Category: Class 10

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 14 Carbon and its Compounds.

AP State Syllabus SSC 10th Class Chemistry Important Questions 14th Carbon and its Compounds

10th Class Chemistry 14th Lesson Carbon and its Compounds 1 Mark Important Questions and Answers

Question 1.
Define Isomerism. (AP March 2016)
Answer:
The phenomenon of possessing same molecular formula but different properties by the compounds is known as “Isomerism”.

Question 2.
Give the names of the functional groups. (AP March 2018)
a) – COOR
b) – OH
Answer:
a) Ester
b) Alcohol

Question 3.
How do you explain the role of Oxygen in combustion process? (TS March 2015)
Answer:
Oxygen helps the combustion (or) No combustion will take place without oxygen.
Ex : C + O₂ → CO₂

Question 4.

Predict and write the products. (TS March 2016)
Answer:

Question 5.
Write two uses of nano tubes. (TS June 2017)
Answer:

1. Nano tubes are used as molecule wires.
2. In intigrated circuits nano tubes are used to connect the components together.
3. Nano tubes are used to incert Bio-molecules into the single cell.

Question 6.
Write two uses of Ethanol in day to day life. (TS March 2018)
Answer:
Ethanol is used in
i) Preparation of Alchoholic drinks
ii) Preparing tincture iodine
iii) Preparing cough syrup and tonics

Question 7.
Write the atomic structure of the following carbon compound. 3, 7-dibromo-4, -6 dichloro – oct-5-ene-l, 2-diol. (TS March 2019)
Answer:

Question 8.
Thanish added acetic acid along with concentrated sulphuric acid to ethanol what would be his observation during the experiment? (AP SCERT: 2019-20)
Answer:

1. He may observe that the resulting mixture is a sweet odoured substance.
2. The substance is ethyl acetate, an ester.

Question 9.
Why do the various micelles present in water do not come together to form a precipitate? Guess the reason. (TS June 2019)
Answer:
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.

Question 10.
Mention any two uses of graphite in day to day life. (TS June 2019)
Answer:
Uses of graphite in day to day life :

1. Pencil lead.
2. Lubricant.

Question 11.
What is “Allotropy”?
Answer:
The property of an element to exist in two or more different forms due to the difference in their atomic arrangement is called “Allotropy” and the different forms are called allotropes.

Question 12.
‘Diamond is a bad conductor of heat.’ Why?
Answer:
Diamond is a bad conductor of heat due to lack of free electrons.

Question 13.
What is ‘cleavage’?
Answer:
Cleavage is a property of splitting of crystals of some minerals in certain directions to produce a flat, even surface.

Question 14.
“Diamond is the hardest natural substance but is brittle.” Why?
Answer:
Diamond is the hardest natural substance but is brittle and can be broken due to the property of cleavage.

Question 15.
Explain about high refractive index of diamond.
Answer:
Diamond has a high refractive index, due to which most of the light that enters the diamond gets reflected back internally. This internally reflected light is responsible for the brilliance of a diamond.

Question 16.
What is catenation?
Answer:
Catenation is the phenomenon in which atoms of same element join together to form long chains.

Question 17.
What is an alkyl group?
Answer:
If one hydrogen is removed from an alkane, it is called alkyl group.
Ex : CH₄ → methane
CH₃ → methyl group

Question 18.
What is polymerization?
Answer:
The reaction in which a large number of identical and simple molecules join together to form a large molecule is called ‘polymerization’.

Question 19.
What do you understand by a ‘Functional group’?
Answer:
A group of atoms in carbon compounds showing characteristic properties is called a functional group.

Question 20.
Name some functional groups.
ANswer:
Alcohol – OH, Aldehyde – CHO, Ketone – > C = O, Carboxylic acid (- GOOH), ester (-COOR), and amine – NH₂ are some important functional groups.

Question 21.
What is pyrolysis?
Answer:
Decomposition of a compound on heating in the absence of air is called pyrolysis.

Question 22.
What is hydrocarbon?
Answer:
Compounds containing only carbon and hydrogen are called ‘hydrocarbons’.
Ex : Alkanes (Saturated hydrocarbons),
Alkenes and Alkynes (Unsaturated hydrocarbons).

Question 23.
What is ‘Saturated hydrocarbon’? (Or) What is an alkane?
Answer:
The valency of carbon is 4, of all the valencies of carbon, are satisfied, the resultant hydrocarbons are referred to as ‘saturated hydrocarbons’ or alkanes. Their general formula is C_nH_2n+2.

Question 24.
What are ‘Unsaturated hydrocarbons’?
Answer:
The hydrocarbons containing one or more double bonds or triple bonds between two carbon atoms are called ‘unsaturated hydrocarbons’.
Ex : C₂H₆ and C₃H₆, etc.

Question 25.
What are alkenes?
Answer:
Alkenes are unsaturated hydrocarbons having at least one (C = C) double bond in their structures, Alkenes are also called olefins. Their general formula is C_nH_2n.
Ex : Ethylene (C₂H4) and propene (C₃H₆), etc.

Question 26.
What are alkynes?
Answer:
Alkynes are unsaturated hydrocarbons having at least one (\(C \equiv C\)) triple bond in their structures. Their general formula is C_nH_2n-2.
Ex: Acetylene (\(\mathrm{HC} \equiv \mathrm{HC}\))

Question 27.
Mention the natural sources of carbon compounds.
Answer:
Plants, wood, natural gas, coal, petroleum, etc. are the natural sources of carbon compounds.

Question 28.
Explain about methanol (or) methyl alcohol.
Answer:
Methanol is the simplest alcohol, It is the first member of the homologous series of alcohol. It is also known as wood alcohol, as it was initially obtained by the destructive distillation of wood.

Question 29.
What is organic chemistry?
Answer:
The chemistry of carbon compounds (excluding the carbonates, bicarbonates, carbides, cyanides, carbon dioxide, and carbon monoxide) is called organic chemistry. The large number of organic compounds necessitated their study in separate branch of chemistry, known as organic chemistry,

Question 30.
What is halogenation?
Answer:
Alkanes react with halogens in the presence of sunlight. For example, when a mixture of methane and chlorine is exposed to sunlight, a hydrogen atom of methane is replaced by a chlorine atom,

Question 31.
How many rings are there in buckminsterfullerene?
Answer:
In buckminsterfullerene, there are 32 rings, of them 12 are pentagonal rings and 20 are hexagonal rings.

Question 32.
Give example for homologous series.
Answer:
CH₄ and C₂H₆ → These differ by a – CH₂ unit.
and C₂H₆ and C₃H₈ → These differ by a – CH₂ unit.

Question 33.
What is hybridisation?
Answer:
The intermixing of orbitals to form equivalent new orbitals is called hybridisation.

Question 34.
What are nanotubes?
Answer:
Nanotubes are allotropic form of carbon.

Question 35.
What are homologous series?
Answer:
The ierles of carbon compound in which successive compounds differ by -CH₂ unit is called homologous series.

Question 36.
Write the molecular formula of the fourth member of the homologous series of alcohols.
Answer:
CH₃ – CH₂ – CH₂ – CH₂ – OH

Question 37.
What is a catalyst?
Answer:
The substance which does not take part in chemical reaction but changes the rate of reaction.

Question 38.
Why are oils liquids at room temperature?
Answer:
Oils are unsaturated compounds so they are in liquid state.

Question 39.
Why are fats solids at room temperature?
Answer:
They are saturated compounds so they are in solid state.

Question 40.
Do you know the police detect whether suspected drivers have consumed alcohol or not? Explain.
Answer:
Orange Cr₂O₇^2- changes bluish green Cr³⁺ during the process of the oxidation of alcohol. The length of die tube that turned into green is the measure of die quantity of alcohol that had been drunk.

Question 41.
What is pka?
Answer:
The negative value of logarithm of dissociation constant of an acid.

Question 42.
What is Saponification?
Answer:
Alkaline hydrolysis of triesters of higher fatty acids producing soaps is called saponification.

Question 43.
What is a soap?
Answer:
Sodium or potassium salt of fatty acid.

Question 44.
What is micelle?
Answer:
A spherical aggregate of soap molecules in water is called micelle.

Question 45.
What change will you observe if you test soap with litmus papers?
Answer:
Red litmus turns into blue.

Question 46.
Write the valency of carbon in CH₃ – CH₃, CH₂ = CH₂ and \(\mathrm{HC} \equiv \mathrm{CH}\)?
Answer:
The valency of carbon in CH₃ – CH₃ is 4.
The valency of carbon in CH₂ = CH₂ is 3.
The valency of carbon in \(\mathrm{HC} \equiv \mathrm{C}\) – H is 2.

Question 47.
Out of butter and groundnut oil which is unsaturated in nature?
Answer:
Groundnut oil is unsaturated in nature.

Question 48.
What are hydrophobic and hydrophilic parts in soap?
Answer:

Question 49.
Name the carboxylic acid used as preservative.
Answer:
Acetic acid is used as preservative.

Question 50.
Why does graphite act as a good conductor of electricity?
Answer:
Graphite is a good conductor of electricity because of delocalized x electron system.

Question 51.
Among objects made of glass and diamond, which one shines more? Why?
Answer:
Diamond shines more because of low conical angle of 24,4° and also high refractive

Question 52.
Write IUPAC names of the following compounds.

Answer:
a) 2, 2, 3, 3 – tetra methyl butane
b) 3-chloro butan-l-oic acid.

Question 53.
What is the difference between combustion and oxidation reaction?
Answer:
Combustion is an oxidation reaction where a compound is burnt in the presence of oxygen, whereas oxidation is addition of oxygen which does not require any burning.

Question 54.
Write the order of priority of functional groups for naming carbon compounds.
Answer:

Question 55.
What is glycerol?
Answer:
The trihydroxy alcohol is called glycerol.

Question 56.
What do you mean by CMC?
Answer:
CMC means Critical Micelle Concentration.

Question 57.
Name the simplest chloride of saturated hydrocarbon.
Answer:
Chloro methane or methyl chloride.

Question 58.
Write the IUPAC name of next homolog of CH₃CH₂CHO.
Answer:
The next homolog of CH₃CH₂CHO is CH₃CH₂CH₂CHO (its IUPAC name is butanol). Since homologs differ by – CH₂.

Question 59.
How do physical properties like boiling point and melting point vary as the number of carbon atoms increases in a homologous series?
Answer:
There is regular gradation in physical properties of homologous series. So the physical properties like boiling point and melting point vary.

Question 60.
What is meant by Hybridisation?
Answer:
Mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals.

Question 61.
Write any two uses of Graphite.
Answer:
i) Conductor
ii) Lubricant

Question 62.
Write any two examples to Amorphous form of carbon.
Answer:
i) Coke
ii) coal
iii) charcoal.

Question 63.
Write any two examples to crystalline forms of carbon.
Answer:
i) Diamond
ii) graphite

Question 64.
What are the applications of Buckminster fullerene?
Answer:
i) Antioxidants
ii) Anti aging and damage agent in cosmetic sector.

Question 65.
What is meant by catenation?
Answer:
Binding of an element to itself through covalent bonds to form chain or ring molecules.

Question 66.
Write any one use of nanotubes.
Answer:
i) Used as molecular wires.
ii) Used in integrated circuits.

Question 67.
On which reason, graphite is used as lubricant and as the lead in pencils?
Answer:
Graphite has free electrons.

Question 68.
How many isotopes are there for C₄H₁₀, what are they?
Answer:
i) n – Butane
ii) Iso – Butane

Question 69.
CH₃ – CH = CH – CH₃, how many sigma bonds are present in the above compound?
Answer:
11

Question 70.
Write the IUPAC name of Ethyle alcohol.
Answer:
Ethanol.

Question 71.
Classify the following into alkanes, alkenes and alkynes.
C₁₂ H₂₂, C₁₀ H₂₂, C₁₁ H₂₂
Answer:
i) C₁₀ H₂₂ – Alkanes
ii) C₁₁ H₂₂ -Alkenes
iii) C₁₂ H₂₂-Alkynes

Question 72.
Hi ……… I am carboxylic acid. I am used in the making vinegar, who am I?
Answer:
Acetic acid.

Question 73.
What does IUPAC represent?
Answer:
International Union of Pure and Applied Chemistry.

Question 74.
Write any one example for esterification reaction.
Answer:
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

Question 75.
A compound with molecules formula C₂H₆O is used in cough syrup. Identify the compound.
Answer:
Ethyl Alcohol.

Question 76.
Which substance is added for the denaturation of ethyl alcohol?
Answer:
Pyridine.

Question 77.
What is the abbreviation of CMC?
Answer:
Critical Micelle Concentration.

Question 78.
Write the names of polar end and non-polar end in a soap.
Answer:
Polar end – COO^– Na⁺, Non-polar end – R.

Question 79.
Write the IUPAC name of the alcohol which one carbon atom.
Answer:
Methanol.

Question 80.
Write the chemical equation which indicates the preparation of ethanol industrially?
Answer:

Question 81.
What is the formula of chloroform? Write its one use.
Answer:
CHCl₃, Anesthetic.

Question 82.
Which type of hydrocarbons are participate in addition reaction?
Answer:
Unsaturate Hydrocarbons.

Question 83.
What are the oxidising agents used in oxidisation of C₂H₅?
Answer:
K₂Cr₂O₇, KMn0₄.

Question 84.
What is meant by catalyst?
Answer:
To change die rate of reaction without itself undergoing any permanent chemical change.

Question 85.
What are the main constituents of LPG?
Answer:
Butane, Methane.

Question 86.
What is the difference between saturated and unsaturated hydrocarbons?
Answer:
Saturated house single bonds, unsaturated have multiple bonds.

Question 87.
Describe a test for carboxylic acid.
Answer:
React with metals liberate hydrogen gas.

Question 88.
What is meant by denatured alcohol?
Answer:
Unfit for human consumption by adding one or more chemicals.

Question 89.
Complete the following equation.
CH₄ + 2O₂ →
Answer:
CH₄ + 2O₂ → CO₂ + 1H₂O

Question 90.

i) In the above substance, what is the hybridisation of 3^rd carbon?
Answer:
sp²

ii) What is the hybridisation of 4^th carbon?
Answer:
sp³

Question 91.
What is the main misuse of Ethanol?
Answer:
Drinking.

Question 92.
What is gasohol?
Answer:
10% Ethyl alcohol with gasoline.

Question 93.
Write any two uses of Ethyle alcohol.
Answer:
i) Good solvent
ii) Additive to automotive gasoline.

Question 94.
Write two IUPAC name

Answer:
3 – Chloro 1 – Butane

Question 95.
Name the following functional groups.
i) – COOR
ii) R – COOH
Answer:
i) – COOR (Ester)
ii) R – COOH (Carboxylic acid)

Question 96.
Name the crystalline allotrope of carbon which conducts electricity.
Answer:
Graphite.

Question 97.
Ravi gets confused while understanding the between R – COOH and R – OH functional groups, ask him one question to classify it.
Answer:
i) What is carboxylic acid?
ii) What is Alcohol?

Question 98.
Formic acid (HCOOH)
Farmaldehyde (HCHO)
Methanol (CH₃OH), then answer the following questions.
i) Which is present in ants?
Answer:
HCOOH (Formic acid).

ii) Which is used to preservation of dead bodies?
Answer:
HCHO (Farmaldehyde).

Question 99.
Write the symbolic representation showing the functional groups.
i) amine
ii) amide
Answer:
i) R – NH₂
ii) R – CONH₂

Question 100.
How many sigma and pi-bonds present in Acetylene?
Answer:
\(\mathrm{HC} \equiv \mathrm{CH}\) ; σ bonds – 3 ; π bonds – 2

Question 101.
Which of the following will give substitution reactions?
CH₄, C₃H₆, C₃H₄, C₅H₁₂, C₄H₈
Answer:
CH₄, C₅ H₁₂

Question 102.
Which of the following will give addition reactions?
CH₄, C₃H₆, C₃H₄, C₅H₁₂, C₄H₁₀
Answer:
C₃H₆, C₃H₄

Question 103.
What is a homologous series?
Answer:
Same functional group, difference between successive members is a simple structural unit – CH₂.

Question 104.
Name the hydrocarbon which is used in the artificial ripening of fruits?
Answer:
C₂H₄

Question 105.
Define fermentation process.
Answer:
Chemical break down of a substance by bacteria, yeast or other microorganisms.

Question 106.
Define functional group.
Answer:
They are specific substituents within molecules that are responsible for die characteristic chemical reactions.

Question 107.
Which hydrocarbons participate in sp² hybridisation?
Answer:
C₂H₄

Question 108.
Name the following compounds,
i) CH₃ – CH₂ – Br
Answer:
1 – Bromo Ethane

ii)

Answer:
Ethanol

iii)

Answer:
2 – Butanone

iv)

Answer:
2, 3 – dichloro Butane

Question 109.
Which constituents are present in tincture Iodine?
Answer:
i) Iodine
ii) Alcohol.

Question 110.
Write the uses of esters in daily life.
Answer:
i) Solvents
ii) Plasticizers

Question 111.
Name the gas evolved when acetic acid reacts with sodium hydrogen carbonate.
Answer:
The gas liberated is carbon dioxide.

Question 112.
Name the organic acid present in vinegar. Write its chemical formula.
Answer:
The acid present in vinegar is acetic acid. Its formula is CH₃COOH.

Question 113.
Why is graphite a good conductors’of electricity?
Answer:
Graphite has free electrons.

Question 114.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Tbtravalency.

Question 115.
Why are alkanes called as paraffins?
Answer:
Low reactivity.

Question 116.
Draw two possible structures with formula C3HgO and what they are called?
Answer:

Question 117.
Draw structure of 3 – methyl pentan-3-ol.
Answer:

Question 118.
Draw the shape of soap molecule.
Answer:

Question 119.
Draw the shape of Micelle.
Answer:

Question 120.
Draw the shape of methane.
Answer:

Question 121.
Draw the structure of pentanoic acid.
Answer:

Question 122.
How do you appreciate the role of diamond in space probes?
Answer:
Since it has the ability to filter out harmful radiations, it is used in making protective windows for space probes.

Question 123.
How do you appreciate the role of acetic acid as a preservative?
Answer:

• Dilute acetic acid is used as a food preservative in the preparation of pickles and sauces,
• As vinegar, it is also used as an appetiser for dressing food dishes.

Question 124.
How do you appreciate the role of diamond in surgery?
Answer:
A sharp edged diamond is used as a tool to remove cataract in eye surgery.

10th Class Chemistry 14th Lesson Carbon and its Compounds 2 Marks Important Questions and Answers

Question 1.
Draw the simple figure of a soap molecule. (AP March 2016)
Answer:
Structure of soap molecule :

Question 2.
Draw the structure of the methane molecule. Write its bond angle. (TS June 2015)
Answer:
The bend angle in methane is 109°2 8′.

Question 3.
a) Why are vegetable oils healthy as compared to vegetable ghee? (TS March 2015)
b) Write the IUPAC name of

Answer:
a) Because vegetable oils contain unsaturated fatty acids or vegetable oils are easily digestible.
b) 3 – Mono chloro butene (or) 3 Chloro butene

Question 4.
What are alkenes? Write the general formula of alkenes. Give an example for alkenes. (TS June 2017)
Answer:

• Unsaturated hydrocarbons those are having carbon * carbon double bond are known as alkenes.
• The general formula of Alkenes is CHH2h.
• Example : Ethelene (C2H4).

Question 5.

Based on the diagram, answer the following.
1) Write the name of the compound.
2) Write the name of functional group in the structure. (AP March 2019)
Answer:

1. The compound is 2, 3-di ethyl-cycle hexan-1-ol.
2. Alcohol (OH) is the functional group in the structure.

Question 6.
Identify the functional groups in the following compounds and write IUPAC names.

Answer:

The IUPAC name of the compound Is 2 – Chloro-Butan 1-ol.

The IUPAC name of the compound is 3 – Methyl-2-Butan-one.

Question 7.
Draw the structure of butanoic acid C₃H₇COOH.
Answer:
Formula of butanoic acid is C₄H₅O₂.

Question 8.
What is ‘Isomerism’?
Answer:
Compounds having same molecular formula but different structures are called isomers, and the phenomenon is called isomerism.
Ex: C₄H₁₀ exists an n-hutane and iso-butane.

Question 9.
How do you detect leakage in the cylinder?
Answer:

• To detect any leakage of gas from die cylinder, a strong-smelling substance like ethyl mercaptan (C₂H₅ SH) is added to die gas.
• Then the leakage can be easily detected by the foul smell of die ethyl mercaptan.

Question 10.
How is LPG gas useful for environment?
Answer:

• Because of its heat producing capacity (calorific value), it is considered to be a good fuel.
• It bums without producing smoke. Hence, it does not cause any pollution.
• It is a dean fuel and can be conveniently handled.

Question 11.
How is ethanol useful in pharmaceutical industry?
Answer:

• Solutions in ethanol are often prepared in pharmaceutical industry, these solutions are known as tinctures.
• For example, a solution of Iodine and potassium iodide in ethanol is called tincture of iodine.
• It is also used as an important raw material for the synthesis of many organic compounds, for example, ethanol, ethanoic acid, ethanoie anhydride, esters, chloroform, etc.

Question 12.
How are synthetic detergents harmful for environment?
Answer:

• Some synthetic detergents resist biodegradation, i.e. they are not decomposed by micro-organisms such as bacteria.
• Hence, they cause water pollution in lakes and rivers.
• They tend to persist for a long time, making the water unfit for aquatic life.

Question 13.
Explain about allotropic forms of carbon.
Answer:

Question 14.
Diamond is considered to be the purest form of carbon. How can we prove it?
Answer:
When diamond is heated in oxygen alone, it bums at about 800° C and forms carbon dioxide leaving no residue. This proves that diamond to be the purest form of carbon.

Question 15.
Why does carbon not form C⁴⁺? Why?
Answer:

• Electronic configuration of carbon is 1s²2s²2p².
• If carbon loses four electrons from the outer shell, it will form C⁴⁺ ions.
• This requires huge amount of energy which is not available normally.
• Therefore C⁴⁺ formation is not possible.

Question 16.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Carbon is unable to form C⁴⁺ ion as well as C^4- ion. So carbon has to satisfy its tetra- valency by sharing electrons with other atoms. So it mainly forms covalent bonding.

Question 17.
Define Allotropy. What are the allotropic forms of carbon?
Answer:
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. The allotropic forms of carbon are graphite, diamond, etc.

Question 18.
Identify the unsaturated compounds of the following.
a) CH₃ – CH₂ – CH₂
b) CH₃ – CH = CH₃
c)

Answer:
a) CH₃ – CH₂ – CH₂ saturated compound.
b) CH₃ – CH = CH₃ unsaturated compound.
c)

Question 19.
Define Isomers. Write structural formula of isomers of butane.
Answer:
Compounds having same molecular formula but different properties are called isomers.
Isomers of butane :
1) CH₃ – CH₂ – CH₂ – CH₃
Butane

Question 20.
What happens when a small piece of sodium is dropped into ethanol?
Answer:
When a small piece of sodium is dropped into ethanol it releases hydrogen gas and forms sodium ethoxide.
2C₂H₅OH + 2 Na → 2C₂H₅ONa + H₂

Question 21.
What type of reaction takes place between ethane and chlorine?
Answer:
Substitution reaction takes place between ethane and chlorine in die presence erf sunlight
CH₄ + Cl₂ → CH₃Cl + HCl
CH₃Cl +Cl₂ → CH₂Cl₂ + HCl
CH₂Cl₂ + Cl₂ → CHCl₃ + HCl
CHCl₃ + Cl₂ → CCl₄ + HCl

Question 22.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:

1. Catenation
2. Isomerism.

Question 23.
How could you name the following compounds?
a) CH₃ – CH₂ – CH₂ – Br
b) CH₃ – CH₂ – CH₂ – CH₂
Answer:
a) Bromo propane
b) Hexyne

Question 24.
Give examples for primary, secondary and tertiary amines.
Answer:
Primary amine – CH₃NH₂
Secondary amine – CH₃ – NH – CH₃

Question 25.
Write the following conversions.
1) Ethanol to Ethene
2) Ethene to Ethanol
3) Methane to carbon tetra chloride.
Answer:
1) Ethanol reacts with cone. H₂SO₄ at about 170°C to give ethene.

2) Ethanol is prepared from ethene by the addition of water vapour in the presence of catalyst P₂O₅.

3) Methane reacts with chlorine in the presence of sunlight. Hydrogen atoms of CH₄ are replaced by chlorine at

Question 26.
Name the following compounds and which one is saturated among them.
a) CH₃ – \(\mathrm{C} \equiv \mathrm{H}\) – CH₃
b) CH₃ – CH = CH – CH₃
c) CH₃ – CH₂ – CH₂ – CH₃
Answer:
a) 2-Butyne
b) 2 – Butene
c) Butane

Butane does not show any double or triple bonds. Its valency is completely satisfied with formation of single bond. So it is a saturated compound.

Question 27.
How do you identify the given organic compound contains carboxylic acid functional group?
Answer:

• On adding carbonates and bicarbonates the compound containing carboxylic acid group evolves carbon dioxide gas.
• When warmed with alcohol and cone. H₂SO₄ a pleasant fruity smell is produced due to formation of ester.

Question 28.
Explain briefly about the structure of “Diamond”.
Answer:

• In a diamond, each carbon atom is surrounded by four other carbon atoms.
• In these carbon atoms, each carbon atom undergoes in its excited state sp3 hybridisation.
• These are placed at the four corners of a regular tetrahedron.
• This results in a 3-dimensional network of carbon atoms.
• So diamond is in three dimensional structure.

Question 29.
Explain briefly about the structure of “Graphite”.
Answer:

• In graphite, each ‘C’ is surrounded by three other ‘C’ atoms.
• The ‘C’ atoms are arranged in layers.
• In the layer structure, the carbon atoms are in trigonal planar environment.
• Each layer consists of a 2-dimensional hexagonal network.

Question 30.
Diamond is an extremely bad conductor of electricity.” Why?
Answer:
1) In diamond, each carbon atom is covalently bonded with four other carbon atoms.
2) So, the four outermost electrons of a carbon atom are engaged or trapped in the covalent bonds, having no free electrons making it a bad conductor of electricity.

Question 31.
Why is diamond hard but graphite is smooth and slippery?
Answer:
Diamond has sp³ hybridisation with tetrahedral environment. As C – C bonds are very strong any attempt to distort the diamond structure requires large amount of energy. Hence diamond is one of the hardest material.

Whereas graphite has sp² hybridisation with layer structure with trigonal planar environment. The layers tend to slide on one another. So graphite is smooth and slippery.

Question 32.
An organic compound X with a molecular formula C₂H₆O undergoes oxidation within presence of alkaline KMnO₄ to form a compound Y. X on heating in presence of con. H₂SO₄ at 443 K gives Z. Which on reaction with Br₂ and decolorizes it? Identify X, Y, and Z and write the reactions involved.
Answer:
X is ethanol.

Question 33.
Complete the following reactions.

Answer:

Question 34.

What are A and B?
Answer:
1) Alkynes undergo addition reaction in the presence of nickel catalyst and hydrogen to form Alkene.

Question 35.
Draw the structure for the following compounds.
a) Propanoic acid
b) Chlorobutane
c) Hexanone
d) Pentanal
Answer:
a) CH₃CH₂COOH
b) CH₃CH₂CH₂CH₂Cl
c) CH₃CH₂CH₂CH₂COCH₃
d) CH₃CH₂CH₂CH₂CHO

Question 36.
Give IUPAC names of the following compounds. If more than one compound is possible, name all of them.
i) A chloride derived from butane.
ii) A ketone derived from pentane.
Answer:
i) The following chlorides are possible for butane.

ii) The following ketones are possible for pentane.

Question 37.
a) What are the various possible structural formulae of a compound having molecular formula C₃H₆?
b) Give IUPAC names of the above possible compounds and represent them in structure.
c) What is the difference between those
Answer:

b) The IUPAC names of compounds are propene and cycle propane.
c) The main difference Is that the first compound Is alkene-an unsaturated compound and second is cyclo alkane-a saturated compound.

Question 38.
Draw isomeric forms of C₆H₁₄.
Answer:
Isomers of hexane :

Question 39.
How do you appreciate the role of carbon in everyday life?
Answer:

• Major components of our daily food have carbohydrates, proteins, fats, etc. which are all made up of carbon compounds.
• The fibres of cloth are made up of cellulose and other types of materials, which are all carbon compounds.
• Cement and steel form the core of any of the modern buildings. Carbon bestows steel with hardness, while limestone (CaCO₃) a major constituent of cement also contains carbon.

Question 40.
How do you appreciate the role of oxygen in combustion process?
Answer:

• When the oxygen supply is insufficient, the fuels burn incompletely producing mainly a yellow flame.
• When the oxygen supply is sufficient, the fuels burn completely producing a blue flame.

Question 41.
How do you appreciate the role of Ethanol as a fuel?
Answer:

• A material which is burnt to obtain heat is called a fuel. Since ethanol burns with a clear flame giving a lot of heat, it is used as a fuel.
• Some countries add ethanol to petrol to be used as a fuel in cars. Thus ethanol is used as an additive in petrol.
• Ethanol alone can also be used as a fuel for cars.

Question 42.
What are the uses of fullerenes?
Answer:
Fullerenes are under study for potential medical use such as specific antibiotics to target resistant bacteria and even target cancer cells such as melanoma.

Question 43.
Write the HJPAC names of the following compounds.
i) CH₃ – CH₀ – CH₂ – CH₂ – CH₂ – CH₂ – CH₂ – CH₂ – CH₂OH
ii) CH₃ – CH₂ – CH = CH- CH₂ – \(\mathrm{C} \equiv \mathrm{CH}\)
iii) CH₃ – CH₂ – CH₂ – CH₂ – CHO
iv) CH₃ – CH₂ – CH₂ – CH₂ – COOH
Answer:

1. nananol
2. 4- ene – 1 heptyne
3. pentanal
4. pentanoic acid

Question 44.
What are the uses of alcohol?
Answer:

• Alcohols are goods solvent for resin and gums.
• Ethanol is used in the thermometers because of its low freezing point.
• One of the products of ethyl alcohol is chloroform, which is used as an aesthetic.
• 10% ethanol in gasoline is a good motor fuel.
• It is used in medicines such as tincture iodine, cough syrups and many tonics.

Question 45.
What are the uses of acetic acid?
Answer:

• 5 to 8% solution of acetic acid in water is called vinegar and is used widely as a preservative in pickles.
• Used as a laboratory reagent.
• Used in the production of perfumes, dyes, esters, etc.
• Used in medicine.

10th Class Chemistry 14th Lesson Carbon and its Compounds 4 Marks Important Questions and Answers

Question 1.
Write IUPAC names for the following carbon compounds.

Answer:
A) 2 – methyle pentane – 3 – ol
B) 3 – chloro, 4 – Methyle hexanoic acid
C) 2 Bromo – Bute – 2 – ene
D) 2, 5 Dimethyle hexane

Question 2.
(AP June 2017)
Observe the given carbon compound and answer the following questions.
a) Give numbering to the carbons in the given compound according to IUPAC rules.
b) Name the functional group present in the given compound.
c) Name the word root for the given carbon compound.
d) Write the IUPAC name of the given compound.
Answer:
a)

b) The given compound contains functional group – OH. It is an alcohol.
c) Word root: The number of carbon atoms present in the molecules is called word root. Here the word root is (C₅) – pent.
d) IUPAC name of the given compound is pent 4 – ene 2 – ol.

Question 3.
Alkanes are considered as Paraffins. So, they undergo substitution reactions but not addition reactions. Explain with suitable example. (AP March 2017)
Answer:

Question 4.

Observe the structure and answer the following.
a) Write the name of principal functional group present in the compound.
b) Identify the parental chain in the compound.
c) What are the substituents in the above compound?
d) Name the above compound as per IUPAC nomenclature. (AP June 2018)
Answer:
a) Ketone
b)

c) Methyl group ; Hydroxy group
d) 7 – hydroxy – S – methyl heptan – 2 – one

Question 5.
In the table given below, fill the information in the empty boxes and give answers to the following questions. (TS June 2015)
a Write the general formula of alkanes from the table.

b) How many a bonds are there in C₃H₆?
c) What sequential order did you notice in the molecular formulae?
d) There exist single bonds between carbon atoms of alkanes. Do you agree with this statement? Give reasons.
Answer:

a) The general formula of Alkanes is C_nH_2n+2
b) The number of o bonds in C₂H₆ are 7.
c) Two successive alkanes are differed by – CH₂ group.
d) Except Methane all other alkanes have single bonds between carbon atmos because it is a saturated hydro carbon.

Question 6.
Why do we call alkanes as paraffins? Explain the substitution reactions of alkanes. (TS June 2016)
Answer:
a) 1. Alkanes are saturated hydrocarbons with least reactivity.
2. Therefore they are called paraffins.
3. Parum = little and affins = affinity.

b) 1. A reaction in which one atom or a group of atoms in a given compound is replaced by other atom or group of atoms is called a substitution reaction.
2. Alkanes have single bonds and undergo substitution reactions.

3. For example :
Methane (CH₄) reacts with chlorine in the presence of sunlight.

Question 7.
Write the types of Allotropes of Carbon. Give any three examples of each. (TS March 2016)
Answer:
The allotropes of carbon are classified into two types. They are
i) Amorphous forms,
ii) Crystalline forms.
Examples :
Amorphous forms :
Coal, coke, wood charcoal, animal charcoal, lamp black, gas carbon, petroleum coke, sugar charcoal, etc.

Crystalline forms :
Diamond, graphite, and buckminsterfullerene.

Question 8.
Write any 4 characteristic features of homologous series of Organic compounds. (TS March 2016)
Answer:
Homologous series :
The series of carbon compounds in which two successive compounds differ by – CH2 unit is called homologous series.

Characteristic features of homologous series :

1. They have one general formula.
   Ex : Alkane (C₄H_{2n + 2}), Alkene (C₄H_2n), Alkyne (C₄H_2n-2)
2. Successive compounds in their series possess a difference of (- CH₂) unit.
3. They possess similar chemical properties due to the same functional group.
4. They show a regular gradation in their physical properties.

Question 9.
List out the materials required to conduct the experiment to understand the esterification reaction. Explain the procedure of the experiment. How can you identify that an ester is formed in this reaction?(TS March 2017)
Answer:
Required Material :
Test tube, beaker, tripod* burner, water, wire guage, ethanol (absolute alcohol), glacial acetic acid, concentrated sulphuric acid.

Procedure :

1. Take 1 ml of ethanol and 1 ml of glacial acetic acid along with a few drops of concentrated sulphuric acid in the test tube.
2. Warm it in a water bath or in a beaker containing water for atleast five (5) minutes.
3. Pour the warm contents into a beaker containing 20-50 ml of water and observe the odour of the resulting mixture.
   If we smell sweet odour from the beaker, we can confirm that ester is formed.

Question 10.
Explain the Isomerism and Catenation properties of carbon. (TS March 2018)
Answer:
Catenation properties of carbon :
i) Carbon has ability to form longest chains with its own atoms. This special property of carbon is called catenation.
ii) Due to catenation property of carbon it can form largest chain containing millions of carbon atoms, branches and cyclic compounds.

Isomerism of carbon :
The phenomenon of possessing some molecular formula but different properties by the compounds is known as isomerism.

Molecular formula of above two molecules is C₄ H₁₀ but they have different structure. These two are isomers.

By there two special properties of carbon it can make number of compounds.

Question 11.

Observe the above table and answer the following questions. (TS March 2019)
1) Write the general formula of Alkanes.
2) Mention the names of unsaturated hydrocarbons.
3) Write the homologous series of Alkynes.
4) Write the formula of Hexyne.
Answer:
1) General formula for Alkanes : C_nH_2n+2.
2) Unsaturated Hydrocarbons in the list are :
Propene C₃H₆, Butene C₄H₆, Pentyne C₅H₈, Hexyne C₆H₁₀.

3) Homologous series of Alkynes is C₂H₂ (Ethyne), C₃H₄ (Propyne), C₄H6 (Butyne), C₅H₈ (Pentyne), C₆H₁₀ (Hexyne).

4) Formula of Hexyne is C₆H₁₀.

Question 12.
Complete the following table based on functional groups of organic compounds, their structural formulas and respective suffixes. (AP SCERT: 2019-20)

Answer:

Question 13.
Explain the occurrence of carbon.
Answer:
Carbon occurs in nature in free state as well as in combined state.

Question 14.
What is sp hybridisation? Explain.
Answer:

• Each carbon is only joining to two other atoms rather than four or three.
• Here the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridise two of the orbitals.
• They use the ‘s’ orbital (2s) and one of the 2p orbitals, but leave the other 2p orbitals unchanged.
• The new hybrid orbitals formed are called sp-hybrid orbitals, because they are made by an s-orbital and a p-orbital reorganizing themselves.

Question 15.
Write the characteristics of homologous series of organic compounds.
Answer:
Characteristics of homologous series :

1. They have one general formula.
   e.g.: Alkanes (C_nH_2n+2)
2. Successive compounds in the series possess a difference of – CH₂ unit.
3. They possess similar chemical properties due to same functional group.
   e.g.: C – OH
4. They show a regular gradation in their physical properties.

Question 16.
What is sp³ hybridisation with diagram? Explain.
Answer:
The excited carbon atom allows its one s-orbital (2s) and three p-orbitals (2p_x, 2p_y, 2p_z) to intermix and reshuffle into four identical orbitals known as sp³ orbitals. Thus, carbon atom undergoes sp³ hybridization. The four electrons enter the new four identical hybrid orbitals known as sp³ hybrid orbitals, one each as per Hu nd’s rule.

1) Since carbon has four unpaired electrons, it is capable of forming bonds with four other atoms.

2) When carbon reacts with hydrogen, four hydrogen atoms allow their ‘s’ orbitals containing one electron each to overlap with four sp³ orbitals of carbon atom which are oriented at an angle of 109°. 28’.

3) Four orbitals of an atom in the outer shell orient along the four corners of a tetrahedron to have minimum repulsion between their electrons. ‘The nucleus of the atom is at the centre of the tetrahedron.

