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AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 14 Carbon and its Compounds Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 14th Lesson Carbon and its Compounds

10th Class Chemistry 14th Lesson Carbon and its Compounds Textbook Questions and Answers

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Question 1.
Name the simplest hydrocarbon. (AS1)
Answer:
The simplest hydrocarbon is alkane called Methane (CH₄). It’s an aliphatic, saturated compound of Hydrogen and Carbon.

Question 2.
What are the general molecular formulae of alkanes, alkenes and alkynes? (AS1)
Answer:
General molecular formula of alkane is C_nH_2n+2.
General molecular formula of alkene is C_nH_2n.
General molecular formula of alkyne is C_nH_2n-2.

Question 3.
Name the carboxylic acid used as a preservative. (AS1)
Answer:
Vinegar with chemical formula CH₃COOH is used as preservative. 5 – 8% of solution of acetic acid or ethanoic acid in water is called vinegar and it is used widely as preservative in pickles.

Question 4.
Name the product other than water formed on burning of ethanol in air. (AS1)
Answer:
C₂H₃OH + 3O₂ → 2CO₂ + 3H₂O + Energy
So, the product other than water formed on burning of ethanol in air is carbon dioxide (CO₂).

Question 5.
Give the IUPAC name of the following compounds. If more than one compound is possible, name all of them. (AS1)
i) An aldehyde derived from ethane.
ii) A ketone derived from butane.
iii) A chloride derived from propane.
iv) An alcohol derived from pentane.
Answer:
i) An aldehyde derived from ethane is ethanal. Its formula is CH₃CHO.
ii) A ketone derived from butane. Its IUPAC name is Butanone.
Its chemical formula is CH₃COCH₂CH₃
It is also known as methyl ethyl ketone. (Its general name)

iii) A chloride derived from propane.
A) 1-Chloro propane. Its formula is CH₃CH₂CH₂Cl.

iv) An alcohol derived from pentane :
A) 1-Pentanol. Its formula is CH₃CH₂CH₂CH₂CH₂OH.
B) 2-Pentanol. Its formula is CH₃CHOH CH₂CH₂CH₃
C) 3-Pentanol. Its formula is CH₃CH₂ CHOH CH₂CH₃

Question 6.
A mixture of oxygen and ethyne is burnt for welding ; can you tell why a mixture of ethyne and air is not used? (AS1)
Answer:

• Ethyne when burnt in the presence of oxygen gives enough heat that can be used for welding.
• Whereas if it is burnt in air which contains nitrogen, CO₂ and other inactive gaseous contents, sufficient oxygen is not available for burning ethyne to give the required heat.

Question 7.
Explain with the help of a chemical equation, how an addition reaction is used in vegetable ghee industry. (AS1)
Answer:

• The addition of hydrogen to an unsaturated hydrocarbon to obtain a saturated hydrocarbon is called hydrogenation. The process of hydrogenation takes place in the presence of nickel or palladium metals as catalyst.
• The process of hydrogenation has an important industrial application. It is used to prepare vegetable ghee (or vanaspati ghee) from vegetable oils.
• Vegetable oils are unsaturated fats having double bonds between some of their carbon atoms.
• When a vegetable oil (like groundnut oil) is heated with hydrogen in the presence of finely divided nickel as catalyst, a saturated oil called vegetable ghee (or vanaspati ghee) is formed. This a reaction is called hydrogenation of oils and it can be represented as follows.

Here vegetable oil is a liquid whereas vegetable ghee is a solid (or a semi solid).

Question 8.
a) What are the various possible structural formulae of a compound having molecular formula C₃H₆O? (AS1)
b) Give the IUPAC names of the above possible compounds and represent them in structures. (AS1)
c) What is the similarity in these compounds? (AS1)
Answer:
a) They are CH₃COCH₃and CH₃ CH₂ CHO

b) i) The IUPAC name of CH₃COCH₃ is propanone.
ii) The IUPAC name of CH₃ CH₂ CHO is propanal.

Question 9.
Name the simplest ketone apfl write its molecular formula. (AS1)
Answer:
Acetone is the simplest ketone. Its molecular formula is CH₃COCH₃ Its IUPAC name is propanone.

Question 10.
What do we call the Self linking property of carbon? (AS1)
Answer:
The property of self combination (or linking) of carbon atoms to form long chains is useful to us because it gives rise to an extremely large number of carbon compounds (or organic compounds). This is known as catenation.

Question 11.
Name the compound formed by heating ethanol at 443 K with excess of cone. H₂SO₄. (AS1)
(OR)
What is the compound formed when ethyhalcohol (Ethanol) is dehydrated ? Write the chemical equation of the reaction.
Answer:
1. When ethanol is heated with excess of cone. H₂SO₄ at 443 K (170° C), it gets dehydrated to form ethene (which is an unsaturated hydrocarbon).

2. During dehydration of ethanol molecules (CH₃ – CH₂OH), H from the CH₃ group and OH from CH₂OH group are removed in the form of a water molecule (H₂O) regulating in the formation of this molecule (CH₂ = CH₂).
3. In this reaction concentrated sulphuric acid acts as a dehydrating agent.

Question 12.
Give an example for esterification reaction. (AS1)
Answer:
The reaction between carboxylic acid and an alcohol in the presence of cone. H₂SO₄ to form a sweet odoured substance, ester with the functional group

is called esterification.

Ex: Ethanoic acid (carboxylic acid) reacts with Ethanol (alcohol) and forms ethyl acetate.

Question 13.
Name the product obtained when ethanol is oxidized by either chromic anhydride or alkaline potassium permanganate. (AS1)
(OR)
If the ethanol is oxidized by either chromic anhydride or alkaline potassium permanganate, what is the product obtained from them?
Answer:
Ethanol (Ethyl alcohol) undergoes oxidation to form the product of Acetaldehyde and finally Acetic acid.

Question 14.
Write the chemical equation representing the reaction of preparation of ethanol from ethane. (AS1)
Answer:
1. Ethane in the absence of air on heating forms ethene

2. Then Ethanol is prepared on large scale from ethene by the addition of water vapour to it in the presence of catalyst like P₂O₅, Tungsten oxide at high pressure and temperature.

Question 15.
Write the IUPAC name of the next homologous of CH₃OHCH₂CH₃. (AS1)
Answer:
The IUPAC name of the next homologous of CH₃OHCH₂CH₃ is HO-CH₃CH₂CH₂CH₃ 1 – butanol.

Question 16.
Define homologous series of carbon compounds. Mention any two characteristics of homologous series. (AS1)
Answer:
1. The series of carbon compounds in which two successive compounds differ by – CH₂ unit is called homologous series.
Ex : 1) CH₄, C₂H₆, C₃H₈, ………………..
2) CH₃OH, C₂H₅OH, C₃H₇OH, ………………..

2. If we observe above series of compounds, we will notice that each compound in the series differs by – CH₂ unit by its successive compound.

3. Characteristics of homologous series :
i) They have one general formula.
Ex : alkanes (C_nH_2n+2), alkynes (C_nH_2n-2), alcohols (C_nH_2n+1) OH, etc.
ii) Successive compounds in the series possess a difference of (-CH₂) unit.
iii) They have similar chemical properties.

Question 17.
Give the names of functional groups
(i) – CHO
(ii) – C = O. (AS1)
(OR)
Write the names of the given functional groups
(i) – CHO
(ii) – C = O
Answer:
i) – CHO → aldehyde
ii) – C = O → ketone

Question 18.
Why does carbon form compounds mainly by covalent bonding? (AS1)
Answer:
Since carbon atoms can achieve the inert gas electron arrangements only by the sharings of electrons, therefore, carbon always forms covalent bonds.

Question 19.
Allotropy is a property shown by which class substance: elements, compounds or mixtures? Explain allotropy with suitable examples. (AS1)
Answer:

1. Allotropy is a property shown by the elements.

2. The property of an element to exist in two or more physical forms having more or less similar chemical properties but different physical properties is called allotropy.

3. The different forms of the element are called allotropes and are formed due to the difference in the arrangement of atoms.

4. Example for allotropes : Allotropes of carbon.

Allotropes of carbon are classified into two types. They are
1) Amorphous forms,
2) Crystalline forms.

5) Amorphous forms of carbon:
Coal, coke, wood, charcoal, animal charcoal, lampblack, gas carbon, petroleum coke, sugar charcoal.

6) Crystalline forms of carbon :
Diamond, graphite and buckminsterfullerene.

Question 20.
Explain how sodium ethoxide is obtained from ethanol. Give chemical equations. (AS1)
Answer:
As ethanol is similar to water molecule (H₂O) with C₂H₅ group in place of hydrogen, it reacts with metallic sodium to liberate hydrogen and form sodium ethoxide.

Question 21.
Describe with chemical equation how ethanoic acid may be obtained from ethanol. (AS1)
Answer:
Ethyl alcohol (Ethanol) undergoes oxidation to form the product Acetaldehyde and finally acetic acid (Ethanoic acid).

Question 22.
Explain the cleansing action of soap. (AS1)
Answer:
When a dirty cloth is put in water containing dissolved soap, the hydrocarbon ends of the soap molecules in the micelle attach to the oil or grease particles present on the surface of dirty clothes.

Question 24.
Explain the structure of graphite in terms of bonding and give one property based on this structure. (AS1)
(OR)
Why does graphite act as lubricant?
Answer:

• Graphite forms a two dimensional layer structure with C – C bonds within the layers.
• There are relatively weak interactions between the layers.
• In the layer structure, the carbon atoms are in a trigonal planar environment.
• This is consistent with each carbon atom in sp² hybridisation.
• Interactions between the sp² orbitals (overlaps) lead to the formation of C – C bonds.
• Each carbon atom is with one unhybridised ‘p’ orbital.
• The unhybridised ‘p’ orbitals interact to form a π system that is delocalised over the whole layer.
• The interactions known as London dispersion forces between the layers which are separated by a distance of 3.35 A° are weakened by the presence of water molecules so that it is easy to cleave graphite.
• For this reason graphite is used as lubricant and as the lead in pencils.

Question 25.
Name the acid present in vinegar. (AS1)
Answer:
1) The acid present in vinegar is Ethenoic acid or acetic acid (CH₃COOH).
2) 5 – 8% solution of acetic acid in water is called vinegar.

Question 26.
What happens when a small piece of sodium is dropped into ethanol? (AS2)
Answer:
Ethanol reacts with sodium to liberate hydrogen and form sodium ethoxide.

Question 27.
Two carbon compounds A and B have molecular formula C₃H₈ and C₃H₆ respectively. Which one of the two is most likely to show addition? Justify your answer. (AS2)
Answer:

• It is a saturated hydrocarbon. It shows substitution reaction.

• This is an unsaturated hydrocarbon. Hence it shows addition to become saturated. During the reactions, addition of reagent takes place at the double bonded carbon atoms.

Justification :
In the following, C₃H₆ undergoes addition reaction.

Question 28.
Suggest a test to find the hardness of water and explain the procedure. (AS3)
(OR)
How do you test whether a given water sample is soft or hard?
Answer:

• Take about 10 ml hard water (well water or hand pump water) in a test tube.
• Add five drops of soap solution to it.
• Shake the test tube vigorously.
• We see that no lather is formed at first.
• Only a dirty white curd like scum is formed on the surface of water.
• From this, we conclude that soap does not form lather easily with hard water.
• We have to add much more soap to obtain lather with hard water.

Question 29.
Suggest a chemical test to distinguish between ethanol and ethanoic acid and explain the procedure. (AS3)
Answer:

1. Take ethanol and ethanoic acid in two different test tubes.
2. Add nearly 18 g of sodium bicarbonate (NaHCO₃) to each test tube.
3. Lots and lots of bubbles and foam will be observed from the test tube containing ethanoic acid. This is due to release of CO₂.
   NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
4. Ethanol will not react with sodium bicarbonate and thus we won’t observe any change in the test tube containing ethanol.
   Thus we can separate ethanol from ethanoic acid.

Question 30.
An organic compound ‘X’ with a molecular formula C₂H₆O undergoes oxidation with alkaline KMnO₄ and forms the compound ‘Y’, that has molecular formula C₂H₄O₂. (AS3)
i) Identify ‘X’ and ‘Y’.
Answer:
X is Ethanol is CH₃CH₂OH and T is Ethanoic acid, i.e., CH₃COOH.

ii) Write your observation regarding the product when the compound X is made to react with compound IT which is used as a preservative for pickles.
Answer:
Ethyl alcohol undergoes oxidation to form the product Acetaldehyde and finally Acetic acid.

Here CH₃COOH is used as preservative for pickles.

When X reacts with Y it forms ethyl acetate and water which is called esterification reaction.
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

Question 31.
Prepare models of methane, ethane, ethene and ethyne molecules using clay balls and matchsticks. (AS4)
Answer:
Stick and ball model :
1) Methane (CH₄) :

2) Ethane (C2H₆):

3) Ethene (C₂H₄):

4) Ethyne (C₂H₂)

Question 32.
Collect information about artificial ripening of fruits by ethylene. (AS4)
Answer:

• Seasonal fruits like mango, banana, papaya, sapota and custard apple are often harvested in nature. But due to unripe condition they are subsequently allowed to ripen by natural release of ripening harmone (ethylene) from the fruit.
• However, natural ripening in some fruits is a slow process, which leads to high weight loss, desiccation of fruits and under ripening. With the rapid development of fruit trade, artificial ripening has become essential and the methods practised earlier by small traders are smoking and calcium carbide treatment.
• Fruits ripened with calcium carbide though seem attractive and colourful are inferior in taste, flavour and spoil faster.
• Government of India has banned the use of calcium carbide for artificial ripening of fruits under PFA Act 8-44AA, 1954.
• Artificial ripening of fruits by using the above steps spoils the health of consumers, so we should not use such type of fruits.
• Government has to take serious action on the fruit sellers who are practising the above said methods.

Question 33.
Draw the electronic dot structure of ethane molecule (C₂H₆). (AS6)
Answer:
C₂H₆:

Question 34.
How do you appreciate the role of esters in everyday life? (AS6)
Answer:

• Esters are usually volatile liquids having sweet or pleasant smell.
• They are also said to have fruity smell.
• Esters are used in making artificial perfumes.
• This is because of the fact that most of the esters have a pleasant smell.
• Esters are also used as flavouring agents.
• This means that esters are used in making artificial flavours and essences used in ice-cream, sweets and cool drinks.
• The alkaline hydrolysis of esters is known as saponification (Soap making).
• That’s why we can appreciate the role of esters in everyday life.

Question 35.
How do you condemn the use of alcohol as a social practice? (AS7)
Answer:

• Consumption of alcohol in the form of beverages is harmful to health.
• It causes severe damage to blood circulation system.
• Addiction to alcohol drinking leads to heart diseases and damages the liver.
• It also causes ulcers in small intestines due to increased acidity and damages the digestive system.
• Alcohol which is consumed in raw form under the names liquor, gudumba which is more harmful to health due to adulteration.
• Alcohol mixed with pyridine is called denatured spirit. Consumption of denatured spirit causes blindness and death.
• Hence use of alcohol is a social evil which harms the society.

Question 36.
An organic compound with molecular formula C₂H₄O₂ produces brisk effervescence on addition of sodium carbonate/bicarbonate.
Answer the following :
a) Identify the organic compound. (AS1)
Answer:
The organic compound is Ethanoic acid (CH₃COOH).

b) Write the chemical equation for the above reaction. (AS1)
Answer:

c) Name the gas evolved. (AS2)
Answer:
CO₂

d) How will you test the gas evolved? (AS3)
Answer:
1) Pass the evolved gas through lime water in a test tube.
2) We will find that lime water turns milky.
3) Only CO₂ gas can turn lime water milky.

e) List two important uses of the above compound. (AS1)
Answer:
1) Dilute ethanoic acid (CH₃COOH) is used as a food preservative in the preparation of pickles and sauces.
2) Ethanoic acid is used for making cellulose acetate which is an important artificial fibre.

Question 37.
1 ml glacial acetic acid and 1 m/of ethanol are mixed together in, a test tube. Few drops of concentrate sulphuric acid is added in the mixture are warmed in a water bath for 5 min.
Answer the following:
a) Name the resultant compound formed.
b) Represent the above change by a chemical equation.
c) What term is given to such a reaction?
d) What are the special characteristics of the compound formed?
Answer:
a) Ethyl acetate.

c) Esterification
d) It has fruity smell or pleasant smell.

Fill In The Blanks

1. Carbon compounds containing double and triple bonds are called ………………….
2. A compound which is basic constituent of many cough syrups ………………………
3. Very dilute solution of ethanoic acid is ………………..
4. A sweet odour substance formed by the reaction of an alcohol and a carboxylic acid is ………………
5. When sodium metal is dropped in ethanol …………………. gas will be released.
6. The functional group present in methanol is …………………….
7. IUPAC name of alkene containing 3 carbon atoms is ………………….
8. The first member of homologous series among alkynes is ……………………
9. The product that is formed by dehydration of ethanol in cone, sulphuric acid is ………………….
10. Number of single covalent bonds in ammonia are ………………..
11. Type of reactions shown by alkanes is ……………….
Answer:

1. unsaturated compounds
2. ethanol
3. vinegar
4. ester
5. H₂
6. – OH (Alcohol)
7. propene
8. ethyne (C₂H₂)
9. ethene (C₂H₄)
10. 3
11. substitutional

Multiple Choice Questions

1. Which of the four test tubes containing the following chemicals shows the brisk effervescence when dilute acetic acid was added to them?
i) KOH
ii) NaHCO₃
iii) K₂CO₃
iv) NaCl
A) i & ii
B) ii & iii
C) i & iv
D) ii & iv
Answer:
B) ii & iii

2. Which of the following solution of acetic acid in water can be used as preservative?
A) 5-10%
B) 10-15%
C) 15-20%
D) 100%
Answer:
A) 5-10%

3. The suffix used for naming an aldehyde is
A) – ol
B) – al
C) – one
D) – ene
Answer:
B) – al

4. Acetic acid, when dissolved in water, it dissociates into ions reversibly because it is a
A) weak acid
B) strong acid
C) weak base
D) strong base
Answer:
A) weak acid

5. Which one of the following hydrocarbons can show isomerism?
A) C₂H₄
B) C₂H₆
C) C₃H₈
D) C₄H₁₀
Answer:
D) C₄H₁₀

6. Combustion of hydrocarbon is generally accompanied by the evolution of
A) Heat
B) Light
C) Both heat and light
D) Electric current
Answer:
C) Both heat and light

7. 2 ml of ethanoic acid was taken in each of the three test tubes A, B and C and 2 ml, 4 ml and 8 ml water was added to them respectively. A clear solution is obtained in:
A) Test tube A only
B) Test tubes A & B only
C) Test tubes B and C only
D) All the test tubes
Answer:
D) All the test tubes

8. If 2 ml of acetic acid was added slowly in drops to 5 ml of water then we will notice
A) The acid forms a separate layer on the top of water
B) Water forms a separate layer on the top of the acid
C) Formation of a clear and homogenous solution
D) Formation of a pink and clear solution
Answer:
C) Formation of a clear and homogenous solution

9. A few drops of ethanoic acid were added to solid sodium carbonate. The possible results of the reactions are
A) A hissing sound was evolved
B) Brown fumes evolved
C) Brisk effervescence occurred
D) A pungent smelling gas evolved
Answer:
C) Brisk effervescence occurred

10. When acetic acid reacts with ethyl alcohol, we add cone. H₂SO₄, it acts as and the process is called
A) Oxidizing agent, saponification
B) Dehydrating agent, esterification
C) Reducing agent, esterification
D) Acid and esterification
Answer:
B) Dehydrating agent, esterification

10th Class Chemistry 14th Lesson Carbon and its Compounds InText Questions and Answers

10th Class Chemistry Textbook Page No. 254

Question 1.
Can carbon get helium configuration by losing four electrons from the outer shell?
Answer:

• If carbon loses four electrons from the outer shell, it has to form C⁴⁺ ions.
• This requires huge amount of energy which is not available normally.
• Therefore C⁴⁺ formation is also a remote possibility.
• Carbon has to satisfy its tetravalency by sharing electrons with other atoms.
• It has to form four covalent bonds either with its own atoms or atoms of other elements.

10th Class Chemistry Textbook Page No. 255

Question 2.
How do carbon atoms form bonds in so many different ways?
Answer:
As per valence bond theory, the four unpaired electrons in a carbon atom is main cause to form many bonds.

Question 3.
Explain the four unpaired electrons in carbon atom through excited state.
Answer:
Electronic configuration of carbon (ground state):

Electronic configuration of carbon (excited state):

10th Class Chemistry Textbook Page No. 256

Question 4.
Where does this energy to excite electron come from?
Answer:

• We have to understand that free carbon atom would not be in excited state under normal conditions.
• When the carbon atom is ready to form bonds with other atoms, the energy required for excitation is taken up from bond energies, which are the liberated energies when bonds are formed between carbon atom and other atoms.

Question 5.
In methane (CH₄) molecule all four carbon – hydrogen bonds are identical and bond angle HCH is 109°28′. How can we explain this?
Answer:
In excited state, carbon atom has three unpaired electrons in p-orbitals and one electron in s-orbital. These four valence electrons are with different energies. These orbitals combine to form four identical orbitals. Four hydrogen atoms form four identical C -H bonds with bond angle 109° 28′. This is called hybridisation.

Question 6.
How do these energetically unequal valence electrons form four equivalent covalent bonds in methane molecule?
Answer:
1) When bonds are formed, energy is released and the system becomes more stable. If carbon forms four bonds rather than two, still more energy is released and so the resulting molecule becomes even more stable.

2) The energy difference between the 2s and 2p orbitals is very small. When carbon atom is ready to form bonds it gets a small amount of energy from bond energies and gets excited to promote an electron from the 2s to the empty 2p to give four unpaired electrons.

3) We have got four unpaired electrons ready for bonding, but these electrons are in two different kinds of orbitals and their energies are different.

4) We are not going to get four identical bonds unless these unpaired electrons are in four identical orbitals.

10th Class Chemistry Textbook Page No. 257

Question 7.
How to explain the four orbitals of carbon containing unpaired electrons as energetically equal?
Answer:
With hybridisation we explai n the four orbitals of carbon containing unpaired electrons are energetically equal.
Ex : Methane (CH₄).

10th Class Chemistry Textbook Page No. 258

Question 8.
How do you explain the ability of C – atom to form two single covalent bonds and one double bond?
Answer:
Ethylene (CH₂ = CH₂) explains the ability of carbon atom to form two single covalent bonds and one double bond.
Ex:

10th Class Chemistry Textbook Page No. 259

Question 9.
How do you explain the ability of carbon atom to form one single bond and one triple bond?
Answer:
Ethyne (HC \(\equiv\) CH) explains the ability of carbon atom to form one single bond between one hydrogen and carbon, and one triple bond between carbon and carbon.
Ex : H – C \(\equiv\) C – H.

10th Class Chemistry Textbook Page No. 260

Question 10.
What are bond angles H\(\widehat{\mathbf{C}}\)H in CH₄, C₂H₄ and C₂H₂ molecules?
Answer:

10th Class Chemistry Textbook Page No. 262

Question 11.
How do you understand the markings (writings) of a pencil on a paper?
Answer:

1. When we write with a pencil, the inter layer attractions breakdown and leave graphite layers on the paper.
2. It is easy to remove pencil marks from paper with an eraser because, the layers do not bind strongly to the paper.

10th Class Chemistry Textbook Page No. 265

Question 12.
Allotting completely one special branch in chemistry to compounds of only one element. Is it justified when there are so many elements and their compounds but not with any special branches?
Answer:

1. We understand that all molecules that make life possible carbohydrates, proteins, nucleic acids, lipids, hormones, and vitamins contain carbon.
2. The chemical reactions that take place in living systems are of carbon compounds.
3. Food that we get From nature, various medicines, cotton, silk and fuels like natural gas and petroleum almost all of them are carbon compounds.
4. Synthetic fabrics, plastics, synthetic rubber are also compounds of carbon.
5. Hence, carbon is a special element with the largest number of compounds:

10th Class Chemistry Textbook Page No. 266

Question 13.
What are hydrocarbons?
Answer:
The compounds containing only carbon and hydrogen in their molecules are called hydrocarbons.

Question 14.
Do all the compounds have equal number of C and H atoms?
Answer:
No. All the compounds do not have equal number of C and H atoms.

10th Class Chemistry Textbook Page No. 269

Question 15.
Observe the following two structures.
a) CH₃ – CH₂ – CH₂ – CH₃
b)

i) How about their structures? Are they same?
Answer:
No, they are not same compounds.

ii) How many carbon and hydrogen atoms are there in (a) and (b) structures?
Answer:
Carbon – 4 ; Hydrogen – 10.

iii) Write the condensed molecular formulae for (a) and (b), do they have same molecular formulae?
Answer:
C₄H₁₀; Yes.

Question 16.
Can carbon form bonds with the atoms of other elements?
Answer:
Carbon forms compounds not only with atoms of hydrogen but also with atoms of other elements like oxygen, nitrogen, sulphur, phosphorus, halogens, etc.

10th Class Chemistry Textbook Page No. 272

Question 17.
What do you mean by nomenclature of Organic componds?
Answer:
Nomenclature of organic chemistry is systematic method of naming organic compound.

Question 18.
What is the basis for nomenclature?
Answer:
The basic of the nomenclature is number of carbons in the parent chain in a compound.

10th Class Chemistry Textbook Page No. 273

Question 19.
What are the word – root and suffix?
Answer:
Word root:
Word root indicates the number of carbon atoms in the longest possible continuous carbon chain also known as parent chain.

Suffix :
Suffix is added immediately after the word root. It is two types

1) Primary Suffix :
It is used to indicate the degree of saturation or unsaturation of the main chain.

2) Secondary Suffix :
It is used to indicate the main functional group in the organic compound.

10th Class Chemistry Textbook Page No. 274

Question 20.
What do you mean by the term ‘alkyl’?
Answer:
Alkyl:
Alkyl is a substituent, that is attached to the molecular fragment.
General formula of alkyl is C_nH_{2n + 1}

10th Class Chemistry Textbook Page No. 278

Question 21.
Can we write the structure of a compound if the name of the compound is given?
Answer:
Yes, we can write the structure of a compound if the name of the compound is given.

10th Class Chemistry Textbook Page No. 279

Question 22.
Why do sometimes cooking vessels get blackened on a gas or kerosene stove?
Answer:
Because of the inlets of air getting closed, the fuel gases do not completely undergo combustion. Hence, it forms a sooty carbon form which gets coated over the vessels.

10th Class Chemistry Textbook Page No. 280

Question 23.
Do you know what is a catalyst?
Answer:
A catalyst is a substance which regulates the rate of a given reaction without itself finally undergoing any chemical change.

10th Class Chemistry Textbook Page No. 281

Question 24.
Do you know how the police detect whether suspected drivers have consumed alcohol or not?
Answer:

1. The police officer asks the suspect to blow air into a plastic bag through a mouth piece of the detecting instrument which contains crystals of potassium-di-chromate (K₂Cr₂O₇).
2. As K₂Cr₂O₇ is a good oxidizing agent, it oxidizes any ethanol in the driver’s breath to ethanal and ethanoic acid.
3. Orange Cr₂O₇^2- changes to bluish green Cr³⁺ during the process of the oxidation of alcohol.
4. The length of the tube that turned into green is the measure of the quantity of alcohol that had been drunk.
5. The police even use the IR Spectra to detect the bonds C – OH and C – H of CH₃ – CH₂OH.

10th Class Chemistry Textbook Page No. 283

Question 25.
What are esters?
Answer:

The compounds which contain the functional group and the general formula R – COO – R’, where R and R’ are alkyl groups or phenyl groups, are known as “Esters”.

10th Class Chemistry Textbook Page No. 284

Question 26.
What is a true solution?
Answer:
A true solution is that in which the solute particles dispersed in the solvent are less than 1 nm in diameter.

