reaction ₉₂U²³⁸ + ₀n¹ → ₉₃NP²³⁹ + X. X is ………….
Answer:
1. Marie Curie and pierre Curie
2. Spontaneous
3. more than 83
4. self-disintegration
5. Becquerel
6. Becquerel
7. β rays
8. a helium nucleus
9. Gamma
10. Alpha, beta
11. alpha
12. beta
13. critical mass
14. 1.602 × 10^-13 J
15. ₂He⁴
16. E = mc²
17. Gamma radiation
18. Radio-gold Au¹⁹⁸
19. 100 mR per week
20. Water, air and helium
21. Breeder
22. Bhaba Atomic Research Centre
23. 22
24. ₁₁Na²⁴
25. _-1e⁰

III. State whether the following statements are true or false: If false, correct the statement.

1. Artificial radioactivity is induced in the elements like boron, aluminum etc.
2. Natural radioactivity can be controlled.
3. 1 Rd = 3.7 × 10¹⁰ disintegrations / second.
4. The S.I units of radioactivity is roentgen.
5. The charge of each beta particle is – e
6. Beta ray travels with velocity of light.
7. Example for alpha decay is ₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴
8. Otto Hann and Strassman discovered nuclear fission.
9. In the mass of the fissile material is more than the critical mass it is termed as supercritical.
10. Hydrogen bomb is based on uncontrolled chain reaction.
11. 1eV = 1.602 × 10^-16 Joule.
12. Hydrogen is based on nuclear fusion.
13. Positrons are emitted in nuclear fusion.
14. To diagnose anaemia P³² is used.
15. Radio phosphorous is used in the treatment of skin diseases.
16. To prevent the leakage of nuclear radiation concrete wall with lead lining is used.
Answer:
1. True
2. False – Artificial radio activity can be controlled.
3. False – 1 curie = 3.7 × 10¹⁰⁶ disintegrations / second.
4. False – The S.l unit of radioactivity is Becquerel.
5. True
6. False – Gamma ray travels with velocity of light.
7. True
8. True
9. True
10. False – Atom bomb is based on uncontrolled chain reaction.
11. False – 1eV = 1.602 × 10^-9 Joule
12. True
13. True
14. False – To diagonise aneamia Fe⁵⁹ is used.
15. True
16. True
17. True

IV. Match the following.

Question 1.
Match the Column I with Column II.

Answer:
(a) – (iii)
(b) – (iv)
(c) – (v)
(d) – (i)

Question 2.
Match the Column I with Column II.

Answer:
(a) – (iv)
(b) – (v)
(c) – (ii)
(d) – (i)

Question 3.
Match the Column I with Column II.

Answer:
(a) – (v)
(b) – (iv)
(c) – (i)
(d) – (ii)

V. Numerical problems.

Question 1.
A radioactive isotope emits radiation of 7.4 × 10⁴ G Bq per second. Convert this disintegration in terms of curie.
Answer:
One curie = 3.7 × 10¹⁰ Becquerel
Becquerel (1 Bq) = One disintegration / second
∴ 1 Bq = \(\frac{1}{3.7×10^{10}}\) curie
∴ 7.4 × 10⁴ G Bq
= 7.4 × 10⁴ × 10⁹ × \(\frac{1}{3.7×10^{10}}\)
= 2 × 10³ curie
= 2,000 curie

Question 2.
The isotope ₉₂U²³⁸ successfully undergoes three α – decays and two β – decays. What is the resulting isotope?
Answer:
Number and nature of decays = 3α and 2β
Atomic number of isotope is Z = 92
Mass number of isotope is A = 238
Resulting isotope:
After the emission of 3α particles atomic number
Z’ = Z = (3 × 2)
= Z – 6
= 92 – 6
Z’ = 86
After the emission of 2β particles atomic number
Z” = Z – 2 (-1) = 86 + 2 = 88
Mass number after the emission of 3α particles
A’ = A – 3α
= 238 – 3(4)
= 238 – 12
= 226
Mass number after the emission of 2β particles
A” = A’ – 2(0)
= 226
∴ Resulting isotope is ₈₈Ra²²⁶

Question 3.
₈₈Ra²²⁶ experiences one α – decay and one β – decay. Calculate the number of neutrons in the final daughter nucleus that is formed.
Answer:
α particle = ₂He⁴
Atomic number of α particle = 2
Mass number of α particle = 4

Number of neutrons = Mass number – atomic number
= 222 – 87 = 135

Question 4.
State the number of protons, electrons neutrons and nucleons in ₄Be⁹ atom.
Answer:
Atomic number of ₄Be⁹ atom Z = 4
Mass number of ₄Be⁹ atom A = 9
Number of protons = Atomic number Z = 4
Number of neutrons = A – Z = 9 – 4 = 5
Number of nucleons = A = 9

Question 5.
Determine X and Y in the following transformations.
(i) ₄Be⁹ + α → X
(ii) ₉₂U²³⁵ + ₀n¹ → ₅₀Ba¹⁴¹ + Y + 3₀n¹
Answer:
(i) ₄Be⁹ + ₂He⁴ → ₆C¹³ [α = ₂He⁴]
∴ X = ₆C¹³

(ii) ₉₂U²³⁵ + ₀n¹ → ₅₀Ba¹⁴¹ + Y + 3₀n¹
Equating atomic number on both sides we get,
235 + 1 = 141 + A + 3
236 = 144 + A
∴ A = 236 – 144
= 92
Equating atomic number on both sides are get
92 + 0 = 56 + 2 + 0
Z = 92 – 56 = 36
∴ Y is ₃₆Kr⁹²

Question 6.
Calculate the amount energy released when a radioactive element undergoes fusion and results in a mass defect of 4g.
Answer:
Velocity of light = 3 × 10⁸ m/s
Mass defect Δm = 4 × 10^-3
By Einstein’s equation, energy released
E = mC²
= (4 × 10^-3) × (3 × 10⁸)²
= 36 × 10¹³ J
= 3.6 × 10¹⁴ J

Question 7.
Identify the radio isotopes in the following reactions.
Answer:
(i) ₁₅P⁸ + ₀n¹ → X + γ
(ii) ₀n¹ → _ZY^A + _-1e⁰ + γ^–
Answer:
(i) ₁₅P³¹ + ₀n¹ → X + γ
Mass number of X = 31 + 1 = 32
Atomic number of Y= 15 + 0 =15
∴ X is ₁₅P³¹

(ii) ₀n¹ → _ZY^A + _-1e⁰ + γ^–
Mass number 1 = A + 0
∴ A = 1
Atomic number 0 = Z – 1
∴ Z = 1
_ZY^A is ₁H¹

VI. Assertion and reason type questions.

Question 1.
Assertion: Natural radioactivity is a spontaneous process.
Reason: It cannot be controlled.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 2.
Assertion: In artificial radioactivity neutron and positron are emitted. Reason: It can be controlled.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.

Question 3.
Assertion: p rays are electrons.
Reason: p rays travel with speed of light.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(c) If the assertion is true, but the reason is false.

Question 4.
Assertion: Gamma rays have maximum ionising power.
Reason: Alpha rays have low penetrating power.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(d) If the assertion is false, but the reason is true.

Question 5.
Assertion: Critical mass of a fissile material is the mass necessary to sustain chain reaction.
Reason: Critical mass depends on the nature, density and size of the fissile material.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(c) If the assertion is true, but the reason is false.

Question 6.
Assertion: Nuclear fusion takes place in the cores of the sun and other stars.
Reason: Nuclear fusion is called as thermonuclear reaction.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 7.
Assertion: An alpha particle carries double the positive charge of protons.
Reason: The charge of alpha particle is equal to charge on the helium nucleus.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

Question 8.
Assertion: Gamma rays are electromagnetic waves.
Reason: Gamma rays travel with speed of light.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If the assertion is true, but the reason is false.
(d) If the assertion is false, but the reason is true.
Answer:
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.

VII. Answer in one or two words (VSA).

Question 1.
Who discovered the nucleus?
Answer:
Rutherford

Question 2.
Define ‘Natural Radioactivity’.
Answer:
This phenomenon of spontaneous emission of radiation from certain elements on their own is called ‘Natural radioactivity’.

Question 3.
State elements exhibiting radioactivity with atomic number less than 83.
Answer:
Uranium and radium

Question 4.
What is the nature of radioactive substances?
Answer:
They are earth metals and transition metals.

Question 5.
Define ‘One Rutherford (Rd)’.
Answer:
It is another unit of radioactivity. It is defined as the quantity of a radioactive substance, which produces 106 disintegrations in one second.
1 Rd = 10⁶ disintegrations per second.

Question 6.
In ₄Be⁹ + ₂He⁴ → X. What is X?
Answer:
X is ₆C¹³

Question 7.
What is the value of 1 curie?
Answer:
1 curie = 3.7 × 10¹⁰ disintegrations / second.

Question 8.
State the value of Rutherford.
Answer:
1 Rutherford = 10⁶ disintegrations / second.

Question 9.
Explain the term ‘fissile material’ Give an example.
Answer:
Fissionable material is a radioactive element, which undergoes fission in a sustained manner when it absorbs a neutron. It is also termed as ‘fissile material’.
E.g. U²³⁵, plutonium (Pu²³⁹ and Pu²⁴¹)

Question 10.
Which ray has greatest ionising power?
Answer:
Alpha ray

Question 11.
Which ray posses minimum ionising power?
Answer:
Gamma ray

Question 12.
Which ray is not deflected by electric and magnetic fields?
Answer:
Gamma ray

Question 13.
Name the ray that travels with speed about 9/10 times that of light?
Answer:
Beta ray

Question 14.
What is the charge of particle?
Answer:
It’s charge is +2e.

Question 15.
Which particle has a charge of -e?
Answer:
Beta particle

Question 16.
State that the ray consisting of protons.
Answer:
Gamma ray

Question 17.
When uranium emits an alpha particle which element is obtained as daughter element.
Answer:
Thorium ₉₀Th²³⁴

Question 18.
Mention the value of 1eV in terms of joule.
Answer:
1eV = 1.602 × 10^-19 joule

Question 19.
What is the value of 1Mev?
Answer:
1MeV = 10⁶eV = 1.602 × 10^-13 joule

Question 20.
Which radio isotope is used in the treatment of skin diseases?
Answer:
Radio Phosphorus P³²

Question 21.
What is the use of radio iodine?
Answer:
Radio iodine is used to cure goiter.

Question 22.
Which isotope is used in airlines?
Answer:
Califorium Cf²³² is used in airlines.

Question 23.
State the isotope used as a smoke detector.
Answer:
Americium Am²⁴¹

Question 24.
Write down the uses of a nuclear reactor.
Answer:
Uses of a nuclear reactor

• Nuclear reactors are widely used in power generation.
• They are also used to produce radioisotopes, which are used in a variety of applications
• Some reactors help us to do research in the field of nuclear physics.
• Breeder reactors are used to convert non-fissionable materials into fissionable materials.

Question 25.
What are gamma rays?
Answer:
Gamma rays are electromagnetic waves consisting of photons.