4) This leads to form four sp³ – s sigma bonds between carbon atom and four hydrogen atoms, All these bonds are of equal energy,

Question 17.
What is sp² hybridisation? Explain.
Answer:
Consider ethene molecule

• In the formation of CH₂ – CH₂ each carbon atom in its excited state undergoes sp² hybridisation by intermixing one s-orbital (2s) and two p-orbitals (say 2p_x and 2p_y) and reshuffling to form three sp² orbitals.
• Mow each carbon atom is left with one ‘p’ orbital (say 2p_z) unhybridised,
• The three sp² orbitals having one electron each get separated around the nucleus of carbon atoms at an angle of 120°.
• When carbon is ready to form bonds one sp² orbital of one carbon atom overlaps the sp² orbital of the other carbon atom to form sp² – sp² sigma (σ) bond,
• The remaining two sp² orbitals of each carbon atom get overlapped by ‘s’ orbitals of two hydrogen atoms containing unpaired electrons.
• The unhybridised p_z orbitals on the two carbon atoms overlap laterally as shown in figure to form a π (pi) bond.
• Hence, there exist a sigma (σ) bond and a pi π (pi) bond between two carbon atoms in ethene molecule. Hence, the molecule ethene (C₂H₄) is

Question 18.
The list of some organic compounds is given below.
Ethanol, ethane, methanol, methane, ethyne and ethene.
From the above list name the compound …………..
a) formed by the dehydration of ethanol by cone. H₂SO₄.
b) which forms methanoic acid on oxidation?
c) which forms chloroform on halogination in the presence of light?
d) which are unsaturated compounds?
e) which have compounds containing alcohol group?
Answer:
a) Dehydration ethanol in the presence of Cone. H₂SO₄ forms ethene,
b) Methanol on oxidation turns to methanoic acid,
c) Methane in the presence of light forms chloroform,
d) Unsaturated compounds are ethene and ethyne.
e) The compounds containing alcohol group are methanol, ethanol,

Question 19.
Give the IUPAC names of the following compounds.

Answer:

1. 1 – propyne
2. 3 – pentanel (or) pentamSml
3. 2 – methyl propane
4. 1, 2 dichloro ethane

Question 20.
Give one example of each of the following.
i) Saturated hydrocarbon
ii) Cyclic compounds
iii) Unsaturated hydrocarbon
iv) Functional group
v) Homologous series
Answer:
i) Saturated hydrocarbons are Alkanes, So the examples are methane (CH₄), Ethane, (C₂H₆).
ii) Cyclic compounds are cycle alkanes, eg : Cyclo propane (C₃H₆), Cycle butane (C₄H₆).
iii) Unsaturated hydrocarbons are Aikynes, eg : Ethene (C₂H₄), Propene
iv) The examples for functional groups are ‘ 1. Aldehyde – CHO, 2. Alcohol = OH
v) A series of carbon compounds that differ by – CH₂ with similar chemical properties is called homologous series.
eg: 1, Alkane, 2, Alkene

Question 21.
Write the differences between saturated and unsaturated hydrocarbons.
Answer:

Saturated hydrocarbons	Unsaturated hydrocarbons
1) All the four valencies of each carbon atom are satisfied by forming single covalent bonds with carbon and hydrogen atoms,	1) The valencies of at least two carbon atoms are not fully satisfied by the hydrogen atoms.
2) Carbon atoms are joined only by single bonds.	2) Carbon atoms are joined by at least one double bond or by a triple bond.
3) They are less reactive due to non­availability of electrons in the single covalent bond therefore they undergo substitution reactions,	3) They are more reactive because of the presence of electrons in the double or triple bond and therefore undergo addition reactions.

Question 22.
Answer the following.
a) What are the first three members of carboxylic acid series?
b) Name the compounds which can be oxidised directly or in stages to produce ethanoic acid.
c) Write one equation each when acetic acid reacts with a metal, a base, and a carbonate.
d) Name the organic compound formed when acetic acid and ethanol react together.
Answer:
a) The first three members of carboxylic acids are :
i) Methanoic acid – HCOOH
ii) Ethanoic acid – CH₃COOH
iii) Propanoic acid – CH₃CH₂COOH

b) Ethanol in stages oxidises to acetic acid whereas ethanol directly oxidises to ethanoic acid.

c) i) 2 CH₃COOH + 2 Na → 2CH₃COONa + H₂
ii) CH₃COOH + NaOH → CH₃COONa + H₂O
iii) CH₃COOH + Na₂CO₃ → CH₃COONa + H₂O + CO₂

d) When ethanol reacts with ethanoic acid it forms an ester namely ethyl acetate.

Question 23.
What are the rules to be followed to name a carbon compound?
Answer:
Rules to be followed
i) Longest carbon chain is selected,
ii) Chain is numbered in such a way that the branched chain or substituent gets the smallest number,
iii) If the functional group is present, it is given the. lowest number,
iv) Substituents are named in the alphabetical order,
v) The position of substituents are prefixed with hyphen,
vi) Multiple substituents are written with numerical prefixes such as di or tri,

Question 24.
Write suffixes and prefixes for some important characteristic functional group in a tabular form.
Answer:

Question 25.
Correct the following statements.
1) Alkenes undergo substitution reactions.
2) Alkanes are polar in nature.
3) When sodium piece is added to ethanol oxygen gas liberates.
4) On complete combustion of carbon compound it gives carbon monoxide and water.
Answer:

1. Alkenes are unsaturated hydrocarbons. So they undergo addition reactions.
2. Alkanes are covalent compounds. So they are non-polar in nature.
3. When sodium piece is added to ethanol it releases hydrogen gas.
4. On complete combustion of carbon compound it forms carbon dioxide and water.

Question 26.
Copy and complete the following table which relates to three homologous series of hydrocarbons.

Answer:

Question 27.
Draw the structures of isomers of butane.
Answer:
Isomers of butane are n-butane, iso butane and cyclo butane :
Structures :

Question 28.
Draw the structures of the following.
a) Ethanoic acid
b) Propanal
c) Propene
d) Chloro propene
Answer:
Structures:

Question 29.
Draw the structures of the following compounds
a) 2 – bromo pentane
b) 2 – methyl propane
c) butanal
d) 1 – hexyne
Answer:

Question 30.
Write the molecular formula of the first four compounds of the homologous series of aldehydes.
Answer:
Homologous series of aldehydes ate Formaldehyde, Acetaldehyde, Propionaldehyde and Butanaldehyde.

Question 31.
How many isomers can be drawn for pentane with molecular formula C-H(2? What are they? Draw their structures and mention theii common names.
Answer:
Isomers of pentane are three. These are
1) Pentane
2) Iso pentane
3) Neo pentane.
Structures :

Question 32.
Draw the Allotropes of Carbon. Diamond
Answer:

Question 33.
Draw the Graphite.
Answer:

Question 34.
Draw the Buckminsterfullerene (₆₀C).
Answer:

Question 35.
Draw the Nanotubes. A.
Answer:

Question 36.
Draw the structures of Methane :
Answer:
Methane :

Question 37.
Draw the structures of Ethyne :
Answer:

Question 38.
Draw structures of the Ethane and electron dot structure of Chlorine.
Answer:
Ethane:

Chlorine:

Question 39.
Draw the electron dot structures of Ethanoic acid arid Ethyne (Acetylene).
Answer:
Ethanoic acid (Acetic acid) :

Structure of Ethyne (Acetylene) :

Question 40.
Draw the electronic dot structure of ethane molecule.
Answer:

Question 41.
Write the structures of the following compounds.
a) prop-l-ene
b) 2, 3-dimethyl butane
c) 3-hexene
d) 2-methyl prop-l-ene
Answer:

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 5 Refraction of Light at Plane Surfaces.

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

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces 1 Mark Important Questions and Answers

Question 1.
Take a bright metal ball and make it black with soot on a candle flame. Immerse it in the water. Mention one observation. (AP June 2015)
Answer:
1) The ball shines.
2) The ball appears to raise up in water.

Question 2.
What is critical angle? (AP March 2015)
Answer:
The angle of incidence at which the light ray propagates from denser to rarer graze along interface is called critical angle of denser medium.

Question 3.
If refractive index of glass is \(\frac{3}{2}\), then what is speed of light in glass? (AP June 2016)
(OR)
Find the speed of light in a transparent medium, whose refractive index is 3/2.
Answer:
The refractive index of glass or transparent medium = \(\frac{3}{2}\)

Question 4.
Write any two questions about the ‘formation of mirages’. (AP June 2017)
Answer:

1. When does a mirage form?
2. How does a mirage form?

Question 5.
Optical Fibre Cable (OFC) are oftenly used in tele-communications. What is the working principle behind the OFC? (AP March 2017)
Answer:
Total Internal Reflection.

Question 6.
Among objects made of glass and diamond, which one shines more? Why? (AP June 2015)
Answer:
Diamond shines more because of low conical angle of 24.4°.

Question 7.
Suggest reasons for the phenomenon associated with the following : Twinkling of stars. (TS March 2015)
Answer:
Refraction of light is the reason for the twinkling of stars.

Question 8.
Draw the diagram showing the path of the ray when it travels from denser medium to rarer medium when the incident angle is more than the critical angle. (TS June 2016)
Answer:

Question 9.
Why does the light ray deviate in refraction? (AP SA-1:2019-20)
Answer:
Light ray always chooses the path of least time to travel. Hence speed of light changes at interface of two media. So, the light ray deviate in refraction.

Question 10.
Name the phenomenon involved in the function of optical fibre. (AP SCERT : 2019-20)
Answer:
Total Internal Reflection.

Question 11.
What is Fermat’s principle?
Answer:
The light ray always travels in a path which needs shortest possible time to cover the distance between the two given points.

Question 12.
What happens when light travels from one medium to another medium?
Answer:
It bends towards or away from normal.

Question 13.
When does speed of light decrease?
Answer:
When it travels from rarer to denser medium.

Question 14.
What do you mean by denser medium?
Answer:
The medium which has more optical density.

Question 15.
What is refraction?
Answer:
The process of changing speed when light travels from one medium to another is called refraction of light.

Question 16.
Which quantity will compare the refractive indices of two media?
Answer:
Relative refractive index.

Question 17.
What is relative refractive index?
Answer:
It is the ratio of refractive index of second medium to refractive index of first medium.

Question 18.
When does a light ray bend away from normal?
Answer:
When a light ray moves from denser to rarer medium it bends away from normal.

Question 19.
When will angle of refraction be equal to 90°?
Answer:
When angle of incidence is equal to critical angle then angle of refraction will be equal to 90°.

Question 20.
When angle of incidence of light ray is greater than critical angle, what happens?
Answer:
Light ray undergoes total internal reflection.

Question 21.
What happens to refractive index of air with height?
Answer:
Refractive index of air increases with height.

Question 22.
Which has greater refractive index between these?
1) cool air at the top
2) hotter air just above the road
Answer:
Cooler air has greater refractive index due to more density.

Question 23.
What is mirage?
Answer:
The virtual images of distant high objects cause the optical illusion called mirage.

Question 24.
What happens to a light ray when it falls perpendicular to one side of the slab surface?
Answer:
It comes out without any duration.

Question 25.
What are the conditions for total internal reflection?
Answer:

1. The rays of light must travel from denser to rarer medium.
2. The angle of incidence of denser medium must be greater than critical angle.

Question 26.
What is meant by a vertical shift?
Answer:
When a ray emerges out of a glass slab, it is parallel to the incident ray but is displaced laterally relative to incident ray. This shift of emergent ray is called vertical shift.

Question 27.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
It bends towards the normal. This is because it travels from an optically rarer to , optically denser medium.

Question 28.
When does Snell’s law fail?
Answer:
Snell’s law fails when light is incident normally on the surface of a refracting medium.

Question 29.
Why does a ray of light bend when it travels into another medium?
Answer:
It bends because its velocity changes when it moves from one medium to the other.

Question 30.
A pencil when dipped in a glass tumbler containing water appears to be bent at the interface of air and water. Explain why.
Answer:

• When light travels obliquely from one transparent medium to another, the direction of propagation of light changes due to refraction of light.
• In this case, light travels from denser medium to rarer medium, hence it bends away from the normal and the pencil appears to be bent.

Question 31.
Why does light travel in vacuum?
Answer:
Light travels in vacuum because it does not require medium for its propagation.

Question 32.
What are the factors on which refractive index depends?
Answer:

1. Nature of material
2. Wavelength of light used.

Question 33.
What does the ratio of sine of angle of incidence and sine of angle of refraction give?
Answer:
The ratio of sine of angle of incidence and sine of angle of refraction gives refractive index.

Question 34.
What is the relationship between critical angle and refractive index?
Answer:
The relationship between critical angle and refractive index

Question 35.
What is the relationship between angle of incidence and shift?
Answer:
As the angle of incidence increases, the shift also increases.

Question 36.
Why does a coin placed in a water appear to be raised?
Answer:
It is due to refraction of light.

Question 37.
Can you guess what happens when light travels from denser medium to rarer medium?
Answer:
The light ray bends away from normal.

Question 38.
A ray of light falls normally on a face of a glass slab. What are the values of angle of incidence and angle of refraction of this ray?
Answer:
Both angles are zero.

Question 39.
When does light ray from slab not undergo any deviation?
Answer:
The light ray that incidents perpendicular to one side of the slab surface comes out without any deviation.

Question 40.
What is the factor on which refraction depends?
Answer:
Refraction depends on optical density.

Question 41.
What is absolute refractive index?
Answer:
It is the ratio of speed of light in vacuum to speed of light in medium.
\(\mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}}\)

Question 42.
In water filled vessel, the coin of the bottom can be seen at a height. Give reasons.
Answer:
Rising of coin when water is poured in a cylindrical transparent vessel.

Question 43.
Write one activity in showing the process of ‘total internal refraction’.
Answer:
Due to refraction of light speed of light changes when it travels from one medium to another medium.

Question 44.
Define “Glass Slab”.
Answer:
A thin glass slab is formed when a medium is isolated from its surroundings by two plane surfaces parallel to each other.

Question 45.
A ray of light is incident normally on a plane glass slab. What will be the angle of refraction and angle of deviation for the ray?
Answer:
The ray is incident normally on a plane glass slab. So there is no deviation of light ray. Therefore the angle of refraction and angle of deviation both have 0° values.

Question 46.
A light ray in passing from water to a medium (a) speeds up, (b) slows down. In each case get one example of the medium.
Answer:
a) Air, because its optical density is less than water,
b) Glass, because its optical density is more than water.

Question 47.
If an angle of refraction is 90°, what is the corresponding angle of incidence called?
Answer:
The angle of incidence is called critical angle.

Question 48.
If the angle of incidence is more than critical angle, what happens to light ray if the light ray travels from denser to rarer medium?
Answer:
The light ray undergoes total internal reflection.

Question 49.
The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?
Answer:
It means that light travels 2.42 times faster in vacuum than in diamond.

Question 50.
For the same angle of incidence 45°, the angle of refraction in two transparent media, I and II is 20° and 30° respectively. Out of I and II, which medium is optically denser and why?
Answer:
Medium I is optically dense as angle of refraction is lesser in it, hence light bends towards normal.

Question 51.
For which colour of white light is the refractive index maximum and for which colour of white light is the refractive index minimum?
Answer:
The refractive index is maximum for violet because its wavelength is least.
The refractive index is minimum for red because its wavelength is maximum.

Question 52.
Correct the statement. “If the angle of incidence is greater than the critical angle the light is refracted when it falls on the surface from a denser medium to rarer medium”.
Answer:
If the angle df incidence is greater than the critical angle the light undergoes total internal reflection when it falls on the surface from a denser medium to rarer medium.

Question 53.
A light ray passes from medium 1 to medium 2. Which of the following quantities of refracted ray will differ from that of the incident ray?
Speed, intensity, frequency, wavelength.
Answer:
Speed, intensity and wavelength will differ from that of incident ray.

Question 54.
The refractive indices of alcohol and turpentine oil with respect to air are 1.36 and 1.47 respectively. Find the refractive index of turpentine oil with respect to alcohol. Which one of these permits the light to travel faster?
Answer:
The refractive index of turpentine oil with respect to alcohol = \(\frac{1.47}{1.36}\) = 1.08.

The refractive index increases when the speed of light decreases. So light travels faster in alcohol as its refractive index is less.

Question 55.
Light enters from air to diamond which has refractive index of 2.42. Calculate the speed of light in diamond, if speed of light in air 3 × 10⁸ ms^-1.
Answer:
Absolute refractive index = \(\frac{c}{v}\)
2.42 = \(\frac{3 \times 10^{8}}{\mathrm{v}}\) ⇒ v = 1.24 × 10⁸ ms^-1.

Question 56.
A glass block 3.0 cm thick is placed over a stamp. Calculate the height through which image of stamp is raised. Refractive index of glass is 1.54.
Answer:

Question 57.
The refractive index of water is 4/3. Calculate the critical angle for water – air interface (sin 49 = 3/4).
Answer:

10th Class Physics 5th Lesson Reflection of Light by Different Surfaces 2 Marks Important Questions and Answers

Question 1.
A ray of light enters from air to a medium X. The speed of light in the medium is 1.5 × 10⁸ m/s and the speed of light in air is 3 × 10⁸ m/s.
Find the refractive index of the medium X. (TS March 2015)
Answer:

Question 2.
What are the applications of optical fibres?
(OR)
Write two uses of fibre optics in daily life. (TS June 2016)
Answer:
Applications oruses of optical fibres :

1. Light pipes using optical fibres may be used to see places which are difficult to reach things such as inside of a human body.
2. The other important application of fibre optics is to transmit communication signals through light pipes.

Question 3.
Focal length of the lens depends on its surrounding medium. What happens, if we use a liquid as surrounding media of refractive index, equal to the refractive index of lens? (TS June 2018)
Answer:

• When the refractive index of surrounding media is equal to the refractive index of lens, the lens looses its characteristics.
• Lens do not diverge or converge the light.
• Light do not get refracted when it passes through that lens.

Question 4.
Why does the light ray travel slowly in diamond when compared to vacuum? (AP SA-I : 2019-20)
Answer:

• Refractive index of diamond (2.42) is greater than that of vacuum (1).
• Speed of light is inversely proportional to refractive index of substances.
• Hence, light ray travel slowly in diamond when compared to vacuum.

Question 5.
Write about laws of refraction.
(OR)
Write the laws of refraction.
Answer:
Laws of refraction :

1. The incident ray, the refractive ray and the normal to interface of two transparent media at a point of incidence lie in the same plane.
2. During refraction light follows Snell’s law, i.e., the ratio of sine of angled of incidence to sine of angle of refraction is constant.
   n₁ sin i = n₂ sin r (OR) \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\) = constant.

Question 6.
What is total internal reflection ? What are the applications of total internal reflection?
Answer:
When the angle of incidence is greater than critical angle, the light ray is reflected into denser medium at interface i.e., light never enters rarer medium. This phenomenon is called total internal reflection.
1) Brilliance of diamonds :
Total internal reflection is the main cause for brilliance of diamonds. The critical angle of diamonds is very low (24.4°). So if a light ray enters a diamond it is very likely to get total internal reflection which makes the diamond shine brilliant.

2) Optical fibres:
Total internal reflection is the basic principle for working of optical fibre.

Question 7.
What are optical fibres? How do they work?
Answer:

• An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer.
• A bunch of such thin fibres forms a light pipe.

Working :

1. Because of the small radius of the fibre, light going into it makes a nearly glancing incidence on the wall.
2. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
3. The light is thus transmitted along the fibre.

Question 8.
How can a patient’s stomach be viewed by using optical fibres?
(OR)
How do you observe patient’s stomach by using a light pipe?
Answer:

• The patient’s stomach can be viewed by inserting one end of a light pipe into the stomach through the mouth.
• Light is sent down through one set of fibres in the pipe.
• This illuminates the inside of the stomach.
• The light from the inside travels back through another set of fibres in the pipe and the viewer gets the image at the outer end.

Question 9.
State four differences between reflection and total internal reflection.
Answer:

Reflection	Total internal reflection
1) Smooth polished surface is required for reflection.	1) No smooth polished surface is required for total internal reflection.
2) It takes place for all angles of incidence.	2) It takes place only, when angle of incidence is greater than critical angle.
3) It takes place when the rays of light travel from rarer to denser medium to an opaque medium.	3) It takes place when rays of light travel from denser to rarer medium.
4) Some amount of light is absorbed by reflecting surface.	4) No light is absorbed by reflecting surface.

Question 10.
The figure shows refraction and emergence of a ray of light incident on a rectangular glass slab. Copy the diagram and mark the lateral displacement of the incident ray. Name the two factors on which the lateral displacement depends.

Answer:
The lateral displacement depends on

1. The angle of incidence of the incident ray PQ, on the slab and
2. The thickness of the glass slab.

The perpendicular distance between the emergent forward.

Question 11.
1) What happens to a ray of light when it travels from one equal refractive indices?
2) State the cause of refraction of light.
Answer:
1) No refraction or bending would take place. The light will travel in a straight line.

2) The refraction occurs due to change in speed of light as it enters from one medium to another.

Question 12.
A coin placed at the bottom of a tank appears to be raised when water is poured into it. Explain.
Answer:

• It happens due to the phenomenon of refraction of light.
• When the rays of light from the coin, in the denser medium fall on the interface separating the two media, the rays of light move away from the normal after refraction.
• The point from which the refracted rays appear to come gives the apparent position of the coin.
• As the rays appear to come from a point above the coin, therefore, the coin seems to be raised.

Question 13.
Define refractive index. Explain the relationship between the refractive index of the medium and to the speed of light in the medium.
Answer:
The ratio of speed of light in vacuum to the speed of light in that medium is defined as refractive index ‘n’ with respect to the vacuum. It is also called absolute refractive index.

When refractive index of a medium is high, then the speed of light is low and vice-versa.

Question 14.
Explain lateral shift and vertical shift.
Answer:
Lateral shift:
The distance between incident and emergent ray is called lateral shift.

Vertical shift :
The perpendicular distance between object and its image is called vertical shift.

Question 15.
During refraction of light, which of the following quantities does not change.
(1) velocity,
(2) wavelength,
(3) frequency,
(4) amplitude.
Answer:
During refraction of light velocity of light changes and also wavelength and amplitude. Frequency does not change during refraction.

Question 16.
The upper surface of water contained in a beaker and held above the eye level appears silvery. Why?
Answer:
Critical angle for water is 48°. The rays of light entering in water from below, suffer refraction. If these rays strike the water-air surface at an angle which is greater than 48°, they get totally internally reflected. These rays on emerging out of water, appear to come from the upper surface of water, which in turn appear silvery.

Question 17.
Why don’t the planets twinkle?
Answer:

• The planets are much closer to the earth, and are thus seen as extended sources.
• We can consider a planet as a collection of a large number of point-sized sources of light.
• The total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying twinkling effect

Question 18.
Why did an empty test tube placed obliquely in water, appears filled with mercury, when seen from above?
Answer:
When the rays of light travelling through water they strike the water glass interface of test tube at an angle, which is more than critical angle for water, they suffer total internal reflection. When these totally reflected rays reach eye, then to the eye they appear as they come from surface of test tube, which in turn appears filled with mercury.

Question 19.
Why are the bubbles rising up the fish tank appear silvery?
Answer:
When the rays of light travelling through water they strike the water air interface of the bubble at an angle, which is greater than critical angle for water, they get totally internally reflected. These reflected rays on reaching the eye appear to come from air bubble, which in turn appears silvery.

Question 20.
Why does a crack in a window pane appear silvery?
Answer:
There is always some amount of air present in the crack. When the rays of light travelling through glass, strike the glass, the glass air interface at an angle, greater than critical angle of glass, they are totally internally reflected. When these reflected rays reach eye, then to the eye they appear to come from the crack, which in turn appears silvery.

Question 21.
Explain why a straight stick appears to be bent when dipped in water.
Answer:

• Suppose two rays originate from the end of the stick in water.
• As these rays get refracted into the air, they bend away from the normal.
• When these two refracted rays are produced backwards they seem to meet at a point higher than the end of stick.
• This point gives the apparent position of the end of the stick. Thus, the stick appears to be bent.

Question 22.
A pond appears to be shallower than it really is when viewed obliquely. Why?
Answer:

• Suppose two rays originate from the bottom of the pond. As these rays get refracted into the air, they bend away from the normal.
• When these two refracted rays are produced backwards they seem to meet at a point higher than the bottom of the pond.
• This point gives the apparent position of the bottom of the pond.
• Thus, the pond appears to be shallower.
• This effect is absent if the pond is viewed normally.

Question 23.
Frame some questions to know about the formation of mirages.
Answer:

1. What are mirages?
2. What is the principle involved in mirages?
3. Can mirages be photographed?
4. Where does the water on the road go?

Question 24.
A glass slab is placed over a piece of paper on which VIBGYOR is printed with each letter into corresponding colours.
1) Will the image of all the letters be in the same place?
2) The letter of which colour appears to be raised maximum and which colour minimum? Explain your answer.
Answer:

1. The image of all letters will not be in the same place.
2. The letter of violet colour appears to be raised maximum, while the letter of red colour appears to be raised minimum because refractive index of glass is most for the violet light while least for the red light, therefore the apparent depth is least for violet and most for red.

Question 25.
Why does sun appear bigger during the sunset or the sunrise?
Answer:

1. We already know that the apparent position of sun is higher than actual position in the horizon.
2. Moreover, due to refraction, the apparent image of sun is closer to eye than the actual position. Since during sunset or sunrise, the rays of light travel through maximum length of atmosphere therefore the refraction is also maximum.
3. Hence apparent image of sun is very much closer to eye. Thus it appears bigger.

Question 26.
Write the material required in finding out the relation between angle of incidence and angle of refraction.
Answer:
Material required :
A plank, white chart, protractor, scale, small black painted plank, a semi-circular glass disc of thickness nearly 2cm, pencil and laser light.

Question 27.
Write the aim and apparatus experiment in finding the refractive index of the glass slab.
Answer:
Aim :
Finding the refractive index of the glass slab.

Apparatus :
Glass slab, white chart, pin.

Question 28.
Observe the following table and answer the following question.

Questions :
1) Find out from the table the medium having highest optical density and the medium with lowest optical density.
2) You are given kerosene, turpentine oil and water. In which of these does the light travel fast? Use the information given in the table.
3) The refractive index of diamond is 2.42. What is the meaning of this statement?
4) When light travels from water to crown glass, what happens?
5) When light travels from diamond to air, what happens?
Answer:
1) The medium with highest optical density is diamond as its refractive index is maximum, i.e. 2.42.
The medium with lowest optical density is air, as its refractive index is minimum, i.e. 1.0003.

2) The refractive index of medium is given by the expression, n = \(\frac{c}{v}\) or v = \(\frac{c}{n}\)
This expression shows that light travels faster in the medium whose refractive index is minimum. From the table, we can find that water has the minimum value of refractive index. Therefore light travels faster in water.

3) This statement means that light travels 2.42 times faster in vacuum than in diamond.

4) The light bends towards normal.

5) The light bends away from the normal.

Question 29.
A ray of light enters from a medium A into a slab made up of a transparent substance B. Refractive indices of medium A and B are 2.42 and 1.65 respectively. Complete the path of ray of light till it emerges out of slab.

Answer:

Question 30.
A glass slab made of material of refractive index n₁ is kept in medium of refractive index n₂. A light ray is incident on the slab. Complete the path of rays of light emerging from glass slab, if a) n₁ > n₂ b) n₁ = n₂ c) n₁ < n₂.
Answer:
a) n₁ > n₂

b) n₁ = n₂
There is no deviation of light ray

c) n₁ < n₂

Question 31.
How do you appreciate the process of total internal reflection in nature?
Answer:

1. Total internal reflection is responsible for brilliance of diamond.
2. Total internal reflection is basic pruxiplo behind working of optical fibres which are used in getting the images ol internal ouaiis and also used in telecommunications. So the role of total internal reflection is thoroughly appreciated.

Question 32.
Write the application of optical fibres in communication.
Answer:

• Optical fibres are used to transmit communication signals through light pipes.
• For example, about 2000 telephone signals, approximately mixed with lightwaves, may be simultaneously transmitted through a typical optical fibre.
• The clarity of the signals transmitted in this way is much better than other conventional methods.

Question 33.
Write the applications of total internal reflection.
Answer:
Application of total internal reflection :

1. Brilliance of diamonds,
2. Optical fibres.

Question 34.
A monochromatic ray of light strikes the surface of transparent medium at an angle of incidence 60° and gets refracted into the medium at an angle of refraction 45°. What is the refractive index of the medium?
Answer:

Question 35.
A light ray enters a liquid at an angle of incidence 45° and it gets refracted on liquid at angle of refraction 30°. Calculate the refractive index of the liquid.
Answer:

Question 36.
Refractive index of water is 4/3. Calculate the speed of light in water.
Answer:

Question 37.
A postage stamp placed under glass appears raised by 8 mm. If refractive index of glass is 1.5, calculate the actual thickness of glass slab.
Answer:
Let real thickness of glass = x.
Vertical shift = 8 mm.

Question 38.
Refractive index of glass is 1.5. Find its critical angle.
Answer:

Question 39.
What is the advantage of using prism in place of plane mirror in periscope or binocular?
Answer:

• When total internal reflection occurs from a prism, the entire incident light is reflected back into the denser medium.
• Whereas in ordinary reflection from a plane mirror, some light is refracted and absorbed. So the reflection is partial.
• This is the reasons why total reflecting prism is used in place of a plane mirror to deviate the light ray by 90° in a periscope and 180° in a binocular.

10th Class Physics 5th Lesson Reflection of Light by Different Surfaces 4 Marks Important Questions and Answers

Question 1.
What is the angle of deviation produced by the glass slab? Explain with ray diagram. (AP June 2015)
(OR)
Which angle of deviation is produced by glass slab? Write your explanation with a ray diagram.
Answer:

1. Angle of deviation is the angle between incident ray and emergent ray.
2. The angle of deviation produced by a glass slab is ‘O’, because the incident ray and emergent ray are parallel to each other that can be seen in the figure.

Question 2.
Explain the phenomenon of total internal reflection with two examples. (AP June 2018)
(OR)
What is total internal reflection? Explain with examples. (AP SA-I:2019-20)
Answer:

• When the angle of incidence is greater than the critical angle, the light ray is reflected into denser medium at interface. This phenomenon is called total internal reflection.
• Total internal reflection is the main reason for brilliance of diamonds. The critical angle of a diamond is very low. So if a light ray enters a diamond it is very likely to undergo total internal reflection which makes the diamond shine.
• Total internal reflection is the basic principle behind working of optical fibre. Because of the small radius of the fibre light going into it makes a nearly glancing incidence is greater than the critical angle and hence total internal reflection takes place. The light is thus transmitted along the fibre.

Question 3.
Explain the relation between angle of incidence and angle of refraction with an experiment. (AP March 2018)
Answer:
Aim :
To verify the relation between angle of incidence and angle of refraction.

Material required :
A plank, white chart, protractor, semicircular glass disc, pencil and leser light.

Procedure :

1. Take a drawing sheet on a cardboard and mark different angles (on both side of MM line)
2. Place a semi circular glass disc, so that its diameter coincides with the line “MM”.
3. Send a laser light along a line with makes 15° with NN.
4. Let it is incident angle.
5. Measure its corresponding angle of refraction by observing light coming from outside of the glass slab.
6. Repeat this experiment with various values of angle of incidence, refraction and not in the table.

7.

8. From the above table we observe that \(\frac{\sin \mathrm{i}}{\sin r}\) = constant.

Question 4.
Give some daily life consequences of refraction of light.
Answer:

• A star appears twinkling in the sky.
• The sun is seen a few minutes before it rises above the horizon in the morning and in the evening few minutes longer after it sets.
• A coin kept in a vessel not visible when seen from just below the edge of the vessel, can be viewed from the same position when water is poured into the vessel.
• A print appears to be raised when a glass block placed over it.
• A piece of paper stuck at the bottom of a glass block appears to be raised when seen from above.
• A tank appears shallow than its actual depth.
• A person’s legs appear to be short when standing in a tank.
• An object placed in a denser medium when viewed from a rarer medium appears to be at a lesser depth.
• An object in a rarer medium, when viewed from a denser medium, appears to be at a greater distance than its real distance.

Question 5.
What are the factors which influence refractive index of material?
Answer:

• Nature of medium, i.e. its optical density. Smaller the speed of light in a medium relative to air, higher is the refractive index of the medium.
• Physical condition such as temperature. With rise in temperature the speed of light in medium increases, so the refractive index of medium decreases.
• The colour or wavelength of light (refractive index increases with decrease in wavelength, eg : µ_v > µ_R).

Question 6.
What is the advantage of total internal reflection over reflection?
Answer:

• In the process of total internal reflection, 100% energy is reflected back.
• No other device such as plane mirror, etc. produces 100% reflection due to absorption and refraction of some part of light.
• Due to this property the phenomenon total internal reflection is of great practical application in the construction of periscope, binocular and certain type of camera.

Question 7.
The diagram below shows a glass block suspended in a liquid. A beam of light of single colour is incident from liquid on one side of block.
1) Draw diagrams to show how light bends when it travels from liquid to glass and then to liquid if (i) the light slows down in glass (ii) the light speeds up in glass.
2) State two conditions under which the light ray moving from liquid to glass passes straight without bending. Will the glass be visible them?
Answer:
1) If light slows down in going from liquid to glass (i.e., µ_glass > µ_liquid), it will bend towards the normal at the point of incidence in passing from liquid to glass at the first surface, while it is bent away from normal at the second surface in passing from glass to liquid. In the ray diagram, the light beam suffers lateral shift.

2) If light speeds up in going from liquid to glass (i.e., µ_glass < µ_liquid). It will bend away from the normal at the point of incidence on the first surface in passing from liquid to glass, while it bends towards the normal at the second surface in passing from glass to liquid. The light beam suffers lateral shift in direction opposite to that
Note that in both cases, the emergent ray is parallel to the incident ray.

Question 8.
A ray of light is incident on a rectangular glass block PQRS, which is silvered at the surface RS. The ray is partly reflected and partly refracted.

1) Trace the path of reflected and refracted rays.
2) Show at least two rays emerging from the surface PQ after reflection from the surface RS.
3) How many images are formed in the above case? Which image is the brightest?
Answer:
1) In the figure OB is reflected ray and OC is the refracted ray for the incident ray AO.

2) Two rays emerging from surface PQ after reflections for the surface RS are labelled as 1 and 2.

3) Multiple (or infinite) images are formed. The second image formed due to first reflection at C at the silvered surface RS is the brightest. It is seen in the direction of ray 1.

Question 9.
What are the factors which affect critical angle? The critical angle for a given pair of media depends on their refractive index which is affected by the following factors.
Answer:
1. Effect of colour of light:
The refractive index of transparent medium is more for violet light and less for red light, therefore the critical angle for pair of media is less for the violet light and more for the red light. Thus critical angle increases with increase in wavelength of light.

2. Effect of temperature :
On increasing the temperature of medium, its refractive index decreases, so the critical angle for that pair of media increases. Thus critical angle increases with increase in temperature.

Question 10.
The table shows the refractive index of some material media.

Material Medium	Refractive Index
Air	1.0003
Ice	1.31
Water	1.33
Kerosene	1.44
Fused quartz	1.46
Turpentine oil	1.47
Crown glass	1.52
Benzene	1.50

Answer the following questions with the help of the above table.
1) Find the speed of light in Benzene.
2) Write the relationship between mass density and optical density of kerosene and water.
3) What are the factors that refractive index depends on?
4) Write the relative refractive index of kerosene with water.
Answer:
1)

Speed of light in benzene = 2 × 10⁸ m/s

2) Optical denser medium may not possess greater mass density. Kerosene with high refractive index is optically denser than water although its mass density is lesser than water.

3) Refractive index depends on
1) nature of material,
2) wavelength of light used.

4) Relative refractive index of kerosene with water = \(\frac{1.44}{1.33}\) = 1.08

Question 11.
Red light of wavelength 6600A travelling in air gets refracted in water. If the speed of light in air is 3 × 108 ms-1 and refractive index of water is 4/3, find the
(i) frequency of light in air,
(ii) the speed of light in water,
(iii) the wavelength of light in water.
Answer:

Question 12.
Draw the ray diagram which shows the ray takes curved path because of total internal reflection.
Answer:

Question 13.
Give some daily life consequences of total internal reflection?
Answer:

• On a hot sunny day, a driver may see a pool of water on the road before him. It is the phenomenon of mirage which is often observed in desert.
• An empty test tube placed in a beaker with mouth outside the water surface shines like a mirror.
• A crack in a glass vessel often shines like a mirror.
• A piece of diamond sparkles when viewed from certain directions.
• An optical fibre is used to transmit a light signal over a long distance with negligible loss of energy.

Question 14.
Light travels from air to water, then the refraction index of water is 1.33. Hence find the refractive index when light travels from water to air.
Answer:
Refractive index of water (n₂₁) = 1.33
Refractive index of air (n₁₂) = \(\frac{1}{1.33}\) = 0.75

Question 15.
The refractive index of diamond is 2.42 and the refractive index of glass is 1.5; compare the critical angle between them. (Diamond 24°, glass 42°)
Answer:
Refractive index of diamond (µ₁) = 2.42

Question 16.
A ray of light travels from an optically denser to rarer medium. The critical angle of the two media is ‘C’. What is the maximum possible deviation of the ray?
Answer:
The relation between angle of deviation and angle of incidence, angle of emergence and angle of prism is given by
Angle of deviation = i₁ + i₂ – A
For maximum deviation, Angle of incidence (i₁) = 90°
Angle of emergence (i₂) = 90°

∴ Maximum deviation = i₁ + i₂ – A = 90 + 90 – 2C = 180 – 2C = n – 2C.

Question 17.
A ray of light strikes a glass slab 5 cm thick making an angle of incidence equal to 30°.
a) Construct the ray diagram showing emergent ray and refracted ray through the glass block. The refractive index of glass is 1.5.
b) Measure the lateral shift of the ray.
Answer:

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 13 Principles of Metallurgy.