10th Class Chemistry Textbook Page No. 286

Question 27.
What is the action of soap particles on the greasy cloth?
Answer:

• Soaps and detergents make oil and dirt present on the cloth come out into water, thereby making the cloth clean.
• Soap has one polar end and one non-polar end.
• The polar end is hydrophilic in nature and this end is attracted towards water.
• The non-polar end is hydrophobic in nature and it is attracted towards grease or * . ; oil on the cloth, but not attracted towards water.
• When soap is dissolved in water, its hydrophobic ends attach themselves to dirt and remove it from the cloth.
• The hydrophobic end of the soap molecules move towards the dirt or grease particles. ’
• The hydrophobic ends attach to the dirt particle and try to pull out.
• The molecules of soap surround the dirt particle at the centre of the cluster and form a spherical structure called micelle.
• These micelles remain suspended in water like particles in a colloidal solution.
• The various micelles present in water do not come together to form a precipitate as each micelle repels the other because of the ion-ion repulsion.
• Thus, the dust particles remain trapped in micelles and are easily rinsed away with water.
• Hence, soap micelles remove dirt by dissolving it in water.

10th Class Chemistry Textbook Page No. 280

Question 28.
Why we are advised not to use animal fats for cooking?
Answer:

• Animal fats have recently been implicated as the cause of heart disease and obesity. So, we are advised not to use animal fats for cooking.
• Excess animal fat is stored in lipocytes, which expand in size until the fat is used for fuel.

Question 29.
Which oil is recommended for cooking? Why?
Answer:
Canola oil :

• A recent entrant into the Indian market Canola is flying off the shelves.
• Canola oil which is made from the crushed seeds of the Canola plant, is said to be amongst the healthiest of cooking oils.
• It has the lowest saturated fat content of any oil.
• It’s seen as a healthy alternative as it’s rich in monosaturated fats and is high in omega-3 and omega a fats.
• It has a medium smoking point and is an oil that works well for fruits, baking, sauteing, etc.

10th Class Chemistry 14th Lesson Carbon and its Compounds Activities

Activity – 1

Question 1.
Observe the structural formula of the following hydro carbons and write their names in your notebook.
Answer:
1) CH₃ – CH₂ – CH = CH₂
Sol. But-l-ene

Sol. 2-Methyl butane

3) CH₃ – CH₂ – CH₂ – CH₂ – CH₂ – CH₃
Sol. Hexane

Sol. 3-Methyl, but-l-ene

5)

Sol. Prop-l-yne

Activity-2

Question 2.
Read the names of the following hydro carbons and draw their structures in your notebook.
1. 2,2-Dimethyl hexane
Sol.

2. But-l-yne
Sol. CH₃ – CH₂ – C = CH

3. 3-Methyl Pent-2-ene
Sol.

4. But-1.2-diene
Sol. CH₃ – CH₃ = c = CH₂

5. Hept-2 en, 4-yne
Sol.

Activity – 3

Question 3.
Write an activity to show esterification reactions.
Answer:
The compound formed is ester. The process is called esterification.

1. Take 1 ml of ethanol and 1 ml of glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
2. Warm it in a water bath or a beaker containing water for at least five minutes.
3. Pour the warm contents into a beaker containing 20-50 ml of water and observe the odour of the resulting mixture.
4. We will notice that the resulting mixture is sweet odoured subatance.
5. This substance is nothing but ethyl acetate, an ester.
6. This reaction is called esterification reaction.

Activity – 4

Question 4.
Write an activity to show soap solution separates oil from water.
Answer:

1. Take about 10 ml of water each in two test tubes.
2. Add a drop of oil to both the test tubes.
3. Label them as A and B.
4. Add a few drops of soap solution to test tube B.
5. Now shake both the test tubes vigorously for the same period of time.
6. We can see the oil and water layers separately in both the test tubes immediately after we stop shaking them.
7. Leave the test tubes undisturbed for sometime and observe.
8. The oil layer separates out first in which test tube we added drops of soap solution.

 

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 10 Chemical Bonding Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 10th Lesson Chemical Bonding

10th Class Chemistry 10th Lesson Chemical Bonding Textbook Questions and Answers

Improve Your Learning

Question 1.
List the factors that determine the type of bond that will be formed between two atoms. (AS1)
(OR)
How can you identify the type of bond formation between two atoms?
Answer:

• The strength of attraction or repulsion between atoms.
• Electrons in valence shell (valence electrons).

Question 2.
Explain the difference between the valence electrons and the covalency of an element. (AS1)
(OR)
How are valence electrons different from the covalency of element? Explain with examples.
Answer:
Valence electrons :

• Number of electrons in the outermost orbit or an atom is called its valence electrons.
• Ex: Na (Z = 11). It has 2e^– in I orbit, 8e^– in II orbit and 1e^– in III orbit.
• So number of valence electrons in Na atom are ‘l’.

Covalency of an element:

• Number of valance electrons which are taking part in covalent bond is called covalency.
• The electron configuration of Boran is 1s² 2s² 2p¹.
• It has three valance electrons.
• So its covalency is 3.

Question 3.
A chemical compound has the following Lewis notation : (AS1)
a) How many valence electrons does element Y have?
b) What is the valency of element Y?
c) What is the valency of element X?
d) How many covalent bonds are there in the molecule?
e) Suggest a name for the elements X and Y.

Answer:
a) 6
b) 2
c) 1
d) two
e) X – is hydrogen and Y – is oxygen. The formed molecule is H₂O.

Question 4.
Why do only valence electrons involve in bond formation? Why not electron of inner shells? Explain. (AS1)
(OR)
Which shell electrons involve in bond formation? Explain. What is the reason behind it?
Answer:

• The nucleus and the electrons in the inner shell remain unaffected when atoms come close together.
• But the electrons in the outermost shell (valence shell) of atoms get affected.
• The inner shell electrons are strongly attracted by the nucleus when compared to the valence electrons.
• So electrons in valence shell (valence electrons) are responsible for the formation of bond between atoms.

Question 5.
Explain the formation of sodium chloride and calcium oxide on the basis of the concept of electron transfer from one atom to another atom. (AS1)
(OR)
Explain the formation of any two compounds according to Kossel’s theory.
Answer:
I. Formation of sodium chloride (NaCl) :
1) Sodium chloride is formed from the elements sodium (Na) and chlorine (Cl).

2) Cation formation:
i) When sodium (Na) atom loses one electron to get octet electron configuration, it forms a cation (Na⁺).
ii) Now Na⁺ gets electron configuration that of Neon (Ne) atom.

3) Anion Formation :
i) Chlorine has shortage of one electron to get octet in its valence shell.
ii) So it gains the electron that was lost by Na to form anion and gets electron configuration of Argon (Ar).

4) Formation of NaCl :
i) Transfer of electrons between ‘Na’ and ‘Cl’ atoms, results in the formation of ‘Na⁺‘ and ‘Cl^–’ ions.
ii) These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).
Na⁺_(g) + Cl^–(g) → Na⁺Cl^–_(s) or NaCl

II. Formation of calcium oxide (CaO) :
1. Calcium (Ca) reacts with oxygen (0) to form an ionic compound calcium oxide (CaO).

2. Atomic number of Calcium is 20. Its electronic configuration is 2, 8, 8, 2.
3.

4. Atomic number of Oxygen is 8. Its electronic configuration is 2, 6.

5.

6. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound calcium oxide (CaO).
Ca²⁺ + O^2- → Ca²⁺O^2- (or) CaO

Question 6.
A, B and C are three elements with atomic number 6, 11 and 17 respectively.
i) Which of these cannot form ionic bond? Why? (AS1)
ii) Which of these cannot form covalent bond? Why? (AS1)
iii) Which of these can form ionic as well as covalent bonds? (AS1)
Answer:
i) ‘A’ cannot form ionic bond. Its valence electrons are 4. It is difficult to lose or gain 4e^– to get octet configuration. So it forms covalent bond [Z of A is 6 so it is carbon (C)].

ii) ‘B’ cannot form covalent bond. Its valence electrons are 1 only. So it is easy to donate for other atom and become an ion. So it can form ionic bond [Z of B is 11, so it is sodium (Na)].

iii) Element C can form ionic as well as covalent bonds. Atomic number of Cl is 17. It is able to participate with Na in ionic bond and with hydrogen in HCl molecule as covalent bond.

Question 7.
How do bond energies and bond lengths of molecule help us in predicting their chemical properties? Explain with examples. (AS1)
(OR)
How can you explain with examples that bond energies and bond lengths are used to recognise chemical properties?
Answer:
1. Bond length :
Bond length or bond distance is the equilibrium distance between the nuclei of two atoms which form a covalent bond.

2. Bond energy :
Bond energy or bond dissociation energy is the energy needed to break a covalent bond between two atoms of a diatomic covalent compound in its gaseous state.

3. If the nature of the bond between the same two atoms changes the bond length also changes. For example, the bond lengths between two carbon atoms are C – C > C = C > C = C.

4. Thus the various bond lengths between the two carbon atoms are in ethane 1.54 Å, ethylene 1.34 Å, acetylene 1.20 Å.

5. The bond lengths between two oxygen atoms are in H₂O₂ (O – O) is 1.48 Å and in O₂ (O = O) is 1.21 Å.
6. Observe the table.

7. When bond length decreases, then bond dissociation energy increases.

8. When bond length increases, then bond dissociation energy decreases.

9. Bond length of H – H in H₂ molecule is 0.74 Å and its bond dissociation energy is 436 KJ/mol, whereas bond length of F – F in F₂ molecule is 1.44 Å and its bond dissociation energy is 159 KJ/mol.

10. Melting and boiling points of substances also can be determined by this bond energies and bond lengths.

Question 8.
Predict the reasons for low melting point for covalent compounds when compared with ionic compounds. (AS2)
(OR)
“Covalent compounds have low melting point.” What Is the reason for this statement? Explain.
Answer:
They are covalent compounds.

• The melting point is low due to the weak Vander Waal’s forces of attractions between the covalent molecules.
• The force of attraction between the molecules of a covalent compound is very weak.
• Only a small amount of heat energy is required to break these weak molecular forces, due to which covalent compounds have low melting points and low boiling points.
• Please note that some of the covalent solids like diamond and graphite have, however very high melting points and boiling points.

Question 9.
Collect the information about properties and uses of covalent compounds and prepare a report. (AS4)
(OR)
Generally these compounds are non-polar in nature. What are those compounds? Explain their properties and uses.
(OR)
Write any two uses and two properties of covalent compounds.
Answer:
The compounds are covalent.
Properties of covalent compounds :

1. Covalent compounds are usually liquids or gases, only some of them are solids.
2. The covalent compounds are usually liquids or gases due to the weak force of attraction between their molecules.
3. Covalent compounds have usually low melting and low boiling points.
4. Covalent compounds are usually in soluble in water but they are soluble in organic solvents.
5. Covalent compounds do not conduct electricity.

Uses of covalent compounds :

1. Covalent compounds form 99% of our body.
2. Water is a covalent compound. We know its many uses.
3. Sugars, food substances, tea and coffee are all covalent compounds.
4. Air we breathe in contains covalent molecules of oxygen and nitrogen.
5. Almost everything on earth other than most simple in organic salts are covalent.

Question 10.
Draw simple diagrams to show how electrons are arranged in the following covalent molecules : (AS5)
a) Calcium oxide (CaO)
b) Water (H₂O)
c) Chlorine (Cl₂)
Answer:
a) Calcium oxide (CaO) :

b) Water (H₂O):

The formation of water molecule can be shown like this also

c) Chlorine (Cl₂):

We can explain the formation of Cl₂ molecule in this way also.

Question 11.
Represent the molecule H₂O using Lewis notation. (AS5)
(OR)
How can you explain the formation of H₂O molecule using dot structure?
Answer:
One atom of oxygen shares its two electrons with two hydrogen atoms to form a water molecule.

Question 12.
Represent each of the following atoms using Lewis notation : (AS5)
a) Beryllium
b) Calcium
c) Lithium
Answer:

Question 13.
Represent each of the following molecules using Lewis notation : (AS5)
a) Bromine gas (Br2)
b) Calcium chloride (CaCl₂)
c) Carbon dioxide (CO₂)
d) Which of the three molecules listed above contains a double bond?
Answer:
a) Bromine gas (Br₂) :

b) Calcium chloride (CaCl₂)

c) Carbon dioxide (CO₂) :

d) CO₂, contains double bond in above list. Its structure is like this : O = C = O.

Question 14.
Two chemical reactions are described below. (AS5)
♦ Nitrogen and hydrogen react to form ammonia (NH₃).
♦ Carbon and hydrogen bond to form a molecule of methane (CH₄).
For each reaction give :
a) The valency of each of the atoms involved in the reaction.
b) The Lewis structure of the product that is formed.
Answer:
a) ♦ Nitrogen and hydrogen react to form ammonia (NH₃):
i) The valency of nitrogen is 3 and hydrogen is 1.
ii) The chemical formula of the product is NH₃

♦ Carbon and hydrogen bond to form a molecule of methane (CH₄):
i) The valency of carbon is 4 and hydrogen is 1.
ii) The chemical formula of the product is CH₄.

b) ♦ The Lewis structure of the product that is formed (: NH₃)

♦ The Lewis structure of the product that is formed (CH₄)

Question 15.
How does Lewis dot structure help in understanding bond formation between atoms? (AS6)
(OR)
What is the use of Lewis dot structure in bond formation? Explain.
Answer:

1. Only the outermost electrons of an atom take part in chemical bonding.
2. They are known as valence electrons.
3. The valence electrons in an atom are represented by putting dots (•) on the symbol of the element, one dot for each valence electron.
4. For example, sodium atom has 1 valence electron in its outermost shell, so we put 1 dot with the symbol of sodium and write Na• for it.
5. Sodium atom loses this 1 electron to form sodium ion.
6. By knowing the valence electrons of two different atoms by Lewis dot structure, we can understand which type of bond is going to establish between them and forms corresponding molecule.

Question 16.
What is octet rule? How do you appreciate role of the ‘octet rule’ in explaining the chemical properties of elements? (AS6)
(OR)
Which rule decides whether given element is chemically stable or not? Appreciate that rule.
Answer:
Octet rule decides whether given element is stable or not.
Octet rule :

• ‘The atoms of elements tend to undergo chemical changes that help to leave their atoms with eight outer shell electrons.”
• It was found that the elements which participate in chemical reaction get octet (or) ns2 np6 configuration similar to that of noble gas elements.

Role of octet in chemical properties of elements :

1. Except He remaining inert gas elements have 8 electrons in their outermost orbit. Since these elements are having stable octet configuration in their outermost orbit, they are very stable.
2. They do not allow the outermost electrons to take part in chemical reactions.
3. So by having octet configuration for these elements we can conclude these are chemically inertial.
4. If any group of elements (take halogens) which contain 7 electrons in their outermost orbit, they require only 1 e^– to get octet configuration.
5. So they try to participate in chemical reaction to get that 1 difference electron for octet configuration.
6. Similarly, Na contains 2, 8, 1 as its electronic configuration.
7. So it loses le^– from its outermost shell; it should have 8e^– in its outer shell and get the octet configuration.
8. Thus the octet rule helps in explaining the chemical properties of elements.

Question 17.
Explain the formation of the following molecules using valence bond theory.
a) N₂ molecule
b) O₂ molecule
(OR)
Write the formation of double bond and triple bond according to valence bond theory.
(OR)
Who proposed Valence Bond Theory? Explain the formation of N₂ molecule by using this theory.
Answer:
Linus Pauling was proposed valence bond theory.
Formation of N₂ molecule :

1. Electronic configuration of Nitrogen is 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹.
2. Suppose that p_x orbital of one Nitrogen atom overlaps the p_x orbital of other ‘N’ atom giving σ p_x – p_x bond along the inter nuclear axis.
3. The p_y and p_z orbitals of one ‘N’ atom overlaps with the p_y and p_z orbital of other ‘N’ atom laterally giving π p_y – p_y and π p_z – p_z bonds.
4. Therefore, N₂ molecule has a triple bond between two Nitrogen atoms.

Formation of O₂ molecule :

1. Electronic configuration of ‘O’ is 1s² 2s² 2p_x² 2p_y¹ 2p_z¹.
2. If the P_y orbital of one ‘O’ atom overlaps the p_y orbital of other ‘O’ atom along inter- nuclear axis, a σ p_y – p_y bond is formed.
3. p_z orbital of oxygen atom overlaps laterally, perpendicular to inter nuclear axis giving a π p_y – p_z bond.
4. So O₂ molecule has a double bond between the two oxygen atoms.

Question 18.
What is hybridisation? Explain the formation of the following molecules using hybridisation.
a) BeCl₂
b) BF₃
Explain the formation of sp and sp² hybridisation using examples.
(OR)
What is the name given to inter mixing of atomic orbitals to form new orbitals. Explain the formation of following molecules by using that process,
a) BeCl₂
b) BF₃
Answer:
This process is called hybridisation.
Hybridisation :
It is a phenomenon of inter mixing of atomic orbitals of almost equal energy which are present in the outer shells of the atom and their reshuffling or redistribution into the same number of orbitals but with equal properties like energy and shape.

a) Formation of BeCl₂ (Beryllium chloride) molecule :

1. ₄Be has electronic configuration 1s² 2s².
2. It has no unpaired electrons.
3. It is expected not to form covalent bonds, but informs two covalent bonds one each with two chlorine atoms. „
4. To explain this, an excited state is suggested for Beryllium in which an electron from ‘2s’ shifts to 2p_x level.
5. Electronic configuration of ₄Be is 1s² 2s¹ 2p_x¹].
6. Electronic configuration of ₁₇Cl is 1s² 2s² 2p⁶ 3s² 3p_x² 3p_y² 3p_z¹.
7. If Be forms two covalent bonds with two chlorine atoms, one bond should be σ 2s-3p due to the overlap of ‘2s’ orbital of Be, the ‘3p_z‘ orbital of one chlorine atom.
8. The other bond should be σ 2p-3p due to the overlap of ‘2p_x’ orbital of Be atom the 3p orbital of the other chlorine atom.
9. As the orbitals overlapping are different, the bond strengths of two Be-Cl must be different.
10. But, both bonds are of same strength and Cl\(\hat{\mathrm{Be}}\) Cl is 180°.

The Hybridisation of BeCl₂ can be explained in this way also :
a) Be atom in its excited state allows its 2s orbital and 2p_x orbital which contain unpaired electrons to intermix and redistribute to two identical orbitals.
b) As per Hund’s rule each orbital gets one electron.
c) The new orbitals based on the types of orbitals that have undergone hybridisation are called sp orbitals.
d) The two sp orbitals of Be get separated by 180°.
e) Now each chlorine atom comes with its 3p_z¹ orbital and overlaps it the sp orbitals of Be forming two identical Be-Cl bonds (σ sp-p bonds).
Cl\(\hat{\mathrm{Be}}\) Cl = 180°.

f) Both the bonds are of same strength.

b) Formation of BF₃ molecule :

1. ₅B has electronic configuration 1s² 2s² 2pxh
2. The excited electronic configuration of ₅B is 1s² 2s¹ 2p_x¹ 2p_y¹
3. As it forms three identical B-F bonds in BF₃.
4. It is suggested that excited ‘B’ atom undergoes hybridisation.
5. There is an intermixing of 2s, 2p_x, 2p_y orbitals and their redistribution into three identical orbitals called sp² hybrid orbitals.
6. For three sp² orbitals to get separated to have minimum repulsion the angle between any two orbitals is 120° at the central atom and each sp² orbital gets one election.
7. Now three fluorine atoms overlap their 2p_z orbitals containing unpaired electrons (F₉ 1s² 2s² 2p_x² 2p_y² 2p_z¹) the three sp² orbitals of ‘B’ that contain unpaired electrons to form three σsp²-p bonds.

Fill in the Blanks

1. Electrons in the outermost orbit are called …………………… .
2. Except …………………… gas all other noble gases have octet in their valence shell.
3. Covalency of elements explains about member of …………………… formed by the atom.
4. Valence bond theory was proposed by …………………… .
5. In …………………… bonding the valence electrons are shared among all the atoms of the metallic elements.
Answer:

1. valence electrons
2. Helium
3. covalent bonds
4. Linus Pauling
5. covalent

Multiple Choice Questions

1. Which of the following elements is electronegative?
A) Sodium
B) Oxygen
C) Magnesium
D) Calcium
Answer:
B) Oxygen

2. An element ₁₁X²³ forms an ionic compound with another element ‘Y’. Then the charge on the ion formed by X is
A) +1
B) +2
C) -l
D) – 2
Answer:
A) +1

3. An element ‘A’ forms a chloride ACl₄. The number of electrons in the valence shell of ‘A’
A) 1
B) 2
C) 3
D) 4
Answer:
D) 4

10th Class Chemistry 10th Lesson Chemical Bonding InText Questions and Answers

10th Class Chemistry Textbook Page No. 153

Question 1.
How do elements usually exist?
Answer:
They may exist as a single atom or as a group of atoms.

Question 2.
Do atoms exist as a single atom or as a group of atoms?
Answer:
Atoms exist as a single atom, sometimes as a group of atoms also.

Question 3.
Are there elements which exist as atoms?
Answer:
Yes. There are elements which exist as atoms.

Question 4.
Why do some elements exist as molecules and some as atoms?
Answer:
By following different laws of chemical combination the chemical compounds take place as a result of combination of atoms of various elements in different ways.

Question 5.
Why do some elements and compounds react vigorously while others are inert?
Answer:
1) Number of electrons in their outermost shell.
2) Bond strength between the atoms in compound.

Question 6.
Why is the chemical formula for water H₂O and for sodium chloride NaCl, why not HO₂ and NaCl₂?
Answee:
Valencies of the atoms participating in the molecules.

Question 7.
Why do some atoms combine dille tl do not?
Answer:
1) Atoms which have 8e“ in their outer shell will not combine.
2) Atoms which have more than or less than 8e“ in their outer shell will combine.

Question 8.
Are elements and compounds simply made up of separate atoms Individually arranged?
Answer:
No. Elements and compounds are not simply made up of separate atoms individually arranged.

Question 9.
Is there any attraction between atoms?
Answer:
Yes. There is some attraction betwen atoms.

Question 10.
What is that holding them together?
Answer:
Force of attraction between them.

10th Class Chemistry Textbook Page No. 155

Question 11.
Why is there absorption of energy in certain chemical reactions and release of energy in other reactions?
Answer:’
Because of bond energy between the atoms in a molecule.

Question 12.
Where does the absorbed energy go?
Answer:
For breaking chemical bonds between atoms in a molecule.

Question 13.
Is there any relation to energy and bond formation between atoms?
Answer:
Yes. There is some relation to energy and bond formation between atoms.

Question 14.
What could be the reason for the change in reactivity of elements?
Answer:
Number of electrons in their outermost orbit.

Question 15.
What could be the reason for this?
Answer:
They have 8 (e^–) electrons in their outermost orbit.

10th Class Chemistry Textbook Page No. 157

Question 16.
What did you notice in Lewis dot structure of noble gases and electronic configurations of the atoms of these elements shown in table – 1?
Answer:
Except He remaining Ne, Ar, Kr have 8 electrons in their outermost orbit.

10th Class Chemistry Textbook Page No. 158

Question 17.
What have you observed from the above conclusions about the main groups?
Answer:

1. Number of gained electrons of non-metals in their valency.
2. Number of lost electrons of metals in their valency.

Question 18.
Why do atoms of elements try to combine and form molecules?
Answer:
To attain stable electronic configuration.

10th Class Chemistry Textbook Page No. 159

Question 19.
Is it accidental that IA to VIIA main group elements during chemical reactions get eight electrons in the outermost shells of their ions, similar to noble gas atoms?
Answer:
No, it cannot be simply accidental.

Question 20.
Explain the formation of ionic compounds NaCl, MgCl₂, Na₂O and AlCl₃ through Lewis electron dot symbols (formulae).
Answer:
1) Lewis electron dot symbol for NaCl:

Formation of sodium chloride (NaCl) :
Sodium chloride is formed from the elements sodium and chlorine. It can be explained as follows.
Na_(s) + ½Cl_2(g) → NaCl₂

Cation formation :
When sodium (Na) atom loses one electron to get octet electron configuration it forms a cation (Na+) and gets electron configuration that of Neon (Ne) atom.

Anion formation :
Chlorine has shortage of one electron to get octet in its valence shell. So it gains the electron from Na atom to form anion and gets electron configuration as that of argon (Ar).

Formation of the compound NaCl from its ions :
Transfer of electrons between ‘Na’ and ‘Cl’ atoms, results in the formation of ‘Na⁺‘ and ‘Cl^–‘ ions. These oppositely charged ions get attracted towards each other due to electrostatic forces and form the compound sodium chloride (NaCl).
Na⁺_(g) + Cl^–_(g) → Na⁺Cl^–_(s) or NaCl

2) Lewis electron dot symbol for MgCl₂:
MgCl₂

Formation of magnesium chloride (MgCl₂):
Magnesium chloride is formed from the elements magnesium and chlorine. The bond formation MgCl₂ in brief using chemical equation is as follows :
Mg_(s) + Cl_2(g) → MgCl_2(g)
Cation formation:

Anion formation :

The compound MgCl₂ formation from its ions :
Mg²⁺ gets ‘Ne’ configuration and
Each Cl^– gets ‘Ar’ configuration
Mg²⁺_(g) + 2 Cl^–_(g) → MgCl_2(s)
One ‘Mg’ atom transfers two electrons one each to two ‘Cl’ atoms and so formed Mg²⁺ and 2Cl^– attract to form MgCl₂.

3) Lewis electron dot symbol for (Na₂O) :

Formation of di sodium monoxide (Na₂O):
Di sodium monoxide formation can be explained as follows:
Cation formation (Na⁺ formation):

Two ‘Na’ atoms transfer one electron each to one oxygen atom to form 2 Na⁺ and O^2-
Each Na⁺ gets ‘Ne’ configuration and O^2- gets ‘Ne’ configuration.
These ions (2Na⁺ and O^2-) attract to form Na₂O.

4) Lewis electron dot symbol for (AlCl₃):

Formation of aluminium chloride (AlCl₃):
Aluminium chloride formation can be explained as follows:
Formation of aluminium ion (Al³⁺), the cation:

Each aluminium atom loses three electrons and three chlorine atoms gain them, one electron each.
The compound AlCl₃ is formed from its component ions by the electrostatic forces of attractions.
Al³⁺_(g) + 3 Cl^–_(g) → AlCl_3(s)

10th Class Chemistry Textbook Page No. 163

Question 21.
How do cations and anions of an ionic compound exist in its solid state?
Answer:
Ionic compounds exist in crystalline state.

Question 22.
Do you think that pairs of Na⁺ Cl^– as units would be present in the solid crystal?
Answer:
Yes. I think that pairs of Na⁺ Cl^– as units would be present in the solid crystal.

10th Class Chemistry Textbook Page No. 164

Question 23.
Can you explain the reasons for all these?
Answer:
Ionic bond is formed between atoms of elements with electronegativity, difference equal to or greater than 1.9.

10th Class Chemistry Textbook Page No. 165

Question 24.
Can you say what type of bond exists between atoms of nitrogen molecule?
Answer:
Triple bond exists between atoms of nitrogen molecule.

10th Class Chemistry Textbook Page No. 168

Question 25.
What do you understand from bond lengths and bond energies?
Answer:
Bond lengths and bond energies are not same when the atoms that form the bond are different.

Question 26.
Are the values not different for the bonds between different types of atoms?
Answer:
Yes. The values are not different for the bonds between different types of atoms.

10th Class Chemistry Textbook Page No. 170

Question 27.
What is the bond angle in a molecule?
Answer:
It is the angle subtended by two imaginary lines that pass from the nuclei of two atoms which form the covalent bonds with the central atom through the nucleus of the central atom at the central atom.

10th Class Chemistry Textbook Page No. 172

Question 28.
How is MCI molecule formed?
Answer:
The ‘1s’ orbital of ‘H’ atom containing unpaired electron overlaps the ‘3p’ orbital of chlorine atom containing unpaired electron of opposite spin.

10th Class Chemistry 10th Lesson Chemical Bonding Activities

Activity – 1

1. Write the Lewis structures of the given elements in the table. Also, consult the periodic table and fill in the group number of the element.
Answer:

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 9 Classification of Elements- The Periodic Table Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 9th Lesson Structure of Atom

10th Class Chemistry 9th Lesson Structure of Atom Textbook Questions and Answers

Improve Your Learning

Question 1.
Newlands proposed the law of octaves. Mendeleeff suggested eight groups for elements in his table. How do you explain these observations in terms of modem periodic classification? (AS1)
(OR)
Correlate various tables proposed on classification of elements.
Answer:

• According to Newlands, every eighth element starting from a given element jsembles in its properties to that of the starting element, when elements are ranged in ascending order of their atomic weights.
• According to Newlands, the properties of fluorine and chlorine are similar and sodium and potassium are similar. Same aspect is given by modern periodic table.
• Mendeleeff divided it into horizontal rows and vertical columns. He called them peribds and groups respectively. Modem periodic table also gives the same.
• According to Mendeleeff, the elements of same group have similar properties. Modern periodic table also proposed the same thing.
• Mendeleeff gave the general formula for first group elements as R,0, and general formula for second group elements as RO. We can find the same thing in modern periodic table.
• The elements of particular group possess same common valency. Same was proposed by modern periodic table.