Question 26.
Who discovered Cathode rays.
Answer:
J.J. Thomson

VIII. Answer the following question in few sentences.

Question 1.
What is radioactivity?
Answer:
The phenomenon of nuclear decay of certain elements with the emission of radiations like alpha, beta, and gamma rays is called ‘radioactivity’.

Question 2.
What is meant by natural radioactivity?
Answer:
This phenomenon of spontaneous emission of radiation from certain elements on their own is called ‘natural radioactivity’.

Question 3.
Define Rutherford.
Answer:
It is defined as the quantity of a radioactive substance, which produces 106 disintegrations in one second.
1 Rd = 10⁶ disintegrations per second.

Question 4.
What is the unit of radioactivity? Define it.
Answer:
The SI unit of radioactivity is becquerel. It is defined as the quantity of one disintegration per second.

Question 5.
Define nuclear fission. State an example.
This process of breaking (splitting) up of a heavier nucleus into two smaller nuclei with the release of a large amount of energy and a few neutrons is called ‘nuclear fission’.
Eg: Nuclear fission of a uranium nucleus (U²³⁵)
₉₂U²³⁵ + ₀n¹ → ₅₆Ba¹⁴¹ + ₃₆Kr⁹² + 3₀n¹ + Q (energy)

Question 6.
How a product nucleus is identified in a nuclear reaction?
Answer:
In a nuclear reaction, the element formed as the product nucleus is identified by the atomic number of the resulting nucleus and not by its mass number.

Question 7.
What is fissile material?
Answer:
A fissionable material is a radioactive element, which undergoes fission in a sustained manner when it absorbs a neutron. It is also termed as ‘fissile material’.
Eg: U²³⁵, plutonium (Pu²³⁹ and Pu²⁴¹)

Question 8.
What are fertile materials? Mention examples.
Answer:
There are some radioactive elements, which can be converted into fissionable material. They are called as fertile materials.
Eg: Uranium – 238, Thorium – 232, Plutonium – 240.

Question 9.
What is the composition of natural uranium?
Answer:
Natural uranium consists of 99.28 % of ₉₂U²³⁸ and 0.72 % of ₉₂U²³⁵.

Question 10.
What is nuclear fusion? State an example.
Answer:
The process in which two lighter nuclei combine to form a heavier nucleus is termed as ‘nuclear fusion’.
Eg: ₁H² + ₁H² → ₂He⁴ + Q (Energy)

Question 11.
What is meant by chain reaction?
Answer:
A chain reaction is a self-propagating process in which the number of neutrons goes on multiplying rapidly almost in a geometrical progression.

Question 12.
What is meant by mass defect?
Answer:
The mass of the daughter nucleus formed during a nuclear reaction (fission and fusion) is lesser than the sum of the masses of the two parent nuclei. This difference in mass is called mass defect.

Question 13.
Nuclear fusion is named as thermonuclear reaction. Why?
Answer:
Nuclear fusion is possible only at an extremely high temperature of the order of 10⁷ to 10⁹ K and a high pressure to push the hydrogen nuclei closer to fuse with each other. Hence, it is named as ‘Thermonuclear reaction’.

Question 14.
What are the uses of radio isotopes in industries?
Answer:

1. An isotope of californium (Cf²⁵²) is used in the airlines to detect the explosives in the luggage.
2. An isotope of Americium (Am²⁴¹) is used in many industries as a smoke detector.

Question 15.
What is the age of Earth?
Answer:
It is nearly 4.54 × 10⁹ years (around 45 Crore 40 lakh years).

Question 16.
What do you know about dosimeter?
Answer:
Dosimeter is a device used to detect the levels of exposure of an ionizing radiation.

Question 17.
What is the function of moderator in a nuclear reactor?
Answer:
A moderator is used to slow down the high energy neutrons to provide slow neutrons. Graphite and heavy water are the commonly used moderators.

Question 18.
How do coolants function in a nuclear reactor?
Answer:
A coolant is used to remove the heat produced in the reactor core, to produce steam. This steam is used to run a turbine in order to produce electricity. Water, air and helium are some of the coolants.

Question 19.
Name some operating reactors in India.
Answer:
Cirus, Dhuruva and Pumima.

Question 20.
Define Curie.
Answer:
Curie is defined as the quantity of a radioactive substance, which undergoes 3.7 × 10¹⁰ disintegrations in one second. This is actually close to the activity of 1 g of radium-226.

IX. Answer the following question in detail.

Question 1.
Write short notes on induced radioactivity.
Answer:
The phenomenon by which even light elements are made radioactive, by artificial or induced methods, is called ‘artificial radioactivity’ or ‘man-made radioactivity’.

This kind of radioactivity was discovered by Irene Curie and F.Joliot in 1934. Artificial radioactivity is induced in certain lighter elements like boron, aluminium etc., by bombarding them with radiations such as ‘alpha particles’ emitted during the natural radioactivity of uranium. This also results in the emission of invisible radiations and elementary” particles. During such a disintegration, the nucleus which undergoes disintegration is called ‘parent nucleus’ and that which is produced after the disintegration is called a ‘daughter nucleus’. The panicle, which is used to induce the artificial disintegration is termed as projectile and the particle which is produced after the disintegration is termed as ejected panicle. When the projectile hits the parent nucleus, it is converted into an unstable nucleus, which in turn decays spontaneously emitting the daughter nucleus along with an ejected particle.

If you denote the parent and daughter nuclei as X and Y respectively, then the nuclear disintegration is represented as follows: X (P,E) Y. Here, P and E represent the projectile particle and ejected particle respectively.
₄Be⁹ + ₂He⁴ → ₆C^13*
₆C^13* → ₆C¹² + ₀n¹
In the above nuclear reaction, ₆C^13* is unstable and is radioactive. This reaction can be represented as ₄Be⁹ (α, n) ₄C¹²
₄Be⁹ + ₂He⁴ → ₆C¹² + ₀n¹

Question 2.
Explain the concept of ‘Hydrogen bomb’.
Answer:
Hydrogen Bomb:

• This releases a huge amount of energy in a very small time interval and leads to a Hydrogen bomb is based on the principle of nuclear fusion.
• A hydrogen bomb is always designed to have an inbuilt atom bomb which creates the high temperature and pressure required for fusion when it explodes.
• Fusion takes place in the hydrogen core and leads to the release of a very large amount of energy in an uncontrolled manner.
• The energy released in a hydrogen bomb (or fusion bomb) is much higher than that released in an atom bomb (or fission bomb).

Question 3.
Explain
(i) alpha decay,
(ii) Beta decay,
(iii) gamma decay
Answer:
(i) Alpha decay : A nuclear reaction in which an unstable parent nucleus emits an alpha particle and forms a stable daughter nucleus, is called, ‘alpha decay’.
Eg: Decay of uranium (U²³⁸) to thorium (Th²³⁴) with the emission of an alpha particle.
₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ (α – decay)
In α – decay, the parent nucleus emits an a particle and so it is clear that for the daughter nucleus, the mass number decreases by four and the atomic number decreases by two.

(ii) Beta decay : A nuclear reaction, in which an unstable parent nucleus emits a beta particle and forms a stable daughter nucleus, is called ‘beta decay’.
Eg: Beta decay of phosphorous.
₁₅P³² → ₁₆S³² + _-1e⁰(β – decay)
In β – decay there is no change in the mass number of the daughter nucleus but the atomic number increases by one. In a nuclear reaction, the element formed as the product nucleus is identified by the atomic number of the resulting nucleus and not by its mass number.

(iii) Gamma decay: In a γ – decay, only the energy level of the nucleus changes. The atomic number and mass number of the radioactive nucleus remain the same.

Question 4.
What is the principle of atom bomb? Describe its structure.
Answer:
The atom bomb is based on the principle of uncontrolled chain reaction. In an uncontrolled chain reaction, the number of neutrons and the number of fission reactions multiply almost in a geometrical progression. This releases a huge amount of energy in a very small time interval and leads to an explosion.

Structure : An atom bomb consists of a piece of fissile material whose mass is subcritical. This piece has a cylindrical void. It has a cylindrical fissile material which can fit into this void and its mass is also subcritical. When the bomb has to be exploded, this cylinder is injected into the void using a conventional explosive. Now, the two pieces of fissile material join to form the supercritical mass, which leads to an explosion. The structure of an atom bomb.

During this explosion tremendous amount of energy in the form of heat, light and radiation is released. A region of very high temperature and pressure is formed in a fraction of a second along with the emission of hazardous radiation like y rays, which adversely affect the living creatures. This type of atom bombs were exploded in 1945 at Hiroshima and Nagasaki in Japan during the World War II.

Question 5.
What precautions are to he took for those, who are working in radiation laboratories?
Answer:

1. Radio – Iodine (I¹³¹) is used to cure goitre.
2. Radioactive materials should be kept in a thick-walled lead container.
3. Lead coated aprons and lead gloves should be used while working with hazardous radioactive materials.
4. We should avoid eating while handling radioactive materials.
5. The radioactive materials should be handled only by tongs or by a remote control device.
6. Dosimeters should be worn by the users to check the level of radiation.

Question 6.
What do you know about hydrogen bomb?
Answer:
Hydrogen bomb is based on the principle of nuclear fusion. A hydrogen bomb is always designed to have an inbuilt atom bomb which creates the high temperature and pressure required for fusion when it explodes. Then, fusion takes place in the hydrogen core and leads to the release of a very large amount of energy in an uncontrolled manner. The energy released in a hydrogen bomb (or fusion bomb) is much higher than that released in an atom bomb (or fission bomb).

Question 7.
What are the applications of radio isotopes in medicine?
Answer:
Medical applications of radio isotopes can be divided into two parts:

1. Diagnosis
2. Therapy

Radio isotopes are used as tracers to diagnose the nature of circulatory disorders of blood, defects of bone metabolism, to locate tumors, etc., Some of the radio isotopes which are used as tracers are: hydrogen, carbon, nitrogen, sulphur, etc.

1. Radio-sodium (Na²⁴) is used for the effective functioning of heart.
2. Radio-Iodine (I¹³¹) is used to cure goiter.
3. Radio-iron is (Fe⁵⁹) is used to diagnose anaemia and also to provide treatment for the same.
4. Radio-phosphorous (P³²) is used in the treatment of skin diseases.
5. Radio-cobalt (Co⁶⁰) and
6. Radio-gold (Au¹⁹⁸) are used in the treatment of skin cancer.
7. Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.

Question 8.
Mention the applications of radio isotopes in agriculture and industries and archeological research.
Answer:
Agriculture : The radio isotope of phosphorous (P-32) helps to increase the productivity of crops. The radiations from the radio isotopes can be used to kill the insects and parasites and prevent the wastage of agricultural products. Certain perishable cereals exposed to radiations remain fresh beyond their normal life, enhancing the storage time. Very small doses of radiation prevent sprouting and spoilage of onions, potatoes and gram.