AP State Syllabus SSC 10th Class Chemistry Important Questions 13th Principles of Metallurgy

10th Class Chemistry 13th Lesson Principles of Metallurgy 1 Mark Important Questions and Answers

Question 1.
Which method is suitable to enrich sulphide ores? (AP June 2016)
Answer:
Froth flotation method is suitable to enrich sulphide ores.

Question 2.
We use P.V.C. pipes for water supply instead of metal pipes. Why? (AP March 2017)
Answer:
PVC pipes do not rust. So they are used as water pipes instead of metal.

Question 3.
Arrange the metals Fe, Na, Ag and Zn in increasing order of their chemical reactivity. (TS March 2017)
Answer:
Ag < Fe < Zn < Na
(OR)
Ag, Fe, Zn, Na

Question 4.
Write the deferences between Roasting and Calcination. (TS June 2018)
Answer:
1) Burning of ore in the presence of air or oxygen is called “Roasting”.
So in the roasting air is present.

2) Burning of ore in the absence of air or oxygen is called “Calcination.”
So in the calcination air is absent.

Question 5.
What are the preventive methods do you take for rusting iron materials? (TS March 2018)
Answer:

1. Covering the surface of iron materials with paint or by some chemicals.
2. Electroplating.

Question 6.
Mention the application of thermite process in daily life. (AP SCERT: 2019-20)
Answer:
1) Joining railing of railway tracks,
2) Joining cracked machine parts.

Question 7.
What are the essential condition that iron articles get rust? (TS June 2019)
Answer:
The essential condition that iron articles get rust is presence of water and air both.

Question 8.
What is metallurgy?
Answer:
The process of extraction of metals from their ores is called metallurgy.

Question 9.
What is the Bronze an alloy of?
Answer:
Bronze is an alloy of copper and tin.

Question 10.
What are ores?
Answer:
The minerals from which the metals are extracted without economical loss are called ores.

Question 11.
What is the percentage of Aluminium oxide in Bauxite?
Answer:
50-70%.

Question 12.
Why is 16th group called chalcogen family?
Answer:
Chaleo means ore and genus means produce. We notice that the ores of many metals are oxides and sulphides. This is why oxygen-sulphur (16th group) group as chalcogen family.

Question 13.
Which metals form oxides, sulphides and carbonates?
Answer:
Moderate reactive metals.

Question 14.
Based on the reactivity arrange the metals.
Answer:
Based on reactivity we can arrange metals in descending order of their reactivity as shown below:

Question 15.
What is gangue?
Answer:
The unwanted material in the ore is called gangue.

Question 16.
What is activity series?
Answer:
Arrangement of the metals in decreasing order of their reactivity is known as activity series.

Question 17.
Why do we add some impurities to ore?
Answer:
We add some suitable impurities to ore in order to decrease its melting point.

Question 18.
What is roasting?
Answer:
Roasting is a pyrochemical process in which ore is heated in the presence of oxygen or air below its melting point.

Question 19.
What is thermite process?
Answer:
The reaction of metal oxides with aluminium is called thermite process.

Question 20.
Write the chemical equations involving thermite reaction.
Answer:
2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe + heat
2 Al + Cr₂O₃ → Al₂O₃ + 2 Cr + heat

Question 21.
How do you convert cinnabar into mercury?
Answer:
When cinnabar (HgS) is heated in air, it is first converted into (HgO), then reduced to mercury on further heating.

Question 22.
What is distillation of metals?
Answer:
The extracted metal in the molten state is distilled to obtain the pure metal as distillate by distillation of metals. Here the impurities are high boiling point metals.

Question 23.
What is poling?
Answer:
The molten metal is stirred with logs (poles) of greenwood and impurities are removed either as gases or they get oxidized and form slag over surface of the molten metal is called poling.

Question 24.
What is liquation?
Answer:
A low melting metal can be made to flow on a slope surface to separate it from high melting impurities is called liquation.

Question 25.
What is calcination?
Answer:
Calcination is a pyrochemical process in which the ore is heated in the absence of air.

Question 26.
What is flux?
Answer:
Flux is a substance added to the ore to remove the gangue from it by reacting with ore.

Question 27.
What are the ores of iron?
Answer:
Haematite (Fe₂O₃), Magnetite (Fe₃O₄).

Question 28.
What are the ores of zinc?
Answer:
Zinc blende (ZnS), Zincite (ZnO).

Question 29.
What is the formula of gypsum and metal present in gypsum?
Answer:
The formula of gypsum is CaSO₄.2H₂O. The metal present in gypsum is calcium.

Question 30.
What is a furnace?
Answer:
The furnace is one which is used to carry out pyrochemical process in metallurgy.

Question 31.
Arrange the following chlorides in ascending order of reactivity of respective metals. MgCl₂, NaCl, PbCl₂, HgCl₂.
Answer:
The ascending order is HgCl₂, PbCl₂, MgCl₂, NaCl.

Question 32.
What are the fuel and flux for haematite ore?
Answer:
The coke is used as fuel and limestone (CaCO₃) is used as flux for haematite ore.

Question 33.
Why can copper not displace zinc from its compound?
Answer:
Copper is less reactive than zinc. So copper cannot displace zinc from its compound or salt.

Question 34.
How do various metals in activity series react with chlorine on heating?
Answer:

1. All the metals react with chlorine on heating to form their respective chlorides.
2. But the reactivity decreases from top to bottom.

Question 35.
How do you know the reactivity of metals with chlorine decreases from top to bottom?
Answer:
We know that the reactivity of metals with chlorine decreases from top to bottom the heat evolved when the metal reacts with one mole of chlorine gas to form chloride.

Question 36.
Give some examples for corrosion.
Answer:
Examples for corrosion :

1. The rusting of iron (Iron oxide)
2. Tarnishing of silver (Silver sulphide)
3. Development of green coating on copper (Copper carbonate) and bronze.

Question 37.
Why are potassium, sodium, calcium never found in free state?
Answer:
The metals potassium, sodium and calcium are so reactive that is why they never exist in free state.

Question 38.
How do you extract metals at the top of activity series?
Answer:
The metals at the top of activity series are extracted by electrolysis of their fused compounds.

Question 39.
What is meant by enrichment of ore?
Answer:
Some physical methods are useful in removing unwanted rocky material from ore. It is called enrichment of ore.

Question 40.
How do you extract metal from the crude metal?
Answer:
To extract metal from enriched ore it is converted into metallic oxide by reduction reaction. Then this metallic oxide is further reduced to get metal with certain impurities.

Question 41.
What are the impurities you get in the refining of copper?
Answer:
Antimony, selenium, tellurium, silver, gold and platinum.

Question 42.
What is slag?
Answer:
The substance formed due to reaction of gangue and flux,
Eg : CaSiO₃, FeSiO₃

Question 43.
What is meant by pyrochemical reactions?
Answer:
Pyre means heat. So the chemical reactions involving heat are called pyrochemical reactions.

Question 44.
Can you mention some articles that are made up of metals?
Answer:
Jewellery, conducting wires and utensils.

Question 45.
Why are we mixing small amount of carbon to iron?
Answer:
To make iron hard and strong.

Question 46.
What is the main difference between steel and stainless steel?
Answer:
Steel will rust whereas stainless steel will not rust.

Question 47.
Give some examples for corrosion.
Answer:
The rusting of iron, tarnishing of silver, development of green coating on copper.

Question 48.
Do metals exist in the same form as that we use in our daily life?
Answer:
No, they exist as ores and minerals and some may exist in the form of metals.

Question 49.
Do you know how metals are obtained?
Answer:
The metals are extracted from their ores.

Question 50.
Have you ever heard the words like ore, mineral and metallurgy?
Answer:
Yes, these words are related to extraction of metal.

Question 51.
Where do we carry out pyrochemical processes in metallurgy?
Answer:
Pyrochemical processes can be carried out inside the furnace.

Question 52.
Which process converts sulphide ore into oxide ore?
Answer:
Roasting is the process which converts sulphide ore into oxide ore.

Question 53.
Is silver mineral or ore? Justify your answer.
Answer:
Silver is neither mineral nor ore. It is a metal.

Question 54.
Give two examples for corrosion.
Answer:
1) Rusting of iron
2) Green coating on copper.

Question 55.
Name the form of carbon used in the blast furnace for the extraction of iron.
Answer:
Carbon is used in the form of coke to reduce iron in blast furnace.

Question 56.
Give name and formulae of sulphide ore of lead and mercury.
Answer:
a) Sulphide ore of lead is Galena. Its formula is PbS.
b) Sulphide ore of mercury is Cinnabar. Its formula is HgS.

Question 57.
What are the various pyrochemical processes used in metallurgy?
Answer:
The various pyrochemical processes used are
a) Smelting,
b) Roasting,
c) Calcination.

Question 58.
What is the gas released at anode when fused sodium chloride is electrolysed?
Answer:
When sodium chloride is electrolysed, sodium metal is formed at cathode and chlorine gas is formed at anode.
NaCl → Na⁺ + Cl^–

Question 59.
Which pyrochemical process is useful to convert zinc blende into oxide ore?
Answer:
Zinc blende is sulphide ore of zinc. Its formula is ZnS. So it can be converted into oxide ore by heating strongly in excess of air known as roasting.
2 ZnS + 3 O₂ → 2 ZnO + 2 SO₂

Question 60.
Which pyrochemical process is useful to convert Magnesite into oxide ore?
Answer:
Magnesite is carbonate ore of magnesium. Its formula is MgCO₃. So it can be converted into oxide ore by heating in the absence of air.
MgCO₃ → MgO + CO₂

Question 61.
What is the main impurity present in iron when it is removed from the blast furnace?
Answer:
The main impurity that can be removed is slag because it is formed when gangue in the ore reacts with flux.
CaO + SiO₂ → CaSiO₃

Question 62.
Name two metals normally manufactured by the electrolysis of fused compounds.
Answer:
Metals with high reactivity can be extracted by electrolysis of their fused compounds.
The examples are potassium, sodium and calcium.

Question 63.
What are the examples of corrosion?
Answer:

1. Rusting of iron.
2. Tarnishing of silver.
3. Development of green coating on copper (CuCO₃).
4. Green coating on Bronze.

Question 64.
What is importance of prevention of corrosion?
Answer:

1. Save the money.
2. Preventing accidents such as a bridge collapse.
3. Failure of a key component.

Question 65.
What is meant by galvanisation?
Answer:
Preventing the rust on metals by using layer of zinc. This phenomena is called galvanisation.

Question 66.
Write the example of electroplating in daily life.
Answer:

1. Rold gold.
2. Copper coating on cookware.

Question 67.
What is formula and name of iron rust?
Answer:
Iron rust is equal to hydrated ferric oxide.
Formula : Fe₂O₃ × H₂O

Question 68.
What is importance of alloying?
Answer:

1. Improving the properties of metal.
2. To avoid the rust.
3. To increase the hardness.

Question 69.
Which one used as flux extracting of iron from heamatite?
Answer:
Limestone or calcium carbonate (CaCO₃).

Question 70.
Marne the two metals which corrode easily?
Answer:
Iron and copper.

Question 71.
Atmospheric air always contains moisture. Then, how can you protect iron articles from the affect of atmosphere?
Answer:
By painting, oiling and greasing, etc.

Question 72.
Explain the terms gangue and flux.
Answer:
The impurity present in the ore is called gangue. The substance added to the ore to remove gangue from it is called flux.

Question 73.
What are the metals are present in carnallite?
Answer:
Potassium (K) and Magnesium (Mg).

Question 74.
Write the elements are present in high reactivity series.
Answer:
Na, Mg, Al, K, Ca
(11 12 13 19 20)

Question 75.
Write the elements that are in moderate reactivity series.
Answer:
Fe, Cu, Zn, Pb.

Question 76.
Name the two metals which do not corrode easily?
Answer:
Gold and platinum.

Question 77.
Mention some important methods of refining.
a) Distillation
b) Poling
c) Liquation
d) Electrolysis
Answer:
d) Electrolysis

Question 78.
What is the role of furnace in metallurgy?
Answer:
Furnace is the one which is used to carry out pyrochemical process in metallurgy.

Question 79.
What is meant by calcination?
Answer:
It is the process of heating the concentrated ore in the absence of air.

Question 80.
Write the equation of heating of one sulphide ore in the process of roasting.
Answer:

Question 81.
Mention two methods which produce very pure metals?
Answer:
a) Electrolytic reduction.
b) Smelting.

Question 82.
What are the applications of thermite reaction in daily life?
Answer:
a) To join railings of railway tracks.
b) To join cracked machine parts.

Question 83.
Arrange the metals Ag, Mg, K in reactivity series.
Answer:
K > Mg > Ag.

Question 84.
How do you extract highly reactive metals?
Answer:
Highlyreactive metals can be extracted by electrolysis of their fused compounds.

Question 85.
What is dressing of an ore?
Answer:
The process of removal of impurities from an ore is called dressing of the ore or concentration of the ore.

Question 86.
Write the equation of example of calcination.
Answer:

Question 87.
Write the some properties of metals.
Answer:
Malleability, Ductility, Sonarity and Electrical conductivity.

Question 88.
Mention the stages involved in extraction of a metal from its ore.
Answer:

1. Dressing or concentration.
2. Extraction of crude metal.
3. Refining or purification of the metal.

Question 89.
How do you extract moderately reactive metals?
Answer:
These metals are generally sulphides and carbonates. They are converted into oxides before reducing them to metals.

Question 90.
Give an example for reduction of metal oxide with carbon.
Answer:
The oxides are reduced by coke in a furnace which gives the metal and carbon monoxide.

Question 91.
Give an example for reduction of oxide ore with CO.
Answer:

Question 92.
What is flux?
Answer:
Flux is a substance added to the ore to remove the gangue from it by reacting with ore. If the impurity is acidic substance, basic substance is used as flux and vice – versa.

Question 93.
How do various metals in activity series react with chlorine on heating?
Answer:

1. All the metals react with chlorine on heating to form their respective chlorides.
2. But the reactivity decreases from top to bottom.

Question 94.
What are the substances to be added if the gangue is acidic or basic?
Answer:
If the gangue (impurity) is acidic substance like SiO₂, basic substance like CaO is used as flux and if the impurity is of basic nature like FeO acidic flux like SiO₂ is added to the gangue.

Question 95.
Why will stainless steel not rust?
Answer:
Stainless steel is prepared by mixing iron with nickel and chromium. Nickel and chromium are less reactive with oxygen. So stainless steel will not rust.

Question 96.
Why is sodium metal stored in kerosene?
Answer:
Sodium is highly reactive with both air (oxygen) and water. So it should be stored in kerosene.

Question 97.
Which metal gets covered with protective film of oxide when exposed to air?
Answer:
The metal is aluminium. When aluminium is exposed to air it forms a protective layer of aluminium oxide (Al₂O₃).

Question 98.
All ores are minerals, but all minerals need not be ores. Why?
Answer:
A mineral from which a metal can be extracted and economical loss is called ore.

Question 99.
Why is carbon not used for reducing aluminium from aluminium oxide?
Answer:
The oxide of Aluminium is very stable and can be reduced by electrolytic process.

Question 100.
Name few metals which occur in native state in nature. Why?
Answer:
Gold, Platinum, Silver are the metals which occur in native state, because of their low chemical reactivity.

Question 101.
Why do we call oxygen – sulphur group is chalcogen family?
Answer:
Chaleo means ore. We know that most of ores of many metals are oxides and sulphides. That’s why oxygen – sulphur group is called chalcogen family.

Question 102.
Aluminium occurs in combined state in nature whereas gold is in free state. Why?
Answer:
Gold has low reactivity and so occurs in free state. Aluminium is electropositive metal and high reactivity. So it is oxide or chloride.

Question 103.
What are the uses of thermite reaction?
Answer:
Thermite reaction is used to join railings of railway tracks or cracked machine parts.

10th Class Chemistry 13th Lesson Principles of Metallurgy 2 Marks Important Questions and Answers

Question 1.
Define mineral. Mention any two ores of ‘magnesium’. (AP June 2017)
Answer:
1) Minerals :
The elements or compounds of the metals which occur in nature in the earth crust are called ‘minerals’.

2) Two ores of magnesium :
Magnesite – MgCO₃
Carnalite – KCl MgCl₂ 6H₂O

Question 2.
Potassium, Sodium, Magnesium are high reactive metals and occur as chlorides in nature. Suggest and explain the suitable method for the extraction of the above metals from their ores. (AP March 2017)
Answer:

• The suitable method to extract these metals from their chlorides is electrolysis of their fused compounds.
• It is not feasible for method of reduction, electrolysis of their aqueous solutions.

Question 3.
Predict, what happens in the field of domestic use of metals if alloys were not discovered. (TS June 2016)
Answer:
If alloys were not discovered,

1. All the vessels/utensils made of single metal like iron, copper, aluminium, etc. may be used for cooking purpose.
2. We may face problems like rusting of iron, tarnishing of silver and copper, etc.
3. We may face the problems of corrosion of home appliances.
4. We may face difficulties in cleaning of the vessels due to rusting and tarnishing.
5. Cost of the utensils may be risen, because of less availability of the metals like copper.
6. Using of the plastic ware may be risen for storage due to lack of steel containers.
7. Brass, steel, bronze, etc. utensils are not available to use.
8. Making of jewellary is also difficult.

Question 4.
Give an example with the chemical equation for the reduction of ores using more reactive metals. (TS March 2017)
Answer:
The reaction of Iron oxide with aluminium.
Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + Heat
(Or)
Reaction of Titanium Chloride with Magnesium.
TiCl₄ + 2Mg → Ti + 2MgCl₂
(Or)
Reaction of Titanium Chloride with Sodium.
HCl₄ + 4Na → Ti + 4NaCl
(Or)
Reaction of Cromium oxide with aluminium.
Cr₂O₃ + 2Al → 2Cr + Al₂O₃ + Heat

Question 5.
Write two precautions to prevent corrosion of metals in your daily life. (TS June 2018)
Answer:
Precautions to be taken to prevent corrosion of metals.
i) Painting the metals.
ii) By keeping the metals in the dry places.
iii) Cover the surface by other metals that are inert or non reactive to the atmosphere.
iv) Applying oil/grease to the metals.
v) Making of alloys.

Question 6.

High reactivity	Moderate reactivity	Low reactivity
K, Na, Ca, Mg, Al	Zn, Fe, Pb, Cu	Ag, Au

Observe the table and answer the following questions. 4jt*y (June 2019
i) Which of the above metals found even in free state in nature ?
ii) Which of the above metal’s ore are concentrated by using magnetic separation?
Answer:
i) Ag, Au.
ii) Fe.

Question 7.
Silicon is a metalloid. How do you support this?
Answer:
Silicon exhibits following properties, so I conclude that it is a metalloid.

1. It is metallic lustre in nature.
2. It exists in several metallic and non-metallic compounds.
3. It has brittle nature.
4. All metalloids usually occur in combined states both metals and non-metals.

Question 8.
Explain the reaction of various metals in activity with cold water.
Answer:
1) From potassium to magnesium displace hydrogen from cold water with decreasing reactivity. Potassium reacts with cold water violently but reaction of Magnesium is very slow. The reactivity order is given below.
Mg < Ca < Na < K

2) From aluminium to gold do not displace hydrogen from cold water.

Question 9.
How do various metals in activity series react with steam?
Answer:

• The metals from potassium to iron displace H₂ (Hydrogen gas) from steam with decreasing reactivity. That means the reaction of potassium with steam is voilent but the reaction of iron is very slow.
• The metals from lead to gold do not displace hydrogen from steam.

Question 10.
How do various metals in activity series react with dilute strong acids?
Answer:
1) The metals from potassium to lead displace hydrogen from dilute strong acids with decreasing reactivity.
a) The reaction of potassium is explosive.
b) The reaction of magnesium is vigorous.
c) The reaction of iron is steady.
d) The reaction of lead is slow.

2) The metals from copper to gold do not displace H₂ from strong dilute acids.

Question 11.
What are the preventive techniques used in corrosion of metals?
Answer:
Prevention of corrosion of metals :

• Covering the surface of metal with paint or by some chemicals like bisphenol which prevent the surface of metallic object to come in contact with atmosphere.
• Covering the surface of metal by other metals like tin or zinc that are inert or react themselves with atmosphere to save the metal.
• An electrochemical method in which a sacrificial electrode of another metal like magnesium and zinc, etc. corrodes itself to save the metal.

Question 12.
What are the chemical reactions that take place inside blast furnace?
Answer:
The chemical reactions that take place inside the blast furnace.

Question 13.
What are the various types of furnaces? Explain.
Various types of furnaces :
1) Blast furnace:
Blast furnace has both fire box and hearth combined in big chamber which accommodates both ore and fuel.

2) Reverberatory furnace :
It has both fire box and hearth separated, but the vapours (flame) obtained due to burning of the fuel touch the ore in the hearth and heat it.

3) Retort furnace :
In this furnace there is no direct contact between the hearth or fire box and even the flames do not touch the ore.

Question 14.
Why is alloying preferred for metals? Explain with examples.
Answer:

• Alloying is a method of improving properties of a metal. We can get desired properties by this method.
• For example, iron is the most widely used metal. But it is never used in its pure state.
• This is because pure iron is very soft and stretches easily when hot.
• But, if it is mixed with a small amount of carbon, it becomes hard and strong.
• When iron is mixed with nickel and chromium we get stainless steel which will not rust.

Question 15.
What is 22 carat gold? Why is it preferred for making jewellery?
Answer:

• Pure gold, known as 24 carat gold is very soft.
• So it is not suitable for making jewellery.
• It is alloyed with either silver or copper to make it hard.
• So they use 22 carat gold in which pure gold is alloyed with 2 parts of either silver or copper for making gold jewellery.

Question 16.
Write about electrolysis of NaCl.
Answer:
1) Fused NaCl is electrolysed with steel cathode and graphite anode.

2) The metal sodium (Na) will be deposited at cathode and chlorine gas liberates at the anode.
At Cathode : 2 Na⁺ + 2e^– → 2 Na
At Anode : 2 Cl^– → Cl₂ + 2e^–

Question 17.
Identify the metal present in the following ores.
i) Epsom Salt
ii) Horn Silver
iii) Cinnabar
iv) Galena
Answer:
i) Magnesium
ii) Silver
iii) Mercury
iv) Lead

Question 18.
What is meant by extraction of metals? Write the main stages of extraction of metals from its Ore.
Answer:
Separation of metals from ores is called extraction of metals. Extraction of metals involves mainly three stages.

1. Concentration or dressing
2. Extraction of crude metal
3. Refining or purification of the metal.

Question 19.
Write differences between roasting and calcination.
Answer:

Roasting	Calcination
1. Ore is heated in the presence of oxygen or air.	1 Ore is heated in the absence of air.
2. Sulphide ore is converted into oxide ore.	2. Carbonate ore is converted into oxide ore.

Question 20.
What are the differences between minerals and ores?
Answer:

Minerals	Ores
1) Minerals contain a low percentage of metal.	1) Ores contain a large percent of metal.
2) Metals cannot be extracted from minerals.	2) Ores can be used for the extraction of metals.
3) All minerals cannot be called ores.	3) All ores are minerals.

Question 21.
What are the different types of reduction?
Answer:
The different types of reduction are
a) Chemical reduction,
b)Auto reduction,
c) Electrolytic reduction.

Question 22.
How do potassium and sodium react with oxygen?
Answer:
a) Potassium and sodium form oxides of type M₂O in limited supply of oxygen.
4 K + O₂ → 2 K₂O
4 Na + O₂ → 2 Na₂O

b) In excess of oxygen they form peroxides of type M₂O₂.
2 Na + O₂ → Na₂O₂
2 K + O₂ → K₂O₂

Question 23.
How does reactivity of chlorine vary in the reactivity series?
Answer:

1. All metals react with chlorine on heating to form their respective chlorides but with decreasing reactivity in the reactivity series.
2. This is understood from the heat evolved when metal reacts with one mole of chlorine gas to form chloride.

Question 24.
Name two ores of calcium and give their formulae.
Answer:
The ores of calcium are

1. Gypsum (CaSO₄ • 2H₂O)
2. Limestone (CaCO₃)

Question 25.
Which method is useful to separate sand from iron? Explain.
Answer:

• Sand can be separated from iron by using magnetic separation method.
• This can be done by using electromagnet. Iron being a magnetic material is attracted by electromagnet whereas sand is not attracted by electromagnet.
• So these materials are separated.

Question 26.
Which metals do not displace hydrogen from dilute strong acids?
Answer:

• Copper, mercury, silver, platinum, gold do not displace hydrogen from dilute strong acids like HCl, H₂SO₄, etc.
• The reactivity of these metals are less than hydrogen. So, they are unable to displace hydrogen from dilute acids.

Question 27.
Which metals are not found in free state? Why?
Answer:

• The metals like potassium, sodium, calcium, magnesium and aluminium are never found in free state in nature.
• The reason is that these metals have high reactivity. So, they exist as compounds.

Question 28.
Why do silver and gold exist even in free state?
Answer:

• Silver and gold are least reactivity metals. So, they are also called noble metals.
• Due to least reactivity they are unable to react with other elements.

Question 29.
How do moderately reactive metals occur in nature?
Answer:

• The metals like zinc, iron, lead are moderately reactive.
• They are found in the earth’s crust mainly as oxides, sulphides and carbonates.

Question 30.
Mention the most important metals and non-metals from the following products.
a) Annapurna salt
b) Liquid used in thermometer
c) Lead of the pencil
d) Chlorophyll
e) Filament in electric bulb
f) Enamel layer on teeth
Answer:
a) Annapurna salt : Iodine, chlorine – Non-metals
b) Liquid used in thermometer : Mercury – Metal
c) Lead of the pencil : Graphite – Non-metal
d) Chlorophyll : Magnesium – Metal
e) Filament in electric bulb : Tungsten – Metal
f) Enamel layer on teeth : Calcium phosphate – Non-metal

Question 31.
What is a furnace? Explain various parts of furnace.
Answer:
Furnace :
Furnace is the one which is used to cany out pyrochemical processes in metallurgy.

Furnace has mainly three parts :
1) Hearth :
Hearth is the place inside the furnace where the ore is kept for heating.

2) Chimney:
Chimney is the outlet through which flue (waste) gases go out of the furnace.

3) Fire box :
Fire box is the part of the furnace where the fuel is kept for burning.

10th Class Chemistry 13th Lesson Principles of Metallurgy 4 Marks Important Questions and Answers

Question 1.
What is a furnace? Draw Reverberatory furnace and label it parts. (AP March 2018)
Answer:
1) Furnace :
Furnace is the one which is used to carry out pyrochemical processes in metallurgy.
2) Diagram of Reverberatory furnace.

Question 2.
What are the various techniques used in purification of the crude metals? Explain.
(OR)
State the methods used for the purification of crude metals. Explain in which context these methods are used. (TS June 2015)
Answer:
1) The process of obtaining the pure metal from the impure metal is called refining of the metal.
2) Some of the processes of refining are
i) Distillation
ii) Poling
iii) Liquation
iv) Electrolytic refining.

3) The process that has to be adopted for purification of a given metal depends on the nature of the metal and its impurities.

Various methods adopted in purification of metals :
1) Distillation :
This method is very useful for purification of low boiling metals like zinc and mercury containing high boiling metals as impurities. The extracted metal in the molten state is distilled to obtain the pure metal as distillate.

2) Poling :
The molten metal is stirred with logs (poles) of greenwood. The impurities are removed either as gases or they get oxidized and form slag (Scum) over the surface of molten metal.

3) Liquation :
Low melting metal like tin can be made to flow on a slopey surface to separate it from high melting impurities.

4) Electrolytic refining :

1. In this method, the impure metal is made to act as anode.
2. A strip of the same metal in pure form is used as cathode.
3. They are put in a suitable electrolytic bath containing soluble salt of the same metal.
4. The required metal gets deposited on the cathode in the pure form.
5. The metal constituting impurity, goes as the anode mud.

The reactions are :
At anode : M → Mn⁺ + ne^–
At cathode : Mn⁺ + ne^– → M. ; (M = pure metal, n = 1, 2, 3, …….)

Question 3.
Four metals A, B, C and D are in turn added to the following solutions one by one. The observations made are tabulated below. (TS March 2015)

Answer the following based on the given information.
i) Which is the most reactive metal? Why?
ii) What would be observed, if ‘B’ is added to a solution of Copper (II) sulphate and why?
iii) Arrange the metals A, B, C and D in order of increasing reactivity.
iv) Which one among A, B, C and D metals can be used to make containers that can be used to store any of the above solutions safely?
Answer:
i) Metal ‘B’ is more reactive.
– Metal ‘B’ is replacing iron from iron sulphate.
– Metal ‘A’ is replacing copper from copper sulphate.
– Metal ‘C’ is replacing silver from silver nitrate.

ii) Metal B displaces Cu from CuS04 solution. Because metal B is more reactive than Cu.

iii) D < C < A < B.

iv) The container made up of metal D can be used to store any solution mentioned above.

Question 4.
Write the physical methods used for the concentration of the ore. Explain the method used for concentration of the sulphide ore. (TS June 2017)
Answer:
Physical methods used for the concentration of the ore is,
i) Hand Picking
ii) Washing
iii) Froth floatation
iv) Magnetic Separation.

Concentration of sulphide ore :

• Sulphide ore is concentrated by using froth floatation Method.
• The ore with impurities is tinely powdered and kept in water taken in a flotation cell.
• Air under pressure is blown to produce froth in water.
• Froth so produced, taken the ore particles to the surface whereas impurities settle at the buttom.
• Froth is separated and washed to get ore particles.

Question 5.
Draw a neat diagram of froth floatation process for the concentration of sulphide ore why we add pine oil to the mixture in this process? (AP SCERT 7201 9-20)
Answer:

Froth floatation process for the concentration of sulphide ores

1.  The mineral particles in the ore are preferentially wetted by the oil and float on the top of the froath.
2. The gangue particles are wetted by water and settle down.
3. Thus, the minerals can be separated from the gangue by adding pine oil.

Question 6.
Describe the reaction of various metals in activity series with oxygen.
Answer:

• The metals which are at the bottom of activity series have very low reactivity and do not burn or oxidase even on surface.
  Eg : Ag, Pt, Au.
• The metals like Pb, Cu and Hg do not burn but only form a surface layer of oxide, i.e., PbO, CuO, HgO.
• The metals like Al, Zn, Fe react with oxygen to form respective oxides.
• The metals like Ca and Mg burn with decreasing vigorousity to form oxides.
• The metals like K, Na burn vigorously to form Na₂O, K₂O in limited supply of oxygen but form peroxides in excess of oxygen.

Question 7.
How do you reduce purified ore to the metal of the top of activity series? Explain.
Answer:
The reduction of ore to particular metal mainly depends on the position of metal in the activity series.

Extraction of metals at the top of activity series :

1. Simple chemical reduction methods like heating C, CO, etc. to reduce the ores of the metals are not possible with metals like K, Na, Ca, Mg and Al.
2. The temperature required for the reduction is too high and more expensive.
3. The only method available is to extract these metals by electrolysis of their fused compounds.

Question 8.
How do you extract metals in the middle of activity series?
Answer:
Extraction of metals in the middle of the activity series :
The ores of these metals are generally present as sulphides or carbonates. Therefore prior to reduction of ores of these metals, they must be converted into metal oxides.

The metal oxides are then reduced to the corresponding metals by using the following methods :
1) Reduction of metal oxides with carbon :
The oxides are reduced by coke in closed furnace which gives the metal and carbon monoxide (CO).

2) Reduction of oxide ores with CO :

3) Auto (Self) reduction of sulphide ores:
In the extraction of copper from its sulphide ore, the ore is subjected to partial roasting in air to give its oxide.
2 Cu₂S + 3O₂ → 2 Cu₂O + 2SO₂

When the supply of air is stopped and temperature is raised, it results in the reaction of rest of the sulphide ore with oxide to form metal and S02. ‘
2 Cu₂O + Cu₂S → 6 Cu + SO₂

4) Reduction of ores (compounds) by more reactive metals :
When highly reactive metals such as sodium, calcium, aluminium, etc. are used as reducing agents, they displace metals of low reactivity from the compound.

Question 9.
How do you extract metals at the bottom of the activity series?
Answer:
1) Metals at the bottom of the activity series are often found in free state.

2) The oxides of these metals can be reduced to metals by heat alone and sometimes by displacement from their aqueous solutions.

3) When cinnabar (HgS) is heated in air, it is converted into HgO, then reduced to mercury on further heating.

4) Displacement from aqueous solution :
When Ag₂S is dissolved in KCN solution, it forms dicyanoargentate ions. When these ions are treated with Zn dust powder then Ag is precipitated.
Eg : Ag₂S + 4 CN^– → 2 [Ag(CN)₂]^– + S^2-
2 [Ag(CN)₂]^–_(aq) + Zn_(s) → [Zn(CN)₄]^2-_(aq) + 2 Ag_(s)

Question 10.
Explain the process involved in corrosion.
Answer:
1) Corrosion is an electrochemical phenomenon.
2) In corrosion, a metal is oxidised by loss of electrons generally to oxygen and results in the formation of oxides.
3) During corrosion at a particular spot on the surface of an object made of iron, oxidation takes place and that spot behaves as anode.
Anode : 2 Fe_(s) → 2 Fe²⁺ + 4e^–
4) Electrons released at this anodic spot move through the metal and go to another spot and reduce oxygen at that spot in the presence of H+.

5) This spot behaves as cathode.
Cathode : O_2(g) + 4 H⁺_(aq) + 4e^– → 2H₂O_(l)
Net reaction : 2 Fe_(s) + O_2(g) + 4 H⁺_(aq) → 2 Fe²⁺_(aq) + 2 H₂O_(l)

6) The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe₂O₃ . XH₂O) and with further production of hydrogen ions.

Question 11.
Explain electrolytic refining with an example.
Answer:

• The impure metal is taken as anode and pure metal is taken as cathode.
• They are put in a suitable electrolytic bath containing soluble salts of same metal.
• The required metal gets deposited on the cathode in the pure form.
• The metal constituting the impurity goes as the anode mud.

Electrode reactions :
At Anode : M → Mⁿ⁺ + ne^–
At Cathode : Mⁿ⁺ + ne^– → M (M pure metal); where n = 1, 2, 3,………….

Examples :

1. In order to refine copper, impure copper is taken as anode and pure copper strips are taken as cathode.
2. The electrolyte is an acidified solution of copper sulphate.
3. As a result of electrolysis copper in pure form is transferred from anode to cathode.
   At Anode : Cu → Cu²⁺ + 2e^–
   At Cathode : Cu²⁺ + 2e^– → Cu
4. The soluble impurities go into the solution, whereas insoluble impurities from the blister copper deposited at the bottom of anode as anode mud.

Question 12.
What is activity series? Give two examples each of them.
i) Low reactivity metals
ii) Moderate reactivity metals
iii) High reactivity metals
Answer:
Arranging metals in descending order of their reactivity is called activity series,
e.g.:
i) Low reactivity : Ag (Silver), Au (Gold)
ii) Moderate reactivity : Zn (Zinc), Fe (Iron)
iii) High reactivity : K (Potassium), Na (sodium)

Question 13.
Write the balanced chemical equations, extraction of iron from haematite in the Blast furnace.
Answer:

Question 14.
How do you prevent corrosion of various metals?
Answer:
Prevention of corrosion :The corrosion can be prev ented by the following methods.
1) Barrier protection :
In this method the metal surface is not allowed to come in contact with moisture, oxygen and carbon dioxide. This can be achieved by the following.
a) By coating iron with oils, paints, coal tar, grease, pitch, etc.
b) By blowing steam over red hot iron to form protective coating of Fe₃O₄.
c) By alloying iron with Ni, Cr, Si, etc.

2) Sacrificial protection:
Sacrificial protection means covering the iron surface with a layer of metal which is more electropositive than iron thus prevents iron from losing electrons. It is done by following methods.
a) By galvanisation (by dipping iron in a bath of molten zinc).
b) By tinning (by dipping iron in molten tin).
c) By the coating of copper.
d) Decorative coating : By using Zn, Mg and A/ powders mixed with paints.

3) Electrical protection:
In this method, the iron object to be protected from corrosion is connected to more active metal eg. magnesium, zinc or aluminium directly or through a wire. The iron object acts as cathode and the protective metal acts as anode. The anode is gradually used up to the oxidation of metal to its ions due to loss of electrons. Hydrogen ions collect at cathode and prevent rust formation.

4) Using anti-rust solution:
On applying alkyl phosphates and alkyl chromates to the iron objects corrosion can be prevented.

Question 15.
What are the salient features of the activity series?
Answer:
Salient features:

1. Any metal which is placed higher up in the series can displace any metal below it in order to from the salt solution of the later metal.
2. The larger the difference in the position of metals in the series, the more rapidly does the displacement take place.
3. Metals which are placed above hydrogen in the series have the ability to reduce ions from dilute sulphuric acid to liberate hydrogen gas.
4. Oxides of metals K, Na, Ca and Mg cannot be reduced by H2, CO or C.
5. Oxides and nitrates of less reactive metals Hg, Ag and Au decompose to give metals on being strongly heated.
6. Metals below copper such as mercury, silver, platinum and gold do not rust easily.
7. Hydrogen though a non-metal, has been included in the series.
   It occupies the position based on its formation of positive ions.

Question 16.
How do you extract metals based on activity series?
Answer:

• Highly active metals like potassium, sodium, calcium, magnesium and aluminium are obtained by the electrolysis of their fused halides or oxides, that is, by electrolytic reduction because their oxides cannot be reduced by common reducing agents like carbon, carbon monoxide and hydrogen.
• Zinc is obtained only by heating its oxide with carbon.
• Iron, lead and copper are obtained by reduction of their oxides with carbon, carbon monoxide and hydrogen.
• Copper is obtained by reducing black copper oxide with carbon or by air reduction.
• Mercury and silver are obtained by heating their respective oxides to temperature above 300°C when they lose oxygen and are reduced to free metals.
• However, less active mercury can also be obtained by merely heating its sulphide in air.
• Silver and Gold are obtained by displacement from solutions containing their ions by more electropositive metal zinc.