Question 2.
What are the limitations of Mendeleeff’s periodic table? How could the modern periodic table overcome the limitations of Mendeleeff’s table? (AS1)
(OR)
How can the limitations of Mendeleeffs table be overcome with the help of modern periodic table?
Answer:
Limitations of Mendeleeffs periodic table :
1) Anomalous pair of elements :
Certain elements of highest atomic weights precede those with lower atomic weights.
Eg : Tellurium (atomic weight 127.6) precedes iodine (atomic weight 126.9).

2) Dissimilar elements placed together :
a) Elements with dissimilar properties were placed in same group as sub-group A and sub-group Bt
Eg : Alkali metals like Li, Na, K, etc. of IA group have little resemblance with coinage metals like Cu, Ag, Au of IB group.

b) Cl of VII A group is a non-metal and Mn of VII B group is a metal.

Method of overcoming the limitations of Mendeleeffs periodic table by modern periodic table :
1. In modern periodic table, elements are arranged in the ascending order of their atomic numbers. So this arrangement eliminated the problem of anomalous series.
Eg : Though Tellurium (Te) has more atomic weight than Iodine (I), its atomic number is one unit less compared to Iodine.

2. The elements with similar outer shell (valence shell) electronic configurations in their atoms are in the same column called group in modern periodic table. So the elements have similar properties overcoming the Mendeleeffs second limitation.

Question 3.
Define the modern periodic law. Discuss the construction of the long form of the periodic table. (AS1)
(OR)
What are the salient features of modern periodic table?
Answer:
Modern periodic law :
‘The physical and chemical properties of elements are the periodic function of the electronic configurations of their atoms”.

Construction of the long form periodic table :

1. Based on the modern periodic law, the modern periodic table is proposed.
2. This periodic table is known as long form of the periodic table.
3. Long form periodic table is the graphical representation of Aufbau principle.
4. The modern periodic table has 18 vertical columns called groups and 7 horizontal rows known as periods.
5. There are 18 groups, represented by using Roman numerals I to VIII, with letters A and B in traditional notation, (or) 1 to 18 by Arabic numerals.
6. There are 7 periods. These periods are represented by Arabic numerals 1 to 7.
7. The number of main shells present in the atom of particular atom decides to which period it belongs.
8. First period consists 2 elements, 2^nd and 3rd periods contains 8 elements each, 4^th and 5^th periods contains 18 elements each, 6 period contains 32 elements and 7^th period is incomplete.
9. The elements are classified into s, p, d and f block elements.
10. Inert gases are placed in 18^th group.

Question 4.
Explain how the elements are classified into s, p, d and f-block elements in the periodic table and give the advantage of this kind of classification. (AS1)
(OR)
How is the periodic table classified based upon the entering of differenciating electron? Explain that classification. What is the advantage of such classification?
Answer:
1) Depending upon which sub-shell the differentiating electron enters, the elements are classified into s, p, d and f-block elements. They are

1. s – block elements,
2. p – block elements,
3. d – block elements,
4. f – block elements.

2) s – block elements :
i) If the differentiating electron enters in s-sub-shell, then the elements are called s-block elements.
ii) IA (1), IIA (2) group elements belong to this block.

3) p – block elements :
i) If the differentiating electron enters in p-sub-shell, then the elements are called p-block elements.
ii) IIIA(13), IV A (14), V A (15), VIA (16), VIIA (17) belong to p-block.

4) d – block elements :
i) If the differentiating electron enters in d-sub-shell, then the elements are called d – block elements.
ii) I B, II B, III B, IV B, V B, VI B, VII B, VIII B belong to d-block elements.
iii) They are also called transition elements.

5) f – block elements :
i) If the differentiating electrons enter in f-sub-shell, then the elements are called f-block elements.
ii) These are divided into two types
a) Lanthanides (41 elements),
b) Actinides (5f elements).
iii) These are also called as inner transition elements.

Advantage of this classification :
1) The systematic grouping of elements into groups made the study simple.
2) Each period begins with the electron entering a new shell and ends with the complete filling of s and p-sub-shells of that shell.

Question 5.
Given below is the electronic configuration of elements A, B, C, D. (AS1)

A) 1s² 2s²	1. Which are the elements coming within the same period?
B) 1s² 2s² 2p6 3s²	2. Which are the elements coming within the same group?
C) 1s² 2s² 2p6 3s² 3p³	3. Which are the noble gas elements?
D) 1s² 2s² 2p6	4. To which group and period does the element ‘C’ belong?

Answer:
According to electronic configuration
A = Be B = Mg C = P D = Ne
1. Which are the elements coming within the same period?
Answer:
A and D i.e. Be and Ne coming within the same period. [They have same valence shell (n = 2)]

2. Which are the ones coming within the same group?
Answer:
A and B i.e., Be and Mg coming within the same group. [They have same valence subshell with same valency (2s² and 3s²)]

3. Which are the noble gas elements?
Answer:
D, i.e. Ne is the noble gas element. [It has valency as ‘O’ and it has ‘8’ electrons in valence shell].

4. To which group and period does the element ‘C’ belong?
Answer:
Element ‘C’ i.e. ‘P’ belongs to 3^rd period and VA group.

Question 6.
Write down the characteristics of the elements having atomic number 17. (AS1)
1) Electronic configuration ___________
2) Period number _____________
3) Group number _____________
4) Element family ____________
5) No. of valence electrons ___________
6) Valency _____________
7) Metal or non-metal ____________
Answer:

1. 1s² 2s² 2p⁶ 3s² 3p⁵
2. 3
3. VII A or 17
4. Halogen family
5. 7
6. 1
7. Non-metal

Question 7.
a) State the number of valence electrons, the group number and the period number of each element given in the following table : (AS1)

Answer:

b) State whether the following elements belong to a Group (G), Period (P) or neither Group nor Period (N). (AS1)

Answer:

Question 8.
Elements in a group generally possess similar properties, but elements along a period have different properties. How do you explain this statement? (AS1)
(OR)
Elements in a group possess similar properties, but elements along a period have different properties. Explain the reason.
Answer:

• Physical and chemical properties of elements are related to their electronic configurations, particularly the outer shell configurations.
• Therefore, all the elements in a group should have similar chemical properties.
• Similarly, across the table from left to right in any period, elements get an increase in the atomic number by one unit between any two successive elements.
• Therefore, the electronic configuration of valence shell of any two elements in a period is not same. Due to this reason, elements along a period possess different chemical properties.

Question 9.
s – block and p – block elements except 18^th group elements are sometimes called as ‘Representative elements’ based on their abundant availability in the nature. Is it justified? Why? (AS1)
(OR)
Which elements are called representative elements? Why?
Answer:

• s, p – block elements are called representative elements because these are the elements which take part in chemical reactions because of incompletely filled outermost shell.
• These elements undergo chemical reactions to acquire the nearest noble gas configuration by losing or gaining or sharing of electrons.
• So they are called representative elements.

Question 10.
Complete the following table using periodic table. (AS1)

Answer:

Question 11.
Complete the following table using the periodic table. (AS1)

Answer:

Question 12.
The electronic configuration of the elements X, Y, and Z are given below.
a) X = 2
b) Y = 2, 6
c) Z = 2, 8, 2
i) Which element belongs to second period?
Answer:
Y belongs to second period.

ii) Which element belongs to second group?
Answer:
Z belongs to second group,

iii) Which element belongs to 18^th group?
Answer:
X belongs to 18^th group.

Question 13.
Identify the element that has the larger atomic radius in each pair of the following and mark it with a symbol (✓). (AS1)
(i) Mg or Ca
(ii) Li or Cs
(iii) N or P
(iv) B or Al
Answer:

Question 14.
Identify the element that has the lower ionization energy in each pair of the, following and mark it with a symbol (✓). (AS1)
(i) Mg or Na (ii) Li or O (iii) Br or F (iv) K or Br
Answer:

Question 15.
In period 2, element X is to the right of element Y. Then, find which ofitheydements have : (AS1)
i) Low nuclear charge
Answer:
Y has low nuclear charge.

ii) Low atomic size
Answer:
X has lower atomic size,

iii) High ionization energy
Answer:
X has higher ionization energy.

iv) High electronegativity
Answer:
Xhas high electronega^vity.

v) More metallic,character
Answer:
Y has more metallic character.

Question 16.
How does metallic character change when we move
i) Down a group?
ii) Across a period?
Answer:
i) Down a group :
When we move from top to bottom in a group, the metallic character increases.

ii) Across a period:
When we move left to right in a period, the metallic character decreases.

Question 17.
Why was the basis of classification of elements changed from the atomic mass to the atomic number? (AS1)
(OR)
Which atomic property is more suitable for classification of elements? Why?
Answer:

• The first attempt to classify elements was made by Dobereiner.
• Dobereiner’s attempt gave a clue that atomic masses could be correlated with properties of elements:
• Newlands’ law of octaves also followed the same basis for classification but this law is not valid for the elements that had atomic masses higher than calcium.
• Mendeleeff’s classification also based on the atomic masses of elements, but it lead to some limitations like Anomalous pair of elements and Dissimilar elements placed together.
• Moseley by analyzing the X-ray patterns of different elements was able to calculate the number of positive charges in the atoms of respective elements.
• With this analysis, Moseley realized that the atomic number is more fundamental
  characteristic of an element than its atomic weight. ,
• So, he arranged the elements in the periodic table according to the increasing order of their atomic number.
• This arrangement eliminated the problem of anomalous series and dissimilar elements placed together in Mendeleeff’s classification.

Question 18.
What is a periodic property? How do the following properties change in a group and period? Explain. (AS1)
I. a) Atomic radius
b) Ionization energy
c) Electron affinity
d) Electronegativity
II. Explain the ionization energy order in the following sets of elements: (AS1)
a) Na, Al, Cl
b) Li, Be, B
c) C, N, O
d) F, Ne, Na
e) Be, Mg, Ca
Answer:
Periodic property:
The property in which there shall be a regular gradation is called periodic property.

I. a) Atomic radius :
Period :
Atomic radius of elements decreases across a period from left to right because the nuclear charge increases due to increase in atomic number.

Group :
Atomic radius increases from top to bottom in a group due to addition of new shell.

b) Ionization energy:
Period :
When we move from left to right it does not follow a regular trend but generally increases due to increase in atomic number.

Group :
In a group from top to bottom, the ionization energy decreases due to increase in atomic size. –

c) Electron affinity:
Period :
Electron affinity values increase from left to right in a period.

Group :
Electron affinity values decrease from top to bottom in a group.

d) Electronegativity :
Period :
Electronegativity increases from left to right in a period.

Group :
Electronegativity decreases from top to bottom in a group.

II. Ionization energy order :
a) Na, Al, Cl
b) Li, Be, B
c) C, N, O
d) F, Ne, Na
e) Be, Mg, Ca
Answer:
a) In a period ionisation energy increases so the order is Na < kl < Cl.
b) Beryllium has stable configuration 1s² 2s². So it has more ionisation energy. So the order is Li < B < Be.
c) Nitrogen has half-filled p-orbitals. So it has greater ionisation energy. So the order is C < O < N.
d) Ne is inert gas right to F. Whereas Na is a metal ion in third period. So, the order is Na < F < Ne. e) In a group ionisation energy decreases. So the order is Be > Mg > Ca.

Question 19.
Name two elements that you would expect to have chemical properties similar to Mg. What is the basis for your choice? (AS2)
Answer:

• The two elements which have chemical properties similar to Magnesium are Beryllium and Calcium.
• The basis for my expectation is that they belong to same group as we know elements belonging to same group have similar properties.

Question 20.
On the basis of atomic numbers predict to which block the elements with atomic number 9, 37, 46 and 64 belong to? (AS2)
Answer:

1. The element with atomic number 9 belongs to p-block.
2. The element with atomic number 37 belongs to s-block.
3. The element with atomic number 46 belongs to d-block.
4. The element with atomic number 64 belongs to f-block.

Question 21.
Using the periodic table, predict the formula of compound formed between and element X of group 13 and another element Y of group 16. (AS2)
Answer:
The valency of 13^th group elements is 3.
The valency of 16^th group elements is 2.
The formula of compound is X₂Y₃.

Question 22.
An element X belongs to 3rd period and group 2 of the periodic table. State (AS2)
a) The no. of valence electrons
b) The valency.
c) Whether it is metal or a non-metal.
Answer:
a) The number of valence electrons are 2.
b) The valency of element is +2.
c) It is a metal.

Question 23.
An element has atomic number 19. Where would you expect this element in the periodic table and why? (AS2)
Answer:
The clement with atomic number 19 is in 4^th period and first group of the periodic table.
Reason :

1. Electronic configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s or [Ar]4s¹
2. The differentiating electron enters into 4^th shell. Hence it belongs to 4^th period.
3. The differentiating electron is in ‘s’ orbital. So it belongs to ‘s’ block.
4. The outermost orbital has only one electron. Hence it belongs to first group.

Question 24.
Aluminium does not react with water at room temperature but reacts with both dil. HCl and NaOH solutions. Verify these statements experimentally. Write your observations with chemical equations. From these observations, can we conclude that Al is a metalloid? (AS3)
Answer:

• Aluminium reacts with dil. HCl and releases hydrogen gas with formation of Aluminium chloride.
  
• Aluminium reacts with NaOH solution and releases hydrogen gas.
• 
• The above two reactions says that Aluminium is amphoteric.
• Aluminium does not react with water at room temperature.
• This concludes that the properties of Aluminium are in between a metal and non¬metal. So it behaves like a metalloid.

Question 25.
Collect the information about reactivity of VIIIA group elements (noble gases) from internet or from your school library and prepare a report on their special character when compared to other elements of periodic table. (AS4)
Answer:
Reactivity of Noble gases :

• The noble gases show extremely low chemical reactivity.
• He and Ne do not form chemical compounds.
• Xenon, krypton and argon show only minor reactivity.
• The reactivity order follows like this : Ne < He < Ar < Kr < Xe < Rn.
• Xenon can form compounds like XeF₂, XeF₄ and XeF₆, etc.

Reasons for low reactivity :

• The extremely low reactivity of noble gases is due to stable electronic configuration.
• But as we move from top to bottom the reactivity increases. So xenon can form some compounds with high electronegative elements.

Question 26.
Collect information regarding metallic character of elements of IA group and prepare report to support the idea of metallic character increases in a group as we move from top ro bottom. (AS4)
Answer:
Metallic character of IA group elements :

1. Alkali metals exhibit many of the physical properties common to metals but their densities are lower than those of other metals.
2. Alkali metals have one electron in their outer shell which is loosely bound.
3. They have largest atomic radii of the elements in their respective periods.
4. The lower ionization energies result in their metallic properties and high reactivities.
5. An alkali metal can easily lose its valence electron to form positive ion.
6. So they have greater metallic character.
7. The metallic character increases as we move from top to bottom in group due to addition of another shell, it is easy to lose electron.

Question 27.
How do you appreciate the role of electronic configuration of the atoms of elements in periodic classification? (AS6)
(OR)
How does electronic configuration help in the classification of elements in modern periodic table?
Answer:
The quantity is electronic configuration.

1. Modern periodic table is based on electronic configuration. So elements are arranged in ascending order of their atomic numbers.
2. The chemical properties of elements depend on valence electrons. The elements in same group have same number of valence electrons. So the elements belonging to same group have similar properties.
3. So the construction of modern periodic table mainly depends on electronic configuration.
4. Thus electronic configuration plays a major role in the preparation of modern periodic table. So its role is thoroughly appreciated.

Question 28.
Without knowing the electronic configurations of the atoms of elements Mendeleeff still could arrange the elements nearly close to the arrangements in the Modern periodic table. How can you appreciate this? (AS6)
Answer:

• Mendeleeff took consideration about chemical properties while arranging the elements. So the arrangement of elements is close to arrangement of elements in Modern periodic table.
• For this, he violated his periodic law.
• He left some gaps for elements, later those elements are discovered.
• So the efforts of Mendeleeff should be thoroughly appreciated.

Question 29.
Comment on the position of hydrogen in periodic table. (AS7)
Answer:

• Hydrogen is the element which has easier atomic structure than any other element.
• Electron configuration of hydrogen is Is1. It has one proton in its is nucleus and one electron in its is orbital.
• Hydrogen combines with halogens, oxygen and sulphur to form compounds having similar formulae just like alkali metals.
• Similarly, just like halogens, hydrogen also exists as diatomic molecule and combine with metals and non-metals to form covalent compounds.
• As alkali metals hydrogen can lose one electron and accept one electron as halogens.
• So in periodic table, its place may be in IA or VIIA group.
• But based on electronic configuration of hydrogen, it is placed in IA group.

Question 30.
How do the positions of elements in the periodic table help you to predict its chemical properties? Explain with an example. (As7)
Answer:
1) The physical and chemical properties of atoms of the elements depend on their electronic configuration, particularly the outer shell configurations.

2) Elements are placed in the periodic table according to the increasing order of their electronic configuration.

3) The elements in a group possess similar electronic configurations. Therefore all the elements in a group should have similar chemical properties.
Ex : Consider K

• It is the element in 4^th period 1^st group.
• Electron configuration : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.
• Differentiating electron enters into s-orbital. Hence it belongs to s-block.
• It is on the left side of the periodic table. Hence it is a metal.
• It is ready to lose one electron to get octet configuration. Hence its reactivity is more.
• It is Alkali metal.
• All alkali metals react with both acids and bases and releases H₂ gas.

Fill In The Blanks

1. Lithium, ……………… and potassium constitute a Dobereiner’s triad.
2. ……………… was the basis of the classifications proposed by Dobereiner, Newlands, and Mendeleeff.
3. Noble gases belong to ……………… group of periodic table.
4. The incomplete period of the modern periodic table is
5. The element at the bottom of a group would be expected to show …………….. metallic character than the element at the top

Answer:

1. Sodium
2. Atomic weight
3. VIIIA or 18 group
4. 7^th
5. higher

Multiple Choice Questions

1. Number of elements present in period – 2 of the long form of periodic table …………
A) 2
B) 8
C) 18
D) 32
Answer:
B) 8

2. Nitrogen (Z = 7) is the element of group V of the periodic table. Which of the following is the atomic number of the next element in the group?
A) 9
B) 14
C) 15
D) 17
Answer:
C) 15

3. Electronic configuration of an atom is 2, 8, 7. To which of the following elements would it be chemically similar?
A) Nitrogen (Z = 7)
B) Fluorine (Z = 9)
C) Phosphorous (Z – 15)
D) Argon (Z = 18)
Answer:
B) Fluorine (Z = 9)

4. Which of the following is the most active metal?
A) lithium
B) sodium
C) potassium
D) rubidium
Answer:
D) rubidium

10th Class Chemistry 9th Lesson Classification of Elements-The Periodic Table InText Questions and Answers

10th Class Chemistry Textbook Page No. 129

Question 1.
Can you establish the same relationship with the set of elements given in the remaining rows?
Answer:
Yes, we can establish the approximately same relationship between other elements given in the table.

Question 2.
Find average atomic weights of the first and third elements in each row and compare it with the atomic weight of the middle element. What do you observe?
Answer:
The atomic weight of middle element is arithmetic mean coverage of first and third elements.

10th Class Chemistry Textbook Page No. 135

Question 3.
What is atomic number?
Answer:
The number of positive charges (protons) in the atom of element is the atomic number of element.

10th Class Chemistry Textbook Page No. 142

Question 4.
How does the valency vary in a period on going from left to right?
Answer:
It does not follow a regular trend when we move from left to right in a period. First, it increases and then decreases and finally ‘O’ for inert gases.

Question 5.
How does the valency vary on going down a group?
Answer:
The valency is constant when we move from top to bottom in a group because the number of valence electrons are same for same group elements.

10th Class Chemistry Textbook Page No. 144

Question 6.
Do the atom of an element and its ion have same size?
Answer:
No, the positive ion has smaller size than neutral atom whereas negative ion has greater size than neutral atom.

Question 7.
Which one between Na and Na⁺ would have more size? Why?
Answer:

• The atomic number of Sodium is 1 and it has 11 protons and 11 electrons with outer electron as 3s¹ whereas Na⁺ ion has 11 protons but only 10 electrons.
• The 3s shell of Na⁺ has no electron in it.
• So the outer shell configuration is 2s²2p⁶.
• As proton number is more than electrons, the nucleus of Na⁺ ion attracts outer shell electrons with strong nuclear force.
• As a result the Na⁺ ion shrinks in size.
• Therefore, the size of Na⁺ ion is less than Na atom.

Question 8.
Which one between Cl and Cl^– would have more size? Why?
Answer:

• The electronic configuration of chlorine (Cl) atom is 1s2 2s² 2p⁶ 3s² 3p⁵ and the electronic configuration of chloride (Cl^–) ion is 1s² 2s² 2p⁶ 3s² 3p⁶.
• Both chlorine and chloride ions have 17 protons each but there are 17 electrons in chlorine atom, whereas 18 electrons in chloride ion.
• Therefore, the nuclear attraction is less in Cl^– ion when compared with chlorine atom.
• Therefore the size of the chlorine (Cl) atom is less size than chloride of Cl^– ion.

Question 9.
Which one in each of the following pairs is larger in size? Why?
a) Na, Al
b) Na, Mg⁺²
c) S^2-, Cl^–
d) Fe²⁺, Fe³⁺
e) C^4-, F^–.
Answer:
a) Na has larger size because Sodium and Aluminium are third period elements in which Na is left to Al. As we move from left to right in a period atomic size decreases.

b) Mg²⁺ has smaller size because Mg²⁺ has 10 electrons and 12 protons whereas Na has 11 electrons and 11 protons. So the distance between nucleus and outermost orbital is less in Mg²⁺ due to greater nuclear attraction.

c) S^2- has, larger size because S^2- has 18 electrons and 16 protons and Cl^– has 18 electrons and 17 protons. So nuclear attraction over outermost orbital is more in Cl^– when compared with S^2-. So S^2- has larger size.

d) In Fe²⁺ it has 26 protons and 24 electrons whereas for Fe³⁺ it has 26 protons and 23 electrons. So nuclear attraction over outermost orbital is more in Fe³⁺. So Fe³⁺ has smaller size (or) Fe²⁺ has larger size.

e) C^4- has 6 protons and 10 electrons whereas F^– has 9 protons and 10 electrons. So nuclear attraction is less in C^4-. So size of C^4- is more than F^–.

10th Class Chemistry Textbook Page No. 129

Question 10.
What relation about elements did Dobereiner want to establish?
Answer:
Dobereiner wanted to give a relationship between the properties of elements and their atomic weights.

Question 11.
The densities of calcium (Ca) and barium (Ba) are 1.55 and 3.51 gem-3 respectively. Based on Dobereiner’s law of triads can you give the approximate density of strontium (Sr)?
Answer:
Molecular weight is directly proportional to density.

So density of strontium is mean of calcium and barium according to Dobereiner.
∴ Density of strontium = \(\frac{1.55+3.51}{2}\) = 2.53.

10th Class Chemistry Textbook Page No. 130

Question 12.
Do you know why Newlands proposed the law of octaves? Explain your answer in terms of the modern structure of the atom.
Answer:

• John Newlands found that when elements were arranged in the ascending order of their atomic weights, they appeared to fall into seven groups.
• Each group contained elements with similar properties.
• If we start with hydrogen and move down, the next eighth element is fluorine, and then next eighth element is chlorine and the properties of these elements are similar.
• Similarly, if we start from Lithium their eighth element is Sodium and next eighth element is potassium. These show similar properties.

Question 13.
Do you think that Newlands’ law of octaves is correct? Justify.
Answer:
No, there are some limitations of Newlands’ model:

• There are instances of two elements fitted into the same slot. Eg : Cobalt and Nickel.
• Certain elements, totally dissimilar in their properties, were fitted into the same group.
• Law of octaves holds good only for the elements up to Calcium.
• Newlands’ periodic table was restricted to only 56 elements and did not leave any room for new elements.
• Newlands had taken consideration about active pattern sometimes without caring the similarities.

10th Class Chemistry Textbook Page No. 134

Question 14.
Why did Mendeleeff have to leave certain blank spaces in his periodic table? What is your explanation for this?
Answer:
1) Mendeleeff predicted that some elements which have similar properties with the elements in a group are missing at that time.
2) So he kept some blanks in the periodic table by writing ’eka’ to the name of the element immediately above the empty space.
3) Later these elements are discovered and they are fitted into those empty spaces.

Question 15.
What is your understanding about Ea₂O₃, EsO₂?
Answer:

• Mendeleeff predicted that after aluminium there was another element namely eka- aluminium (Ea) and after silicon, there was another element namely eka-silicon (Es).
• He also gave the formulae of those oxides as Ea203 and Es02.
• Later those elements are discovered namely gallium and germanium and Ea₂O₃ and EsO₂ as Ga₂O₃ and GeO₂.

10th Class Chemistry Textbook Page No. 135

Question 16.
All alkali metals are solids but hydrogen is a gas with diatomic molecules. Do you justify the inclusion of hydrogen in first group with alkali metals?
Answer:
No, hydrogen shows the properties of both alkali metals and halogens. Still the position of hydrogen has some questions. So it was kept just above alkali metals in first group.

10th Class Chemistry Textbook Page No. 141

Question 17.
Why are lanthanoids and actinoids placed separately at the bottom of the periodic table?
Answer:
The properties of these elements do not coincide with other elements because the valence electron enters 4f and 5f orbitals respectively. So they are placed separately at the bottom of the periodic table.

Question 18.
If lanthanoids and actinoids are inserted within the table, imagine how the table would be?
Answer:
It looks very big in size, and it is difficult to identify, as these elements have similar properties.

10th Class Chemistry Textbook Page No. 145

Question 19.
Second ionization energy of an element is higher than its first ionization energy. Why?
Answer:

• The energy required to remove an electron from unipositive ion is called second ionisation energy.
• It is difficult to remove an electron from unipositive ion when compared with neutral atom due to an increase in nuclear attraction.
• So always second ionisation energy is higher than first ionisation energy.

10th Class Chemistry Textbook Page No. 146

Question 20.
The calculated electron gain enthalpy values for alkaline earth metals and noble gases are positive. How can you explain this?
Answer:

• Generally alkaline earth metals having one, two or three valence electrons prefer to lose electrons in order to get inert gas configuration. So it is difficult to add electron to alkaline earth metals. So they have positive electron gain enthalpy values.
• Noble gases are stable. So they do not prefer to take electrons. So they have positive electron gain enthalpy.

Question 21.
The second period element, for example, ‘F’ has less electron gain enthalpy than the third period element of the same group for example ‘Cl’. Why?
Answer:

• Electron gain enthalpy values decrease in a group as we go down and increase from left to right along a period.
• But the size of Fluorine is small compared chlorine.
• So it is difficult to add electron to fluorine.
• So fluorine has less electron gain enthalpy.

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table Activities

Activity – 1

Question 1.
Observe the following table. Establish the relationship of other elements given in the table.

Answer:

Activity – 2

Question 2.
Some main group elements of s-block and p-block have family names as given in the following table.
Observe the long form of a periodic table and complete the table with proper information.

Answer:

Activity – 3

Question 3.
Collect valencies of first 20 elements.
Answer:

AP State Board Syllabus AP SSC 10th Class Chemistry Solutions Chapter 8 Structure of Atom Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Chemistry Solutions 8th Lesson Structure of Atom

10th Class Chemistry 8th Lesson Structure of Atom Textbook Questions and Answers

Improve Your Learning

Question 1.
What information does the electronic configuration of an atom provide? (AS1)
Answer:

• The distribution of electrons in shells, sub-shells and orbital in an atom is known as electronic configuration.
• It provides the information of position of an electron in the space of atom.
• The distribution of electrons in various atomic orbitals provides an understanding of the electronic behaviour of the atom and in turn its reactivity.
• The short hand notation is as shown below.