Industries : In industries, radioactive isotopes are used as tracers to detect any manufacturing defects such as cracks and leaks. Packaging faults can also be identified through radio activity. Gauges, which have radioactive sources are used in many industries to check the level of gases, liquids and solids.

1. An isotope of californium (Cf ²⁵²) is used in the airlines to detect the explosives in the luggage.
2. An isotope of Americium (Am²⁴¹) is used in many industries as a smoke detector.

Archeological research : Using the technique of radio carbon dating, the age of the Earth, fossils, old paintings and monuments can be determined. In radio carbon dating, the existing amount of radio carbon is determined and this gives an estimate about the age of these things.

Question 9.
Compare the features of nuclear fission with nuclear fusion.
Answer:

Question 10.
Mention the preventive measures to be followed while working with radioactive materials.
Answer:
Radioactive materials should be kept in a thick walled lead container.

1. Lead coated aprons and lead gloves should be used while working with hazardous radioactive materials.
2. You should avoid eating while handling radioactive materials.
3. The radioactive materials should be handled only by tongs or by a remote control device.
4. Dosimeters should be worn by the users to check the level of radiation.

Question 11.
Write a note on nuclear power plants in India.
Answer:
Indian Atomic Energy Commission (AEC) was established in August 1948 by the Department of Indian Scientific Research committee at Bombay (now Mumbai) in Maharashtra. It is the nodal agency for all the research done in the field of atomic energy. Dr. Homi Jahangir Bhaba was the first chairman of Indian Atomic Energy Commission. Now, it is known as Bhaba Atomic Research Centre (BARC).

Nuclear power is the fifth largest source of power in India. Tarapur Atomic Power Station is India’s first nuclear power station. Now, there are a total of seven power stations, one each in Maharashtra, Rajasthan, Gujarat, Uttar Pradesh and two in Tamilnadu. In Tamilnadu, we have nuclear power stations in Kalpakkam and Kudankulam. Apsara was the first nuclear reactor built in India and Asia. Now, there are 22 nuclear reactors which are operating in India. Some other operating reactors are

1. Cirus
2. Dhuruva
3. Pumima

Question 12.
Distinguish (i) atom bomb from Nuclear reactor,
(ii) Nuclear fission from Nuclear fusion.
Answer:
(i) atom bomb from nuclear reactor:

(ii) Nuclear fission from Nuclear fusion

X. Hot Questions.

Question 1.
A neutron is most effective as bulet in nuclear reactions. Why?
Answer:
This is because a neutron carries no charge. It can hit the nucleus directly without being repelled or attracted by the nucleus or electrons.

Question 2.
Why is \({ _{ 92 } }U^{ 238 }\) not suitable for chain reaction?
Answer:
Only fast-moving neutrons of 12 MeV can cause fission of \({ _{ 92 } }U^{ 238 }\) nuclei. But such neutron has fewer chances of interaction. They escape the fissionable material without causing fission.

Question 3.
Why cadmium is used as control rods in the nuclear reactor?
Answer:
As Cadmium has a high cross section for absorption of neutrons, cadmium is used as control rod.

Question 4.
Why is nuclear fusion not possible in a laboratory (or) Why is nuclear fusion difficult to carry out?
Answer:
Nuclear fusion requires a very high temperature of 10⁶ – 10⁷ K. This temperature is attained by causing an explosion due to the fission process. Moreover, no solid container can withstand such a high temperature.

Question 5.
Why is the penetrating power of a – particle is less than that of β – particle?
Answer:
Ionising power is directly proportional to kinetic energy that varies directly as mass of particle and square of its velocity. Hence ionising power of an alpha particle is more than that of β particle.

Question 6.
Why heavy water is used to slow down the fast moving neutrons?
Answer:
Since heavy water (D₂O) is rich in hydrogen, it is used to slow down the fast moving neutrons.

Question 7.
The penetrating power of α – particle is less than that of β – particle – Why?
Answer:
(i) Penetrating power varies inversely as mass of the particle. But mass of alpha particles is very much greater than mass of beta particles.
(ii) Hence, penetrating power of α – particle is less than that of β – particles.

Question 8.
What are the units of activity of radio active elements?
Answer:
1 curie = 3.7 × 10¹⁰ disintegrations / second
1 rutherford = 10⁶ disintegrations / second

Students can download 10th Social Science History Chapter 10 Social Transformation in Tamil Nadu 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 History Chapter 10 Social Transformation in Tamil Nadu

Samacheer Kalvi 10th Social Science Social Transformation in Tamil Nadu Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
…………….. was the pioneer of social Reformers in India.
(a) C. W. Damotharanar
(b) Periyar
(c) Raja Rammohan Roy
(d) Maraimalai Adigal
Answer:
(c) Raja Rammohan Roy

Question 2.
…………… established a full-fledged printing press in 1709, at Tranquebar.
(a) Caldwell
(b) F.W.Ellis
(c) Ziegenbalg
(d) Meenakshisundaram
Answer:
(c) Ziegenbalg

Question 3.
…………….. was the official newspaper of the Self Respect Movement.
(a) Kudi Arasu
(b) Puratchi
(c) Viduthalai
(d) Paguththarivu
Answer:
(a) Kudi Arasu

Question 4.
Periyar wanted religion to be replaced by ………………
(a) Nationalism
(b) Iconoclasm
(c) Rationalism
(d) Spiritualism
Answer:
(c) Rationalism

Question 5.
…………….. founded Adi Dravida Mahajana Sabha in 1893.
(a) Rettaimalai Srinivasan
(b) B.R. Ambedkar
(c) Rajaji
(d) M.C. Rajah
Answer:
(a) Rettaimalai Srinivasan

Question 6.
India’s first organised trade union, the Madras Labour Union was formed in ………….
(a) 1918
(b) 1917
(c) 1916
(d) 1914
Answer:
(a) 1918

Question 7.
…………….. was established by the Justice Party Government for the selection of Government officials.
(a) Staff Selection Board
(b) Public Service Commission
(c) Provincial Staff Recruitment Board
(d) Staff Selection Commission
Answer:
(a) Staff Selection Board

Question 8.
……………. was the first elected Legislative Council Member from the depressed class in Madras Province.
(a) M. C. Rajah
(b) Rettaimalai Srinivasan
(c) T.M. Nair
(d) P. Varadarajulu
Answer:
(a) M. C. Rajah

II. Fill in the blanks

1. ……………… was the first non-European language that went into print.
2. The College of Fort St. George was founded by ………………
3. ……………… is considered the father of Tamil linguistic purism.
4. ……………… was the first to approve participation of women in the electoral politics.
5. The name Suriyanarayana Sastri changed in Tamil as ………………
6. ……………… gave prominence to Tamil music.
7. The first Woman Legislator in India was

Answers:

1. Tamil
2. F.W. Ellis
3. Maraimalai Adigal
4. Justice Party Government
5. Parithimar Kalaignar
6. Abraham Pandithar
7. Muthulakshmi Ammaiyar

III. Choose the correct statement

Question 1.
(i) Thirukkural was one of the earliest Tamil literary texts to be published in 1812.
(ii) Maraimalai Adigal collected and edited different palm leaf manuscripts of the Tamil grammars and literature.
(iii) Robert Caldwell established the close affinity between the Dravidian languages in contrast with Sanskrit and also established the antiquity of Tamil.
(iv) Thiru. V. Kalyanasundaram was an early pioneer in Trade union movement.
(a) (i) and (1%) are correct
(b) (i) and (iii) are correct
(c) (iv) is correct
(d) (ii) anA (iii) are correct
Answer:
(b) (i) and (iii) are correct

Question 2.
(i) Margret Cousin was one of the founders of Women’s India Association.
(ii) Periyar spent his entire life campaigning against superstitions through Thinkers or Rationalists Forums he had formed.
(iii) Singaravelar was a staunch supporter of the Hindu Mahasabha.
(iv) Periyar emphasised that the caste system in South India is linked with the arrival of Brahmins from the North.
(a) (iii) and (iv) are correct
(b) (ii), (iii) and (iv) are correct
(c) (i), (ii) and (iv) are correct
(d) (ii) and (iii) are correct
Answer:
(c) (i), (ii) and (iv) are correct

Question 3.
Assertion (A): The Justice Party continued to remain in Government from 192CE1937 in Madras Presidency.
Reason (R): The Congress Party boycotted the Madras Legislature during this period of Dyarchy.
(a) Both A and R are correct
(b) A is correct, but R is not the correct explanation
(c) Both A and R are wrong
(d) R is correct, but it has no relevance to A
Answer:
(b) A is correct, but R is not the correct explanation

Question 4.
Assertion (A): As World War I was in progress, the British Government was considering the introduction of representative institutions for Indians after the war.
Reason (R): In 1920, Dyarchy as a form of Government was introduced in the provinces.
(a) A is correct, but R is not the correct reason
(b) Both A and R are wrong
(c) Both A and R are right
(d) A is wrong and R has no relevance to A
Answer:
(c) Both A and R are right

IV. Match the following

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

V. Answer briefly

Question 1.
Write a note on Tamil Renaissance.
Answer:
C.W. Damotharanar collected and edited different palm leaf manuscripts of the Tamil grammars and literature. U.V. Swaminathar took efforts to publish the classical texts. These triggered an intellectual revolution and it was popularly known as Tamil Renaissance. Tamil language and culture played a significant role in their identity construction. The introduction of printing press, linguistic research on Dravidian languages, etc., underpinned the process of Tamil renaissance.

Tamil was the first non-European language that went into print. In the nineteenth century, Tamil scholars like C.W. Damotharanar, and U.V. Swaminathar spent their lifetime in the rediscovery of the Tamil classics. This provided the Tamil people with a revelation about their heritage. Therefore, the rediscovery of ancient classics and their publication is considered the foundation of Tamil renaissance.

Question 2.
Highlight the contribution of Caldwell for the cause of South Indian languages.
Answer:

1. In 1816, F. W. Ellis who founded the College of Fort St. George, formulated the theory that the south Indian languages belonged to a separate family which was unrelated to the Indo – Aryan family of languages.
2. Robert Caldwell (1814-1891) expanded this argument in a book titled. A comparative grammar of the Dravidian (or) South Indian family of languages in 1856.
3. He established the close affinity between the Dravidian languages in contrast with Sanskrit and also established the antiquity of Tamil.

Question 3.
List out the personalities who contributed to the revival of Tamil literature through their writings.
Answer:
C.W. Damotharanar, U.V. Swaminathar, F.W. Ellis, Robert Caldwell, Subramania Bharati, Thiru Vi. Kalyanasundaram, Singaraveiar, PanditViar lyotheethassar, Sundaram Pillar and Maraimalai Adigal.

Question 4.
Discuss the importance of Hindu Religious Endowment Act passed by the Justicite Ministry.
Answer:

1. The Justice Party further concentrated on reforms in religious institutions.
2. Tamil Nadu has a large number of temples and these commanded huge resources.
3. In general, the resources were monopolised and exploited by the ‘ dominant caste in the society and led to mismanagement of public resources.
4. The Justice Party introduced the Hindu Religious Endowment (HRE) Act in 1926 and enabled any individual irrespective of their caste affiliation to become a member of the temple committee and govern the resources of the religious institutions.

