Category: Class 10

MP Board Class 10th Science Solutions Chapter 4 Carbon and Its Compounds

MP Board Class 10th Science Chapter 4 Intext Questions

Intext Questions Page No. 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:

Intext Questions Page No. 68,69

Question 1.
How many structural isomers can you draw for pentane?
Answer:
Three structural isomers are possible for pentane.

Question 2.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:
The two features of carbon that give rise to a large number of compounds are as follows:

1. Carbon has the unique ability to form bonds with other atoms of carbon, giving rise to large molecules. This property is called catenation.
2. Since carbon has a valency of four, it is capable of bonding with four other atoms of carbon or atoms of some other mono-valent element.

Question 3.
What will be the formula and electron dot structure of cyclopentane?
Answer:
The formula for cyclopentane is C₅H₁₀. Its electron dot structure is given below:

Question 4.
Draw the structures for the following compounds:
(i) Ethanoic acid
(ii) Bromopentane
(iii) Butanone
(iv) Hexanal
Are structural isomers possible for bromo-pentane?
Answer:

Yes, there are many structural isomers possible for bromo-pentane. Among them, the structures of the three isomers are given.

Question 5.
How would you name the following compounds?

Answer:
(i) Bromoethane
(ii) Methanal (formaldehyde)
(iii) Hexyne.

Intext Questions Page No. 71

Question 1.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Answer:
Addition reaction means adding oxygen. Adding ethanol to potassium permanganate, we get ethanoic acid. Hence this reaction is called oxidation reaction.

Since in this reaction one oxygen is added to ethanol, hence it 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:
If a mixture of oxygen and ethyne is burnt, then ethyne burns completely producing a blue flame. The oxygen ethyne flame is extremely hot and produces a very high temperature which is used for welding metals. A mixture of ethyne and air is not used for welding because the burning of ethyne in air produces a sooty flame due to incomplete combination which is not enough to melt metals for welding.

Intext Questions Page No. 74

Question 1.
How would you distinguish experimentally between an alcohol and a carboxylic acid?
Answer:

1. We can distinguish between an alcohol and a carboxylic acid on the basis of their reaction with carbonates and hydrogen carbonates. The acid reacts with carbonate and hydrogen carbonate to evolve CO₂ gas that turns lime-water milky.
2. Metal carbonate/Metal hydrogen carbonate + Carboxylic acid → Salt + Water + Carbon dioxide.
3. In the litmus test, alcohol shows no change in colour whereas carboxylic acid turns blue litmus red.
   With sodium metal, alcohol gives effervescence but carboxylic acid does not give it. Alcohols, on the other hand, do not react with carbonates and hydrogen carbonates.

Question 2.
What are oxidising agents?
Answer:
some substances are capable of adding oxygen to others. These substances are known as oxidising agents.

Intext Questions Page No. 76

Question 1.
Would you be able to check if the water is hard by using a detergent?
Answer:
Margents are remaining effective in hardwater. Because of this reason, we can check if water is hard by using a detergent.

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:
The soap molecules form structures called micelles in water, where one end of the molecules is towards the oil droplet while the ionic end-faces outside. This forms emulsion in water and we can wash our clothes clean.

MP Board Class 10th Science Chapter 4 Ncert Textbook Exercises

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 that.
(a) The food is not cooked completely.
(b) The fuel is not burning completely.
(c) The fuel is wet.
(d) The fuel is burning.
Answer:
(b) the fuel is not burning completely

Question 4.
Explain the nature of the covalent bond using the bond formation in CH₃Cl.
Answer:
The structure of CH₃Cl is given below:

Carbon has four valence electrons. It shares one electron each with three hydrogen atoms and one electron with chlorine. The bond between C and Cl atoms is covalent but due to higher value of electro-negativity of Cl, the C-Cl bond is polar in nature.

Question 5.
Draw the electron dot structures for:
(a) Ethanoic acid
(b) H₂S
(c) Propanone
(d) F₂

Question 6.
What is a homologous series? Explain with an example?
Answer:
A series of compounds in which the same functional group substitutes for hydrogen in a carbon chain is called a homologous series.
Eg: CH₄ and C₂H₆ – These differ by a CH₂ unit.
C₂H₆ and C₃H₈ – these differ by a CH₂ unit.

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

1. Ethanol is a liquid at room temperature with a pleasant smell. Ethanoic acid has a melting point of 17°C. Since it is below the room temperature so, it freezes during winter. Moreover, ethanoic acid has a smell like vinegar.
2. Ethanol does not react with metal carbonates while, ethanoic acid reacts with metal carbonates to form a salt, water and carbon dioxide.
   For example:
   2CH₃COOH + Na₂CO₃ → 2CH₃COONa + CO₂ +H₂O
3. Ethanol does not react with NaOH while ethanoic acid reacts with NaOH to form sodium ethanoate and water.
   For example,
   CH₃COOH + NaOH → CH₃COONa + H₂O
4. Ethanol is oxidized to give ethanoic acid in the presence of acidified KMnO₄ while no reaction takes place with ethanoic acid in the presence of acidified KMnO₄.

Difference in physical properties:

Question 8.
Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?
Answer:
Soaps are molecules in which the two ends have differing properties, one is hydrophilic that is, it interacts with water, while the other end is hydrophobic, that is it interacts with hydrocarbons. When soap is at the surface of water, the hydrophobic 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. Thus, 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. A micelle will not be formed in other solvents such as ethanol.

Question 9.
Why are carbon and its compounds used as fuels for most applications?
Answer:
Carbon and its compounds give large amount of heat on combustion due to the high percentage of carbon and hydrogen. Carbon compounds used as fuel have optimum ignition temperature with high calorific values and are easy to handle. Their combustion can be controlled. Therefore, carbon and its compounds are used as fuels.

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
When soap reacts with water, calcium and magnesium salts are formed which causes hardness for water. Ionic ends of soap interacts with water while the carbon chain interacts with oil. The soap molecules, thus form structures called micelles where one end of the molecules is towards the oil droplet while the ionic-end faces outside. This forms an emulsion in water.

Question 11.
What change will you observe if you test soap with litmus paper (red and blue)?
Answer:
Since soap is basic in nature, it will turn red litmus blue. However, the colour of the blue litmus will remain blue.

Question 12.
What is hydrogenation? What is its industrial application?
Answer:
Unsaturated Hydrocarbons react with Hydrogen, in presence of catalysts such as palledium or Nickel and forms saturated Hydrocarbons. This is called Hydrogenation of oils.
This process is useful in hydrogenation of oils derived from plants.

Question 13.
Which of the following hydrocarbons undergo addition reactions:
C₂H₆, C₃H₈, C₃H₆, C₂H₂ and CH₄.
Answer:
Unsaturated hydrocarbons undergo addition reactions. Being unsaturated hydrocarbons, C₃H₆ and C₂H₂ undergo addition reactions.

Question 14.
Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.
Answer:
Saturated Hydrocarbons will not react with Bromine, but unsaturated hydrocarbons change the colour of Bromine.

Question 15.
Explain the mechanism of the cleaning action of soaps.
Answer:
The dirt present on clothes is organic in nature and insoluble in water. Therefore, it cannot be removed by washing with water only. When soap is dissolved in water, its hydrophobic ends attach themselves to the dirt and remove it from the cloth. Then, the molecules of soap arrange themselves in micelle formation and trap the dirt at the centre of the cluster. These micelles remain suspended in the water. Hence, the dust particles are easily rinsed away by water.

MP Board Class 10th Science Chapter 4 Additional Questions

MP Board Class 10th Science Chapter 4 Multiple Choice Questions

Question 1.
Which of the following is a three-carbon compound?
(a) Ethene
(b) Ethane
(c) Propane
(d) Acetylene
Answer:
(c) Propane

Question 2.
Which one of the following is an unsaturated hydrocarbon?
(a) Acetylene
(b) Butane
(c) Propane
(d) Decane
Answer:
(a) Acetylene

Question 3.
Two neighbours of homologous series differ by:
(a) -CH
(b) -CH₂
(c) -CH₃
(d) -CH₄
Answer:
(b) -CH₂

Question 4.
General formula of alkanes is –
(a) C_nH_2n+2
(b) C_nH_2n
(c) C_nH_2n-2
(d) C_nH_n
Answer:
(a) C_nH_2n+2

Question 5.
Which of the following represents alkynes?
(a) -C – C-
(b) -C = C-
(c) -C ≡ C-
(d) None of these
Answer:
(c) -C ≡ C-

Question 6.
Which of the following represents ketones?
(a) -C = O
(b) -OH
(c) -CHO
(d) COOH
Answer:
(a) -C = O

Question 7.
Which of the following is not an aliphatic hydrocarbon?
(a) ethene
(b) ethane
(c) propyne
(d) benzene
Answer:
(d) benzene

Question 8.
Complete combustion of a hydrocarbon gives:
(a) CO + H₂O
(b) CO₂ + H₂O
(c) CO + H₂
(d) CO₂ + H₂
Answer:
(b) CO₂ + H₂O

Question 9.
Which is NOT correct for isomers of a compound?
(a) They differ in physical properties.
(b) They differ in chemical properties.
(c) They have the same molecular formula.
(d) They have the same structural formula.
Answer:
(d) They have the same structural formula.

Question 10.
Buckminsterfullerene is an example of ………….. of carbon.
(a) an isomer
(b) an isotope
(c) an allotrope
(d) a functional group
Answer:
(c) an allotrope

Question 11.
Who prepared urea for the first time by heating ammonium cyanate?
(a) Wohler
(b) Lavoisier
(c) Fuller
(d) Haber
Answer:
(a) Wohler

Question 12.
Hexanone is a four-carbon compound with the functional group:
(a) Carboxylic acid
(b) Aldehyde
(c) Ketone
(d) Alcohol
Answer:
(c) Ketone

Question 13.
Major constituent of LPG is ………….
(a) Ethene
(b) Butane
(c) Propane
(d) Pentane
Answer:
(b) Butane

Question 14.
The gas used in welding and cutting metals is:
(a) Ethyne
(b) Ethene
(c) Ethane
(d) Propane
Answer:
(a) Ethyne

Question 15.
How is carbon atoms arranged in Buckminster fullerenes?
(a) Triangle shape
(b) Hexagonal array
(c) Football shape
(d) None
Answer:
(c) Football shape

Question 16.
Vinegar is a solution of –
(a) 40%-45% acetic acid.
(b) 90%-95% acetic acid.
(c) 5-20% acetic acid and water.
(d) 35-40% acetic acid and water.
Answer:
(c) 5-20% acetic acid and water.

Question 17.
How many covalent bonds are there in Bromoethane?
(a) 4
(b) 6
(c) 10
(d) 7
Answer:
(d) 7

Question 18.
Which functional group is present in propane?
(a) Aldehyde
(b) No group
(c) Ketone
(d) Alcohol
Answer:
(b) No group

Question 19.
Which compound/molecule is being presented by the following formula: H: C:: C: H
(a) Ethane
(b) Ethene
(c) Ethyne
(d) None of these
Answer:
(b) Ethene

Question 20.
Next homologous to C₂H₅OH will be:
(a) CH₄
(b) C₂H₆
(c) C₃H₅
(d) C₃H₇OH
Answer:
(d) C₃H₇OH

Question 21.
When we burn naphthalene it produces:
(a) Smoky flame
(b) Non-sooty flame
(c) Colourless flame
(d) No flame
Answer:
(a) Smoky flame

Question 22.
Bunsen burner is used for:
(a) making food.
(b) study flames type.
(c) low heating work.
(d) all the above.
Answer:
(c) low heating work.