Question 2.
a) How many maximum number of electrons that can be accommodated in a principal energy shell?
Answer:
The maximum number of electrons that can be accommodated in a principal energy shell is 2n². Here n is principal quantum number.

b) How many maximum number of electrons that can be accommodated in a sub-shell?
Answer:
The maximum number of electrons that can be accommodated in a sub-shell is 2(2l +1) (where l is orbital quantum number).

c) How many maximum number of electrons can that be accommodated in an orbital?
Answer:
The maximum number of electrons that can be accommodated in an orbital is 2.

d) How many sub-shells are present in a principal energy shell?
Answer:
The number of sub-shells in a principal energy shell is n (n is principal quantum number).

e) How many spin orientations are possible for an electron in an orbital?
Answeer:
The spin orientations possible for an electron in an orbital are 2.

Question 3.
In an atom the number of electrons in M-shell is equal to the number of electrons in the K and L-shell. Answer the following questions. (AS1)
a) Which is the outermost shell?
Answer:
The outermost shell is N shell.

b) How many electrons are there in its outermost shell?
Answer:
Two electrons are there in outermost shell.

c) What is the atomic number of element?
Answer:
The atomic number of element is 22.

d) Write the electronic configuration of the element.
Answer:
The element is Ti (Titanium). Its electronic configuration is 1s²2s²2p⁶3s²3p⁶4s²3d².

Reason :

• Electrons enter M shell after completion of K and L.
• So the number of electrons in M shell is 10.
• But after completion of 3p orbital electron enters 4s before entering to 3d.
• So outermost orbit or shell is N shell.
• So the atomic number of element is 22.
• Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d².

Question 4.
Rainbow is an example for continuous spectrum – explain. (AS1)
(OR)
Which is naturally occurring continuous spectrum ? Explain.
Answer:

• Rainbow is a spectrum of different colours (VIBGYOR) with different wavelengths.
• These colours are continuously distributed.
• There is no fixed boundary for each colour.
• Hence, rainbow is a continuous spectrum.

Question 5.
How many elliptical orbits are added by Sommerfeld in third Bohr’s orbit ? What was the purpose of adding these elliptical orbits? (AS1)
Answer:
Sommerfeld added two elliptical orbits to Bohr’s third orbit.

Purpose of adding elliptical orbits :

• Bohr’s model failed to account for splitting of line spectra and line spectrum.
• In an attempt to account for the structure of line spectrum, Sommerfeld modified Bohr’s atomic model by adding elliptical orbits.

Question 6.
What is absorption spectrum?
Answer:
Absorption spectrum: The spectrum formed by the absorption of energy when electron jumps from lower energy level to higher energy level is called absorption spectrum. It contains dark lines on bright background.

Question 7.
What is an orbital? How it is different from Bohr’s orbit? (AS1)
(OR)
Comparison between orbit and orbital.
Answer:
The region of space around the nucleus where the probability of finding electron is maximum is called orbital. Whereas orbit is the path of the electron around the nucleus.

These two are differentiated like this.

Orbit	Orbital
1. Path of electron around nucleus.	1) Probability of finding electron around nucleus.
2. Orbits are represented by letters K, L, M, N, 0, …….etc.	2. Orbitals are represented by letters s, p, d, f, g, …….etc.
3) Its information is given by principal	3) Its information is given by orbital quantum number.
4) It is two dimensional.	4) It is three dimensional.
5) It does not satisfy Heisenberg’s uncertainty principle.	5) It satisfies the Heisenberg’s principle of uncertainty.

Question 8.
Explain the significance of three quantum numbers in predicting the positions of an electron in an orbit. (AS1)
(OR)
How are quantum numbers helpful to understand the atomic structure?
Answer:
Significance of three quantum numbers in predicting the positions of an electron in an orbit.

1) Principal quantum number (n) :
The principal quantum number explains about the size and energy of shells (or) orbitals. It is denoted by n.

As ‘n’ increases, the orbitals become larger and the electrons in those orbitals are farther from the nucleus.

It takes values 1, 2, 3, 4, ……………. for that the shells are represented by letters K, L, M, N, ……….

The number of electrons in a shell is limited to 2n².

2) The Angular – momentum quantum number (l) :
The angular momentum quantum number defines the shape of the orbital occupied by the electron and the orbital angular momentum of the electron, is in motion.

l takes values from 0 to n – 1 for these values the orbitals are designated by letters s, p, d, f, ………….. etc.

l also governs the degree with which the electron is attached to nucleus. The larger the value of l, the smaller is the bond with which it is maintained with the nucleus.

3) Magnetic orbital quantum number (m_l) :
The orientation of orbital with external magnetic field determines magnetic orbital quantum number.

m_l has integer values between – l and l including zero.

The number of values for m, are 2l + l, which give the number of orbitals per sub-shell. The maximum number of electrons in orbitals in the sub-shell is 2 (2l + l).

Question 9.
What is nl^x method? How is it useful? (AS1)
(OR)
What is nl^x method? How is it useful in electronic configuration?
Answer:
The shorthand notation consists of the principal energy level (n value) the letter representing sub – level (l value), and the number of electrons (x) in the sub-shell is written as superscript nl^x.

It is useful in writing electron configuration of elements. For example, in Hydrogen (H), the set of quantum numbers is n = 1, l = 0, m_l = 0, m_s = ½ or – ½. The electronic configuration is

Question 10.
Following orbital diagram shows the electronic configuration of nitrogen atom. Which rule does not support this? (AS1)

(OR)
Write the correct electronic configuration of the given nitrogen atom with the help of Hund’s rule.
Answer:

• This electron configuration does not support Hund’s rule.
• According to Hund’s rule, the orbitals of equal energy are occupied with one elec-tron each before pairing of electrons starts.
• Here, pairing of electrons in 2p_x orbital was taken place without filling of an elec-tron in 2p_z orbital.
• Hence the correct electron configuration is as follows.

Question 11.
Which rule is violated in the electronic configuration 1s⁰ 2s² 2p⁴?
Answer:

• Aufbau principle is violated in this electronic configuration because according to Aufbau principle, electron enters orbital of lowest energy.
• Among 1s, 2s and 2p, Is has least energy.
• So Is orbital must be filled before the electron should enter 2s.

Question 12.
Write the four quantum numbers for the differentiating electron of sodium (Na) atom. (AS1)
Answer:
The electronic configuration of sodium (Na) is 1s² 2s² 2p⁶ 3s¹. So the differentiating electron enters 3s. Therefore the four quantum numbers are

Question 13.
What is emission spectrum?
(OR)
When radiation is emitted what is the name given to such spectrum? Explain such spectrum.
Answer:

• The spectrum produced by the emitted radiation is known as emission spectrum.
• This spectrum corresponds to liberation of energy when an excited electron returns back to ground state.

Emission spectrum is of two types :

1) Continuous spectrum :
When white light passes through a prism it dissociates into seven colours. This spectrum is called continuous spectrum.

2) Discontinuous spectrum :
Discontinuous spectrum is of two types.

a) Line spectrum :
The spectrum with sharp and distinct lines. It is given by gaseous atoms.

b) Band spectrum :
The spectrum very closely spaced lines is known as band spectrum. It is given by molecule.

Question 14.
i) An electron in an atom has the following set of four quantum numbers to which orbital it belong to : (AS2)
Answer:

This electron belongs to 2s orbital.
Spin is in clockwise direction. ⇒ 2s¹

ii) Write the fojur quantum numbers for Is1 electron. (AS1)
Answer:
The four quantum numbers for Is1 electron are

Question 15.
Which electronic shell is at a higher energy level K or L? (AS2)
Answer:
L – shell is at higher energy level, because it is far from nucleus than K shell.

Question 16.
Collect the information regarding wavelengths and corresponding frequencies of three primary colours red, blue and green. (AS4)
Answer:
The wavelengths and corresponding frequencies of three primary colours red, blue and green are given below.

Primary colours	Wavelength in nm (1 nm = 10-9m)	Frequency in Hz (Hertz)
Red	700	4.29 × 1014
Green	530	5.66 × 1014
Blue	470	6.38 × 1014

Question 17.
The wavelength of a radio wave is 1.0 m. Find its frequency. (AS7)
Answer:
c = 3 × 10⁸ m/s ; λ = 1m ; c = vλ ⇒ v = \(\frac{\mathrm{c}}{\lambda}=\frac{3 \times 10^{8}}{1}\) = 3 × 10⁸ Hz.

Question 18.
Why are there exemptions in writing the electronic configurations of Chromium and Copper?
Answer:
1. Elements which have half-filled or completely filled orbitals have greater stability.

2. So in chromium and copper the electrons in 4s and 3d redistributes their energies to attain stability by acquiring half-filled and completely filled d-orbitals.

3. Hence the actual electronic configuration of chromium and copper are as follows.

Fill In The Blanks

1. If n = 1, then angular momention quantum number (l) = …………………
2. If a sub-shell is denoted as 2p, then its magnetic quantum number values are …………………, …………………, …………………
3. Maximum number of electrons that an M-shell contain is / are …………………
4. For ‘n’, the minimum value is ………………… and the maximum value is …………………
5. For?, the minimum value is ………………… and the maximum value is …………………
6. For’m/ the minimum value is ………………… and the maximum value is …………………
7. The value of ‘ms’ for an electron spinning in clockwise direction is ………………… and for anti-clockwise direction is …………………
Answer:

1. 0
2. – 1, 0, + 1
3. 18
4. 1, – ∞
5. 0, (n – 1)
6. – l, + l
7. + ½, – ½

Multiple Choice Questions

1. An emission spectrum consists of bright spectral lines on a dark back ground. Which one of the following does not correspond to the bright spectral lines?
A) Frequency of emitted radiation
B) Wavelength of emitted radiation
C) Energy of emitted radiations
D) Velocity of light
Answer:
D) Velocity of light

2. The maximum number of electrons that can be accommodated in the L-shell of an atom is
A) 2
B) 4
C) 8
D) 16
Answer:
C) 8

3. If l = 1 for an atom, then the number of orbitals in its sub-shell is
A) 1
B) 2
C) 3
D) 0
Answer:
C) 3

4. The quantum number which explains about size and energy of the orbit or shell is
A) n
B) l
C) m_l
D) m_s
Answer:
A) n

10th Class Chemistry 8th Lesson Structure of Atom InText Questions and Answers

10th Class Chemistry Textbook Page No. 112

Question 1.
How many colours are there in a rainbow? What are they?
Answer:
There are seven colours in a rainbow. They are Violet, Indigo, Blue, Green, Yellow,Orange and Red.

Question 2.
What are the characteristics of electromagnetic waves?
A.nswer:
Electromagnetic energy is characterised by wavelength (l) and frequency (u).

10th Class Chemistry Textbook Page No. 113

Question 3.
Can we apply this equation c = uA, to a sound wave?
Answer:
Yes. It is a universal relationship and applies to all waves.

10th Class Chemistry Textbook Page No. 114

Question 4.
What happens when you heat an iron rod on a flame? Do you find any change in colour while heating an iron rod?
Answer:

• When we heat an iron rod, some amount of heat energy that was absorbed by iron rod is emitted as light.
• First iron turns into red (lower energy corresponding to higher wavelength) and as the temperature rises it glows and turns into orange, yellow, blue or even white respectively (higher energy and lower wavelength).
• If we go on heating the rod, it turns into white light which includes all visible wavelengths.
• So we find some changes in colour while heating an iron rod.

Question 5.
Do you observe any other colour at the same time when one colour is emitted?
Answer:
While heating the rod if the temperature is high enough, other colours will also be emitted, but due to higher intensity of one particular emitted colour (eg.: red), others cannot be observed.

Question 6.
How do various colours come from fire works?
(OR)
Do you enjoy Deepavali fire works? Variety of colours is seen from fire works. How do these colours come from fire works?
Answer:
Yes. The electrons present in atoms of elements absorb energy and move to excited states and they return to ground state with emission of energy in visible spectrum. So the colours observed during fire works are the emitted energy by various elements in different fire works.

10th Class Chemistry Textbook Page No. 115

Question 7.
Do you observe yellow light in street lamps? Which will produce yellow light?
Answer:
Yes, sodium vapours produce yellow light in street lamps.

Question 8.
Why do different elements emit different flame colours when heated by the same non-luminous flame?
Answer:

• All the materials are made up of atoms and molecules. These atoms and molecules possess certain fixed energy.
• An atom or molecule having lowest possible energy is said to be in ground state.
• When we heat the materials the electrons of these atoms gain energy and move to excited states (higher energy state).
• An atom of molecule in excited state can emit light to lower its energy in order to get stability and come back to ground state.
• Light emitted in such process has certain fixed wavelength for one kind of atoms.
• The light emitted by different kinds of atoms is different because the excited states electrons will go are different. So different elements produce different flame colours.

Question 9.
What happens when an electron gains energy?
Answer:
The electron moves to higher energy level called the excited state.

10th Class Chemistry Textbook Page No. 116

Question 10.
Does the electron retain the energy forever?
Answer:
The electron loses the energy and comes back to its ground state. The energy emitted by the electron is seen in the form of electromagnetic energy.

Question 11.
Did Bohr’s model account for the splitting of line spectra of a hydrogen atom into finer lines?
Answer:
No, Bohr’s model failed to account for splitting of line spectra.

Question 12.
Why is the electron in an atom restricted to revolve around the nucleus at certain fixed distances?
Answer:
In order to explain the atomic spectra, Bohr-Sommerfeld model proposed that the electrons are restricted to revolve around the nucleus at certain fixed distances.

10th Class Chemistry Textbook Page No. 117

Question 13.
Do the electrons follow defined paths around the nucleus?
Answer:
No, they revolve around the nucleus in a region called orbital.

Question 14.
What is the velocity of the electron?
Answer:
It is very close to light.

Question 15.
Is it possible to find exact position of electron? How do you find the position and velocity of an electron?
Answer:
No, as the electrons are very small, light of very short wavelength is required for this task.

This short wavelength light interacts with the electron and disturbs the motion of electron. So it is not possible to find exact position and velocity of electron simultaneously. Whereas we can find the region where the probability of finding electron is more.

Question 16.
Do atoms have a definite boundary, as suggested by Bohr’s model?
Answer:
Yes, atoms have definite boundary.

Question 17.
What do we call the region of space where the electron might be, at a given time?
Answer:
The region of space around the nucleus where the probability of finding an electron is maximum, called an orbital.

10th Class Chemistry Textbook Page No. 118

Question 18.
What information do the quantum numbers provide?
Answer:
The quantum numbers describe the space around the nucleus where the electrons are found and also their energies.

Question 19.
What does each quantum number signify?
Answer:
The quantum numbers signify the probability of finding electron in the space around nucleus.

10th Class Chemistry Textbook Page No. 119

Question 20.
What is the maximum value of/for n = 4?
Answer:
The maximum value of / for n = 4 is 3.

Question 21.
How many values can l have for n = 4?
Answer:
l takes values from 0 to n – 1. So l has 4 values for n = 4. Those values are 0, 1,2, 3.

Question 22.
Do all the p-orbitals have the same energy? A. Orbitals in the sub-shell belonging to same shell possess same energy but they differ in their orientations.

10th Class Chemistry Textbook Page No. 121

Question 23.
How are two electrons in the Helium atom arranged?
Answer:
They are arranged in pair in Is orbital and the electronic configuration is 1s².

10th Class Chemistry Textbook Page No. 122

Question 24.
What are the spins of two electrons in an orbital?
Answer:
The two electrons in an orbital have opposite spins. If one is clockwise spin, then other electron has anti-clockwise spin.

Question 25.
How many electrons can occupy an orbital?
Answer:
An orbital can hold only two electrons.

10th Class Chemistry 8th Lesson Structure of Atom Activities

Activity – 1

Question 1.
Explain the wave nature of light.
(OR)
How does light behave? Explain.
Answer:

• Light is an electromagnetic wave.
• Electromagnetic waves are produced when an electric charge vibrates.
• This vibrating electric charge creates a change in the electric field. The changing electric field creates a changing magnetic field.
• This process continues with both the created fields being perpendicular to each other and at right angles to the direction of propagation of the wave.
• This electromagnetic wave is produced.

Activity – 2

Question 2.
Write an activity which shows metal produces colour in flame.
(OR)
‘Metal produces colour in a flame.’ Prove the statement by giving examples.
Answer:
A)

• Take a pinch of cupric chloride in a watch glass and make a paste with concentrated hydrochloric acid.
• Take this paste on a platinum loop and introduce it into a non-luminous flame.
• Cupric chloride produces a green colour flame.

B)

• Take a pinch of strontium chloride in a watch glass and make a paste with concentrated hydrochloric acid.
• Take this paste on a platinum loop and introduce it into a non-luminous flame.
• Strontium chloride produces a crimson red flame.

Activity – 3

Question 3.
Complete the electronic configuration of the following elements.

Answer:

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 7 Human Eye and Colourful World Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 7th Lesson Human Eye and Colourful World

10th Class Physics 7th Lesson Human Eye and Colourful World Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
What is the function of lens in human eye?
Answer:
The eye lens forms a real image on retina then we can see the objects.

Question 2.
How does it help to see objects at long distances and short distances?
Answer:
Eye lens can adjust itself in shape, so that it helps to see the objects at long and short distances.

Question 3.
How is it possible to get the image at the same distance on the retina?
Answer:
When the eye is focussed at distant object, the ciliary muscles are relaxed. So the focal length of eye lens adjusted itself which is equal to the distance of object from the retina. Then we can see the object clearly.
When the eye is focussed on a closer object the ciliary muscles adjust the focal length in such a way that the image is formed on retina and we see the object clearly.

Question 4.
Are we able to see all objects in front of our eye clearly?
Answer:
Yes. By maintaining the 25 cm distance of the object from our eye, we can see the objects in front of our eye.

Question 5.
How do the lenses used in spectacles correct defects of vision?
Answer:
To form image on retina.
To correct
myopia- Concave lens
hypermetropia i- Convex lens

Improve Your Learning

Question 1.
How do you correct the eye defect Myopia? (AS1)
Answer:

• Some people cannot see objects at long distances but can see nearby objects clearly. This type of defect in vision is called ‘Myopia’ or ‘Near sightedness’.
• Myopia is corrected by using a con-cave lens of focal length equal to the distance of the far point F from the eye.
• This lens diverges the parallel rays from distant object as if they are coming from the far point.
• Finally the eye lens forms a clear im-age at the retina.

Question 2.
Explain the correction of the eye defect Hypermetropia. (AS1)
Answer:

• A person with hypermetropia can see distant objects clearly but cannot see objects at near distances. This is also known as ‘far sightedness’.
• Eye lens can form a clear image on the retina when any object is placed beyond near point.
• To correct the defect of hypermetropia, we need to use a lens which forms an image of an object beyond near point at H, when the object is between H and L. This is possible only when a double con¬vex lens is used.
• The image acts like an object for the eye lens. Hence final image due to eye is formed at retina.

Question 3.
How do you find experimentally the refractive index of material of a prism? (AS1)
(OR)
Write the experimental procedure in finding the refractive index of a prism.
(OR)
Which quantity will decide whether a given medium is denser or rarer? How do you find that quantity of prism experimentally?
Answer:
Refractive index decides whether a medium is denser or rarer.
Aim :
To find the refractive index of a prism.

Materials required :
Glass prism, white chart of size 20 x 20 cm, pencil, pins, scale and protractor.

Procedure :

1. Keep a prism on white chart.
2. Draw the triangular base of the prism with pencil.
3. Remove the prism.
4. The shape of the outline drawn prism is triangle and name its vertices as P, Q and R.
5. PQ and PR be the refracting surfaces.
6. Find the angle between PQ and PR. This is the angle of the prism (A) (or) Refracting angle.
7. Mark M on PQ and draw a perpendicular line to PQ at M.
8. Place the centre of the protractor at M, along the normal and mark an angle of 30° and then draw a line up to M. This line denotes incident ray. This angle is called angle of incidence.
9. Place the prism in its position (triangle) again.
10. Now fix two pins vertically on the line at points A and B .
11. See the images of pins through the 2nd refracting side (PR).
12. Fix another two pins at points C and D such that all the four pins appear to lie along the same line.
13. Remove the pins and prism, join the pin-holes. Draw the incident and emergent rays.
14. The angle between the normal and the emergent ray at N is the angle of emergence.
15. The line passing through the points A, B, M, N, C and D represents the path of light when it suffers refraction through prism.

The angle of deviation :

• Extend incident and emergent rays are intercept at a point ‘O’.
• The angle between these two rays is the angle of deviation (d).
• Note the emergent deviation angles for different values of i, in the given table.

• Draw the graph between angle of incidence on X – axis and the angle of deviation on Y – axis.
• We notice that the angle of deviation decreases first and then increases with increase of the angle of incidence.
• Mark points on a graph paper and join the points to obtain a graph (smooth curve).
• Draw a tangent line to the curve, which parallel to X-axis, at the lowest point of the graph.

• The point where this line cuts the Y- axis gives the angle of minimum deviation. It is denoted by D.
• From the graph, we notice that, at angle of minimum deviation, the angle of incidence is equal to the angle of emergence.
• By finding A and D we can find refractive index of prism by using formula .

Question 4.
Explain the formation of rainbow. (AS1)
(OR)
What is the natural spectrum occuring in sky? Explain the formation of that spectrum.
Answer:

• A rainbow is a natural spectrum of sunlight in the form of bows appearing in the sky when the sun shines on rain drops.
• It is combined result of reflection, refraction and dispersion of sunlight from water droplets, in atmosphere.
• It always forms in the direction opposite to the sun.
• To see a rainbow, the sun be must behind us and the water droplets falls in front of us.
• When a sunlight enters a spherical raindrop, it is refracted and dispersed. The different colours of light are bent in different angles.
• When different colours of light falls on the back inner surface of drop, it (Water drop) reflects (different colours of light) interwnally (total internal reflection).
• The water drops again refract the different colours, when it comes out from the raindrop.
• After leaving this different colours from the raindrop as rainbow, reach our eye. Thus, we see a rainbow.

Question 5.
Explain briefly the reason for the blue of the sky. (AS1)
Answer:

• The reason for blue sky is due to the molecules N₂ and O₂.
• The size of these molecules are comparable to the wavelength of blue light.
• These molecules act as scattering centres for scattering of blue light.
• So scattering of blue light by molecules of N₂ and O₂ is responsible for blue of the sky.

Question 6.
Explain two activities for the formation of artificial rainbow. (AS1)
(OR)
Give two activities for the formation of the artificial rainbow? And explain it.
(OR)
Suggest an experiment to create a rainbow in your classroom and explain the procedure.
Answer:
Activity -1 :

• Select a white coated wall on which the sun rays fdll.
• Stand in front of a wall in such a way that the sun rays fall on your back.
• Hold a tube through which water is flowing.
• Place your finger in the tube to obstruct the flow of water.
• Water comes out from small gaps between the tube and finger like a fountain.
• Observe the changes on wall while showering the water.

Activity – II : (Activity – 4)

• Take a metal tray and fill with water.
• Place a mirror in water such that it makes an angle to the water surface.
• Now focus white light on the mirror through the water.
• Keep a white card board sheet above the water surface.
• We may observe the colours VIBGYOR on the board.
• The splitting of white light into different colours (VIBGYOR) is called dispersion.
• So consider a white light is a collection of waves with different wavelengths.
• Violet has shortest wavelength, and red has longest wavelength.

Question 7.
Derive an expression for the refractive index of the material of a prism. (AS1)
Answer:
Derivation of formula for refractive index of a prism :
PQ and PR are refracting surfaces of prism, i, is an angle of incidence, i2 is an angle of emergence, is an angle of refraction of first surface and r2 is an angle of incidence on second surface.

1. From triangle OMN, we get
d = i₁ – r₁ + i₂ – r₂
∴ d = (i₁ + i₂)-(r₁+ r₂) ……… (1)
2. From triangle PMN, we have
∠A + (90° – r₁) + (90° – r₂) = 180°
⇒ r₁ + r₂ = A ……… (2)
3. From (1) and (2),
⇒ d = (i₁ + i₂) – A
⇒ A + d = i₁ + i₂
4. This is the relation between angle of incidence, angle of emergence, angle of deviation and angle of prism.

5. From Snell’s law, we know that
n₁ sin i = n₂ sin r

6. Let ‘n’ be the refractive index of the prism.
7. Using Snell’s law at M, refractive index of air n₁ = 1; i = i₁; n₂ = n; r = r₁.
⇒ sin i₁ = n sin r₁ …………. (4)
8. Similarly, at N, n₁ = n; i = r₂; n₂ = 1; i₁ = i₂
⇒ n sin r₂ = n sin i₂ ………… (5)
9. We know that at the angle of minimum deviation position (D), i.e. i₁ = i₂
10. We will notice that MN is parallel to the base of prism QR.

14. This is the formula for the refractive index of the prism.

Question 8.
Light of wavelength λ₁ enters a medium with refractive index n₂ from a medium with refractive index n₁ What is the wavelength of light in second medium? (AS1)
Answer:
Wavelength of first medium =λ₁
Refractive index of first medium = n₁
Let the wavelength of second medium = λ₂,
Refractive index of second medium = n₂
Light enters from first medium to second medium = \(\Rightarrow \frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{\lambda_{1}}{\lambda_{2}}\) (∵ υ = const)
\(\Rightarrow \lambda_{2}=\frac{\lambda_{1} \mathrm{n}_{1}}{\mathrm{n}_{2}}\)

Note:
For the below questions the following options are given. Choose the correct option by making hypothesis based on given assertion and reason. Give an explanation.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true and R is not the correct explanation of A.
c) A is true but R is false.
d) Both A and R are false.
e) A is false but R is true.

Question 9.
Assertion (A) : The refractive index of a prism depends only on the kind of glass of which it is made of and the colour of light. (AS2)
Reason (R) : The refractive index of a prism depends on the refracting angle of the prism and the angle of minimum deviation.
Answer:
(b) Both A and R are true and R is not the correct explanation of A.

Question 10.
Assertion (A) : Blue colour of sky appears due to scattering of light.
Reason (R) : Blue colour has shortest wavelength among all colours of white light. (AS2)
Answer:
(C) A is true but R is false.

Question 11.
Suggest an experiment to produce a rainbow in your classroom and explain the procedure. (AS3)
Answer:
Activity to produce a rainbow in classroom :

• Take a prism and place it on the table near a vertical white wall.
• Take a thin wooden plank, make a small hole in it and fix it vertically on the table.
• Place the prism between the wooden plank and wall.
• Place a white light source behind the hole of the wooden plank. Switch on the light.
• Adjust the height of the prism such that the light falls on one of the lateral surfaces.
• Observe the changes in the emerged ray of the prism.
• Adjust the prism by slightly rotating it till you get an image on the wall.
• We observe a band of different colours on the wall.
• These colours are nearly equal to the colours of the rainbow, i.e., VIBGYOR.

Question 12.
Prisms are used in binoculars. Collect information why prisms are used in binoculars. (AS4)
Answer:

• Binoculars consists of a pair of identical or mirror symmetrical telescope mounted side by side and aligned to point accurately in the same direction, allowing the viewer to use both eyes, when viewing distant objects.
• The size of binoculars is reduced by using prisms.
• We get good image with more brightness.
• Objective size and optical quality should be increased by using prisms in binoculars.

Question 13.
Incident ray on one of the face (AB) of prism and emergent ray from the face AC are given in figure. Complete the ray diagram. (AS5)

Answer:

Question 14.
How do you appreciate the role of molecules in the atmosphere for the blue colour of the sky? (AS6)
(OR)
How do you appreciate the nature of molecules, responsible for the blue of the sky?
Answer:

• The sky appear blue due to atmospheric refraction and scattering of light through molecules.
• Molecules are scattering centres.
• The reason to blue sky is due to the molecules N2 and 02.
• The sizes of these molecules are comparable to the wavelength of blue light.
• In the absence of molecules there will be no scattering of sunlight and the sky will appear dark.
• We should appreciate the molecules which are scattering centres.