Question 5.
What do you know of the Cheranmahadevi Gurukulam incident?
Answer:
A Gurukulam was established in Cheranmahadevi by V.V.S. Iyer with the financial support of the Tamil Nadu Congress Committee. However, students were discriminated on the basis of caste. Brahmin and non-Brahmin students were made to dine separately. Periyar was disturbed to see the discrimination. He questioned the practice, and severely criticized it. Despite his objections, Congress continued to support the iniquitous practice in the Gurukulam. This disappointed Periyar.

Question 6.
Name the newspapers published by the South Indian Liberal Foundation.
Answer:
Dravidian in Tamil, Justice in English and Andra Prakasika in Telugu to propagate ideals of the Party.

Question 7.
Estimate Periyar as a feminist
Answer:
Periyar was a great feminist. He believed in gender equality and raised voice over the plight of women. He emphasized on women’s right to divorce and property. He objected the terms like “giving in marriage”. This, he said, treats woman as a thing. He wanted it to substitute by “valkaithunai,” (companion) a word for marriage taken from the Tirukkural. Periyar’s most important work on this subject is Why the Woman is Enslaved ? He believed that property rights for women would provide them a social status and protection. He welcomed equal rights for males and females in property, guardianship and adoption.

Question 8.
Explain the proceedings of All India Trade Union Congress Conference held in 1920.
Answer:

1. The first All India Trade Union Conference (AITUC) was held on 31st October 1920 in Bombay. The delegates discussed several resolutions.
2. These included a demand for protection from police interference in labour disputes, the maintenance of an unemployment register restriction on exporting food stuffs, compensation for injuries and health insurance.
3. In addition, the delegates demanded that Indian workers be given some representation in the Government just as empolyers had representatives on legislative councils.

VI. Answer the questions given under each caption

Question 1.
Periyar e.v.r

(a) When did Periyar founded the Dravidar Kazhagam?
Answer:
Periyar founded the Dravidar Kazhagam in the year 1944.

(b) What were the Newspapers and Journals run by Periyar?
Answer:
Kudi Arasu (Democracy) (1925) Revolt (1928) Puratchi Revolution (1933) Paguth tharivu (Rationalism) (1934) and Viduthalai (Liberation) (1935).

(c) Why was Periyar known as Vaikom hero?
Answer:

1. In the name of caste dharma the “lower caste” people were denied access to the temples and the streets surrounding the temple.
2. In Vaikom Catown in the then princely state of Travancore and in present day Kerala, people protest against this practice.
3. People hailed him as Vaikom Virar (Hero of Vaikom) because Periyar led the movement and was imprisoned.

(d) Which was the most important work of Periyar?
Answer:
Periyar’s most important work on this subject is why the women is enslaved?

Question 2.
Labour Movement in Tamil Nadu

(a) Highlight the factors that caused the birth of Trade Union Movement in Madras.
Answer:
The factors that caused the birth of Trade Union Movement in Madras are

• Retrenchment of workers at the end of the First World War.
• Nationalists’ support to the cause of labour.

(b) Identify the three prominent persons associated with the Madras Labour Union.
Answer:
Three prominent persons associated with the Madras Labour Union are B.P. Walia, M. Singaravelar and Thiru.Vi. Kalyanasundaram.

(c) Where was the first conference of All India Trade Union Congress held?
Answer:
The first All India Trade Union Conference was held in Bombay.

(d) Who organized the first ever celebration of May Day in Madras and which year?
Answer:
M. Singaravelar organised the first ever celebration of May Day in Madras in 1923.

Question 3.
Maraimalai adigal

(a) Name the Sangam texts for which Maraimalai Adigai wrote commentaries.
Answer:
Maraimalai Adigai wrote commentaries on the sangam texts Pattinappalai and Mullaipattu.

(b) Name the Journal where he worked as a young man.
Answer:
Siddhanta Deepika.

(c) Why did he oppose imposition of Hindi?
Answer:
He promoted the use of pure Tamil words, impact on Tamil culture especially in language and literature.

(d) Who were the key influences in Maraimalai Adigal’s life?
Answer:
Maraimalai Adaigal’s teachers such as P. Sundaranar and Somasundra Nayagar were key influences in his life.

VII. Answer in detail

Question 1.
Attempt an essay on foundation and development of Tamil Renaissance in the 19fh Century.
Answer:
(i) In Tamil Nadu, the proliferation of the printing press acted as a catalyst for the publication and spread of secular ancient Tamil literature.

(ii) In the nineteenth century, Tamil scholars like C.W. Damotharanar, and U.V. Swaminathar spent their lifetime in the rediscovery of the Tamil classics.

(iii) C.W. Damotharanar collected and edited different palm leaf manuscripts of the Tamil grammars and literature. U.V. Swaminathar took efforts to publish the classical texts. These triggered an intellectual revolution and it was popularly known as Tamil Renaissance.

(iv) The transformation not only revived Tamil language and literature but also Tamil culture. It challenged the prevailed caste hierarchy and influenced the rise of Dravidian consciousness and established Tamil as a language of the Dravidian people.

(v) Tamil renaissance questioned the cultural hegemony of Brahminism. These developments were reflected in art, literature, religion, etc.

Question 2.
Describe the background for the formation of the Justice Party and point put its contribution to the cause of social justice.
Answer:

1. As World War -1 was in progress the British Government was considering the introduction of representative institutions for Indians after the War.
2. Fearing that such political reforms would further strengthen the political power of Brahmins, educated Non – Brahmins decided to organise themselves politically.
3. On 20th, November 1916 around 30 prominent Non – Brahmin leaders including Dr. C. Natesanar Sir, Pitti Theyagarayar, T.M. Nair and Alamelu Mangai Thayarammal came together to form the South Indian Liberation Federation (SILF).

Programme and Activities:

1. The Justice Party is the fountain head of the Non – Brahmin Movement in the country.
2. The Justicite removed the legal hindrances restricting inter – caste marriages and broke the barriers that prevented Depressed Classes from the use of public wells and tanks.
3. Hostels were established for the students belonging to this social group in 1923.
4. The Justice Party Government was the first to approve participation of women in the electoral politics in (1921)
5. The Justice Party worked towards legislating provisions for communal representation reservations for various communities.
6. The Justice Party rule established the Staff Selection Board in 1924 for the selection of Government officials and encouraged all the communities to share the administrative powers.
7. In 1929, the Government of British India adopted the pattern and established the Public Service Commission.
8. The Justice Party introduced the Hindu Religious Endowment (HRE) Act in 1926 and enabled any individual, irrespective of their caste affiliation, to become a member of the temple committee and govern the resources of the religious institutions.

Question 3.
Estimate Periyar E.V.R’s decisive contribution to the social transformation of Tamil Nadu?
Answer:
Periyar was a great social reformer. His contribution to the social transformation of Tamil Nadu is really praise worthy.
(i) He launched ‘Temple entry’ movement to provide the lower caste people easy access to the temples.

(ii) He also started the Self-Respect movement in 1925, with the determination that there ought to be ‘no God; no religion; no Gandhi; no Congress; and no Brahmins’.

(iii) Periyar was critical of patriarchy. He condemned child marriage and the Devadasi system. Right from 1929, when the Self-Respect conference began to voice its concern over the plight of women, Periyar had been emphasising on women’s right to divorce and property.

(iv) He advocated atheism as a mode of critique to deconstruct the established practices of faith, culture and custom. Periyar wanted religion to be replaced by rationalism.

(v) He welcomed equal rights for males and females in property, guardianship and adoption. He was strong champion of birth control and contraception.

VIII. Students Activity

Question 1.
Students can be taught to distinguish between Labour Movement and Trade Union Movement. Project work on the activities of local trade union organisations may be done by students.
Answer:
Distinguish between Labour Movement and Trade Movement.

Labour Movement:

1. It is a political movement campaigning to improve the rights and conditions of the working people.
2. If originated in U.S.A focussing on rights, interests and protection of employees.
3. Labour movement generally work for better wages, reasonable working hours and safer working conditions.

Trade union Movement:

1. It is an association of workers forming a legal unit to improve their status by united action.
2. It originated in Great Britain focussing to protect and promote the welfare of the employees.
3. Trade Union Movement regulate employer – employee relationship and generally work for better wages, reasonable working hours and safer working conditions.

Activities of a local trade union: Students should visit with few of their classmates (with prior permission from the subject teacher and school authorities and parents) to visit a nearby industry where trade unions are organised by the workers.

They have to meet the trade, union leader and gather information on the activities performed by them in the factory. The collected information should be written down neatly and submit to the subject teacher.

Eg: In general, the following are the few activities performed by the trade union.

1. Negotiate wages.
2. Regulate relationship between workers (employees) and the employers.
3. Raise demand for new uniform, or safer working condition, pure drinking water, sanitary facilities etc.,
4. Protect the workers against discrimination.
5. Protect the workers from exploitation.

Question 2.
Students can compile the activities of the local writers’ associations or women’s collectives.
Answer:
Local writers Association/women collectives: Students should find people who are writers and they write scripts for audio formats, digital media (or) T.V or screenwriters for films. Later, students can collect information about their.

1. Name of their Association.
2. Location.
3. Job site.
4. Nature of work.
5. Year of joining the Association
6. Membership fee
7. Total no. of members
8. Usefulness being the member

and submit in this format to the subject teacher. If students can find women organised themselves as a group, in the above-mentioned format can be used for that also. Eg: Women collectives (self-help group)

They work for their betterment together as a group (Sua Udavi Kuzhu).