Question 23.
See the figure carefully.

Choose the suitable name of isomer:
(a) Neo-pentane
(b) n-pentane
(c) Iso-pentane
(d) All
Answer:
(c) Iso-pentane

Question 24.
Which of the following is a structure of ethanoic acid?

Answer:

Question 25.
What is the name of CH₃-CH₂-Br? Choose from the following:
(a) Hex-1-one
(b) Hexanal
(c) Ethanoic acid
(d) None
Answer:
(d) None

Question 26.
What happens on the litmus test of soap?
(a) No change
(b) Red litmus turns blue
(c) Red litmus turn purple
(d) Red litmus turn green
Answer:
(b) Red litmus turns blue

MP Board Class 10th Science Chapter 4 Very Short Answer Type Questions

Question 1.
Name two groups which can have the same general formula.
Answer:
Both alkenes and cyclo-alkanes can be represented by the same general formula.

Question 2.
Which group of compounds have general formula C₂H_2n?
Answer:
The general formula C_nH_2n represents alkenes group of compounds.

Question 3.
What is the common name and IUPAC name for CH₃COCH₃?
Answer:
Acetone is the common name and propanone is the IUPAC name for CH₃COCH₃.

Question 4.
Do isomers always show same chemical properties?
Answer:
No, isomers always do not show the same chemical properties.

Question 5.
What is the common name and formula for ethanol?
Answer:
Alcohol, CH₃CH₂OH.

Question 6.
What are the products of complete combustion of a hydrocarbon?
Answer:
Carbon dioxide and water.

Question 7.
What is next homologue of C₃H₇OH is called?
Answer:
The next homologue of C₃H₇OH is called butanol C₄H₉OH.

Question 8.
What are isomers?
Answer:
The compounds which have the same molecular formula but different structures and chemical properties are called isomers.

Question 9.
Which one are more reactive unsaturated hydrocarbons or saturated hydrocarbons? Give reason.
Answer:
Unsaturated hydrocarbons: The Presence of double and triple covalent bonds make them more reactive.

Question 10.
Discuss the general nature of covalent compounds in water.
Answer:
Generally, they are insoluble in water.

Question 11.
What type of hydrocarbons takes part in an addition reaction?
Answer:
Unsaturated hydrocarbons.

Question 12.
Which carboxylic acid freezes during winter or under cold climate conditions?
Answer:
Acetic acid and hence, known as glacial acetic acid.

Question 13.
What is the difference in molecular masses of any two successive homologous alkanes?
Answer:
14 units.

Question 14.
What is the molecular formula of the alcohol which can be derived from propane?
Answer:
Alcohol obtained from propane is propanol -1 and the molecular formula is C₃H₇OH.

Question 15.
Give the names of the functional groups: (CBSE 2007)

1. -CHO
2. -COOH

Answer:

1. Aldehydic group.
2. Carboxylic acid group.

Question 16.
Give the names of the following functional groups: (CBSE 2007)

1. -OH
2. -CO

Answer:

1. Alcoholic.
2. Ketonic.

MP Board Class 10th Science Chapter 4 Short Answer Type Questions

Question 1.
What is meant by the term functional group?
Answer:
An atom or a group of atoms, which makes a carbon compound reactive and decides its properties, is called a functional group.
For example aldehyde, ketone etc.

Question 2.
Which R functional groups always occur at the terminal position of a carbon chain?
Answer:
Aldehydic Group, R-CHO (R is the alkyl group),
Carboxyl Group, R-COOH (R is the alkyl group)

Question 3.
Why a candle flame burns yellow, while a highly-oxygenated gas fuel flame burns blue?
Answer:
The most important factor determining the colour of the flame is oxygen supply and the extent of fuel-oxygen, which determines the rate of combustion and thus, the temperature and reaction paths, thereby producing different colour hues. In case of a candle, it is incomplete combustion and the flame temperature is not high. This gives a yellow flame, while a highly-oxygenated gas (e.g., ethyne) flame burns blue because of complete combustion raising a very high temperature.

Question 4.
Why is the reaction between methane and chlorine considered a substitution reaction? (CBSE 2008)
Answer:
Methane reacts with chlorine in the presence of sunlight to form chloromethane and hydrogen chloride. Since chlorine substitutes or replaces hydrogen of methane to form chloromethane, it is considered as substitution reaction.
CH₄ + Cl₂ → CH₃Cl + HCl
With the excess of chlorine, four hydrogen atoms of methane are replaced by chlorine atoms to form carbon tetrachloride (CCl₄).

Question 5.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Being tetravalent carbon atom, it is neither capable of losing all of its four valence electrons nor it can easily accept four electrons to complete its octet. Both of these are requirements of ionic bond formation and are energetically less favourable. Carbon completes its octet by sharing electrons and hence, covalent bonding is preferred.

Question 6.
What do you mean by Octane rating?
Answer:
Gasoline is rated on a scale known as octane rating, which is based on the way they burn in an engine. The higher the octane rating, the greater the percentage of complex-structured hydrocarbons that are present in the mixture, the more uniformly the gasoline burns, and the less knocking there is in the automobile engine. Thus, a gasoline rated 92 octane will burn more smoothly than one rated 87 octanes.

Question 7.
What is covalent bonding?
Answer:
The chemical bonding that takes place due to the mutual sharing of electron pairs of two or more atoms of different elements is called covalent bonding. By mutual sharing of electron pairs, atom attains noble gas configuration, e.g., hydrogen molecule (H₂), the two H-atoms combine by covalent bonding (H-H).

Question 8.
What are hydrocarbons? Give examples.
Answer:
Compounds of carbon and hydrogen are called hydrocarbons. Methane, ethane, butane, ethyne, propane, benzene, petroleum products – all are examples of hydrocarbons.

Question 9.
What are saturated hydrocarbons? (CBSE 2011)
Answer:
The hydrocarbons in which valency of carbon is satisfied by a single covalent bond are called saturated hydrocarbons. Alkanes like methane (CH₄), ethane(C₂H₆), propane (C₃H₈) etc. are examples of saturated hydrocarbons. Saturated hydrocarbons will generally give a clean flame.

Question 10.
Why do ionic compounds have high melting points? (HOTS)
Answer:
Ions have strong electrostatic forces of attraction among them forming ionic compounds. It requires a lot of energy to break these ionic bonds or forces. That’s why ionic bonds have high melting points.

Question 11.
What are homologous series? (HOTS)
Answer:
Homologous series are:

1. Compounds with the same formula.
2. Belong to the same functional group.
3. Have general methods of separation.
4. Have similar chemical properties.

Show similar gradation of physical properties, e.g., boiling points of alcohol increase with an increase in their molecular weights. Similarly, solubility decreases with increase in molecular weights.

Question 12.
What is a heteroatom? What is the heteroatom in the alcohol functional group? (HOTS)
Answer:
In a hydrocarbon chain, one or more hydrogen atoms can be replaced by other atoms according to their valencies. The element wh replaces hydrogen in the chain is called a heteroatom, e.g., in alcohol (-OH) functional group, oxygen is the heteroatom.

MP Board Class 10th Science Chapter 4 Long Answer Type Questions

Question 1.
Distinguish between saturated and unsaturated hydrocarbons by the way of their burning in air and bromine test inferences.
Answer:
1. Saturated compounds are burnt in air, to give a clear (blue) flame but the burning of unsaturated compounds (alkenes and alkynes) give a sooty (yellowish) flame because saturated compounds contain comparatively less percentage of carbon which is completely oxidized by the oxygen present in the air.

On the other hand, the percentage of carbon in unsaturated compounds is more and it requires more oxygen to get completely oxidized that is not fulfilled by air. So, due to incomplete oxidation, they burn with a sooty flame.

2. Bromine-water test: Br₂ water is a brown coloured liquid:

1. Unsaturated hydrocarbons give addition reaction with Br₂. So, the colour of Br₂ water gets decolourised.
   
2. Saturated hydrocarbons do not react with Br₂ water, so the colour of B₂-water does not get decolourised.

Question 2.
Two compounds A and B react with each other in the presence of a dehydrating agent to produce an ester. Both react with Na to evolve hydrogen gas. On reaction with Na₂CO_3, only A evolves CO₂. Identify the functional groups present in A and B giving the reason for your answer.
Answer:
Compound A contains -COOH group while compound B contains -OH group. Since, carboxylic acids and alcohols react with each other to form an ester, out of A and B, one is an alcohol and the other is a carboxylic acid. This is further strengthened by the reaction of both with Na to evolve hydrogen gas. Only carboxylic acids react with Na₂CO₃ to evolve CO₂, A contains -COOH group while B contains -OH group.

Question 3.
An organic compound ‘X’ is widely used as a preservative in pickles and has a molecular formula C₂H₂O₂. This compound reacts with ethanol to form a sweet-smelling compound ‘Y’.

1. Identify the compound ‘X’.
2. Write the chemical equation for its reaction with ethanol to form compound ‘Y’.
3. How can we get compound ‘X’ back from ‘Y’?
4. Name the process and write a corresponding chemical equation.
5. Which gas is produced when compound ‘X’ reacts with washing soda? Write the chemical equation.
6. Answer:
7. Compound X is ethanoic acid which gives and ester (Y) when reacts with ethanol.
8. CH₃COOH + CH₃CH₂OH → CH₃COOC₂H₅.
9. Esters give back alcohol and carboxylic acid in the presence of acid or base.
10. Saponification reaction: CH₃COOC₂H₅ + NaOH → C₂H₅OH + CH₃COOH + Na.
11. CO₂ gas is released,
    CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂.

Question 4.
“Saturated hydrocarbons burn with a blue flame while unsaturated hydrocarbons burn with a sooty flame.” Why?
Answer:
Saturated hydrocarbons have only C-C and C-H single bonds and thus, contain the maximum possible number of hydrogen atoms per carbon atom. With sufficient oxygen, saturated hydrocarbons burn completely and give blue flame,
CH₄ + 2O₂ → CO₂ + 2H₂O
Unsaturated hydrocarbons contain a carbon-carbon double bond (C=C) or triple bond (C=C). Hence, they contain less number of hydrogen than carbon. Unsaturated hydrocarbons undergo incomplete combustion and give yellow flame along with black sooty carbon.
C₂H₄ + O₂ → CO₂ + 2H₂O + C(s)

Question 5.
What makes some molecular formula compound different? (HOTS)
Answer:
The arrangement makes them different compounds with identical molecular formula but different structures are called structural isomers. Organic compounds show a great level of isomerism. Isomers may be structural (due to difference in the arrangement of C atoms forming chain) or stereo (due to arrangement of bonds in a chain). With the increase in the number of carbon atoms in molecular formula, it leads to an increase in the number of isomers.
For example:

MP Board Class 10th Science Chapter 4 Textbook Activities

Class 10 Science Activity 4.1 Page No. 58

1. Make a list of ten things you have used or consumed since the morning.
2. Compile this list with the lists made by your classmates and then sort the items into the following table.
3. If there are items which are made up of more than one material, put them into both the relevant columns.
   
4. (C) Indicates carbon. Most substances contain carbon in it.

Class 10 Science Activity 4.2 Page No. 67

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

Difference: 70 – 60 = 14U
All three groups given above are homologous.