Question 17.
X is an element in the form of a powder. X burns in oxygen and the product is soluble in water. The solution is tested with litmus.
Write down the answers for the following questions from the above information and give reasons.
i) If X is a metal, then which colour will litmus turn ?
ii) If X is a non-metal, then which colour will litmus turn ?
iii) If X is a reactive metal, what gas will be released with dilute sulphuric acid ?
Answer:
i) If X is a metal, then the litmus turns into blue because metal reacts with oxygen and forms metallic oxide and aqueous solution of metallic oxide ore basic in nature.

ii) Mf X is a non-metal, then the litmus turns into red because non-metal reacts with oxygen and forms non-metallic oxide and aqueous solution of non-metallic oxide ore acidic in nature.

iii) If X is a reactive metal, then it releases hydrogen gas from sulphuric acid because more reactive metal displaces hydrogen from acid.

Question 18.
Complete the missing statements and give reasons.
i) Metals are ……………………….., while non-metals are poor conductors of heat.
ii) Metals are malleable, while non-metals are ……………………….. .
iii) Metals form positive ions, while non-metals form ……………………….. .
iv) Non-metals form acidic oxides, while metals form ……………………….. .
Answer:
i) Good conductors.
Reason :
Metals containing free electrons are very good conductors of electricity whereas non-metals are bad conductors of electricity because they do not have free electrons.

ii) Non-malleable.
Reason :
Metals are hard. So, they can be beaten into sheets whereas non-metals are soft, so they are non-malleable.

iii) Negative ions.
Reason :
Metals are electropositive in nature. They easily lose electrons to form positive ions, whereas non-metals are electronegative in nature. So, they gain electrons to form negative ions.

iv) Basic oxides.
Reason :
Non-metallic oxide solutions turn blue litmus into red. They are acidic in nature. So, they are called acidic oxides whereas metallic oxide solutions turn red litmus into blue. They are basic in nature. So, they are called basic oxides.

Question 19.
Answer the following questions.
a) i) Name two naturally occurring compounds of zinc other than carbonate and give their formulae.
ii) Give equations for the extraction of zinc from zinc carbonate.
b) Write equations for the action of zinc on the following.
i) dil. H₂SO₄
ii) Copper (II) sulphate solution.
Answer:
a) i) The ores of zinc other than carbonate ore are zinc blende (ZnS) and Zincite (ZnO).

b) i) Zn + dil. H₂SO₄ → ZnSO₄ + H₂
Zr_(s) + CuSO_4(aq) → ZnSO_4(aq) + Cu_(s)

Question 20.
i) The ore zinc blende is an important source of the metal zinc. What is the name of zinc compound in zinc blende?
ii) What is the compound obtained by roasting zinc blende?
iii) What is the type of chemical reaction carried out after roasting in order to obtain zinc?
iv) What is the name of the alloy formed between zinc and copper?
Answer:
i) The zinc compound in zinc blende is ZnS (zinc sulphide).
ii) By roasting zinc blende it converts into zinc oxide.
2 ZnS + 3O₂ → 2 ZnO + 2 SO₂
iii) The chemical reaction carried out to convert zinc oxide to zinc metal is reduction in the presence of coke.

iv) The alloy of copper and zinc is bronze.

Question 21.
The basic material used for the production of iron in the blast furnace are limestone, coke and air in addition to iron ore.
a) Name one iron ore and write its formula.
b) Hot air is blown at the base of furnace where it reacts with coke. Give the chemical equations for the reactions that take place.
c) Higher up in the furnace the iron ore is reduced to iron by one of the gases produced in the furnace. Give the chemical equation for the reaction by which the gas is produced and give a balanced equation to show how the ore is reduced to iron.
d) Which compound produced from limestone takes part in forming the slag?
Answer:
a) The iron ore is Haematite (Fe₂O₃).
b) Coke bums partially to produce carbon monoxide gas.
2 C_(s) +O_2(g) → 2 C0_(g)

c) Iron oxide reacts with carbon monoxide gas and reduces to iron.
Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂

d) Calcium carbonate (limestone) undergoes calcination to produce calcium oxide which takes part in the reaction to form slag.
CaCO_3(s) → CaO_(s) + CO₂
CaC_(s) + SiO_2(s) → CaSiO_3(l)

Question 22.
What information do you get from metal activity series given below.
K > Na > Ca > Mg > Al > Zn > Fe > Pb > [H] > Cu > Ag > Pt > Au
Answer:

• Metals below hydrogen [H] cannot displace hydrogen from acids and above hydrogen can displace hydrogen from acids.
• Metals which are higher in the series, can displace metals below it from the salt solution.
• The higher the position, the more active is the metal.
• Hydrogen has electropositive character, so it is placed among the metals.

Question 23.
The results of reactions of metals A, B, C, D, E with different solutions are given in the table below. Observe the table and write answers.

1) Which is the highly reactive metal? Why?
2) Which is the least reactive metal? Why?
3) Which metals form brown layer?
4) Arrange the metals A, B, C, D, E in the order of their reactivity?
5) Among these identify the silver, copper, iron, zinc and aluminium.
Answer:

1. Metal ‘E’ is more reactive among all the metals given because it displaces all the elements from the compounds given in the table,
2. Metal ‘C’ is the least reactive metal because it does not displace any other metal from the compounds given in the table.
3. Metals B and E will form brown layer.
4. The ascending order is as follows C < A < D < B < E.
5. C is silver, A is copper, D is iron, B is zinc and E is aluminium.

Question 24.
Draw the diagram of blast furnace and label its parts.
Answer:

AP State Board Syllabus AP SSC 10th Class Biology Solutions Chapter 8 Heredity Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Biology Solutions 8th Lesson Heredity

10th Class Biology 8th Lesson Heredity Textbook Questions and Answers

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Question 1.
What are variations? How do they help organisms?
Answer:

1. Differences in characters within very closely related groups of organisms are referred to as variations.
2. Variations develop during reproduction in organisms.
3. Variations are passed from parent to offspring through heredity.
4. Beneficial variations are selected by the nature in evolution.
5. Variations increase the survival chance of the organisms.
6. These variations help the organisms to adapt to their environments.
7. For example, green colour in the beetles is a variation that gave a survival advantage to the beetles as they cannot be seen by the crows.
8. Some variations do not help organisms to survive. For example, colour variation occurs in red beetles and some blue beetles are produced instead of red beetles as they are eaten by crows easily.

Question 2.
One student (researcher) wants to cross pure tall plant (TT) with pure dwarf (tt) plant, what would be the Fj and F2 generations? Explain.
Answer:

1. Pure tall plant has both the factors of the same type ‘TT’.
2. Pure dwarf plant has both the factors of the same type ‘tt’.
3. When a pure tall plant (TT) is crossed with pure dwarf plant (tt), all the offsprings in Fj generation are tall (Tt).
4. So all the plants are heterozygous tall, as ‘T’ is the dominating factor.
   
5. On self pollination of these F₁ generation plants the new breed can have any combination of T and t like TT, Tt, Tt or tt.
   
6. So in F₂ generation 75% of plants are tall and 25% of plants are dwarf. Thus the phenotype ratio is 3 : 1.
7. Among 75% of tall plants 25% are pure tall (TT) or homozygous tall, remaining 50% are heterozygous tall (Tt, tT).
8. The remaining 25% dwarf plants are pure or homozygous dwarf (tt).
9. So the genotype ratio is 1 : 2 : 1.

Question 3.
One experimenter cut the tails of parent rats, what could be the traits in offsprings? Do the daughter rats contain tails or not? Explain your argument.
Answer:

1. If the tails of parent rats were cut, their offsprings will have normal tails.
2. Daughter rats do not contain tails because the bodily changes are not inherited.
3. So the change would not be passed to their offsprings.
4. This was proved experimentally by Augustus Weisemann and rejected the theory ‘inheritance of acquired characters’ proposed by Lamarck.

Question 4.
In a mango garden a farmer saw one mango tree with full of mango fruits but with a lot of pests. He also saw another mango tree without pests but with few mangoes. But the farmer wants the mango tree with full of mango fruits and pests free. Is it possible to create new mango tree which the farmer wants? Can you explain how it is possible?
Answer:

1. Yes, it is possible to create new mango tree which one the former wants with full of mango fruits and pests free.
2. The former can cross two plants one with full of mangoes and pests and another plant with less mango fruits and without pests,
3. In F₁ generation he may get plants with full of mango fruits and without pests. Such plants are called hybrid plants.
4. The F₁ generation plants can be self pollinated and desired plants can be selected from the mixed population of F₂ generation.
5. The plant with desirable characters can be vegetatively propagated to get required number of plants.

Question 5.
ExplaIn monohybrid experiment with an example. Which law of inheritance can we understand? Explain.
Answer:

1. We can understand the law of inheritance with an example of monohybrid cross between pure yellow pea seeds with pure green pea seeds.
2. A pure breed (parental) yellow will have both the factors which denote them by ‘YY’ and pure breed (parental) green seed will have both the factors denote them by ‘yy’.
3. During reproduction one factor (genes) from each parent is taken to form a new pair in the progeny (off spring).
4. In F₁ generation all pea plants are Yellow.
   
5. F₁ generation pea plants are self pollinated.
6. In F₂ generation 75% of the plants produced were Yellow seeds and the remaining 25% produced were green seed. It can be represented as
   
   In F₂ generation the produced plants are YY, Yy; yY or yy.
   From the above example of monohybrid cross we can understand the following laws of inheritance.

1) When pure breed Yellow (YY) and green (yy) seeds were crossed, only Yellow seeds were expressed phenotypically in the F₁ generation. It indicates that Yellow seed character is dominent over green seed characters.
This is “LAW OF DOMINANCE”.

2) When F₁ plants are self pollinated each parent passes randomly selected allele (Y or y) of one of these factor to offsprings. This is segregation of alleles or genes during production of gametes.
This is “LAW OF SEGREGATION”.

Question 6.
What is the law of independent assortment? Explain with an example.
Answer:

1. In the inheritance of more than one pair of characters (traits), the factors for each pair of characters assorts independently of the other pairs. This is known as “Law of independent assortment”.
2. If pea plants with two different pairs of characteristics (eg. : Round / yellow and green wrinkled) are breed with each other, the F₁ progeny plants would have all round and yellow seeds.
3. This implies that round and yellow seeds are dominant characters over green and wrinkled seeds.
4. In F₂ progeny there would be some plants with round and yellow seeds and some with green and wrinkled seeds.
5. However, there would be some plants with mixed characters – yellow and wrinkled seeds and green and round seeds.
6. This depicts that round/wrinkled trait and yellow / green trait are inherited independent of each other (law of independent assortment).
   The following punnet square explains this.
   
7. The different combination of characters resulted from dihybrid cross.
   a) RRYY, RRYy, RrYy, RrYY, RRYy, RrYy, RrYy, RrYY, RrYy are having round and yellow seeds.
   b) RRyy, Rryy, Rryy have round and green seeds.
   c) rrYy, rrYy, rrYY have wrinkled and yellow seeds.
   d) rryy have wrinkled and green seeds.
8. From the result, it can be concluded that the factors for each character or trait remains separate and maintain its identity in the gametes. This is known as “Law of independent assortment”.

Question 7.
How does sex determination take place in human?
(OR)
Explain sex determination in humans with the help of flow chart.
Answer:

1. Each human cell contains 23 pairs (46) of chromosomes.
2. Out of 23 pairs, 22 pairs of chromosomes are called autosomes.
3. Remaining one pair is called allosomes or sex chromosomes.
4. There are two types of sex chromosomes – one is ‘X’ and the other is ‘Y’.
5. These two chromosomes determine the sex of an individual.
6. Females have two ‘X’ chromosomes in their cells (XX).
7. Males have one ‘X’ and one ‘Y’ chromosomes in their cells (XY).
   
8. All the gametes produced by women (ovum) will be with only X chromosomes.
9. The gametes produced by man (sperm) will be of two types, one with X chromosomes and other with Y chromosomes.
10. If the sperm carries X chromosome and fertilizes with the ovum, the resultant baby will have XX condition. So the baby will be a girl.
11. If the sperm carries Y chromosome and fertilises with the ovum, the resultant baby will have XY condition. So the baby will be a boy.

Question 8.
Explain Darwin’s theory of ‘Natural selection’ with an example.
(OR)
What do you understand by the term Natural selection? Write Darwin’s theory of evolution.
Answer:

1. Darwin proposed the theory of Natural selection.
2. Nature only selects or decides which organism should survive or perish in nature.
3. The organism with useful traits will survive and the organisms having harmful traits are going to perished or eliminated from its environment.
4. For example, a group of twelve red beetles live in a bush of green leaves.
5. They will grow their population by sexual reproduction.
6. So they generate variations in their population. Let us assume crows eat the red beetles more the population of red beetles slowly reduced.
7. Crows eat these red beetles and their population slowly reduces.
8. During this time a colour variation arises by the sexual reproduction.
9. So that there appears one beetle that is green in colour instead of red.
10. Moreover this green colour beetle passes its colour to its offsprings; so that all its progeny are green.
11. Crows cannot see the green coloured beetles on green leaves of the bushes and therefore crows cannot eat them.
12. The crows can see the red beetles and eat them as a result, there are more and more green beetles than red ones which decrease in their number.
13. The variation of green colour beetle gave a survival advantage to green beetles’ than red beetles. They were naturally selected.

Question 9.
What are variations? Explain with a suitable example.
Answer:

1. Differences in characters within very closely related groups of organisms are referred to as variations.
2. Often a new character in a group may lead to variations that are inherited.
3. If we observe parents and offsprings, there will be some similar features in the offspring of the parents.
4. At the same time we find differences between parents and offspring in their features.
5. These differences are an example of variations.
6. Variations are quite apparent among closely related groups of organisms.
7. If we take roses as an another example, we observe number of varieties in them.
8. But we can still find some characters similar to all plants.
9. Thus rose plants have similar physical features, at the same time they have differences in characters like flower colour, number of petals, leaf size, stem, spines, etc.
10. These differences in features are variations.

Question 10.
What variations generally have you observed in the species of cow?
Answer:
In the species of cow the following contrasting variations can be observed:

1. White coloured – spotted
2. Longhorns – short horns
3. Height – dwarf
4. Long-tail – short tail
5. Elongated face – stunted face
6. More milk giving – less milk giving, etc.

Question 11.
What are the characters that Mendel selected for his experiments on pea plants
(OR)
Write the seven pairs of contrasting characters in pea plant identified by Mendel and mention their traits.
Answer:
Mendel selected the following characters on pea plants for his experiment. They are:

Character	Description
1. Colour of the flower	1. Purple or white.
2. Position of the flower	2. Axial or terminal in position
3. Colour of the seed	3. Either yellow or green.
4. Shape of the seed	4. Either round or wrinkled.
5. Shape of the pod	5. Inflated and constricted
6. Colour of the pod	6. Yellow or green.
7. Length of the stem	7. Tall and dwarf.

Question 12.
In what way Mendel used the word ‘Traits’? Explain with an example.
Answer:

1. Trait is a separate variant of an organism.
2. Mendel hypothesized that characters were carried as traits.
3. An organism always carried a pair of factors for a character.
4. He also hypothesized that distinguishing traits of the same character were present in the population of an organism.
5. He assumed that the traits shown by the pea plants must be in the seeds that produce them.
6. The seeds must have obtained these traits from the parent plants.
7. The factors which are responsible for the character or trait of an organism, are now named as “genes”.
8. By all these we can assume that Mendel used the word ‘traits’ for indicating the variant of an organism expressed by a pair of factors or genes.
9. For example, height is a character of pea plant while the tallness is a trait expressed by a pair of factors either TT or Tt and dwarfness is another trait expressed by a pair of factors tt.

Question 13.
What are the differences that Mendel observed between parent and F₂ generation?
Answer:
Mendel identified the following differences between parent and F₂ generation.

Parent	F2 Generation
1. They are pure breeds.	1. They consist of mixed population.
2. They consist of homozygous alleles.	2. They consist of homozygous alleles in some plants and heterozygous alleles in some other plants.
3. They have some fixed characteristic features.	3. New combination of characters will appear.

Question 14.
Male is responsible for sex determination of baby – do you agree? If so write your answer with a flow chart.
Answer:

1. Yes, I agree with the statement that male is responsible for sex determination of baby.
2. There are two types of sex chromosomes in human beings, one is ‘X’ and other is ‘Y’.
   
3. Females have two ‘X’ chromosomes in their cells (XX) whereas males have one ‘X’ and one ‘Y’ chromosomes in their cells (XY).
4. All the gametes produced by woman (ovum) will be with only X chromosomes.
5. The gametes produced by man (sperm) will be of two types one with X chromosomes and other Y chromosomes.
6. If the sperm carrying X chromosome fertilizes the ovum, the resultant baby will have XX condition. So the baby will be a girl.
7. If the sperm carrying Y chromosome fertilizes the ovum, the resultant baby will have XY condition. So the baby will be a boy.
8. So the gamete produced by the male is the deciding factor for sex determination of the baby.

Question 15.
Write a brief note on analogous organs.
(OR)
What are analogous organs?
Answer:

1. The organs which are structurally different but functionally similar are known as ‘Analogous organs’.
2. Wings of birds and bats is the example for analogous organs.
3. The wings of bats are skin folds stretched mainly between elongated fingers.
4. But the wings of birds are a feathery covering all along the arm.
5. The designs of the two wings, their structure and components are different.
6. They look similar because they have common use for flying, but their origins are not common.
7. This makes the ‘analogous’ characteristics.
8. This type of evolution is called convergent evolution.

Question 16.
How do scientists utilise information about fossils?
(OR)
“Fossils are valuable material that nature had preserved to know about ancient organisms.” Write the information you have collected about fossils.
Answer:

1. Fossils are evidence of ancient life forms or ancient habitats which have been preserved by natural processes.
2. The scientific study of fossils is called ‘Palaeontology’.
3. Scientists utilise information about fossils to understand the evolutionary history of life.
4. This information is also useful to study ecology and environmental history, such as ancient climates.
5. This also helps to find out how old that certain layer of earth is.
6. This information is also utilized as indicators of possible fossil fuel deposits which are of great interest to humanity.
7. Thus scientists utilize the information on fossils to learn more about the earth’s past.

Question 17.
Mendel selected a pea plant for his experiments. Mention the reasons for the selection as these plants.
(OR)
Why did Mendel select the pea plant for his experiment? (OR)
Which characters in the pea plant are selected by Mendel, for his experiments?
What are the reasons for selecting pea plant by Mendel to conduct his experiments?
Answer:
Mendel chose the pea plant (Pisum sativam) for his breeding experiments for the following reasons.

1. It is sexually reproducing.
2. Flowers are bisexual.
3. Predominantly self-pollinated.
4. Predominantly self-fertilization.
5. Well developed characters.
6. Early hybridization.
7. It is an annual plant.
8. These plants have short maturity and can produce large number of seeds in a single generation.
9. Pea plants have short life cycle.
10. These plants can easy to grow either on the ground or in pots.

Question 18.
If the theory of inheritance of acquired characters proposed by Lamarck was true, how will the world be?
Answer:
If the theory of inheritance of acquired characters proposed by Lamarck was correct,

1. All the organisms which lost some of their body parts should give birth to the offsprings without the lost parts.
2. Rat which lost their tail should give birth to tail less rats.
3. A handicapped who lost their legs in an accident should give birth to babies v without legs.
4. A body builder’s children should be body builders.
5. But all these are not happening because bodily changes won’t be passed to its offspring.

Question 19.
Collect information on the inherited traits in your family members and write a note on it.
Answer:

1. My grandfather and father- had curling hair. I too have curling hair. So it’s an inherited trait in family.
2. My mother and I both have long noses which appear similar. It’s another inherited trait.
3. Eyes of my grandmother, my brother and mine are similar. It’s another inherited trait.
4. Ear lobes of my father, brother and mine are similar. This is another inherited trait.

Question 20.
With the help of given information write your comment on evidences of evolution.

Mammals have fore limbs as do birds, reptiles and amphibians. The basic structure of the limbs is similar, though it has been modified to perform different functions.

Answer:

1. The given information gives the evidences of evolution.
2. Mammals. birds, reptiles and amphibians all these have forelimbs which have similar basic structure.
3. But they are modified to perform different functions.
4. This indicates that all the vertebrates have evolved from a common ancestor. These organs are called homologous organs. This type of evolution is called divergent evolution.
5. In case of bat (mammal) and bird the designs of the two wings, their structure and components are different.
6. They look similar because they have common use for flying, but their origins are not common.
7. These organs which are structurally different but functionally similar are known as ‘Analogous organs’. This type of evolution is called “convergent evolution”.
8. There are remarkable similarities in the embryos of above mentioned animals even in their limb formation. These are called embryological evidences.

Question 21.
Collect information about carbon dating method. Discuss with your physical science teacher.
(OR)
Write about the carbon dating method from the information collected by you.
Answer:

1. Carbon dating is the method used to calculate the age of rocks, minerals or fossils.
2. The breakdown of radioactive isotopes of certain elements such as carbon, uranium and potassium takes place at a known rate. So the age of rock or mineral containing isotopes can be calculated.
3. Archaeologists use the exponential, radioactive decay of carbon 14 to estimate the death dates of organic material.
4. The earth’s atmosphere contains various isotopes of carbon, roughly in constant proportions.
5. These include the main stable isotope ¹²C and an unstable istope ¹⁴C.
6. Through photosynthesis, plants absorb both forms from carbon dioxide in the atmosphere.
7. When an organism dies, it contains the standard ratio of ¹⁴C to ¹²C.
8. But as the ¹⁴C decays with no possibility of replenishment, the proportion of carbon 14 decreases at a known constant rate.
9. The time taken for it to reduce by half is known as the half-life of ¹⁴C, which is 5730.
10. The measurement of the remaining proportion of ¹⁴C in organic matter thus an estimation of its age.
11. As the half life of carbon – 14 is 5,700 years, it is useful for dating objects up (o about 60,000 years old.

Question 22.
Draw a checker board, show the law of independent assortment with a flowchart and explain the ratio.
Answer:

1. The phenotypic ratio is 9 : 3 : 3 : 1. i.e., 9 round and yellow seeds 3 round and green seeds, 3 wrinkled and green seeds and 1 wrinkled and green seed.
2. RRYY, RRYy, RrYy, RrYy, RRYy, RrYy, RrYy, RrYY and RrYy are round and yellow seeds.
3. RRyy, Rryy, Rryy are round and green.
4. rrYY, rrYy, rrYy are wrinkled and yellow.
5. rryy are wrinkled and green.

From the above result, it can be concluded that factors for each character or trait remains separate and maintains its identity in the gametes. Thus in the inheritance of more than one pair of characters, the factors for each pair of characters assort independently of the other pairs. This is known as “Law of independent assortment”.

Question 23.
Explain the process to understand the monohybrid cross of Mendel experiment with a checker board.
Answer:

Question 24.
Prepare a chart showing the evolution of man through ages.
Answer:

Question 25.
Nature selects only desirable characters. Prepare a cartoon.
Answer:
Nature selects only desirable characters

Question 26.
What is your understanding about survival of the fittest? Give some situations or examples that you observe in your surroundings.
Answer:

1. Nature favours only useful variations.
2. Each species tends to produce a large number of offspring.
3. They compete with each other for food, space, mating and other species.
4. In this struggle for existence only the fittest can survive.
5. When cat tries to catch some rats, the rats which can run fast and hide in its hole can survive and which is slow can become prey for the cat.
6. When we spray some insecticide on insects, most of them will die but few which can withstand that chemical will escape.
7. When a pest attacks our garden plants, most of them may die but Which can withstand the pest can survive.
8. When the dog tries to catch chickens, the chickens which will run fast and escape can survive but the slower ones will become food for the dog.

Question 27.
Write a monologue on the evolution of a human to perform a stage show on the theatre day in your school.
Answer:

1. Hai, I am a human being. I am going to recall what had happened to me so far, how I had evolved, simply my journey from my origin to till now.
2. Nearly 1.6 – 2.5 million years ago, during the gelasian pleistocene period, I used to wander in the forest. It is belived that, I evolved from apes.
3. Between 1-1.8 million years ago, I gradually evolved into Homo erectus. I lived in this stage throughout most of the pleistocene. I used more diverse and sophisticated stone tools than my predecessors and it is belived that I travelled over oceans using rafts.
4. Around 1,00,000 – 40,000 thousand years ago I evolved into Homo sapiens neanderthalensis. I was stronger than present in those days. I made advanced tools. I had language to communicate.
5. Around 40 thousand years ago, I reached the present form of human being, the modern humans known as Homo sapiens. I learnt cultivation, construction of houses, cooking, etc. I had invented various things that help me to live comfortably.
6. But my journey did not stop. It is still continuing. Let us see what may happen? Where can I reach? What changes may come in me? Hope for the best.
   Thank you.

Fill in the blanks.

1. The process of acquiring change is called ———–.
2. Mendel’s experiment explains about ———–.
3. The four characters observed in the experiments on law of independent assortment are ———–.
4. If we cross pollinate red flower plant with white flower we will get percent of ———– recessive trait plants.
5. TT or YY, Tt or Yy are responsible for a ———– character.
6. Female baby having 23 pairs of autosomes at the age of 18 years, has ———– pair autosomes and ———– of sex chromosomes.
7. The population grows in ———– progression whereas food sources grow in ———– progression.
8. A goat which walks properly can’t live for a long time. According to Darwin, this represents ———–.
9. Forelimb of whale is for swimming whereas in horse it is used for ———–.
10. The study of fossils is called ———–.

Answer:

1. evolution
2. heredity
3. Round, wrinkled, yellow, green
4. 100
5. dominant
6. 22, one pair
7. geometrical, arithmetic
8. survival of the fittest
9. running
10. palaeontology

Choose the correct answer.

1. Which of the following is not a variation in rose plant?  [ ]
   A) Coloured petals
   B) Spines
   C) Tendrils
   D) Leaf margin
   Answer: C
2. According to Mendel, alleles are  [ ]
   A) Pair of genes , Responsible for character
   B) gene
   C) Production of Gametes
   D) Recessive factors
   Answer: B
3. Natural selection means  [ ]
   A) Nature selects desirable characters
   B) Nature rejects undesirable characters
   C) Nature reacts with an organism
   D) A, B
   Answer: A
4. Palaeontologists deal with  [ ]
   A) Fossilised Embryological evidences
   B) Fossil evidences
   C) Fossilised Vestigial organ evidences
   D) All
   Answer: D

10th Class Biology 8th Lesson Heredity InText Questions and Answers

10th Class Biology Textbook Page No. 166

Question 1.
How does evolution take place?
Answer:

1. Evolution takes place through the accumulation of new characters or variations in a species of organisms.
2. Accumulation of variations occurs only when new characters are passed on from one generation to other and much more new characters are added to the pre-existing once.
3. So this happens oVer a kirig period of time, sometimes several generations may pass.
4. Hence it happens in a slow and steady manner.
5. It is not just about change but producing something new and different.
6. It is about the formation of new species and their adaptation to their environments.

10th Class Biology Textbook Page No. 168

Question 2.
Is variation all about apparent differences? Or is it about some subtle differences as well that we most often overlook?
Answer:

1. Variations are not always apparent differences.
2. Sometimes these may be subtle differences that we most often overlook.
3. When these subtle differences accumulate together they may become apparent.

10th Class Biology Textbook Page No. 171

Question 3.
How do parent plants pass on their traits to the seeds?
Answer:

1. Every character or trait is controlled by a pair of factors called genes.
2. At the time of sexual reproduction, one factor or each trait will pass to the gametes.
3. By the fussion of male and female gametes zygote will form in which factors from both male and female parents get paired again.
4. This zygote will develop into seed in the later stages.
5. Thus parent plants pass on their traits to the seeds.

Question 4.
Will the seeds from tall plants always produce new tall plants?
Answer:

1. No. Tall plants may or may not produce tall plants again.
2. This is because tallness is a dominant character in most of the plants, especially in peas.
3. So tall plant may be homozygous tall (TT) or heterozygous tall (Tt).
4. If the parental plant is homozygous tall (pure breed), then they always produce new tall plants.
5. If the parental plant is a heterozygous tall plant, then they produce the tall and dwarf plants in the ratio of 3 : 1.

10th Class Biology Textbook Page No. 175

Question 5.
What should be the percentage of each type of plants in F₂ generation produced in dihybrid cross between pea plants with yellow, smooth seeds and green wrinkled seeds?
Answer:

1. In F₂ generation of dihybrid cross between pea plants with yellow, smooth seeds and green wrinkled seeds, new plants will produce with the following combination.
   i) Round and yellow
   ii) Round and green and
   iii) Wrinkled and green iv) Wrinkled and yellow
2. They will produce in the ratio of 9 : 3 : 3 : 1 respectively.
3. So 56 (56.25%) of plants should be with round and yellow seeds. 19 (18.75%) of plants should be with wrinkled and yellow seeds. 19 (18.75 %) of plants should be with round and green seeds and 6% (6. 25%) of plants should be with wrinkled and green seeds.

10th Class Biology Textbook Page No. 178

Question 6.
What will happen if the sperm containing X chromosomes fertilizes the ovum?
Answer:

1. If the sperm containing X chromosomes fertilizes the ovum which has X chromosome, the baby will have XX condition.
2. So the baby will be a girl.

Question 7.
Who decides the sex of the baby – mother or father?
Answer:
Father decides the sex of the baby.

Question 8.
Is the sex also a character or trait? Does It follow Mendels’ law of dominance?
Answer:

1. Yes, sex is also a character or a trait.
2. It has two contrasting characters male and female.
3. Male character is represented by a pair of allosomes ‘XY‘ (heterozygous).
4. In this, we can consider Y as dominant and X as recessive.
5. In this, recessive character is expressed only when it is homozygous recessive, i.e. female.
6. Homozygous dominant is not existing as reproduction occurs between male (heterozygous dominent XY) and female (homozygous recessive XX) only.
7. As X is not exhibiting its nature when Y is present along with it. it follows Mendel’s law of dominance.

Question 9.
Were all your traits similar to that of your parents?
Answer:

1. No, all my traits are not similar to my parents.
2. There are certain traits which differ from my parents.

10th Class Biology Textbook Page No. 185

Question 10.
How does the evolution of organisms have taken place?
Answer:

1. Variations which are beneficial are selected by nature and passed from parents to offspring through heredity.
2. The same process happens with every new generation until the variation becomes common feature.
3. As the environment changes, the organism within environment adopts and changes to the new living conditions.
4. Over a long period of time, each species of organisms can acuumulate so many changes that it becomes a new species.
5. Thus evolution of organisms took place from common pre-existing ancestors.

Question 11.
Are birds and bats more closely related to each other than to squirrels or lizards?
Answer:

1. No, bats are mammals whereas birds belong to aves.
2. Squirrels are mammals and lizards belong to reptiles.
3. So bats and birds are not closely related to each other as they belong to two different groups.
4. Both bats and birds have wings.

10th Class Biology Textbook Page No. 186

Question 12.
Do embryological evidences indicate that frogs have evolved from ancestors of fish?
Answer:
Yes the embryological evidences indicate that frogs have evolved from ancestors of fish.

Question 13.
Does the life history of every individual exhibit the structural features of their ancestors?
Answer:

1. Yes. The life history of every individual exhibit the structural features of its ancestors.
2. The resemblance is so close at an early stage.

10th Class Biology Textbook Page No. 189

Question 14.
Think why did ancient human beings travelled from one place to other and how did they travelled?
Answer:

1. Ancient human beings travelled from one place to other in search of better living conditions such as availability of food, water shelter and other facilities.
2. They did not travel in a single line.
3. They went forwards and backwards with groups, sometimes separating from each other.
4. This travel is responsible for the formation of races.

10th Class Biology Textbook Page No. 183

Question 15.
In a forest there are two types of deer, in which one type of deer can run very fast. Whereas second type of deer can not run as fast as the first one. Lions, Tigers hunt deer for their food. Imagine which type of deer are going to survive in the forest and which type of deer population is going to be eliminated? And why?
Answer:

1. Deer that can run fast can survive in the forest. Because they can escape easily from lions and tigers, when compared to second type.
2. Deer that run slowly are going to be eliminated. Because they can be caught easily by its predators. So the survival chance will decrease.

10th Class Biology 8th Lesson Heredity Activities

Activity – 1

Think of your own family, what similarities do you share with your father and mother? Draw a table to represent the similarities of some characters like colour of eye (cornea), colour of hair, shape of nose, shape of face, type of earlobe (attached or free), inner thumb markings, etc. Write your characters in one column and that of your parents in the other columns.
Table – 1

Characters	In me	In my Mother/Father	In my Brother/Sister	In my grandma/grandpa

Answer:

S.No.	Characters	In me	In my Mother/Father	In my Brother/Sister	In my grandma/grandpa
1.	Colour of eye	Black	Black	Black	Black
2.	Colour of hair	Black	Black	Black	Black
3.	Shape of nose	Long	Long	Short	Short
4.	Shape of face	Oval	Round	Oval	Oval
5.	Type of earlobe	Free	Free	Free	Free
6.	Type of hair	Curling	Curling	Straight	Straight
7.	Inner thumb marking	Conical	Round	Round	Conical
8.	Skin colour	Fair	Fair	Fair	Fair

1. Is there any character in you similar to that of your mother as well as your grandma?
Answer:
There are four characters in me similar to my mother as well as my grandma. They are

1. Colour of eye
2. Colour of hair
3. Type of earlobe and
4. Skin colour.

2. Is there any character in you similar only to that of your grandma?
Answer:
Two characters are similar in me and in my grandma. They are

1. Shape of face and
2. Inner thumb marking.

3. How do you think these characters may have been inherited by you from grandma?
Answer:
These characters are hereditary from parent to child.

4. Is there any character that is not present in grandma but present in your mother and you?
Answer:
Two characters are not present in grandma which are only present in me and my mother. They are

1. Shape of nose
2. Type of hair

5. Think where from your mother got that character?
Answer:
This character is the result of inherited traits transmitted from parent to progeny.

Activity – 2
Observe some of your friends and note their characters in the following table. Fill in yours as well.
Table – 2

Name of your friend	Colour of skin	Earlobes Free / Attached	Marking on inner side of thumb	Length of forehead	Colour of eyes (Cornea)	Any other features

Answer:

Name of your friend	Colour of skin	Earlobes Free / Attached	Marking on inner side of thumb	Length of forehead	Colour of eyes (Cornea)	Any other features
Ravi	Black	Free	Round	Broad	Blue	Straight hair, long nose, and face, etc.
Ganesh	Black	Attached	Conical	Narrow	Black	Straight hair, short nose, oval face, etc.
Vi jay	Fair	Free	Conical	Broad	Black	Curling hair, short nose, round face, etc.
Karthik	Fair	Free	Round	Broad	Black	Straight hair, long nose, round face, etc.

1. Compare your characters to that of any one of your friend. How many characters did you find were similar among you and your friend?
Answer:
Only few characters such as black hair and black eye were similar among me and my friend.

2. Do you share more similar characters with your parents or with your friends?
Answer:
I share more similar characters with my parents than my friends.

3. Do you think that your differences from parents are same as differences from friends? Why / Why not?
Answer:
My differences from parents are not same as differences from friend. This is because the differences from parents are subtle as there is more genetic relation with parents but the differences from friends are apparent.

Activity – 3

Observe seeds in a pea or bean pod. You may observe several parts to arrive at a generalisation.

1. Can you find two similar seeds there?
Answer:
No, all the seeds are not similar. They had certain variations.

2. What makes them vary? even though they are in the same pod. (Hint: You know that seeds are formed from ovules).
Answer:

1. They vary from one another because they are produced from different ovules.
2. Ovules of a plant are female gametes.
3. These gametes carry different factors (genes) for different characters randomly.

3. Why variations are important? How are variations useful for an organism or a population?
Answer:

1. Variations perhaps help a certain group of organisms in a community when conditions would otherwise be unfavourable for other groups.
2. Desirable variations can be selected by nature.
3. Desirable variations increase the chance of survival of an organism.
4. Accumulation of variations after a long period leads to formation of new species.

Activity – 4

Let us do the following activity to understand the Mendelian principles of Heredity. Materials required :
a) 3 cm length and 1cm breadth chart pieces – 4
b) 2 cm length and 1cm breadth chart pieces – 4
c) Red buttons – 4
d) White buttons – 4
e) Chart, scale, sketch pen, pencil, 2 bags.

Method: Prepare a chart with 2×2 boxes along with number and symbol as shown in the figure.

Game 1: Monohybrid cross (starting with hybrid parents)
To start with take 1, 2 or 3, 4 . In case you start 1, 2 pick all the 16 long and short pieces and prepare such pairs in each of which you have a long and short piece.

Take 4 pairs each of long and short strips and put them in two separate bags. Now each bag contains 8 strips (4 long and 4 short).One bag say ‘A’ represents male and the bag ‘B’ represents female. Now randomly pick one strip each from bag A and B and put them together in the 1 on the chart. Keep picking out the strips and arrange them in the same manner till your bags are empty. Same time your boxes in the chart are filled with pairs of strips. You might have got the following combinations, two long strips, one long and one short strip, two short strips.

1. What is the number of long strip pairs?
Answer:
There are four long strip pairs.

2. What is the number of one long and one short pairs?
Answer:
There are eight, one long and one short strip pairs.

3. What is the number of short strips pairs?
Answer:
There are four short strip pairs.

4. What is the percentage of each type? Also find their ratios.
Answer:
The percentage of long strip pairs, one long and one short strip pairs and short strip pairs are 25%, 50% and 25% respectively and the ratio is 1 : 2 : 1.

5. What can you conclude from this game?
Answer:
From this game I have concluded that:

1. Every individual possesses a pair of alleles, for any particular trait.
2. Each parent passes a randomly selected copy (allele) of these to an offspring.
3. The offspring then receives its own pair of alleles for that trait one each from both parents.
4. If the long strip is considered as dominant 75% exhibit dominant and 25% exhibit recessive character. Thus the phenotype ratio is 3 : 1 in monohybrid cross.
5. The genotype ratio is 1 : 2 : 1.