Samacheer Kalvi 10th Social Science Social Transformation in Tamil Nadu Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
………………. is also identified as the Indian renaissance.
(a) Historical development
(b) Political development
(c) Economic development
(d) None
Answer:
(a) Historical development

Question 2
Periyar transformed the justice Party into ………….
(a) Akali Dal
(b) Dravidar kazhagam
(c) Swarajya
Answer:
(b) Dravidar kazhagam

Question 3.
The basic inspiration of renaissance is:
(a) Knowledge
(b) Humanism
(c) Painting
(d) Transition
Answer:
(b) Humanism

Question 4.
Vaikom is a place in ……………
(a) Tamil Nadu
(b) Andhra Pradesh
(c) Kerala
Answer:
(c) Kerala

Question 5.
The introduction of ………………. on Dravidian languages are underpinned the Process of tamil renaissance
(a) Printing press
(b) linguistic research
(c) both (a) and (b)
(d) knowledge
Answer:
(c) both (a) and (b)

Question 6.
The Justice Party was established in 1916 by ……………
(a) Periyar
(b) Anna
(c) C. Natesanar, Pitti Theyagarayar, T.M. Nair and Alamelu Mangai Thayarammal
Answer:
(c) C. Natesanar, Pitti Theyagarayar, T.M. Nair and Alamelu Mangai Thayarammal

Question 7.
In which year Thirukkural, Tamil literary text to be published?
(a) 1810
(b) 1805
(c) 1808
(d) 1812
Answer:
(d) 1812

Question 8.
Self Respect Movement was started in the year …………….
(a) 1927
(b) 1925
(c) 1930
Answer:
(b) 1925

Question 9.
Who was collected and edited different palm – leaf manuscripts of the Tamil grammerand literature?
(a) U.V. Swaminathar
(b) C.W. Damotharanar
(c) Meenakshi Sundaranar
(d) P. Sundaranar
Answer:
(b) C.W. Damotharanar

Question 10.
Devadasi system was abolished by …………..
(a) Dr. Muthulakshmi Reddy
(b) Ramamirdhan
(c) Dr. S. Dharmambal
Answer:
(a) Dr. Muthulakshmi Reddy

Question 11.
………………. was popularly known as Vallalar.
(a) Ramalinga Adigal
(b) Parithimar Kalaignar
(c) Thiru.Vi. Ka
(d) Maraimalai Adigal
Answer:
(a) Ramalinga Adigal

Question 12.
Whose name is associated with the ‘Self-Respect’ movement?
(a) Muthulakshmi Reddy
(b) Rajaji
(c) Rettimalai Srinivasan
(d) Periyar
Answer:
(d) Periyar

Question 13.
Who was the first to argue that Tamil is a classical languages?
(a) Parithimar Kalaignar
(b) Maraimalai Adigal
(с) E.V. Ramasamy
(d) M. Singaravelar
Answer:
(a) Parithimar Kalaignar

Question 14.
Whose efforts led to the abolition of the devadasi system?
(a) Muthulakshmi Reddy
(b) Muvalur Ramamirtham
(c) M.C. Rajah
(d) Periyar
Answer:
(a) Muthulakshmi Reddy

Question 15.
Which organisation focused on educating and supporting Non – Brahmin graduates?
(a) Madras Dravidian Association
(b) Tani Tamil Iyakkam
(c) Tamil Isai Iyakkam
(d) None
Answer:
(a) Madras Dravidian Association

Question 16.
Who was the founder of Tani Tamil Iyakkam? (Pure Tamil Movement).
(a) M. Singaravelar
(b) Parithimar Kalaignar
(c) Ramalinga Adigal
(d) Maraimalai Adigal
Answer:
(d) Maraimalai Adigal

Question 17.
In which year Non – Brahmin Manifesto was released?
(a) 1905
(b) 1910
(c) 1912
(d) 1916
Answer:
(d) 1916

Question 18.
In which year the Congress contested the electron for the first time?
(a) 1930
(b) 1933
(c) 1935
(d) 1937
Answer:
(d) 1937

Question 19.
In which year Justice Party Government was the firstto approval participation of women in the electrol politics?
(a) 1910
(b) 1915
(c) 1919
(d) 1921
Answer:
(d) 1921

Question 20.
Which Party introduced the Hindu Religious Endowment? (HRE)
(a) The Justice Party
(b) The Congress Party
(c) both (a) and (b)
(d) none
Answer:
(a) The Justice Party

Question 21.
Who was the founder of the Self – Respect Movement?
(a) E.V. Ramasamy
(b) Rajaji
(c) Rettaimalai
(d) none
Answer:
(a) E.V. Ramasamy

Question 22.
In which year Rettaimalai Srinivasan was born?
(a) 1849
(b) 1852
(c) 1855
(d) 1859
Answer:
(d) 1859

Question 23.
Who was one of the prominent leaders from the “depressed class”?
(a) B.R. Ambedkar
(b) M.C. Rajah
(c) Rajaji
(d) none
Answer:
(b) M.C. Rajah

Question 24.
Who was a pioneer in the labour movement activities?
(a) M. Singaravelar
(b) M.C. Rajah .
(c) B.R. Ambedkar
(d) Rajaji
Answer:
(a) M. Singaravelar

Question 25.
Who was in the forefront of the campaign pressing for legislation to abolish Devadasi system?
(a) Nagammai
(b) Dr. Muthulakshmi
(c) Rukmani
(d) Neelambikai
Answer:
(b) Dr. Muthulakshmi

II. Fill in the blanks

1. ………………. traditions were treated as primitive.
2. ………………. was portrayed as conservative.
3. ………………. brought changes in the socio – cultural life of the Indian sub – continent.
4. Manonmaniam written by ……………….
5. Tamil renaissance questioned the cultural hegemony of ……………….
6. ………………. an early pioneer in Buddhist revival promoted communism and socialism.
7. ………………. worked in a Journal Siddhanta Deepika.
8. ………………. held high in the radical philosophy.
9. ………………. complied a dictionary that provide pure Tamil equivalent to Sanskirt.
10. The Madras Non – Brahmins Association was founded in ……………….
11. ………………. founded the Madras United League.
12. ………………. won the election and formed the first ever Indian Cabinet in Madras.
13. ………………. later renamed as Madras Dravidian Association to support Dravidian uplift.
14. ………………. became the chief minister of the Madras Presidency during (1920-26).
15. ………………. is the fountain head of the Non – Brahmin Movement in the country.
16. In 1929, the Government of ………………. adopted the pattern and established the public service commission.
17. ………………. proposed a resolution regarding the rights of “Untouchables” to temple entry.
18. ………………. demand for a separate electorates for scheduled castes.
19. ………………. The Chief Minister of Madras State (1952 – 54) introduced to vocational training.
20. ………………. wanted religion to be replaced by rationalism.
21. ………………. who had worked for the the progress of the deprived castes.
22. The first All India Trade Union (AITUC) was held in ……………….
23. ………………. demanded abolition of untouchability.
24. ………………. was a pioneer in the Labour Movement Activities.
25. The first Tamil Isai conference was held in ……………….
26. ………………. was one of the important objectives of the Self – Respect Movement.
27. ………………. formed the All India Women’s Conference (AlWC).
28. The Madras ………………. was enacted by the Government.
29. ………………. founded the Tanjore Sangitha Vidhya Mahajana Sangam.
30.  ………………. published a Tamil Newspaper Thozhilalan (Worker).

Answers:

1. Indigenous
2. Indian society
3. The rise of Humanism
4. P. Sundaranar
5. Brahminism
6. M. Singaravelar
7. Maraimalai Adigal
8. E.V. Ramasamy
9. Neelambikai
10. 1909
11. C. Natesanar
12. The Justice Party
13. Madras United League
14. A. Subbarayalu
15. The Justice Party
16. British India
17. Periyar
18. Dr. B.R. Ambedkar
19. Rajaji
20. Periyar
21. Rettaimalai Srinivasan
22. Bombay
23. M.C. Rajah
24. M. Singaravelar
25. 1943
26. Women’s Liberation
27. Women’s India Association (WIA)
28. Devadasis Act
29. Abraham Pandithar
30. M. Singaravelar

III. Choose the correct statement

Question 1.
(i) C.W. Damotharanar collected and edited different palm-leaf manuscripts of the Tamil grammar.
(ii) Ramalinga Adigal gave prominence to Tamil music.
(iii) Maraimalai Adigal promoted the use of pure Tamil words and removal of the Sanskirt influence from the Tamil language.
(iv) Muvalur Ramamirtham to became the first women legislator in India in 1926.
(a) (i) and (ii) are correct
(b) (i) and (iii) are correct
(c) (iv) is correct
(d) (ii) and (iii) are correct
Answer:
(b) (i) and (iii) are correct

Question 2.
(i) The Justice Party Rule established the Staff Selection Board in 1924 for the selection of Government officials.
(ii) Periyar had a close relationship with Singaravelar who is considered the first communist of South India.
(iii) The anti – Hindi agitation (1932 -1936) had a big impact on Tamil Nadu’s politics.
(iv) Rettaimalai Srinivasan criticised it as Kula Kalvi Thittam and opposed it tooth and nail.
(a) (i) and (ii) are correct
(b) (i) and (iv) are correct
(c) (iii) is correct
(d) (ii) and (iv) are correct
Answer:
(a) (i) and (ii) are correct

Question 3.
(i) Periyar was keen on the introduction of reservation in representative bodies such as the legislative council for Non – Brahmins.
(ii) All India Women’s Conference (AIWC) in 1937 to address the problem of Non – Brahmins.
(iii) The First World War (1912 -1920) provided stimulus to industrial growth ih India.
(iv) Women’s Liberation was one of the important objectives of Self – Respect Movement.
(a) (i) and (ii) are correct
(b) (i) and (iii) are correct
(c) (iv) is correct
(d) (i) and (iv) are correct
Answer:
(d) (i) and (iv) are correct

Question 4.
Assertion (A): Rettaimalai Srinivasan who had experienced the horrors of untouchability worked for the progress of the deprive castes.
Reason (R): He served as president of the scheduled castes federations and the Madras Provincial Depressed Classes.
(a) Both A and R are correct.
(b) A is correct but R is not the correct explanation of A.
(c) Both A and R wrong.
(d) R is correct but it has no relevance to A.
Answer:
(a) Both A and R are correct.

Question 5.
Assertion (A): Periyar E.V. Ramasamy was the founder of the Self – Respect movement.
Reason (R): The movement was concerned with the marginalised sections of the society and criticised Brahminism and it advocated a casteless society.
(a) Both A and R are wrong.
(b) A is correct and R is correct explanation of A.
(c) A is correct and R is wrong.
(d) Both A and R are correct but R is not the correct explanation of A.
Answer:
(b) A is correct and R is correct explanation of A.

IV. Match the following

Question 1.
Match the Column I with Column II.

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

Question 2.
Match the Column I with Column II.

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

Question 3.
Match the Column I with Column II.

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

Question 4.
Match the Column I with Column II.

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

Question 5.
Match the Column I with Column II.

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

Question 6.
Match the Column I with Column II.

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

Question 7.
Match the Column I with Column II.

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

Question 8.
Match the Column I with Column II.

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

V. Answer briefly

Question 1.
What are the aims of Self Respect Movement?
Answer:

1. The Self Respect Movement condemned and fought against Brahmin domination over other castes, society, politics and religion.
2. It fought to abolish traditionalism and superstitions.
3. It advocated women education, widow remarriage, inter-caste marriages and opposed child marriage.

Question 2.
Explain about the basic inspiration of renaissance.
Answer:

1. The basic inspiration of renaissance is humanism and questioning the fundamentalist religious practices that denied humans their dignity.
2. This ideology of humanism stimulated creative energy in all spheres of social life and knowledge such as language, literature, philosophy, music, painting, architecture etc.

Question 3.
What were the causes for the failure of the Justice Party?
Answer:

1. The Justice Party came to power when the world was in economic depression.
2. The Nationalist Movement was popular and heading for freedom.
3. Caste and wealth played vital role. Hence the party lost its hold.
4. The rise of congress under the leadership of Ghandhi became a formidable challenge to Jus¬tice Party.

