Author name: Prasanna

These NCERT Solutions for Class 10 Science Chapter 4 Carbon and Its Compounds Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Carbon and Its Compounds NCERT Solutions for Class 10 Science Chapter 4

Class 10 Science Chapter 4 Carbon and Its Compounds InText Questions and Answers

In-text Questions (Page 61)

Question 1.
What would be the electron dot structure of carbon dioxide which has the formula CO₂ ?
Answer:

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur ? (Hint: The eight atoms of sulphur are Joined together in the form of a ring)
Answer:

In-text Questions (Page 68-69)

Question 1.
How many structural isomers can you draw for pentane ?
Answer:
There chain isomers or structural isomers of pentane are possible namely n-pentane, 2-methyl butane and 2, 2-Dimethyl propane.

Question 2.
What are two properties of carbon which lead to the huge number of carbon compounds we see around us ?
Answer:
The two main properties of carbon are :
(i) Tetravalency : Carbon has atomic number six. Its electronic configuration, 2, 4.
Carbon has four electrons in its valence shell and, therefore, it can attain a noble gas electronic configuration either by losing or gaining or sharing four electrons. But the loss or gain of four electrons by the carbon atom to form highly charged C⁺⁴ or C^-4 ions would require a very large amount of energy which is not generally available during a chemical reaction. So it is unable to form ionic bonds and as such it can participate only in the formation of covalent bond. Carbon is capable to form covalent bonds with Oxygen, Hydrogen, Nitrogen, Sulphur, Chlorine and many other elements giving rise to a large number of compounds.

(ii) Catenation : Carbon has the unique property to form bonds with atoms of carbon. This property of self linking of carbon atoms is called catenation. Due to catenation carbon atoms can form various types of straight chain, branched chain and cyclic chains, thus giving rise to a large no, of compounds.

Question 3.
What will be the formula and electron do not structure of cyclopentane ?
Answer:
Formula of cyclopentane : C₅H₁₀

Question 4.
Draw the structure for the following compounds.
(i) Ethanoic acid
(ii) Bromo pentane
(iii) Butanone
(iv) Hexanol
Are structural isomers possible for bromopentane.
Answer:
(i) Ethanoic acid : Molecular Formula, C₂H₅OH

(ii) Bromo pentane : Molecular Formula, C₅H₁₁-Br

(iii) Butanone, Molecular Formula, C₄H₈O

(iv) Hexanal: Molecular Formula: C₆H₁₂O
Structure :

Yes, structural isomers of Bromo pentane are possible. Some of them are as follows :

Question 5.
How would you name the following compounds ?

Answer:
(i) CH₃ – CH₂ – Br
1 – Bromo ethane

In-text Questions (Page 71)

Question 1.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction ?
Answer:
A reaction in which a substance gain oxygen or loose hydrogen is called oxidation reaction. During the conversion of ethanol to ethanoic acid, oxidising agent like alkaline potassium permanganate or acidic potassium permanganate release oxygen atoms which are gained by ethonol and converts to ethanoic acid. So the conversion of ethanol to ethanoic acid is an oxidation reaction.

Question 2.
A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used ?
Answer:
A high temperature is required for welding. Ethyne burn in oxygen and produce a large amount of heat and H₂O and CO₂ as by product. But if we used a mixture of ethyne and air it produce oxides of nitrogen with H₂O and CO₂. The oxides of nitrogen are the major air pollutants so they pollute the environment, that is why a mixture of air and ethylene is not used for welding.

In-text Questions (Page 74)

Question 1.
How would you distinguish experimentally between an alcohol and a carboxylic add ?
Answer:
Carboxylic acid turns blue litmus paper or blue litmus solution to red, but alcohols do not. Carboxylic acids reacts with sodium carbonate and liberates carbondioxide, when the evolved gas passes through the lime water, lime water turns milky. Alcohol does not give this test.

Question 2.
What are oxidising agent.
Answer;
A substance which provides oxygen for oxidation and reduce itself in the reaction is called an oxidising agent e.g. Alkaline KMnO₄.

In-text Questions (Page 76)

Question 1.
Would you be able to check if water is hard by using a detergent?
Answer:
No, because detergents do not form any precipitate with hard water and form equal amount of foam with hard and soft water.

Question 2.
People use a variety of methods to wash clothes, usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Answer:
Soaps or detergents forms micelles with the greasy or oily material associated with the clothes. To remove oily or greasy material it is necessary to mix soaps with clothes in such a way so that it combine with greasy/oily material to form a micelles. On agitation the oily dirt tends to lift the dirty surface.

So agitation is necessary for the throughout mixing of a soap with the clothes and it give rises to the formation of micelles and we can get clean clothes.

Class 10 Science Chapter 4 Carbon and Its Compounds Textbook Questions and Answers

Page No. 77

Question 1.
Ethane with the molecular formula – C₂H₆ – has
(a) 6 covalent bonds
(b) 7 covalent bonds
(c) 8 covalent bonds
(d) 9 covalent bonds
Answer:
(b) 7 covalent bonds

Question 2.
Butanone is a four-carbon compound with the functional group
(a) Carboxylic acid
(b) aldehyde
(c) ketone
(d) alcohol
Answer:
(c) Ketone

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means.
(a) the food is not cooked completely
(b) the fuel is not burning completely.
(c) the fuel is wet
(d) the fuel is burning completely.
Answer:
(b) the fuel is not burning completely.

Question 4.
Explain the nature of the covalent bond using the bond formation in CH₃Cl.
Answer:
Carbon has four electrons in its valence shell. So it require four more electrons to complete its octet. Hydrogen has one electron in its valence shell so it form one covalent bond with carbon. CH₃Cl has three hydrogen atoms and each atom will form a covalent bond with carbon. The nature of ‘C-H’ bond is purely covalent. So CH₃Cl has three pure covalent bonds. Chlorine has seven electrons in its valence shell, so it require only one electron to complete its octect. Chlorine form one covalent bond with carbon and complete its octect. But the nature of ‘C-Cl’ bond is not purely covalent, it has some polar character due to the large difference in the electronegativity of carbon and chlorine. The four covalent bonds of carbon are diverted towards the four corners of a tetrahedron because the four valency of carbon is diverted towards the four corners of tetrahedron. So the geometry of CH₃Cl will be tetrahedral.

Question 5.
Draw electron dot structures for:
(a) Ethanoic acid
(b) H₂S
(c) Propanone
(d) F₂
Answer:
(a) Ethanoic acid M.F = CH₃COOH

(b) H₂S

(c) Propane (CH₃ – C – CH₃)

(d) F₂

Question 6.
Which is a homologous series ? Explain with an example.
Answer:
It is series of similarly constituted compounds in which the members have the same functional group and have similar characteristics properties and the two consecutive members differ in their molecular formula by – CH₂.

The different members of the homologous series are called homologoues.
For example homologous series of alcohols:
CH₃-OH
CH₃-CH₂-OH
CH₃-CH₂-CH₂-OH
CH₃-CH₂-CH₂-CH₂-OH
CH₃-CH₂-CH₂-CH₂-CH₂-OH

In the above series all the members have same functional group, alcoholic group (-OH) and the two consecutive members differ in their molecular formula by – CH₂. So the given series is a homologous series.

Question 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties ?
Answer:

Question 8.
Why does micelle formation place when soap is added water ? Will a micelle be formed in other solvents such as ethanol also ?
Answer:
Soaps molecules have two ends, one is Hydrophillic, i.e., water soluble while the other end is Hydrophobic i.e., water repelling. This unique nature of soaps leads to the formation of a micelle. When soap is dissolved in water, the hydrophobic ‘tail’ of soap will not be soluble in water and the hydrocarbon ‘tail’ protruding out of water. Inside water, these molecules have a unique orientation that keeps the hydrocarbon portion out of the water. Thus a cluster is formed which is known as micelle. Ethanol also form a micelle.

Question 9.
Why are carbon and its compounds used as fuels for most applications ?
Answer:
Carbon, in all its allotropic forms, burns in oxygen to give carbonoxide along with the release of heat and light on combustion. Carbon and its compounds have a very high calorific value.
C + O₂ → CO₂ + heat + light
CH₄ + 2O₂ → CO₂ + 2H₂O + heat + light
CH₂CH₂OH + 3O₂ → 2CO₂ + 3H₂O + heat + light
There are the oxidation reactions.
Due to high calorific value, carbon and its compounds are used as a fuel.

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
Hardness of water is due to the dissolved impurities of the salts like bicarbonates, chlorides and sulphates of calcium and magnesium. Hard water does not produce lather with soap solution readily because the cations (Ca²⁺ and Mg²⁺) present in hard water react with soap to form scum (precipitates) of calcium and magnesium salts of fatty acids.

Thus scum is formed when soap is treated with hard water.

Question 11.
What change will you observe if you test soap with litmus paper (red and blue) ?
Answer:
Soap is the Sodium of Potassium salt of long chain fatty acids e.g, Sodium stearate (C₁₇H₃₅COONa). When soaps are dissolved in water, they are dissociated in C₁₇H₃₅COO^– and Na⁺ ions. Water contains H⁺ and OH^– ions. Water contains H⁺ and OH^– ions. So C₁₇H₃₅COO combine with H⁺ to form C₁₇H₃₅COOH and Na⁺ ions combine with OH^– ions to form NaOH. So soap changes blue litmus to red due to acidic character and red litmus to blue due to basic character.