Question 15.
Eye is the only organ to visualise the colourful world around us. This is possible due to accommodation of eye lens. Prepare a six line stanza expressing your wonderful feelings. (AS6)
Answer:
“Many people simply see
The world in black and white
But through my eyes the world’s alive
So colourful and bright
Don’t close your mind to the sights
That light up the night and day
There’s so much to see here on this earth
And not a rupee do you have to pay”
The most obvious things aren’t the ones
There is beauty in the unknown
with willing eyes and an open mind
The true wonders you will be shown

Question 16.
How do you appreciate the working of ciliary muscles in the eye? (AS6)
(OR)
Which muscles are helpful in changing the focal length of eye lens? How do you appreciate those muscles?
Answer:

• Ciliary muscles are helpful in changing focal length of eye lens.
• The ciliary muscles which attached with eye lens help to change the focal length of eye lens.
• When the eye is focussed on a distant object, these are relaxed. So the focal length of eye lens increases to its maximum value.
• The parallel rays coming into the eye are focussed on the retina then we can see the object clearly.
• When the eye is focussed on a nearer object the muscles are strained so the focal length of eye-lens decreases. The ciliary muscles adjust the focal length and the image is formed on retina then we can see the object.
• Their process of adjusting focal length of eve lens is called accomodation.
• So ciliary muscles must be appreciated for its accomodation of eye lens.

Question 17.
Why does the sky sometimes appear white? (AS7)
(OR)
The sky sometimes appears white. What is the reason behind it?
Answer:

• Our atmosphere contains atoms and molecules of different sizes.
• According to their sizes, they are able to scatter different wavelengths of light.
• For example, the size of the water molecule is greater than the size of the N₂ or O₂.
• It acts as a scattering centre for light other frequencies which are lower than the frequency of blue light.
• On a hot day due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere. These water molecules scatter the colours of other frequencies (other than blue).
• All such colours of other frequencies reach our eye and white colour is appeared to us.

Question 18.
Glass is known to be transparent material. But ground glass is opaque and white in colour. Why? (AS7)
Answer:

• Ground glass is glass whose surface has been ground that produces a flat but rough finish.
• Ground glass has the effect of rendering the glass translucent by scattering of light during transmission thus blurring visibility while still transmitting light.
• To get more permanent frost, the glass may be ground by rubbing with some gritty substance.

Question 19.
If a white sheet of paper is stained with oil, the paper turns transparent. Why? (AS7)
(OR)
What is the reason behind the paper turns transparent when it is immersed in oil?
Answer:

• The paper fibres have higher index of refraction probably much greater than 1.5.
• The oil or fat also has a high index of refraction so that it nearly matches the index of refraction of the paper fibres and it reduces the scattering significantly.
• The fat adhering to the cellulose fibers lowers the index of refraction of the cellulose and also fills in air voids, so that visible light passes through the bag with significantly less scattering.
• The oil connects the fibres in the paper with a liquid which can transmit by refraction (rather than scatter) light that falls upon it. As a result, the paper stained with oil is turned transparent.

Question 20.
A light ray falls on one of the faces of a prism at an angle 40° so that it suffers angle of minimum deviation of 30°. Find the angle of prism and angle of refraction at the given surface. (AS7)
Answer:
Given that, incident ray on one of the prisms (i₁) = 40°
Angle of minimum deviation (D) = 30°
Angle of prism (A) = ?
A+D = 2i₁
⇒ A = 2i₁ – D
⇒ A = 2(40°) -30°
= 80° – 30°
= 50°
∴ Angle of prims (A) = 50°
Angle of refraction (r₁) = ?

Question 21.
The focal length of a lens suggested to a person with Hypermetropia is 100 cm. Find the distance of near point and power of the lens. (AS7)
Answer:
i) The distance of near point :
If the distance of near point is ‘d’ and focal length is ‘f’ then the relation between

Question 22.
A person is viewing an extended object. If a converging lens is placed in front of his eye, will he feel that the size of object has increased? Why? (AS7)
Answer:

• A simple magnifier allows us to put the object closer to the eye than we could normally focus and forms an enlarged virtual images.
• The principle behind this is angular magnification.
• The magnification is Ma = \(\frac{25}{f}\) for the close focus point, but since that causes eye strain, it is usually desirable to put the images at infinity giving Ma = \(\frac{25}{f}\)
• So he fees the size of object is increased. The reason is mentioned.
• Angular magnification: ‘The ratio of the angle substended at the eye by the image formed by an optical instrument to the angle substended at the eye by the object being viewed.”

Fill In The Blanks

1. The value of least distance of distinct vision is about ……………………… .
2. The distance between the eye lens and retina is about ……………………… .
3. The maximum focal length of the eye lens is about ……………………… .
4. The eye lens can change its focal length due to working of ……………………… muscles.
5. The power of lens is ID then focal length is ……………………… .
6. Myopia can be corrected by using ……………………… lens.
7. Hypermetropia can be corrected by using ……………………… lens.
8. In minimum deviation position of prism, the angle of incidence is equal to angle of ……………………… .
9. The splitting of white light into different colours (VIBGYOR) is called ……………………… .
10. During refraction of light, the character of light which does not change is ……………………… .
Answer:

1. 25 cm
2. 2.5 cm
3. 2.5 cm
4. ciliary
5. 100 cm
6. bi-concave
7. bi-convex
8. emergence
9. dispersion of light
10. frequency

Multiple Choice Questions

1. The size of an object as perceived by an eye depends primarily on
A) actual size of the object
B) distance of the object from the eye
C) aperture of the pupil
D) size if the image formed on the retina
Answer:
B) distance of the object from the eye

2. When objects at different distances are seen by the eye which of the following remains constant?
A) focal length of eye-lens
B) object distance from eye-lens
C) the radii of curvature of eye-lens
D) image distance from eye-lens
Answer:
D) image distance from eye-lens

3. During refraction, …………….. will not change.
A) wavelength
B) frequency
C) speed of light
D) all the above
Answer:
B) frequency

4. A ray of light falls on one of the lateral surfaces of an equilateral glass prism placed on the horizontal surface of a table as shown in figure. For minimum deviation of ray, which of the following is true?

A) PQ is horizontal
B) ‘QR’ is horizontal
C) ‘RS’ is horizontal
D) Either ‘PQ’ or ‘RS’ is horizontal
Answer:
B) ‘QR’ is horizontal

5. Far point of a person is 5 m. In order that he has normal vision what ki nd of spectacles should he use?
A) concave lens with focal length 5 m
B) concave lens with focal length 10 m
C) convex lens with focal length 5 m
D) convex lens with focal length 2.5 mm
Answer:
A) concave lens with focal length 5 m

6. The process of re-emission of absorbed light in all directions with different intensi¬ties by the atom or molecule is called ……………
A) scattering of light
B) dispersion of light
C) reflection of light
D) refraction of light
Answer:
A) scattering of light

10th Class Physics 7th Lesson Human Eye and Colourful World InText Questions and Answers

10th Class Physics Textbook Page No. 104

Question 1.
Can you imagine the shape of rainbow when observed during travel in an airplane?
Answer:

The shape of rainbow, when observed during travel in an airplane is circular as shown in the following figure.

10th Class Physics Textbook Page No. 88

Question 2.
Why do the values of least distance of distinct vision and angle of vision change with person and age?
Answer:

• The ciliary muscle helps the eye lens to change its focal length by changing radii of curvature of eye lens.
• When the eye is focussed on a distant object, the ciliary muscles are relaxed so that the focal length of eye lens has its maximum value which is equal to its distance from the retina.
• The working of ciliary muscle in eye changes from person to person.
• So, the values of least distance of distinct vision and angle of vision changes with person and age.

10th Class Physics Textbook Page No. 89

Question 3.
How can we get same image distance for various positions of objects?
Answer:
For different positions of object, the image distance remains constant only when focal length of lens changes.

Question 4.
Can you answer above question using concepts of refraction through lenses?
Answer:
The focal length of a lens depends on the material by which it has been made and radii of curvatures of lens. We need to change focal length of eye lens to get same image distance for various positions of object.

10th Class Physics Textbook Page No. 90

Question 5.
How does eye lens change its focal length?
(OR)
What is the role of ciliary muscles in the eye ? Write the answer in one or two sentences only.
Answer:
The ciliary muscles to which eye lens is attached help the eye lens to change its focal length by changing the radii of curvature of eye lens.

Question 6.
How does this change (eye lens changes its focal lengths) take place in eye ball?
Answer:
When the eye is focussed on a distant object the ciliary muscles are relaxed. So the focal length of eye lens has its maximum value which is equal to its distance from the retina. The parallel rays coming into the eye are then focussed on to the retina and we see the object clearly.

Question 7.
Does eye lens form a real image or virtual image?
Answer:
The eye lens forms a real and inverted image of an object on the retina.

Question 8.
How does the image formed on retina help us to perceive the object without change in its shape, size and colour?
Answer:

• The eye lens forms a real and inverted image of an object on the retina.
• This retina is a delicate membrane, which contains large number of receptors (125 million) called ‘rods’ and ‘cones’.
• They receive the light signals and identify the colour, and the intensity of light.
• These signals are transmitted to the brain through optic-nerve fibres.
• The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 9.
Is there any limit to change of focal length of eye lens?
Answer:
Yes. When the object is at infinity, the parallel rays from the object falling on the eye lens are refracted. They form a point-sized image on retina. In this situation, eye lens has a maximum focal length.

Question 10.
What are the maximum and minimum focal lengths of the eye lens?
Answer:
Maximum focal length is 2.5 cm and minimum focal length is 2.27 cm.

Question 11.
How can we find the maximum and minimum focal lengths of the eye lens?
(OR)
Calculate the maximum and minimum focal lengths of the eye lens.
Answer:’
a) When the object is at infinity,
u = – ∞ ; v = 2.5 cm (image distance which is equal to distance between eye lens and retina)

b) Object at a distance of 25 cm from eye. In this case, eye has minimum focal length.
Here u = -25 cm ; v 2.5 cm

10th Class Physics Textbook Page No. 91

Question 12.
What happens if eye lens is not able to adjust its focal length?
Answer:
In this case the person cannot see the object clearly and comfortably.

Question 13.
What happens if the focal length of eye lens is beyond the range of 2.5 cm to 2.27 cm?
Answer:
The vision (image) becomes blurred due to defects of eye lens.

10th Class Physics Textbook Page No. 92

Question 14.
What can we do to correct myopia?
Answer:
To correct myopia, we use concave lens.

10th Class Physics Textbook Page No. 93

Question 15.
How can you decide the focal length of the lens to be used to correct myopia?
Answer:
Assume that the object distance (u) is infinity and image distance (v) is equal to distance of far point.
u = – ∞ ; v = distance of far point = – D
Let ‘f be the focal length of bi-concave lens.

Here ‘f is negative showing that it is a concave lens.

Question 16.
What happens when the eye has a minimum focal length greater than 2.27 cm?
Answer:
In this case, the rays coming from the nearby object after refraction at eye lens, forms image beyond the retina.

10th Class Physics Textbook Page No. 94

Question 17.
How can you decide the focal length of convex lens to be used?
Answer:
Here object distance (u) = -25 cm
Image distance (v) = – d (distance of near point)
Let ‘f be the focal length of bi-convex lens.

If d > 25 cm, then ‘f’ becomes positive then we need to use bi-convex lens to correct hypermetropia.

10th Class Physics Textbook Page No. 95

Question 18.
Have you ever observed details in the prescription?
Answer:
A prescription that contains some information regarding type of lens to be used to correct vision.

Question 19.
What does it (“my sight is increased or decreased”) mean?
Answer:
Usually doctors, after testing the defects of vision prescribe corrective lenses indicating
their power which determines the type of lens to be used and its focal length.

Question 20.
What do you mean by power of lens ? flftorH)
Answer:
The reciprocal of focal length is called power of lens.

Question 21.
Doctor advised to use 2D lens. What is its focal length?
Answer:
Given power of lens P = 2D

∴ Focal length of lens (f) = 50 cm.

10th Class Physics Textbook Page No. 96

Question 22.
How could the white light of the sun give us various colours of the rainbow?
Answer:
Due to reflection, refraction and dispersion of sunlight.

Question 23.
What happens to a light ray when it passes through a transparent medium bounded by plane surfaces which are inclined to each other?
Answer:
When light incident on one of the plane surfaces and emerges from the other.

Question 24.
What is a prism?
Answer:
A prism is a transparent medium separated from the surrounding medium by consisting two refracting plane surfaces which are inclined.

10th Class Physics Textbook Page No. 97

Question 25.
What is the shape of the outline drawn for a prism?
Answer:
The outline drawn for a prism is in a triangle shape.

10th Class Physics Textbook Page No. 98

Question 26.
How do you find the angle of deviation?
Answer:
The angle between the extended incident and emergent rays is called angle of deviation.
(OR)
Extend both incident and emergent rays till they meet at a point ‘O’. Measure the angle between these two rays. This is the angle of deviation.

Question 27.
What do you notice from the angle of deviation?
Answer:
The angle of deviation decreases first and then increases with increase in the angle of incidence.

Question 28.
Can you draw a graph between angle of incidence and angle of deviation?
Answer:

Yes, we can draw the graph between angle of incidence and angle of deviation.

Question 29.
From the graph, can you find the minimum of the angle of deviation?
Answer:
Yes, we can. Draw a tangent line to the curve, parallel to X – axis, at the lowest point of the graph. The point where this line cuts Y – axis gives the angle of minimum deviation.

Question 30.
Is there any relation between the angle of incidence (i) angle of emergence (r) and piangle of deviation (d)?
Answer:
(i₁ + i₂)= A + D
i + r = A + D

10th Class Physics Textbook Page No. 101

Question 31.
In activity-3,we noticed that light has chosen different paths. Does this mean that the refractive index of the prism varies from colour to colour?
Answer:
Yes, refractive index of the prism varies from colour to colour.

Question 32.
Is the speed of light of each colour different?
Answer:
In vacuum – No, the speed of each colour is constant.
In medium – Speed is different for different colours.

Question 33.
Can you guess now, why light splits into different colours when it passes through a prism?
Answer:
Due to dispersion of light and different wavelength of colours in medium.

10th Class Physics Textbook Page No. 102

Question 34.
Does it (light passing through a prism) split into more colours? Why?
Answer:
We know the frequency of light is the properly of the source and it is equal to number of waves leaving the source per second. This cannot be changed by any medium. Hence frequency doesn’t change due to refraction. The coloured light passing through any transparent medium retains its colour.

Question 35.
Can you give an example in nature, where you observe colours as seen in activity 3?
Answer:
Yes, in rainbow. It is a good example of dispersion of light.

Question 36.
When do you see a rainbow in the sky?
Answer:
Dut to the refraction, reflection and dispersion of sunlight, when the sunlight passes through the rain drops, we can see the rainbow in the sky.

Question 37.
Can we create a rainbow artificially?
Answer:
Yes, we can create a rainbow artificially.

10th Class Physics Textbook Page No. 104

Question 38.
Why does the light dispersed by the raindrops appear as a bow?
Answer:

• A rainbow is not the flat two dimensional arc as it appears to us.
• The rainbow we see is actually a three dimensional cone with the tip of our eye.
• All the drops that disperse the light towards us lie in the shape of the cone – a cone of different layers.
• The drops that disperse red colour to our eye are on the outermost layer of the cone, similarly, the drops that disperse orange colour to our eye are on the layer of the cone beneath the red colour cone.
• In this way the cone responsible for yellow lies beneath orange and so on it till the violet colour cone becomes the innermost cone.

 

Question 39.
Why is the sky blue?
Answer:
A clear cloudless day – time sky is blue because molecules in the air scatter blue light from the sun more than thev scatter red liuht.

Question 40.
What is scattering?
A. Atoms or molecules which are exposed to absorb light energy and emit some part of the light energy in different directions is called scattering of light.

10th Class Physics Textbook Page No. 106

Question 41.
Why is that the sky appears white sometimes when you view it in certain direction on hot days?
Answer:

• On a hot day, due to rise in the temperature, water vapour enters atmosphere which leads to abundant presence of water molecules in atmosphere.
• These water molecules scatter the colours of other frequencies (other than blue).
• All such colours of other frequencies reach our eye and the sky appears white.

Question 42.
Do you know the reasons for appearance the red colour of sun during sunrise and at sunset?
(OR)
Sun appears red in colour during sunrise and sunset. Give reason.
(OR)
Why does sky appear red at Sunshine and Sunset?
Answer:

• The light rays from the sun travel more distance in atmosphere to reach our eye in morning and evening times.
• During sunrise and sunset except red light all colours scatter more and vanish before they reach us.
• Since scattering of red light is very less, it reaches us.
• As a result sun appears red in colour during sunrise and sunset.

Question 43.
Can you guess the reason why sun does not appear red during noon hours?
Answer:
During noon hours, the distance to be travelled by the sun rays in atmosphere is less than when compared to morning and evening hours. Therefore all colours reach our eye without scattering. Hence the sunlight appears white in noon hours.

10th Class Physics 7th Lesson Human Eye and Colourful World Activities

Activity – 1

Question 1.
How do you find least distance of distinct vision?
(OR)
What is the least distance a person can see an object comfortably and distinctly known as ? Write an activity to find that (least, distance of distinct vision) distance.
Answer:

• The least distance a person can see an object comfortably and distinctly is known as least distance of distinct vision.
• Hold the textbook at certain distance with your hands.
• Try to read the contents on the page.
• Gradually move the book towards eye, till it reaches very close to your eyes.
• You may see that printed letters on the page appear blurred or you feel strain to read.
• Now move the book backwards to a position where you can see clear printed letters without strain.
• Ask your friend to measure distance between your eye and textbook at this position.
• Note down its value.
• Repeat this activity with other friends and note down the distances for distinct vision in each case.
• Find the average of all these distances of clear vision.
• You will notice that to see an object comfortably and distinctly, you must hold it at a distance about 25 cm from your eyes.
• This 25 cm distance is called least distance of distinct vision.
• This value varies from person to person and with age.

Activity – 2

Question 2.
How can you find angle of vision?
(OR)
What is maximun angle a person is able to see whole object? Write an activity to find that angle.
Answer:

• The maximum angle, at which we are able to see the whole object is called angle of vision.
• Arrange a retort stand.
• Collect a few wooden sticks (or) PVC pipes that are used for electric wiring.
• Prepare sticks or pipes of 20 cm, 30 cm, 35 cm, 40 cm, 50 cm from them.
• Keep the retort stand on a table and stand near the table such that vour head is beside the vertical stand.
• Adjust the clamp on horizontal rod and fix at a distance of 25 cm from the eyes.
• Ask one of your friends to fix a wooden stick of 30 cm height to the clamp in vertical position.
• Now keeping your vision parallel to horizontal rod of the stand, try to see the top and bottom of wooden stick kept in vertical position.
• If you are not able to see both ends of stick at this distance (25 cm), adjust the vertical stick on the horizontal rod till you are able to see both ends of the stick at nearest possible distance from your eye.
• Fix the clamp to the vertical stick at this position.
• Without changing the position of the clamp on horizontal rod, replace this stick of 30 cm length.
• Try to see the top and bottom of the stick simultaneously without any change in the position of eye.
• Try the same activity with various lengths of the sticks.
• You can see the whole object AB which is at a distance of 25 cm (least distance of clear vision) because the rays coming from the ends A and B of the object will enter the eye.
• Similarly we can also see complete object CD with eye as explained above.
• Let us assume that AB is moving closer to eye to a position A B .
• You notice that you will be able to see only the part (EF) of the object A B because the rays coming from E and F enter your eye.

• The rays coming from A and B cannot enter your eye.
• The rays coming from the extreme ends of an object forms an angle at the eye.
• If this angle is below 60°, we can see whole object.
• If this angle is above 60°, then we see only the part of the object.
• This maximum angle, at which we are able to see the whole object is called angle of vision.
• The angle of vision for a healthy human being is about 60°.
• It varies from person to person and with age.

Activity – 3

Question 3.
Describe an activity for dispersion of light.
(OR)
What is the name given to process when white light passes through a prism it splits into different colours ? Explain the process with an activity.
Answer:

• The splitting of white light into different colours is called despersion of light.
• Do this experiment in the dark room.
• Take a prism and place it on the table near a vertical white wall.
• Take a thin wooden plank.
• Make a small hole in it and fix it vertically on the table.
• Place the prism between the wooden plank and wall.
• Place a white light source behind the hole of wooden plank.
• Switch on the light.
• The rays coming out of the hole of plank become a narrow beam of light.
• Adjust the height of the prism such that the light falls on one of the lateral surfaces.
• Observe the changes in emerged rays of the prism.
• Adjust the prism by slightly rotating it till you get an image on the wall.
• You will observe that white light is splitting into certain different colours.

Activity – 6

Question 4.
Describe an experiment for scattering of light.
(OR)
What is the principle involved in blue of the sky ? Explain the principle with an experiment?
Answer:

• Take the solution of sodium-thio-Sulphate (hypo) with sulphuric acid in a glass beaker.
• Place the beaker in which reaction is taking place in an open place where abundant sunlight is available.
• Watch the formation of grains of sulphur and observe changes in beaker.
• You will notice that sulphur precipitates as the reaction is in progress.
• At the beginning, the grains of sulphur are smaller in size as the reaction progresses, their size increases due to precipitation.
• Sulphur grains appear blue in colour at beginning and slowly their colour becomes white as their size increases.
• The reason for this is scattering of light.
• At the beginning, the size of grains is small and almost comparable to the wavelength of blue light.
• Hence they appear blue in the beginning.
• As the size of grains increases, their size becomes comparable to the wavelengths of other colours.
• As a result of this, they act as scattering centres for other colours.
• The combination of all these colours appears as white.

 

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 6 Refraction of Light at Curved Surfaces Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 6th Lesson Refraction of Light at Curved Surfaces

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Textbook Questions and Answers

Review of Your Previous Knowledge

Question 1.
Have you ever touched a magnifying glass with your hand?
Answer:
Yes.

Question 2.
Have you touched the glass in the spectacles used for reading with your hand?
A. Yes.

Question 3.
Is it a plane or curved surface?
Answer:
Curved surface.

Question 4.
Is it thicker in the middle or at the edge?
Answeer:
Magnifying glass and some spectacle are thicker in middle whereas some spectacles are thicker at edge.

Improve Your Learning

Question 1.
A man wants to get a picture of a zebra. He photographed a white donkey after fitting a glass, with black stripes, on to the lens of his camera. What photo will he get? Explain. (AS1)
(OR)
A person wants to get a picture of zebra and he photographed a white donkey fitting a glass with black stripes. Does he get photo of zebra? Explain.
Answer:
The person was unable to gel the picture <>l zebra because only two rays are enough to form complete image after convergence. So he will get the image of white donkey but the intensity may be less.
(OR)
He will get a picture of while donkey because e\ery part of lens forms an image so if you cover lens with stripes still it forms a complete image. However, the intensity of the image will be reduced.

Question 2.
Two converging lenses are to be placed in the path of parallel rays so that the rays remain parallel after passing through both lenses. How should the lenses be arranged? Explain with a neat ray diagram. (AS1)
Answer:

• Two lenses are placed in the path of parallel rays as shown in figure.
• The first lens is placed in the direction of parallel lines, which converges at focus.
• The second lens is arranged so that it is the focus of 2nd then emerging rays will be parallel.

Question 3.
The focal length of a converging lens is 20 cm. An object is 60 cm from the lens. Where will the image be formed and what kind of image is it? (AS1)
Answer:
f = 20 cm (by sign conversion f = + 20 cm)
u = 60 cm (by sign conversion u = – 60 cm)

Image will be formed at 30cm in between F₁, and 2F1. Image is real, inverted and diminished.

Question 4.
A double convex lens has two surfaces of equal radii ‘R’ and refractive index n = 1.5. Find the focal length ‘f’. (AS1)
(OR)
What is the focal length ‘f, when its double convex lens has two surfaces of equal radii ‘R’ and refractive index n = 1.5?
Answer:
R₁ = R₂ = R (suppose)
Focal length (f) = ?; Refractive index (n) = 1.5

∴ Focal length of lens = Radius of curvature of surface.

Question 5.
Write the lens maker’s formula and explain the terms in it. (AS1)
(OR)
Ravi wants to make a lens. Which formula he has to follow ? Write the formula and explain the terms in it.
(OR)
Write lens formula.
Answer:
Lens maker’s formula:
\(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
n = Refractive index of the medium
R₁ = Radius of curvature of 1 st surface
R₂ = Radius of curvature of 2nd surface
f = Focal length

Question 6.
How do you verify experimentally that the focal length of a convex lens is increased when it is kept in water? (AS1)
(OR)
Write an activity to show that the focal length of a lens depends on its surrounding medium.
Answer:
Aim :
To prove focal length of convex lens is increased when it is kept in water.

Apparatus :
Convex lens, water, cylindrical vessel, circular lens holder, stone.

Procedure :

1. Take a cylindrical vessel like glass tumbler.
2. Its height must be greater than the focal length of lens, (the around four times focal length of lens).
3. Keep a black stone inside the vessel at its bottom.
4. Pour the water into the vessel such that the height of the water level from the top of the stone is greater than the focal length of lens.
5. Now dip the lens horizontally using a circular lens holder.
6. Set the distance between stone and lens that is equal to or less than focal length of lens.
7. Now see the stone through the lens.
8. We can see the image of the stone.
9. If we dip the lens to a certain height which is greater than the focal length of lens in air, still we can see the image.
10. This shows that the focal length of convex lens has increased in water.
11. Thus we conclude that the focal length of lens depends upon the surrounding medium.

Note : For convenience, use 5 or 10 cm focal length convex lens.

Question 7.
How do you find the focal length of a lens experimentally? (AS1)
Answer:

• Take the lens (Ex : Convex), which focused towards the distant object.
• A white coated screen (Ex : White paper) is placed on the other side of the lens.
• Adjust the screen till you get a clear image of the object.
• At this position measure the distance between the lens and screen which is equal to the focal length of the lens.

Question 8.
Harsha tells Siddhu that the double convex lens always behaves like a convergent lens. But Siddhu knows that Harsha’s assertion is wrong and corrected Harsha by asking some questions. What are the questions asked by Siddhu? (As2)
Answer:
The questions asked by Siddhu :

1. Is the object placed beyond 2f point?
2. Is the object located at 2f point?
3. Is the object located in between the 2f and the focal point?
4. Is the object located at the focal point?
5. Is the object located in front of the focal point?
6. Is the lens kept in a medium with refractive index less than lens or more than lens?

Question 9.
Assertion (A): A person standing on the land appears taller than his actual height to a fish inside a pond. (AS2)
Reason (R) : Light bends away from the normal as it enters air from water.
Which of the following is correct? Explain.
a) Both A and R are true and R is the correct explanation of A.
b) Both A and R are true and R is not the correct explanation of A.
c) A is true but R is false.
d) Both A and R are false.
e) A is false but R is true.
Answer:
Answer a is correct.
Explanation :
Because the light travelling from water to air it bends away from the normal so the fish observes the apparent image of the person, appears taller than his original.

Question 10.
A convex lens is made up of three different materials as shown in the figure Q-10. How many of images does it form? (AS2)

Answer:

• A lens made of three different materials of refractive indices say n₁, n₂ and n₃.
• These three materials will have three different refractive indices. Thus for a given object it forms three images.

Question 11.
Can a virtual image be photographed by a camera? (AS2)
Answer:
Yes, we can.
Ex : – A plane mirror forms a virtual image, we can able to take photograph of that image in plane mirror.

Question 12.
You have a lens. Suggest an experiment to find out the focal length of the lens. (AS3)
(OR)
Through an experiment, find out the focal length of the lens.
Answer:
Aim :
To find focal length of given lens.

Apparatus :
Object (candle), convex lens, v – stand, screen.

Procedure :

• Take a v-stand and place it on a long table at the middle.
  Place a convex lens on the v-stand. Imagine the principal axis of the lens.
• Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
• Adjust a screen (a sheet of white paper placed perpendicular to the axis) which is on other side of the lens until you get an image on it.
• Measure the distance of the image from the v-stand of lens (image distance V) and also measure the distance between the candle and stand of lens (object distance ‘u’). Record the values in the table.

• Now place the candle at a distance of 60 cm from the lens, try to get an image of the candle flame on the other side on a screen. Adjust the screen till you get a clear image.
• Measure the image distance V and object distance ‘u’ and record the values in table.
• Repeat the experiment lor various object distances like 50 cm, 40 cm, 30 cm, etc. Measure the image distances in all cases and note them in table.
• Using the formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\), find f in all the cases. We will observe the value ‘f is equal in all cases. This value off is the focal length of the given lens.

Question 13.
Let us assume a system that consists of two lenses with focal length f₁, and f₂ respectively. How do you find the focal length of the system experimentally, when
i) two lenses are touching each other
ii) they are separated by a distance ‘d’ with common principal axis? (AS3)
Answer:
Experimental Proof:
i) Two lenses are touching each other :
Aim :
To find focal length of combination of two convex lenses, touching each other. Material required : Convex lenses – 2 (with known focal lengths say f, and f2); V-stands – 2, candle, screen scale.