Class 10 Science Activity 4.3 Page No. 69

Caution:

1. This Activity needs the teacher’s assistance.
2. Take some carbon compounds (naphthalene, camphor, alcohol) one by one on a spatula and burn them.
3. Observe the nature of the flame and note whether smoke is produced.
4. Place a metal plate above the flame. Is there a deposition on the plate in case of any of the compounds?

Observations:

Class 10 Science Activity 4.4 Page No. 69

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

Observations:

1. Yellow, Sooty flame is formed – when the hole is closed.
2. A blue flame is observed – when the hole is open.

Class 10 Science Activity 4.5 Page No. 70

1. Take about 3 ml of ethanol in a test tube and warm it gently in a water bath.
2. Add a 5% solution of alkaline potassium permanganate drop by drop to this solution.
3. Does the colour of potassium permanganate persist when it is added initially?
4. Why does the colour of potassium permanganate not disappear when excess is added?

Observations:
Doing the above activities we found that potassium permanganate act here as oxidising agents only and their colour do not change at,

Class 10 Science Activity 4.6 Page No. 72

Teacher’s demonstration:

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

Observations:
Sodium is an inflammable substance hence, it should be handled very carefully. When we place it in alcohol, hydrogen gas is evolved and sodium ethoxide is formed,
2Na + 2CH₃CH₂OH → 2CH₃CH₂ONa⁺ + H₂ ↑

Class 10 Science Activity 4.7 Page No. 73

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

Observations:
The litmus test and pH test show the acidity and alkalinity of substance or chemical:

Class 10 Science Activity 4.8 Page No. 73

1. Take 1 ml ethanol (absolute alcohol) and 1 ml glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
2. Warm in a water-bath for at least five inutes as shown in Figure.
3. Pour into a beaker containing 20-50 ml of water and smell the resulting mixture.
   

Observations:
When acetic acid reacts with alcohol a new compound with an ester functional group is formed. It has fruit like smell. This reaction is called esterification reaction.

Class 10 Science Activity 4.9 Page No. 74

1. Take a spatula full of sodium carbonate in a test tube and add 2 ml of dilute ethanoic acid.
2. What do you observe?
3. Pass the gas produced through freshly prepared lime-water. What do you observe?
4. Can the gas produced by the reaction between ethanoic acid and carbonate be identified by this test?
5. Repeat this Activity with sodium hydrogen carbonate instead of sodium carbonate.

Observations:
Sodium acetate is produced when we add carbonate or hydrogen carbonate to acetic acid.
2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂
CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂

Class 10 Science Activity 4.10 Page No. 74

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

Observations:
Yes, a layer of oil separates out by reacting with the soap solution. Dirt has an oily nature. It happens first in test tube B.

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 9 Heredity and Evolution

MP Board Class 10th Science Chapter 9 Intext Questions

Class 10th Science Chapter 9 Intext Questions Page No.143

Question 1.
If a trait Aexists in 10% of a population of an asexually reproducing species and trait B exists in 60% of the same population, which trait is likely to have arisen earlier?
Answer:
Trait ‘B’ is likely to have arisen earlier. Because in asexually reproducing species, small differences are seen due to DNA replication.

Question 2.
How does the creation of variations in a species promote survival?
Answer:
Depending on the nature of variations, different individuals would have different kinds of advantages. Bacteria that can withstand heat will better in a heat wave, as selection of variants by environmental factors forms the basis for evolutionary process.

Class 10th Science Chapter 9 Intext Questions Page No. 147

Question 1.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
Mendel used a number 0 contrasting visible characters of garden peas-round/wrinkled seeds, tall/short plants, white/violet flowers and so on. He took pea plants with different characteristics a tall plant and a shot plant, produced progeny by crossing then, and calculated the percentages of tall or short progeny.

In the first place, there were no halfway characteristics in this first generation, or F₁ progeny no medium-height plants. All plants were tall. This means that only one of the parental traits was seen, not some mixture of the two, so the next question was were the tall plants in the F₁ generation exactly the same as the tall plants of the parent generation? Mendelian experiments test this by getting both the parental plans and these F₁ tall plants to reproduce by self-pollination.

The progeny of the parental plants are of course, all tall. However the second generation, or F₂ progeny of the F₁ tall plants are not all tall. Instead, one quarter of them are short. This indicates that both the tallness and shortness traits were inherited in the F₁ plants, but only the tallness trait was expressed. This led Mendel to propose that two copies of factor (now called genes) controlling traits are present in sexually reproducing organism. These two may be identified or may be different, depending on the percentage. A pattern of inheritance can be worked out with this assumption as shown in figure.

In this explanation, both TT and Tt are tall plants, while only it is a short plant. Traits like T are called dominant traits, while those that behave like ‘t’ are called recessive traits.

Question 2.

How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel crossed

1. pure breeding tall plants having round seeds and
2. pure breeding short plants having wrinkled seeds.

The plants of F1 generation were all tall with round seeds indicating that the traits of tallness and round seeds were dominant.

While self breeding, of F1 yielded plants with characters of 9 tall round seeded, 3 tall wrinkled seeded, 3 short round seeded and one short wrinkled seeded. Tall wrinkled and short round seeded plants are new combinations which can develop only when the traits are inherited independently.

Question 3.
A man with blood group A marries a woman with blood group C and their daughter has blood group O. Is this information enougl to tell you which of the traits – blood group A or O – is dominant? Why or why not?
Answer:
From this information, it is not possible to tell which of the traits-blood group A or O is dominant. AA group becomes AO. So this information is complete.

Question 4.
How is the sex of the child determined in human beings?
Answer:
Pair of sex chromosomes determines the particular sex of a child. In human, the males have one X and one Y chromosome and the females have two X chromosomes therefore, the females are XX and the males are XY. The gametes receive half of the chromosomes. The child gametes have 22 autosomes and either X or Y sex chromosome in males while X in females.
Type of male gametes: 22 + X Or 22 + Y.
Type of female gamete: 22 + X.
This is the basis of sex determination in human beings.

Class 10th Science Chapter 9 Intext questions Page No. 150

Question 1.
What are the different ways in which individuals with a particular trait may increase in a population?
Answer:
A big reason could be accurate copying of DNA and limited variations. Variation causes in generation of new traits and non preserving of existing parental traits. Individuals with a particular trait may also increase due to natural selection that is trait offers some survival advantage. Genetic drift which is caused by genes governing that trait become common in a population.

Question 2.
Why are traits acquired during the life-time of an individual not inherited?
Answer:
Variation is not hereditary from generation to generation. In case of Asexual reproduction DNA will not transfer to germ cells. So experiences of an individual during its lifetime cannot be passed on to its progeny and cannot direct evolution.

Question 3.
Why are the small numbers of survivingjigers a cause of worry from the point of view of genetics?
Answer:
Tigers are adopted to their environment as per genes. If tigers number is decreasing number of genes also decrease. So its generation is becoming less.

Class 10th Science Chapter 9 Intext Questions Page No. 151

Question 1.
What factors could lead to the rise of a new species?
Answer:
Variations, non copying of DNA, natural selection, genetic drift and acquisition of traits during the life time of an individual can give rise to new species.

Question 2.
Will geographical isolation be a major factor in the speciation of a self-pollinating plant species? Why or why not?
Answer:
Geographical isolation will not a major factor in the speciation of a self pollinating plant species, because self pollination is taking place in one plant. In cross pollination it is a major factor.

Question 3.
Will geographical isolation be a major factor in the speciation of an organism that reproduces asexually? Why or why not?
Answer:
Geographical isolation will not be a major factor in the speciation of an organism that reproduces asexually because it is a major factor in organisms that reproduces sexually.

Class 10th Science Chapter 9 Intext Questions Page No. 156

Question 1.
Give an example of characteristics being used to determine how close two species are in evolutionary terms.
Answer:
The characteristics in different organisms would be similar because they are inherited from a common ancestor. As an example, consider the fact that mammals have four limbs, as do birds, reptiles and amphibians. The basic structure of the limbs is similar through it has been modified to perform different functions in various vertebrates. Such a homologous characteristic helps to identify an evolutionary relationship between apparently different species.

Question 2.
Can the wing of a butterfly and the wing of a bat be considered homologous organs? Why or why not?
Answer:
The wing of a butterfly and the wing of a bat be is not considered homologous organs because the designs of the two wings, their structure and components are very different.

Question 3.
What are fossils? What do they tell us about the process of evolution?
Answer:
Fossils are the remains of organisms that once existed on earth.

We can gather information about the development of the structures from simple structured to complex structured organisms. They tell us about the phases of evolutions through which they must have undergone in order to sustain themselves in the competitive environment.

Class 10th Science Chapter 9 Intext Questions Page No. 158

Question 1.
Why are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species?
Ans.
A species is a group of organisms that are capable of interbreeding to produce a fertile offspring. Skin colour, looks and size are all variety of features present in human beings. These features are genetic but also environmentally controlled. Various human races are formed based on these features. All human races have more than enough similarities to be classified as same species. Therefore, all human beings are a single species as humans of different colour, size and looks are capable of reproduction and can produce a fertile off spring.

Question 2.
In evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a ‘better’ body design? Why or why not?
Answer:
Bacteria is a better body design, because even though it a simple organism, it can survive in hot springs, at the bottom of sea and even in coldest ice covered place such as Antarctica.

MP Board Class 10th Science Chapter 9 NCERT Textbook Exercises

Question 1.
A Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic make-up of the tall parent can be depicted as:
(a) TTWW
(b) TTww
(c) TtWW
(d) TtWw
Answer:
(c) TtWW. Genetic make up of tall plant.

Question 2.
An example of homologous organs is:
(a) our arm and a dog’s fore-leg.
(b) our teeth and an elephant’s tusks.
(c) potato and runners of grass.
(d) all of the above.
Answer:
(d) Both organs in all option have same basic structural design but have different functions and appearance.

Question 3.
In evolutionary terms, we have more in common with:
(a) a Chinese school-boy
(b) a chimpanzee
(c) a spider
(d) a bacterium
Answer:
(a) a Chinese school-boy.

Question 4.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
No, we cannot say anything about whether the light eye colour trait is dominant or recessive. As this information is not sufficient. For considering a trait as dominant or recessive, we need data of at least last three generations.

Question 5.
How are the areas of study – evolution and classification – interlinked?
Answer:
An example will help with this A brother and a sister are closely related. They have common ancestors in the first generation before them, namely their parents. A girl and her first cousin are also related, but less than the girl and her brother. This is because cousins have common ancestors, their grandparents, in the second generation before them, not in the first one. We can now appreciate that classification of species is in fact a reflection of their evolutionary relationship.

Question 6.
Explain the terms analogous and homologous organs with examples.
Answer:

Consider the fact that mammals have four limbs, as do birds, reptiles and amphibians. The basic structure of the limbs is similar though it has been modified to perform different functions in various vertebrates. This is an example of homologous characteristic.

We find that the wings of bats are skin folds stretched mainly between elongated fingers. But the wings of birds are feathery covering all along the arm. The design of the two wings, their structure and components, are thus very different. They look similar because they have a common use for flying, but their origins are not common. This makes them analogous characteristics, rather than homologous characteristics.

Question 7.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
There are variety of genes that govern coat colour of a dog. At least eleven identified gene series (A, B, C, D, E, F, G, M, P, S, T) that influence coat colour in dog.

A dog inherits one gene from each of its parents. The dominant gene gets expressed in the phenotype. For example, in the B series, a dog can be genetically black or brown.