Question 12.
What is hydrogenation ? What is its industrial application ?
Answer:
Unsaturated hydrocarbons (alkane and alkynes) combine readily with hydrogen in the presence of finely-divided nickel, platinum or palladium as catalysts. This process is known as Hydrogenation. The ultimate products of Hydrogenation are alkanes.
e.g.,

Hydrogenation is used in the manufacture of vanaspati ghee from vegetable oils.

Question 13.
Which of the following hydrocarbons undergo addition reaction; C₂H₆, C₃H₈, C₃H₆, C₂H₂, and CH₄.
Answer:
Unsaturated hydrocarbons undergo addition reactions.
So C₃H₆ i.e. CH₃ – CH = CH₂ and
C₂H₂ i.e. CH = CH will give addition
reations but C₂H₆, C₃H₈ and CH₄ will not give addition reactions because they are saturated.

Question 14.
Give a test that can be used to differentiate chemically between butter and cooking oil.
Answer:
When the butter is heated above 250°C they decompose with production of acrolein which has intense smell but oil does not give this test.

Question 15.
Explain the mechanism of the cleaning actions of soaps.
Answer:
Soaps molecules have two different ends, one is Hydrophillic (water soluble) while the other end is hydrophobic (water repelling), When soap is at the surface, the Hydorpholbic tail of soap will not be soluble in water and the soap will align along the surface of water with the ionic end in water and the hydrocarbon ‘tail’ protruding out of water.

Inside water, those molecules have unique orientation that keeps the hydrophobic end out of the water. This is achieved by forming clusters of molecules in which the hydrophobic tails are in the interior of the duster and the ionic ends are on the surface of the duster. This formation is called a micelle. Soap in the form of a micelle is able to dean, since the oily dirt will be collected in the centre of the micelle. The micelle stay in solution as a colloid and will not come together to precipitate because of ion-ion repulsion. So the dirt suspended in the micelles is also easily rinsed away.

Class 10 Science 4 Carbon and Its Compounds Textbook Activities

Activity 4.1 (Page 58)

• Make a list of ten things you have used or consumed since the morning.
• Compile this list with the lists made by your classmates and then sort the items into the following table.
• If there are items which are made up of more than one material put them into bot the relevant columns.

Activity 4.2 (Page 67)

• Calculate the difference in the formulae and molecular masses for (a) CH₃OH and C₂H₅OH and C₃H₇OH and (c) C₃H₇OH and C₄H₉OH.
• Is there any similarity in these three?
• Arrange these alcohols in the order of increasing carbon atoms to get a family. Can we call this family a homologous series?
• Generate the homologous series for compounds containing upto four carbons for the other functional groups given in table 4.3

Answer:
(a) CH₃OH and C₂H₃OH
Difference in the molecular formula = CH₂
The molecular mass of CH₃OH
= C + 3 × H + 1 × O × 1 × H
= 12 + 3 × 1 + 16 + 1 × 1
= 12 + 3 + 16 + 1 = 32

The molecular mass of C₂H₅OH
= 2 × C + 5 × H + 1 × O + 1 × H
= 2 × 12 + 5 × 1 + 1 × 16 + 1 × 1
= 24 + 5 + 16 + 1 = 46
The difference in molecular mass = 46 – 32 = 14

(b) C₂H₅OH and C₃H₇OH
Difference in the molecular formula = CH₂
The molecular mass of C₂H₅OH
= 2 × C + 5 + H + 1 × O + 1 × H
= 2 × 12 + 5 × 1 + 1 × 16 – 1 × 1
= 24 + 5 + 16 + 1
= 46
The molecular mass of C₃H₇OH
= 3 × C + 7 × H + 1 × O + 1 × H
= 3 × 12 + 7 × 1 + 1 × 16 + 1 × 1
= 36 + 7 + 16 + 1
= 60
Difference in the molecular mass = 60 – 46 = 14

(c) C₃H₇OH and C₄H₉OH
Difference in the molecular formula = CH₂
The molecular mass of C₃H₇OH
= 3 × C + 7 × H + 1 × O + 1 × H
= 3 × 12 + 7 × 1 + 1 × 16 + 1 × 1
= 36 + 7 + 16 + 1 = 60
The molecular mass of C₄H₉OH
= 4 × C + 9 × H + 1 × O + 1 × H
= 4 × 12 + 9 × 1 + 1 × 16 + 1 × 1
= 48 + 9 + 16 + 1
= 74
The difference in molecular mass = 74 – 60 = 14

• Yes there is a similarity in the three organic compouds. They all have same functional group.
• CH₃OH
• C₂H₅OH
• C₃H₇OH
• Yes we can call this family a homologous I series because the difference between the molecular formula of the two consecutive members differ by -“CH₂” and they have same functional group. (OH)
• Homologous series of halo-alkane

Homologous series of alcohol
CH₃-OH
CH₃-CH₂-OH
CH₃-CH₂-CH₂-OH₂-OH
CH₃-CH₂-CH₂-CH₂-OH
Homologous series of aldehyde
H-CHO
CH₃-CHO
CH₃CH₂-CHO
CH₃CH₂CH₂-CHO

Homologous series of Ketons:

Homologous series of carboxylic acid :
H-COOH
CH₃-COOH
CH₃-CH₂-COOH
CH₃-CH₂-CH₂-COOH

Activity 4.3 (Page 68)

Caution : This activity needs the teacher’s assitance.

• Take some carbon compounds (naphthalene, camphor, alcohol) one by one on a spatual and burn them.
• Observe the nature of the flame and note whether smoke is produced.
• Place a metal plate above the flame. If there a deposition on the plate in case of any of the compounds ?

Observations : Saturated hydrocarbon generally give a clean flame while unsaturated carbon compounds give a yellow flame with lots of black smoke. It results in a sooty deposit on the metal plate. Smoke is also produced in all cases and a sooty deposit on metal place.

Activity 4.4 (Page 68)

• Light a bunsen burner and adjust the air hole at the base to get different types of flames/presence of smoke.
• When do you get a yellow, sootry flame ?
• When do you yet a blue flame.

Observation: During incomplete combustion a yellow sooty flame is observed and when a sufficient oxygen is supplied it gives a clean blue flame.

Activity 4.5 (Page 70)

• Take about 3 ml of ethanol in a test tube and warm it gently in a water bath.
• Add a 5% solution of alkaline potassium permanganate drop by drop to this solution.
• Does the colour of potassium permanganate persist when it is added initially ?
• Why does the colour of potassium permanganate not disappear when excess is added ?
• Alkaline potassium permanganate is an oxidising agent so it oxidises alcohols to acids. An excess potassium permanganate, contains a lot of permanganate ions which a pink colour to the solution because all the permanganate ions are not reduced into MnO₂.

Observation : Yes the colour of alkaline potassium permanganate persist initially but is disappears immediately.

Activity 4.6 (Page 72)

Teacher Demonstration:

• Drop a small piece of sodium, about the size of a couple of grains of rice, into ethanol (absolute alcohol).
• What do you observe ?
• How will you test the gas evolved ?

Observation : Alcohols react with sodium metal and produce hydrogen sodium ethoxide. Put a candle near the mouth of the test tube taken, the evolved gas burn with a pop sound, it indicated the presence of hydrogen.

Activity 4.7 (Page 73)

• Compare the pH of dilute acetic acid and dilute hydrochloric acid using both litmus paper and universal indicator.
• Are both acids indicated by the litmus test ?
• Does the universal indicator show them as equally strong acids?

Observation:

• pH of oil HCl : 1.0
• pH of dil. acetic acid : 3.5
• Yes both the acids indicated by the litmus test. They turn blue litmus to red.
• No, Universal indicator shows that dilute hydrochloric acid is stronger than acetic acid.

Activity 4.8 (Page 73)

• Take I mL ethanol (absolute alcohol) and 1 mL glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
• Warm in a water-bath for at kast five minutes as shown in Fig.
• Pour into a beaker containing 20-50 mL of water and smell the resulting mixture.

Observation : Ethanol reacts with glacial acetic acid in presence of concentrated sulphuric acid to form esters. Esters have fruity smell or sweet smell.

Activity 4.9 (Page 74)

• Set up the apparatus as shown in chapter 2. activity 2.5
• Take a spatula-full of sodium, carbonate in a test tube and add 2 mL of dilute ethanoic acid.
• What do you observe?
• Pass the gas produced through freshly prepared lime-water. What do you observe?
• Can the gas produced by the reaction between ethanoic acid and sodium carbonate be identified by this test?
• Repeat this activity with sodium hydrogen carbonate instead of sodium carbonate.

Observation:
Ethanoic acid reacts with sodium carbonate to give rise to a salt, carbondioxide and water.

It turns lime-water milky.

Yes the gas evolved can be identified by this test.

Ethanoic acid reacts with sodium hydrogen carbonate to give rise to salt. CO2, and water.

Same result is observed with lime-water.

Activity 4.10 (Page 74)

• Take about 10 mL of water each in two test tubes.
• Add a drop of oil (cooking oil) to both the test tubes and label them as A and B.
• To test tube B, add a few drops of soap solution.
• Now shake both the test tubes vigourously for the same period of time.
• Can you see the oil and water layers separately in both the test tubes immediately after you stop shaking them ?
• Leave the test tubes undisturbed for some time and observe. Does the oil layer separate out ? In which test tube does this happen first ?