Question 4.
List out a few manuscripts of the Tamil grammar edited by C.W. Damotharanar and U.V. Swaminathar.
Answer:

1. C.W. Domotharanar collected and edited different palm – leaf manuscripts of the Tamil grammar and literature.
2. His edition included such texts as Tholkappiyam, Viracholiyam, Iraiyanar Akapporul, Illakkana Villakkam, Kaliththokai, and Culamani.
3. U.V. Swaminathar took efforts to publish the classical texts such as Civakachinthamani, Paththupattu, Chilapathikaram, Purananuru, Purapporul – Venpa – Malai, Manimekalai, Ainkurunuru, and Pathitrupathu.

Question 5.
What was the different official positions hold by Periyar?
Answer:
Periyar held different official positions of Erode that included the chairmanship of Municipal Council, President of the Tamil Nadu Congress and President of the Justice Party.

Question 6.
What was the impact of Tamil language movement in the 20th century?
Answer:
In the 20th century Tamil language movement such as Tani Tamil Iyakkam and Tamil Isai Iyakkam made a significant cultural impact in creating a pure Tamil free from the influence of Sanskrit.

Question 7.
Discuss the work done by the Justice Party towards legislating provisions for communal representation?
Answer:
The Justice Party rule established the Staff Selection Board in 1924 for the selection of government officials and encouraged ail the communities to share the administrative powers. This ensured equitable distribution in appointments among various castes and communities as a part of achieving social justice.

Question 8.
Whom was Women’s India Association (WIA) started and what was its work?
Answer:

1. Women’s India Association was started in 1917 by Annie Besant, Dorothy JinaraJadasa and Margaret cousins at Adyar Madras.
2. The Association published pamphlets and bulletins in different languages to detail the problems of personal hygiene, marriage laws, voting rights, child care and women’s role in the public.

Question 9.
Who initiated the Self-Respect movement? What was it about?
Answer:
The Self-Respect movement was initiated by Periyar E.V. Ramaswamy. The movement was concerned with the marginalised sections of the society and criticised Brahminism and the cultural hegemony of the Brahmins. It advocated a casteless society devoid of rituals and difference based on birth. The movement declared rationality and self-respect as the birthright of all human beings and held these as more important than self-rule.

Question 10.
Who worked hard to abolish the Devadasi system?
Answer:
Dr. Muthulakshmi ammaiyar, was in the forefront of the campaign pressing for a legislation to abolish the devadasis system.

The Madras Devasasis (prevention of Dedication) Act 1947 was enacted by the Government.

Question 11.
What was the devadasi system? Why was it abolished?
Answer:
The devadasi system was a custom of dedicating young girls to the Hindu temples as a servant of God. Though intended as a service to God, it soon got corrupted leading to extensive immorality and abuse of the women.

Muthulakshmi Reddi started a campaign pressing for a legislation to abolish this system. Finally, the Madras Devadasis (Prevention of Dedication) was enacted by the government in 1947.

Question 12.
What were the titles given to Rettaimalai Srinivasan?
Answer:
Rettaimalai Srinivasan was honoured with such titles as Rao Sahib (1926), Rao Bahadur (1930) and Divan Bahadur (1936) for his selfless social services.

VI. Answer the question given under each caption

Question 1.
Rise Of the Dravidian Movement

(a) In what situation the Dravidian movement emerged?
Answer:
The Dravidian movement emerged as a defence of the Non – Brahmins against the Brahmin dominance.

(b) When was Madras Non – Brahmin Association formed?
Answer:
In 1909.

(c) Who was founded the Madras United League?
Answer:
C. Natesanar, a medical doctor founded the Madras United League.

(d) What is the main aim of the organisation?
Answer:
The organisation focused on educating and supporting Non – Brahmin graduates and conducting regular meetings to share their grievances.

Question 2.
Self Respect Movement:

(a) Who started the Self Respect Movement?
Answer:
E.V. Ramaswamy Periyar started the self respect movement.

(b) Why did he start?
Answer:
He started it in order to spread and execute his ideas and policies.

(c) When was it started?
Answer:
It was started in 1925.

(d) Name the laws passed by the Government due to the constant struggle of Self Respect Movement.
Answer:
Widow Remarriage Act, Women’s Right to property Act and abolition of Devadasi Act.

Question 3.
M.G. rajah

(a) Who was popularly known as M.C. Rajah?
Answer:
Mylar Chinnathambi Raja, popularly known as M.C. Rajah.

(b) What was his demand?
Answer:
He demanded abolition of untouchability and organised a number of political meetings for the “Untouchable castes” to have access to public wells and pathways to burial grounds.

(c) In which year All India Depressed Classes Association was formed?
Answer:
In 1928.

(d) What was the terms he advocated?
Answer:
He advocated the use of such as terms Adi – Dravida and Adi – Andhra.

VII. Answer in detail

Question 1.
Explain briefly about the self- respect movement.
Answer:

1. The Self- Respect Movement introduced a programme of Non – Brahmin uplift in Tamil Nadu.
2. The movement was concerned with the marginalised sections the society and Criticised Brahminism and the cultural hegemony of the Brahmin.
3. It advocated a casteless society devoid of rituals and diffemeces based on birth.
4. The movement declared rationality and Self – Respect as the birth right of all human beings and held these as more important than self – rule.
5. The movement demanded women’s emancipations, deplored superstition and emphasised rationality.
6. The movement also advocated self – respect marriage.
7. The Self – Respect Movement championed not only the cause of the Non – Brahmin Hindus, but also that of the Muslims.
8. The Self Respect Movement extrolled the lofty principles of Islam such as equity and brotherhood.
9. Muslim elite considered the Tamil Muslims as Dravidian.
10. He wanted the Dravidian Muslim to follow Mustapha Kemal Pasha of Turkey and Amanullah of Afghanistan who initiated reforms in Islamic society.

Question 2.
Give an account of Labour Movements in Tamil Nadu.
Answer:

1. The first World War (1914 – 18) provided stimulus to industrial growth India.
2. These industries catering to war time needs, had employed a huge number of workers.
3. At the end of the War there were retrenchments across the industries as the War time requirements receded.
4. Combined with high prices, this gave a momentum to the labour movement.
5. The nationalist realising the value of organised labour power began to supported the cause of labour,
6. B.P. Wadia, M. Singaravelar, ThiruVi. Ka and others initiated the formation of labour unions in the Madras presidency.
7. In 1918 India’s first organised trade union the Madras Labour Union, was formed.
8. The first All India Trade Union Conference (AITUC) was held on 31st October 1920 in Bombay.
9. These included a demand for protection from police interference in labour disputes, the maintenance of an unemployment register restriction on exporting food stuffs, compensation for injuries and health Insurance.
10. In addition, the delegates demanded that Indian workers be given some representation in the Government, just as employers, had representatives on legislative councils.
11. M. Singaravelar (1860 – 1946) was a pioneer in the Labour Movement activities in the Madras Presidency.
12. He organised the first over celebration of May Day in 1923.
13. He published a Tamil newspaper Thozhilalan (workers) to address the problem of the working class.

VIII. Draw a time line for the following

Question 1.
Write any five important events between 1905 -1927.
Answer:
Given period: 1905 – 1927

Question 2.
Write any five important events between 1920 -1930.
Answer:
Given period: 1920 – 1930

Question 3.
Write any five important events between 1930 -1947.
Answer:
Given period: 1930 – 1947

Question 4.
Write any five important events between 1931 -1945.
Answer:
Given period: 1931 – 1945

Question 5.
Write any five important events between 1937 -1947.
Answer:
Given period 1937 – 1947

Question 6.
Write any five important events between 1906 -1925.
Answer:
Given period: 1906 – 1925

Students can download 10th Social Science Geography Chapter 1 India: Location, Relief and Drainage 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 1 India: Location, Relief and Drainage

Samacheer Kalvi 10th Social Science India: Location, Relief and Drainage Text Book Back Questions and Answers

I. Choose the correct answer

Question 1.
The north-south extent of India is:
(a) 2,500km
(b) 292,933 km
(c) 3,214km
(d) 2,814km
Answer:
(c) 3,214km

Question 2.
The Southern most point of India is ……….
(a) Andaman
(b) Kanyakumari
(c) Indira Point
(d) kavaratti
Answer:
(c) Indira Point

Question 3.
The extent of Himalayas in the east-west is about:
(a) 2,500 km
(b) 2,400 km
(c) 800 km
(d) 2,200 km
Answer:
(a) 2,500 km

Question 4.
……… River is known as ‘Sorrow of Bihar’.
(a) Narmada
(b) Godavari
(c) Kosi
(d) Damodar
Answer:
(c) Kosi

Question 5.
Deccan Plateau covers an area of about sq.km.
(a) 8 lakh
(b) 6 lakh
(c) 5 lakh
(d) 7 lakh
Answer:
(d) 7 lakh

Question 6.
A landmass bounded by sea on three sides is referred to as ………
(a) Coast
(b) Island
(c) Peninsula
(d) Strait
Answer:
(c) Peninsula

Question 7.
The Palk Strait and Gulf of Mannar separates India from:
(a) Goa
(b) West Bengal
(c) Sri Lanka
(d) Maldives
Answer:
(c) Sri Lanka

Question 8.
The highest peak in South India is …………
(a) Ooty
(b) Kodaikanal
(c) Anaimudi
(d) Jindhagada
Answer:
(c) Anaimudi

Question 9.
Plains are formed by the older alluviums.
(a) Bhabar
(b) Tarai
(c) Bhangar
(d) Khadar
Answer:
(c) Bhangar

Question 10.
Pulicat Lake is located between the states of ………….
(a) West Bengal and Odisha
(b) Karnataka and Kerala
(c) Odisha and Andhra Pradesh
(d) Tamil Nadu and Andhra Pradesh
Answer:
(d) Tamil Nadu and Andhra Pradesh

II. Match the following

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

III. Give Reasons

Question 1.
The Himalayas are called young fold mountains.
Answer:
The Himalayas are called young fold mountains. Because they have been formed only few million years ago and also they were formed because of the folding of the Earth crust due to tectonic activity’.
[OR]
High pointed peaks and parallel ranges is the features of young fold mountains. Since the Himalayas have these features they are called young fold mountains.

Question 2.
North Indian Rivers are perennial.
Answer:
Most of the North Indian Rivers are perennial. It means that they have water throughout the year. These rivers receive water from rain as well as from melted snow from the lofty mountains.

Question 3.
Chottanagpur Plateau is rich in mineral resources.
Answer:
This region is a part of the ancient Gondwana region. So, Chotanagpur plateau is rich in mineral resources like iron ore and coal.

Question 4.
The great ludian desert ii called Marusthall.
Answer:
The Thar desert, also known as the Great Indian desert is a large arid region in the north
western part of the Indian Subcontinent that covers an area of 2,00,000 Km² and forms a natural boundary between India and Pakistan.
Marusthali means sand-dune. It covered eastern portion of the Great indian Thar Desert in western Rajasthan. It extends over about 24,000 square miles north of the Luni River.

Question 5.
The Eastern states are called seven sisters.
Answer:
The Eastern states in India are the continuous seven states. Assam, Arunachal Pradesh, Nagaland, Mizoram, Manipur, Tripura and Meghalaya. “A paradise unexplored” named by a journalist Jyothi Prakash Salkia in 1972 during the inaugural foundation of North East states.