Let us assume that one parent is homozygous black (BB), while the other parent is homozygous brown (bb)

In this case, all the off springs will be heterozygous (Bb).
Since black (B) is dominant, all the offsprings will be black. However, they will have both B and b alleles.
If such heterozygous pups are crossed, they will produce 25% homozygous black (BB), 50% heterozygous black (Bb), and 25% homozygous brown (bb) offsprings.

Question 8.
Explain the importance of fossils in deciding evolutionary relationships.
Answer:
Analysis of the organ structure in fossils allows us to make estimates of how far back evolutionary relationships go. The wild cabbage plant is a good example. Humans have over more two thousand years, cultivated wild cabbage as a food plant, and generated different vegetables from it by selection. This is of course, artificial selection rather than natural selection. Kale, cauliflower. Broccoli, cabbage, Red cabbage and Kohl rabi all these have same ancestor.

Question 9.
What evidence do we have for the origin of life from inanimate matter?
Answer:
As we all know, that there occurs a time when our planet was lifeless, at that time intial matter to develop life was water, sand and some atmospheric gases as CO₂, CH4 and nitrogen. The evidence for the origin of life from inanimate matter, was provided through an experiment, conducted in 1953, by Stanley L. Miller and Harold C. Urey. In experiment, they assembled an atmosphere containing molecules like ammonia, methane and hydrogen sulphide, but not oxygen.

This was similar to atmosphere that existed on early earth . This was maintained at a temperature just below 100°C and sparks were passed through the mixture of gases to simulate lightning. At the end of a week, 15% of the carbon from methane, had been converted to simple compounds of carbon including amino acids which make up protein molecules and support the life in basic form.

Question 10.
Explain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually?
Answer:

Change in on-reproductive tissues cannot be passed on to the DNA of the germ cell. Therefore the experiences of an individual during its lifetime cannot be passed on to its progeny and cannot direct evolution.

Ex: If we breed a group of mice, all their progeny will have tails, as expected. Now, if the tails of these mice are removed by surgery in each generation, do these tailless mice have tailless progeny? The answer is no and it makes sense because removal of the tail cannot change the genes of the germ cells of the mice. Hence sexual reproduction gives rise to more viable variations than asexual reproduction.

Question 11.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
Genetic inheritance begins at the time of conception, progeny inherited 23 chromosomes from female parent and 23 from male parent. Together it form 22 pairs of autosomal chromosomes and a pair of sex chromosomes (either XX in case of female, or XY in male). Homologous chromosomes have the same genes in the same positions, but may have different alleles (varieties) of those genes. An individual has two copies of alleles, and that can be homozygous (both copies the same) or heterozygous (the two copies are different) for given gene.

Hence, in human beings, equal genetic contribution of male and female parents is ensured in the progeny through inheritance of equal number of chromosomes from both parents. Females have a equal pair of two X sex chromosomes and males have a pair of one X and one Y sex chromosome. As fertilisation takes place, the male gamete (haploid) fuses with the female gamete (haploid) resulting in formation of the diploid zygote. The zygote in the progeny receive an equal contribution of genetic traits from the parental generations.

Question 12.
Only variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement? Why or why not?
Answer:
Only variations that confer an advantage to an individual organism will survive in a population we agree to this statement variation is convenient for survival. This provides diversity for organisms.

MP Board Class 10th Science Chapter 9 Additional Important Questions

MP Board Class 10th Science Chapter 9 Multiple Choice Questions

Question 1.
Fossil archaeopteryx exhibits connection between:
(a) Amphibian and fish
(b) Reptiles and fish
(c) Reptile and birds
(d) Birds and mammals
Answer:
(c) Reptile and birds

Question 2.
The sex of the human child depends on the sex chromosome present in the:
(a) Egg
(b) Spenn
(c) Both (a) and (b)
(d) None of these
Answer:
(b) Spenn

Question 3.
Genetic information is carried out by long chain of molecules made up of:
(a) Enzymes
(b) DNA
(c) Amino acids
(d) Proteins
Answer:
(b) DNA

Question 4.
Which one of the following represents a ratio of monohybrid cross?
(a) 9 : 7
(b) 3 : 1
(c) 1 : 1 : 1 : 1
(d) 9 : 3 : 3 : 1
Answer:
(b) 3 : 1

Question 5.
On which plant Mendel carried his experiments of inheritance?
(a) Cow pea
(b) Wild pea
(c) Garden pea
(d) Pigeon pea
Answer:
(c) Garden pea

Question 6.
A gamete certains which of the following?
(a) Both alleles of a gene
(b) Only one allele of a gene
(c) All alleles of a gene
(d) No allele of a gene
Answer:
(b) Only one allele of a gene

Question 7.
Chromosomes are made up of
(a) Proteins
(b) DNA
(c) RNA
(d) All of these.
Answer:
(d) All of these.

Question 8.
Pea plants were more suitable than cats for Mendel’s experiments because:
(a) Cats have many genetic traits
(b) No pedigree record of cats
(c) Pea plants can be self-pollinated or fertilised
(d) Pea plants favour cross pollination.
Answer:
(c) Pea plants can be self-pollinated or fertilised

Question 9.
In a cross Tt × Tt, the percentage of offsprings produced having same phenotype as the parents would be:
(a) 50%
(b) 100%
(c) 25%
(d) 0%
Answer:
(a) 50%

Question 10.
Who proposed the laws of heredity?
(a) Darwin
(b) Mendel
(c) Morgan
(d) Dalton
Answer:
(b) Mendel

MP Board Class 10th Science Chapter 9 Very Short Answer Type Questions

Question 1.
Define heredity.
Answer:
The process by which traits and characteristics are reliably inherited or passed from the parents to the offspring is called heredity.

Question 2.
What is a gene?
Answer:
Gene is a functional segment of DNA on a chromosome occupying specific position, which carries out a specific biological function.

Question 3.
Name the plant on which Mendel performed his experiments.
Answer:
Garden pea (Pisum sativum).

Question 4.
Define the term variation.
Answer:
Variation: There are differences found in structure, function, behaviour and genetic make up of different individuals of the same parentage, variety, race and species. These differences refer to variation.

Question 5.
Write the expanded form of DNA.
Answer:
Deoxyribonucleic acid (DNA).

Question 6.
Define genetics.
Answer:
The branch of biology which deals with heredity and variations, is known as genetics.

Question 7.
Define the term offspring.
Answer:
Offspring is an individual formed as a result of sexual reproduction involving the formation and fusion of two gametes. The genotype of an offspring is different from either of the parents due to shuffling of chromosomes and their genes.

Question 8.
What are reciprocal crosses?
Answer:
They are two types of crosses involving two groups of individuals where the male of one group is crossed with the female of the other and vice versa.

Question 9.
Where are the genes located? What is the chemical nature of gene?
Answer:
Genes are located at a specific position on a chromosome. Chemical Nature of Gene: Chemically, gene is a segment of deoxyribonucleic acid (DNA) consisting of specific sequence of the nucleotides. The sequence of the constituent nucleotides determines the functional property of a gene.

MP Board Class 10th Science Chapter 9 Short Answer Type Questions

Question 1.
Define genetics. What is the contribution of Mendel in this branch of Biology?
Answer:
Genetics is the branch of science of heredity and variations which deals with the study of the transmission of traits from parents to the offsprings and the occurrence of differences among the individuals.

Contribution of Mendel: Mendel did his experiments on garden pea (Pisum sativum) and discovered the scientific principles, which govern patterns of inheritance i.e., the principle of inheritance. He explained that contrasting characters are controlled by units which he called ‘Factors Today, these factors are called genes.

Question 2.
Differentiate between inherited and acquired traits.
Answer:
Inherited traits:

1. The traits which are inherited from the parents (Father and Mother) to the offsprings (progeny) are called inherited traits.
2. These traits are due to genetic make up of the progeny.

Acquired characters

1. These traits cannot be passed on to their future generations.
2. These traits develop in response to the environment.

Question 3.
What are Mendel’s laws of inheritance?
Answer:
Law of dominance: When two homozygous individuals with one or more sets of contrasting characters are crossed the characters that appear in the F, hybrids are dominant characters.

Law of segregation: Contrasting characters brought together in hybrid remain together without being contaminated and when gametes are formed from the hybrid, the two separate out from each other and only one enters each gamete.

Law of independent assortment: In inheritance of more than one pair of contrasting characters simultaneously, the factors for each pair of characters assort independently of other pair’s.

Question 4.
How did life originate on earth?
Answer:
Life originated on earth from inorganic elements and compounds under extreme atmospheric conditions (such as very high temperature, electric discharges, reducing atmosphere etc.) by formation of complex organic compounds such as amino acids.

Question 5.
Why did Mendel choose garden pea for his experiments?
Answer:
Due to the following reasons, Mendel selected garden pea for his experiment:

1. Garden pea flowers are normally self-pollinated but can be easily cross-pollinated.
2. Many varieties with distinguished contrasting characters e.g., smooth seed coat, wrinkled seed coat are available.
3. A large number of progeny can be produced in a short duration.
4. Its flowers can be easily handled for experimentation.

Question 6.
What are the factors which help in speciation?

1. Genetic drift: Due to genetic drift, there will be accumulation of # different changes in each sub-populations. The levels of gene flow ’ between them will decrease if they are further isolated, it will be more on a small sub-population.
2. Over generations, genetic drift will accumulate, causing different changes in the populations.
3. Natural selection may also operate differently in the different geographical location.
4. Together, genetic and natural selection will make the population more and more different from each other. As a result, members will be incapable of reproducing with each other. Changes may be due to change in DNA or number of chromosomes.

Question 7.
Does geographical isolation of individuals of a species lead to formation of a new species? Provide a suitable explanation.
Answer:
Yes, geographical isolation of sub-populations of a population of a species leads to genetic drift. This may impose limitations to.sexual reproduction of the separated population. Slowly, the separated individuals will reproduce among themselves and generate new variations. Continuous accumulation of those variations through a few generations may ultimately lead to the formation of a new species.

Question 8.
What tools have been used to study human evolution?
Answer:
The tools used for tracing evolutionary line are:

1. Excavating time – dating and study of fossils.
2. Determining DNA sequences.

MP Board Class 10th Science Chapter 9 Long Answer Type Questions

Question 1.
A husband have 46 chromosomes, his wife has 46 chromosomes. Then, why don’t their offspring have 46 pairs of chromosomes, which is obtained by the fusion of male and female gametes? Support your answer with a neat illustration.
Answer:
At the stage of gamete formation, meiosis division (reduction division) occurs. As a result, each gamete receives half number of chromosomes of the parent. So, when male gamete (sperm) fuses with egg, original number of chromosomes of the parent is received by the zygote.

Fertilisation

Question 2.
Name the characters studied by Mendel in garden pea.
Answer:

Question 3.
Explain the terms:
Monohybrid cross, dihybrid cross, monohybrid ratio and dihybrid ratio.
Answer:
Monohybrid cross: Monohybrid cross is that cross which is made to study the inheritance of a single pair of genes or factors of a character.

Dihybrid cross: It is a cross which is made to study the inheritance of two pairs of genes or two characters.

Monohybrid ratio: It is the ratio which is obtained in the F2 generation when a monohybrid cross is made. It is usually 3 : 1 (Phenotypic ratio) or 1 : 2 : 1 (genotypic ratio).

Dihybrid ratio: It is the ratio, which is obtained in the F2 generation when a dihybrid cross is studied. It is usually 9 : 3 : 3 : 1 (phenotypic ratio).