Procedure:

• Place two V-stands with two convex lenses as they touch each other on a table.
• Place a candle (object) far away from the lenses.
• Adjust a screen, which is placed other side of the lenses until we get a clear image on it.
• At that position, measure the image distance (v) and object distance (u).
• Do this experiment for several object distances and record in the given table.

ii) They are separated by a distance of ‘d’ :
Procedure :

• Now place v-stands along with lenses with distance’d’.
• Do the same procedure again.
• Record the observations in the given table.
• Find the average of the ‘f’_comb.

Question 14.
Collect the information about the lenses available in an optical shop. Find out how the focal length of a lens may be determined by the given power’ of the lens. (AS4)
Answer:
I had collected the information regarding different lenses available at optical shops.
The relationship between power and focal length is power (D) = \(\frac{1}{f}\). f is in meters.

Power of lens in diopters	Type of lens	Focal length
0.25	Convex	400 cm
0.5	Convex	200 cm
1	Convex	100 cm
-2	Concave	50 cm
– 1	Concave	– 100 cm
-0.5	Concave	– 200 cm
-0.25	Concave	– 400 cm

Question 15.
Collect the information about lenses used by Galileo in his telescope. (AS4)
(OR)
What lenses are used by Galileo in his telescope?
Answer:

A Galilean telescope is defined as having one convex lens and one concave lens. The concave lens serves as the ocular lens or the eye piece, while the convex lens serves as the objective. The lens are situated on either side of a tube such that the focal point of the ocular lens is the same as the focal point for the objective lens.

Question 16.
Use the data obtained by activity – 2 in table-1 of this lesson and draw the graphs of u vs v and \(\frac{1}{u}\) vs \(\frac{1}{v}\) (AS5)
(OR)
By obtaining data from activity – 2 in table – 1 of this lesson, draw the graphs of u vs v and \(\frac{1}{u}\) vs \(\frac{1}{v}\)
Answer:
Graph of u – v using data obtained by activity – 2. Take lens with focal length 30 cm.

Object distance (u)	Image distance (v)	Focal length (f)
60 cm	60 cm	30 cm
50 cm	75 cm	30 cm
40 cm	120 cm	30 cm

The graph looks like this

The shape of the graph is rectangular hyperbola.

Graph of \(\frac{1}{u}\) – \(\frac{1}{v}\)

For these values the graph is straight line which touches the axis as shown in figure.

Question 17.
Figure shows ray AB that has passed through a divergent lens. Construct the path of the ray up to the lens if the F position of its foci is known. (AS5)

Answer:

The path of the ray up to the lens if the position of foci is known for ray AB is diverging lens or concave lens path.

Question 18.
Figure shows a point light source and its image produced by a lens with an optical axis N₁, N₂. Find the position of the lens and its foci using a ray diagram. (AS5)

Answer:

1. The object is in between focus and optic centre.
2. The image is virtual, erect and magnified. Nv
3. l is the lens, ‘O’ is the object and T is the image.

Question 19.
Find the focus by drawing a ray diagram using the position of source S and the image S’ given in the figure. (AS5)

Answer:

1.  Image is real.
2. l’ is lens, ‘O’ is object and T is image.
3.  Lens is convex.

(Or)

1. Image is real.
2. l’ is lens, ‘O’ is object and ‘I’ is image.
3. Lens is convex.

Question 20.
A parallel beam of rays is incident on a convergent lens with a focal length of 40 cm. Where should a divergent lens with a focal length of 15 cm be placed for the beam of rays to remain parallel after passing through the two lenses? Draw a ray diagram. (AS5)
Answer:

1. A parallel beam of rays when incident on a convergent lens, after refraction they meet at the focus of the lens.

2. A beam of rays which is incident on a divergent lens, after refraction, pass parallel to the principal axis. If we extend these incident rays, they seems to meet at focus of the lens.

3. Hence the divergent lens should be kept at 25 cm distance from convergent lens (40 – 15 = 25 cm) as shown in the figure.

PF = 40 cm (Focal length of convergent lens)
P’F = 15 cm (Focal length of divergent lens)
PP’ = 40 – 15 = 25 cm (Position of divergent lens)

Question 21.
Draw ray diagrams for the following positions and explain the nature and position of image.
i) Object is placed at 2F₂
ii) Object is placed between F₂ and optic centre P. (AS5)
Answer:
i) Object is placed at 2F₂:

Nature : Real, inverted and diminished.
Position : Image is formed on the principal axis between the points F₁, and 2F₁.

ii) Object is placed between F₂ and optic centre P :

Nature : Virtual, erect and magnified.
Position : Same side of the lens where object is placed.

Question 22.
How do you appreciate the coincidence of the experimental facts with the results obtained by a ray diagram in terms of behaviour of images formed by lenses? (AS6)
Answer:

• Ray diagrams are very useful in optics.
• By the ray diagrams, we can easily find the values of image distance, object distance, focal length, radius of curvature, magnification, etc.
• These results are exactly equal to the result gotten by an experiment.
• For example : In the experiment, with a convex lens, we get clear image of an object, on a screen by adjusting the screen.

Then, we measure the image distane (v) practically. This takes more time and requires equipped lab also.

But, by simply draw a ray diagram on a paper, we can get exact image distance (v) very easily, without lab.

• So, ray diagrams are very useful in the construction of microscopes, telescopes, etc.
• Hence, one can trust and depend on the result of ray diagrams instead of several lab experiments.
• So, I appreciate the ray diagrams.

Question 23.
Find the refractive index of the glass which is a symmetrical convergent lens if its focal length is equal to the radius of curvature of its surface. (AS7)
Answer:
Given that lens is convergent symmetrical

Question 24.
Find the radii of curvature of a convexo – concave convergent lens made of glass with refractive index n = 1.5 having focal length of 24 cm. One of the radii of curvature is double the other. (AS7)
Answer:

Question 25.
The distance between two point sources of light is 24 cm. Where should a convergent lens with a focal length of f = 9 cm be placed between them to obtain the images of both sources at the same point? (AS7)
Answer:
For Source S₁ :

∴ The convex lens may be placed between the two sources, such that a distance of 18 cm from one source, and 6 cm from other source.

Question 26.
Suppose you are inside the water in a swimming pool near an edge. A friend is standing on the edge. Do you find your friend taller or shorter than his usual height? Why? (AS7)
(OR)
If your friend is standing near an edge of the swimming pool and you are in the water, do you find he is taller or shorter than his usual height?
Answer:

1. My friend appears to be taller because the light is travelling from rarer to denser.
2. The rays bend in such away that they seems to be coming from long distance.
3. So it is actually apparent image of my friend which appears to be taller due to refraction.

Fill in the Blanks

1. The rays from the distant object, falling on the convex lens pass through ……………….. .
2. The ray passing through the ……………….. of the lens is not deviated.
3. Lens formula is given by ……………….. .
4. The focal length of the plano-convex lens is 2R where R is the radius of curvature of the surface. Then the refractive index of the material of the lens is ……………….. .
5. The lens which can form real and virtual images is ……………….. .
Answer:

1. Tocus
2. optical centre
3. \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
4. 1.5
5. convex lens

Multiple Choice Questions

1. Which one of the following materials cannot be used to make a lens?
A) water
B) glass
C) plastic
D) clay
Answer:
D) clay

2. Which of the following is true?
A) The distance of virtual image is always greater than the object distance for convex lens.
B) The distance of virtual image is not greater than the object distance for convex lens.
C) Convex lens always forms a real image.
D) Convex lens always forms a virtual image.
Answer:
B) The distance of virtual image is not greater than the object distance for convex lens.

3. Focal length of the plano-convex lens is when its radius of curvature of the surface is R and n is the refractive index of the lens.

4. The value of the focal length of the lens is equal to the value of the image distance when the rays are
A) passing through the optic centre
B) parallel to the principal axis
C) passing through the focus
D) in all the cases
Answer:
D) in all the cases

5. Which of the following is the lens maker’s formula?

Answer:
C

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Additional Questions and Answers

Question 1.
Derive a relation between refractive indices of two media (n₁, n₂), object distance (u), image distance (v) and radius of curvature (R) for a curved surface.
(OR)
Derive \(\frac{\mathbf{n}_{2}}{\mathbf{v}}-\frac{\mathbf{n}_{1}}{\mathbf{u}}=\frac{\mathbf{n}_{2}-\mathbf{n}_{1}}{\mathbf{R}}\)
(OR)
Derive curved surface formula.
Answer:

• Consider a curved surface separating two media of refractive indices n₁, and n₂.
• A point object is placed on the principal axis at point ‘O’.
• The ray which travels along the principal axis passes through the pole undeviated.
• The second ray, which forms an angle with a princi¬pal axis, meets the interface at A. The angle of incidence is Q₁. The ray bends and passes through the second medium along the line AI. The angle of refraction is Q₂.
• The two refracted rays meet at I and the image is formed there.
• 6) Let the angle made by the second refracted ray with principal axis be γ and the angle between the normal and principal axis be β.
• From figure,

PO = u (object distance), PI = v (Image distance),
PC = R (radius of curvature) and n₁, n₂ are refractive indices of the media.
From ∆ACO, θ₁ = α + β
∆ACI, β = θ₂ + γ
⇒ θ₂ = β – γ
According to Snell’s law, n₁sin θ₁ = n₂ sin θ₂.
∴ n₁ sin (α + β) = n₂ sin (β – γ) …………….. (1)
As per paraxial approximation,
sin (α + β) = α + β and sin (β – γ) = β – γ.
∴ (1) ⇒ n₁(α + β) = n₂ (β – γ)
⇒ n₁ α + n₁β = n₂ β – n₂ γ — (2)
Since all angles are small, we can write

∴ This is the required relation for curved surfaces.

Question 2.
Derive expression for lens maker’s formula.
(OR)
Prove \(\frac{1}{\mathbf{f}}=(\mathbf{n}-\mathbf{1})\left(\frac{1}{\mathbf{R}_{1}}-\frac{1}{\mathbf{R}_{2}}\right)\).
Answer:

Procedure :

• Imagine a point object ‘O’ placed on the principal axis of the thin lens
• Let this lens be placed in a medium of refractive index n_a and let refractive index of lens be n_b.
• Consider a ray, from ‘O’ which is incident on the convex surface of the lens with radius of curvature R₁ at A.
• The incident ray refracts at A.
• It forms image at Q, if there were no concave surface.
• From figure Object distance PO = – u;

Image distance PQ = v = x
Radius of curvature R = R₁
n₁ = n_a and n₂ = n_b.

• But the ray that has refracted at A suffers another refraction at B on the concave surface with radius of curvature (R₂).
• At B the ray is refracted and reaches I.
• The image Q of the object due to the convex surface. So I is the image of Q for concave surface.
• Object distance u = PQ = + x
  Image distance PI = v
  Radius of curvature R = – R₂
• The refraction of the concave surface of lens is medium -1 and surrounding is medium – 2.
  ∴ n₁ = n_b and n₂ = n_a

Question 3.
Derive the lens formula.
Answer:
1. Consider an object 00′ placed on the principal axis in front of a convex lens as shown in the figure. Let II’ be the real image formed by the lens, i.e. the other side of it.

2. From the figure : PO, PI, PFt are the object distance, image distance and focal length respectively.

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 64

Question 1.
What happens to a ray that is incident on a curved interface separating the two media? Are the laws of refraction still valid?
Answer:
It undergoes deviation from its path. Yes, the laws of reflection are still valid.

Question 2.
How do rays betid when they are incident on a curved surface?
Answer:
A ray will bend towards the normal when it travels from rarer to denser medium and bends away from the normal when it travels from denser to a rarer medium.

10th Class Physics Textbook Page No. 65

Question 3.
What happens to ray that travels along the principal axis? Similarly, a ray that travels through the centre of curvature?

Answer:
According to Snell’s law the ray which travels along the normal drawn to the surface does not deviate from its path. Hence both rays in the given condition travel along normal, so they do not deviate.

Question 4.
What difference do you notice in the refracted rays in 4 (a) and 4 (b)? What could be the reason for that difference?

Answer:

• In figure 4 (a) ray travelling parallel to the principal axis strikes a convex surface and passes from a rarer medium to a denser medium.
• In figure 4 (b) a ray travelling parallel to the principal axis strikes a convex surface passes from a denser medium to a rarer medium.
• Figure 4 (a) : The refracted ray moves towards the normal.
• Figure 4 (b) : The refracted ray moves away from the normal.
  Reason : The main reason is that light passes through different media.

10th Class Physics Textbook Page No. 66

Question 5.
What difference do you notice in refracted rays in 4 (c) and 4 (d)? What could be the reasons for that difference?
(OR)
Draw the ray diagrams when the incident ray passes through the curved surfaces.
a) Rarer medium to denser medium.
b) Denser medium to rarer medium.

Answer:

• In figure 4 (c) a ray travelling parallel to the principal axis strikes a concave surface and passes from a denser medium to a rarer medium.
• In figure 4 (d) a ray travelling parallel to the principal axis strikes a concave surface and passes from a rarer medium to a denser medium.

Reasons :

• Figure 4 (c) :The refracted ray reaches a particular point on the principal axis.
• Figure 4 (d) : The refracted ray moves away from the principal axis.
• The main reason is that light passes through different media.

Question 6.
You might have observed that a lemon in the water of a glass tumbler appears bigger than its actual size, when viewed from the sides of tumbler.
1) How can you explain this (appeared) change in size of lemon?
Answer:
It can be explained by using refraction. When light travels from one medium to another medium it undergoes refraction.

2) Is the lemon that appears bigger in size an image of lemon or is it the real lemon?
Answer:
That is image of lemon.

3) Can you draw a ray diagram to explain this phenomenon?
Answer:

10th Class Physics Textbook Page No. 70

Question 7.
What happens to the light ray when a transparent material with two curved surfaces is placed in its path?
Answer:
The light ray undergoes refraction.

Question 8.
Have you heard about lenses?
Answer:
Yes, we have heard about lenses. A transparent material bounded by two spherical v surfaces is called lens.

Question 9.
How does a light ray behave when it is passed through a lens?
Answer:
A light ray will deviate from its path in some cases and does not deviate in some other cases.

10th Class Physics Textbook Page No. 72

Question 10.
How does the lens form an image?
Answer:
Lens forms an image through converging light rays or diverging light rays.

Question 11.
If we allow a light ray to pass through the focus, which path does it take?
Answer:
The ray passing through the focus takes a parallel path to principal axis after refraction.

10th Class Physics Textbook Page No. 73

Question 12.
What happens when parallel rays of light fall on a lens making some angle with the principal axis?
Answer:’
The rays converge at a point (or) appear to diverge from a point lying on the focal plane.

Question 13.
What do you mean by an object at infinity? What type of rays fall on the lens?
Answer:
The distance between the lens and the object is very much greater than when compared to object size is known as object at infinity. Parallel rays fall on the lens.
The object at infinity means distant object. The rays falling on the lens from an object at infinity are parallel to principal axis.

10th Class Physics Textbook Page No. 77

Question 14.
Could you get an image on the screen for every object distance with a convex lens?
Answer:
No, when the object is placed between pole and focus we will get virtual, erect and enlarged image on the other side of the- object.

Question 15.
Why don’t you get an image for certain object distances?
Answer:
Because at those distances the light rays diverge each other.

Question 16.
Can you find the minimum limiting object distance for obtaining a real image? What do you call this minimum limiting object distance?
Answer:
Yes, this minimum limiting object distance is called focal length.

Question 17.
When you do not get an image on the screen, try to see the image with your eye directly from the place of the screen. Could you see the image? What type of image do you see?
Answer:
Yes, we can see the image. This is a virtual image which we cannot capture on screen.

Question 18.
Can you find the image distance of a virtual image? How could you do it?
Answer:
We can find the image distance of virtual image by using lens formula \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}\) (if we know the focal length of lens and object distance.)

10th Class Physics 6th Lesson Refraction of Light at Curved Surfaces Activities

Activity – 1

Question 1.
Write an activity to observe the light refraction at curved surface.
Answer:
Procedure and observation :

• Draw an arrow of length 4 cm usfng a black sketch pen on a thick sheet of paper.
• Take an empty cylindrical-shaped transparent vessel.
• Keep it on the table.
• Ask your friend to bring the sheet of paper on which arrow was drawn behind the vessel while you look at it from the other side.
• We will see a diminished image of the arrow.
• Ask your friend to fill vessel with water.
• Look at the arrow from the same position as before.
• We can observe an inverted image.

Explanation :

• In the first case, when the vessel is empty, light from the arrow refracts at the curved interface, moves through the glass, enters in to air then it again undergoes refraction on the opposite curved surface of vessel and comes out into the air.
• In this way light travels through two media, comes out of the vessel and forms a diminished image.
• In the second case, light enters the curved surface, moves through water, comes out of the glass and forms an inverted image.

Lab Activity

Question 2.
Write an activity to know the characteristics of image due to convex lens at various distances.
Answer:
Aim:
Determination of focal length of bi-convex lens using UV method.

Material Required :
V Stand, convex lens, light source, screen, meter scale. Take a V-stand and place it on a long (nearly 2m) table at the middle. Place a convex lens on the v-stand. Imagine the principal axis of the lens. Light a candle and ask your friend to take the candle far away from the lens along the principal axis. Adjust a screen (a sheet of white paper placed perpendicular to the axis) which is on other side of the lens until you get an image on it.

Procedure :

1. Take a V-stand and place a convex lens on this stand.
2. Imagine the principal axis of the lens.
3. Light a candle and ask your friend to take the candle far away from the lens along the principal axis.
4. We use a screen because it forms a real image generally which will form on a screen. Real images cannot be seen with an eye.
5. Adjust the screen, on other side of lens until clear image forms on it.
6. Measure the distance of the image from the stand and also measure the distance between the candle and stand of lens.
7. Now place the candle at a distance of 60 cm from the lens such as the flame of the candle lies on the principal axis of the lens.
8. Try to get an image of candle flame on the other side on a screen.
9. Adjust the screen till you get a clear image.
10. Measure the distance of image (v) from lens and record the value of’u’ and V in the table.
11. Repeat this for various distances of images; in all cases note them in the table.

Observation :

Conclusion : From this we conclude that a convex lens forms both real and virtual images when object is placed at various positions.

AP State Board Syllabus AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Physics Solutions 5th Lesson Refraction of Light at Plane Surfaces

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Textbook Questions and Answers

Improve Your Learning

Question 1.
Why is it difficult to shoot a fish swimming in water? (AS1)
(OR)
If the fish is swimming in water, why it is difficult to shoot?
(OR)
A shooter finds it difficult to shoot a fish swimming in water. Why?
Answer:
Due to refraction of light, it is difficult to shoot a fish swimming in water.

Reason :
The light rays coming from the fish towards shooter, bend at water-air interface. So, shooter sees only image of the fish, but not actual fish.

Question 2.
The speed of light in a diamond is 1,24,000 km/s. Find the refractive index of diamond if the speed of light in air is 3,00,000 km/s. (AS1)
Answer:
Speed of light in diamond = 1,24,000 km/s
Speed of light in vacuum = 3,00,000 km/s

Question 3.
Refractive index of glass relative to water is 9/8. What is the refractive index of water relative to glass? (AS1)
Answer:
Refractive index of glass relative to water = \(\frac{n_{g}}{n_{w}}=\frac{9}{8}\)
∴ Refractive index of water relative to glass = \(\frac{\mathrm{n}_{\mathrm{w}}}{\mathrm{n}_{\mathrm{g}}}=\frac{8}{9} \cdot\left[\because \mathrm{n}_{12}=\frac{1}{\mathrm{n}_{21}}\right]\)

Question 4.
The absolute refractive index of water is 4/3. What is the critical angle ? (AS1)
Answer:
Absolute refractive index of water = 4/3

Critical angle of water = C = 48°5′ = 48.5°.

Question 5.
Determine the refractive index of benzene if the critical angle is 42°. (AS1)
Answer:
Critical angle of benzene = 42°.

Question 6.
Explain the formation of mirage. (AS1)
(OR)
How is the mirage formed? Explain.
(OR)
A person walking on a road observed some water being present on the road but when he went there actually he did not find water. What is that actually formed called? Explain that process.
(OR)
Sometimes during the hot summer at noon time on tar roads, it appears that there is water on the road, but there would really be no water. What do you call this phenomenon? Explain why it happens.
(OR)
Why do you see a mirage the road on a hot summer day?
Answer:

• During hot summer day, air just above the road surface is very hot and the air at higher altitudes is cool.
• We know that refractive index of air increases with density.
• So, the cooler air at the top has greater refractive index than hotter air just above the road.

• Light travels faster through the thinner hot air than the denser cool air above it.
• On hot days, the temperature decreases with height.
• Thus the refractive index of air increases with height.
• When the light from a tall object such as tree or from the sky passes, through a medium just above the road whose refractive index decreases towards ground, suffers refraction and takes a curved path because of total internal reflection.
• This refracted light reaches the observer in a direction shown as in second figure.
• This appears to the observer that the ray is reflected from ground.
• Hence we will see water on road, which is the virtual image of sky and an inverted image of tree on the road.
• Such virtual images of distant high objects cause the optical illusion called ‘mirage’.

Question 7.
How do you verify experimentally that \(\frac{\sin i}{\sin r}\) is a constant? (AS1)
(OR)
Explain the experiment that shows the relation between angle of incidence and angle of refraction through figure.
(OR)
Write an experiment to obtain the relation between angle of incidence and angle of refraction.
Answer:
Aim:
Identifying relation between angle of incidence and angle of refraction.

Materials required :
A plank, white chart, protractor, scale, small black painted plank, a semi-circular glass disc of thickness nearly 2 cm pencil and laser light.

Procedure :

1. Take a wooden plank which is covered with white chart.
2. Draw two perpendicular lines, passing through the middle of the paper as shown in the figure (a).
3. Let the intersecting point be O.
4. Mark one line as NN which is normal to the another line marked as MM.
5. Here MM represents the line drawn along the interface of two media and NN represents the normal drawn to this line at ‘O’.
6. Take a protractor and place it along NN in such a way that its centre coincides with ‘O’ as shown in figure (b).
7. Then mark the angles from 0° to 90° on both sides of the line NN.
8. Repeat the same on the other side of the line NN.
9. The angles should be represented on circular line.
10. Now place semi circular glass disc so that its diameter coincides with the interface line (MM) and its centre coincides with the point O.
11. Take the laser light and send it along NN in such a way that the laser propagates from air to glass through the interface at point O and observe the way of laser light coming from other side of disc.
12. There is no deviation.
13. Send Laser light along a line which makes 15° (angle of incidence) with NN and see that it must pass through point O.
14. Measure its corresponding angle of refraction.
15. Repeat the experiment with angle of incidences of 20°, 30°, 40°, 50° and 60° and note the corresponding angles of refraction.

Observation :

• Find sin i, sin r for every i and r note down the values in table.

• Evaluate \(\frac{\sin i}{\sin r}\) for every incident angle i.
• We get \(\frac{\sin i}{\sin r}\) as constant.
• That is the relationship between angle of incidence and angle of refraction.
• The ratio of sin i and sin r is called refractive index.

Question 8.
Explain the phenomenon of total internal reflection with one or two activities. (AS1)
Answer:
Procedure :

1. Place the semi-circular glass disc in such a way that its diameter coincides with interlace line MM and its center coincides with point O’.
2. Now send light from the curved side of the semicircular glass disc.
3. The light travels from denser medium to rarer medium.
4. Start with angle of incidence (i), equals to 0° and observe for refracted on other side of the disc.
5. It does not deviate into its path when entering rarer medium.
6. Send laser light along angles of incidence 5°, 10°, 15°, etc. and measure the angle of refraction.
7. And tabulate the results in the given table.

Observation :

• Make a table shown below and note the values ‘i’ and ‘r’.

• At particular angle of incidence, the refracted ray does not come out but grazes the interface separating the air and glass. This angle is called critical angle.
• When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e. light never enters rarer medium. This phenomenon is called total internal reflection.

Question 9.
How do you verify experimentally that the angle of refraction is more than angle of incidence when light rays travel from denser to rarer medium? (AS1)
(OR)
When the light rays travel from denser to rarer medium, how can you explain, the angle of refraction is more than angle of incidence experimentally?
Answer:
Procedure :

• Take a metal disc. Use a protractor and mark angles along its edge as shown in the figure.
• Arrange two straws at the centre of the disc, in such a way that they can be rotated freely about the centre of the disc.
• Adjust one of the straws to make an angle 10°.
• Immerse half of the disc vertically into the water, filled in a transparent vessel. While dipping, verify that the straw at 10° must be inside the water.
• From the top of the vessel, try to view the straw which is inside the water as shown in the figure.
• Then adjust the other straw which is outside of the water until both straws look like they are in a single straight line.
• Then take the disc out of the water and observe the two straws on it. You will find that they are not in a single straight line.
• Measure the angle between the normal and second straw. Note the values in the following table.

• Do the same for various angles and find the corresponding angles of refraction and note them in the table.

Observation :
We will find the angle of refraction is more than angle of incidence.
i. e., r > i.

Conclusion :
When light travels from denser (water) to rarer (air) it bends away from the normal.

Question 10.
Take a bright metal ball and make it black with soot in a candle flame. Immerse it in water. How does it appear and why? (Make hypothesis and do the above experiment) (AS2)
Answer:

• The black metallic .ball appears to be raised up in the water because the path of the ray changes its direction at the interface, separating the two media, i.e., water and air.
• This path is chosen by light ray so as to minimize time of travel between ball and eye.
• This can be possible only when the speed of light changes at interface of two media.
• In another way the speed of light is different in different media.

Hypothesis :
Speed of light changes when it travels from one medium to another medium.

Question 11.
Take a glass vessel and pour some glycerine into it and then pour water up to the brim. Take a quartz glass rod. Keep it in the vessel. Observe the glass rod from the sides of the glass vessel.
1) What changes do you notice?
2) What could be the reasons for these changes? (AS2)
Answer:

1. We cannot see the glass rod in glycerine but we can see the rod in water.
2. We can also observe an apparent image of glass rod in water.
3. Reasons:
   i) Glycerine has essentially same refractive index as glass.
   ii) So, any light passing through these is bent equally.
   iii) Since both are transparent, it is not possible for our eye to distinguish the boundary by a change in the angle of reflection, and the glass seems to vanish.
   iv) But, the refractive index of glass and water are different.
   v) So the glass rod is visible to our eye in water. .

Question 12.
Do Activity-7 again. How can you find critical angle of water? Explain your steps briefly. (AS3)
Answer:

Procedure:

1. Take a cylindrical transparent vessel.
2. Place a coin at the bottom of the vessel.
3. Now pour water until you get the image of the coin on the water surface.
4. This is due to total internal reflection.

Critical angle of water :

1. Refractive index of water = 1.33
2. The sine of critical angle of water = \(\frac{1}{\text { Refractive index }}\)
3. Sin C = \(\frac{1}{\text { 1.33 }}\) ⇒ sin C = 0.7518.
   ∴ C = 8.7°
4. ∴ The critical angle of water = 48.7°.

Question 13.
Collect the values of refractive index of the following media. (AS4)

Water, coconut oil, flint glass, crown glass, diamond, benzene and hydrogen gas.

Answer:

Medium	Refractive Index
1. Water	1.33
2. Coconut oil	1.445
3. Flint glass	1.65
4. Crown glass	1.52
5. Diamond	2.42
6. Benzene	1.50
7. Hydrogen gas	1.000132

Question 14.
Collect information on working of optical fibres. Prepare a report about various uses of optical fibres in our daily life. (AS4)
(OR)
What do you know about the working of optical fibres and make a report of various uses of optical fibres in our daily life?
(OR)
How are the optical fibres working? What are the various uses of optical fibres in our daily life?
Answer:
1) An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer (10-6 m).
2) A bunch of such thin fibres form a light pipe.

Working :
1. Optical fibre having three parts-namely core (n = 1, 7), clading (n = 1, 6) and shielding.
2. The ray of light AB gets refracted at point ‘B’ into core and incident at ‘C’ with angle of incidence i (i > c).

3. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
4. The light is thus transmitted along the fibre.
5. The optical fibre is also based on ‘Fermat’s principle.