Question 6.
The river Godavari is often referred to as Vridha Ganga.
Answer:

1. The Godavari is India’s second longest river after Ganga.
2. It rises from the slopes of the Western Ghats in the Nasik district of Maharashtra.
3. It covers a large number of states as the Ganga does. Because of its large size and extend among the peninsular rivers, the river Godavari is often referred to as Vridha Ganga.

IV. Distinguish between the following

Question 1.
Himalayan rivers and Peninsular rivers.
Answer:
Himalayan Rivers:

1. Originate from the Himalayas.
2. Perennial in nature.
3. Long and Wide.
4. Middle and Lower courses are navigable.
5. Unsuitable for hydropower generation.

Peninsular Rivers:

1. Originate from Western ghats.
2. Non-perennial in nature.
3. Short and narrow.
4. Not useful for navigation.
5. Suitable for hydro-power generation.

Question 2.
Western Ghats and Eastern Ghats.
Answer:
Western ghats:

1. Western ghats forms the Western edge of the peninsular plateau.
2. It runs parallel to the Arabian sea coast.
3. They are continuous range.
4. Anaimudi is the highest peak.

Eastern ghats:

1. Eastern ghats form the Eastern edge of the peninsular plateau.
2. It runs parallel to the Bay of Bengal coast.
3. They are not continuous dissected by east flowing rivers.
4. Mahendragiri is the highest peak.

Question 3.
Himadri and Himachal.
Answer:
Himadri:

1. Northern most range of the Himalayas.
2. Known as the Greater Himalayas.
3. Lofty peaks like Mt.Everest, Kanchenjunga etc., are located in this range.
4. Region of permanent snow cover.

Himachal:

1. Middle range of the Himalayas.
2. Known as the lesser Himalayas.
3. Major hill stations like Shimla, Daijeeling etc., are located in this region.
4. Subjected to extensive erosion.

Question 4.
Western Coastal Plains and Eastern Coastal Plains.
Answer:
Western Coastal plains:

1. Lies between Western ghats and Arabian Sea.
2. It extends from Rann of Kutch in the North to Kanyakumari in the South.
3. Dotted with sandy beaches, lagoons and estuary.
4. It is known as Konkan plains in the North Kanara in the middle and Malabar in the South.

Eastern Coastal plains:

1. Lies between Eastern ghats and the Bay of Bengal.
2. It stretches along the states of West Bengal, Odisha, Andhra Pradesh and Tamil Nadu.
3. Has regular shoreline and composed of alluvial deposits.
4. Northern part (Mahanadhi-Krishna) called as Northern circars and Southern part (Krishna-Kaveri) is called Coromandal coast.

V. Answer in brief

Question 1.
Name the neighbouring countries of India.
Answer:
The neighbouring countries of India are Pakistan in the West, Afghanistan in the North-West, China, Nepal and Bhutan in the North and Bangladesh and Myanmar in the East and Srilanka in the South.

Question 2.
Give the importance of IST.
Answer:
To avoid the confusion of time in different states of India. One meridian is taken to have a uniform time for a whole country. Therefore 80° 30′ E has been chosen as the Standard Meridian of India which is almost passing from the centre of India.

Question 3.
Write a short note on Deccan Plateau.
Answer:

1. Deccan plateau is the largest part of the peninsular plateau.
2. It is roughly triangular in shape.
3. It covers an area of 7 lakh sq km.
4. The height ranges from 500 to 1000m above sea level.
5. It extends from Rajmahal hills on the East to Kanyakumari in the South. Satpura range in the West to Kanyakumari in the South. Bordered by Western Ghats on the West and Eastern ghats on the East.

Question 4.
State the west-flowing rivers of India.
Answer:

• The Narmada and the Tapi are the west flowing rivers of the Peninsular India.
• These rivers drain into the Arabian Sea.
• These rivers form estuaries on the west coast.
• These rivers are devoid of an large tributary system.
• Narmada river is the largest among the west flowing rivers of peninsular India. Its principal tributaries are Bushner, Halon, Heran, Banjar, Dudhi, Shakkar, Tawa, Bama and Kolar.
• Tapti river is one of only the three rivers in peninsular India that run from east to west. The others being the Narmada and the Mahi.
• The major tributaries are Vaki, Gomai, Arunavati, Aner, Nesu, Buray Panjhra and Bori. It outfall into the Arabian sea through the Gulf of Cambay.

Question 5.
Write a brief note on the island group of Lakshadweep.
Answer:

1. The Lakshadweep islands of India are located off the West coast of India.
2. These islands are of coral origin.
3. It covers an area of 32 sq km and consists of 27 islands.
4. Kavaratti is the administrative capital of Lakshadweep.
5. Earlier it had three divisions namely Laccadives, Minicoy and Aminidivi and was named as Lakshadweep in 1973.
6. “Pitt Island” of this group has a bird sanctuary and it is uninhabited.

VI. Answer In a paragraph

Question 1.
Explain the divisions of the Northern Mountains and its importance to India.
Answer:

• The Himalayas geologically young and structurally fold mountains stretch over the northern borders of India.
• The Himalayas represent the loftiest and one of the most rugged mountain barriers of the world.
• The mountain ranges run in a west-east direction from India to the Brahmaputra.

Question 2.
Give an account on the major peninsular rivers of India.
Answer:

1. The rivers in South India are called the Peninsular rivers.
2. Most of these rivers originate from the Western ghats.
3. These rivers are seasonal and are mainly fed by rain.
4. Based on the direction of flow the Peninsular rivers are divided into East flowing rivers and West flowing rivers.

East flowing rivers:

1. Mahanadi: Originates near Sihawa in Raipur district of Chattisgarh.
   • Flows through Odisha, Jharkand, Chattisgarh (851 km in length)
   • Major tributaries: Seonath, Telen, Sandur and lb.
   • Distributaries: Paika, Birupa, Chitartala, Genguti and Nun.
   • These distributaries forms the Delta of Mahanadhi, one of the largest delta in India.
   • This river flows into Bay of Bengal.

(ii) Godavari:

• It is the longest river (1,465km) among the peninsular rivers covers 3.13 lakh km2 area.
• Originates in Nasik district of Maharashtra.
• Flows through the States of Telangana and Andhra Pradesh ends into Bay of Bengal.
• Major distributaries: Puma, Penganga, Pranitha, Indravathi, Tal and Salami.
• Distributaries: Near Rajahmundry divides into two channels. Vasishta and Gautami and form one of the largest delta.
• Fresh water lake: Kolleru is located in the deltaic region of Godavari.

(iii) Krishna:

• It is the second longest peninsular river (1400 km) covers 2.58 lakh sq km area.
• Originates from a spring in Mahabaleshwar in the Western ghats of Maharashtra.
• Flows through Andhra Pradesh.
• Major tributaries: Bhima, Peddavagu, Musi, Koyna and Thungabhadra.
• Joins in Bay of Bengal at Hamasaladevi.

(iv) Kaveri:

• Originates at Talakaveri – Kudagu hills of Karnataka (800km).
• It is also known as Dakshin Ganga or Ganga of the South.
• Main Tributaries: Hemavathi, Kabini, Bhavani, Arkavathy, Amaravathi, Harangi etc.
• Distributary – Coleroon
• Sacred lslands made by river Kaveri: Sri ranagpattinam(Kamataka) Srirangam (TamilNadu) Waterfalls – Sivasamudram.
• Flows through Karnataka and TamilNadu.
• Enters TamilNadu through Hoghenekkal Falls (Dharmapuri).
• Drains into Bay of Bengal at Poompuhar.

West Flowing Rivers:

(i) Narmadha:

• Origin: Araarkantak Plateau (Madhya Pradesh)
• Length: 1312 km covers an area of 98,796 sq km.
• Largest among the West flowing rivers.
• Tributaries: Burhner, Halon, Heran, Banjar, Dudhi, Shakkar, Tawa, Bama, Kolar.
• States benefited: Madhya Pradesh, Maharashtra.
• Ends into Arabian Sea through the Gulf of Cambay.
• Forms 27 km long estuary.

(ii) Tapti:

• Rises near Multani in the Betul district of Madhya Pradesh.
• Length: 724 km. Covers an area of 65,145 sq km.
• Tributaries: Vaki, Gomai, Arunnavati, Aner, Nesu, Buray, Panjhra and Bori.
• Outfalls into the Arabian Sea through Gulf of Cambay.
• States: Flows through Madhya Pradesh and Maharashtra.
• Since all these rivers flow in the plateau region they are swift and are useful for hydro power generation.

Question 3.
Give a detailed account on the basin of the Ganga.
Answer:

• On the basis of deposition of Sediments by various rivers and topographical characteristics, the Northern plains of India is divided into the four major regions. Ganga plains is one among them.
• It extends from the Yamuna River in the West to Bangladesh in the East.
• The extensive plain is the largest plain of India covers 3.75 lakh sq. km.
• It is watered by the River Ganga and its tributaries such as Ghaghra, Gandak, Kosi, Yamuna, Chambal, Betwa etc.
• It is formed by the sediments of these rivers and is fertile in nature.
• The general slope of the entire plain is towards East and South-East.
• Ganga plains covers the states of Uttar Pradesh, Bihar and West Bengal.

VII. Map exercises

Mark the following in the outline map of India

Question 1.
Major mountain ranges – Karakoram,Ladakh, Zaskar, Aravaili, Western Ghats,Eastern Ghats.
Answer:

Question2.
Major rivers – Indus, Ganga, Brahmaputra, Narmada, Tapti, Mahanadi, Godavari, Krishna and Kaveri.
Answer:

Question 3.
Major plateaus – Malwa, Chotanagpur, Deccan.
Answer:

VIII. Activities

Question 1.
Observe the Peninsular Plateau map of India and mark the major plateau divisions of India.
Answer:

Question 2.
Prepare a table showing the major West flowing and East flowing rivers of peninsular India.
Answer:
West flowing rivers:

1. Narmada, Tapti, Mahi.
2. Flows into Arabian Sea through Gulf of cambay.
3. Do not make deltas but form estuary.

East flowing rivers:

1. Mahanadhi, Godhavari, Krishna and Kaveri.
2. Flows into Bay of Bengal.
3. Make deltas at their lower courses.

Question 3.
Assume that you are travelling from West Bengal to Gujarat along the beautiful coasts of India. Find out the states which you would pass through?
Answer:
From West Bengal, if we travel along the East coast we could pass through Odisha, Andhra Pradesh, Tamil Nadu states from North to South and from South to North along the West Coast we would pass through the states of Kerala, Karnataka, Goa, Maharashtra and reach Gujarat.

Question 4.
Find out the states through which the river Ganga flows.
Answer:
River Ganga flows through the States of Uttar Pradesh, Bihar and West Bengal.