Question 4.
Describe any three methods of tracing evolutionary relationships among organisms.
Answer:
The following methods help us in tracing evolutionary relationships:

(i) Study of homologous organs: Organs which have similar structure and origin are called homologous organs. For example: limbs of birds, frog, human may look different but they have similar structure and origin. Such homologous organs help to identify an evolutionary relationship between apparently diffejent species.

(ii) Study of analogous organs: Analogous organs are similar in function but differ in structure and origin. For example’. forelimbs of birds and bats are used for flying but their origins and components are not common. Thus, study of analogous organs reveals difference in their ancestry and their evolutionary relationship.

(iii) Study of fossils: All impressions, casting of body or hard remains of ancient life in the sedimentary rocks are called fossils. Study of fossils helps in finding out:
(a) Interrelationship of ancient life.
(b) Correlation of forms of life existing today and their line of evolution from ancient life.

Question 5.
A tall pea plant bearing violet flowers is crossed with short pea plant bearing white flowers. Work out the F1 and F2 generations. Give F2 ratio.
Answer:
Parents: Tall pea plant with violet flower × Short pea plant with white flower

Question 6.
Given below is the experiment carried out by Mendel to study inheritance of two traits in garden pea:
(а) What do A, B, C, D, E, F and G represent in these boxes?
(b) State the objective for which Mendel performed this experiment.

Independent inheritance of two separate traits, shape, and colour of seeds.
Answer:
(a) A = gamete (Ry) of round green plant.
B = gamete (rY) of wrinkled yellow plant.
C = (RrYy).
D = 9, E = 3, F = 3, G = 1.
(b) To show independent inheritance of traits or to prove law of independent assortment.

MP Board Class 10th Science Chapter 9 NCERT Textbook Activities

Class 10 Science Activity 9.1 Page No. 143

Observe the ears of all the students in the class. Prepare a list of students having free or attached earlobes and calculate the percentage of students having each (Fig. 9.1). Find out about the earlobes of the parents of each student in the class. Correlate the earlobe type of each student with that of their parents. Based on this evidence, suggest a possible rule for the inheritance of earlobe types.

(a) Free and (b) attached earlobes. The lowest part of the ear, called the earlobe, is closely attached to the side of the head in some of us, and not in others. Free and attached earlobes are two variants found in human populations.

Observations:

• Earlobes can be free or attached. The genes for earlobe inheritance consists of two alleles. Both the alleles for attached and free earlobes can be present in a single human being, the one which is dominant shows and the recessive one do not express itself,

Class 10 Science Activity 9.2 Page No. 144

• In Fig. 9.2, what experiment would we do to confirm that the F2 generation did in fact have a 1 : 2 : 1 ratio of TT, Tt and tt trait combinations?

Inheritance of traits over two generations.

Observations:
The cross fertilisation of pea plants showing different traits can be done at F2 stage and the number of plants for particular trait (height here) can be studied and used to confirm 1 : 2 : 1 ratio of TT, Tt and it

MP Board Class 10th Science Solutions

MP Board Class 10th Science Solutions Chapter 14 Sources of Energy

MP Board Class 10th Science Chapter 14 Intext Questions

Class 10th Science Chapter 14 Intext Questions Page No. 243

Question 1.
What is a good source of energy?
Answer:
We could then say that a good source of energy would be one.

1. Which would do a large amount of work per unit volume or mass.
2. be easily accessible.
3. be easy to store and transport, and
4. perhaps most importantly, be economical.

Question 2.
What is a good fuel?
Answer:
A good fuel is one which

• produces more heat per unit mass. It has high calorific value.
• produces less harmful gases on combustion.
• is cheap and easily available.
• is every to handle safe to transport and convenient to store.

Question 3.
If you could use any source of energy for heating your food, which one would you use and why?
Answer:
We should select which is easily available and it should be cheaper. Bio-gas is an excellent fuel as it contains. It burns without some. Its heating capacity is high. This gas is convenient for consumption and transportation.

Class 10th Science Chapter 14 Intext questions Page No. 248

Question 1.
What are the disadvantages of fossil fuels?
Answer:
Disadvantages of fossil fuels are as following:

1. Fossil fuels are limited source and we can not use them for more than 110-120 years. From not so designing or making machines dependent on them will be a big failure after their loss.
2. Fossil fuel generates big amount of pollution which will destroy our atmosphere and increase temperature of earth which lead to destruction of our ecosystem.

Question 2.
Why are we looking at alternate sources of energy?
Answer:
Fossil fuels are a non-renewable source of energy. So we need to conserve them. If we were to continue consuming these sources at such alarming rates, we would soon run out of energy. In order to avoid this, alternate sources of energy were explored.

Question 3.
How has the traditional use of wind and water energy been modified for our convenience?
Answer:
(1) Wind energy: The kinetic energy of the wind can be used to do work. This energy was harnessed by wind mills in the past to do mechanical work. For example in a water lifting pump, the rotatory motion of windmill is utilized to lift water from a well. Today wind energy is also used to generate electricity. A windmill essentially consists of a structure similar to a large electric fan that is erected at some height on a rigid support.

A number of windmills are erected over a large area, which is known as wind energy farm. The energy output of each windmill in a farm is coupled together to get electricity on a commercial scale wind energy farms can be established only at those places where wind blows for the greater part of a year. The wind speed should also be higher than 15 km/h to maintain the required speed of the turbine, since, the tower and blades are exposed to the vagaries of nature like rain, sun, storm and cyclone, they need a high level of maintenance.

(2) Water energy: In order to produce hydel electricity, high rise dams are constructed on the river to obstruct the flow of water and thereby collect water in larger reservoirs. The water level rises and in this process the kinetic energy of flowing water gets transformed into potential energy. The water from the high level in the dam is carried through pipes, to the turbine, at the bottom of the dam. Sine the water in the reservoir would be refilled each time. It rains (hydropower is a renewable source of energy) we would not have to worry about hydro electricity sources getting used up the way fossil fuels would get finished one day.

Class 10th Science Chapter 14 Intext Questions Page No. 253

Question 1.
Can any source of energy be pollution free? Why or why not?
Answer:
Yes, nature show itself some examples of energy conversion which arg pollution free as photosynthesis, as we know in this process photo energy is converted to chemical energy. Hence, solar energy is best way to produce pollution free energy.

Question 2.
Hydrogen has been used as rocket fuel. Would you consider it a cleaner fuel than CNG? Why or why not?
Answer:
Yes, it is cleaner because it does not create any residual hazardous product or chemicals which pollute environment or disbalance the ecosystem. But, handling such big amount of energy properly is required.

Class 10th Science Chapter 14 Intext Questions Page No. 254

Question 1.
Name two energy sources that you would consider to be renewable. Give reasons for your choices.
Answer:
Energy derived from water, wind, sun and ocean all are renewable. All these energies can be harnessed into usable form as long as the solar system exists.

Question 2.
Give the names of two energy sources that you would consider to be exhaustible. Give reasons for your choices.
Answer:

• fossil fuels
• Nuclear fuels

Fossil fuels are present in a limited. amount in the earth. Once exhausted, they will not be available to us again. It takes millions of years for fossil fuel to be formed. The nuclear materials which can be conveniently extracted from earth 7. are limited and hence they will get exhausted one day.

Class 10th Science Chapter 14 NCERT Textbook Exercises

Question 1.
A solar water heater cannot be used to get hot water on:
(a) A sunny day
(b) A cloudy day
(c) A hot day
(d) A windy day
Answer:
(b) A cloudy day

Question 2.
Which of the following is not an example of a bio-mass energy source?
(a) Wood
(b) Gobar-gas
(c) Nuclear energy
(d) Coal
Answer:
(c) Nuclear energy

Question 3.
Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the sun’s energy?
(a) Geothermal energy
(b) Wind energy
(c) Nuclear energy
(d) Bio-mass
Answer:
(c) Nuclear energy

Question 4.
Compare and contrast fossil fuels and the sun as direct sources of energy.
Answer:

Question 5.
Compare and contrast bio-mass and hydro electricity as sources of energy.
Answer:

Question 6.
What are the limitations of extracting energy from—
(a) the wind?
(b) waves?
(c) tides?
Answer:
(a) The wind:

1. Wind energy farms can be established only at those places where wind blows for the greater part of a year.
2. The wind speed should also be higher than 15 km/h to maintain the required speed of the turbine.
3. Establishment of wind energy farms require large are of land.

These are the limitations of extracting energy from the wind.

(b) Limitations of extracting waves energy: The waves are generated by strong winds blowing across the sea. Wave energy would be a viable proposition only where waves are very strong.

(c) Limitations of extracting tidal energy: The locations where such dams can be built are limited.

Question 7.
On what basis would you classify energy sources as
(a) renewable and non-renewable?
(b) exhaustible and inexhaustible?
Are the options given in (a) and (b) the same?
Answer:
The options given in (a) and (b) are the same.

Question 8.
What are the qualities of an ideal source of energy?
Answer:
The qualities of an ideal source of energy are as follows:

1. Which would do a large amount of work per unit volume or mass.
2. Be easily accessible.
3. Be easy to store and transport, and
4. Perhaps most importantly, be economical.

Question 9.
What are the advantages and disadvantages of using a solar cooker? Are there places where solar cookers would have limited utility?
Answer:
Solar cooker is a device used to trap solar energy and utilize it to cook food. It consists of a box painted black from inside to absorb heat of the Sun (black colour is the best absorber of heat). A thick glass lid is placed over the box to trap heat energy of the sun. A plane mirror reflector is also attached to the box so that a strong beam of sunlight falls over the cooker’s top.

The reflector of the solar cooker sends strong beams of sunlight over the top of the cooker. Sunlight consists of about 1/3 rd infra-red rays which have a heating effect. These infra-red rays are of shorter wavelength as these are produced by a very hot source of heat. The glass lid over the cooker allows these infra-red rays of short wavelength into the cooker but does not allow the infra-red rays which are emitted by the black surface of the cooker to escape as these are of longer wavelength. Thus, heat energy of the sun gets trapped in the black box of the solar cooker. This heat cooks the food material kept in the black box.

Fig. 14.1: Solar heating device (solar cooker).

Advantages of solar cooker:

1. It is used to cook food and saves precious fossil fuel.
2. It does not cause any pollution.
3. No smoke is produced during the working of a solar cooker.
4. Nutrients of food material, which is to be cooked in the solar cooker, do not get destroyed.
5. Four food items can be cooked at the same time.

Disadvantages:

1. It cannot be used to cook food during the night.
2. It cannot be used to cook food on a cloudy day.
3. The direction of the reflector has to be changed after a small interval of time with the change of position of the sun.

Question 10.
What are the environmental consequences of the increasing demand for energy? What steps would you suggest to reduce energy consumption?
Answer:
They are:

• Burning of fossil fuels to meet the increasing demand for energy causes air pollution.
• Construction of dams and rivers to generate hydroelectricity destroys large ecosystems which get submerged underwater in the dams further, a large amount of methane [which is a green house gas] is produced when submerged vegetation rots under anaerobic conditions.

In order to reduce energy consumption

• Fossil fuel should be used with care and caution to derive maximum benefit out of them.
• Fuel saving devices such as pressure cookers etc should be used.
• Efficiency of energy sources should be maintained be getting them regularly serviced.
• And last of all, we should be economical in our energy consumption as energy saved is energy produced.