Observation:

• No, oil and water layer are not separated immediately.
• After some time water and oil layers are separated. In test tube ‘A’ oil and water layers are separated first.

Activity 4.11 (Page 76)

• Take about 10 mL of distilled water (or rain water) and 10 mL of hard water (from a tubewell or hand-pump in separate test tubes.
• Add a couple of drops of soap solution to both.
• Shake the test tubes vigourously for an equal period of time and observe the amount of foam formed.
• In which test tube do you get more foam?
• In which test tube do you observe a white curdy precipitate.

Note for teacher : If hardwater is not available in your locality, prepare some hard water by dissolving hydrogen carbonates/ sulphates/ chlorides of calcium or magnesium in water.

Observation : The test tube which contains distilled water has more foam than the test tube which contains hard water.

Activity 4.12 (Page 76)

• Take two test tubes with about 10 mL of hard water in each.
• Add five drops of soap solution to one and five drops of detergent solution to the other.
• Shake both test tubes for the same period.
• Do both test tubes is a curdy solid formed ?

Observation:

• Both the test tube have different amount of foam.
• A while curdy solid is formed in soap solution test tube.

Class 10 Science Chapter 4 Carbon and Its Compounds Additional Important Questions and Answers

Very Short Answers Type Questions

Question 1.
What is the co-valency of Carbon ?
Answer:
Four.

Question 2.
What is Vinegar ?
Answer:
5-8% solution of acetic acid in water in called vinegar.

Question 3.
What is the formula of chloroform ?
Answer:
CHCl₃.

Question 4.
How many covalent bonds are present in N₂?
Answer:
Three

Question 5.
Give the electron dot structure of Hydrogen ?
Answer:

Short Answer Type Questions

Question 1.
What are the characteristics of a homologous series?
Answer:
The general characteristics of a homologous series are:

• All the members of a series can be represented by the same general formula.
  e.g. Alcohol sereis : C_nH^– _2n+1 OH.
• Any two consecutive members have a difference of CH₂.
• All the members of a series have same functional group.
• All the members of a series have almost similar chemical properties.
• All the members show a regular gradation in the physical properties.
• All the member of a series can be synthesized by the identical methods.

Question 2.
What is longest carbon chain rule ? Explain with an example.
Answer:
During naming of organic compound certain rules are followed. Longest carbon chain rule is as follows select the longest continous chain of carbon atoms as the parent chain. If some carbon-carbon double or triple bond is present, the parent chain must contain the carbon atoms involved in it.

In this compound the three continuous chain of carbon atoms are present. One of them contain four carbon atoms, another one contain six and third chain contain seven carbon atoms. According to this rule you select the longest carbon chain which contain seven carbon atoms.

Question 3.
What is lowest number or lowest sum rule ? Explain.
Answer:
The selected longest chain is numbered using Arabic numerals and the positions of substituent groups are indicated by the number of carbon atom to which the alkyl group is attached.

“The numbering is done in the parent chain in such a way that the substituted carbon atoms have the lowest possible numbers. When series of locants containing the same number of terms are compared term by term, that series is lowest which contains the lowest number on the occasion of first difference.”

Long Answer Type Question

Question 1.
Discuss the micelles formation and cleaning action of soap with figures.
Answer:
Soaps are molecules in which the two ends have differing properties, one is Hydrophilic, that is, it dissolved in water, while the other end is Hydrophobic, that t dissolves in Hydrocarbons. When soaps is at the surface of water, the Hydrophobic ‘tall’ of soap

will not be soluble in water and the soap will align along the surface of water with the ionic end in water and the Hydrocarbon ‘tail’ protruding out of water. Inside water, these molecules have a unique orientation that keeps and Hydrocarbons portion out of the water. This is achieved by forming clusters of molecules in which the Hydrophobic tails are in the interior of the cluster and the ionic ends are on the surface of the cluster. This formation is called a micelle.

Soap in the form of a micelle is able to clean, since the oily dirt will be collected in the centre of the micelle. The micelles stay in solution as a colloid and will not come together to precipitate because of ion-ion repulsion. Thus, the dirt suspended in the micelles is also easily rinsed away. The soap micelles are large enough to scatter light. Hence a soap solution appears cloudy.

Multiple Choice Questions

Question 1.
The IUPAC name of Acetone
(a) Propanone
(b) Butanone
(c) Alkanone
(d) Propanal
Answer:
(a) Propanone

Question 2.
Aldehydes and Ketones commonly have
(a) Carboxyl group
(b) smell
(c) Melting point
(d) Boiling point
Answer:
(a) Carboxyl group

Question 3.
Sodium/Potassium salts of fatty acid are called
(a) Soaps
(b) Detergents
(c) Micelles
(d) Hydrocarbons
Answer:
(a) Soaps

Question 4.
In porpane the no. of Carbon-Carbon single bonds is/are
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 5.
The molecular formula of benzene is
(a) C₆H₁₂
(b) C₆H₆
(c) C₆H₁₀
(d) C₆H₈
Answer:
(b) C₆H₆

These NCERT Solutions for Class 10 Science Chapter 5 Periodic Classification of Elements Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Periodic Classification of Elements NCERT Solutions for Class 10 Science Chapter 5

Class 10 Science Chapter 5 Periodic Classification of Elements InText Questions and Answers

In-text Questions (Page 81)

Question 1.
Did Dobereiner’s triads also exist in the columns of Newlands’ Octaves? Compare and find out.
Answer:

Question 2.
What were the limitations of Dobereiner’s classification?
Answer:

• Concept of triads could be applied only to a limited number of elements.
• Dobereiner could identify only three triads from the elements known at that time. So this system was not found to be useful.

Question 3.
What were the limitations of Newlands’ Law of Octaves?
Answer:

1. It was found that Law of Octaves was applicable only, up to calcium.
2. It was assumed by Newland that only 56 elements existed in nature and no more elements would be discovered in the future. But, later on, a no. of new elements were discovered, whose properties did not fit into the Law of Octaves.
3. In order to fit elements into his table, Newlands adjusted two elements in the same slot, but put also some unlike elements under the same note. For example iron, which resembles cobalt and nickel in properties, had been placed far away from these elements.

In-text Questions (Page 85)

Question 1.
Use Mendeleev’s Periodic Table to predict the formulae for the oxides of the following elements: K, C, AI, Si, Ba.
Answer:
KO₂, CO₂, Al₂O₃, SiO₂, Ba₂O

Question 2.
Besides gallium, which other elements have since been discovered that were left by Mendeleev in his Periodic Table?
Answer:
Scandium, germanium.

Question 3.
What were the criteria used by Mendeleev in creating his Periodic Table?
Answer:

• Mendeleev proposed that the physical and chemical proportion of an elements are the periodic function of then atomic masses.
• The formulae of the hydrides and oxides formed by element were treated as one of the basic properties of an element for its classification.

Question 4.
Why do you think the noble gases are placed In a separate group?
Answer:
Inert gases like He, Ne, Ar, Kr, Xe were discovered very late because they are very invert and present in extremely low concentrations in our atmosphere. One of the strengths of Mendeleev’s periodic table was that, when these gases were discovered they could be placed in a new group without disturbing the existing order. Because inert gases have 2, or 8 electrons in their outer most shell and do not reach with other elements in normal conditions. So a new group can only justify their properties.

In-text Questions (Page 90)

Question 1.
How could the Modem Periodic Table remove various anomalies of Mendeleev’s Periodic Table?
Answer:
The Modern Periodic Table is based on the atomic number which is more fundamental property of the elements. The Modern Periodic Table rectify the various anomalies of Mendeleev’s Periodic Table; which are given below.

• Since this classification is based on the atomic number, the position of placing isotopes at one place is justified.
• It separate metals and non-metals and metalloids clearly.
• The position of some elements which were misfit on the basis of atomic mass is justified on the basis of atomic no. For example, argon proceeds potassium because argon has atomic number 18 and potassium has 19.
• The lanthanoids and actinoids which have properties different from other groups are placed separately at the bottom of the periodic table.
• It so removed the other anomalies found in Mendeleev’s periodic table such as grouping of the chemically dissimilar elements, and separation of the chemically similar elements.
• It separates very soft elements (Na, K…) from very hard elements (Cu, Ag, Au).
• This periodic table does not has sub-groups but it has only groups starting from 1 to 18.

Question 2.
Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis of your choice.
Answer:
Calcium and Stronsium. Because they belongs to same group. Group no. 2.

Question 3.
Name
(a) three elements that have a single electron in their outermost shells.
(b) two elements that have two electrons in their outermost shells.
(c) three elements with filled outermost shells.
Answer:
(a) Li (3) = 2, 1
Na (11) = 2, 8, 1
K (19) = 2, 8, 8, 1

(b) Be (4) = 2, 2
Mg (12) = 2, 8, 2

(c) He (2) = 2
Ne (10) = 2, 8
Ar (18) = 2, 8, 8

Question 4.
(a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?
(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything do their atoms have in common?
Answer:
(a) Atoms of lithium, sodium, and potassium has one electron in their outermost or valence shell.
(b) The atoms of ‘He’ and ‘Ne’ have complete octet. That is why they are inert.