Activity – 5

Observe the below diagram showing variation in beetle population and its impact.
Let us consider a group of twelve beetles. They live in bushes on green leaves. Their population will grow by sexual reproduction. So they were able to generate variations in population. Let us assume crows eat these red beetles. If the crows eat more Red beetles, their population is slowly reduced. Let us discuss the above three different situations in detail.
Answer:
Situation-1: In this situation, a colour variation arises during reproduction. So that there appears one beetle that is green in colour instead of red.
Moreover, this green coloured beetle passes its colour to Its offspring (Progeny). So that all its progeny are green. Crows cannot see the green coloured beetles on green leaves of the bushes and therefore crows cannot eat them. But crows can see the red beetles and eat them. As a result there are more and more green beetles than red ones which decrease in their number.

The variation of colour in beetle ‘green’ gave a survival advantage to’green beetles’ than red beetles. In other words it was naturally selected. We can see that the ‘natural selection’ was exerted by the crows. The more crows there are, the more red beetles would be eaten and the more number of green beetles in the population would be. Thus the natural selection is directing evolution in the beetle population. It results in adaptation in the beetle population to fit in their environment better.

Let us think of another situation.
Situation-2: In this situation a colour variation occurs again in its progeny during reproduction, but now it results in ‘blue’ colour beetles instead of ‘red’colour beetle. This blue colour beetled can pass its colour to its progeny. So that all its progeny are blue.
Crows can see blue coloured beetles on the green leaves of the bushes and the red ones as well. And therefore crows can eat both red and blue coloured beetles. In this case there is no survival advantage for blue coloured beetles as we have seen in case of green coloured beetles.

What happens initially in the population, there are a few blue beetles,but most are red. Imagine at this point an elephant comes by and stamps on the bushes where the beetles live. This kills most of the beetles. By chance the few beetles survived are mostly blue. Again the beetle population slowly increases. But in the beetle population most of them are in blue colour.

Thus sometimes accidents may also result in changes in certain characters of the population. Characters as we know are governed by genes. Thus there is change in the frequency of genes in small populations. This is known as “Genetic drift’, which provides diversity in the population.

Let us think of another situation :
Situation-3: In this case beetles population is increasing, but suddenly bushes were affected by a plant disease in which leaf material were destroyed or in which leaves are affected by this beetles got less food material.
So beetles are poorly nourished. So the weight of beetles decrease but no changes take place in their genetic material (DNA). After a few years the plant disease are eliminated. Bushes are healthy with plenty of leaves.

What do you think will be condition of the beetles?
Answer:
The weight of beetles will increase once again as they get plenty of food material again.

Activity – 6

Let us observe different stages of development of vertebrate embryos. Try to find out similarities and differences and discuss with your friends.
(OR)
What do you infer about the embroyological evidences of various organisms?
Answer:

1. There are remarkable similarities in the embryos of different animals from fish to man.
2. The resemblance is so close at an early stage.
3. Gradually the similarities are decreased when they become babies.
4. The embryological evidences give us an idea that all the organisms have evolved from a common ancestors.

 

AP State Board Syllabus AP SSC 10th Class Biology Important Questions Chapter 6 Reproduction.

AP State Syllabus SSC 10th Class Biology Important Questions 6th Lesson Reproduction

10th Class Biology 6th Lesson Reproduction 1 Mark Important Questions and Answers

Question 1.
What questions you ask the doctor, who visited your school on World AIDS day?
Answer:

1. How does AIDS disease occurs?
2. How does the AIDS transmit?
3. What are the symptoms of AIDS?
4. What are the precautions to be taken to prevent AIDS?

Question 2.
What is colostrum?
Answer:
The first secretion from the Mammary glands, after giving birth, rich in antibodies.
During the end of pregnancy, a watery yellowish lymph like fluid accumulates in mammary glands. It is known as colostrum.

Question 3.
Name the types of asexual reproduction in the following organisms:
a) Paramoecium b) Yeast
Answer:
a) Paramoecium : Paramoecium reproduces by splitting into two. (Transverse binary fission)
b) Yeast: Yeast reproduces by Budding.

Question 4.
What are the advantages of grafting method in plants.
Answer:

1. Grafting is used to obtain a plant with desirable characters.
2. It can be used to produce varieties in seedless fruits.

Question 5.
What measures can be taken to avoid sexually transmitted diseases?
Answer:

1. Avoid sex with unknown or multiple partners.
2. Sex with life partners only.
3. Follow ethical and healthy life practices because contraceptives always cannot prevent STD’s.
4. In case of any doubt, consult a qualified doctor for early detection if diagonised with disease take complete treatment.

Question 6.
What is parthenogenesis?
Answer:
The process of developing zygote from female gametes without fertilization is known as parthenogenesis.

Question 7.
In flowering plants, I am formed as the result of double fertilization. The cotyledons digest and absorb me. Who am I?
Answer:
Endosperm.

Question 8.
In what way does mitotic division help the living organism?
Answer:

1. growth
2. cell repair
3. healing wounds.

Question 9.
Give any two suggestions to create awareness to stop female foeticide.
Answer:

1. Preparing relevant slogans
2. Organising rallies
3. Awareness campaign by using electronic and print media

Question 10.
Write two precautions you take, while observing Rhizopus in the laboratory.
Answer:

1. Don’t touch the experimental bread with hand.
2. If you touch the bread, thoroughly wash your hands.
3. Leave the bread in the open air for about an hour.
4. Avoid opening of the plastic bag as much as you can.
5. Sprinkle water over bread.
6. Place the bag in a dark and warm place.

Question 11.
Mention two materials you have used to observe Rhizopus on bread mould.
Answer:
Bread mould sample, plain glass slide, cover slip, water, disposable gloves.

Question 12.
What type of reproduction occurs in paramoecium during favourable conditions?
Answer:
During favourable conditions paramoecium reproduce asexually by fission.

Question 13.
What type of reproduction occurs in paramoecium during unfavourable conditions?
Answer:
During unfavourable conditions paramoecium reproduce sexually by conjugation.

Question 14.
Which bacteria is responsible for formation of curd from milk?
Answer:
Lactobacillus bacteria is responsible for formation of curd from milk.

Question 15.
What is asexual reproduction?
Answer:
The reproduction in which a single parent is involved, without formation of gametes is known as asexual reproduction.

Question 16.
What is fission?
Answer:
Splitting of organisms into two or more offsprings in a symmetrical manner is known as fission. Ex: Paramoecium and bacteria.

Question 17.
How budding occurs in yeast?
Answer:

1. A bud develops as an outgrowth due to repeated cell division at specific site.
2. These buds develop when fully mature, detach from the parent body and become new independent individuals.

Question 18.
Which animals reproduce through fragmentation?
Answer:
Fragmentation is a common mode of reproduction in Algae, Fungi and many land plants.

Question 19.
What is Regeneration?
Answer:

1. Many fully differentiated organisms have the ability to give rise to new individual organism from their body parts.
2. If the individual is some how cut or broken up into many pieces, many of these pieces grow into separate individuals. Ex: Hydra and planaria.

Question 20.
In which plant, small plants grow at the edge of leaves?
Answer:
In Bryophyllum, small plants grow at the edge of leaves.

Question 21.
By means of which plants propagate vegetatively through stem?
Answer:
Plants propagate vegetatively through stem by means of stolons, bulbs, corms, tuber etc.

Question 22.
Through which the Vallisneria, Strawberry propagate vegetatively?
Answer:
Vallisneria, Strawberry propagate vegetatively through stolons.

Question 23.
Which plants produce the new plants through roots?
Answer:
New plants are produced from the roots of Dahlia, radish, carrot etc.

Question 24.
What are the artificial propagation methods in plants?
Answer:
Cutting, Layering and Grafting are the artificial propagation methods in plants.

Question 25.
Which method is used to obtain a plant with desirable characters?
Answer:
Grafting is used to obtain a plant with desirable characters.

Question 26.
Which method will you adopt to get two desirable characters from two different plants in a single plant?
Answer:
I will adopt grafting method to get two desirable characters from two different plants in a single plant.

Question 27.
Which fungus is commonly called as bread mould?
Answer:
Rhizopus is commonly called bread mould.

Question 28.
How bread mould appears when you observe it under microspore?
Answer:
The common bread mould consists of fine thread like projections called hyphae and thin knob like structures called Sporangia.

Question 29.
In which plants leaf is known as Sporophyll? Why?
Answer:
In fern plants leaf is known as Sporophyll. Because on the lower surface of the leaf sporangia are present.

Question 30.
What is fertilisation?
Answer:
Union of male and female gametes is known as fertilisation.

Question 31.
What is external fertilisation?
Answer:
If the fertilisation occurs outside the body of the organism then it is known as external fertilisation. Eg : Frog and fish.

Question 32.
What is internal fertilisation?
Answer:
If the fertilisation occurs inside the body of the female organism then it is known as internal fertilisation. Eg : Terrestrial animals (Reptiles, Aves, Mammals).

Question 33.
What are the parts present in male reproductive system of man?
Answer:
A pair of testis, Accessory glands and System of ducts.

Question 34.
The male reproductive organ testis produces?
Answer:
Testis produces male reproductive cells or spermatozoa.

Question 35.
Sperms are temporarily stored in which part of duct system?
Answer:
Sperms are temporarily stored in epididymis of duct system.

Question 36.
What are the accessory glands present in male reproductive system?
Answer:
The accessory glands present in male reproductive system are one prostrate gland . and two cow cowper glands.

Question 37.
The fluid secreted by accessory glands is
Answer:
The fluid secreted by the accessory glands is semen.

Question 38.
What is the function of semen?
Answer:
Semen provide nutrients for sperm to keep alive and helps as a medium for the movement of sperms.

Question 39.
Which hormone regulates the development of the male reproductive organs?
Answer:
The hormone testosterone regulates the development of the male reproductive organs.

Question 40.
How are the secondary sexual characters are controlled in males?
Answer:
Secondary sexual characters in males are controlled by the male sex hormone testosterone.

Question 41.
Men produce sperm from the age of about?
Answer:
Men produce sperm from the age of about 13 or 14 years and can go on doing so most their lives.

Question 42.
Which are capable of changing the sex of the organism in which they grow like wasp?
Answer:
Some bacteria and other micro-organisms have been found capable of changing the sex of the organism of wasp in which they live.

Question 43.
The female gamete ovum is produced by
Answer:
The female gamete ovum is produced by graffian follicles of Ovary.

Question 44.
What is ovulation?
Answer:
The release of ovum from graffian follicle is known as ovulation.

Question 45.
Fertilisation of ovum occurs in which part of female reproductive system?
Answer:
Fertilisation of ovum occurs in fallopian tube or oviduct of female reproductive system.

Question 46.
What is placenta?
Answer:
Placenta is the nourishment tissue formed by the outer membrane of the embryo called chorion and the part of the uterine tissue.

Question 47.
When do placenta is formed during the development of embryo?
Answer:
Placenta is formed at around 12 weeks of pregnancy or during the embryonic development.

Question 48.
What keeps embryo moist and protects it from minor mechanical injury?
Answer:
The embryo develop in amniotic fluid filled cavity which keeps it moist and protects it from minor mechanical injury.

Question 49.
Which membrane forms umbilical cord?
Answer:
Allantois membrane which originates from the digestive canal of the embryo forms the major part of tube like structure called umbilical cord.

Question 50.
What is foetus?
Answer:
From the third month of pregnancy the embryo is called foetus.

Question 51.
What is gestation period?
Answer:
Total time required for the development of embryo and foetus is called gestation period.

Question 52.
What is the gestation period in human beings?
Answer:
The gestation period in human beings is 9 months or 280 days.

Question 53.
Collect the information about gestation periods in different animals.
Answer:
Gestation period in different animals:

Animal	Gestation period
Cat and dog	63 days
Horse	330 days
Cow	280 days
Rat and mouse	20-22 days

Question 54.
What is after birth?
Answer:
The muscular contractions of the uterus continue until they push out the tissues of the placenta, which are commonly called the ‘after birth’.

Question 55.
What are labour pains?
Answer:
The rhythmic contraction and relaxation of muscle layers of the uterus is known as labour pains.

Question 56.
What is colostrum?
Answer:
Colostrum: During the end of pregnancy a watery yellowish lymph like fluid accumulates in the mammary glands. It is known as colostrum.

Question 57.
What is the importance of feeding colostrum to new born baby?
Answer:
It is very important to feed colostrum to the new born baby because it helps in developing the immune system of the child.

Question 58.
What is the need of sexual reproduction?
Answer:
Sexual reproduction help organisms to develop characters that would be help them to adapt better to their surroundings.

Question 59.
In which mountain regions can Sal trees grow?
Answer:
Sal trees grow in the Himalayan mountains.

Question 60.
What are the different parts of a flower?
Answer:
Sepals, petals, stamens and carpels are the different parts of a flower.

Question 61.
What are stamens and carpels?
Answer:
The reproductive parts of a flower which possess the sex cells or germ cells are called stamens and carpels.

Question 62.
What are unisexual flowers? Give examples.
Answer:
Flowers having either stamens or carpels are called unisexual flowers.
Eg: Bottlegourd, papaya.

Question 63.
What are Bisexual flowers? Give some examples.
Answer:
Flowers having both the stamen and carpel are called bisexual flowers. Eg: Datura.

Question 64.
What are the three parts of carpel or gynoecium?
Answer:
The three parts of carpel or gynoecium are ovary, style and stigma.

Question 65.
What is self pollination?
Answer:
Plants having flowers. Where reproductive cells of stamen of the flower fertilise the female reproductive cells of the carpel of the same flower is called self pollination.
Eg: Plants of pea family.

Question 66.
How cross fertilisation occurs?
Answer:
If the male cells of flower of a plant fertilise the female cells of flowers on the same or different plants of the same species, the type of pollination is called cross pollination.

Question 67.
What did Darwin showed regarding fertilization of plants?
Answer:
Darwin in 1876 showed that plants when isolated had the greatest tendency to self fertilize while when surrounded by varieties of the same flower, they readily cross fertilize.

Question 68.
Which cells are composed the embryosac of ovule?
Answer:
The embryosac of ovule composed of gametophyte cells.

Question 69.
How many cells and nuclei does an embryosac consisting in majority of flowering plants?
Answer:
The majority of flowering plants have an embryosac consisting of seven cells and eight nuclei.

Question 70.
What is double fertilisation?
Answer:
Double fertilisation: Union of one male nucleus with an egg and the second male nucleus with the fusion nucleus is called double fertilisation.

Question 71.
What is germination?
Answer:
The seed produced after fertilisation contains the future plant or embryo that develops into a seedling under appropriate conditions. This process is called germination.

Question 72.
Who gave the phrase “omnis cellula de cellula”? What does it mean?
Answer:
The ‘phrase omnis cellula de cellula’ means cells arise from pre-existing cells. It was given by Rudolph Virchow who discovered cell division.

Question 73.
Who stated that the animals can reproduce through binary fission of cells?
Answer:
In 1852 Robert Remak of Germany stated that animals can reproduce through binary fission of cells.

Question 74.
Who discovered the process of mitosis?
Answer:
Mitosis was discovered by Walther Flemming in 1879.

Question 75.
What is the most important discovery of Walther Flemming regarding chromosomes?
Answer:
Walther Flemming’s most important discovery was chromosomes appear double in nature.

Question 76.
Who proposed that chromosomes carried a different set of heritable elements?
Answer:
Wilhelm Roux proposed that chromosomes carried a different set of heritable elements.

Question 77.
What are the hypothesis made by August Weiseman on chromosomes?
Answer:

1. In successive generations, individuals of the same species have the same number of chromosomes.
2. In successive cell division the number of chromosomes always remain constant.

Question 78.
Who confirmed the scheme of mitotic division?
Answer:
The scheme of mitotic division was confirmed in 1904 by Theodor Boveri.

Question 79.
Who discovered the structure of DNA?
Answer:
The structure of (DNA) deoxy ribonucleic acid was discovered in 1953 by James Watson and Francis Crick.

Question 80.
The cells in which organ do not divide?
Answer:
Cells present in organs such as heart and brain of an individual never divide.

Question 81.
What is time required for completion of mitosis?
Answer:
The process of mitosis is completed in 40 to 60 minutes.

Question 82.
What is interphase?
Answer:
The period between two cell divisions is called interphase.

Question 83.
Into how many phases the interphase can be divided?
Answer:
Interphase can be divided into three phases. They are G₁ phase, S phase and G₂ phase.

Question 84.
What is G₁ phase of interphase?
Answer:
G₁ phase is the linking period between the completion of mitosis and the begining of DNA replication (Gap 1 phase).

Question 85.
What is S phase of interphase?
Answer:
S phase is the period of DNA synthesis leading to duplication of chromosomes.

Question 86.
What is G₂ phase of interphase?
Answer:
G₂ phase is the time between the end of DNA replication and the beginning of mitosis (Gap 2 phase).

Question 87.
Who conducted some experiments using the cell fusion technique on phases of interphase?
Answer:
Potu Narasimha Rao and Johnson conducted some experiments using the cell fusion technique to understand the functional relationship between the phases of interphase.

Question 88.
What is cytokinesis?
Answer:
Division of cytoplasm is called cytokinesis.

Question 89.
What are the different stages present in mitosis?
Answer:
Prophase, Metaphase, Anaphase and Telophase are the different stages present in mitosis.

Question 90.
In which phase of the mitosis chromosomes split lengthwise to form chromatids?
Answer:
In prophase of the mitosis chromosomes split lengthwise to form chromatids.

Question 91.
During which phase of mitosis chromatids are pulled towards poles?
Answer:
During anaphase of mitosis chromatids are pulled towards poles.

Question 92.
How many haploid daughter cells are formed after meiosis?
Answer:
Four haploid daughter cells are formed after meiosis.

Question 93.
What are the diseases that can be sexually transmitted?
Answer:
Sexually transmitted diseases include bacterial infections such as Gonorrhoea and Syphilis and Viral infections such as AIDS.

Question 94.
In what way the sexually transmitted diseases spread from person to person?
Answer:
Sexually transmitted diseases spread by unsafe sexual contacts, using infected devices, infected blood transfusion, from an infected mother to child.

Question 95.
Which state has the highest number of HIV patients in the country?
Answer:
Andhra Pradesh and Telangana has the highest number of HIV patients in the country.

Question 96.
Which factors are contributing to the spread of HIV in Andhra Pradesh?
Answer:
Illiteracy, poor health, unemployment, migration, non-traditional sex practise, unethical contacts and trafficking are some of the factors contributing to the spread of HIV in Andhra Pradesh.

Question 97.
Expand “ASHA”.
Answer:
Accredited Social Health Activist.

Question 98.
What is Red ribbon express?
Answer:
Red Ribbon express is an AIDS/HIV awareness campaign train by the Indian Railways. The motto of the Red ribbon express is “Embarking on the Journey of Life”.

Question 99.
What is contraception?
Answer:
The prevention of pregnancy in women by preventing fertilisation is called contraception.

Question 100.
Which device not only prevents fertilisation but also transmitting some sexually transmitted diseases?
Answer:
Condoms and diaphragm (cap) prevents fertilisation and also useful to not transmitting some sexually transmitted diseases like gonorrhoea, syphilis, AIDS.

Question 101.
What are spermicides?
Answer:
Spermicides are the pills used for killing sperms.

Question 102.
What are the surgical methods to birth control in males and females?
Answer:
Vasectomy for males and Tubectomy for female are the surgical birth control methods in human beings.

Question 103.
What is Vasectomy?
Answer:
In males, a small portion of vas deferens is removed by surgical operation ami both ends are tied properly. This method is called vasectomy.

Question 104.
What is Tubectomy?
Answer:
In females, a small portion of oviducts (fallopian tube) is removed by surgical operation and the cut ends are tied. This prevents the ovum from entering into the oviducts. This method is called Tubectomy.

Question 105.
What is the marriage age for girls in India?
Answer:
The marriage age for girls in India is 18 years.

Question 106.
What is foeticide?
Answer:
Foeticide is the act of destruction or aborting a foetus because it is female.

10th Class Biology 6th Lesson Reproduction 2 Marks Important Questions and Answers

Question 1.
What are the questions you asked the doctor who visited your school to know “the ways of transmission of HIV”?
Answer:
I shall ask the following questions to the doctor.

1. What are the ways of transmission of HIV?
2. How can we prevent the spread of HIV?
3. What precautions should we take while doing transfusion of blood:
4. How does HIV transmit from mother to baby?
5. Why should we use disposable syrenges?

Question 2.
The chromosomal number is reduced to half in the daughter cells produced by meiosis. What happens if the number is not reduced to half in daughter cells?
(OR)
In Meiosis, the chromosome number in the daughter cells are reduced to half that of their parent cells. Guess, what would happen, if the reduction of chromosome number is not done.
Answer:

1. If the reduction of chromosomes number is not done, the chromosomal number is doubled in the offsprings.
2. The change in chromosomal number leads to development of abnormal characters in the individual.
3. The offspring differs from parental generation.
4. Abnormal characters will be formed in new generation, which are not useful for the existence of individual.

Question 3.
What questions do you ask a doctor to know about different birth control methods?
Answer:

1. What is family planning?
2. What is meant by contraception?
3. How many types of contraceptive methods are there?
4. What are the contraceptive devices used for female?
5. What are the contraceptive devices used for male?
6. What is tubectomy?
7. What is vasectomy?
8. What are surgical methods of birth controls?

Question 4.
Apparao and Ramulamma are a newly married illiterate couple. They don’t want children for few years. Suggest some birth control methods for them.
(OR)
Mention any four birth control methods.
Answer:
a) condoms
b) diaphragm (Cap)
c) pills
d) copper – T
e) loop

Question 5.
Why is it important for gametes to have half the number of chromosomes?
Answer:

1. If gametes have 2 sets of chromosomes, the number of chromosomes will be 4 sets in zygote after fertilization because of this the chromosomal number will be doubled in each generation. This results in abnormalities in off-spring.
2. Hence, to maintain a constant number of chromosomes, garnets should have half set of chromosomes.

Question 6.
Identify the flower parts a, b, c, d and write their main function.
Answer:
a) Ovary: Female reproductive organ in flower. It produces female gametes called ovules.
b) Style: Ovary has a pipe like structure called style. It allows the pollen tube to enter the ovary for fertilization.
c) Stamen: These are male parts called androecium. It has two parts. They are filament and Anther.
d) Anther : Produces male gametes called pollen grain.

Question 7.
Draw and label the diagram of human sperm cell.
Answer:

Question 8.
How can we get the desired useful triats with the help of two selected triats by using grafting method?
Answer:

1. Two plants are joined together in such a way that two stems join and grow as a single plant.
2. One which is attached to soil is called stock and the cut stem of another plant without roots is called scion.
3. Both stock and scion are tied with the help of a twine thread and covered by a polythene cover.
4. Grafting is used to obtain a plant with desirable characters.
5. This technique is very useful in propagating improved varieties of plants with various flowers and fruits. Ex: Mango, citrus, apple, rose.

Question 9.
Draw the labelled diagram of Embryo-sac A.
Answer:

Question 10.
Observe the diagram and answer the following questions.
i) Which phases take same time duration?
Answer:
G₁ phase and S phase.
ii) In which phase, DMA synthesis takes place?
Answer:
S Phase.

Question 11.
Write the process involved in seedless fruit development with two suitable examples.
Answer:
In some plants ovary directly develops into fruit without the process of fertilization, this phenomenon is called as parthenocarypy.
Ex: Grapes, water melon.

Question 12.
What precautions will you take to keep away from diseases like AIDS and other sexually transmitted diseases?
Answer:

1. Avoid sex with unknown partners or multiple partners.
2. Use condom every time.
3. Use disposable syringes and needles.
4. Transfusion of safe blood to the patients.
5. HIV mother can have child with doctor’s advice only.

Question 13.
Observe the diagram and answer the following questions.
i) Name male and female reproductive parts of the above figure.
Answer:
Male reproductive parts – anther / pollen grain / stamen
Female reproductive parts – ovary / ovule / style / stigma.

ii) Write the names of (1) and (2) in the diagram.
Answer:

1. Sepal or calyx
2. Petal or corolla

Question 14.
When does Parthenogenesis occur? Write names of two animals in which parthenogenesis takes place.
Answer:
a) Parthenogenesis is a process of reproduction where there is a shift from sexual to asexual mode of reproduction.
b) In this process generally the female garnets develops into zygote without fertilization.
c) This strange kind of reproduction occur in bees, ants and wasps.
d) The parthenocarpic zygote develop into male (Monoploid) while the fertilized one developed into female (Diploid)

Question 15.
Draw the figure of metaphase in mitosis, and write about it.
Answer:

1. Chromosomes move to spindle equator, centromeres attached to spindle fibres.
   
2. Centromeres split, separating the chromatids.

Question 16.
Prepare 4 questions on meiosis, to conduct a Quiz programme.
Answer:

1. Where does meiosis occur in?
2. How many daughter cells are produced at the end of meiosis?
3. In which phase of meiosis karyokinesis takes place?
4. Name the scientist who discovered meiosis for the first time.

Question 17.
Write slogans on ‘Child marriages – a social evil’.
Answer:

1. Child marriage, a loosing game.
2. She is a child herself, why burden her with another child?
3. My childhood, my right.
4. A child should call ‘mother’ but a child should not be called mother.
5. Good marriages take place slowly. Go slow with children’s marriage.
6. Say no to child marriage.

Question 18.
Write 5 slogans on the prevention of HIV/AIDS.
Answer:

1. Open your eyes before AIDS closes them.
2. Hate the disease but not the diseased.
3. Spread the knowledge not the virus.
4. Wear protection to prevent infection.
5. AIDS brings pain! Girls please obstain.

Question 19.
What is fission? Give examples.
Answer:

1. Fission is a method of asexual reproduction in which a single-celled organism splits into two or more offsprings.
2. This splitting usually occurs in a symmetrical manner.
3. When an organism is split into two offsprings it is called binary fission.
4. When an organism is split into more offsprings, it is called multiple fission.
5. This is often the only mode of reproduction for single celled organisms.
   Ex : Paramoecium and bacteria.

Question 20.
Write a short notes on fragmentation.
Answer:

1. Fragmentation is a reproductive method in multicellular organisms with relatively simple body organisation.
2. Some can grow from a separate piece of parent organism. This can be from any part of the body.
3. This happens only in the simplest such as some flat-worms, moulds, lichens, spirogyra, etc.
4. Fragmentation is a common mode of reproduction in algae, fungi and many land plants.

Question 21.
What do you know about parthenogenesis? Explain with examples.
Answer:

1. Parthenogenesis is an asexual reproduction in which unfertilized eggs develop into offsprings.
2. In this process generally egg develops into new individual without meiosis and fertilization. So the offsprings are diploid.
3. In some species of animals reproduction occurs only through parthenogenesis. There are no males known in these species. Ex: Rotifers.
4. In another type of parthenogenesis meiosis does occur and the egg can develop whether fertilized or not.
5. The monoploid offsprings develop into males and diploid into females.
   Ex: Bees, Ants and Wasps.
6. Nowadays we are able to develop seed less fruits like watermelon, grapes, pomegranate etc.

Question 22.
Describe the vegetative propagation through the stem with examples.
Answer:

1. Production of new plants from the vegetative parts such as stem, root, leaves of the existing plant is called vegetative propagation.
2. Aerial weak stems like runners and stolons, when they touch the ground, give off adventitious roots.
3. When the connection with the parent plant is broken, the portion with the newly struck roots develops into an independent plant.
4. Some examples for propagation by stem are from stolons, bulbs, corms and tubers as follows.
   a) Stolons – Vallisneria, Strawberry
   b) Bulbs – Alliumcepa or onion
   c) Corms – Colacasia
   d) Tuber – Potato

Question 23.
Write short note on artificial propagation method cutting.
Answer:

1. Cutting is an artificial method of vegetative propagation in which new plants are developed from the cut portion of existing plant.
   
2. Some plants grow individually when a piece of the parent plant having bud is cut from the existing plant.
3. The lower part of this cutting is buried in moist soil.
4. After few days the cut parts having buds grow as an individual plant.
   Ex: Rose, Hibiscus.

Question 24.
What is layering? Explain briefly about it.
Answer:

1. Stems that form roots while still attached to the parent plants are called layers. Propagating the plants in this method is layering.
2. A branch of the plant with at least one node is bent towards the ground and a part of it is covered with moist soil leaving the tip of the branch exposed above the ground.
3. After sometime, new roots develop from the part of the branch hurried in the soil.
4. The branch is then cut off from the parent plant, later it develops roots and grows to become a new plant. Ex: Nerium.

Question 25.
Write a short note on Grafting.
Answer:

1. Grafting is a method of artificial vegetative propagation in which two plants are joined together in such a way that two stems join and grow as a single plant.
   
2. One which is attached to soil is called stock and the cut stem of another plant without roots is called scion.
3. Both stock and scion are tied with help of a twine thread and covered by a polythene cover.
4. After few days both will unite by forming new tissue and grow as a single one.
5. Grafting is used to obtain a plant with desirable characters.
6. Plants in which grafting is done more in mango, apple, citrus, plants.

Question 26.
What are the advantages of grafting?
Answer:

1. Grafting enables us to combine the most desirable characteristics of the two plants (scion and stock) in its flower and fruits.
2. By grafting method, a very young scion can be made to flower and produce fruits quite fast when it is grafted to the stock.
3. Grafting can be used to produce varieties of seedless fruits.

Question 27.
How is tissue culture more beneficial than other traditional methods for the artificial propagation of plants? (OR)
What is tissue culture? What are its uses?
Answer:

1. The traditional methods for the artificial propagation of plants are being replaced by the modern methods of artificial propagation of plants involving tissue culture, as it is more beneficial than the traditional methods.
2. In tissue culture, a few plant cells or plant tissue are placed in a growth medium with plant hormones in it and it grows into new plants.
3. Thousands of plants can be grown in very short interval of time.
4. There will be no climatic impact on the propagation, so multiplication can be done throughout the year.
5. It is possible to obtain plants that are free from pathogens.

Question 28.
How does the Rhizopus propagate?
Answer:

1. Rhizopus propagates by means of spores.
   
2. The Rhizopus parent plant produces hundreds of microscopic reproductive units called spores.
3. When the spore case of the plant bursts, the spores spread into air.
4. These air borne spores fall on food or soil, under favourable conditions like damp and warm conditions, they germinate and produce new plants.

Question 29.
Write a short note on spore formation. (OR)
How spores are produced in sporangia of fungi?
Answer:

1. Spore formation is a method of asexual reproduction which occurs through microscopic reproductive units called spores.
2. Most of the fungi like rhizopus, mucor etc., bacteria and non-flowering plants such as ferns and mosses reproduce by the method of spore formation.
3. In fungi like rhizopus spores are produced in some specialised structures called sporangia which bursts and spreads the spores into air. These spores when fall on food or soil under favourable conditions germinate and produce new plants.
4. In non-flowering plants like fern, the leaves called sporophyll bears clusters of sporangia on their lower side. These sporangia produce the spores which produce the new plant when it falls on ground under favourable conditions.

Question 30.
How is external fertilisation different from internal fertilisation? (OR)
What are the differences between external and internal fertilisation?
Answer:

1. Fertilisation that takes place outside the body of mother is called external fertilisation. This is most common in animals like fishes and amphibians. As the chance of fertilisation is controlled by nature it becomes necessary to give rise to vast number of eggs and sperms by these animals.
2. Fertilisation that takes place inside the body of mother is called internal fertilisation. This is common in most of the land animals. As the chance of fertilisation is not controlled by the nature, these animals generally produce less number of eggs.

Question 31.
Write a short note on ovulation. (OR)
What is ovulation? How it occurs?
Answer:

1. Release of the egg or ovum is called ovulation.
2. The ova develop in tiny cellular structures in ovary called follicles, which at first look like cellular bubbles.
3. As a follicle grows, it develops a cavity filled with fluid.
4. Each follicle contains a single ovum.
5. When an ovum is mature, the follicle ruptures at the surface of the ovary and the tiny ovum is flushed out.
6. This release of ovum is called ovulation.

Question 32.
How does the uterus get adapted to receive the embryo?
Answer:

1. The uterus at the time of fertilization is beautifully adapted to receive the developing embryo, providing it with food and disposing of its wastes.
2. A few days prior to this time, the uterus was small, its tissues were thin, and its supply of blood vessels was poor.
3. When the fertilized egg or zygote is about to enter the uterus become much larger, its inner wall becomes thick, soft and moist with fluid, its blood supply is greatly increased and waiting for an embryonic occupant.

Question 33.
What is colostrum? What is its importance?
Answer:

1. During the end of pregnancy, a watery lymph like fluid accumulates in the mammary glands.
2. This is called colostrum.
3. For the first few days after the baby is born, the mammary glands secrete only colostrum.
4. It is very important to feed the new born baby with colostrum because it helps in developing the immune system of the child.

Question 34.
What is the importance of mitosis in human beings?
Answer:

1. Mitosis is the cell division that transforms a human fertilized egg into a baby in nine months and into an adult in the next 20 years.
2. The bone marrow cells actively divide by mitosis to produce red blood cells.
3. Mitosis helps in replacing the worn out cells in the skin.
4. Mitotic divisions in the cells surrounding the wound helps in cease the wound and healing.

Question 35.
Collect the information about the significance of the experiments done by Dr Potu Narasimha Rao and Johnson.
Answer:

1. Nearly 4 decades back Dr.P.N. Rao and Johnson did some elegant experiments using the cell fusion technique to understand the functional relationship between the phases of cell cycle.
2. These experiments have, for the first time provided evidence that the progression of cells through the cell cycle is sequential and unidirectional and are controlled by a series of chemical signals that can diffuse freely between nucleus and cytoplasm.
3. These experiments revealed for the first time the structure of interphase chromosomes that are not ordinarily visible under the microscope.
4. These experiments are considered to be a ‘milestone’ in the cell cycle studies.

Question 36.
Ramu said that it is very essential to create more awareness in Andhra Pradesh on the risk of HIV infection and AIDS. Do you support him? If so, how can you support his statement?
Answer:
Yes, what Ramu said is right. I support his statement with the following reasons.

1. Andhra Pradesh has the highest number of HIV patients in the country.
2. According to official statistics, the state had 5 lakh of the 24 lakh HIV positive patients
   in the country during 2011-12.
3. While one in every 300 adults is suffering from HIV elsewhere, in Andhra Pradesh one in every 100 adults is a HIV patient, that is almost one per cent.
4. The prevalence of HIV is 1.07 per cent among males and 0.73 among females in the state, which again is higher than in other states.

Question 37.
Briefly explain about the contraception and contraceptive methods.
Answer:
The prevention of pregnancy in the woman by preventing fertilisation is called contraception. Any device or chemical which prevents pregnancy in a woman is called a contraceptive. Contraceptive methods are of various types and used by any of the partners as preferable. Some of the contraceptive methods are:

1. Use of physical devices such as condoms and diaphragm (cap).
2. Use of hormonal pills which stop the ovaries from releasing ovum into oviduct.
   These pills can be induced either orally or inserting into female reproductive organ vagina.
3. Use of spermicides that kills the sperms.
4. Use of intra-uterine device called copper – T, loop, etc.
5. Use of surgical methods such as vasectomy for male and tubectomy for female.

Question 38.
Classify the given organisms basing on the type of reproduction.
Man, Flatworm, Mould, Dog, Bacteria, Frog, Fern, Datura, Hen, Yeast.
Answer:

Sexually reproducing organisms	Asexually reproducing organisms
Man	Flat worm
Dog	Mould
Frog	Bacteria
Datura	Fern
Hen	Yeast

Question 39.
What will happen if the amnion is ruptured before the foetus is developed completely?
Answer:

1. Amnion is the embryonic membrane that grows around the embryo itself.
2. The cavity within the amnion is filled with a fluid called amniotic fluid, which keeps the growing embryo moist and protects it from minor mechanical injury.
3. If the amnion ruptures by accident before the foetus developed completely, the amniotic fluid is released out through vagina.
4. As there is no protective fluid around the foetus, it starts getting damaged.
5. So if possible delivery must done immediately by surgerical method, otherwise abortion must be done.
6. If baby dies inside the uterus which leads to infections in uterus causing problems
   to mother that leads to death.

Question 40.
How will you appreciate the contribution of August Weiseman to the cell biology?
Answer:

1. Science is not advanced only by the collection of data. Someone must think about and interpret the data. August Weiseman belongs to this category who think and interpret the data.
2. Even though his poor eyesight not allowed him to use a microscope to study cells, he made great contribution to the cell biology making use of his thinking capacity and interpretation skills.
3. He hypothesised that
   a) In successive generations, individuals of the same species have the same number of chromosomes.
   b) In successive cell division, the number of chromosomes remains constant.
4. His hypothesis proved right in case of mitosis.
5. We should take such a great person who overcame his defect with his will as our role model.

Question 41.
How will you appreciate the contribution of Dr. P.N. Rao to the ceil biology?
Answer:

1. Dr. Potu Narasimha Rao, a renowned scholar and eminent cytologist came from a poor family in Muppalla village of Guntur district.
2. He did his research work on the cytogenetics of tobacco plant and cancer cells in culture medium.
3. He conducted research in cell kinematics and triggering factor of cell division i.e., mitosis.
4. He observed the interphase and its three phases.
5. To understand the functional relationship between these phases he did elegant experiments on cell fusion technique along with his research associate Dr.Johnson.
6. His researches revealed that the cell cycle is sequential, unidirectional and controlled by a series of chemical signals.
7. His experiments are considered to be a milestone in the cell cycle.
8. He is an exemplary person who proved that poverty is not a barrier to the talent and wisdom.

Question 42.
Write briefly about natural vegetative propagation in plants.
Answer:

1. In natural vegetative propagation new plants are produced from stem, root, leaves of old plants without the help of any reproductive organs.
2. In bryophyllum small plants grow at the edge of leaves.
3. Aerial weak stems like runners stolons, when they touch the ground give it adventitious roots.
4. When the connection with the parent plant is broken the stem portion with the adventitious roots develops into an independent plant.
5. Some examples for propagation by stem are from stolons, bulbs, corms, tuber etc.
   