Uses :

1. Optical fibres are used in ‘endoscopy’ to see the internal organs like throat, stomach, intestines, etc.
2. Optical fibres are used in transmitting communication signals through light pipes.
3. Optical fibres are used in international telephone cables laid under the sea, in large computer networks, etc.
4. In optical fibre about 2000 telephone signals appropriately mixed with light waves may be simultaneously transmitted through a typical optical fibre.

Question 15.
Take a thin thermocol sheet. Cut it in circular discs of different radii like 2 cm, 3 cm, 4 cm, 4.5 cm, 5 cm etc. and mark centers with sketch pen. Now take needles of length nearly 6 cm. Pin a needle to each disc at its centre vertically. Take water in a large opaque tray and place the disc with 2 cm radius in such a way that the needle is inside the water as shown in figure. Now try to view the free end (head) of the needle from surface of the water.

1) Are you able to see the head of the needle?
Now do the same with other discs of different radii. Try to see the head of the needle, each time.
Note : The position of your eye and the position of the disc on water surface should not be changed while repeating the Activity with’other discs.
2) At what maximum radius of disc, were you not able to see the free end of the needle ?
3) Why were you not able to view the head of the nail for certain radii of the discs ?
4) Does this Activity help you to find the critical angle of the medium (water) ?
5) Draw a diagram to show the passage of light ray from the head of the nail in different situations. (AS4)
Answer:

1. Yes, we can see head of the needle.

2. Height of the pin = 6 cm
Radius of the disc = x cm
Critical angle of water = C = 48.7°
Tan C = 48.7°
\(\frac{x}{6}\) = 1.138 ⇒ x = 6.828 cm
So, at radius of 6.8 cm we cannot see the free end of the needle.

3. Because the light rays coming from object undergoing total internal reflection by touching the surface of disc.

4. Yes, we can find critical angle.
Refractive index of air (n₂) = 1.003 ; Refractive index of water (n₁) = 1.33
Sin C = \(\frac{n_{2}}{n_{1}}=\frac{1.003}{1.33}\) = 0.7541 ⇒ C = 48.7°

5.

Question 16.
Explain the refraction of light through the glass slab with a neat ray diagram. (AS5)
(OR)
Draw a glass slab diagram and explain the refraction of light through glass slab.
(OR)
Write the procedure of a lab Activity to understand lateral shift of light rays through a glass slab.
(OR)
How can you find lateral shift using glass slab?
Answer:
Aim :
A) Determination of position and nature of image formed by a glass slab.
B) Understanding lateral and vertical shift.
C) Determination of refractive index of given glass slab.

Materials required :
Plank, chart paper, clamps, scale, pencil, thin glass slab and pins.

Procedure :

1. Place a piece of chart on a plank. Clamp it. Place a glass slab in the middle of the paper.
2. Draw border line along the edges of the slab by using a pencil. Remove it. You will get a figure of a rectangle.
3. Name the vertices of the rectangle as A, B, C and D.
4. Draw a perpendicular at a point on the longer sides (AB) of the rectangle.
5. Now draw a line, from the point of intersection where side AB of rectangle and perpendicular meet, in such a way that it makes 30° angle with the normal.
6. This line represents the incident ray falling on the slab and the angle it makes with normal represents angle of incidence.
7. Now place the slab on the paper in such a way that it fits in the rectangle drawn. Fix two identical pins on the line making 30° angle with normal, such that they stand vertically with same height.
8. By looking at the two pins from the other side of the slab, fix two pins in such a way that all pins appear to be along a straight line.
9. Remove the slab and take out pins. Draw a straight line by joining the dots formed by the pins up to the edge CD of the rectangle.
10. This line represents emergent ray of the light.
11. Draw a perpendicular to the line CD where our last line drawn meets the line CD.
12. Measure the angle between emergent ray and normal.
13. This is called angle of emergence.
14. The angle of incidence and angle of emergence are equal.
15. Incident emergent rays are parallel.
16. The distance between the parallel rays is called shift.

Question 17.
Place an object on the table. Look at the object through the transparent glass slab. You will observe that it will appear closer to you. Draw a ray diagram to show the passage of light ray in this situation. (AS5)
Answer:

Question 18.
What is the reason behind the shining of diamond and how do you appreciate it? (AS6)
(OR)
For which reason is the diamond shining and how is it appreciable?
Answer:

• The critical angle of diamonds is very low, i.e., 24.4°.
• So if a light ray enters diamond, it undergoes total internal reflection.
• It makes the diamond shine brilliant.
• So total internal reflection is main cause of brilliance of diamonds.
• Majority of people are attracted towards diamonds due to this property.
• So we have to thoroughly appreciate total internal reflection for brilliance of diamonds.

Question 19.
How do you appreciate the role of Fermat’s principle in drawing ray diagrams? (AS6)
(OR)
How do you admire the role of Fermat’s principle in drawing ray diagrams? Fermat’s principle: The light ray always travels in a path which needs shortest possible time to cover distance between two points.
This principle has lot of importance on optics. This is used in

1. Laws of reflection (i.e., angle of incidence = angle of reflection)
2. Laws of refraction (Snell’s law)
3. To derive refractive index of a medium.
4. To derive refractive index of glass slab.
   So, I appreciate the Fermat’s principle.

Question 20.
A light ray is incident on air-liquid interface at 45° and is refracted at 30°. What is the refractive index of the liquid? For what angle of incidence will the angle between reflected ray and refracted ray be 90°? (AS7)
Answer:
i) Given that angle of incidence (i) = 4S°
angle of refraction (r) = 30°

ii) Given that angle between reflected and refracted ray is 90°.
We know angle of incidence = angle of reflection
∴ Angle of refraction (r) = 90 – angle of incidence
= 90 – i

Critical angle = 54.7°. This angle is also known as polarising angle.

Question 21.
Explain why a test tube immersed at a certain angle in a tumbler of water appears to have a mirror surface from a certain viewing position. (AS7)
Answer:
When a test tube is immersed at a certain angle in a tumbler of water appears to have
a mirror surface from a certain viewing positions due to total internal reflection.

Explanation :

• The critical angle for glass is 42°.
• The glass and air in test tube works as denser and rarer mediums.
• The rays of light while travelling through water strike glass – air interface of test tube at an angle of more than 42° (i > c) they get totally internal reflected as shown figure.
• When these reflected rays reach the eye, they appear to come from the surface of test tube itself.
• Now the test tube appears like silvary.

Question 23.
In what cases does a light ray not deviate at the interface of two media? (AS7)
Answer:

1. When a light ray incident is perpendicular to the interface of surface, it does not undergo deviation.
2. When a light ray incident is more than critical angle, it does not undergo deviation (refraction) but it undergoes reflection to come back into the original medium.

Question 25.
When we sit at camp fire, objects beyond the fire seen swaying. Give the reason for it. (AS7)
(OR)
What are the reasons for the objects beyond the fire seen swaying, when we sit at camp fire?
Answer:

• The temperature of the surrounding air changes due to convection of heat by the camp fire.
• This leads to chang in density and refractive index of air, continuously.
• The continuous change in refractive index of air changes the refracted path of the light ray.
• This is the cause for swaying of an object.

Question 26.
Why do stars appear twinkling? (AS7)
(OR)
What is the reason for the appearance of stars like twinkling?
Answer:

• The twinkling of a star is due to atmospheric refraction of star light.
• The atmosphere consists of a number of layers of varying densities.
• When light rays coming from a star pass through this layers and undergo refraction for several times.
• Thats why stars appear twinkling.

Question 27.
Why does a diamond shine more than a glass piece cut to the same shape? (AS7)
(OR)
What is the reason for shining of diamond brightly as compared to glass piece cut?
Answer:

• The critical angle of a diamond is very low (i.e., 24.4°).
• So if a light ray enters a diamond it definitely undergoes total internal reflection.
• Whereas it is not possible with glass piece cut to the same shape.
• So diamond shines more than a glass piece.

Fill In The Blanks

1. At critical angle of incidence, the angle of refraction is ……………… .
2. n₁ sin i = n₂ sin r, is called ……………… .
3. Speed of light in vacuum is ……………… .
4. Total internal reflection takes place when a light ray propagates from …………. to …………… medium.
5. The refractive index of a transparent material is 3/2. The speed of the light in that medium is …………… .
6. Mirage is an example of ……………… .
Answer:

1. 90°
2. Snell’s law
3. 3 × 10⁸ m/s
4. denser, rarer
5. 2 × 10⁸ m/s
6. optical illusion / total internal reflection

Multiple Choice Questions

1. Which of the following is Snell’s law?

Answer:
B)

2. The refractive index of glass with respect to air is 2. Then the critical angle of glass air interface is ………………….
A) 0°
B) 45°
C) 30°
D) 60°
Answer:
C) 30°

3. Total internal reflection takes place when the light ray travels from …………….. .
A) rarer to denser medium
B) rarer to rarer medium
C) denser to rarer medium
D) denser to denser medium
Answer:
C) denser to rarer medium

4. The angle of deviation produced by the glass slab is …………… .
A) 0°
B) 20°
C) 90°
D) depends on the angle formed by the light ray and normal to the slab
Answer:
D) depends on the angle formed by the light ray and normal to the slab

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Additional Questions and Answers

Question 1.
Derive Snell’s law.
(OR)
Prove n₁ sin i = n₂ sin r.
(OR)
Derive the Snell’s formula from Fermat’s principle.
(OR)
Derive the formula in realtion with and of incidence and angle of refraction.
Answer:
Let X be the path and A be the point above X and B be the point below X.
Now, we have to find the way from A to B.

1. Let us try to calculate how long it would take to go from A to B by the two paths through point D and another through point C.
2. If we draw a perpendicular DE, between two paths at D, we see that the path on line is shortened by the amount EC.
3. On the other hand, in the water, by drawing corresponding perpendicular CF we find that we have to go to the extra distance DF in water. These times must be equal since we assumed there was no change in time between two paths.
4. Let the time taken by the man to travel from E to C and D to F be ∆t and v₁ and v₂ be the speeds of the running and swimming. From figure we get,
   EC = v₁ ∆t and DF = v₂ ∆t
   ⇒ \(\frac{\mathrm{EC}}{\mathrm{DF}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}\) ………….. (1)
5. Let i and r be the angles measured between the path ACB and normal NN, perpendicular to shore line X.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces InText Questions and Answers

10th Class Physics Textbook Page No. 56

Question 1.
Why should you see a mirage as a flowing water?
Answer:

• A mirage is a naturally occuring optical phenomenon, in which light rays are bent to produce a displaced image of distant objects or the sky.
• As light passes from colder air (higher place) to warmer air (lower place), the light ray bends away from the direction of the temperature gradient.
• Once the rays reach the viewer’s eye, the visual cortex interprets it as if it traces back along a perfectly straight “line of sight”. However this line is at a tangent to the path the ray takes at the point it reaches the eye.
• The result is that an “inferior image ” of the sky above appears on the ground.
• The viewer may incorrectly interprets this sight as water that is reflecting the sky, which is to the brain, a more reasonable and common occurrence.

Question 2.
Can you take a photo of a mirage?
Answer:

• Yes, I can take a photo of a mirage.
• Our eye can catches the total internal reflected rays.
• So, camera lens also catches the same.

10th Class Physics Textbook Page No. 46

Question 3.
What difference do you notice in fig 2(a) and Fig 2(b) with the respect to refracted rays?
(OR)
Draw the ray diagram of refraction in between denser and rarer medium.

Answer:
In figure 2(a) the light ray bends towards normal whereas in 2(b) the light ray bends away from the normal.

Question 4.
Is there any relation between behaviour of refracted rays and speed of the light?
Answer:
Yes. The speed of light changes when it travels from one medium to another medium. So the light may bend towards normal or away from normal.

10th Class Physics Textbook Page No. 47

Question 5.
Why db different material media possess different values of refractive indices?
Answer:
Refractive index depends on nature of material. So different media have different values of refractive indices.

10th Class Physics Textbook Page No. 48

Question 6.
On what factors does the refractive index of a medium depend?
Answer:
Refractive index depends on (1) Nature of material and (2) Wavelength of light used.

10th Class Physics Textbook Page No. 49

Question 7.
Can we derive the relation between the angle of incidence and the angles of refraction theoretically?
Answer:
Yes, we can derive the relation between angle of incidence and angles of refraction theoretically. We can get nt sin i = n2 sin r.

10th Class Physics Textbook Page No. 53

Question 8.
Is there any chance that angle of refraction is equal to 90° ? When does this happen?
Answer:
Yes, when angle of incidence is equal to critical angle then angle of refraction is equal to 90°.

10th Class Physics Textbook Page No. 54

Question 9.
What happens to light when the angle of incidence is greater than critical angle?
Answer:
When the angle of incidence is greater than critical angle, the light ray gets reflected into denser medium at the interface, i.e., light never enters rarer medium. This phenomenon is called total internal reflection.

10th Class Physics Textbook Page No. 57

Question 10.
How does light behave when a glass slab is introduced in its path?
Answer:
The light ray undergoes refraction two times.

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces Activities

Activity – 1

Question 1.
Procedure :
Take some water in a glass tumbler. Keep a pencil in it. See the pencil from one side of glass and also from the top of the glass.
Observation:

1. How does it look?
Answer:
From the side it appears to be bent. From the top it appears as it is raising.

2. Do you find any difference between two views?
Answer:
Yes, the position of pencil is different.

Activity – 2

Question 2.
Procedure :

1. Go to a long wall (of length of 30 feet) facing the Sun. Go to the one end of a wall and ask someone to bring a bright metal object near the other end of the wall.
2. When the object is few inches from the wall, it will distort and we will see a reflected image on the wall as though the wall were a mirror.

Observation:

Why is there an image of the object on the wall?
Answer:
The image is due to refraction of light.

Activity-3 Refraction

Question 3.
Procedure: –

1. Take a shallow vessel with opaque walls such as a mug, a tin or a pan.
2. Place a coin at the bottom of the vessel.
3. Move away from the vessel until we cannot see the coin (fig. 2). Ask someone to fill the vessel with water. When the vessel is filled with water the coin comes back into view (fig. 3).

1. Why are you able to see the coin when the vessel is filled with water?
Answer:
The ray of light originated from the coin does not reach your eye when the vessel is empty. Hence you are not able to see the coin. But the coin becomes visible after the vessel is filled with water.

2. How is it possible? Do you think that the ray reaches your eye when the vessel is filled with water?
Answer:
Yes, it reaches the second instance.

3. What happens to the light ray at interface between water and air?
Answer:
It bends towards the normal.

4. What could be the reason for this bending of the light ray in the second instance?
Answer:
It is based on Fermat’s principle, which states that the light ray always travels in a path which needs shortest possible time to cover the distance between the two points.

Activity – 4

Question 4.
Prove that when light ray travels from denser to rarer medium it bends away from the normal.
Answer:


Procedure :

1. Take a metal disc. Use protractor and mark angles along its edge as shown in the figure.
2. Arrange two straws from the centre of the disk.
3. Adjust one of the straws to the angle 10°.
4. Immerse half of the disc vertically into the water, filled in a transparent vessel.
5. Inside the water the angle of straw should be at 10°.
6. From the top of the vessel try to view the straw which is inside the water.
7. Then adjust the other straw which is outside the water until both straws are in a single straight line.
8. Then take the disc out of the water and observe the two straws on it.
9. We will find that they are not in a single straight line.
10. It could be seen from the side view while half of the disc is inside the water.
11. Measure the angle between the normal and second straw. Draw table for various angles and corresponding angles of refraction.

Observation :
We observe that ‘r’ is greater than ‘i’ in all cases and when light travels from
denser to rarer medium it bends away from the normal.

Activity – 6

Question 5.
Why can we not see a coin placed in water from the side of glass?
Answer:
Procedure :

1. Take a transparent glass tumbler and coin.
2. Place a coin on a table and place glass on the coin.
3. Observe the coin from the side of the glass. We can see the coin.
4. Now fill the glass with water and observe the coin from the side of the glass tumber.
5. Now we cannot see the coin because the coin rises up due to refraction.

Activity – 7

Question 7.
Why can we see the coin in water from top? What is the phenomenon behind that?
Answer:

Procedure :

1. Take a cylindrical transparent vessel. Place a coin at the bottom of the vessel.
2. Now pour water until we will get the image of the coin on the water surface.
3. This is due to total internal reflection.

Activity – 8

Question 8.
Write an Activity to find refractive index of glass slab by calculating vertical shift.
(OR)
Explain the experiment with glass slab in determination of refraction through vertical shift.
Answer:

Procedure :

1. Take a glass slab and measure the thickness of the slab.
2. Take a white chart and fix it on the table.
3. Place the slab in the middle of the chart.
4. Draw line around it.
5. Remove the slab from its place.
6. The lines form a rectangle. Name the vertices of it as A. B, C and D. ‘
7. Draw a perpendicular to the longer line AB of the rectangle at any point on it.
8. Place slab again in the rectangle ABCD.
9. Take a needle. Place at a point P in such a way that its length is parallel to the AB on the perpendicular line at a distance of 20 cm from the slab.
10. Now take another needle and by seeing at the first needle from the other side of the slab, try to keep the needle so that it forms a straight line with the first needle. 1 1)
11. Remove the slab and observe the positions of the needles.
12. They are not in same line.
13. Draw a perpendicular line from the second needle to the line on which the first needle is placed.
14. Take the intersection point as Q.
15. The distance between P and Q is vertical shift.
16. We will get the same vertical shift placing needle at different distances.
17. 

SCERT AP Board 9th Class Social Solutions 24th Lesson Traffic Education Textbook Questions and Answers.

AP State Syllabus 9th Class Social Studies Solutions 24th Traffic Education

9th Class Social Studies 24th Lesson Traffic Education Textbook Questions and Answers

Improve Your Learning

Question 1.
What documents should a driver carry while driving and what skills are needed to drive safely?
Answer:
The following documents should be carried while driving :

1. Driving licence
2. Registration certificate
3. Taxation certificate
4. P.U.C. certificate
5. Insurance certificate
6. Fitness certificate and permit

The following is a basic description of the skills and abilities a driver needs before getting in the driver’s seat. This applies to people of all ages.

Physical skills and abilities :
Driving requires physical strength. It takes a lot of muscle work too.

1. Hold the body upright to use and control the steering wheel,
2. Maintain sitting balance,
3. Control the head, neck, arms and legs, feet and hands.
4. To operate a vehicle

Physical and mental stamina and muscle flexibility are needed to :

1. sit and drive
2. focus constantly on the task of driving.
3. twist and turn
4. move the head and’ neck side to side, up and down, back and forth.

In addition to these driving requires a clear and alert mind.

Question 2.
What will happen if someone jumps the traffic signal?
Answer:
Traffic signals help to control traffic. It someone jumps the traffic signal, there is a chance of accident.

Question 3.
Suggest a few steps that are taken in your area for road safety.
Answer:
Road safety is a result of contributing efforts from all the sections the society including both civilians and government officials. In addition to the human sufferings, the estimated costs of the road injuries are a noticeable amount in GNP per annum. So some steps can be fruitful in this direction.
A few important road safety steps :

1. Don’t use mobile phone whilst driving.
2. Belt up in the back.
3. Don’t drink and drive.
4. Always adhere to speed limits.
5. Take special care about children, senior citizens and pedestrians.
6. Don’t drive if tired.
7. Pedestrians should walk cautiously.
8. Always observe and anticipate other road users.
9. Keep your distance and
10. Always wear helmets and seat belts.

We follow all these road safety rules in our area.

Question 4.
Explain mandatory, caution and information traffic signs with examples.
Answer:
1) Mandatory signs :
Mandatory signs are indicated in a circular form. In accordance with the motor vehicle Act 1988, every driver of a motor vehicle shall drive the vehicle in conformity with any indication given by the mandatory signs and not obeying there signs is an offence.
Ex : 1. Stop
2. No Parking
3. Overtaking prohibited

2) Cautionary /Warning signs :
Cautionary signs are meant for cautioning the driver about the hazards lying ahead on the road. Drivers must obey there signs for safety. These signs are indicated in a triangular form
Ex : 1. T -inter section
2. Right hand curve
3. School ahead

3) Information signs:
Informatory signs are erected on the road to provide information on direction, destination, road side facilities etc., to the road users.
Ex : 1. Park this side
2. First aid post
3. Public telephone

Question 5.
Kamala wants to purchase a new vehicle. Explain her what are the steps to be taken and what documents are to be produced for the registration of a vehicle?
Answer:
The steps to be taken by Kamala in purchasing a car or something else :

1. Starting out
2. Using incentives and rebates
3. Pricing the car
4. Finding the exact car you want to buy
5. Test driving the car salesman
6. If you are trading in your old car
7. Negotiating for the best finance options
8. Closing the deal
9. Reviewing and signing the paper work,
10. Inspecting and taking possession of your new car.

Documents to be produced for the registration of a vehicle :

1. R.T.O. forms – a) Form 20 b) Form 34
2. Pan card , copy of sales certificate
3. Address proof
4. Insurance cover note
5. Person’s authorized signature
6. Copy of road worthiness certificate
7. Pollution under control certificate

Question 6.
Ramu wants to interchange his vehicle number to other vehicle. Is it correct or not?
Explain, why.
Answer:
No. Ramu cannot interchange his vehicle number to other vehicle.
Reason : Any vehicle registered for its particular registration mark will remain its
identifications and interchange of it is not allowed.

Question 7.
Explain the need of road safety.
Answer:
India loses more than 1,00,000 lives due to road traffic crashes every year. It has a road traffic fatality rate of 16.8 deaths per, 1,00,000 population. Approximately half of all deaths on the country’s roads are among vulnerable road users – motorcyclists, pedestrians and cyclists.

Hence road safety is very important to avoid the accidents and control loss of lives. Road safety ensures that every road user follow traffic rules and thereby avoid traffic jams and deaths due to accidents. For systematic regulation of vehicular traffic, road safety is needed.

Question 8.
Read the table of page ‘Accident Victims Age’ on page 286, identify the age group for which more number of cases registered and draw a bar graph.

Age	Cases	Age	Cases
00-05	24	50-55	207
05-10	58	55-60	138
10-15	40	60-65	113
15-20	152	65-70	57
20-25	345	70-75	49
25-30	380	75-80	13
30-35	254	80-85	12
35-40	294	85-90	0
40-45	226	90-95	0
45-50	215	95-100	0

Answer:
More number of cases registered age group 35 – 40 years age group.

Question 9.
Read the paras under the title ‘Traffic Chaos’ of page 287 and comment on them.
Traffic Chaos:
You have to go to school on the morning. If you are late, you may miss classes. You are struck in a traffic jam. What will you do?
Students, employees, labourers, teachers, doctors and all are affected by traffic jams. Foot paths (Side walks) are considered a boon for pedestrains. Sometimes motorists drive on these side walks also.

Stray animals, fruit and vegetable seller, private vehicles like cars, autorickshaws are parking at No Parking Zones are the main causes for traffic jams. As there is an increase in population and use of automobiles, there has been a rapid increase in the volume of traffic on roads. To avoid the accidents, one must know the prescribed rules and regulations.
Answer:

Road sense on Indian streets is often completely missing. The Indian traffic conditions are chaotic, the drivers are reckless, and the roads are poor repair conditions.

There is a pecking order for right of way – cows / buffaloes are at the top, trucks and buses are second, and dogs and pedestrians are at the bottom. Two wheelers are pretty low down. Pot holes and speed breaker bends are common and rarely marked. Pedestrians, . animals, bicycles, ox carts and tractors all use the roads.

Question 10.
Collect the data from the traffic police /RTA officials who are nearest to you.
Month : Place :
No. of cases booked :

Analyse the data and discuss in your classroom regarding traffic situations in your area.
Answer:
Month : September Place : Vijayawada
No. of cases booked : 1986

The above data is revealing us the negligence of the vehicle riders and road users. They are to be strictly punished.

9th Class Social Studies 24th Lesson Traffic Education InText Questions and Answers

Question 1.
Why is it compulsory to have a driving license? (Text Book Page No. 288)
Answer:
Yes, it is compulsory to have a driving license. As per Motor Vehicle Act 1988, a valid driving license is necessary to drive any motor vehicle on public roads.

Question 2.
Observe the given table and answer the questions. (Text Book Page No. 286)

Age	Cases	Age	Cases
00-05	24	50-55	207
05-10	58	55-60	138
10-15	40	60-65	113
15-20	152	65-70	57
20-25	345	70-75	49
25-30	380	75-80	13
30-35	254	80-85	12
35-40	294	85-90	0
40-45	226	90-95	0
45-50	215	95-100	0

1) In which group do you find more cases? Can you say, why.
Answer:
25 – 30. As the people in this age become more independent, they are exposed to risks.

2) How many cases are there in the age group of both 20 – 25 and 25 – 30?
Answer:
345; 380

Question 3.
Observe the following pie-chart and answer the following questions. (Text Book Page No. 287)
1) Discuss the data relating to the accidents – accused vehicles in your classroom.

Answer:
Group discussion in classroom

2) Which type of vehicles are accused of more accidents? Can say why?
Answeer:
Two wheelers. Because they are in more number.

3) What are Traffic Rules and regulations? Discuss in your classroom.
Answer:
Traffic rules and regulations of the road are both the laws and the informal rules that may have been developed to facilitate the orderly and timely flow of traffic.
Note : Students should discuss in the classroom.

Question 4.
With the help of your teacher collect the road surface markings from RTA/Traffic police and discuss the uses of various markings in the classroom. (Text Book Page No. 290)
Answer:
Uses of various markings :

1. Road surface markings are used on paved roadways to provide guidance and information to drivers and pedestrians.
2. These markings promote road safety.
3. These are used to supplement the message of road signs and other devices.

Note : Students should collect road surface markings and discuss in the classroom.

SCERT AP Board 9th Class Social Solutions 23rd Lesson Disaster Management Textbook Questions and Answers.

AP State Syllabus 9th Class Social Studies Solutions 23rd Disaster Management

9th Class Social Studies 23rd Lesson Disaster Management Textbook Questions and Answers

Improve Your Learning

Question 1.
Explain how a natural harzard becomes a disaster.
Answer:

• A natural hazard is a natural event of unusual magnitude that people don’t expect and cannot control.
• Natural hazards threaten people’s lives and their activities and can forever change their ways of living.
• A natural hazard event can become a natural disaster when causes the destruction of people’s property or their injury and/or death.

Question 2.
What is Terrorism? What are the motives of terrorists?
Answer:
A common definition of terrorism is the systematic use or threatened use of violence to intimidate a population or government and thereby effect political, religious or ideological change.
Motives of terrorism :

• Terrorism is a type of violence used tactically in peace, conflict and war.
• The motivation of terrorists are quiet various depending on what they hope to accomplish.
• The majority of the motives can be broken down into three broad areas: rational, psychological and cultural.
• Although some motives are difficult for the majority of the world to comprehend given a motive, a terrorist group has a purpose for violence.

Question 3.
What safety measures should be followed to avoid fire accidents?
Answer:
Safety measures to be followed to avoid fire accidents:

• Prevention of fire is necessary for avoiding accidents.
• Never leave the kitchen during cooking, no matter what the circumstance.
• All the inflammable things or liquids should not be kept near the fire.
• Electrical appliances should have incorporated in an automatic cut off feature for some time. Timers can be used for these.
• Wear suitable dresses while cooking.
• Installation of fire detectors in the house will alert automatically the fire department.
• One should not smoke in bed/throw burning buds here and there.
• One should not play crackers etc., near the house.
• One should not keep match boxes, lighter etc., at the reach of the children.
• One should not keep agarbattis carelessly.
• One should not make fire when wind is blowing.
• One should close the regulator of the gas cylinder properly etc.

Question 4.
What are the main causes of road accidents? How can we reduce the road accidents?
Answer:
The main causes of road accidents:

1. Alcohol & drunk – driving.
2. Speeding
3. Weather related
4. Hydroplanning
5. Animals in road.
6. Street racing.
7. Cell phones.
8. Music, CD players, Radio distractions
9. Driver fatigue
10. Suicide car accidents.

Steps to reduce road accidents :

1. Strict enforcement of speed limits.
2. Heavy penalty should be imposed on all those who cross speed limits.
3. Lower age limit for 2 wheeler and heavy vehicle licence should be raised to 21.
4. Driving tests for issue of driving license is to be made more stringent and fool proof.
5. Helmet should be made compulsory by law.
6. Existing traffic rules should be strictly enforced, etc.