Question 5.
In the Modem Periodic Table which are the metals among the first ten elements?
Answer:
Lithium (Li)
Beryllium (Be)

Question 6.
By considering their position in the periodic table, which one of the following elements would you expect to have maximum metallic characteristics? Ga, Ge, As, Se, Be
Answer:
Be.

Class 10 Science Chapter 5 Periodic Classification of Elements Textbook Questions and Answers

Page No. 91

Question 1.
Which of the following statements is not a correct statement about the trends when going from left to right across the periods of periodic table.
(a) The elements become less metallic in nature.
(b) The number of valence electrons increases.
(c) The atoms lose their electrons more easily.
(d) The oxides becomes more acidic.
Answer:
(c) The atoms lose their electrons more easily.

Question 2.
Element X forms a chloride with the formula XCI₂, which is a solid with a high melting point. X would most likely be in the same group of the periodic table as.
(a) Na
(b) Mg
(c) Al
(d) Si
Answer:
(b) Mg

Question 3.
Which element has
(a) two shells, both of which are completely filled with electrons?
(b) the electrons configuration is 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second shell as in its first shell?
Answer:

Question 4.
(a) What property do all elements in the same column of the periodic table as boron have in common?
(b) What property do all elements in the same column of periodic table as fluorine have in common?
Answer:
(a) Boron belongs to group no. 13 of the modern periodic table. All the elements of this group has similar valency as boron has. The valency of Boron in Z, because it has three electrons in its valency shell. So all the elements of G-13 have three valency.

(b)Valency of all the elements of the same group will be same. So all the elements of this group shows one valence like fluorine.

Question 5.
An atom has electronic configuration 2, 8, 7.
(a) What is the atomic number of this element?
(b) To which of the following elements would it be chemically similar? Atomic number are given in parentheses).
N (7), F (9), F (15), Ar (18)
Answer:
(a) 17
(b) F (9)

Question 6.
The position of three elements A, B, and C in the periodic table are shown below:

(a) State whether ‘A’ is a metal or non-metal.
(b) State whether ‘C’ is more reactive or less reactive than A.
(c) Will C be larger or smaller in size than B?
(d) Which type of ion, cation or anion, will be formed by element A?
Answer:
(a) ‘A’ is a non-metal because group no. 17 is present in the right side of periodic table and in right side of the periodic. Table non-metals are present.
(b) ‘A’ is more reactive than ‘C’ because ‘A’ has higher position in the group so it has high electronegativity than C
(c) ‘C’ will be smaller in size than B because across the period from left to right (G-16 to G-17) size decreases.
(d) A form anion (A) due to its high el ectronegativity.

Question 7.
Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the periodic table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?
Answer:
N (7) = 2, 5
P (15) = 2, 8, 5
Nitrogen is more electronegative than phosphorus because the size of nitrogen is smaller than phosphorus. So nitrogen has high effective nuclear charge. The electron negativity depend upon size and effective nuclear charge. Smaller the size higher will be the electronegativity and higher the effective nuclear charge, again in favour of high electronegative.

Question 8.
How does the electronic configuration of an atom relate to its position in the Modem Periodic Table?
Answer:
Modern Periodic Table is based on the atomic number or electronic configuration of elements. The elements of a group have same electronic configuration of the outer most valence shell. After a certain interval repetition of similar outer electronic configuration of elements of atoms causes periodicity.
e.g.,
Li (3) = 2, 1
Na (11) = 2, 8, 1
K (19) = 2, 8, 8, 1
Rb (37) = 2, 8, 18, 8, 1
Cs (55) = 2, 8, 18, 18, 8, 1
Fr (87) = 2, 8, 18, 18, 32, 8, 1
Thus, it is because of similarity in electronic configuration that all the elements have similar properties, and all the atom have the same no. of electrons in their valence shell so they belong to the same group. So the electronic configuration of an atom justify its position in the periodic table.

Question 9.
In the Modem Periodic Table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12,19,21 And 38. Which of these have physical and chemical properties Resembling calcium?
Answer:
The electronic configuration of ‘Ca’ (20) = 2, 8, 8, 2
The electronic configuration of elements with atomic numbers 12, 19, 21 and 38 are as follows:
= 2, 8, 2
= 2, 8, 8, 1
= 2, 8, 8, 3
= 2, 8, 18, 8, 2
The electronic configuration of elements with atomic numbers 12 and 38 is similar to calcium. So the physical and chemical properties of these elements is similar to calcium.

Question 10.
Compare and contrast the arrangement of elements in mendeleev’s periodic table and the Modem Periodic Table.
Answer:
Comparison in Mendeleev’s Periodic Table and the modern Periodic Table:

• Both the tables gives a systematic study of elements.
• In both the table elements are arranged in rows and columns.
• The columns are known as groups and the rows are known as periods.
• Both the tables have metals in left hand side and non-metals in right hand side.
• In both the tables after a certain interval the elements of similar properties are arranged in a same group and dissimilar elements are separated.
• The variation in the valence of elements with respect to hydrogen and oxygen across the period remains same for both the periodic tables.
• The metallic character decreases across the period from left to right and increases down the group in a similar way in both the periodic tables.

Contrast in Mendeleev’s and Modem Periodic Table.

Class 10 Science Chapter 5 Periodic Classification of Elements Textbook Activities

Activity 5.1 (Page 84)

• Looking at its resemblance to alkali metals and the halogen family, try to assign hydrogen a correct position in Mendeleev’s Periodic Table.

Question 1.
To which group and period should hydrogen be assigned?
Answer:
Electronic configuration of hydrogen resemblances of that alkali metals. Like alkali metals hydrogen combines with halogens (F, Cl, Br, …….) Oxygen and sulphur to form compounds having similar formula i.g. HCl, H₂O, H₂S like NaCl, Na₂O, Na₂S.

On the other hand. just, like halogens, hydrogen also exists as diatomic molecules and it combines with metals and non-metals to from covalent compounds.

Certainly, no fixed position can be given to hydrogen in the periodic table. This was the first limitation of Mendelev’s periodic table. He could not assign a correct position to hydrogen in his table.

Activity 5.2 (Page 85)

• Consider the isotopes of chlorine, Cl-35 and Cl-37

Question 1.
Would you place them in different slots because their atomic masses are different?
Or would you place them in the same position because their chemical properties are the same?
Answer:
According to Mendeelev’s periodic table elements of different atomic masses should be placed at different positions.

Because the chemical properties of Cl-35 and Cl-37 are same so we would like to place them in the same position in the periodic table.

Activity 5.3 (Page 85)

Question 1.
How were the positions of cobalt and nickel resolved in the Modern Periodic Table?
Answer:
The Mendeleev’s Periodic Table contain eight groups. In eighth group, there are there rows of elements namely Fe, Co, Ne and Ru, Rh, Pd and Os, Ir, Pt. In modern periodic table these three rows has been converted into three groups. The modern periodic table is based upon atomic no. not upon atomic mass. So the problem of atomic mass of Co, and Ni resolved by considering their atomic number.

Question 2.
How were the positions of isotopes of various elements decided in the Modem Periodic liable?
Answer:
The positions of isotopes of various elements decided in the periodic table by considering their atomic no. in place of atomic mass because isotopes of an element have same atomic no. So they are placed at the same position in the periodic table.

Question 3.
Is it possible to have an element with atomic number 1.5 placed between hydrogen and helium?
Answer:
No, it is not possible that an element with atomic mass 1.5 be placed between hydrogen and helium.

Question 4.
Where do you think should hydrogen be placed in the Modem Periodic Table?
Answer:
The modem periodic table is based on the atomic number of the element. Hydrogen shows resemblance with first group elements (Li, Na, K……) and 17 group elements (F, Cl, Br. ) also. So it is better that hydrogen should be placed with alkali metals on the basis of its electronic configuration. But the best way in this, that it should be placed at a separate portion in the periodic table neither with alkali metals nor with halogens.

Activity 5.4 (Page 87)

Question 1.
Look at the group 1 of the Modem Periodic Table, and name the elements present in it.
Answer:
H, Li, Na, K, Rb, Cs

Question 2.
Write down the electronic configuration of the first three elements of group J.
Answer:
H = 1
Li = 2, 1
Na = 2, 8, 1

Question 3.
What similarity do you find in their electronic configurations?
Answer:
All the elements have equal no. of electrons m their outer most shell.

Question 4.
How many valence electrons are present in these three elements?
Answer:
One.

Activity 5.5 (Page 87)

Question 1.
If you look at the Modern Periodic table, you will find that the elements Li, Be, B, C, N, O, E, and Ne are present in the second period. Write down their electronic configurations.
Answer:
Li (3) = 2, 1
Be (4) = 2, 2
B (5) = 2, 3
C (6) = 2, 4
N (7) = 2, 5
O (8) = 2, 6
F (9) = 2, 7
Ne (10) = 2, 8

Question 2.
Do these elements also contain the same number of valence electrons?
Answer:
No, these elements contain different no. of valence electrons.

Question 3.
Do they contain the same number of shells?
Answer:
Yes they contain the same no. of shells, two.