6. Stolons – Vallisneria, strawberry.
   Bulbs – Onion (Alliumcepa)
   Corms – Colacasia
   Tuber – Potato

Question 43.
What are sexually transmitted diseases and mention the ways to prevent them?
Answer:

1. A disease which can be transmitted through sexual contact is called sexually transmitted disease or STD.
2. These include bacterial infections such as gonorrhoea, syphilis, Herpis and viral infections such as herpes and AIDS.
3. Lack of hygiene is usually a major factor in providing conditions for spread of STDs.
4. But unprotected sex with multiple and unknown partners is the highest reason for the spread of STDs.
5. Some of the ways to prevent STD are as follows.
   a) Being faithful to one’s life partner.
   b) Avoid sexual contact with unknown person.
   c) Using condom during sexual intercourse.
   d) Maintaining personal hygiene.

Question 44.
Why more complex organisms cannot give rise to new individual through regeneration ?
Answer:

1. Many organisms have the ability to give rise to new individual organisms from their body parts.
2. Regeneration happens through mitosis and a particular type of tissue can give rise to its own kind only.
3. In complex organisms, different tissues and organs have altogether different structures.
4. Regenerating a different kind of tissue from another kind is not possible.
5. Hence complex organisms are not able to give rise to new individuals through regeneration.

Question 45.
How an organism will be benefited if it reproduces through spores?
Answer:

1. Reproduction through spores gives several advantages to an organism like they are produced in very large numbers and it helps in propagation of species.
2. Spores can remain dormant till favourable conditions become available.
3. Spores help an organism to overcome unfavourable conditions.
4. Spores can be spread through water, air or animals and thus is good for the spread of an organism to more places.

Question 46.
What is the role of the placenta in embryo development?
Answer:

1. Placenta is a tissue formed by the cells from the embryo and the mother.
2. It is formed around 12 weeks of pregnancy and becomes an important structure for nourishment of the embryo.
3. Placenta is a disc which is embedded in the uterine wall. It contains villi on the embryo’s side of the tissue.
4. On the other side mother’s blood spaces are present.
5. This provides a large surface area for diffusion of glucose, oxygen and other nutrients from the mother of the embryo.

Question 47.
Why do we practise vegetative propagation for growing some types of plants?
(OR)
Why vegetative propagation is adopted over other types of propagation?
Answer:
Vegetative propagation is practised in some plants because

1. It is the only method of reproduction in seed less plants.
2. We get more number of matured plants in a very short time.
3. Thousands of plants can be grown in very short time.
4. This method can help the breeder in preserving the characters he need.
5. It is very easy and economical method for the multiplication of ornamental plants.

Question 48.
What is Mitosis? Which type of cells it occurs in organisms? Write about the different stages of it.
Answer:

1. Mitosis is a method of cell division, in which the nucleus divides into two daughter nuclei.
2. Each containing the same number of chromosomes as the parent nucleus.
3. Mitosis takes place in all body cells which retains same number of chromosomes.
4. Different stages of mitosis:
   1. Prophase
   2. Metaphase
   3. Anaphase
   4. Telophase

10th Class Biology 6th Lesson Reproduction 4 Marks Important Questions and Answers

Question 1.
Explain the changes involved in the formation of seed from Ovule.
(OR)
Pollen grain reached the stigma of a flower. Explain the changes that occurs up to the formation of seeds in a sequence.
Answer:
Process of double fertilization:

1. At the time of fertilization there will be a total of 7 cells arranged in three groups in a mature embryo sac.
2. They are one egg (female garnet) two synergids, one central cell (secondary or polar nucleus) and three antipodals.
3. While all the cells are in haploid (n) condition only the polar nucleus is diploid (2n). This is due to the fusion of two nuclei.
4. The synergids are also known as helper cells.
5. Fertilization is the process of fusion of male and female gametes. For the fusion pollen grains have to reach the surface of the stigma. This is called pollination.
   
6. Pollen grain received by the stigma, germinate and give rise to pollen tubes. The pollen tube has two male nuclei.
7. Usually the pollen tube enters the ovule through microphyle and discharges the two male gametes into the embryo sac.
8. One male nucleus (garnet) approaches the egg and fuses with it to form diploid (2n) zygote this is called first fertilization.
9. The other male nucleus reaches the secondary nucleus (2n) (polar nucleus) and fuses with it to form endosperm nucleus which will be triploid. This is second fertilization. Thus double fertilization occurs in embryosac.
   Changes after double fertilization:
10. After double fertilization, the ovule increases in size rapidly as a result of formation of endosperm tissue by mitosis and the development of new embryo.
11. The embryo consists of cotyledons an epicotyl and a hypocotyl. The cotyledons become greatly enlarged because of stored food for the seedling.
12. The zygote divides several times to form an embryo within the ovule. The ovule develops a tough coat and is converted into a seed. The ovary grows to form a fruit.

Question 2.
Observe the given diagram and answer the following questions.
i) What are the four main parts of a flower?
Answer:
Calyx, Corolla, Androecium and Gynoecium are the main parts of a flower.

ii) Which parts of the flower produces gametes?
Answer:
Androecium and gynoecium produces gametes.

iii) Which parts of the flower help in pollination?
Answer:
Petals or corolla help flower in pollination.

iv) Which part protect the flower during its bud stage?
Answer:
Sepals or calyx protect flower in bud stage.

v) Which part of the flower will turn into a fruit in the future?
Answer:
Ovary of the flower will change into fruit.

Question 3.
Organisms reproduce asexually in many ways. Some of the organisms are given below. Fill the below table based on the collected information about the organism and mode of asexual reproduction in it.
a) Onion b) Spirogyra c) Strawberry d) Ginger e) Honey-bee f) Paramoecium g) Planaria h) Yeast

Name of the organism	Mode of Asexual reproduction

Answer:

Name of the organism	Mode of Asexual reproduction
a) Onion	Bulb
b) Spirogyra	Fragmentation
c) Strawberry	Stolons
d) Ginger	Rhizome
e) Honey – bee	Parthenogenesis
f) Paramoecium	Binary fission
g) Planaria	Regeneration
h) Yeast	Budding

Question 4.
i) Draw a neat labelled diagram of L.S. of flower.
ii) What are the sexual parts in the flower ?
Answer:
i)
ii) A. Androecium or Stamen
B. Gynoecium or Pistil

Question 5.
Read carefully and answer the following questions.

According to Weismann prediction, every organism undergoes two kinds of cell divisions. In Mitosis, there is no change in chromosomal number (2n) and in Meiosis, chromosomal number is reduced to half (n).

i) What does ‘n’ and ‘2n’ indicate?
Answer:
‘n’ indicates haploid state. ‘2n’ indicates diploid state.

ii) In which cells, Meiosis takes place?
Answer:
Meiosis occurs in sex cells during the formation of gametes.

iii) What happens, if chromosomal number is not reduced in Meiosis?
Answer:
The chromosomal number not constant in successive generations.

iv) Which type of cell division occurs in the skin cells?
Answer:
Mitosis

Question 6.
Observe the diagram and answer the following.

i) Which part produce the female gamete?
Answer:
Ovary

ii) Where does the fertilization takes place in female reproductive system?
Answer:
Fallopian tube

iii) Where does the embryo develops until it is ready to born?
Answer:
Uterus

iv) In some cases doctor’s cut and tie the cut ends of the fallopian tubes. What is the name of surgery?
Answer:
Tubectomy

Question 7.
Briefly explain the stages of cell cycle.
Answer:
The process of cell division is called “mitosis”. The period between two cell divisions is called “Interphase”.
This is actually the period when the genetic material makes it’s copy so that it is equally distributed to the daughter cells during mitosis. Interphase can be devided into three phases.
G₁ Phase: This is the linking period between the completion of mitosis and the beginning of DNA replication (GAP-1 Phase). The cell size increase during this period.
S Phase: This is the period of DNA synthesis (Synthesis phase) leading duplication of chromosomes.
G₂ Phase: This is the time between the end of DNA replication and the beginning of mitosis. Cell organells devide and prepare chromosome for mitosis.

Question 8.
i) Draw a labelled diagram of the human male reproductive system.
ii) What is the function of testosterone?
Answer:
i) Male reproductive system:
ii) The function of testosterone hormone is maintaining of secondary sexual chracters in males.

Question 9.
Describe the life cycle of a flowering plant with a help of neat labelled diagrams. (OR) Draw the life cycle of a flowering plant.
Answer:

1. Adult plant produces flowers:
   When the plant matures and is ready to reproduce, it develops flowers. Flowers are special structures involved in sexual reproduction, which includes pollination and fertilisation.
   
2. Pollination: The transfer of pollen grains from the anther of a stamen to the stigma of a carpel is called pollination.
3. Fertilisation:
   i) After pollen grains falls on the stigma fertilization occurs when the male gamete present in pollen grains joins with the female gametes present in the ovule.
   ii) In the ovary the male nucleus of pollen combines with the nucleus of female gamete or egg present to form zygote.
4. Formation of fruit and seed: After fertilisation, a combined cell i.e. zygote grows into an embryo within a seed formed by the ovule.
5. Each seed contains a tiny plant called an embryo which has root, stem and leaf parts ready to grow into a new plant when conditions are favourable.
6. Another part of the flower (the ovary) grows to form fruit, which protects the seeds and helps them spread away from the parent plant to continue the cycle.

Question 10.
Analyze the following information and answer the following questions.

S.No.	Name of the plant	Method of propagation
1.	Mango	Grafting
2.	Rose, Hibiscus	Cutting
3.	Jasmine	Layering
4.	Bryophyllum	Small plants grow on edges of leaves
5.	Colacasia	Cor ms
6.	Onions	Bulbs

i) What do you call the given reproduction methods?
Answer:
Given reproduction methods are called ‘vegetative propagation’.

ii) What is the major difference between sexual reproduction and vegetative reproduction in plants?
Answer:
In sexual reproduction gametes form zygote. Plant parts like root, stem and leaf are used in vegetative reproduction. It is one of asexual method.

iii) Potato plants do not produce seeds. How can you propagate this plant?
Answer:
Potato plants propagates through the ‘eyes’.

iv) What are the advantages of propagating plants with the above given methods?
Answer:
In vegetative propagation

1. More plants are produced in less time
2. Characters are not changed.
3. It would be possible to develop new varieties with useful characters.

Question 11.
Explain the methods of artificial propagation in various plants.
Answer:
Artificial propagation:

1. Cutting: Some plants can grow individually when a piece of the parent plant having bud is cut off from the existing plant. The lower part of this cutting is buried in moist soil.
   
   After few days the cut parts having buds grow as an individual plant after developing roots. E.g. Rose, Hibiscus.
2. Layering: A branch of the plant with atleast one node is bent towards the ground and part of it is covered with moist soil. After a few days new roots develop from the part of the branch buried in the soil. The branch is then cut off from the parent plant.
   E.g: Nerium, Jasmine
   
3. Grafting: Two plants are joined together in such a way that two stems join and grow as a single plant. This technique is very useful in propagating improved varieties of various flowers and fruits. Grafting is used to obtain a plant with desirable character. E.g: Mango, citrus, apple, rose.
   

Question 12.
Observe the following figures and find the stages of cell division and explain.
Answer:
In the mitotic cell division, the division of nucleus (karyokinesis) followed by the division of cytoplasm (cytokinesis). Finally brings about the formation of two daughter cells. There are four stages in mitosis division.
They are

1. Prophase
2. Metaphase
3. Anaphase
4. Telophase

1) Prophase	1) Chromosomes condense and get coiled. They become visible even in light microscope. Nucleoli become smaller. 2) Chromosomes split lengthwise to form chromatids, connected by centromeres. 3) Nuclear membrane disappears. 4) Centrosome, containing rod-like centrioles, divide and form ends of spindle
2) Metaphase	1) Centrosomes move to spindle equator, spindle fibres attached to centromeres.
3) Anaphase	1) Centromeres split, separating the chromatids. 2) Spindle fibres attached to centromeres contract, pulling chromatids towards poles.
4) Telophase	1) Chromatids elongate, replication at this stage to become chromosomes and become invisible. 2) Nuclear membrane form round daughter nuclei. 3) Cell membranes pinches into form daughter cells (animals) or new cell wall material becomes laid down across spindle equator (plants) 4) Nucleus divides into two and division of cytoplasm starts. Two cells are form.

Question 13.
Mention the stages of Mitosis with the help of diagrams. Explain the changes that takes place in Prophase.
Answer:
Mitosis is a method of cell division, in which the nucleus divides into two daughter nuclei each containing the same number of chromosomes as the parent nucleus. Mitosis takes place in all body cells which retains same number of chromosomes.
Different stages of mitosis:
1) Prophase 2) Metaphase 3) Anaphase 4) Telophase

1) Prophase

1. In this phase chromosomes condense and get coiled.
2. They become visible even in light microscope.
3. Nucleoli becomes smaller.
4. Chromosome split lengthwise to form chromatids, connected by centromeres.
5. Nuclear membrane breaks down.
6. Centrosome containing rod like centrioles, divide and form ends of spindle.

Question 14.
Describe the process of double fertilization in plants. Explain the uses of endosperm that is formed.
Answer:
Double fertilization:

1. In flowering plant germinated pollen grain forms pollen tube.
2. The end of the pollen tube ruptures and two male garnets are released in the Embryosac.
3. Out of two male garnets one male garnet fuses with female garnet which is called fertilization.
4. Another male garnet fuses with the secondary nucleus and forms endosperm.
5. So in flowering plant fertilization occures twice hence it is called double fertilization.

Uses of Endosperm:

1. Cotyledons develops by utilizing endosperm.
2. The Cotyledons utilizes the stored food in the endosperm.
3. Some of the plants utilizes the endosperm completely and changes in to seed.
4. Because of the stored food the size of the cotyledons increases.

Question 15.
Explain any two natural and two artificial vegetative propagation methods to produce more number of plants in less time period with examples.
Answer:
Natural propagation:
i) Leaves – Small plant grow at the edge of the leaves. Ex: Bryophyllum
ii) Stems:
a) Stolon – Ex: Jasmine, strawberry b) Bulbs – Ex: Onion
c) Corns – Ex: Colocasia d) Rhizome – Ex: Ginger e) Tuber – Ex: Potato
iii) Root – Ex: Roots of murayya, guava
Artificial propagation:
Cutting: Some plants can grow individual when a piece of parent plant having bud is cut off from the existing plants. Ex: Rose, Hibiscus.
Layering: A branch of the plant with at least one node is bent towards the ground and a part of it is covered with moist soil leaving the tip of the branch exposed above the ground. Ex: Nerium, Jasmine.
Grafting: Two plants are joint together in such a way that stems join and grow as a single plant one which is attached to soil is called stock and stem of another plant without roots is called scion. Both stock and scion are tied with a twine thread and cover by a polythene cover. Ex: Mango, citrus, apple, rose.

Question 16.
Read the following table and answer the following questions.

SI. No.	Structure	Location
1.	Tricuspid valve	Right auriculo-ventricular aperture
2.	Guard cells	Epidermis of leaves
3.	Glomerulus	Nephron
4.	Alveoli	Lungs
5.	Acrosome	Above the head of a sperm.

i) Name the structure concerned to the heart.
Answer:
Tricuspid valve

ii) What is the function of acrosome?
Answer:
It helps the sperm in penetrating into ovum.

iii) Name the structures which are helpful for gaseous exchange.
Answer:
Alveoli and guard cells

iv) Name the part performing Excretion.
Answer:
Glomerulus

Question 17.
a) Draw a neat and labelled diagram of Human female reproductive system.
b) What happens when the Fallopian tubes are closed?
Answer:
a) Female reproductive system
b) If fallopian tubes are closed the sperm can not reach the ova, fertilization will not happen and zygote will not form.

Question 18.
Observe the following table.

Reproduction system	Organisms
Fission	Paramoecium, Bacteria
Budding	Yeast, Hydra
Fragmentation	Flatworms, Spirogyra
Rhizome	Ginger, Turmeric
Cutting	Rose, Hibiscus
Grafting	Citrus, Apple

On the basis of information given in the table write- the answers to the following questions.
i) Write the names of two organisms that show Asexual reproduction.
Answer:
Yeast, Hydra, Bacteria, Paramoecium (any two you may write)

ii) Write two artificial vegetative propagation methods mentioned in the table.
Answer:
Cutting, Grafting

iii) Write the names of two plants, which undergo natural vegetative propagation mentioned in the table.
Answer:
Ginger, Turmeric

iv) In fission, how many organisms can we get from one organism?
Answer:
Two

Question 19.
Among the following organisms can we see asexual reproduction? Write about the method of asexual reproduction in any of the two organisms.
Answer:
а) Paramoecium b) Yeast c) Spirogyra d) Amoeba e) Planaria
Yes, we can see asexual reproduction in all the following organisms.

Method of asexual reproduction – Organism
Binary fission                                – Paramoecium, amoeba
Budding                                        – Yeast
Fragmentation                              – Spirogyra
Regeneration                                – Planaria

1) Binary fission in Paramoecium: A single cell divides into two equal daughter cells. First the cytoplasm divides into two parts followed by nuclear division.
2) Asexual reproduction in Yeast: Budding is the common method of asexual reproduction in yeast. In this method, yeast cell wall at a particular region becomes soft and bulges into an outgrowth called bud. Cytoplasm enters into this bulge and then nucleus divides mitotically into two nuclei, one moves into the bud. Finally bud is detached from the parent cell and grows into an independent yeast cell.

Question 20.
See the adjacent picture. Which type of pollination will occur in this ? Why do you think so?
Answer:

1. Self-pollination occurs if stamens and carpels matures at the same time.
2. If they mature at different times, cross pollination occurs.
3. Cross pollination occurs in this plant.
4. For cross pollination the pollen grains are carried from other plants belonging to the same species.
5. The mechanism of dispersal of pollen grains from one plant to other plant is facilitated mostly by wind and insects.
6. Cross pollination is believed to be advantageous for the plant.
7. The seeds produced by the flower will contain another source of genetic material
8. Which may contain genes which are advantageous to the survival of the seedlings.

Question 21.
What are the consequences if meiosis do not happen in the body cells of the organism?
Answer:

1. Each organism has a fixed number of chromosomes.
2. This number has to be maintained in its offspring.
3. Any sudden change in the number of chromosomes will be harmful to the offspring. Assume parent has 10 chromosomes.
4. In the absence of meiosis during sexual reproduction gametes will also have the same number of chromosomes as parent i.e., 10 chromosomes.
5. Union of female and male gametes occur forming zygote during sexual reproduction. The number of chromosomes doubled in zygote will have 10+10 chromosomes.
6. In the next generation, the offspring will have forty chromosomes. If this continues cells in the offsprings will have thousands of chromosomes within few generation.
7. This results in formation of abnormalities in each generation. Hence by way of meiotic division, the chromosome number is maintained constant from generation to generation.

Question 22.
Describe different artificial vegetative methods to produce large scale production of plants.
Answer:

1. Different artificial vegetative propagation methods are cutting, layering, grafting and tissue culture methods.
2. Cutting: Some plants grow individually when a piece of parent plant having bud is cut from the existing plant. After burying in the soil the cut parts having buds grow as an individual plant after developing roots. E.g. Rose.
3. Layering: A branch of the plant with at least one node is bent towards the ground and part of it is covered with moist soil. After sometime, new roots develop from the part of the branch hurried in the soil. The branch is then cut off from the parent plant. E.g: Nerium.
4. Grafting: Two plants are joined together in such a way that two stems join and grow as a single plant. This technique is very useful in propagating improved vari¬eties of various flower and fruits. Grafting is used to obtain a plant with desirable character. E.g: Mango, citrus, apple, rose.
5. Tissue culture: In this method, few plant cells or plant tissues are placed in a growth medium with plant hormones in it and it grows into new plants. Thousands of plants can be grown in very short interval of time.

Question 23.
i) Labelled parts of A, B, C, D above drawn Human female reproductive system.

ii) In which part fertilization takes place?
iii) Which part is in connection with implantation?
iv) What is ovulation?
Answer:
i) A: Fallopian tube
B: Ovary
C: Uterus
D: Vagina
ii) Fertilization takes place in fallopian tube.
iii) Uterus
iv) Release of ovum from graffian follicle of ovary is known as ovulation.

Question 24.
Write some programmes conducted by you to bring awareness in the people about health and hygeine and family planning?
Answer:

1. Organising Health camps on World Health day to people of the village.
2. Conducting immunisation programs for every three months.
3. Supplying tablets on the deworming day.
4. Organising seminars by expert doctors on individual health and cleanliness programs.
5. Propagating small family norms conducting camps for family planning operations.
6. Educating the masses through pamplets on the needs of taking balanced diet.
7. Need of using toilets and washing hands and legs before and after meals.
8. Educating the people by conducting adult education centres. This is basically required for enlightening the people on health aspects.

Question 25.
Government made an act on determining sex through ultrasound scanning and telling it as crime. What do you do to tell this to others?
Answer:

1. I will educate people knowing the sex of foetus inside mother’s womb is a severe crime as per the act made by government.
2. The purpose of ultrasound tests are to know the growing condition of the foetus and also to see whether it is suffering with severe ailments.
3. By knowing the sex of the foetus, if it is female people are ready for aborting it.
4. This leads to reduction in male female ratio in the country.
5. Children either male or female are equal to parents.
6. We should see proper development of girl child after her birth.

Question 26.
Write about the embryonic membranes that nourish, protect and support to the embryo?
Answer:

1. The growing embryo form two membranes – Chorion and Amnion.
2. Chorion establishes connection with the walls of the uterus and helps in the supply of nutrients to the embryo and in the removal of wastes from the embryo.
3. Amnion forms a sac like structure around the embryo and amniotic fluid is present between layers of Amnion.
4. Amnion and Amniotic fluid give protection to the embryo against mechanical shocks.
5. Placenta is a tissue formed around 12 weeks of pregnancy by the cells from the embryo and mother.
6. Embryo receives all the required nutrients and oxygen for its metabolism from the mother through the blood vessels present in the placenta.
7. Another membrane called allantois, which originates from the digestive canal of the embryo forms the major part of a tube like structure called umbilical cord.
8. Umbilical cord contains very important blood vessels that connect the embryo with the placenta.

Question 27.
Write brief history of cell division.
Answer:

1. In 1852 a German scientist, Robert Remak published his observations on cell division and stated that the binary fission of cells was the means of reproduction of animal cells.
2. This view was widely publicized by Rudolf Virchow who gave the phrase “Omnis cellulade cellula” means all cells arising from pre existing cells.
3. In 1879 Walther Flemming reported that there were string like structures in the nucleus which split longitudinally during cell division. He named the process as mitosis means fine threads as the dividing structures resembled threads.
4. Wilhelm Roux proposed that each chromosome carried a different set of heritable elements and suggested that the longitudinal splitting observed by Flemming ensured the equal division of these elements.
5. Combined with the rediscovery of Gregor Mendel’s 1866 paper on heritable elements in peas, these results highlighted the central role of the chromosomes in carrying heritable material or genetic material.
6. The scheme of mitotic division was confirmed in 1904 by Theodor Boveri.
7. The chemical nature of the genetic material was determined in a series of experiments over the next fifty years.
8. The structure of DNA – the constituent of the genetic material was determined in 1953 by James Watson and Francis Crick.

Question 28.
Explain briefly about child birth. (OR) How child birth occurs after gestation period?
Answer:

1. Total time required for the embryonic and foetal development is about 9 months or 280 days.
2. After this time, foetus is expelled from the uterus by the mother. This is child birth.
3. Child birth is a complicated process and involves the participation of child and mother.
4. The foetal hormones produced inside, stimulate the contraction of the muscles present in the walls of uterus.
5. These contractions called labour pains help in the expulsion of the foetus from the uterus.
6. During this process the amnion ruptures, placenta is separated from the walls of J the uterus.
7. At child birth the head usually comes out first.
8. The foetus is still attached to the mother’s uterus through the umbilical cord, which is later separated by the doctors.

Question 29.
Draw the life history of flowering plant in the form of block diagram.
Answer:
Life history of a flowering plant:

Question 30.
In a flower self fertilization takes place. Write the process, the flower organs which involve in self fertilization.
Answer:

1. Fusion of male and female gametes produced by the same individual is called self fertilization.
2. Self ferlization occurs in bisexual flowering plants.
3. The flower organs which involve in self fertilization are stamens (androecium) and carpels (Gynoecium).
4. Majority of flowering plants have an embryo sac consisting of seven cells and eight nuclei.
   
5. The pollen grains produced by anther of stamen are transferred to the stigma of the same flower by wind or insects.
6. The stigma of the carpel secretes a sticky substance which promotes the growth of pollen grains.
7. Under favourable conditions pollen grains germinate on the stigma and give rise to pollen tubes.
   Only one pollen tube finally reaches the embryo sac.
8. This pollen tube will have two male nuclei, which migrate to the tip of the pollen tube at the time of fertilization. Usually the pollen tube enters the ovule through micropyle and discharges the two male gametes into its embryo-sac.
9. One male nucleus (gamete) approaches the egg and fuses with it to form a diploid zygote. This is first fertilization.
10. The other male nucleus reaches the secondary nucleus (2n) and fuses with it to form the endosperm nucleus which will be triploid. This is second fertilization in the embryo sac.
11. Thus double fertilization occurs in embryo sac which is unique in flowering plants.

Question 31.
Describe the structure of flower with a neatly labelled diagram.
Answer:

1. A typical flower consists of an outer whorl of green sepals (calyx) which protects the parts with in.
2. The second whorl has petals (corolla) which are usually brightly coloured. They sometimes emit fragrance also.
3. Petals are soft and are useful to attract insects to facilitate cross pollination.
   
4. The third whorl of the flower consists of stamens (Androecium) which are the male reproductive organs.
5. Each stamen is made up of a filament and an anther.
6. Each anther usually has two anther lobes. The anther produces pollen grains (microspores).
7. The inner most fourth whorl is gynoecium or pistil. It consists of ovary, style and stigma.
8. Ovary occupies central portion on the thalamus. A swollen ovary is present on the thalamus.
9. Inside the ovary future seeds, known as ovules are present.
10. Ovary has a pipe like extension called style. The tip of the style ends in stigma. The stigma receive the pollen grains.

Question 32.
Write a brief note on male reproductive system of human beings.
Answer:

1. The male reproductive system of human beings consists of a pair of testis, accessory glands and a system of ducts.
2. Testis are male reproductive organs and produces spermotozoa or sperms and also secretes male sex hormone Testosterone.
   
3. Inside each testis several lobules are present. Each lobule has several tubules called seminiferous tubules.
4. Germinal epithelial cells in the seminiferous tubules undergo meiotic division to produce sperms.
5. The accessory glands include one prostrate gland and two cowper glands. Secretion of these glands produce semen.
6. The duct system consists of vasa efferentia.
   They collect spermatozoa from seminiferous tubules.
7. Vasefferentia continue as epididymis where sperms are stored temporarily.
8. From epididymis sperms moved into tubule called vas deference and then into urethra.

Question 33.
Describe the female reproductive system in human beings.
Answer:

1. A pair of ovaries, oviducts, uterus and vagina are the parts present in female reproductive system.
2. Ovaries are present just below the Kidneys in the abdominal cavity.
3. Each ovary has several sac like structures called ovarian follicles or Graffian follicles.
   
4. Every time only one follicle matures and release one ovum into the body cavity.
5. Ovaries secrete two female sex hormones called oestrogen and progesterone which control the development of female reproductive organs, ovulation and menstruation.
6. Just above the ovaries are the tubes called oviducts or fallopian tubes where fertilisation takes place.
7. The two oviducts connect to a bag like organ called uterus at their other ends.
8. The uterus is connected through a narrow opening called cervix to another tube called vagina which opens to the outside of the body.
9. Vagina is a tubular structure and is also called birth canal because it is through this passage that the baby is born after the completion of development inside the uterus of the mother.

Question 34.
Describe briefly about the reduction division or meiosis.
(OR)
Why meiosis is also known as reduction division? Comment on it.
Answer:

1. Meiosis occurs only during the formation of gametes in sexual reproduction.
2. During meiosis only one set of chromosomes are passed on to the daughter cells. Hence daughter cells have half the number of chromosomes of the mother cells.
   
3. In meiosis karyokinesis and cytokinesis occur two times.
4. During first phase of meiosis the parent cell divides twice, though the chromosomes divide only once.
5. The second phase meiosis is similar to normal mitosis, but chromosomes do not duplicate more over the chromosome number distributed equally to each cells.
6. Thus the four daughter cells have just half the number of chromosomes of the parent cells.
7. These are haploid (containing only one set of chromosomes).
8. Thus meiotic division is also called reduction division.

Question 35.
Describe the developmental stages of human embryo after fertilization with the help of neatly labelled diagrams.
Answer:

1. During fertilization, chromosomes of the ovum and the chromosomes of the sperm make up into pairs and the resulting cell is called zygote.
2. Fertilization takes place in the oviduct or fallopian tube.
   
3. The zygote which is diploid travels down the fallopian tube. As it moves it undergoes several mitotic divisions forming the embryonic stage called blastocyst.
4. Blastocyst moves towards the wall of the uterus and finally gets attached and embedded in the wall of the uterus. This is called implantation.
5. The growing embryo forms two membranes Chorion and Amnion.
6. Chorion establishes connection with the walls of the uterus and helps in the supply of nutrients to the embryo and removal of wastes from the embryo.
7. Amnion forms a sac like structure around the embryo. The space between the amnion and embryo is filled with a fluid called amniotic fluid.
8. Amnion and amniotic fluid give protection to the embryo against minor mechanical injury.
9. Placenta is a tissue formed by the cells from the embryo and the mother. It is formed around 12 weeks of pregnancy.
10. Placenta nourishes the growing embryo.
    
11. A tough cord called umbilical cord is also formed by the embryo which is connected to the walls of the uterus through the placenta.
12. From 3 months of pregnancy, the embryo is called foetus.
13. Pregnancy lasts on an average 9 months or 280 days. This period is called gestation period.
14. After this time foetus is expelled from the uterus by the mother – this is child birth.
15. This process is complicated and involves the participation of foetus and mother.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 1 Heat.

AP State Syllabus SSC 10th Class Physics Important Questions 1st Lesson Heat

10th Class Physics 1st Lesson Heat 1 Mark Important Questions and Answers

Question 1.
What is humidity? (AP March 2015)
Answer:
Humidity :
The amount of water vapour present in air is called humidity.

Question 2.
Define latent heat of Fusion. (AP Morch 2016)
Answer:
Latent heat of Fusion :
At constant temperature, the heat energy required to convert one gram of solid completely into liquid is called latent heat of Fusion.

Question 3.
While drinking water, Rama spilled some water on the floor. After some time, the water disappeared from the floor. What happened to the water? (TS June 2015)
Answer:

• The water disappeared due to evaporation because we know that as the surface area increases rate of evaporation also increases.
• So water molecules escape from the floor to air.

Question 4.
Give an example to explain that evaporation is a cooling process. (TS March 2016)
Answer:
The examples to explain that evaporation is a cooling process are

1. Drying of wet clothes
2. When the floor is washed with water, the water on the floor disappears.
3. Sweating, etc.

Question 5.
Let heat is not lost by any other process between two objects in thermal contact, “Net heat lost (by hot body) = Net heat gain (by cold body).” above statement indicates a principle. Write the name of that principle. (AP March 2019)
Answer:
Principle of method of mixtures.

Question 6.
Convert 25°C into Kelvin scale. (AP SCERT: 2019-20)
Answer:
25°C = (273 + 25) K = 298 K

Question 7.
Given a beaker with water, a thermometer and a stand, draw the arrangement of an experiment to measure boiling point of water. (AP SA-1:2019-20)
Answer:

Question 8.
Define heat.
Answer:
Heat: Heat is a form of energy which is transferred from one body to the other body due to the difference in their temperature.

Question 9.
What is meant by thermal equilibrium?
Answer:
Thermal Equilibrium :
Two bodies are at the same temperature then they are said to be in thermal equilibrium.

Question 10.
Define dew.
Answer:
Dew :
The water droplets condensed on surface are known as dew.

Question 11.
What is boiling?
Answer:
Boiling is a process in which the substance changes from liquid to gas.

Question 12.
What is melting?
Answer:
Melting :
The process in which the substance changes from solid to liquid state is called melting.

Question 13.
What are the different energies possessed by system (body or material)?
Answer:

1. Linear kinetic energy
2. Rotational kinetic energy
3. Vibrational energy
4. Potential energy and Internal energy (I.E).

Question 14.
Why does samosa seem to be cool but hot when we eat?
Answer:
Because the curry inside samosa contains ingredients with higher specific heats.

Question 15.
On which factors does rate of evaporation depend?
Answer:

1. Surface area
2. temperature
3. the amount of vapour already present in the surrounding air.

Question 16.
What is the value of latent heat of vapourization of water?
Answer:
Latent heat of vapourization of water = 540 cal/gm. (or) 2.26 × 10⁶ J/kg

Question 17.
What is the value of latent heat of fusion of ice?
Answer:
Latent heat of fusion of ice is 80 cal/gm. (or) 3.26 × 10⁵ J/kg

Question 18.
Give some liquids which solidify (convert into solid) in winter season.
Answer:
Coconut oil, ghee are some liquids which solidify in winter season.

Question 19.
What is freezing?
Answer:
Freezing :
The process in which the substance changes from liquid to solid state by losing some energy from it is called freezing.

Question 20.
Which will have lower temperature when we take out a wooden piece and a metal piece from a fridge?
Answer:
The metal piece will have lower temperature as compared to the wooden piece when they are taken out of the fridge.

Question 21.
When do you say there is thermal equilibrium between two bodies?
Answer:
It is said that there is thermal equilibrium between two bodies when there is no transfer of heat energy between them.
(OR)
When temperature between two bodies is same it is said that there is thermal. equilibrium between them.

Question 22.
What is absolute temperature?
Answer:
Temperature measured in Kelvin scale is called absolute temperature.

Question 23.
What is latent heat of vapourisation?
Answer:
At constant temperature the heat energy required to change one gram of liquid into gaseous state.

Question 24.
What is boiling point?
Answer:
The temperature at which the substance changes from liquid to gaseous state at the fixed temperature is called boiling point.

Question 25.
What is melting point?
Answer:
Melting point :
The temperature at which the substance changes from solid to liquid state at constant temperature is called melting point.

Question 26.
How is aquatic animal able to live at poles?
Answer:
The ice has less density compared to water. So it forms a layer on the top of water which prevents the solidification of water.

Question 27.
What are the phases of water present at 0° C?
Answer:
Two phases namely, ice and water.

Question 28.
What happens if external pressure of liquid increases?
Answer:
The boiling point of the liquid will increase.

Question 29.
Does ice melt below 0° C?
Answer:
Yes, if the external pressure increases it melts at low temperature.

Question 30.
What happens when two objects of same temperature are in contact with each other?
Answer:
Heat does not flow between two objectives.

Question 31.
What is the principle involved in pressure cooker?
Answer:
Boiling point of liquid increases with external pressure.

Question 32.
What happens to kinetic energy of particles if we increase the temperature?
Answer:
Kinetic energy of particles increases with increase of temperature.

Question 33.
Why does transfer of heat energy take place between systems?
Answer:
When heat energy gives to the system, internal energy increases. Similarly, internal energy decreases when heat energy flows out of the system.

Question 34.
What is internal energy?
Answer:
Internal energy :
It is the energy possessed by the system by virtue of its molecular motion and molecular configuration. It is a stored energy. It depends on the temperature of the system.

Question 35.
What is transit energy?
Answer:
Transit energy:
Energy possessed by a system which can cross its boundary is called transit energy. Heat and work are transit energies.

Question 36.
Where does air get? vapour from?
Answer:
The vapour may come from evaporation of water from the surfaces of rivers, lakes, ponds, and from the drying of wet clothes, sweat, and so on.

Question 37.
Why do pigs toil in the mud during hot summer days?
Answer:
They do not have sweat glands for evaporation process. So pigs toil in the mud.

Question 38.
Why is it easy to cook food in a pressure cooker?
Answer:
We know as the atmospheric pressure increases the boiling point of water increases. So we can increase the boiling point of water to 120°C in a pressure cooker. So it is \ easy to cook in a pressure cooker.

Question 39.
Why is water used as coolant?
Answer:
Water has the highest specific heat. So it takes lot of time to become hot. So it is used as coolant.

Question 40.
How is fog formed?
Answer:
The water molecules present in vapour condense on the dust particles in air and form small droplets of water which form a thick mist called fog.

Question 41.
Equal amounts of water is kept in a cup and in a disc. Which will evaporate faster? Why?
Answer:
The water present disc evaporates faster because of greater surface area.

Question 42.
Explain why dogs pant during hot summer days using the concept of evaporation.
Answer:
Dog does not have pores on its body. The only place where a dog can sweat is on its foot pads and the rest of the body is covered in a fur coat. So it cannot sweat; that’s why dogs pant to keep cool themselves.

Question 43.
Same amount of heat is supplied to two liquids A and B. The liquid A shows a greater rise in temperature. What can you say about the specific heat of A?
Answer:
The specific heat of A is less than that of B because rise in temperature is inversely proportional to temperature.

Question 44.
What is the specific heat of-water at boiling point?
Answer:

1. Generally, the specific heat of water is 1. (or) 4.187 KJ / Kg K
2. Specific heat of water at 100° C = 4.219 KJ / KgK

Question 45.
What is the equation of heat energy when change the state?
Answer:
Q = mL
Where m = mass of body, L = latent heat.

Question 46.
Convert 212°F into Kelvin scale.
Answer:
212°F= 100°C. So 100 + 273 = 373 K.

Question 47.
Convert 310 K into centigrade system.
Answer:
310-273 = 37°C.

Question 48.
Are the processes of evaporation and boiling the same?
Answer:
No. Evaporation takes place at any temperature, while boiling occurs at a definite temperature called the boiling point.

Question 49.
Define latent heat of vaporization?
Answer:
The heat energy required to change one unit mass of liquid to gas at constant temperature is called latent heat of vaporization.
\(L=\frac{Q}{m}\)
The value of latent heat of vapourization of water is 540 cal/gm.

Question 50.
What is meant by internal energy?
Answer:
Combination of linear kinetic energy, rotational kinetic energy, vibrational energy, and potential energy of molecules is known as internal energy of the system.