Question 5.
Terrorist attacks cause loss of life and wealth. What are the precautions to be taken to prevent these activities?
Answer:

• Devastating acts have left many concerned about the possibility of future incidents.
• Taking preparatory action can reassure us that we can exert a measure of control even in the face of such events.
• Finding out what can happen is the first step. Then develop a disaster plan.
• Create an emergency communications plan.
• Establish a meeting place.
• Assemble a disaster supplies kit.
• If disaster strikes :
  a) Remain calm and be patient.
  b) Follow the instructions of local emergency officials.
  c) If the disaster occurs near you, check for injuries. Give first aid and get help for seriously injured people.

Thus we can protect from terrorist attacks.

Question 6.
Identify the causes of rail accidents.
Answer:
Causes of rail accidents :

1. Lack of proper maintenance.
2. Human error or sabotage.

Question 7.
List out the reasons and risk inducing factors in your home, school or village.
Answer:
Three risk inducing factors :

1. In my home: Nothing.
2. In my school:
   a. There is a big pit in our school. I think it is dangerous.
   b. We have 3 storeyed building as our school. The little children are in the third stair. I think it is also a risk factor.
3. In my village : My village is in earthquake zone.

Question 8.
Locate the following places attacked by terrorists on the India map.
a) Mumbai
b) Hyderabad
c) Bhagalpur
d) Kumbakonam
e) Bengaluru
Answer:

Question 9.
Write about an incident related disaster you know.
Answer:
Some years ago we went to Marina beach in Chennai. We enjoyed a lot by playing there. We built some structures in the sand. We ate peanuts, groundnuts, mango pieces, etc. on our return we walked far away from the sea. Suddenly we heard the noises of the people in the beach. Then we saw a big wave. Many were drowned and lost their lives. We ran very fast from the place. By god’s grace we are still alive.

Question 10.
Read the para under the title ‘Fire Accidents’ on page 281 and comment on it.
Fire Accidents :
The 2004 fire in a school in Kumbakonam, Tamilnadu sparked ofT debates and arguments on the safety of schools in the*ountry. 93 innocent people were charred to death. The main causes of this fire were lack of awareness amongst children and teachers as to what to do in case of a fire, lack of preparedness, kitchen close to the classroom etc. However, incidents like these are not new.
Answer:
Fire accidents can be due to various reasons. They include open flames such as a candle being accidentally being knocked over or through electrical faults. It is important to know the fire safety rules because fire accidents can result in disastrous personal injury and distressing damage. Fire safety plans are to be planned.

Question 11.
Collect information from newspapers and magazines about the human induced disaster that have taken place in the past few years in your state. Find out, what measures are being taken in your area to reduce such risks in the future.
Answer:
Nowadays road accidents became a common scenario in India. The main reasons behind the road accidents are not obeying speed limits, poor roads, drinking and driving, rash driving, aggressive driving etc.

Information about road accidents in Telangana and Andhra Pradesh :

Year	No. of accidents	No. of deaths
2004	39,390	10,621
2005	38,913	11,098
2006	42,867	12,606
2007	45,163	13,791
2008	46,389	14,529
2009	45,977	15,203
2010	44,570	15,696
2011	42,869	15,100
2012	41,712	14,975

On an average of last 6 years the number of accidents are 44,446 and the number of deaths are 14,882 in Telangana and Andhra Pradesh.
The following measures are taken to control the road accidents in Andhra Pradesh :

1. Increased traffic fines in India.
2. Suspending licence for drunken drivers.
3. Established traffic courts.
4. Started traffic awareness programmes for drivers.

Question 12.
List the various losses incurred due to terrorism.
Answer:

1. Fear in people
2. Psychological problems in people
3. Negative effect on tourism
4. Financial or economic losses
5. Unstable governments and
6. Loss of human lives, etc.

9th Class Social Studies 23rd Lesson Disaster Management Activities

Question 1.
From the nearest fire station, find out what are the other simple Do’s and Don’ts that you can follow and awareness to reduce fire accidents.
Answer:
I live in Vijayawada. We have fire station in Krishnalanka. I visited that and learnt the following. Do’s :

• Do have an adult always present when cooking is going on the kitchen. Children should not be allowed alone.
  Do keep hair tied back and do not wear synthetic clothes when you are cooking.
  Do make sure that the curtains on the window near the stove are tied back and will not blow on the flame or burner.
  Do check to make sure that the gas burner is turned off immediately if the fire is not ignited and also switched off immediately after cooking.
  Do keep matches out of the reach of children, etc.

Don’ts :

• Don’t put any clothes near the stove.
• Don’t wear loose fitting clothes or chunnies, etc.
• Don’t store spray cans or cans carrying inflammable items near the stove.
• Don’t let small children near an open oven door.
• Don’t lean against the stove to keep warm, etc.

Question 2.
Identify recent terrorist acts that have taken place in India, and discuss the possible effects of these acts on children.
Answer:
Terrorist attack at Amarnath on July 11, 2017 is recent attack that have taken place in India.

Many children are exposed to trauma and life-threatening solution, duringthe last few decades, thousands of children have been exposed to terrorism. The attacks significantly, affect the mental health of the children. Terrorist attacks and their aftermath have had a powerful impact on children and their families. Media and television exposure of terrorist events throughout the world has increased during the past few years. There is increasing concern about the effects of this exposure on children who witness these violent images. Recent studies have examined the effect of remote exposure of terrorist attacks and have shown a significant clinical impact on children and families.

SCERT AP Board 9th Class Social Solutions 22nd Lesson Women Protection Acts Textbook Questions and Answers.

AP State Syllabus 9th Class Social Studies Solutions 22nd Women Protection Acts

9th Class Social Studies 22nd Lesson Women Protection Acts Textbook Questions and Answers

Improve Your Learning

Question 1.
What are the disadvantages of child marriage?
Answer:
Disadvantages of child marriage :

• Underage pregnancy.
• Provision for trafficking and sale of girls.
• Pushing the unmatured into family system.
• Increased abortions, premature births which results not only infant mortality but also the death of mothers.
• Handicapped child births or dead child births.
• Mental as well as physical health problems.
• Obstacle to education.
• Hurdle for physical growth (especially regarding female).
• Become child labourers for family maintenance.

Question 2.
Why has domestic violence become a common practice? In what forms domestic violence is seen? Find out the reasons.
Answer:
Reasons for the domestic violence which became common practice :

1. The family system in the society is on the edge of a blade.
2. No moral values are taught to the children.
3. Lack of understanding capacity in the people.
4. Changing cultures in the society.
5. Hurting manner.
6. Lack of adjustment between family members.
7. Lack of respect towards women.
8. Uncontrolled anger, i.e. mental imbalance.

Question 3.
You have read about various problems girls and women. Have you noticed any sort of problems in your village/town? Specify them and what is to be done.
Answer:
I observe the following in our area.

1. An estimated 85 to 90 percent of domestic violence victims are females.
2. Females are victims of intimate partner violence at a rate about five times that of males.
3. Females between the ages of 16 and 24 are most vulnerable to domestic violence.

To control this domestic violence –

1. Proper counselling should be given to them.
2. Value oriented education should be taught to the children.
3. The nature of thinking from others’ point of view’ should be developed from childhood.

Question 4.
Many acts are made by the government. What do you suggest for better implementation?
Answer:
The Acts cause changes physically. But there should be a change in the minds of the people. The government should take necessary steps for this change.

Question 5.
Write an essay on general problems, the women facing.
Answer:
A Thomson Reuters Foundation expert poll last year ranked India as the world’s fourth most dangerous country for a woman, behind only Afganisthan, Congo, and Pakisthan.

Even though the practice is outlawed 300,000 to 600,000 female foetuses are aborted every year in India because of the preference for boys.

From the time they are born or not born and continuing till late in life when they become wives or mothers, it’s a vicious cycle of discrimination and violence keeps on continuing.

Nearly half of Indian girls are married off before the age of 18. Girls are also seen as a financial liability. The practice of dowry is banned by the government, but it’s still as common as ever.

A 2012 of UNICEF study found more than half of Indian males think it is justifiable to beat a wife under certain circumstances.

Outside the household, crimes against women in India are also on the rise, and the evidence is shocking. According to government data, more than 24,000 cases of rape were reported in 2011.

Indian women in some ways, have also made some strides. Literacy rates have gone up, maternal mortality rates have gone down, and millions of women have joined the workforce.

Authorities acknowledge that action is needed and say they are taking steps to try to better protect women.

Helplines have been set up. A number of fast track courts have also been established because of the December 16 gang rape, exclusively for cases of sexual assault and rape. But women’s rights activists say that when discrimination begins even before birth, change will not come easily.

Question 6.
Imagine that you are Tehsildar. How would you prevent child marriages?
Answer:
As a Tehsildar –

1. I will raise the awareness in parents and their relatives.
2. If they don’t listen to me, I will request the police department to take them into custody. Thus I will prevent child marriages.

Question 7.
Prepare a pamphlet to understand the problems faced by the women.
Answer:

Women in India

Women in Independent India are comparitively in a respectable position. Some of the problems which had been haunting the community of women for centuries are not found now.
But still, for a woman, her life is a battle for survival and dignity from her birth to death. Female infanticide, eve-teasing, sexual exploitation, sexual harassment, rapes are still common. There are still many areas where women have not equal rights and are not encouraged to take up education. Dowry practice is still rampant, eventhough it termed illegal. Women still have to make a lot of sacrifices in every areas of their lives.

Question 8.
Read the paragraph under the title ‘Sexual Assault and Torturing’ of page 275 and comment on it.

Sexual Asssault an Torturing :
Recently, central government has appointed Justice J.S. Verma Committee asking suggestions to prevent sexual assault and abuse. Based on the suggestions, Hon. President of India issued an ordinance on 2nd February, 2013. Some important features of the ordinance, are;

• Minimum 20 years of imprisonment.
• Recruitment of women police to address the complaints from the victims.
• It is not necessary for the victim to attend before the police officers individually.
• There will be no punishment if the accused dies in struggle at the time of attack with acid on women.
• There is scope to short video at the time of complaining and trial on the request of the victims.

Answer:

1. Only legislation and law enforcement agencies cannot prevent the incident of crime against women.
2. There is a need of social awakening and change in the attitude of masses, so that due respect and equal status is given to women.

Question 9.
What facilities does the government provide to the poor for seeking judicial justice?
Consult a lawyer and get the information.
Answer:

• Lok Adalat is a relatively recent product in the justice for the poor.
• Next to that there are Legal Aid Committees, Legal Services Authority.

Services offered by the legal services authority :

1. Payment of court and other process fee,
2. Changes for preparing, drafting and filing of any legal proceedings.
3. Changes of a legal practitioner or legal advisor.
4. Costs of obtaining decrees, judgements, orders or any other documents in a legal proceeding.
5. Costs of paper work, including printing, translation, etc.

These are the facilities provided by the government to the poor for seeking judicial justice.

9th Class Social Studies 22nd Lesson Women Protection Acts InText Questions and Answers

Question 1.
Have you ever seen child marriages? (Text Book Page No. 269)
Answer:
No, I have never seen child marriages.

Question 2.
Have you ever observed harassment and violence made to women and girls? (Text Book Page No. 269)
Answer:
Yes, I have observed in many of my neighbouring houses.

Question 3.
Have you ever observed the various types of domestic violence threatening women in day to day life? (Text Book Page No. 269)
Answer:
I have observed the following,

1. Persisting denial of food,
2. Insisting on perverse sexual conduct
3. Constantly locking a woman out of the house.
4. Denying the woman access to children, thereby causing mental torture.
5. Physical violence.
6. Taunting, demoralising and putting down the woman with the intention of causing mental torture.
7. Abusing children in their mother’s presence with the intention of causing her mental torture.
8. Mental abuse (threaten to harm, stopping from jobs, force to marry whom she doesn’t like).
9. Threatening divorce unless dowry is given.

Question 4.
What shall we do to stop such violence caused to girls and women? (Text Book Page No. 269)
Answer:
We should take the following steps to stop such violence to girls and women.

1. Value oriented education should be given to the children in schools.
2. The persons, who caused violence, should be punished.

Question 5.
Have you ever seen or heard about women being tortured for dowry in your locality? How are they being tortured? What sort of suggestions do you make to prevent? Who will held responsible? (Text Book Page No. 273)
Answer:
Yes, I have seen such a case. My neighbour ‘Vasu’ is harassing his wife for dowry. He has his mothers support also. They usually beat her, abuse her. I suggest her to proceed women protection cell authorities. Then the project officer will hold responsible.

Question 6.
Domestic violence takes place now and then and gradually it becomes a bad habit pattern. More violence generates from violence. How do we stop this? Who will help? (Text Book Page No. 274)
Answer:
In order to put a stop to the domestic violence, there are various things which should be followed.
1) Awareness:
To make the people aware about the drawbacks and harms of the domestic violence, setthe rules againstthe practice and impose stringent punishments to the attacker.

2) Need for stringent laws:
It is very important that the law against domestic violence has to be imposed stringently.

3) Empowered non-government organizations:
Individuals can also seek the help of NGOs.

4) Seek police help:
In case of severe violence, individuals can seek legal help from the police and put an end to domestic violence.

5) Be aware of the domestic violence facts:
The facts of the domestic violence should be learnt by all.

6) Encourage and not threaten:
It is very important to organize a meeting and encourage people to come up with the solutions.

7) Counselling :
It is very important to have counselling in terms of danger.

Question 7.
Sometimes child marriages are performed for the under aged i.e., less than 15 years without their consent. How do we stop such marriages? Who will help us? (Text Book Page No. 271)
Answer:
We can approach the higher officials in the area – to stop the child marriages.

1. District Collector at district level.
2. R.D.O at division level
3. Tehsildar or Mandal level ICDS officer.
4. Gram Panchayat Secretary at village level are the incharges in stopping the child marriages. They will help us.

 

SCERT AP Board 9th Class Social Solutions 21st Lesson Human Rights and Fundamental Rights Textbook Questions and Answers.

AP State Syllabus 9th Class Social Studies Solutions 21st Human Rights and Fundamental Rights

9th Class Social Studies 21st Lesson Human Rights and Fundamental Rights Textbook Questions and Answers

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Question 1.
Which of the following is not an instance of an exercise of a fundamental right
a) Workers from Bihar go to Punjab to work on the farms
b) Religious minority set up a chain of schools
c) Men and women government employees get the same salary
d) Parents’ property is inherited by their children
Answer:
d) Parents’ property is inherited by their children

Question 2.
Which of the following freedoms is not available to an Indian citizen?
a) Freedom to criticise the government
b) Freedom to participate in armed revolution
c) Freedom to start a movement to change the government
d) Freedom to oppose the central values of the Constitution
Answer:
b) Freedom to participate in armed revolution

Question 3.
Which of these statements about the relationship between democracy and rights is more valid? Give reasons for your preference.
a) Every country that is a democracy gives rights to its citizens.
b) Every country that gives rights to its citizens is a democracy.
c) Giving rights is good, but it is not necessary for a democracy.
Answer:
I prefer the first sentence. The reasons are –
a)

1. All democratic countries ensure certain rights to its citizens.
2. Rights are reasonable claims of the people.
3. Democratic governments strive to preserve equal ground for all.

b)

1. During the early days of modern history, all despotic governments granted certain rights to the people. That was done under great pressure.
2. Therefore every country that gives rights to its citizens is not a democratic.

c) The rights are so important that they are also expressed by many democratic countries and codified by the UNO and find first place in the universal declaration of human rights. Hence it is necessary for democracy to give rights.

Question 4.
Are these restrictions on the right to freedom justified? Give reasons for your answer.
a) Indian citizens need permission to visit some border areas of the country for reasons of security.
b) Outsiders are not allowed to buy property in some areas to protect the interest of the local population.
c) The government bans the publication of a book that can go against the ruling party in the next elections.
Answer:
a)

1. Yes, it is justifiable.
2. It is the responsibility of the government to protect the life of the people.
3. Border areas are high risk areas and tension always persists there.
4. Hence permission is necessary.

b)

1. No, not justifiable.
2. People in democracy have freedom to reside in any part of India.
3. Hence, this statement is against the fundamental rules.
4. But, there is one clause in our Constitution itself, that we cannot buy property in the state of Jammu and Kashmir.
5. So Jammu and Kashmir is an exemption to the fundamental rule.

c)

1. No, not justifiable.
2. Democracies grant civil liberties to its citizens.
3. Under civil liberties, we can express our ideas through media or books.
4. But generally, government bans certain books in order to avoid internal tensions.

Question 5.
Look through this chapter and the previous one and make a list of the six Fundamental Rights in the Constitution.
(OR)
What are fundamental rights? How are these helping us to live better?
(R)
Write any four fundamental rights enjoyed by the citizens of India.
Answer:
There are six fundamental rights. They are :

1. Right to equality
2. Right to freedom
3. Right to religious freedom
4. Right against exploitation
5. Right to education and culture
6. Right to constitutional remedies.

Fundamental rights protect the liberties and freedom of the citizens against any invasion by the state, prevent the establishment of the authoritarian and dictatorial rule in the country. They are very essential for the all-round development of the individuals and the country.

Question 6.
Are the Fundamental Rights being violated in each of the following cases? If so, which Fundamental Right or Rights? Discuss with your classmates.
a) Suppose a person is kept in a police station for 4 days without being told the reasons, which law was broken?
b) Suppose your neighbor tries to claim some of your land as her own.
c) Suppose your parents do not allow you to go to school. They make you take up a job in a match factory instead because they cannot afford to feed you properly.
d) Suppose your brother refuses to give you the land that you have inherited from your father.
Answer:
a)

1. In the first case, arresting a person without proper reason and keeping in a police station for 4 days is against the fundamental rights.
2. “Rights to life” and personal liberty ensures that “no one can be arrested without being told the grounds for his arrest.
3. “Hebeaus Corpus” writ protects the individuals from the arrest.

b)

1. Occupation of one’s land by another is not a violation of fundamental rights.
2. Right to property is a legal right.

c)

1. Not allowing a child to go to school is definitely violation of fundamental right.
2. “Right to education” is a part of “right to life”.
3. Government is responsible for providing free and compulsory education to all the children up to the age of 6 to 14 years.

d)

1. Refusal on part of your brother to give you land that you inherited is not violation of fundamental rights.
2. Right to property is a legal right and civil courts will solve the problem.

Question 7.
Suppose you are an advocate. How would you argue the case for a group of people who come to you with the following request:
“The river in our area is getting very polluted by the factories upstream. We get our drinking water from the river. People in our villages keep falling ill because of the polluted water. We have complained to the government but there has been no action from their side. This is surely a violation of our Fundamental Rights.” ;
Answer:

• Being an advocate I would like to file a writ in the court for the interests of the public.
• This is surely a violation of our fundamental rights.
• Hence I argue the case in such a way that immediately stay order would be issued to close down the factory.
• When the government did not respond to our complaints, courts would definitely safeguard our interests.

Question 8.
Read the paragraph under the heading ‘Abolition of Title’ and answer the following question:
Abolition of Title:
In another move to remove arbitrary and unequal classification of the aristocratic class and the bourgeoise, the Constitution prohibits the State from conferring any titles. The British government had created an aristocratic class known as Rao Bahadurs and Khan Bahadurs in India -these titles were also abolished. Citizens of India cannot accept titles from a foreign State. However, military and academic distinctions can be conferred on the citizens of India. The awards like the Bharat Ratna, the Paramveer Chakra, and the Padma Vibhushan cannot be used by the recipient as a title and do not, accordingly, come within the constitutional prohibition.
The awards can’t be used by the recipient as a title. Why?
Answer:

• In order to remove arbitrary and unequal classification of the aristocracy and middle class, the constitution prohibits the state from conferring any titles.
• Hence the awards like the Bharat Ratna, the Padma Vibhushan cannot be used by the recipients as a title.

Question 9.
Analyse an incident you know about where the Fundamental Rights are violated.
Answer:

• The 1984 Anti-Sikh Riots was a four-day period during which sikhs were massacred by members of the secular-centrist Congress party of India, some estimates that more than 2000 were killed. (Religious violation)
• Dalits and indigeneous peoples continue to face discrimination, exclusion and acts of communal violence.
• Narco analysis test (against to the Indian constitution), “nobody may be made a witness against himself, etc.

Question 10.
Invite a senior advocate into your classroom and collect the following information by conducting an interview.
– violation of fundamental rights and its consequences
– violation of children rights
– ways of struggle for rights in democracy
– any other related
Answer:
Students : Good morning sir.
Advocate : Good morning children.
Students : Sir, today we are going to known about the fundamental rights and importance of other rights from you sir.
Advocate : Yes, children, I will explain. What do you know about.
Students : Sir what will happen, if we violate fundamental rights.
Advocate : Courts will punish us.
Students : Sir please explain one example?
Advocate : If any person is created a nusence in the public places, he created inconvience to the freedom of other people. Then the police arrested that person and kept in the prision.
Students : What will happen when violate the children’s rights?
Advocate : Children are the tomorrow’s citizens generally 6-14 years age is considred as children. So that age children should be in school. But if they did not go to school and work in any where the owner will punish by government / court. Parents should provide education to their children. That is their fundamental right.
Students : Sir what are the ways to struggle for rights in democracy?
Advocate : Students in a democracy always we are fighting for our rights. We will achieve our rights in a peaceful manner. So movements will be in a democratic manner not in a violent manner. These are in through petitions, strikes etc.
Students : Sir please explain any other related issues.
Advocate : Children fundamental rights are provided by our constitution. We will enjoy that not violate and not create any inconvienient to others it will we create any we will punish by government and lost our valuable future also.
Students : Thank you sir.
Advocate : Ok children. Bye.

9th Class Social Studies 21st Lesson Human Rights and Fundamental Rights InText Questions and Answers

Question 1.
Write a few important features of Preamble you studied last year. (Text Book Page No. 256)
Answer:
The Preamble is the heart and soul of our constitution. The important features are –

1. The Preamble starts with the words “We the people of India”. This ensures that sovereignty vests with the people.
2. It also confirms or ensures justice, equality, liberty, and fraternity to all its citizens.
3. It declares our country as sovereign, socialistic, secular, democratic, republic. Each of the words have different meaning.

Question 2.
What kinds of rights to equality does the Constitution ensure? Give examples. (Text Book Page No. 259)
Answer:
The Constitution ensures the following rights to equality to its citizens.

1. Equal protection of law * The laws apply to all in the same manner, regardless of a person’s income, status, background, etc.
2. Social Equality – The state condemns any sort of discriminations of human beings.
3. Equality of opportunity – The constitution guarantees equality of opportunity for all citizens regarding education or employment.
4. Abolition of untouchability.
5. Abolition of titles – In order to remove inequalities “Titles” of any sort are abolished.

Question 3.
What would happen if the Fundamental Right to Equality was not in the Constitution? Discuss. (Text Book Page No. 259)
Answer:

1. Democratic systems preserve equal grounds for all.
2. Democracies work on the principle of equality which is also known as “rule of law”.
3. If the fundamental “Right to equality” was not in the Constitution, the very essence of democracy would be lost.

Question 4.
What associations are there in your area? (Text Book Page No. 261)
Answer:
There are so many associations in my area. Some of them are –

1. Teachers Associations
2. Workers Association
3. Foremen’s Association
4. Students’ Associations
5. Auto – Rickshaw Association
6. Trade Union Associations
7. Rice Millers’ Associations
8. Fishermen’s Associations, etc.

Question 5.
Why are workers’ unions formed? What problems do they face? (Text Book Page No. 261)
Answer:
Workers unions are formed to protect the rights of workers and to solve their problems. These unions hold meetings to discuss their problems and take their demands to the officers of the factory. The following are their problems.
a) Their working conditions will not be healthy.
b) Salaries, dearness allowances, pensions, etc., will not be paid to them in time.
c) Sometimes their services will not be regularised.
d) They will not provide any educational facilities to their children of the factory workers.
e) Medical reimbursement, generally, not given to them.

Question 6.
Why do people want to move and settle in various parts of the country? (Text Book Page No. 261)
Answer:
In search of job opportunities people move and settle in various parts of the country.

Question 7.
What do you remember about the difference between the role of the police and that of the court? (Text Book Page No. 262)
Answer:

• Police generally file a case on any person who had committed a crime.
• He has to submit the accused in the case before the court.
• The court decides whether a person is guilty or not. And finally gives judgement.

Question 8.
What are the different types of schools you see in your area? Why do you think are there such different types of schools? (Text Book Page No. 262)
Answer:
There are the following types of schools
Basing on the management –

1. Zilla Parishad School
2. Mandal Parishad Schools
3. Government Schools
4. Social Welfare Schools
5. Tribal Welfare Schools
6. Government aided schools
7. Private Schools

Basing on the classes –

1. Primary Schools (I class to 5th class)
2. Upper Primary Schools (1st class to 7th class)
3. High Schools (6th class to 10th class)

The children, whose parents are rich and can afford private school, go to private schools. And rest of the children generally go to government schools.

Question 9.
Can anyone not follow any religion if he/she wishes? (Text Book Page No. 263)
Answer:
No, everyone can follow whatever religion they want. “Right to religious freedom” is our fundamental right.

Question 10.
State some instances of violations of Human Rights. (Text Book Page No. 266)
Answer:

1. Arbitrary arrests
2. Denial of right to information and corruption
3. Sexual cruelty
4. Rape of women
5. Delay in investigation of crimes
6. Female infanticide
7. Kidnapping for ransom
8. Deplorable conditions of women, children and downtrodden people
9. Discrimination against women in the family
10. Cruelty to domestic servants

The above are some instances of the violations of human rights. These violations continue unchecked because people are not aware of their human rights.

Question 11.
Read the passage and answer the following questions. (Text Book Page No. 262)
The Constitution states, “no child below the age of 14 shall be employed to work in any factory or mines or engaged in any other hazardous employment.” Accordingly, laws have been made that prohibit children from making matches, crackers, beedis, and carpets, or doing printing and dyeing, etc.
Do you think this right has been made available to children in the villages and cities in your area?
Answer:
There are number of children who are still working in small scale industries like dyeing, printing, beedi making etc.

Question 12.
Discuss whether you think each of the following is a violation of the Fundamental Right to Equality, Also discuss whether you think it is constitutionally right or wrong to do such things. (Text Book Page No. 259)

• While filling water from a public source, some people object if the vessel of another person touches their pots.
• Some communities are never provided a place to live within the village but always outside.
• In some schools, certain children are not allowed to serve water because they belong to a particular caste.
• Members of some communities do not go to many places of worship because they fear that they will be ill treated or beaten up.

Answer:

• The above examples are the clear evidences of practice of untouchability.
• The practice of untouchability is a crime.
• Anyone doing so is punishable under law.

Question 13.
With the help ofthe teacher find out the minimum wages in your state. (Text Book Page No. 262)
Answer:
Minimum wages were received by the people in different sector in our state.
Male – Rs. 200 (per day) – Female – Rs. 150 (per day)
Teacher – Rs. 300 to 2000 (per day).

Agriculture labour
Male – Rs. 180 (per day)
Female – Rs. 120 (per day)

Mason
Male – Rs. 300 (per day)
Female – Rs. 180 (per day)

Question 14.
How does the practice of “Sati” violate fundamental rights? (Text Book Page No. 263)
Answer:

• No citizen can be denied his life and liberty except by law.
• “The right to life” does not include “the right to die”, and hence suicide, forced death or sati, etc. are offences.
• “Sati”, hence, is violation of fundamental right of “the right to live”.

Question 15.
What is the responsibility of the government towards workers who are able to find some work, in the city but don’t have a proper place to live? (Text Book Page No. 261)
Answer:

• Our Government has introduced subsidised housing schemes to the people of below poverty line.
• Under urban basic schemes, Indira Aavas Yojana, Rajiv Gruha Yojana, etc., so many people acquired houses.
• They are constructed in the prime localities of the cities.
• Most of the slums in the cities were removed.
• Development is still going on in this regard.

Question 16.
Write a petition to the NHRC if you know any instances of human rights violation in your area. (Text Book Page No. 266)
Answer:

Question 17.
Is there a State Human Rights Commission in our state? Find out about its activities. (Text Book Page No. 266)
Answer:
Yes, there is a State Human Rights Commission in our state.
A Human Rights Commission, also known as a Human Relations Commission is a body set up to investigate, promote or protect human rights.

Its activities are

1. Inquire into any violation of human rights.
2. Look into negligence in the prevention of human rights violation by a public servant.
3. They can take cognizance either sumotu or on a petition presented to it or on an order of a court.
4. They intervene in any proceeding involving allegation of violation of human rights pending before a court etc.