MP Board Class 10th Science Chapter 14 Additional Important Questions

MP Board Class 10th Science Chapter 14 Multiple Choice Questions

Question 1.
Can we convert any form of energy to its other form?
(a) Yes
(b) No
(c) May be Yes
(d) None
Answer:
(a) Yes

Question 2.
Photosynthesis represents what kind of conversion of energy:
(a) Conversion of photo-energy to chemical energy
(b) Conversion of chemical energy to photo-energy
(c) Conversion of physical energy to photo-energy
(d) Conversion of chemical energy to electrical energy
Answer:
(a) Conversion of photo-energy to chemical energy

Question 3.
Which one of the following is good source of energy?
(a) Burning wood
(b) Burning cooking oil
(c) Burning wax
(d) Burning petroleum
Answer:
(d) Burning petroleum

Question 4.
Fuel not used for cooking is:
(a) LPG
(b) Coal
(c) Petrol
(d) PNG
Answer:
(c) Petrol

Question 5.
Renewable source of energy is:
(a) Diesel
(b) Water
(c) Coal
(d) Petrol
Answer:
(b) Water

Question 6.
Physical energy is changed to electrical energy in:
(a) Thermal power plant
(b) Hydro power plant
(c) Photosynthesis
(d) None
Answer:
(a) Thermal power plant

Question 7.
Which one among following is not a property of biomass fuel?
(a) It contains 75% methane
(b) Bum without smoke
(c) Leaves no residue
(d) Easy to install
Answer:
(c) Leaves no residue

Question 8.
What kind of energy is used in wind energy form?
(a) Potential energy
(b) Kinetic energy
(c) Chemical energy
(d) Photo-energy
Answer:
(b) Kinetic energy

Question 9.
Which one is among alternative non-conventional source of energy?
(a) Fossil fuel
(b) Biomass
(c) Solar energy
(d) None of these
Answer:
(c) Solar energy

Question 10.
What percent of total of solar energy is absorbed on earth:
(a) 10%
(b) 50%
(c) 100%
(d) Can’t calculate
Answer:
(a) 10%

Question 11.
Ultimate source of energy on earth is:
(a) Electricity
(b) Nuclear energy
(c) Fossil fuel
(d) Sun
Answer:
(d) Sun

Question 12.
To maintain a wind energy farm, the wind speed should be:
(a) 1-2 km/h
(b) 180-200 km/h
(c) 15-20 km/h
(d) Can’t be calculated
Answer:
(c) 15-20 km/h

Question 13.
Which of the following element is chosen for generating nuclear energy?
(a) Uranium
(b) Silicon
(c) Germanium
(d) Carbon
Answer:
(c) Germanium

Question 14.
How bio gas is generated in biomass plant?
(a) By distillation
(b) By anaerobic fermenting
(c) By reduction
(d) By simple burning
Answer:
(b) By anaerobic fermenting

Question 15.
Natural gas contains:
(a) CO₂
(b) H₂O
(c) CH₄
(d) NH₃
Answer:
(c) CH₄

Question 16
…………. is used in a solar cell for storing energy.
(a) Gold
(b) Silver
(c) Germanium
(d) Silicon
Answer:
(d) Silicon

Question 17.
The stored heat in the earth is harnessed as:
(a) Fuel
(b) Geothermal energy
(c) Solar energy
(d) Biomass
Answer:
(b) Geothermal energy

Question 18.
Main constituent of LPG is:
(a) Butane
(b) Methane
(c) Propane
(d) Ethane
Answer:
(c) Propane

Question 19.
High calorific value of a material represents:
(a) A good manure
(b) A good fuel
(c) Protein rich material
(d) A good conductor of electricity
Answer:
(b) A good fuel

Question 20.
…………………… is a coal with highest carbon content
(a) Bituminous
(b) Peat
(c) Anthracite
(d) None
Answer:
(c) Anthracite

MP Board Class 10th Science Chapter 14 Very Short Answer Type Questions

Question 1.
Name two things which are considered as good source of energy.
Answer:
Coal and petrol.

Question 2.
Write a main characteristic of a good fuel.
Answer:
It should release a large amount of energy per unit volume.

Question 3.
Which food product can be used to produce light energy?
Answer:
Cooking oil and fodder.

Question 4.
What kind of energy source can be used with the help of concave mirror?
Answer:
Solar energy when being used in solar appliances.

Question 5.
What kind of energy is heat energy of molten rock present inside earth core?
Answer:
Geothermal energy.

Question 6.
Name two energy sources which are conventional and renewable in nature.
Answer:
Biomass and hydroelectricity.

Question 7.
Write one limitation of use of fossil fuels.
Answer:
Pollution.

Question 8.
How much percent of CH₄ is present in Bio gas?
Answer:
75%.

Question 9.
Write two kind of solar energy manifestation to oceans.
Answer:
Wave energy and ocean thermal energy.

Question 10.
What percentage of nuclear energy is contributed in total energy production in India?
Answer:
Approximately 3-4%.

Question 11.
What is main source of solar energy?
Answer:
Sun is the main source of energy.

Question 12.
Give one cause of limitation of solar energy.
Answer:
Solar energy is costly.

Question 13.
What is nuclear energy?
Answer:
Nuclear fission is the process during which two nucleus fuse to form one nucleus. The energy which is produced in this process called nuclear energy.

Question 14.
Give three examples of energy which is produced from sea.
Answer:
Tidal Energy, wave energy and Ocean thermal energy.

MP Board Class 10th Science Chapter 14 Short Answer Type Questions

Question 1.
Why it is important to consume energy wisely?
Answer:
When we use energy in its usable form we convert the form of energy and get our work done during the process. Since, we cannot reverse . the change involved in this process so we cannot get back the original usable form of energy. Due to this, it becomes important to think about energy shortage and the related energy crisis.

Question 2.
Write characteristics of a good source of energy.
Answer:
Characteristics of a good source of energy are:

• It should be effective i.e., able to do large amount of work in mass or volume.
• It should be easily accessible.
• It should be easily transported from one place to other.
• It should be economical.

Question 3.
What are conventional sources of energy?
Answer:
The sources of energy which have been in knowledge since a long time are conventional sources of energy. Firewood, coal, petroleum, natural gas, hydel energy, wind energy and nuclear energy are considered to be the conventional sources of energy.

Question 4.
Give two examples of fossil fuels.
Answer:
Two examples of fossil fuels used commercially for producing energy are:

Coal: The plants buried under swamps after death for very long time and due to high pressure and high temperature inside the earth; they are converted into coal. Coal is the highest used energy source in India.

Petroleum: The animals body after death get buried under the ocean surface and due to high pressure and high tempera lure inside the water were converted into petroleum; in due course of time Petroleum is among the major source of energy. Petroleum products are used as automobile fire! and also in the chemical industries.

Question 5.
Differentiate between non-renewable and renewable sources of Energy.
Answer:
Non-renewable sources of Energy: It takes millions of years for the formation of fossils fuels. Since, they cannot be replenished in the foreseeable future, they are known as non-renewable sources of energy.

Renewable sources of Energy: Those sources of energy which can be replenished quickly are called renewable sources of energy. Hydel energy, wind energy and solar energy are examples of renewable sources of energy.

Question 6.
What is hydel energy?
Answer:
Hydel Energy: Hydel energy is produced by utilizing the kinetic energy of flowing water. Huge dams are built over a source of water. Water is collected behind ‘the dam and released. When the water falls on the turbine; the turbine moves; because of kinetic energy of water. Electricity is generated by the turbine. Electricity; thus generated is called hydel energy of hydroelectricity. Water in the reservoir is replenished with rainwater and so, availability of water is, not a problem for hydroelectricity.

Question 7.
How electricity is generated in thermal power plant?
Answer:
In a thermal power plant, coal or petroleum is used for converting water into steam. The steam is used to run the turbine and thus, electricity in generated.

Question 8.
Define biomass.
Answer:
Biomass: The plants and animals constitute the biomass. Farm waste; such as stalks of harvested plants and dung of cattle; can be used to generate methane. The decomposition of biomass produces methane; which can be channelized for useful purposes.

Question 9.
What is bio-gas plant? How it is channelized?
Answer:
Bio-gas plant: Bio-gas plant can be very useful in solving the energy need of rural areas. A bio gas plant is a dome like structure which is usually built from bricks and concrete. In the mixing tank; the slurry is made from cow dung and water. The slurry then goes to the digester; which is a closed chamber. Since oxygen is absent in the digester, the anaerobes carry on their work of decomposition. The process of decomposition produces bio gas. Bio gas has about 70% of methane and the rest is composed of other gases.

The bio gas is channelized through a pipe and can be utilitzed as kitchen fuel and also as fuel for getting light. The slurry; left behind; is removed. It is used as manure, once it dries.

Question 10.
What is wind energy?
Answer:
Wind energy has been in use since ages. The sail boats of the pre-industrialisation era used to run on wind power. Windmills have been in use; especially in Holland; since the medieval period. Nowadays, windmills are being used to generate electricity. The kinetic energy of wind is utilized to run the turbines; which generate electricity.

Question 11.
What is non-conventional sources of energy?
Answer:
Energy sources which are relatively new are called non-conventional sources of energy, e.g., nuclear power and solar energy.

Question 12.
Explain solar energy.
Answer:
The sun is the main source of energy for all living beings on this earth. Even the energy in the fossil fuels has come from the sun. The sun is an endless reservoir of energy which would be available as long as the solar system is in existence. Technologies for harnessing the solar energy have been developed in recent times.

Question 13.
What is solar cooker and how food is cooked in it?
Answer:
Solar cooker is very simple in design and mode of function. It is usually made from mirrors. Plane mirrors are placed inside a rectangular box. The light reflected from the plane mirrors concentrates the solar energy inside the solar cooker which generates enough heat to cook food.

Question 14.
What are solar cells?
Answer:
Solar cells: Solar cells are made from silicon. The solar panel converts solar energy into electrical energy which is stored in a battery; for later use.

Question 15.
What is tidal energy?
Answer:
Due to the gravitational pull of the moon, tides happen near seashores. Water rushes up near the seashore during a high tide and goes down during a low tide. Dams are built near seashores to collect the water which comes during a high tide. When the water runs back to the ocean, the flow of water can be utilized to generate electricity.

Question 16.
How energy is generated from molten rocks?
Answer:
The molten rocks from the inside the earth are pushed in certain regions of the earth. Such regions are called the hot spots of the earth. When groundwater comes in contact with such hot spots, lot of steam is generated. This steam can be harnessed to produce energy which is called Geothermal Energy.

Question 17.
What is nuclear energy?
Answer:
Nuclear fission is the process during which two nucleus fuse to form one . nucleus. The process generates a huge amount of energy. This phenomenon is utilized in nuclear power plants. Nuclear power is safest for the environment but the risk of damage due to accidental leaks of radiation is pretty high. Further, storage of nuclear waste is a big problem because of potential risk of radiation involved. Nonetheless, many countries are using nuclear power in a big way.

MP Board Class 10th Science Chapter 14 Long Answer Type Questions

Question 1.
Define biomass. What is Bio gas plant? How it is channelized?
Answer:
Biomass: The plants and animals constitute the biomass. Farm waste; such as stalks of harvested plants and dung of cattle; can be used to generate methane. The decomposition of biomass produces methane; which can be channelized for useful purposes.