Question 51.
Write the formula for resultant temperatures of a mixture, when V₁ ml of water at T₁°C is mixed with V₂ ml of water at T₂° C.
Answer:
Resultant temperature \(T=\frac{V_{1} T_{1}+V_{2} T_{2}}{V_{1}+V_{2}}\)

Question 52.
Write the equation of heat energy when change the temperature.
Q = mS∆T
m == mass, S = specific heat, AT = change in temperature

Question 53.
The figure shows change in state of ice from – 5°C to 110°C with temperature. What are the melting and vaporization curves?

Answer:
BC = melting
DE = vaporization curve.

Question 54.
Write principle of method of mixtures.
Answer:
When two or more bodies are brought into thermal contact, then heat lost by hot body is equal to heat gain by cold body. Until they attain thermal equilibrium.

Question 55.
Evaporation is a cooling process. Why?
Answer:
During evaporation process, the energy of the molecules inside the liquid decreases and they slow down.

Question 56.
Which factors are influence the rate of evaporation of a liquid?
Answer:
Rate of evaporation of a liquid depends on surface area, temperature, pressure, and amount of vapour present in surrounding air.

Question 57.
What is meant by fog?
Answer:
The droplets keep floating in the air and form a thick mist which restricts visibility. This thick mist is called fog.

Question 58.
Why do we sweat while doing a work?
Answer:
When we work our body produces heat. As a result, the temperature of the skin becomes higher and the water in the sweat glands starts evaporating. This evaporation cools the body.

Question 59.
A samosa appears to be cool when touched outside but it is hot when we eat it. Why?
Answer:
A samosa appears to be cool outside but it is hot when we eat it because the curry inside the samosa contain ingredients with higher specific heats. Hence they remain hot for a long time.

Question 60.
Equal amounts of water are kept in a cup and in a dish. Which will evaporates faster? Why?
Answer:
Dish evaporates faster, because dish has large surface area. Evaporation of liquids depends on surface area.

Question 61.
Why water is used as coolant in the cooling system of automobile engines?
Answer:
Due to high specific heat, water absorbs large amount of heat and temperature does not rise quickly. So water used as coolant in cooling system of automobile engines.

Question 62.
Why do pigs toil around in the mud?
Answer:
Pigs do not have sweat glands. Water in the mud evaporates and helps the pig to be cool from heat. So pigs toil in the mud during summer.

Question 63.
Take small glass bottle with a tight lid. Fill it with water completely without any gaps and fix the lid tightly in such a way that water should not come out of it. Put the bottle into the deep freezer for a few hours. Take it out from the fridge. You observe the glass bottle is broken. Why?
Answer:
We know, the volume of the water poured into the glass bottle is equal to the volume of the bottle. When the water freezes to ice, the bottle is broken. Because the volume of the ice is greater than the volume of the water filled in bottle.

Question 64.
From the given figure, in which the thermometer mercury level is increases and decreases?

Answer:
Thermometer A(in oil) is increases.
Thermometer B(water) is decreases.

Question 65.
What are the materials are used in to find the specific heat of solid?
Answer:
Calorimeter, thermometer, stirrer, water steam heater, wooden box, and lead shots.

Question 66.
What is the value of following temperatures in Kelvin scale?
(a) 30° C b) 70° C
Answer:
a) 30° C = 30 + 273 = 303 K
b) 70° C = 70 + 273 = 343 K

Question 67.
How much heat energy is required to raise the temperature of unit mass of material by 1° C?
Answer:
1 cal/g – °C = lk cal / kg – K = 4.2 x J/kg – K = 4.2 kJ/kg – K.

Question 68.
How much energy is required to turn 1 g of ice of 0°C into 1 gm of water at 0°C?
Answer:
The energy required to convert 1 g of ice at 0°C into lg of water at 0°C is latent heat of fusion that is 80 cal/g.

Question 69.
What is the temperature of mixture if 10 g of steam at 100°C is mixed with 50 g of ice at 0°C?
Answer:

Question 70.
Boiling water at 100°C and cold water at t°C are mixed in the ratio of 3 : 5 and the resultant temperature is 40°C. Find the value of t.
Answer:
Suppose the quantities of water is 3x and 5x.
Given that

Question 71.
What amount of ice can be melted by 4000 cal of heat?
Answer:
Latent heat of fusion of ice L_f = 80 cal/g
Given that Q = 4000 cal
Q = mL_f ⇒ 4000 = m × 80
∴ m = \(\frac{4000}{80}\) = 50 g

Question 72.
5 gm of ice is at (J°C. It is converted into water at same temperature. How much heat energy is required?
Answer:
In change the state Q = mL
m = 5 gm, L = Latent heat of fusion = 80 Cal/gm
Q = mL = 5 × 80 = 400 cal.

Question 73.
What would be the final temperature of mixture 50 g. of water at 20°C and 50 gm of water at 40°C?
Answer:
If masses are equal, then resultant temperature of mixture = \(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}\)
∴ T = \(\frac{20+40}{2}\) ⇒ T =30

10th Class Physics 1st Lesson Heat 2 Marks Important Questions and Answers

Question 1.
Observe the following table regarding the values of specific heat of substances and answer the following questions : (AP SA-I: 2018-19)

i) Which material is suitable as the base of the cooking vessel?
Answer:
Copper. Because, it has low specific heat value.

ii) Why do we prefer water as a coolant?
Answer:
Due to high specific heat value of water, it can take more time to heat itself and acts as a coolent.

Question 2.
Why do water drops (dew) form on flowers and grass during morning hours of winter season? (AP March 2016)
Answer:
During winter nights, the atmospheric temperature goes down. The flowers, grass, etc. become still colder. The air near them becomes saturated with vapour and condensation begins. The water droplets condensed on such surfaces are known as dew.

Question 3.
Temperatures of two cities at different times are given as follows : (AP March 2019)

On the basis of above table, answer the following questions.
1) In which city, the morning temperature at 6 O’clock is relatively high?
Answer:
In ‘B’ city, the morning temperature at 6 O’clock is relatively high.

2) At what time, both cities are having the equal temperature?
Answer:
At 11 : 30 AM, both cities are having the equal temperature.

Question 4.
A student took the same quantity of water and petrol in two different tumblers. He kept them on a table. When observed after a day there was water in a glass but petrol was completely evaporated. Give reasons why water was not evaporated completely but petrol was completely evaporated.
Answer:
Petrol has a lower vapour point than water. And evaporation depends on the nature of the liquid so petrol evaporates quickly than water at room temperature.

Question 5.
What are the differences between dew and fog?
Answer:

Dew	Fog
1) Dew is the droplets that appear on the exposed objects in the morning or evening.	1) Fog is nothing but cloud on ground.
2) Dew does not effect visibility.	2) Due to fog visibility is greatly effected.
3) Dew is formed when relative humidity higher than temperature.	3) Fog is formed when island area is warmer than the ocean or large body of water.

Question 6.
Why is spirit evaporated in petri dish quickly under a fan when compared to that kept in closed room?
Answer:

• The blowing air increases the rate of evaporation.
• This is because any molecule escaping from the surface is blown away from the vicinity of the liquid.
• This increases the rate of evaporation.
• This is the reason why the spirit in petri dish evaporates quickly when compared to that kept in closed room.

Question 7.
Does the temperature of water rise continuously if heat is supplied continuously?
Answer:
Yes. If heat is supplied to water its temperature rises continuously till it reaches 100°C. At 100°C there would be no further rise of temperature, because the heat is sterilized to convert water to water vapour. So if heat is supplied beyond 100°C, all the water is converted into vapour.

Question 8.
Why does the mercury level of thermometer rise up when it is placed in hot water and fall down when it is placed in cold water?
Answer:

• We know that bodies which are in contact achieve thermal equilibrium due to transfer of heat energy.
• If we keep thermometer in hot water, its mercury level rises because heat is transformed from hot body to cold body.
• Similarly, we observe that mercury level comes down when it is placed in cold water.

Question 9.
When we place thermometer in hot water, there is a rise in mercury level, thereafter it stops. What is the reason for steadiness of mercury level? What does reading of thermometer give at that time?
Answer:

• The steadiness of the mercury column of the thermometer indicates that, flow of heat between the thermometer liquid (mercury) and water has stopped and thermal equilibrium has been attained between the water and thermometer liquid.
• The reading of thermometer gives thermal equilibrium state that is temperature.

Question 10.
What is the relationship between temperature and kinetic energy?
Answer:

• The average kinetic energy of molecules/particles of the hotter body is more than the colder body.
• So we can say that the temperature of a body is an indicator of the average kinetic energy of molecule of that body.
• So the average kinetic energy of molecules is directly proportional to the absolute temperature. [KE_avg ∝ T]

Question 11.
What is the relationship between rise in temperature and specific heat of material?
Answer:

• Temperature depends on nature of the material, hence the specific heat depends on its nature.
• If the specific heat, is high, the rise in temperature is low. [Q = mSΔt]
• It gives us an idea of degree of reluctance of a material to rise in temperature.

Question 12.
What is the principle of method of mixtures?
Answer:
When two or more bodies at different temperatures are mixed with each other, then net heat lost by the hot bodies, is equal to net heat gained by the cold bodies until they attain thermal equilibrium or equal temperature.

Net heat lost = Net heat gained

This is known as principle of method of mixtures.

Question 13.
Why is evaporation of a liquid faster under a fan?
Answer:

• If air is blown over the liquid surface in an open petri dish, a number of molecules evaporate from the surface of liquid.
• Because any molecule escaping from the surface is blown away from the vicinity of liquid.
• This increases the rate of evaporation.
• So evaporation of a liquid is faster under a fan.

Question 14.
Why do we get dew on the surface of a cold soft drink bottle kept in open air?
Answer:

• The temperature of surrounding air is higher than the temperature of cold soft drink.
• Air contains water molecules in the form of water vapour which tend to condense by losing their kinetic energy to form water droplets.

Question 15.
Why does water take more time to become hot and take more time to become cool?
Answer:
Water has the highest specific heat among all liquids. So it takes more time to become hot and takes more time to become cool.

Question 16.
What is calorimeter? Name the material of which it is made of. Give two reasons for using the material stated by you.
Answer:

• The vessel used for measurement of heat is calorimeter.
• It is made of thin sheet of copper.
• The reason is that the specific heat of copper is low and by making the vessel thin, its thermal capacity becomes low so that it takes a negligible amount of heat from its contents to attain the temperature of contents.

Question 17.
Why are burns caused by steam at 100°C more painful than that of water at 100°C?
Answer:

• Water at 100°C takes additional heat energy to convert from liquid state to vapour (steam) state. This energy is called latent heat of vapourisation.
• Hence, steam at 100°C contains more heat energy than that of water at 100°C.
• So, burns caused by steam at 100°C are more painful than that of water at 100°C.

Question 18.
Why is cooking fast in a pressure cooker compared to open vessel?
Answer:

• Boiling point of liquid increases with external pressure.
• Pressure cooker gives external pressure to the water in it.
• So, while cooking in pressure cooker it increases boiling point of water more than 100°C.
• So that, cooking is fast in pressure cooker compared to open vessel.

Question 19.
What happens to the water when wet clothes dry?
Answer:
When wet clothes dry, the water molecules from wet clothes, after evaporation, change into water vapour and mix with water molecules present in surrounding air, in the form of water vapour.

Question 20.
Why do we use hot water bottles for fomentation?
Answer:
The reason is that water does not cool quickly due to its large specific heat, so hot water bottle provides heat energy for fomentation for a long time.

Question 21.
Why do all plants and animals have a high content of water in their bodies?
Answer:
All plants and animals have nearly 80% to 90% of water in their bodies. So it helps in maintaining the body temperature in all seasons due to high specific heat.

Question 22.
Water is used as an effective coolant. Give reason.
Answer:
1) Water is used as an effective coolant because by allowing water to flow in pipes around the heated parts of machine, heat energy from such parts is removed (e.g. radiators in car and generator are filled with water).
2) Water in pipes extracts more heat from surroundings without much rise in its temperature because of its large specific heat.

Question 23.
Why is the base of cooking pan made thick?
Answer:

• By making the base of cooking pan thick, its thermal capacity becomes large and it imparts sufficient heat energy at a lower temperature to the food for its proper cooking.
• Further, it keeps the food warm for a long time, after cooking.

Question 24.
Water in lakes and ponds in cold countries does not freeze all at once. Give reason.
Answer:

• The latent heat of fusion of ice is sufficiently high.
• So to freeze water, a large quantity of heat has to be withdrawn, hence it freezes slowly and thus keeps the surroundings moderate.

Question 25.
Why do drinks get more quickly cooled by adding pieces of ice at 0°C than ice-cold water at 0°C?
Answer:

• This is because 1 g of ice at 0°C takes 336 J of heat energy from the drink to melt into water at 0°C.
• Thus drink loses an additional 336 J of heat energy for 1 g of ice at 0°C than for 1 g ice cold water at 0°C. Therefore cooling produced by 1 g of ice at 0°C is more than that by 1 g of water at 0°C.

Question 26.
When ice in a frozen lake starts melting, its surroundings become very cold? Why?
Answer:
The reason is that the heat energy required for melting the frozen lake is absorbed from the surrounding atmosphere. As a result, the temperature of surroundings falls and it becomes very cold.

Question 27.
Why is it more cold after the hail-storm than during or before the hail-storm?
Answer:
The reason is that after the hail-storm, the ice absorbs the heat energy required for melting from the surroundings, so the temperature of the surroundings falls further down and we feel more cold.

Question 28.
Which of the substances A, B, and C has the least specific heat? The temperature versus time graph as shown below.
Answer:

The substance ‘A’ has least specific heat because the rise in temperature is more for substance. We know that specific heat and rise in temperature are inversely proportional. So A has the least specific heat.

Question 29.
Why do we need pressure cooker to cook food at higher altitudes?
Answer:

• At higher altitudes, such as hills and mountains, atmospheric pressure is low, therefore water boils at a temperature lower than 100°C and so it does not provide the required heat energy for cooking.
• Thus cooking there becomes very difficult and it takes a much longer time.
• So we require a pressure cooker to cook the food at a faster rate.

Question 30.
Lalitha wants to determine the specific heat of Aluminium shots. What apparatus of material is required to do this experiment?
Answer:
The apparatus required is calorie meter, thermometer, stirrer, water, steam heater, wooden box, and aluminium shots.

Question 31.
What are the material required in order to find specific heat of soild?
Answer:
Calorimeter, thermometer, stirrer, water, steam, heater, wooden box, and lead shots.

Question 32.
Your teacher made an experiment to show the formation of dew and frost. Explain how you show the formation of dew and frost.
Answer:
Place a water bottle in a deep fridge of refrigerator. After some time remove bottle from the refrigerator. We can observe ice in the bottle and water droplets are formed outside the bottle. This experiment is useful in formation of dew and frost.

Question 33.
Why a bottle completely filled with water and closed with a tight cap break after freezing?
Answer:

• Density of ice is less than that of water.
• This means water expands on freezing and converts into ice.
• So, excess space is required to expand the water.
• The bottle completely filled with water and closed with a tight cap has no excess space to expand water.
• So, bottle breaks while freezing water in it.

Question 34.
Which of the following substances take more time to raise its temperature for a certain degree Celsius? Give reason.

Answer:
Water takes more time to raise temperature because it has greater specific heat is, i. e., 1 cal/g-°C. As the specific heat of substance increases, it takes more time to raise its temperature.

Question 35.
The graph shows variations of temperature (T) of one kilogram of material with the heat (H) supplied to it.
At ‘O’ the substance is in the solid state. From the graph can conclude that

i) The melting point of the solid is …………….
Answer:
The melting point of the solid is T₁

ii) The latent heat of fusion is …………….
Answer:
The latent heat of fusion is (H₂ – H₁)

iii) The latent heat of vaporisation is …………….
Answer:
The latent heat of vaporisation is (H₄ – H₃)

iv) The boiling point of the substance is ……………
Answer:
The boiling point of the substance is T₃

Question 36.
Which of the substances A, B, and C has the highest specific heat? The temperature versus time graph as shown given below.

Answer:

• The substance ‘C’ has the highest specific heat.
• Because ‘C’ does not rise its temperature quickly. In other words ‘C’ takes more time to rise its temperature.

Question 37.
Can the average kinetic energy of a body be even zero?
Answer:
The average kinetic energy of a body can be zero.
Reason :
K.E = \(\frac{1}{2}\) mv²
m is never zero
if v = 0; then body is not in the motion.

Question 38.
A slab of ice at -50°C is constantly heated till the steam attains a temperature of 150°C. Draw a graph showing the change in temperature with time. Label the various parts of the graph properly.
Answer:

Question 39.
Iron of weight 2 kg was supplied with 12000 calories of heat. Initial temperature of iron was 20°C. Its specific heat is 0.1 cal/g-°C. What is the final temperature of iron?
Answer:
Mass of iron (m) = 2 kg = 2 × 1000 g. = 2000 g.
Quantity of heat supplied (Q) = 12,000 cal.
Initial temperature = θ_(i) = 20°C
Final temperature = θ_f = ?
Specific heat of iron (s) = 0.1 cal / g / °C.
Heat = Q = ms∆θ
Q = ms (θ_f – θ_i)
θ_f – θ_i = Q/ms
θ_f – 20 = \(\frac{12000}{2000 \times 0.1}=\frac{12}{2 \times 0.1}\) = 60
θ_f = 60 + 20 = 80°C
∴ The final temperature of iron = θ_f = 80°C

Question 40.
What is the heat energy required to rise 20 kg of water from 25° C to 75° C?
Answer:
Given m = 20 kg = 20,000 gm
t₁ = 25° C ; t₂ = 75° C ;
S = 1 cal/gm °C.
Q = mS∆T = 20000 × 1 × (75 – 25) = 20000 × 50
Q = 1000000 calories

Question 41.
If you drink 200 ml of water at 20° C, what is the heat gained by water from your body? (Body temperature is 37° C)
Answer:
m = 200 g (1 ml of water = 1 gm of water)
S = 1 cal/gm °C ;
t₁ = 20° C; t₂ = 37° C
Q = mS∆T (∆T = t₂ – t₁)
= 200 × 1 × (37 – 20)
= 200 × 17
Q = 3400 calories

Question 42.
What would be the final temperature of a mixture of 60 gm of water at 30°C temperature and 60 gm of water at 60°C temperature?
Answer:
m₁ = 60 g ;
T₁ = 30°C ;
m₂ = 60 g ;
T₂ = 60°C

Question 43.
The quantity of heat which can rise the temperature of ‘x’ gram of a substance through t₁°C and the quantity of heat which can rise the temperature of ‘y’ grams of water through t₂°C is same. What is ratio of specific heats? What is ratio of specific heats if rise in temperatures are same and if amount of substances are same?
Answer:
Suppose specific heats of substance and water are s₁ and s₂ respectively.
Heat absorbed by x gram of substance to rise its temperature to t₁°C.
Q₁ = ms∆T = x × s₁ × t₁
Heat absorbed by y gram of water to rise its temperature to t₂°C.
Q₂ = ms∆T₂ = y × s₂ × t2

10th Class Physics 1st Lesson Heat 4 Marks Important Questions and Answers

Question 1.
Write the factors that effect the process of evaporation. Explain with suitable examples. (AP March 2017)
Answer:
Process of evaporation is effected by surface area, wind speed, humidity, and temperature.
Ex:

• The water kept in a china dish evaporates faster than in a cup because of more surface area.
• Water in wet the clothes are kept under fan evaporates faster than in normal conditions.
• Water in wet clothes evaporates faster on a less humid day than on a more humid day.

Question 2.
A) Write the principle of method of Mixtures.
B) What would be the final temperature of a mixture of 60 gms of water at 50°C and 50 gms of water at 70°C? (AP March 2018)
Answer:
A) Principle of method of mixtures :
Net heat lost by the hot body = Net heat gain by the cold body.

B) m₁ = 60 gms.,
T₁ = 50°C ;
m₂ = 50 gms.,
T₂ = 70°C

Question 3.
Answer the following questions by using the data given in the table. (AP March 2018)

Substance	Specific heat (cal / g°C)
Lead	0.031
Aluminium	0.21
Copper	0.095
Water	1.00
Iron	0.115

a) Write SI units for specific heat.
Answer:
Joule / Kilogram-Kelvin

b) Based on specific heat values, arrange the substances given in the table in ascending order.
Answer:
Lead, Copper, Iron, Aluminium, Water

c) If we supply same quantity of heat, which substance will heat up faster?
Answer:
Lead

d) Calculate the amount of heat required to raise the temperature of 1 kg of Iron through 10°C.
Answer:
Q = mS∆T = 1000 × 0.115 × 10 = 1150 cal.

Question 4.
Suggest an experiment to show that when ice is converted into water, its temperature does not change. How much heat is required to convert 5 grams of ice at 0°C to water, at the same temperature? (Latent heat of fusion of ice is 80 cal/gram). (TS June 2015)
Answer:
Procedure :

1. Take small ice cube in a beaker. Insert the thermometer in the beaker.
2. Now start heating the beaker and note down readings of thermometer every one minute till the ice completely melts and gets converted into water.
3. Before heating the temperature of ice is 0° C or less than 0° C.

Observation :

1. We will observe that the temperature of ice at the beginning is equal to or below 0°C.
2. If the temperature of ice is below 0°C, it goes on changing till it reaches 0° C.
3. When ice starts melting, we will observe no change in temperature though you are supplying heat continuously.

Explanation :

1. Given heat energy is used to break the bonds (H₂O) in ice and melts.
2. So, temperature is constant while melting.

Conclusion :

1. This process is called melting. In this process heat converts solid phase to liquid phase.
2. The temperature of the substance does not change until all the ice melts and converts into water.
3. The heat given to melting is called latent heat of fusion.
4. The heat required to convert 1 gm of solid completely to liquid at constant temperature is called “latent heat of fusion”.
   m = 5 gm; L_f = 80 cal/g

The amount of heat absorbed Q = ML_f = 5 × 80 = 400 cal /g.

Question 5.
The graph shows the values of temperature, when ice is heated till it becomes water vapour. Observe the graph and answer the following questions. (TS March 2016)
(Note that the figure is not completely quantitative and also not to the scale. It is purely qualitative)

a) At what temperature, ice converts into water?
Answer:
Ice converts into water at 0°C and above.

b) What does \(\overline{\mathrm{DE}}\) represent?
Answer:
\(\overline{\mathrm{DE}}\) represents the latent heat of vapourisation.

c) What is the range of temperature of liquid water?
Answer:
The range of temperature of liquid water is 0°C to 100°C

d) Which part of the graph represents change of state from ice to water?
Answer:
\(\overline{\mathrm{BC}}\) represents the change of state of ice to water.

Question 6.
Write the differences between heat and temperature.
Answer:

Heat	Temperature
1) It is a thermal energy.	1) It is the measurement of hotness or coldness.
2) Heat is an extensive property, means it does not depend on amount of the substance that is present.	2) Temperature is an intensive property means that the substance present will not change the specific characteristic.
3) Heat is the amount of energy of the system.	3) Temperature is the measure of the average molecular motions in a system.
4) Its S.I unit is Joules.	4) Its S.I. unit is degrees C or K.
5) Heat is energy itself that flows.	5) It decides the direction of heat flow.

Question 7.
Why is the specific heat different for different substances?
Answer:

• We know that the temperature of a body is directly proportional to the average kinetic energy of particles of the body.
• The molecules of the system have different forms of energies. The total energy of the system is called internal energy of the system.
• When we supply heat energy to the system, the heat energy given to it will be shared by the molecules among the various forms of energy.
• This sharing will vary from substance to substance.
• The rise in temperature is high for a substance, if the maximum share of heat energy is utilised for its linear K.E.
• This sharing of heat energy also varies with temperature. That is why the specific heat is different for different substances.

Question 8.
Explain the process of evaporation.
Answer:

• The molecules of a liquid that kept in a dish, continuously move with random speeds in various directions. As a result, these molecules collide with other molecules.
• During this collision they transfer energy to other molecules. Hence, the molecules at the surface acquire energy and may fly off from the surface.
• Some of these escaping molecules may be directed back into liquid when they collide with the particles of air.
• If the number of escaping molecules is greater than the number returned, then the number of molecules in the liquid decreases.
• Thus when a liquid is exposed to air, the molecules at the surface keep on escaping from the surface till the entire liquid disappears into air. This process is called evaporation.

Question 9.
Define evaporation. Explain what are the affecting factors of evaporation and how they effect the rate of evaporation.
Answer:
Evaporation :
The process of escaping of molecules from the surface of liquid at any temperature is called evaporation.

The affecting factors of evaporation :

1. Temperature,
2. Surface area,
3. Wind speed,
4. Humidity.

The affection on the rate of evaporation :
1) Temperature :
As the temperature increases evaporation increases.

2) Surface area :
As the surface area of liquid increases, more molecules tend to leave the surface. So rate of evaporation increases.

3) Wind speed :
As the wind speed increases rate of evaporation increases.

4) Humidity :
As the humidity increases rate of evaporation decreases.

Question 10.
Why is climate near the seashore moderate?
Answer:

1. The specific heat of water is very high. It is about five times as that of sand.
2. Hence the heat energy required for the same rise in temperature by certain mass pf water will be nearly five times than that required by same mass of sand.
3. Similarly, a certain mass of water will give out nearly five times more heat energy than that given by sand of the same mass for the same fall in temperature.
4. As such, sand (or earth) gets heated or cooled more rapidly as compared to water under similar conditions.
5. Thus, a large difference in temperature is developed between the land and sea due to which land and sea breezes are formed.
6. These breezes make the climate near seashore moderate.

Question 10.
Why do farmers fill their fields with water on a cold winter night?
Answer:

• In the absence of water, if on a cold winter night, the atmospheric temperature falls below 0°C, the water in the fine capillaries of plants will freeze, so the veins will burst due to the increase in volume of water on freezing.
• As a result, plants will die and the crop will be destroyed.
• In order to save crop on such cold nights, farmers fill their field with water because water has high specific heat, so it does not allow the temperature in the plants of surrounding area to fall up to 0°C.

Question 11.
Explain the factors effecting boiling.
Answer:
The factors effecting boiling are
I) Pressure :

1. The boiling point of pure water at one atmospheric pressure is 100°C.
2. Water boils at a temperature higher than 100°C, if the atmospheric pressure is higher than one atmosphere pressure, and boils at a temperature lower than 100°C, if the atmospheric pressure is less than 1 atmosphere.

II) Impurities :
The boiling point of liquid increases by the addition of impurities to it. If a little common salt is added to water, the water boils at a temperature higher than 100°C.

Question 12.
A, B and C are the three liquids at 20°C, 30°C and 40°C respectively. If equal masses of A and B are mixed, the resultant temperature is 26°C. If equal masses of A and C are mixed, the resultant temperature is 33°C. Find the ratio of specific heats of A, B and C.
Answer:

Question 13.
A refrigerator converts 5kg of water at 40°C into ice at 0°C in 20 minutes. Find the power of refrigerator.
Answer:
m = 5kg = 5000gr
∆t = 40°C – 0°C = 40°C
S = 1 (water) .
Q = m.s.∆t = 5000.1.40 = 200000 cal.
We know the relation between heat and work done as

Question 14.
Snow on mountains does not melt all at once. Why?
Answer:

• Snow on mountains does not melt all at once because the ice has a high specific latent heat of fusion.
• It is due to this fact that it changes into water slowly as it gets heat energy from the sun.
• If latent heat would not have been so high, the snow would have melted quickly even with a small amount of heat energy and there would have been floods in rivers.

Question 15.
Collect specific heats of various substances.
Answer:

Substance	Specific heat
	In cal/g – °C	In J/kg – K
Lead	0.031	130
Mercury	0.033	139
Brass	0.092	380
Zinc	0.093	391
Copper	0.095	399
Iron	0.115	483
Glass(flint)	0.12	504
Aluminium	0.21	882
Kerosene oil	0.50	2100
Ice	0.50	2100
Water	1	4180
Sea water	0.95	3900

Question 16.
The graph given below represents a cooling curve for a substance being cooled from a higher temperature to a lower temperature.

a) What is the boiling point of the substance?
Answer:
The boiling point of the substance is 150°C (because the part BC represents condensation where the vapour changes into the liquid without the change in temperature).

b) What happens in the region DE?
Answer:
The region DE represents freezing of the substance where the liquid changes into solid at a constant temperature that is 100°C.

c) What is the melting point of the substance?
Answer:
The melting point of substance is 100°C.

Question 17.
You’ve taken water in vessel at 0°C and closed it with a glass vessel as shown in the figure. You used and created a vacuum inside.

a) Explain what happens.
b) A part of water condenses; what is the amount of water that gest condensed?
Answer:
a) At 0°C also water is available in liquid state (generally at 0°C ice is also available) because the air in vacuum rise the temperature. Here evacuation is possible so it allows evaporation.

b) Let y ml of water is taken at 0°C.
’x’ ml of water is evaporated
Latent heat of vapourisation = L_steam = 540 Cal/g.
Latent heat of ice = L_ice = 80 Cal/g.
After sometime conversion process stops. So equilibrium is possible.

Question 18.
What are the applications of specific heat capacity?
Answer:
1. The oceans behave like heat store houses for the earth. They absorb large amounts of heat at the equator without rise in temperature due to high specific heat capacity of water. So, oceans moderate the surrounding temperature near the equator. Ocean water transports the heat away from the equator to areas closer to the north and south poles. This transported heat helps moderate the climates in parts of the Earth that are far from the equator.

2. Watermelon brought out from the refrigerator retains its coolness for a long time than any other fruit because it contains a large amount of water (water has greater specific heat).

3. The samosa seems to be cool outside but it is hot when we eat it because the curry inside the samosa contains ingredients with higher specific heats.

Question 19.
Some hot water Is added to three times its mass of cold water at 10°C. The resulting temperature is found to be 20°C. Find the initial temperature of hot water.
Answer:
Let the initial temperature of hot water be t°C.
Mass of hot water = mg
Mass of cold water = 3 mg
Initial temperature of cold water = 10°C
And resultant temperature = 20°C

∴ Initial temperature of hot water = 50°C.

Question 20.
40 g of water at 60°C is poured into vessel of 200 g mass containing 50 g of water at 20°C. The final temperature of mixture is 30°C. Calculate the specific heat of vessel.
Answer:
Mass of hot water at 60°C = 40 g
Mass of cold water at 20°C = 50 g
Mass of vessel = 200 g
Suppose the specific heat of vessel is S_v
Heat energy given by hot water = mS_w∆T
= 40 × 1 × (60 – 30) [∵ T₁ = 60°C, T₂ = 30°C]
= 40 × 30 = 1200 cal
Heat energy taken by cold water = mS_w∆T
= 50 × 1 × (30 – 20) = 50 × 10 = 500 cal
Heat energy taken by vessel = mS_v∆T = 200 × S_v × (30 – 20) = 2000 S_v
According to the principle of method of mixtures,
Heat lost by hot water = heat gained by cold water + heat gained by vessel 1200 = 500 + 2000 S_v

Question 21.
A, B and C are three liquids at 20°C, 30°C and 40°C respectively. If equal masses of A and B are mixed, the resultant temperature is 29°C. The equal masses of A and C are mixed, the resultant temperature is 33°C. Find the specific heats of A, B and C.
Answer:
Suppose specific heats of liquids A, B and C are s₁, s₂ and s₃ respectively.
Given that the temperatures of liquids are 20°C, 30°C and 40°C.
Given that equal mass of A and B are mixed, the resultant temperature is 29°C.

Further given that equal masses of A and C are mixed, the resultant temperature is 33°C.

Question 22.
A refrigerator converts 5 kg of water at 40°C into ice at 0°C. Find the total energy released in Joules.
Answer:
m = 5 kg = 5000 g
The heat energy released to convert 5 kg of water at 40°C to 5 kg of water at 0°C.
Q₁ = ms∆T
= 5000 × 1 × (40 – 0) = 5000 × 40 = 200000 cal [s = 1 cal/g-°C for water]
The heat energy released to convert 5 kg of water at 0°C to 5 kg of ice at 0°C.
Q₂ = mL_f
= 5000 × 80 = 400000 cal [∵ L_f = 80 cal/g]
Total energy released = 200000 + 400000
= 600000 cal
= 600 kcal
= 142.86 kJ.

Question 23.
The quantity of heat which can rise the temperature ‘x’ grams of a substance through t₁°C can rise the temperature of ‘y’ grams of water through t₂°C is same.
What is the ratio of specific heats?
Answer:
Given,

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 9 Classification of Elements- The Periodic Table.

AP State Syllabus SSC 10th Class Chemistry Important Questions 9th Classification of Elements- The Periodic Table

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table 1 Mark Important Questions and Answers

Question 1.
What is modern periodic law? (AP June 2015)
Answer:
Modern periodic law :
The physical and chemical properties of the elements are periodic functions of their electronic configurations.

Question 2.
Define Moseley’s periodic law. (AP June 2015)
Answer:
Moseley’s periodic law: The physical and chemical properties of elements are periodic functions of their atomic numbers.

Question 3.
Which group elements are called Carbon family? (AP Mareh 2016)
Answer:
14 (or) IVA Group of elements are called Carbon family.

Question 4.
Which atom is bigger in size, Ne or Ar? Why? (AP June 2018)
Answer:
Ar. In groups as we go down number of shells increases due to the formation of new shell.

“O Group”

He

Ne

Ar

Kr

Xe

Rn

Question 5.
A and B are two elements. The compound formed with A and B is A₂ B. What are the valencies of A and B. (TS March 2018)
Answer:
The valency of A is 1 and B is 2.

Question 6.
A teacher asked to give an example for Dobereiner’s triad. Ramu wrote them as “Li, Na, Mg”. In these three, identify which element does not belongs to this triad? (AP March 2019)
Answer:
Mg or Magnesium do not belongs to this triad.

Question 7.
Write the difference between Mendeleeff’s periodic law and modern periodic law. (AP SCERT: 2019-20)
Answer:
Mendeleeff’s periodic table is prepared based on atomic mass whereas modem periodic table is prepared based on atomic number (electronic configuration).

Question 8.
What is Dobereiner Triad? Give two examples to it.
Answer:
A group of three elements in which atomic weight of middle element is average of first and third element is called Dobereiner triad with similar propertion.
Eg: 1) U, Na, K
2) Cl, Br, I

Question 9.
What is Newlands’ law of octaves?
Answer:
When elements are arranged in the ascending order of their atomic weights, every eighth element starting from a given element resembles in its properties to that of starting element. This is called Newlands’ law of octaves.

Question 10.
What is MendeleefFs periodic law?
Answer:
MendeleefFs periodic law:
The physical and chemical properties of the elements are the periodic functions of their atomic weight.

Question 11.
What is the name given to horizontal rows and vertical columns in MendeleefFs periodic table?
Answer:
Horizontal rows are periods and vertical columns are groups.

Question 12.
What is the property on which MendeleefFs periodic table depends upon?
Answer:
Mendeleeff’s periodic table depends upon atomic weight.

Question 13.
What is the name given to I(A) group elements?
Answer:
Alkali metal family, because aliquili = plant ashes. Na, K, etc. were obtained from plant ash.

Question 14.
Why are VI A group elements called chalcogens?
Answer:
Chalcogeneous = Ore product. As the elements in group 16 (VI A) form ores with metals, they are called chalcogeneous family.

Question 15.
Why are VII A group elements called halogens?
Answer:
Halos – sea salt, genus – produced. So VII A (17) are obtained from nature as sea salt. So they are called halogen family.

Question 16.
What are halogens?
Answer:
Fluorine, Chlorine, Bromine, Iodine, and Astatine of VIIA group elements are called halogens, which are obtained from sea salt.

Question 17.
What are noble gases? What is the general electronic configuration of noble gases?
Answer:
The elements of group VIII A (18) are chemically least reactive so they are called noble gases. Their group electronic configuration is ns²np⁶ (except) for helium it is 1s².

Question 18.
What are Lanthanides?
Answer:
Elements acquiring same properties are called lanthanides, i.e. 4f elements. They are from ₅₈Ce (Cerium) to ₇₁Lu (Lutetium).

Question 19.
What are Actinides?
Elements acquiring different properties are called actinides, i.e. 5f elements. They are from ₉₀Th (Thorium) to ₁₀₃Lr (Lawrensium).

Question 20.
What are metals and non-metals?
Answer:
The elements with three or less electrons in the outer shell are considered to be metals and the ore with five or more electrons in the outer shell are considered to be non-metals.

Question 21.
What are metalloids?
Answer:
The properties of elements which are intermediate between the properties of metals and non-metals are called metalloids.

Question 22.
Which will behave like semi-conductors?
Answer:
Metalloids or semi-metals behave like semi-conductors.

Question 23.
What is valency?
Answer:
The combining power of element with respect to hydrogen, oxygen or indirectly any other element through hydrogen and oxygen is called valency.

Question 24.
What is the latest definition of valency?
Answer:
The number of electrons lost or gained or shared during a chemical reaction.

Question 25.
How do we measure atomic radius of solids?
Answer:
It is half of the distance of radius of each atom.

Question 26.
What is covalent radius?
Answer:
Half of the distance between length of covalent bond is called covalent radius.

Question 27.
In which units is atomic radius measured?
Answer:
Atomic radius is measured in pico meter (pm) units.
1 pm = 10^-12 m.

Question 28.
What is the method given by Milliken to calculate electronegativity of an element?
Answer:
According to Milliken, electronegativity of element is average value of its ionization energy and electron affinity.

Question 29.
What is electropositive character?
Answer:
The tendency of metals to remain positive ions in compounds is called electropositive character. (OR) The tendency of an atom to lose electrons to form positive ions.

Question 30.
What is screening effect or shielding effect?
Answer:
More the shells with electrons between the nucleus and the valence shell, they act as screens to decrease nuclear attraction over valence electron. This is called screening effect or shielding effect.

Question 31.
What do you mean by negative or positive electron gain enthalpy?
Answer:
The negative sign indicates that energy is liberated or lost, and the positive sign indicates that the energy is gained or absorbed.

Question 32.
What is a triad?
Answer:
A group of three elements with similar properties in which atomic weight of middle element is average of other two elements.