Activity 5.6 (Page 88)

• How do you calculate the valency of an element from its electronic configuration?
• What is the valency of magnesium with atomic number 12 and sulphur with atomic number 16?
• Similarly find out the valencies of the first twenty elements.
• How does the valency vary in a period on going from left to right?
• How does the valency vary in going down a group?
• If the no. of electrons in the valence shell is upto four, then its valence is equal to the no. of valence shell electrons. If the no. of valence electrons is more than five then its valence = 8 – No. of valence electrons.
• Mg (12) = 2, 8, 2
  So the valence of Mg = 2
  S (16) = 2, 8, 6
  So the valence of ‘S’ = 8 – 6 = 2
• Valency of the first twenty elements are as follows:
• Valency remain constant in going down a group.

Activity 5.7 (Page 88)

Atomic radii of the elements of the second period are given below:

Question 1.
Arrange them in decreasing order of their atomic radii.
Answer:
Li > Be > B > C > N > O

Question 2.
Are the elements now arranged in the pattern of a period in the Periodic Table?
Answer:
Yes the elements now are arranged in the pattern of a period in the periodic table.

Question 3.
Which elements have the largest and the smallest atoms?
Answer:
Lithium largest and oxygen has the smallest atoms.

Question 4.
How does the atomic radius change as you go from left to right in a period?
Answer:
Atomic radii decreases going from left to right in a period.

Activity 5.8 (Page 89)

Study the variation in the atomic radii of first group elements given below and arrange them in an increasing order.

Question 1.
Name the elements which have the smallest and the largest atoms.
Answer:
Na < Li < K < Rb < Cs
(Na) Sodium – Smallest atoms.
(Cs) Ceisium – Largest atoms.

Question 2.
How does the atomic size vary as you go down a group?
Answer:
Generally atomic size increases, down a group because the no. of shells increases.

Activity 5.9 (Page 89)

Question 1.
Examine elements of the third period and classify them as metals and non-metals.
Answer:
Elements of third period.
Na, Mg, Al, Si, P, S, Cl, Ar
Metals : Na, Mg, Al
Metalloids : Si
Non-metals : P, S, Cl, Ar

Question 2.
On which side of the Periodic Table do you find the metals?
Answer:
In the left hand side of the periodic table, we find metals.

Question 3.
On which side of the Periodic Table do you find the non-metals?
Answer:
In the right hand side of the periodic table, we find non-metals.

Activity 5.10 (Page 89)

Question 1.
How do you think the tendency to lose electrons changes in a group?
Answer:
Down the group, the effective nuclear charge experienced by valence electrons is decreasing because the outermost electrons are farther away from the nucleus. So the tendency to lose electrons decrease down the group.

Question 2.
How will this tendency change in a period?
Answer:
As the effective nuclear charge acting on the valence shell electrons increases across a period, the tendency to loss electrons will decreases.

Activity 5.11 (Page 90)

Question 1.
How would the tendency to gain electrons change as you go from left to right across a period?
How would the tendency to gain electrons change as you go down a group?
Answer:
The electronegativity of the elements increases across the period from left to right so the tendency to gain electrons increase from left to right across a period. The electronegativity of the elements decreases down the group. So the tendency to gain electrons decreases down the group.

Class 10 Science Chapter 5 Periodic Classification of Elements Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
Who gave (i) Law of triads (ii) Law of octaves?
Answer:
(1) Dobereiner
(2) Newland.

Question 2.
To which elements Mendeleev gave the name Eka-aluminium and Eka-silicon?
Answer:
Galium and Germanium respectively.

Question 3.
Which important property did Mendeleev use to classify the elements in his periodic table?
Answer:
Atomic masses

Question 4.
What are horizontal rows and vertical columns of the periodic table called?
Answer:
Horizontal rows are called periods and vertical rows are called groups

Question 5.
What is the basis of long form of the periodic table called?
Answer:
Atomic number.

Short Answer Type Questions

Question 1.
What was the need for classification of elements?
Answer:
When a large number of elements discovered, the detailed study of each and every element became very time consuming and Highly reactive elements formed a large number of compounds. So, it would be very difficult to study the chemistry after all these elements and their large number of compounds individually. For the systematic study of elements, it led to the need for classifying elements into families or groups having somewhat similarities in their chemical and physical properties.

Question 2.
What is the basis of Mendeleev’s Periodic Table?
Answer:
Mendeleev’s Periodic Table is based on the fundamental periodic law which was proposed by Mendeleev. This law is stated as The physical and chemical properties of the elements are the periodic function of their atomic masses. It suggested that when the elements are arranged in order of increasing atomic masses, the elements with similar properties are placed in the same group.

Question 3.
What is periodic table? What do you mean by classification of elements?
Answer:
It is defined as the arrangement of all the known elements starting from atomic no. 1 to 114, according to their physical and chemical properties in a tabular form. This method of arranging similar elements and separating them from dissimilar elements is called classification of elements. The main aim of classification is to make the study of the elements and their compounds systematic and easier. The elements are arranged in groups and periods. The chemical and physical properties of an element can be predicted by knowing its position in periodic table.

Long Answer Type Question

Question 1.
Give a brief account in periodicity of valence.
Answer:
The electrons present in the outermost shell or valence shell of the elements are called valence electrons and they normally represent the valency of the element.

The valence is either equal to the number of electrons in the outermost shell or eight minus the number of outermost electrons.

Variation along a period; The number of electrons in the valence shell increases from one to eight as we move from left to right in the periodic table. The valency with respect to Hydrogen initially increases from 1 to 4 and then decreases from 4 to zero.

In a similar manner, the valency with respect to chlorine also Initially increases from 1 to 4 and then decreases from 4 zero.

Variation along a group; The valency of the element depends upon its valence shell configuration. Since a group, the valence shell configuration of the elements remains almost the same, they exhibit a common valency. For example, all the elements of group 1 have valency one and those of group 2 have valency two.

Multiple Choice Questions

Question 1.
The Dobereiner’s Triad is
(a) Li, Na, K
(b) Li, Na, Ca
(c) Na, K, Ca
(d) Li, Na, Mg
Answer:
(a) Li, Na, K

Question 2.
The valency of alkali metals is/are
(a) 2
(b) 3
(c) 4
(4) 1
Answer:
(4) 1

Question 3.
The Modem periodic table is based on
(a) Atomic no.
(b) Atomic mass
(c) Valency
(d) Inner shell electrons
Answer:
(a) Atomic no.

Question 4.
The Octave rule is given by
(a) Newland
(b) Dobereiner
(c) Moselay
(d) Mendeleev
Answer:
(a) Newland

Question 5.
The Mendeleev’s periodic table is based on
(a) Atomic no.
(b) Atomic mass
(c) Valency
(d) Inner shell electrons
Answer:
(b) Atomic mass

These NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2

Question 1.
Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.
Solution:

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Solution:
Let P (x₁, y₁) and Q (x₂, y₂) be the points of trisection of A (4, -1) and B (- 2, – 3).

Therefore, P divides AB internally in the ratio of 1.2. Therefore, by section formula coordinates of P are

Now, Q also divides AB internally in the ratio 2 : 1, so the coordinates of Q are

Therefore, the coordinates of the points of trisection of the line segment joining A and B are (2, \(\frac { 5 }{ 3 }\)) and (0, \(\frac { 7 }{ 3 }\))

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in given figure below. Niharika runs \(\frac { 1 }{ 4 }\) the distance AD on the 2nd line and posts a green flag. Preet runs \(\frac { 1 }{ 5 }\) th distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Solution:
If the position of Niharika be M and the position of Preet by N the coordinate of point M are (2, 25) and the coordinate of point N are (8, 20) Distance between M and N are

Rashmi has to post a blue flag exactly halfway between the line MN. Suppose this point is 0. Then O divides the line MN at the ratio 1 :1
So, the coordinate of point O (x₁, x₂)

The coordinate of O are (5, 22.5)
∴ The position of the blue flag is 5th line at a distance of 22.5 m

Question 4.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Let the required ratio be k : 1

Question 5.
Find the ratio in which line segment joining A (1, -5) and B (-4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
Solution:

Question 6.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution:

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Solution:

Question 8.
If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = \(\frac { 3 }{ 7 }\) AB and P lies on the line segment AB.
Solution:

Question 9.
Find the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts.
Solution:

Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
[Hint: Area of a rhombus = \(\frac { 1 }{ 2 }\) (product of its diagonals)]
Solution:

These NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3

Question 1.
Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Solution:

Question 2.
In each of the following find the value of ‘k’ for which the points are collinear.
(i) (7, -2), (5, 1), (3, k)
(ii) (8, 1), (k, -4), (2, -5)
Solution:

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Solution:

Question 5.
You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4, -6), B (3, -2) and C (5, 2).
Solution:

These NCERT Solutions for Class 10 Science Chapter 6 Life Processes Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Life Processes NCERT Solutions for Class 10 Science Chapter 6

Class 10 Science Chapter 6 Life Processes InText Questions and Answers

In-text Questions (Page 95)

Question 1.
Why is diffusion insufficient to meet the oxygen requirement of multicellular organisms like humans?
Answer:
In multicellular organisms like man, hydra, spirogyra, the transport of oxygen and nutrients takes place by the process of diffusion but in chain smokers, walls between alveoli of lungs gets ruptured, resulting into abnormal alveoli having less surface for gaseous exchange. So gaseous exchange is reduced. In such cases heart pumps more blood to provide adequate oxygen. Thus, simple diffusion will not meet requirements of all the cells.