Bio gas plant: Bio gas plant can be very useful in solving the energy need of rural areas. A bio gas plant is a dome like structure which is usually built from bricks and concrete. In the mixing tank; the slurry is made from cow dung and water. The slurry then goes to the digester; which is a closed chamber. Since oxygen is absent in the digester, the anaerobes carry on their work of decomposition. The process of decomposition produces bio gas, has about 70% of methane and the rest ‘is composed of other gases. The bio gas is channelized through a pipe and can be utilized as kitchen fuel and also as fuel for getting light. The slurry; left behind; is removed. It is used as manure, once it dries.

Question 2.
What is wind energy? Write limitations of wind energy.
Answer:
Wind energy has been in use since ages. The sail boats of the re-industrialization era used to run on wind power. Windmills have been in use; especially in Holland; since the medieval period.

Nowadays, windmills are being used to generate electricity. The kinetic energy of wind is utilized to run the turbines; which generate electricity.

Limitations of wind Energy: Wind farms can only be established at those places where the wind speed is high enough and is more than 15 km/hr for most parts of the year. Wind farms need to be established on large tracts of land. The fan of the windmill has many moving parts; so cost of maintenance and repair is quite high. The fact, that it has to suffer the vagaries of the nature, further compounds the problem. Initial cost of establishing a wind farm is very high.

Question 3.
Briefly explain energy from sea.
Answer:
Tidal energy: Due to the gravitational pull of the moon, tides happen at seashores. Water rushes up near the seashore during a high tide and goes down during a low tide. Dams are built near seashores to collect the water which comes during a high tide. When the water runs back to the ocean, the flow of water can be utilized to generate electricity.

Wave Energy: Waves can also be a good source of energy. Many devices are being designed and tested to produce wave energy. For example; a hollow tower is built near the seashore. When water gushes in the tube because of wave, it forces the air upwards. The kinetic energy of air in the tube is used to run a turbine. When the wave goes down; air from up goes down the tube which is also used in running the turbine.

Ocean Thermal Energy. The water at sea surface is hot during daytime, while the water at lower level is cold. The temperature differential in water levels can be utilized to generate energy. If the temperature differential is more than 20°C, then ocean thermal energy can be utilized from that place. For this, a volatile liquid; like ammonia; is boiled using the heat from the hot water at the surface. The steam of the volatile liquid is utilized to run the turbine to generate electricity. Colder water from the surface below is utilized to condense ammonia vapour which is then channelized to the surface to repeat the cycle.

Question 4.
Explain Solar Energy. How is food cooked with help of solar energy and also explain its limitations.
Answer:
The sun is the main source of energy for all living beings on this earth. Even the energy in the fossil fuels has come from the sun. The sun has an endless reservoir of energy which would be available as long as the solar system is in existence. Technologies for harnessing the solar energy hav been developed in recent times.

Solar cooker is very simple in design and mode of function. It is usually made from mirrors. Plane mirrors are placed inside a rectangular box. The light reflected from the plane mirrors concentrates the solar energy inside the solar cooker which generates enough heat to cook food.

Limitations of solar Energy: The technologies for harnessing solar energy are at a nascent stage. At present, the cost benefit ratio for using solar energy is not conducive. Using solar energy is exhorbitantly costly.

MP Board Class 10th Science Chapter 14 NCERT Textbook Activities

Class 10 Science Activity 14.1 Page No. 242

• List four forms of energy that you use from morning, when you wake up, till you reach the school.
• From where do we get these different forms of energy?
• Can we call these ‘sources’ of energy? Why or why not?

Observations:

• Forms of energy that we use are electrical energy, mechanical energy, chemical energy for vehicles, chemical energy from food.
• Energy can neither created nor destroyed. It is just transferred from one form to the other.
• The ‘sources’ of energy are the one which releases energy to be used in such forms.

Class 10 Science Activity 14.2 Page No. 243

• Consider the various options we have when we choose a fuel for cooking our food.
• What are the criteria you would consider when trying to categories something as a good fuel?
• Would your choice be different if you lived:

(a) in a forest?
(b) in a remote mountain village or small island?
(c) in New Delhi?
(d) lived five centuries ago?

• How are the factors different in each case?

Observations:

• The criteria for good fuel inducts amount of energy released upon combustion, smoke produced or not, easily accessible, economical and easy be store and transport.
• In forest or remote village, wood from forests can be used as fuel.
• In new Delhi, electrical energy would be a choice.
• Five centuries ago, may be mechanical energy would have been used.
• Factors includes availability of the sunrises and case in utilizing them.

Class 10 Science Activity 14.3 Pages No. 244-245

• Take a table-tennis ball and make three slits into it.
• put semicircular fins cut out a metal sheet into these slits.
• Pivot the tennis ball on an axle through its centre with a straight metal wire fixed to a rigid support. Ensure that the tennis ball rotates freely about the axle.
• Now connect a cycle dynamo to this,
• Connect a bulb in series.
• Direct a jet of water or steam produced in a pressure cooker at the fins (Fig. 14.2). What do you observe?

Fig. 14.2: A model to demonstrate the process – of thermo – electric production.

Observations:

• We observe that fan starts morning. The rotor blade also moves with speed that turn the shaft of the dynamo and concert the mechanical energy into electrical energy.

Class 10 Science Activity 14.4 Page No. 248

• Find out from your grand-parents or other elders:
  • (a) how did they go to school?
  • (b) how did they get water for their daily needs when they were young?
  • (c) what means of entertainment did they use?
• Compare the above answers with how you do these tasks now.
• Is there a difference? If yes, in which case more energy from external sources is consumed? _

Observations:

• Earlier, people used to go school on-foot.
• The water were drawn from wells from far off places and carried on head in pots to the home.
• Means of entertainment includes folk dances, songs etc.
• The scenario has totally changed, nowadays, people are being separately and generally, not in noses as were done decades ago. More energy from internal sources is consumed these days to fulfill the demands of energy which is inversed a lot.

Class 10 Science Activity 14.5 Page No. 249

• Take two conical flasks and paint one white and the other black. Fill both with water.
• Place the conical flasks in direct sunlight for half an hour to one hour.
• Touch the conical flasks. Which one is hotter? You could also measure the temperature of the water in the two conical flasks with a thermometer.
• Can you think of ways in which this finding could be used in your daily life?

Observations:

• The one with black colour is hotter as black absorbs more heat as compared to white one.
• In daily life, we apply this on the colour we wear. The days which are that, we avoid black colours as they absorb more heat. Sincerity for cooling effect, white colours used.

Class 10 Science Activity 14.6 Pages No. 249-250

• Study the structure and working of a solar cooker and/or a solar water- heater, particularly with regard to how it is insulated and maximum heat absorption is ensured.
• Design and build a solar cooker or water-heater using low-cost material available and check what temperatures are achieved in vour system.
• Discuss what would be the advantages and limitations of using the solar cooker or water-heater.

Observations:

• The minimum heat absorption is ensured by painting it black in colour. The glass used on the lop traps the infrared rays from the sun and do not allow them the escape.
• Advantages includes no wastage of energy as solar energy is a trapped to be used further. It is a renewable some of energy that do not create any pollution.
• Limitations includes plausibility of solar rays at certain times of day only. It will not work on sunny days.

Class 10 Science Activity 14.7 Page No. 252

• Discuss in class the question of what is the ultimate source of energy for bio-mass, wind and ocean thermal energy.
• Is geothermal energy and nuclear energy different in this respect? Why ?
• Where would you place hydro electricity and wave energy?

Observations:

• The cleaner source of energy is the sun. Yes, cleaner energy is obtained from fusion or fission of molecules whereas geothermal energy is the energy present in the earth. The energy from geological changes is harnessed to be used in various ways.
• Hydro electricity and wave energy are also form of renewable energy.

Class 10 Science Activity 14.8 Page No. 253

• Gather information about various energy sources and how each one affects the environment.
• Debate the merits and demerits of each source and select the best source of energy on this basis.

Observations:

• Among the various source of energy, renewable sources of energy are the best as they do not harm the environment and this energy can be reused and when required. Non-renewable resources of’energy are harmful to the environment and causes pollution.
• Best sources of energy arc renewable sources of energy like solar, hydro, tidal, wave energy etc.

Class 10 Science Activity 14.9 Page No. 254

• Debate the following two issues in class.

(a) The estimated coal reserves are said to be enough to last us for another two hundred years. Do you think we need to worry about coal getting depleted in this case? Why or why not?
(b) It is estimated that the Sun will last for another five billion years. Do we have to worry about solar energy getting exhausted? Why or why not?

• One the basis of the debate, decide which energy sources can be considered

1. exhaustible
2. inexhaustible
3. renewable
4. non-renewable.

Give your reasons for each choice.

Observations:

• Yes, we need to worry about availability of coal as resources are limited they are made from fossil fuels which take millions of years to be formed. Also, they cause pollution and harm the environment.
• Sun will least for another five billion years but we should use this to the maximum as there are still so many years in which it can be utilized, harness eat and stored.
• Exhaustible are those energy sources which will get exhausted soon from fossil fuel inexhaustible are the one like solar, tidal etc. Renewable can be used whereas non-renewable are not used again and again.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Take π = \(\frac{22}{7}\), unless stated otheriwise

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Radius of the sphere (r₁) = 4.2 cm
∴ Volume of the sphere = \(\frac{4}{3}\) πr₁³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{42}{10} \times \frac{42}{10} \times \frac{42}{10} \mathrm{cm}^{3}\)
Radius of the cylinder (r₂) = 6 cm
Let h be the height of the cylinder.
∴ Volume of the cylinder = πr²h
= \(\frac{22}{7}\) × 6 × 6 × h cm³
Since volume of the metallic sphere = Volume of the cylinder

Hence, the height of the cylinder is 2.744 cm

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radii of the given spheres are
r₁ = 6 cm, r₂ = 8 cm and r₃ = 10 cm
⇒ Volume of the given spheres are

= \(\frac{4}{3} \times \frac{22}{7}\) × [1728] cm³
Let the radius of the new big sphere be R. Volume of the new sphere
= \(\frac{4}{3}\) × π × R₃ = \(\frac{4}{3} \times \frac{22}{7}\) × R₃
Since, the two volumes must be equal.
∴ \(\frac{4}{3} \times \frac{22}{7} \times R^{3}=\frac{4}{3} \times \frac{22}{7} \times 1728\)
⇒ R₃ = 1728 ⇒ R = 12 cm
Thus, the required radius of the resulting sphere is 12 cm.

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Diameter of the cylindrical well = 7 m
⇒ Radius of the cylindrical well (r) = \(\frac{7}{2}\) m
Depth of the well (h) = 20 m
∴ Volume = πr²h = \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) × 20 m³
= 22 × 7 × 5 m³
⇒ Volume of the earth taken out = 22 × 7 × 5 m³
Now this earth is spread out to form a cuboidal platform having length = 22 m, breadth = 14 m
Let h be the height of the platform.
∴ Volume of the platform = 22 × 14 × h m³
Since, the two volumes must be equal
∴ 22 × 14 × h = 22 × 7 × 5
Thus, the required height of the platform is 2.5 m.