Question 33.
Chlorine, bromine, iodine are Dobereiner’s triads. How do you justify?
Answer:
Chlorine, bromine and iodine have similar properties and atomic weight of bromine is average of chlorine and iodine.

Question 34.
Why are lanthanides and actinides placed separately at the bottom of the periodic table?
Answer:
Lanthanides and actinides belong to f – block elements with different properties so they are placed at the bottom of periodic table.

Question 35.
Lithium, Sodium, and Potassium were put in one group on the basis of their similar properties.
1) What are those similar properties?
2) What is the common name of this group of family?
Answer:

1. They have same number of valence electrons that is 1 and valency 1. So they have similar chemical properties.
2. They are called alkali metals.

Question 36.
What are the following groups known as?
1) group VIIA elements
2) Zero group elements.
Answer:

1. Group VII A elements are called Halogens.
2. Zero group elements are called Noble gases.

Question 37.
An element Barium lies in 2^nd group; then answer the following.
1) What is its valency?
2) What will be the formula of its Phosphate?
Answer:

1. The element lies in second group. So its valency is 2.
2. The formula of Phosphate is Ba₃(PO₄)₂ [since the valency of Phosphate is 3],

Question 38.
A, B, C are three elements having their atomic numbers equal to 2,10 and 5 respectively.
a) Which of these elements belong to same period?
b) Which of these elements belong to same group?
Answer:
The electronic configurations of A, B, C are as follows
A – 2, B – 2, 8, C – 2, 3.
a) So, B and C belong to same period because valence electron enters same orbit.
b) A and C belong to same group because both are noble gases.

Question 39.
Which element of 3^rd period will form a chloride of Cl₄?
Answer:
It would be Silicon because its electronic configuration is 2, 8, 4. So, it lies in third period and its valency is 4.

Question 40.
Which two elements of 3^rd period will form a covalent compound?
Answer:
The two elements are phosporous and chlorine.

Question 41.
An element has atomic number 12. State whether it is metal or non-metal. Why?
Answer:
Its electronic configuration is 2, 8, 2. It lies in 2^nd group. The elements towards left of periodic table are generally metals. So the element is metal.

Question 42.
Elements X, Y, and Z belong to IA group of the periodic table. Their atomic radii are as follows.
X → 1.33 Å,
Y → 0.95 Å,
Z → 0.60 Å.
Arrange the elements in the increasing order of atomic number by giving reason.
Answer:
As we move from top to bottom in a group, atomic size increases and atomic number also increases.
So the correct increasing order is Z, Y, X.

Question 43.
An element has an atomic number 16. State
i) period to which it belongs
ii) the number of valence electrons.
Answer:
Its electronic configuration is 2, 8, 6.
i) So it belongs to 3^rd period (orbit number).
ii) The number of valence electrons is 6.

Question 44.
Why is energy absorbed when electron is added to uni-negative ion?
Answer:
It is difficult to add an electron to uni-negative ion. In order to overcome the repulsion between the electrons, actually energy should be supplied to add another electron to uni-negative ion.

Question 45.
When do you observe liberation of energy?
Answer:
Atoms of some elements gain electrons while forming ionic compounds. An atom is able to gain electron when the electron is attracted by the nucleus. Attraction involves the liberation of energy.

Question 46.
Why does nitrogen have less electron affinity value compared to oxygen?
Answer:
The electron affinity of nitrogen is less than oxygen because of stable configuration of nitrogen (i.e., 2p³ configuration).

Question 47.
Which one between Na and Na⁺ would have more size? Why?
Answer:
Na has more size because when one electron is removed from Sodium atom the nucleus attraction over outermost electron increases so atomic size decreases.

Question 48.
Second ionization energy of an element is higher than its first ionization energy. Why?
Answer:
It is difficult to remove an electron from unipositive ion when compared with neutral atom. So second ionization energy is always greater than first ionization energy.

Question 49.
Hydrogen can be placed in group’1 and group 7 periodic table. Why?
Answer:
Hydrogen has both +1 as well as -1 oxidation states. So still there is some ambiguity in position of hydrogen.

Question 50.
Why do inert gases have zero valency value?
Answer:
Inert gases show zero valency because they do not take part in chemical reactions due to stable configuration.

Question 51.
Element ’Z’ belongs to (second) 2nd group in the periodical table. Write the formula of oxide.
Answer:
The formula of oxide of the element is ZO.

Question 52.
Do the atom of an element and its ion have same atomic size?
Answer:
No, generally cation has smaller size and anion has greater size.

Question 53.
The electronegativities of the elements in period 3 of the periodic table are as. follows.

Arrange the elements in which they occur in the periodic table from left to right.
Answer:
Na Mg Al Si P S Cl

10th Class Chemistry 9th Lesson Classification of Elements – The Periodic Table 2 Marks Important Questions and Answers

Question 1.
An element has atomic number 17. Where would you expect this element in the Periodic Table? Why? (AP June 2018)
Answer:

• Electronic configuration of the given element is 1s² 2s² 2p⁶ 3s² 3p⁵.
• So, it is in 3^rd period and 17^th group of periodic table.
• Due to the valency electronic configuration of 3s² 3p⁵ it belongs to 3^rd period and 17^th group.

Question 2.
How do you appreciate the special nature of inert gases?
Answer:
I appreciate the special nature of inert gases because it helps us in explaining the formation of chemical bonds among the atoms of elements and their stability.

Question 3.
The atomic number of an element is 35. Where would you expect the position of this element in the periodic table? Why? (TS June 2015)
Answer:

• The Electronic configuration of element with atomic number 35 is 2, 8, 18, 7.
• So it has seven valence electrons.
• That’s why it is present in 17^th group or VII A group and 4^th period.
• The element is Bromine.

Question 4.
Why were Dobereiner, Newlands and Mendeleeff not 100% successful in their classification of elements? Why is the modern table relatively a better classification? (TS March 2016)
Predict the reason.
Answer:

• All the known elements at the time of Dobereiner could not be arranged in the form of triads.
• Newlands’ periodic table was restricted only for 56 elements.
• As Mendeleeffs classification is based on atomic weight, his classification led to two defects like anomalous pair of elements and dissimilar elements placed together.
• Modern periodic table was prepared on the basis of atomic number. So the periods and groups are clearly defined.

Hence Dobereiner, Newlands, and Mendeleeffs classifications were not 100% successful, but modern classification is successful.

Question 5.
Observe the electronic configurations given below and write the group and period numbers of those elements. (TS March 2016)

Answer:
a) The period number is 3 and group number is 1.
b) The period number is 3 and group number is 15.

Question 6.
Observe the information provided in the table and answer the questions given below it. (TS June 2017)

i) What are the s-block elements in the table?
ii) What are the ‘p’ block and ‘d’ block elements in the table?
Answer:
i) s-block elements : Na, Ca

ii) p-block elements : C, P
d-block elements : Ti, Ni.

Question 7.
Imagine, which one in each of the following pairs is large in size relatively with other? Explain. (AP March 2019)
(X) Na, Al (Y) Na, Mg⁺²
Answer:
(X) 1. Na is large in size than Al.
2. Atomic size gradually decreases from left to right in a period.

(Y) 1. Na is large in size than Mg²⁺.
2. Na is larger than Mg and Mg is larger than Mg²⁺.

Question 8.
What are the limitations of Dobereiner triad?
Answer:

• All the known elements could not be arranged in the form of triads.
• The law failed for very low mass or for very high mass elements.
  Eg : In case of F, Cl, Br the atomic mass of Cl is not an arithmetic mean of atomic masses of F and Br.
• As the techniques improved for measuring atomic masses accurately, the law was unable to remain strictly valid.

Question 9.
Distinguish between electron affinity and electronegativity.
Answer:

Electron affinity	Electronegativity
1. It is the property of an isolated gaseous atom.	1. It is the property of a bonded atom.
2. It is the energy released and is measured in ev/atom or kJ/mole.	2. It is relative quantity and has no units.
3. It is the attraction of an atom for a single electron.	3. It is the attraction of an atom for a pair of electrons.

Question 10.
What is electronegativity? What are the various methods used to determine electronegativity? Explain.
Answer:
Electronegativity :
The electronegativity of an element is defined as the relative tendency of its atom to attract electrons towards itself when it is bounded to the atom of another elements.

Various methods to calculate Electronegativity :
1) Milliken Scale :
According to Milliken, the electronegativity of an element is the average value of its ionization energy and electron affinity.

2) Pauling Scale :
Pauling scale is based on bond energies. The electronegativity of hydrogen is assumed as 2.20. Electronegativity of other elements is calculated with respect to hydrogen.

Question 11.
Give the electronic configurations of following elements. What do say about these elements by writing their electronic configurations?
a) Na
b) Al
c) Sc
d) Ce
Answer:
a) Na : 1s² 2s² 2p⁶ 3s¹
b) M : 1s² 2s²2 2p⁶ 3s² 3p¹
c) Sc : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹
d) Ce : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 6s² 4f²

Inference :
The valence electron enters different orbitals. So these elements belong to different blocks in modern periodic table, i.e. s, p, d, and f respectively.

Question 12.
Do you think that Newlands’ law of octaves is correct? Justify.
Answer:
No, Newlands’ law of octaves was restricted to only 56 elements and did not leave any room for new elements. Elements that were discovered later could not be filled into Newlands’ table in accordance with properties.

Question 13.
Why did Mendeleeff have to leave certine blank spaces in his periodic table?
Answer:
Mendeleeff predicted that some elements were missing in the table so he left blank spaces at the appropriate places in the table.

Question 14.
Give reason for the need of classification of elements.
Answer:
Classification is necessary because it is difficult to remember the properties of all the elements separately. It is easy to identify the properties of elements by making them groups with similar properties.

Question 15.
x, y, and z are the elements of a Dobereiner’s triad. If the atomic mass of ‘x’ is 7 and that of ‘z’ is 39, what should be the atomic mass of ‘y’?
Answer:
The atomic mass of x = 7 ;
The atomic mass of z = 39
x, y, z form Dobereiner triad
∴ Atomic mass oi y = average of x and z = \(\frac{7+39}{2}\) = \(\frac{46}{2}\) = 23

Question 16.
Name the two elements that would expect to have chemical properties similar to element with atomic number 11. What is the base for your choice?
Answer:
The element with atomic number 11 is sodium and its electronic configuration is 1s² 2s² 2p⁶ 3s¹ or 2, 8, 1.

So it has one valence electron, i.e. present in I group. We know that the elements present in same group have same valence electrons. So they show similar properties.

Therefore the other two elements are Lithium and Potassium.

Question 17.
An element X belongs to 3^rd period and group 14 of the periodic table. State
a) the number of valence electrons
b) the valency
c) the name of the element.
Answer:
a) The number of valence electrons are 4.
b) Its valency = 8 – 4 = 4.
c) The electronic configuration of element is 2, 8, 4.
(Because 3^rd period means third orbit, group 14 has 4 valence electrons). So, the element with atomic number 14 is Silicon.

Question 18.
Why is it easier to remove 4f electron than 4s?
Answer:
Orbitals belonging to the same main shell have different penetration power towards the nucleus. In fourth main shell the order of penetration is like this 4s > 4p > 4d > 4f. So, it is easier to remove 4f electron than 4s.

Question 19.
How is screening effect responsible for low ionization of cesium?
Answer:

• More the shells with electrons between the nuclear and the valence shell, they act as screens and decrease nuclear attraction over valence electron. This is called the screening effect.
• More the screening effect, less is the ionization energy.
• Cesium with more inner shells has less ionization energy.

Question 20.
Why does Boron have less ionization energy when compared with Beryllium?
Answer:

• The electronic configuration of Be and B are 1s² 2s² and 1s² 2p² 2p¹.
• The element Boron has less ionization energy due to less penetration power of 2p compared to 2s.

Question 21.
We know that as we move from left to right ionization energy increases. But ionization energy Nitrogen is more than Oxygen. Why?
Answer:

• It is easier to remove an electron from Oxygen when compared to Nitrogen.
• This is because Nitrogen has stable 1s² 2s² 2p³ electronic configuration which contains half filled 2p orbitals whereas Oxygen has 1s² 2s² 2p⁴ configuration.

Question 22.
Why is it difficult to remove an electron from Mg⁺ when compared with Mg?
Answer:

• The energy required to remove the first electron outermost orbit of a neutral gaseous atom of the element is called first ionization energy.
• The energy required to remove from unipositive ion of the element is called second ionization energy.
• Second ionization energy is always more than first ionization energy because it is difficult to move electron from unipositive ion due to greater nuclear attraction.
• So it is difficult to remove an electron from Mg⁺ when compared with Mg.

Question 23.
Using the periodic table predict formula of compound formed between an element ‘X’ of group 2 and another element of group 17.
Answer:

• The element X belongs to group 2. So, the number of valence electrons are 2 and its valency is 2.
• The element Y belongs to group 17 or VII. So, the number of valence electrons are 7 and its valency = 8 – 7= 1.

During formation compound elements exchange their valencies.
∴ The formula of compound is XY₂.

Question 24.
How do electronegativity values vary in period and group?
Answer:
Period :
When we move from left to right in period, the electronegativity increases due to decrease in atomic size.

Group:
When we move from top to bottom in a group, the electronegativity decreases due to increase in atomic size.

Question 25.
How does metallic and non-metallic characters vary in a period and group?
Answer:
Period:
When we move from left to right in a period, the metallic character decreases and non-metallic character increases.

Group :
When we move form top to bottom in a group, non-metallic character decreases and metallic character increases.

Question 26.
How do valency vary in period and group?
Answer:
Period :
When we move from left to right in a period, the valency does not follow a regular trend. For example, in second period the valency starts from 1 and increases to 4, then thereafter decreases to ‘O’.

Group :
When we move top to bottom in a group, the valency remains the same because in a group the valence electrons are same.

Question 27.
How does electron affinity vary in a period and group?
Answer:
Period :
When we move from left to right in period, electron affinity increases due to greater nuclear attraction over electron.

Group :
When we move from top to bottom, the electron affinity decreases in atomic size. As the size of the atom increases, the nuclear attraction over outermost electron decreases. So electron affinity decreases.

Question 28.
An element has atomic number 19. Where would you expect this element in the periodic table anti why?
Answer:
The electronic configuration of element is 1s²2s²2p⁶3s²3p⁶4s¹. So the element is in 4^th period and I group.

Question 29.
The electronic configuration of the element X, Y, and Z are given below,
a) X = 2, 5
b) Y = 2, 8, 1
c) Z = 2, 8
i) Which element belongs to 18^th group?
ii) Which element belongs to 15^th or V group?
iii) Which element belongs to third period?
Answer:
i) Z belongs to 18^th group because it is a noble gase (i.e. Ne).
ii) X belongs to 15^th or V group because it has 5 valence electrons.
iii) Y belongs to 3^rd period because the valence electron is present in 3^rd orbit.

Question 30.
Referring the part of periodic table given below answer the questions that follow.

1) What happens to the atomic size if moved from left to right? Support your answer.
2) What changes do you observe in the metallic properties of the elements when moved from left to right?
Answer:
1) When we move from left to right in a periodic table atomic radii of elements decrease, as a result the size of the,atom decreases,

2) When we move from left to right in a periodic table electronegativity values of elements increase, as a result the metallic properties of the elements decrease.

Question 31.
State the name of element, number of valence electrons, valency, the group number and the period number of each element given in the following table.

Answer:

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table 4 Marks Important Questions and Answers

Question 1.
What is Ionization Energy? Explain the factors that effect Ionization Energy. (AP June 2017)
(OR)
What is ionization energy? What are the factors which influence ionization energy? Explain.
(OR)
Write the factors that influence ionization energy and explain any three of them. (TS March 2019)
Answer:
Ionization energy :
The energy required to remove an electron from the outermost orbit or shell of a neutral gaseous atom is called ionization energy.
Factors influencing ionization energy :
1) Nuclear charge :
As nuclear charge increases, ionization energy increases.

2) Screening effect or shielding effect:
More the screening effect, less is the ionization energy.

3) Penetrating power of the orbitals :
If the orbitals have less penetrating power, then the ionisation energy is less. Generally, the penetrating power of orbits are like this : s > p > d > f.

4) Stable configuration :
The elements having half-filled or completely filled orbitals have more stability. So the ionization energy is more when the element has stable configuration.

5) Atomic size :
As the atomic size increases, the nucleus attraction over outermost electron decreases. So ionization energy decreases.

Question 2.
Elements of one short period of the Periodic Table are given below in the order from left to right. (AP March 2017)
Li, Be, B, C, N, F, Ne
Answer the following:
(i) To which period, do these elements belong?
(ii) One element of this period is missing. Which is the missing element and where it should be placed?
(iii) Which of the above elements belong to the family of halogens? What is its electronegativity value?
(iv) How does the metallic character varies in the Period?
Answer:
(i) 2nd period.
(ii) Oxygen.
It should be placed between Nitrogen and Flourine.

(iii) Flourine
Electro negativity 4.0

(iv) Decreases from left to right.

Question 3.
In the table given below, names of some elements of families are given. Based on this, fill the information in the empty boxes. (TS June 2015)

Answer:

Question 4.
Two elements X and Y belong to Groups 1 and 2 respectively in the same period of the Periodic Table. Compare these elements with respect to : (TS March 2015)
i) number of electrons in their outermost orbit.
Answer:
The number of electrons in the outermost orbit of element X = 1
The number of electrons in the outermost orbit of element Y = 2

ii) their atomic size and their valencies.
Answer:
The atomic size of the Y is lesser than X
Valence of X = 1 ; Valence of Y = 2

iii) their ionisation energy and metallic character.
Answer:
The ionization energy of Y is greater than X, X has higher metallic character than Y.

iv) formulae of their chlorides and sulphates.
Answer:
Chloride of X …. XCl
Chloride of Y …. YCl₂
Sulphate of X …. X₂SO₄
Sulphate of Y …. YSO₄

Question 5.
How are the elements arranged into groups and periods in the Modern Periodic Table? Elements in a group possess similar properties, but elements in a period do not show similarities in their properties. Why? (TS June 2017)
Answer:

• The Modern periodic table is arranged in groups and periods based on the electronic configuration of the atoms of elements.
• Physical and Chemical Properties of elements are related to their electronic con-figurations particularly the outershell configurations.
• The atoms of the elements in a group posses similar electronic configurations.
• The elements in a group should have similar chemical properties and there should be regular gradation in their physical properties from top to bottom.
• Across the table from left to right in any period, elements gets an increase in the atomic number by 1 unit between any two successive elements.
• Therefore the electronic configuration of valence shell of any two elements in a given period is not same.
• Due to this reason elements along a period posses different chemical properties with regular gradation in their physical properties from left to right.

Question 6.
Explain any four factors which influence the electron affinity (Electron Gain Enthalpy). (TS March 2017)
Answer:
Factors effecting of electron affinity
1. Nuclear Charge :
If nuclear charge increases electron affinity increases, similarly it decreases if nuclear charge decreases.

2. Screening effect:
If screening effect value increases electron affinity increases, if it decreases electron affinity decreases.

3. Penetration power of the orbitals :
If penetration power of the orbitals increases electron affinity increases. If it decreases electron affinity decreases.

4. Stable configuration :
If an atom has stable electron configuration electron affinity will decreases.

5. Atomic radius :
If atomic radius increases electronic affinity will be increases. If atomic radius decreases electron affinity decreases.

6. Metallic property :
If metallic property increases electron affinity decreases.

7. Non-Metallic property :
Non-Metallic property increases electron affinity value increases.

Question 7.
Observe the information and answer the following questions. (TS June 2018)

Name of the Element	Atomic Number	Electronic Configuration
Sodium	11	[Ne] 3s1
Magnesium	12	[Ne] 3s2
Potassium	19	[Ar] 4s1
Calcium	20	[Ar] 4s2

1) What is valency of Magnesium?
Answer:
Valency of magnesium is two.

2) Which element has more electro-positivity?
Answer:
Potassium (K) has more electro-positivity.

3) Write the elements which belongs to (third) 3^rd Period.
Answer:
The elements which belongs to 3^rd period are Sodium (Na), Magnesium (Mg).

4) Write the elements which belongs to 1^st Group.
Answer:
Sodium (Na), Potassium (K) belong to 1^st Group.

Question 8.

Answer the following from the above in brmation. (TS March 2018)
i) Which element posses the higher atomic radius in the above table?
Answer:
The element having higher atomic radius is ‘K’ (Potassium)

ii) Mention two plair of element which forms ionic bond.
Answer:
Na, Cl Mg, CL

iii) Name the two elements having valency 2.
Answer:
Elements having valency 2 are Be, Mg, Ca, 0, S, Se.

iv) Which element has electronic configuration of 1s² 2s² 2p⁴.
Answer:
Oxygen.

Question 9.
Explain the significance of three quantum numbers in predicting the position of an electron in an atom. (AP SCERT: 2019-20)
Answer:
Each electron in an atom is described by a set of three quantum numbers n, 1 and ml.
1. Principal quantum number (n):
The principal quantum number is used to describe the size and energy of the main shell. It is denoted by ‘n’. ‘n’ has positive integer values of 1, 2, 3, It is used to know the number of orbitals (n²) and electrons in an orbit. (2n²).

As ‘n’ increases the shells becomes larger and the electrons in those shells are farther from the nucleus and their energies increases.

2. The angular – momentum quantum number (l) :
‘l’ has integer values from O’ to n – 1, for each value of ‘n’. Each ‘l’ value represents one sub-shell. It is used to describe the shape of an orbit.

3. The magnetic quantum number (m_l) :
The magnetic quantum number (m_l) has integer values between -l and +l including zero.

If l = 0, the possible ml value is 1.
l = 1, the possible ml value is -1, 0 and 1.
Thus for a certain value of 1, there are (2l + 1) integer values of ml.

These values describe the orientation of the orbital in space relative to the other orbitals in the atom.

Ex: When l = 1, (2l + 1) = 3, that means ml has 3 values namely -1, 0, 1 or three p orbitals, with different orientations along x, y, z axes, labelled as p_x, p_y and p_z orbitals.

Predicting the position of an electron in an atom :
If the values of n, l, and ml are 2, 1,-1 respectively the electron is present in 2px orbital in L – shell.

Question 10.
Answer the following question based on the values of the atomic radii of the elements of one of the periods in modern periodic table (AP SCERT: 2019-20)
Li (152), Be (111), B (88), C (77), N (74), O (66) and F (64)
a) What is the trend of atomic radii of given elements?
b) In the numerical listing of periods in the modern periodic table, what number was given to above elements?
c) Mention the unit of atomic radius.
d) Why the values of atomic radius varied along the period?
Answer:
a) Atomic radii of elements decrease while going left to right in the periodic table.

b) Period – 2
c) Unit of atomic radius is ‘pm’ (picometer).
d) There should be no change in distance between nucleus and outer most shell for the elements in one period.

But, nuclear charge increases because of the increase in the atomic number of elements in a period.

Hence, the nuclear attraction on the outer shell electrons increases.

As a result the size of the atoms decreases while going left to right in a period.

Question 10.
Mendeleeff classified the then known 63 elements in the form of a periodic table. Mention any two things that benefitted study of chemistry, to support the above statement.
Answer:

• Mendeleeff accepted minor inversions in the order of increasing atomic weights as these inversions resulted in elements being placed in the correct group.
• It was the extraordinary thinking of Mendeleeff that made the chemists to accept the periodic table and recognise Mendeleeff more than anyone else as the originator of the periodic law.
• At the time when Mendeleeff introduced his periodic table even electrons were not discovered.
• Even then the periodic table was prepared to provide a scientific base for the study of chemistry of elements.
• In his honour the 101 element was named “Mendelevium”.

Question 11.
How do these properties vary in period and group?
1) Valency
2) Atomic radius
3) Ionisation energy
4) Electron affinity
5) Electronegativity
6) Electropositivity
7) Metallic nature
8) Non-Metallic nature.
Answer:

Question 12.
Explain the salient features and achievements of the Mendeleeffs periodic table.
Answer:
Mendeleeffs periodic table is based on atomic weight.
1) Periodic law:
The physical and chemical properties of the elements are the periodic functions of their atomic weights.

2) Groups and sub-groups :
The vertical columns in Mendeleeffs periodic table are called groups. There are eight groups and elements in each group have similar properties. Each group is divided into sub-groups A and B.

3) Periods :
The horizontal rows are called periods. There are ‘seven’ periods in Mendeleeffs periodic table.

4) Predicting the properties of missing elements :
Based on the arrangement of elements in table, Mendeleeff predicted that some elements were missing and left blank spaces at appropriate places in the table. Later they were discovered.

5) Correction of atomic weight :
It is useful in correcting atomic weights of elements.

6) Anomalous series :
More atomic weight element like Tellurium (Ti) is placed before the less atomic weight element like Iodine in order to place these elements in the correct group.

Question 13.
How does atomic radius vary la period and group? Explain.
Answer:
Period :

1. As we move from left to right the atomic radius decreases because the electrons enter the same main shell.
2. The nuclear charge increases because of increase in atomic number of elements in period.
3. Hence, the nuclear attraction on the outer shell electron increases. As a result, the size of atom decreases.

Group:

1. Atomic radius increases from top to bottom in a group of the periodic table.
2. As we go down in a group, the atomic number of element increases. In order to accommodate more number of electrons, there are more additional shells.
3. As a result, the distance between the nucleus and the outer shell of atom increases.
4. So atomic size increases.

Question 14.
What is electron affinity? What are the factors which influence electron affinity?
Answer:
Electron affinity :

1. The electron affinity of an element is defined as the energy liberated when an electron is added to its neutral gaseous atom.
2. Electron affinity of an element is also called electron gain enthalpy of that element.
3. M_(g) + e^– → M^–_(g) + EA₁ (M = Atom of element, EA₁ = First Electron affinity)
   M^–_(g) + e^– → M^-2_(g) + EA₂ (EA₂ = Second Electron affinity)

Factors influencing Electron affinity :
1) Nuclear charge :
Greater the nuclear charge, greater the electron affinity value because of greater attraction for incoming electron.

2) Atomic size :
As the atomic size increases, the attractive force of the nucleus on the electron decreases. So electron affinity decreases.

3) Electronic configuration :
The elements having stable electronic configurations of half filled or completely filled valence sub-shells show very small tendency to accept additional electron. So the electron affinity is low or almost zero for these elements.

4) Penetrating power of the orbitals :
As the penetrating power of the orbitals increases, the electron affinity increases.

5) Screening effect or shielding effect:
More the screening effect of orbitals, less is the electron affinity value.

Question 15.
How did Mendeleeff correct atomic weights of various elements?
Answer:

• Atomic weight = Equivalent weight x Valency
• By using the formula, the atomic weight of Beryllium was calculated as 13.5 (Equivalent weight of Be = 4.5, valency = 3)
• With this atomic weight the element should be placed in wrong group.
• So Mendeleeff predicted its valency is only 2. From that he calculated the atomic weight of Beryllium as 9.
• Now it fitted into correct group.
• Similarly, Mendeleeff corrected atomic weights of Indium and Gold.

Question 16.
Answer the following questions if atomic number of element is 15.
1) What is the name of the element?
2) What is the electronic configuration of element?
3) Which period and group does it belong to?
4) How many valence electrons are there in the element?
5) What is the valency of the element?
Answer:

1. The element is phosporous,
2. The electronic configuration of element is 1s² 2s² 2p⁶ 3s² 3p³ or 2, 8, 5.
3. It belongs to 3^rd period (orbit number is 3) and V or 15 group (Number of electrons in valence orbit is 5.)
4. Number of valence electrons are 5.
5. Its valency is 8 – 5 = 3.

Question 17.
If an element belongs to 3^rd period and 17^th group, then answer the following questions.
1) What is its electronic configuration?
2) How many valence electrons are there in the element?
3) What is the valency of element?
4) What is atomic number of element?
5) What is the name of the element?
6) Give two more elements which have similar properties as this element?
Answer:

• The element belongs to 3rd period and 17^th group. So the valence orbit is 3^rd and number of valence electrons in that orbit is 7. So its electron configuration is 2, 8, 7.
• The number of valence electrons are 7.
• The valency of element = 8 – 7 = 1.
• The atomic number of element is 17.
• Name of the element is chlorine.
• Chlorine belongs to Halogen family. So Fluorine, Bromine, Iodine, and Astatine have similar properties as chlorine.

Question 18.
The elements of a periodic table are given below in the order from left to right.
Li Be B C O F Ne
1) To which period do these elements belong?
2) One element of this period is missing. Which is the missing element and where should it be placed?
3) Which one of the elements in this period shows the property of catenation?
4) Place the three elements fluorine, beryllium, and oxygen in the order of increasing electronegativity.
5) Which one of the above elements belongs to halogen series?
Answer:

1. The elements belong to 2^nd period.
2. The element which is missing is Nitrogen which is placed in between carbon and oxygen.
3. Carbon shows the property of catenation.
4. The ascending order of electronegativity for these element is Beryllium < Oxygen < Fluorine.
5. Fluorine belongs to halogen family.

Question 19.
A group of elements in periodic table is given below.
Boron, Aluminium, Gallium, Indium, and Thallium.
(Boron is the first element and Thallium is the last element)
Answer the following questions in relation to the above group of elements.
1) Which element has the most metallic character?
2) Which element would be expected to have the highest electronegativity?
3) If the electronic configuration of Aluminium is 2, 8, 3, how many electrons are there in outer shell of thallium?
4) The atomic number of Boron is 5. Write the chemical formula of the compound formed when Boron reacts with Chlorine.
5) Do the elements in the group to the right of this Boron group have more metallic or less metallic character? Justify your answer.
Answer:
1) Thallium has the most metallic character because as we move from top to bottom in a group the metallic character increases.

2) Boron has the highest electronegativity because as we move from top to bottom in a group electronegativity decreases.

3) Thallium is in the same group as Boron. So, the number of electrons in outermost shell of Thallium is 3.

4) The atomic number of Boron is 5. So, its electronic configuration is 2, 3. Therefore its valency is 3.

Whereas the atomic number of Chlorine is 17. So, its electronic configuration is 2, 8, 7. Therefore its valency is 1.

The formula of compound formed between Boron and Chlorine is BCl₃.

5) The elements in the group right to Boron group have lesser metallic character because as we move from left to right in a period metallic character decreases.

Question 20.
The following questions refer to the periodic table.
1) Name the first and the last element in period 2.
2) What happens to the atomic size of elements moving from top to bottom of a group?
3) Which of the elements has the highest electron affinity among the halogens?
4) What is common feature of the electronic configurations of the elements in group 16?
Answer:
1) The first and the last elements of 2^nd period or Lithium and Neon.

2) The atomic size decreases as we move from top to bottom in a group because there is an addition of shell each time as we move down the group.

3) Chlorine has the highest electron affinity. We know as move from top to bottom the electron affinity values decrease. But due to small size of fluorine there would be more electron-electron repulsions, if we add electron. So Chlorine has more electron affinity.

4) All these have same general outermost electronic configuration that is ns² np⁴.

Question 21.
Answer the following.
1) Elements of which groups have low ionization energy?
2) What is your guess about atomic size of an element with seven electrons among all the elements in the same period?
3) Which element has the highest electronegativity? Why?
4) Which element has the highest electropositivity? Why?
Answer:
1) Group IA, IIA elements have lower ionization energy values because they have metallic character.

2) As we move from left to right in a period atomic size decreases. So element with seven outermost electrons has least size among all the elements in the same period.

3) Fluorine has the highest electronegativity because when we move from left to right in a period atomic size decreases and electronegativity values increase. So Fluorine has the highest electronegativity.

4) Cesium has the highest electropositivity or positive character because when we move from top to bottom in a group atomic size decreases. So electropositive character increases. Therefore Cesium has the highest electropositive character.

Question 22.
Given below is the electronic configuration of A, B, C, D.

A) 1s2 2s2 2p¹	a) Which are the elements coming within the same period?
B) 1s2 2s2 2p6	b) Which are the elements coming within the same group?
C) 1s2 2s2 2p6 3s2 3p6	c) Which are the noble gas elements?
D) 1s2	d) Which group and period does the element C belong to?

Answer:
a) A and B belong to same period because the valence electrons of both the elements lie in the same orbit.
b) Elements A and C and elements B and D.
c) B and D are noble gas elements.
d) C belongs to 3^rd period (orbit number) and III group (Number of valence electrons).

Question 23.
Write down the characteristics of the element having atomic number 16.
i) Electronic configuration
ii) Period number
iii) Group number
iv) Element family
v) Number of valence electrons
vi) Valency
vii) Metal or non-metal
viii) Name of the element
Answer:
i) Electronic configuration of element is 1s² 2s² 2p⁶ 3s² 3p⁴ or 2, 8, 6.
ii) Period number is 3 because valence electron lies in 3^rd orbit.
iii) Group number is 6 because the number of valence electrons are 6.
iv) Element belongs to chalcogen family.
v) Number of valence electrons are 6.
vi) Valency = 8 – 6 = 2.
vii) It is a non-metal because in a period when we move from left to right non-metallic character increases.
viii) Name of the element is sulphur.

Question 24.
The second period element ‘F has electron gain enthalpy than the third period elements of same group ‘Cl’. Why?
Answer:

• In a group of elements, the electron gain enthalpy decreases from top to bottom.
• But in general the second element in a group, i.e. 3^rd period element has greater electron gain enthalpy than the first element, i.e. 2^nd period element.
  Ex : E.A of F < E.A of Cl.
• This is because Fluorine atom is smaller in size than Chlorine atom.
• F₂ also has strong inter electronic repulsions.
• In the addition of an electron to fluorine atom, the electronic repulsions overcome at the expense of a part of the energy liberated.

Hence the overall energy liberated is less than that of Chlorine atom.

Question 25.
Differentiate the metals and non-metals.
Answer:

Metals	Non-Metals
1. Metals have lustrous surface.	1. Non-metals do not have lustrous surface.
2. They show malleability.	2. They do not show malleability.
3. They show ductility.	3. They do not show ductility.
4. They produce sonorous sound.	4. They do not produce sonorous sound.
5. Generally they are hard.	5. Generally they are soft.
6. They are good conductors of electricity.	6. They are bad conductors of electricity.
7. Generally they liberate hydrogen gas when they are treated with acids.	7. They do not liberate hydrogen gas.

Question 26.
The electronic configuration of atom A is 2, 8, 6.
a) What is the atomic number of element A?
b) State whether the atomic size of element A is bigger or smaller than the atom having atomic number 14. Why?
c) Which of the elements exhibits similarity in chemical properties as element A 0(8), C(6), N(7), AV(18). Why?
d) How does the element form inert gas configuration?
Answer:
The electronic configuration of atom – A is 2, 8, 6.
a) Atomic number of element ‘A’ is 16, i.e. Sulphur.

b) The atom which has atomic number – 14 is Silicon (Si).

Atomic size of element decreases across period from left to right. So the atomic size of element ‘A’ is smaller than the atom having atomic number 14.

c) Element oxygen O₈ – exhibits similarity in chemical properties as element A, because they belong to the same group.

d) Given element – A becomes inert gas, i.e. Argon configuration by gaining ‘2’ electrons.

Question 27.
Select the correct answers from the choices A, B, C, D which are given with reference to the variation of properties in the periodic table. Which of the following is generally true?
A : Atomic size increases from left to right across a period.
B : Ionisation energy increases from left to right across a period.
C : Electropositive character increases going down a group.
D : Electronegativity increases going down a group.
Answer:
1) A is wrong because when we move from left to right the atomic number increases. So, the nuclear attraction over outermost orbital increases. Therefore the atomic size decreases.

2) B is correct but it does not follow a regular trend in a period.

3) C is correct. As move from top to bottom in a group atomic size increases. Therefore it is easy to lose electrons. So electropositive character increases.

4) D is wrong because as we move from top to bottom in a group atomic size increases. So electronegativity decreases.

Question 28.
Some elements belonging to second period of periodic table, and their atomic radii are given below. Observe them and write answers.

1) Write the elements in the ascending order of their atomic radii.
2) Which of the 2nd period elements closer to the configuration of inert gas?
3) Which is the outermost orbit of all these elements?
4) Which element’s atomic size is bigger, Beryllium or Carbon? Why?
Answer:

1. The ascending order of atomic sizes is O, N, C, B, Be and Li.
2. Lithium has closest inert gas configuration, i.e. 1s² 2s¹. Its nearest inert gas is Helium.
3. The outermost orbit for all these elements is second orbit.
4. Beryllium has more atomic size than Carbon. Because when we move across a period the atomic number increases. So nuclear attraction over outermost orbit increases and atomic size decreases. So carbon has lesser size than Beryllium.

Question 29.

Refer the above part of periodic table and answer the following questions.
a) Element with the least atomic size.
b) Write the electronic configuration of the elements B and E.
c) Identify the elements that have similar physical and chemical properties as the element Y.
d) Arranged elements increasing order of their electronegativity values.
Answer:
a) The element with least atomic size is E. Because when we move from left to right in a period the atomic size decreases,

b) Electronic configuration of B is 1s² 2s² 2p⁶ 3s² 3p¹ Because the element belongs to 13th group its general configuration is ns² np¹ and the element belongs to third period and its atomic number is 13. Similarly electronic configuration of E is 1s² 2s² 2p⁶ 3s² 3p¹. Because the element belongs to 16th group. Its general configuration is ns2np4 and it is in third period. So its atomic number is 16.

c) The elements which have similar physical and chemical properties with Y are X and Z. Because they lie in a single group, i.e. 1st group. In a group, elements are having similar physical and chemical properties.
d) Z < Y < X < B < C < D < E.

Question 30.
Consider the section of the periodic table given below.

1) Which is the most electronegative?
2) How many valence electrons are present in G?
3) Write the formula of the compound between B and H.
4) Which element has similar properties as J?
5) Which element has greater size-either D or E?
Answer:

1. J is the most electronegative. In a period electronegative values increase.
2. G is present in V group. So the number of valence electrons is 5.
3. B is present in first group. So its valency is 1 and hydrogen also has valency 1. Therefore the compound is BH.
4. K lies in same group as J. Elements belonging to same group have similar properties. So, K has similar properties as J.
5. E has greater size because as we move from top to bottom the atomic size increases.