Question 2.
What criteria do we use to decide whether something is alive?
Answer:
Respiration, digestion, excretion, movement, reproduction etc. are the criteria to decide whether something is alive.

Question 3.
What are outside raw materials used for by an organism?
Answer:
The outside sources of energy could be quite varied; since the environment is not under the control of the individual organism. The process of acquiring oxygen from outside the body, and to use it, is what we commonly call respiration.

Question 4.
What processes would you consider essential for maintaining life?
Answer:
The following processes would be essential for maintain Life are –
(i) Respiration
(ii) Nutrition
(iii) Diggestion
(iv) Excretion
(v) Transportation

In-text Questions (Page 101)

Question 1.
What are the differences between autotrophic nutrition and heterotrophic nutrition ?
Answer:

Question 2.
Where do plant get each of the raw materials required for photosynthesis ?
Answer:
Break-down of pryuvate using (CO₂) takes place in the mitochondria. Exchange of gases takes place through stomata and water from soil through root hairs and light/temperature from the sun and green colour from chlorophyll and the photosynthesis occurs in the grana of chloroplast and respiration in the stroma of chloroplast. So the required raw material for photosynthesis are :
(i) Sunlight, (ii) CO₂, (iii) H₂O (iv) Chlorophyll

Question 3.
What is the role of the acid in our stomach ?
Answer:
HCL
(a) It makes the medium acidic.
(b) Softens the food.
(c) Kill germs i.e., bacteria/any foreign organism.

Question 4.
What is the function of digestion enzymes ?
Answer:
In Mouth (i) Salivary Amylase: Starch (complex sugars) → Maltose
In stomach (ii) (a) Pepsin – Proteins → peptones and proteoses
(b) Renin – Curdles milk (-nt in adults)

In Duodenum: 1. Bile juice – Alkaline – Emulsifies fat.
2. Pancreatic juice – do –
(a) Trypsin : Proteins→ Peptones
Proteoses → Amino Acid and Peptides
(b) Amylosin : Complex sugars → monosaccharides
(c) Lipase : Fats → Fatty acids and glycerol

In Small Intestine – Intestinal juice – Alkaline –
(i) Erepsin : Peptides → Amino Acids
(ii) Invertase : Cane sugar → Glucose + Fructose
(iii) Lactase : Lactose → glucose + galactose

Question 5.
How is the small intestine designed to absorb digested food ?
Answer:
Small Intestine: Small intestine is about 6 metres in length and 2.5 cm in diameter. It remains coiled in the abdominal cavity. The small intestine consists of three parts.

• Duodenum : It is U shaped in appearance. The openings of common bile duct and pancreatic duct opens in it.
• Jejunum : It is about 2.5 metres long.
• Ileum : It is about 3.5 metres long. It is distal in position.

In-text Questions (Page 105)

Question 1.
What advantage over an aquatic organism does a terrestrial organism have with regards to obtaining oxygen for respiration ?
Answer:
The organisms that live in water use the oxygen fairly low compared to the amount of oxygen in the air. So the rate of breathing is aquatic organisms is much faster than that seen in terrestrial organisms

Question 2.
What are the different ways in which glucose is oxidised to provide energy in various organisms ?
Answer:
A biochemical process of stepwise oxidation or breaking down of organic compounds (simple carbohydrate like glucose) to release energy inside the hying cell at body temperature. The energy released in respiration sets stored in Adenosine Triphosphate [ATP] molecules.

Question 3.
How is oxygen and carbon dioxide transported in human beings?
Answer:
When haemoglobin combines with oxygen it forms oxyhaemoglobin.

• Carbon dioxide it forms carbamino haemoglobrn.
• Carbon monoxide it forms carboxy haemoglobin.

Question 4.
How are the lungs designed in human beings to maximize the area tor exchange of gases ?
Answer:
Mechanism of Breathing : Breathing is a mechanical process. It is complete in two steps.
1. Inspiration [Inhalation]: It is a process by which fresh air enters the lungs. During inspiration the volume of thoracic cavity and lungs increases due to

• Contraction of rib muscules, which pull the ribs upwards.
• Contraction of diaphragm muscles causes flatening of dome-shaped diaphragm.
• The air pressure in lungs decreases and air rushes in lungs.

2. Expiration [Exhalation] : It is process by which foul air is expelled out from the lungs. During expiration volume of thoracic cavity decreases due to

• Relaxation of rib muscles which pushes the ribs inward and
• Due to relaxation of diaphragm muscles diaphragm becomes dome shaped. This presses the lungs and air is expelled out.

In-text Questions (Page 110)

Question 1.
What are the components of transport system in human beings? What are the functions of these components?
Answer:
Blood transports food, oxygen and waste materials in our bodies. It is a connective tissue. It consists of a fluid called plasma in which RBCs, WBCs, Blood platelets are suspended. Plasma transports food, carbondioxide and nitrogenous wastes indissolved forms. (O₂) is carried by RBCs and like salts and other substances transported by the blood. So we need a pumping organ to push blood around the body. A network of tubes to reach all the tissues and a system in place to ensure that this network can be repaired if damaged.

Question 2.
Why is it necessary to separate oxygenated and Deoxygenated blood in mammals and birds?
Answer:
It is necessary to separate oxygenated and deoxygenated blood in mammals and birds because while we take food due to the oxidation of food in respiration, energy is released in the form of glucose and CO₂. This CO₂ comes from different organs and tissues to the heart in the form of impure blood and then transferred to the lungs and comes out in the form of expiration (breathing) and intake of (O₂) in the form of inspiration to the lungs and then to the heart as a pureblood. This process takes place though haemoglobin hence it is necessary.

Question 3.
What are the components of transport system in highly organized plants?
Answer:
Xylem and phloem are the main components of transport system in highly organised plants.

Question 4.
How are water and minerals transported in plants?
Answer:
Water and minerals are transported in plants through Xylem components.

Question 5.
How is food transported in plants?
Answer:
Food is transported in plants by the phloem and its components.

In-text Questions (Page 112)

Question 1.
Describe the structure and functioning of nephrons ?
Answer:
Each nephron is made up of two main

Malpighian body (renal corpusle) Renal tubule Malphigian body comprised of cup-shaped Bowman’s capsule and network of blood capillaries the glomerulus. The Bowman’s capsule is lined by a single layer of Squamous epithelial cells. An different arteriole enters a Bowman’s capsule to form glomerulus. The different arteriole leaves the glomerulus. The Bowman’s capsule leads into the renal tubule, is highly coiled

looped and it consists of Proximal convoluted tubule (PCT) loop of Henle (U shaped) and Distal convoluted tubule (DCT). The collecting tubules of many nephrons join to form the collecting duct. It collect urine from nephrons and transport it into renal pelvis through renal pyramids.

Question 2.
What are the methods used by plants to get rid of excretory products?
Answer:
The methods used by plants to get rid of excretory products are excess water by transpiration waste products generated during photosynthesis and are stored as resins and gums. Plants also excrete some waste substances into the soil around them.

Question 3.
How is the amount of urine produced regulated?
Answer:
In a normal healthy adult the initial filtrate in the kidneys about 180 L daily. However, the volume actually excreted is only a litre or two a day, because the remaining filtrate is reabsorbed in the kidney tubules.

Class 10 Science Chapter 6 Life Processes Textbook Questions and Answers

Page No. 113

Question 1.
The kidneys in human beings are a part of the system for
Answer:
(c) Excretion

Question 2.
The xylem in plants are responsible for
Answer:
(a) Transport of water

Question 3.
The autotrophic mode of nutrition requires-
Answer:
(d) All of the above

Question 4.
The break down of pyruvate to give CO₂, H₂O and energy takes place in
Answer:
(b) Mitochondria.

Question 5.
How are fats digested in our bodies? Where does this process take place ?
Answer:
Fats are digested in our body in the liver, which is the largest gland in the human body. It secrets bile juice, which is stored in gall bladder before it is poured into duodenum through a common bile duct. Bile is alkaline due to the presence of bile salts and bile pigments. Bile helps in-

• Emulsification of fat (emulsification means breaking the fat into small globules)
• Activates the pancreatic and intestinal enzymes by making the food alkaline.

Question 6.
What is the role of saliva in the digestion of food?
Answer:
Saliva performs the following functions in the digestion of food. It moistens and lubricates, food and helps in swallowing. It contains an enzyme, salivary amylase, which carries out partial digestion of starch into maltose.

Question 7.
What are the necessary conditions for autotrophic nutrition and what are its by-products?
Answer:
Autotrophic organisms synthesize their own food from simple in organic raw materials i. e. CO₂, and H₂O in the presence of sunlight and chlorophyll. This is known as photosynthesis glucose is the main product of photosynthesis, oxygen is released as a by product of photosynthesis, ATP and NADPH synthesized during light reaction are used in dark reaction for Fixation of CO₂.

Question 8.
What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration?
Answer:

Question 9.
How are the aveoli designed to maximise the exchange of gases ?
Answer:

The alveoli provide a surface where the exchange of gases can take place. The walls of the alveoli contain an extensive network of blood vessels.

Question 10.
What would be the consequences of a deficiency of blood vessels, haemoglobin in our bodies?
Answer:
Function of haemoglobin in our bodies is transportation of (O₂) and (CO₂). Oxygen combines with haemoglobin of red blood cells in lungs to form oxyhaemoglobin (bright red) In tissues Oxyhaemoglobin loses O₂ to form haemoglobin due to low concentration of (O₂) and high concentration of CO₂, water and Energy in the form of ATP. Transport of CO₂, liberated during metabolic activities in the cells. It is transported to lungs in two forms.