Question 5.
Prepare a table showing the major rivers in India and findout it’s tributaries, origin, length and area.
Answer:

TB. PNo: 83
Find Out

Question 1.
The number of Union Territories along the western coast and eastern coast.
Answer:
Union Territories along the western coast are:

1. Diu and Daman
2. Dadra and Nagar Haveli
3. Mahe and Lakshadweep.

Union Territories along eastern coast are:

• Pondicherry and Andaman and Nicobar Islands.

Question 2.
Area wise which is the largest and the smallest state?
Answer:

1. Area wise Rajasthan is the largest state (3,42,239 sq.km).
2. The smallest state is Goa (3,702 sq km).

Question 3.
The states which do not have an international border or lie on the largest state.
Answer:
Haryana, Madhya Pradesh, Jharkhand, Chattisgarh.

Question 4.
Classify into four groups each having common frontiers with

1. Pakistan
2. China
3. Myanmar
4. Bangladesh.

Answer:

1. Pakistan – Jammu and Kashmir, Punjab, Rajasthan, Gujarat.
2. China – Jammu and Kashmir, Himachal Pradesh, Uttarkhand, Sikkim, Arunachal Pradesh.
3. Myanmar – Arunachal Pradesh, Nagaland Manipur, Mizoram.
4. Bangladesh – Meghalaya, Assam, West Bengal, Tripura.

TB. PNo: 85
Find Out

Question 5.
Find the Hill stations which are located in the Himalayan Mountains.
Answer:
Major hill stations of the Himalayas are Shimla, Mussoorie, Nainital, Almora, Ranikhet and Darjeeling.

TB. PNo: 95
Find Out

Question 6.
In which river the Gerosappa (jog) fall is found?
Answer:
In the river Sharavathi is the Gerosappa (jog) falls (830m). Second highest waterfalls in India.

Samacheer Kalvi 10th Social Science India: Location, Relief and Drainage Additional Important Questions and Answers

I. Choose the correct answer

Question 1.
India is the ………………… largest country in Asia.
(a) first
(b) second
(c) fourth
(d) third
Answer:
(b) second

Question 2.
Along with Hindukush and Karakoram range ……… provide a natural boundary in the
north.
(a) Aravalli
(b) Satpura
(c) Himalayas
Answer:
(c) Himalayas

Question 3.
The ………………… latitude passes through the middle of the country.
(a) 0°
(b) 23°.30’N
(c) 66°.30’N
(d) 23°.30’S
Answer:
(b) 23°.30’N

Question 4.
The coconut bearing coastal plains are on the ………
(a) east
(b) west
(c) south
Answer:
(c) south

Question 5.
The ………………… is known as the “Roof of the World”.
(a) Armenian knot
(b) Anaimudi
(c) Pamirknot
(d) Mt.Everest
Answer:
(c) Pamirknot

Question 6.
The …….. land gets abundance sunshine from the sun.
(a) mediterranean
(b) tropical
(c) equatorial
Answer:
(b) tropical

Question 7.
The Great Northern plains covers an area of over sq.km.
(a) 5 lakh
(b) 6 lakh
(c) 7 lakh
(d) 8 lakh
Answer:
(c) 7 lakh

Question 8.
The …… divides India into two halves.
(a) tropic of Cancer 231/2° N
(b) equator 0°
(c) tropic of Capricorn 231/2° S
Answer:
(a) tropic of Cancer 231/2° N

Question 9.
The ………………… desert is the world 9th largest sub tropical desert.
(a) Gobi
(b) Atacama
(c) Kalahari
(d) Thar
Answer:
(d) Thar

Question 10.
India is …….. times bigger than Pakistan.
(a) two
(b) five
(c) four
Answer:
(c) four

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

III. Give Reasons

Question 1.
Peninsular Rivers are Non-Perennial.
Answer:
Peninsular rivers originate from the Western Ghats. They have a large seasonal fluctuation in the volume of water as they are solely fed by the rain. So these Rivers are seasonal and non-perennial.

Question 2.
Andaman and Nicobar islands have dense forests.
Answer:
Since these islands lie close to the equator, the climate remains hot and wet throughout the year and has dense forests.

Question 3.
India has close contact with the countries of the West and the East.
Answer:
India’s Southward extension along the Indian ocean and the trans Indian . ocean routes provide a strategic position to India connecting the countries of Europe in the West and the countries of East Asia. Thus, India has close contact with the countries of the West and the East.

Question 4.
India is geographically vibrant and incredible country.
Answer:
The majestic Himalayan peaks in the North, beautiful beaches in the South, the Great Indian desert in the West and the breathtaking natural heritage in the East make India geographically vibrant and an incredible country.

IV. Distinguish between the following

Question 1.
Andaman and Nicobar Islands and Lakshadweep Islands.
Answer:
Andaman and Nicobar Islands:

1. Located in Bay of Bengal.
2. These islands are of volcanic origin.
3. Port Blair is the administrative capital.
4. Consists of nearly 572 islands.

Lakshadweep Islands:

1. Located in Arabian Sea off the West coast of India.
2. These islands are mainly coral origin.
3. Kavaratti is the administrative capital.
4. Consists of nearly 27 islands.

Question 2.
Bhangar plains and Khadar plains.
Answer:
Bhangar plains:

1. Upland alluvial tracts of Great plains.
2. Formed by the older alluvium.
3. Soil is dark, rich in humus.
4. Well drained and fit for agriculture.

Khadar plains:

1. Plains along the courses of the rivers of Great plains.
2. Formed by the newer alluvium.
3. Soil is sandy, silt, clay and mud.
4. Highly fertile.

Question 3.
West flowing rivers and East flowing rivers.
Answer:
West flowing rivers:

1. Flow into Gulf of Cambay.
2. Form estuaries
3. Narmadha, Tapti and Mahi are the major west flowing rivers.

East flowing rivers of Peninsular plateau or South India:

1. Flow into Bay of Bengal.
2. Form Deltas
3. Mahanadhi, Godavari, Krishna and Kaveri are the major east flowing rivers.

Question 4.
Tributary and Distributary.
Answer:
Tributary:

1. A river or stream that joins with the main river.
2. Contributes water to the main river.
3. Eg: Yamuna tributary of river Ganga.

Distributary:

1. A branch or outlet that leaves the main river.
2. Carries water to the sea or lake.
3. Eg: River Coleroon a distributary of river Kaveri.

V. Answer in brief

Question 1.
Mention the tributaries of Godavari.
Answer:
Ghatprabha, Malaprabha, Manjira, Penganga and Neingarga.

Question 2.
Name the boundaries of India.
Answer:
India is bounded by Bay of Bengal on the East, Arabian sea on the West, Indian ocean in the South and the lofty Himalayas in the North

Question 3.
Why do we need a Standard Meridian for India?
Answer:
To avoid time gap of two hours between east and west part of India which may create confusion in working, it is essential to have a Standard Meridian.

Question 4.
How India is politically divided?
Answer:
India is politically divided into 29 states and 7 union territories (Delhi being the National Capital territory) for administrative convenience.

Question 5.
Why is India called a Sub-continent? How it is separated?
Answer:

1. India is called a Sub-continent because it is big landmass, which stands out as a distinct geographical unit from the rest of the continent.
2. It is separated by natural features like mountains and rivers.
3. India is also separated from rest of the continent by the mighty Himalayas.

Question 6.
Name the three major divisions of the Northern Mountains.
Answer:
The three major divisions of the Northern mountains are:

1. The Trans-Himalayas (or) The Western Himalayas
2. The Himalayas
3. Eastern Hills or Purvanchal Hills

Question 7.
Why is Northern Plain most productive part of India?
Answer:
With a rich soil cover combined with an adequate water supply and favourable climate Northern plain is agriculturally a very productive part of India.

Question 8.
How has Northern Plain become a fertile Plain?
Answer:
The plain is formed of alluvial soil. The deposition of alluvium in a large basis lying at the foothills of the Himalayan over millions of years make northern plains fertile.

VI. Answer in a paragraph

Question 1.
India is a Sub-Continent. Prove it.
Answer:

1. Physical features
2. Climate conditions
3. Natural vegetation
4. Cultural norms
5. Ancient ethic and linguistic groups
6. Huge area

All these distinctive continental characteristic are found in India. Hence we consider India as a Sub-Continent.

Question 2.
Name the old fold mountain in India and write a brief note on it.
Answer:

1. Aravalli range is the old fold mountain in India.
2. These mountains are the relies and are reduced into hills due to erosional features.
3. The Aravalli hills form the North-Western and Western boundary of the Central Highlands a sub division of Peninsular plateau.
4. These hills extends from Gujarat through Rajasthan to Delhi in the North westerly direction (700km).
5. The height of these hills is about 1500m in South west and near Delhi hardly 400m.
6. Gurushikar (1722m) is the highest peak of this range.

Question 3.
Write a note on Purvanchal Hills.
Answer:

1. They are the eastern off-shoot of Himalayas.
2. It extended in the north-eastern states of India.
3. Most of these hills are located along the border of India and Myanmar while others are inside India.
4. Dafla Hills, Abor Hills, Mishmi Hills, Patkai Bum Hills, Naga Hills, Manipur Hills, Mizo Hills, Tripura Hills, Mikir Hills, Garo Hills, Khasi Hills and Jaintia Hills are the hills which are collectively known as Purvanchal Hills.

Question 4.
Write in general about the structure of the Peninsular plateau.
Answer:

1. The peninsular plateau lies to the south of the Great Northern plains.
2. This is the largest physiographic division covers an area of about 16 lakh sq.km.
3. It is made up of old rocks and consists of series of plateaus and hill ranges interspersed with river valleys.
4. The average height of a large portion of this plateau is more than 600m.
5. Anaimudi is the highest point of this place located in Anaimalai.
6. The plateau slopes towards east and is a part of ancient Gondwana.
7. River Narmada divides the Plateau region into two. The region lying to the north of Narmada is called Central Highlands and South of Narmada is called Deccan plateau.

VII. Map exercises

Question 1.
Mark the following in the outline map of India.

Set 1:

1. Mark the boundaries of India
2. Latitudinal Extent
3. Longitudinal Extent
4. Tropic of cancer
5. Lakshadweep Islands
6. Andaman and Nicobar Islands
   
7. Central Meridian of India
   

Set 2:
Peaks: Anaimudi, Mt. Everest, Mt.K2 Mahendragiri, Gurushikhar. NamchaBarwa, Nanga parbat, Kanchenjunga. Dhaulagiri, Makalu. Nandadevi, Dodabetta.

Hill stations: Shimla, Mussourie Ranikhet, Nainital, Almora, Darjeeling.

Hills: Aravalli, Vindhya. Satpura, Naga hills, Mizo hills, Garo hills, Khasi hills, Jaintia hills, Nilgiri hills, Anaimalai hills, Cardomom hills, Raj Mahal hills, Gir hills.

Gulf: Gulf of Mannar, Gulf of Kutch, Rann of Kutch, Palk Strait, Gulf of Cambay.

Set 3:
Seven Union Territories:

1. Delhi
2. Chandigarh
3. Diu and Daman
4. Dadra and Nagar Haveli
5. Pondicherry
6. Andaman and Nicobar Islands
7. Lakshadweep.