(Hb) is called the respiratory pigment found in red blood cells of mammals including humans. But in earthworm it is found dissolved in plasma.

Note: Due to the deficiency of haemoglobin in our body exchange of gases will not take place (reduced). In such cases heart pumps more blood to provide adequate oxygen.

Question 11.
Describe double circulation In human beings. Why is It necessary?
Answer:
Double circulation: “The flow of blood twice through the heart before it is pumped throughout the body is known as double circulation.”

Question 12.
What are the differences between the j transport of materials in xylem and phyloem?
Answer:
Transport of materials in plants: In higher plants the transport system is made up of tube-like structures called xylem and phyloem. The water and minerals absorbed by roots are transported upward through the xylem tissue, while food prepared by the leaves is transported downward through sieve tubes of the phyolem.

Question 13.
Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning?
Answer:
Alveoli in lungs: Oxygen diffuses from alveoli to the blood capillaries where as CO₂, diffuses from blood capillaries into the lungs alveoli. Each branchiole ends into a cluster of tiny air chambers called the air sacs/alveoli. Alveoli are lined by a layer of moist flat epithelial cells and surrounded by network of blood capillaries. It is the functional unit of the lungs. Alveoli are the actual site of gaseous exchange.

Nephrons the Kidneys: Each kidney has large number of filtration units called nephrons. Their main function is to filter the urine.

Class 10 Science Chapter 6 Life Processes Textbook Activities

Activity 6.1 (Page 96)

• Take a potted plant with variegated leaves for example, money plant or crotons.
• Keep the plant in a dark room for three days so that all the starch gets used up.
• Now keep the plant in sunlight for about six hours.
• Pluck a leaf from the plant. Mark the green areas in it and trace them on a sheet of paper.
• Dip the leaf in boiling water for a few minutes.
• After this, immerse it in a beaker containing alcohol.
• Carefully place the above beaker in a water-bath and heat till the alcohol begins to boil.
• What happens to the colour of the leaf?
  What is the colour of the solution?

• Now dip the leaf in a dilute solution of iodine for a few minutes.
• Take out the leaf and rinse off the iodine solution.
• Observe the colour of the leaf and compare this with the tracing of the leaf done in the beginning.
• What can you conclude about the presence of starch in various areas of the leaf?

Answer:
(i) The leaves are kept in dark place for about three days to de-starch, then placed in bright sunshine for three to four days.
(ii) The leaf is boiled in alcohol; so the green parts of the leaf get de-colourised. After washing the leaf in water; pour some iodine solution over the colourless leaf and observe the change in colour of the leaf .
(iii) The outer part of the leal that was originally white does not turn blue-black, which shows that no starch is present in this outer part of the leaf.
(iv) The inner part of the leaf turns blue-black on adding iodine solution to it. This shows that starch is present in the inner part. This is illustrated by the figure.
We conclude that photosynthesis does not take place without chlorophyll.

Activity 6.2 (Page 97)

• Take two healthy pottéd plants which are nearly the same size.
• Keep them in a dark room for three days.
• Now place each plant on separate glass plates. Place a watch-glass containing potassium hydroxide by the side of one of the plants. The potassium hydroxide is used to absorb carbon dioxide.

• Cover both plants with separate bell-jars as shown in Fig.
• Use vaseline to seal the bottom of the jars to the glass plates so that the set-up is air-tight.
• Keep the plants in sunlight for about two hours.
• Pluck a leaf from each plant and check for the presence of starch as in the above activity.

Question 1.
Do both the leaves show the presence of the same amount of starch?
Answer:
(a) With potassium hydroxide : [KOH] – KOH will absorb all the CO₂ present in set-up apparatus. So the leaves of this experiment show very less amount of starch.
(b) Without potassium hydroxide: The leaves shows the presence of the large amount of starch in this set up.

Question 2.
What can you conclude from this activity?
Answer:
We conclude that during photosynthesis synthesize of starch in the leaf does not take place without carbondioxide.

In other words CO₂, is necessary for the process of photosynthesis to occur.

Activity 6.3 (Page 99)

• Take 1 mL starch solution (1%) in two test tubes (A and B).
• Add 1 mL saliva to test tube A and leave both test tubes undisturbed for 20-30 minutes.
• Now add a few drops of dilute iodine solution to the test tubes.

Question 1.
In which test tube do you observe a colour change?
Answer:
In Test Tube (A) it contains an enzyme; salivary amylase, which carries out partial digestion of starch into maltose due to this the colour of this tube will be yellowish.

Question 2.
What does this indicate about the presence or absence of starch in the two test tubes?
Answer:
In Test-Tube (B) after adding few drops of Iodine solution the colour will be blue-black it indicate the presence of starch in T.T. (A) and absence (-nee) in T.T. (B) does not indicate the digestion of starch.

Question 3.
What does this tell us about the action of saliva on starch?
Answer:
This tells us that it is done with the help of biological catalyst called enzymes which contains saliva called salivary amylase that breaks down starch a complex mole and to give sugar.

Activity 6.4 (Page 101)

• Take some freshly prepared lime water in a test tube.
• Blow air through this lime water.
• Note how long it takes for the lime water to turn milky.
• Use a syringe or pichkari to pass air through some fresh lime water taken in another test tube (Fig.)
• Note how long it takes for this lime water to turn milky.

Solution of Test-Tube (a) The less amount of water turns milky because there is very less quantity CO₂ in air. It shows the aerobic respiration.
Solution of Test-Tube (b) Air being exhaled into lime water will turn milky quickly because we realease CO₂ in our breath. It shows the anaerobic respiration.

Question 1.
What does this tell us about the amount of carbon dioxide in the air that we breathe out ?
Answer:
It means we give (CO₂) and intake (O₂) the breathed out CO₂ turns the lime water milky. It tells us about Aerobic respiration.

Activity 6.5 (Page 101)

• Take some fruit juice or sugar solution and add some yeast to this. Take this mixture in a test tube fitted with a one-holed cork.
• Fit the cork with a bent glass tube. Dip the free end of the gbss tube ìnto a test tube containing freshly prepared lime water.

Question 1.
What change is observed in the lime water and how long does it take for this change to occur?
Answer:
Accumulation of lactic acid causes muscle fatigue It depend upon the kind of bacteria. It is called fermentation. Lime water change into milky.

Question 2.
What does this tell us about the products of fermentation ?
Answer:
In Anaerobic respiration, carbondioxide and lactic acid/ethyl alcohol is produced, it is called fermentation.

Activity 6.6 (Page 103)

• Observe fish in an aquarium. They open and close their mouths and the gill-slits (or the operculurn which covers the gill-slits) behind their eyes also open and close. Are the timings of the opening and closing of the mouth and gill-slits coordinated in some manner?
• Count the number of times the fish opens and closes its mouth in a minute.
• Compare this to the number of times you breathe in and out in a minute.
• Fishes take in water through. their mouths and in the gills the dissolved oxygen is taken up by the blood.
• The rate, of breathing is aquatic organisms is much faster than that seen in terrestrial organisms.

Activity 6.7 (Page 105)

• Visit a health centre in your locality and find out what is the normal range of haemoglobin content in human beings.
• It is the same for chiIdrtn and adults?
• Is there any difference in the haemoglobin levels for men and women?
• Visit a veterinary clinic in you locality. Find out what is the normal range of haemoglobiri content In an animal like the buffalo or cow.
• Is this content different in calves, male and female animals ?
• Compare the difference seen in male and female human beings and animals.
• How would the difference, if any, by explained ?

Normal Range of haemoglobin
Hb Level’s for Animals
for buffalo → 12.9 –
for cow → 11.3 –
for calves male → 6 – 11
for calved female → 6 – 11

Activity 6.8 (Page 109)

• Take two small pots of approximately the same size and having the same amount of soil. One should have a plant in it. Place a stick of the same height as the plant in the other pot.
• Cover the soil in both pots with a plastic sheet so that moisture cannot escape by evaporation.
• Cover both sets, one with the plant and the other with the sick1 with plastic sheets and place in bright sunlight for half an hour.

Question 1.
Do your observe any difference in the two cases?
Answer:
From both experiments it is clear that transpiration takes place only by aerial parts of the plants (leaves) not from the pot soil.
Note : The rate of Transpiration by Atmospheric condition.

• modification of leaf into spines.
• having thick cuticle
• If wary coating is made on the sun faced leaf.

Class 10 Science Chapter 6 Life Processes Additional Important Questions and Answers

Very Short Answer Type Questions

Question 1.
What kind of nutrition occurs in fungi?
Answer:
Heterotrophic.

Question 2.
Name the cellular organelles where the following process occurs:
(i) Photosynthesis (ii) Cellular aerobic respiration
Answer:
(i) Chloroplast
(ii) Mitochondria

Question 3.
Which raw material is primarily responsible for the release of molecular (O₂) in photosynthesis?
Answer:
In photosynthesis, the molecular of (O₂) is released from water.

Question 4.
Name the cartilaginous flap which closes the glottis to check the entry of food into It during swallowing.
Answer:
Epiglottis

Question 5.
Give one example each of the following-
(i) Autotroph
(ii) Herbivore
(iii) Carnivore
(iv) Omnivore
Answer:
(i) Green plants,
(ii) Cow,
(iii) Lion,
(iv) Man.

Short Answer Type Questions

Question 1.
Why are plants green?
Answer:
Plants possess green chloroplastic pigments called chlorophyll which absorb most of the blue and red regions of the incident light. They reflect green light so the plants look green in colour.

Question 2.
What is emulsification? Name the organ where fat is emulsified in the alimentary canal of human beings.
Answer:
Emulsification is the breakdown of the large fat droplets into smaller ones. It occurs in the small intestine.

Question 3.
What is the role of hydrochloric acid in human digestive system ?
Answer:
Hydrochloric acid (HCl) secreted inside the stomach makes the medium acidic for digestion of protein. HCl also causes disinfection of the food.

Long Answer Type Question

Question 1.
Explain the following aspects of photosynthesis in plants
(i) The role of chlorophyll
(ii) Dark reaction
(iii) Calvin-Benson cycle.
Answer:
Mechanism of Photosynthesis: G.K., phota = light,
Synthesis = putting together

(i) The role of chlorophyll: Chlorophyll are the green coloured photosynthetic pigments present in all the green plants. They mainly occur in the leaves, through small amount of these pigments may also be present in stems and green branches. There are mainly five spectral forms of chlorophylls chlorophyll a, b, c, d and e.

The chlorophyll (a) and chlorophyll (b) occur in all green plants, whereas chlorophyll (c), (d) and chlorophyll (e)are distributed only in some algae. The chlorophylls give of white light except green light. They reflect green light. This is the reason that photosynthesis does not occur in green light.

(ii) Dark reaction : The dark reaction of photosynthesis utilizes the product of light reaction ie. Assimilatory power (NADPH and ATP) to incorporate carbon from carbon dioxide into carbohydrate. Although, the reaction itself does not require light but it occurs in light and continues for a very brief period after a plant is kept in the dark. The dark reaction occurs inside the stroma of chloroplast where all the necessary enzymes are located.

(iii) Calvin-Benson cycle: Fixation of CO₂, into carbohydrate is not one step process. It involves several reactions which occur in a cyclic manner. Carbon dioxide enters at the starting point of the cycle where it reacts with a pentose sugar, called ribulose bisphosphate (KUBP). The first sugar synthesised from carbon dioxide is a triose sugar- 3 (Phosphoglyceric Acid). A large number of reactions occurring in a cyclic manner synthesise hexose and regenerate ribulose bisphosphate. This cycle was discovered by Melvin-Calvin and Andy Benson. Therefore, this cycle is known as Calvin- Benson cycle/Calvin cycle or C₃, cycle.

Multiple Choice Questions

Question 1.
The matrix of chloroplast is known as
(a) Grana
(b) Stroma
(c) Thylakoid
(d) Lamella
Answer:
(b) Stroma

Question 2.
The green colour of plants is due to the presence of
(a) Chlorophyll
(b) Carotene
(c) Xanthoplyll
(d) Starch
Answer:
(a) Chlorophyll

Question 3.
The word ‘sapro’ means
(a) Cell sap
(b) Dead
(c) Other
(d) Rotten
Answer:
(d) Rotten

Question 4.
The visible part of electromagnetic spectrum lies in between
(a) X-ray and Ultra-violet
(b) Ultraviolet and infrared
(c) Infrared and microwave
(d) Gamma-ray and X-ray
Answer:
(b) Ultraviolet and infrared

Question 5.
Holozoic nutrition occurs in
(a) Amoeba
(b) Plasmodium
(c) Roundworm
(d) Bacteria
Answer:
(a) Amoeba

These NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i) 7 cm, 24 cm,-25 cm
(7)² + (24)² = 49 + 576 = 625 = (25)² = 25
∴ The given sides make a right angled triangle with hypotenuse 25 cm

(ii) 3 cm, 8 cm, 6 cm(8)² = 64
(3)² + (6)² = 9 + 36 = 45
64 ≠ 45
The square of larger side is not equal to the sum of squares of other two sides.
∴ The given triangle is not a right angled.

(iii) 50 cm, 80 cm, 100 cm
(100)²= 10000
(80)² + (50)² = 6400 + 2500
= 8900
The square of larger side is not equal to the sum of squares of other two sides.
∴The given triangle is not a right angled.

(iv) 13 cm, 12 cm, 5 cm
(13)² = 169
(12)² + (5)²= 144 + 25 = 169
= (13)² = 13
Sides make a right angled triangle with hypotenuse 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM . MR.
Solution:
We have PQR is a right triangle and PM ⊥ QR.

Question 3.
In the given figure, ABD is a triangle right angled at A and AC ⊥. BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
Solution:

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:

Given: In ∆ABC, ∠C = 90° and AC = BC
To Prove: AB² = 2AC²
Proof: In ∆ABC,
AB²= BC² + AC²
AB² = AC² + AC² [Pythagoras theorem]
= 2AC²

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC² , Prove that ABC is a right triangle.
Solution:
Given that ABC is an isosceles triangle with AC = BC and given that AB² = 2AC²
Now we have AB² = 2AC²
AB² = AC² + AC²
But AC= BC (Given)
AB² = AC² + BC²
Hence by Pythagoras theorem ∆ABC is a right triangle where AB is the hypotenuse of ∆ABC.

Question 6.
ABC is an equilateral triangle of side la. Find each of its altitudes.
Solution:
Given: In ∆ABC, AB = BC = AC = 2a

We have to find length of AD
In ∆ABC,
AB = BC = AC = 2a
and AD ⊥ BC
BD = \(\frac { 1 }{ 2 }\) x 2 a = a
In right angled triangle ADB,
AD² + BD² = AB²
⇒ AD² = AB² – BD²= (2a)² – (a)² = 4a²– a²= 3a²
AD = √3a

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Given: ABCD is a rhombus. Diagonals AC and BD intersect at O.
To Prove: AB²+ BC²+ CD²+ DA² = AC²+ BD²

Question 8.
In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF².
Solution:
(i) Given: ∆ABC, O is any point inside it,
OD, OE and OF are perpendiculars to BC, CA and AB respectively.
To Prove:

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. ind the distance of the foot of the ladder from base of the wall.
Solution:
By Pythagoras theorem

Hence the distance of the foot of the ladder from base of the wall in 6 m.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Let AB i.e., the vertical pole of height 18 m and AC be the guy wire of 24 m long. BC is the distance from the vertical pole to where the wire will be staked.
By Pythagoras theorem

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1\(\frac { 1 }{ 2 }\) hours?
Solution:

Question 12.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:
We have two poles.
We have
BC = 12 m
AB = 11 – 6
AB = 5 m

By Pythagoras theorem in right triangle ABC
AC² = AB² + BC²
AC² = (12)² +(5)²
AC² = 144 + 25
AC² = 169
AC = 13 m
Hence the distance between the tops is 13 m

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
Solution:
Given: ∆ABC is a right angled at C D and E are the points on the side CA and CB.
To Prove: AE² + BD² = AB² + DE²
Proof : ∆ACE is right angled at C
AE² = AC² + CE²… (i)
(Pythagoras theorem)

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see the figure). Prove that 2AB² = 2AC² + BC².

Solution:
We have
BD = 3CD
∴ BC = BD + DC
⇒ BC = 3CD + CD
BC = 4CD
⇒ CD = \(\frac { 1 }{ 4 }\)BC … (i)
And BD = 3CD
⇒ BD = \(\frac { 3 }{ 4 }\)BC …(ii)
Since ∆ABD is a right triangle, right angled at
AB²= AD² + BD² …(iii)
Similarly, ∆ACD is right angled at D.
AC² = AD² + CD² …(iv)
Substracting (iv) from (iii)
We get

Question 15.
In an equilateral triangle ABC, D is a point on side BC, such that BD = \(\frac { 1 }{ 3 }\)BC. Prove that 9AD² = 7AB².
Solution:

Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
Let ABC be an equailateral traingle of each side x. AD be its altitude.
So, AB = BC = CA = x
and BD = DC = \(\frac { 1 }{ 2 }\) BC = \(\frac { x }{ 2 }\)
In right triangle ADC in which ∠D = 90°
AD = perpendicular,
DC = base and AC = hypotenuse.
Apply Pythagorus theorem, we get

So, we get length of each side is x an length of altitude is \(\frac{\sqrt{3} x}{2}/latex]
Then, three times the square of each side = 3 x (x)² = 3x² … (i)
and four times, the square of its altitudes = 4 x [latex]\frac { 3 }{ 4 }\)x² = 3x² … (ii)
It shows that equations (i) and (ii) are same. Hence times the square of one side an equilateral triangle is equal to four times the square of its altitude.

Question 17.
Tick the correct answer and justify : In ∆ABC, AB = 6\(\sqrt{3}\) cm, AC = 12 cm and BC = 6 cm. The angle B is:
(a) 120°
(b) 60°
(c) 90°
(d) 45
Solution:
(C) is the correct answer.
When the triangle is right angled then by Pythagoras theorem
(12)= (6\(\sqrt{3}\))² + (6)²
144 = 36 x 3 + 36
144 = 144
L.H.S. = R.H.S.
Hence the result is (C)