Question 4.
A well of diameter 3 m is dug 14m deep. The earth taken out of it has been spread evenly all around it in shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Diameter of cylindrical well = 3 m
⇒ Radius of the cylindrical well = \(\frac{3}{2}\) m = 1.5 m
Depth of well (h) = 14 m
∴ Volume of cylindrical well

Let the height of the embankment = H m.
Internal radius of the embankment (r) = 1.5 m.
External radius of the embankment (R)
= (4 + 1.5) m = 5.5 m.
∴ Volume of the embankment
= πR²H – πr²H = πH [R² – r²]
= πH (R + r) (R – r)
= \(\frac{22}{7}\) × H (5.5 + 1.5)(5.5 – 1.5)
= \(\frac{22}{7}\) × H × 7 × 4m³
Since, Volume of the embankment=Volume of the cylindrical well
⇒ \(\frac{22}{7}\) × H × 7 × 4 = 99
⇒ H = 99 × \(\frac{7}{22} \times \frac{1}{7} \times \frac{1}{4} m=\frac{9}{8} m\) = 1.125 m
So, the required height of the embankment = 1.125 m.

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and 6. diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
For the circular cylinder:

Diameter = 12 cm
⇒ Radius (r) = \(\frac{12}{2}\) = 6cm and height (h) = 15 cm
∴ Volume of circular cylinder
= πr²h = \(\frac{12}{2}\) × 6 × 6 × 15 cm³
For conical and hemispherical part of icecream :
Diameter = 6 cm ⇒ radius (R) = 3 cm
Height of conical part (H) = 12 cm

Volume of ice cream cone = (Volume of the conical part) + (Volume of the hemispherical part)

Thus, the required number of cones is 10.

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Solution:
For a circular coin:

Diameter = 1.75 cm
⇒ Radius (r) = \(\frac{175}{200}\) cm
Thickness (h) = 2mm = \(\frac{2}{10}\) cm
∴ Volume of one coin = πr²h = \(\frac{22}{7} \times\left(\frac{175}{200}\right)^{2} \times \frac{2}{10} \mathrm{cm}^{3}\)
For a cuboid:

Length (l) = 10 cm,
Breadth (b) = 5.5 cm
and height (h) = 3.5 cm
∴ Volume = l × b × h = 10 × \(\frac{55}{10} \times \frac{35}{10}\) cm³

Thus, the required number of coins = 400.

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
For the cylindrical bucket:
Radius (r) = 18 cm and height (h) = 32 cm
Volume of cylindrical bucket = πr²h
= \(\frac{22}{7}\) × (18)² × 32 cm³
⇒ Volume of the sand = (\(\frac{22}{7}\) × 18 × 18 × 32) cm³
For the conical heap:
Height (H) = 24 cm
Let radius of the base be R.
∴ Volume of conical heap

Thus, the required radius = 36 cm and slant height = \(12 \sqrt{13}\) cm.

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Width of the canal = 6 m,
Depth of the canal = 1.5 m
Length of the water column in 1 hr = 10 km
∴ Length of the water column in 30 minutes
(i.e., \(\frac{1}{2}\)hr) = \(\frac{10}{2}\) km = 5 km = 5000 m
∴ Volume of water flown in \(\frac{1}{2}\) hr
= 6 × 1.5 × 5000 m³ = 6 × \(\frac{15}{10}\) × 5000 m³
= 45000 m³
Since, the above amount (volume) of water is spread in the form of a cuboid of height
8 cm (= \(\frac{8}{100}\) m)
Let the area of the cuboid = a
∴ Volume of the cuboid = Area × Height

= 562500 m² = 56.25 hectares
Thus, the required area is 56.25 hectares.

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Diameter of the pipe = 20 cm
⇒ Radius of the pipe (r) = \(\frac{20}{2}\) cm = 10 cm
Since, the water flows through the pipe at 3 km/hr.
∴ Length of water column per hour(h) = 3 km
= 3 × 1000 m = 3000 × 100 cm = 300000 cm.
Length of water column per hour(h) = 3 km
Volume of water flown in one hour = πr²h
= π × 10² × 300000 cm³ = π × 30000000 cm²
Now, for the cylindrical tank :
Diameter = 10 m
⇒ Radius (R) = \(\frac{10}{2}\) m = 5 × 100 cm = 500 cm
Height (H) = 2 m = 2 × 100 cm = 200 cm
∴ Volume of the cylindrical tank = πR²H
= π × (500)² × 200 cm³
Now, time required to fill the tank

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 15 Probability Ex 15.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.1

Question 1.
Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = _________
(ii) The probability of an event that cannot happen is ______.Such an event is called _________.
(iii) The probability of an event that is certain to happen is ______ Such an event is called ______
(iv) The sum of the probabilities of all the elementary events of an experiment is ______.
(v) The probability of an event is greater than or equal to _______ and less than or equal to ______.
Solution:
(i) 1 : Probability of an event E + Probability of the event ‘not E’ = 1.
(ii) 0, impossible: The probability of an event that cannot happen is 0. Such an event is called impossible event.
(iii) 1, certain: The probability of an event that is certain to happen is 1. Such an event is called a sure or certain event.
(iv) 1: The sum of the probabilities of all the elementary events of an experiment is 1.
(v) 0, 1: The probability of an event is greater than or equal to 0 and less than or equal to 1.

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.
Solution:
(i) It depends on various factors such as whether the car will start or not. So, the probability of car will start does not equal to the probability of car will not start.
∴ The outcomes are not equally likely.
(ii) It depends on the player’s ability. So, probability that the player shot the ball is not the same as the probability that the player misses the shot.
(iii) The outcomes are equally likely as the probability of answer either right or wrong is \(\frac{1}{2}\)
(iv) The outcomes are equally likely as the probability of ‘newly born baby to be either bay or girls’ is \(\frac{1}{2}\) .

Question 3.
Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?
Solution:
Since on tossing a coin, the outcomes ‘head’ and ‘tail’ are equally likely, the result of tossing a coin is completely unpredictable and so it is a fairway.

Question 4.
Which of the following cannot be the probability of an event?
(A) \(\frac{2}{3}\)
(B) -1.5
(C) 15%
(D) 0.7
Solution:
Since, the probability of an event cannot be negative.
∴ -1.5 cannot be the probability of an event.

Question 5.
If P(E) = 0.05, what is the probability of ‘not E’?
Solution:
∵ P(E) + P(not E) = 1
∴ 0.05 + P(not E) = 1 ⇒ P(not E) = 0.95
Thus, probability of ‘not E’ = 0.95.

Question 6.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
Solution:
(i) Since there are only lemon flavoured candies in the bag.
∴ Taking out orange flavoured candy is not possible.
⇒ Probability of taking out an orange flavoured candy = 0.

(ii) Probability of taking out a lemon flavoured candy = 1.

Question 7.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?
Solution:
Let the probability of 2 students having same birthday = P(SB)
And the probability of 2 students not having the same birthday = P(NSB)
∴ P(SB) + P(NSB) = 1
⇒ P(SB) + 0.992 = 1 ⇒ P(SB) = 1 – 0.992 = 0.008

Question 8.
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is
(i) red?
(ii) not red?
Solution:
Total number of balls = 3 + 5 = 8
∴ umber of possible outcomes = 8
(i) ∵ There are 3 red balls.
∴ Number of favourable outcomes = 3
∴ P (red) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}\)
= \(\frac{3}{8}\)
(ii) Probability of the ball drawn which is not red = 1 – P(red) = \(1-\frac{3}{8}=\frac{8-3}{8}=\frac{5}{8}\)

Question 9.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red?
(ii) white?
(iii) not green?
Solution:
Total number of marbles = 5 + 8 + 4 = 17
∴ Number of all possible outcomes = 17
(i) ∵ Number of red marbles = 5
∴ Number of favourable outcomes = 5
∴ Probability of red marbles, P(red) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{5}{17}\)

(ii) Number of white marbles = 8
∴ Probability of white marbles, P(white) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{8}{17}\)
_ Number of favourable outcomes _ 8 Number of all possible outcomes 17

(iii) Number of green marbles = 4 Number of marbles which are not green
= 17-4 = 13
i.e., Favourable outcomes = 13
∴ Probability of marbles ‘not green’, P(not greeen)
\(\frac{\text { Number of favourable outcomes }}{\text { Number of all possible outcomes }}=\frac{13}{17}\)

Question 10.
A piggy bank contains hundred 50p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin
(i) will be a 50p coin?
(ii) will not be a ₹ 5 coin?
Solution:
Number of coins 50 p = 100, ₹ 1 = 50 ₹ 2 = 20, ₹ 5 = 10
Total number of coins = 100 + 50 + 20 +10 = 180
∴ Total possible outcomes = 180

(i) For a 50 p coin:
Favourable outcomes = 100

(ii) For not a ₹ 5 coin:
Y Number of ₹ 5 coins = 10
∴ Number of ‘not ₹ 5’ coins = 180 – 10 = 170
⇒ Favourable outcomes = 170

Question 11.
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?
Solution:
Number of male fishes = 5
Number of female fishes = 8
∴ Total number of fishes = 5 + 8 = 13
⇒ Total number of outcomes = 13
For a male fish:
Number of favourable outcomes = 5

Question 12.
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see figure), and these are equally likely outcomes. What is the probability that it will point at
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?

Solution:
Total number marked = 8
∴ Total number of possible outcomes = 8
(i) When pointer points at 8:
Number of favourable outcomes = 1

(ii) When pointer points at an odd number:
∵ Odd numbers are 1, 3, 5 and 7
∴ Total odd numbers from 1 to 8 = 4
⇒ Number of favourable outcomes = 4

(iii) When pointer points at a number greater than 2:
∵ The numbers 3, 4, 5, 6, 7 and 8 are greater than 2
∴ Total numbers greater than 2 = 6
⇒ Number of favourable outcomes = 6

(iv) When pointer points at a number less than 9:
∵ The numbers 1, 2, 3, 4, 5, 6, 7 and 8 are less than 9.
∴ Total numbers less than 9 = 8
∴ Number of favourable outcomes = 8

Question 13.
A die is thrown once. Find the probability of getting:
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Solution:
Since, numbers on a die are 1, 2, 3, 4, 5 and 6.
∴ Total number of possible outcomes = 6
(i) Since 2, 3 and 5 are prime number.
∴ Favourable outcomes = 3

(ii) Since the numbers between 2 and 6 are 3, 4 and 5
∴ Favourable outcomes = 3

(iii) Since 1, 3 and 5 are odd numbers.
⇒ Favourable outcomes = 3

Question 14.
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) a red face card
(iv) the jack of hearts
(v) a spade
(vi) the queen of diamonds
Solution:
Number of cards in deck = 52
∴ Total number of possible outcomes = 52
(i) ∵ Number of red colour kings = 2
[∵ King of diamond and heart is red]
Number of favourable outcomes = 2

(ii) For a face card:
∵ 4 kings, 4 queens and 4 jacks are face cards
∴ Number of face cards = 12
⇒ Number of favourable outcomes = 12

(iii) Since, cards of diamond and heart are red
∴ There are 2 kings, 2 queens, 2 jacks i.e., 6 cards are red face cards.
∴ Number of favorable outcomes = 6

(iv) Since, there is only 1 jack of hearts.
∴ Number of favourable outcomes = 1

(v) There are 13 spades in a pack of 52 cards.
∴ Number of favourable outcomes = 13

(vi) ∵ There is only one queen of diamond.
∴ Number of favourable outcomes = 1

Question 15.
Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
We have five cards.
∴ Total number of possible outcomes = 5
(i) ∵ Number of queen = 1
∴ Number of favourable outcomes = 1

(ii) The queen is drawn and put aside.
∴ Only 5 – 1 = 4 cards are left.
∴ Total number of possible outcomes = 4
(a) ∵ There is only one ace.
∴ Number of favourable outcomes = 1

(b) Since, the only queen has been put aside already.
∴ Number of favourable outcomes = 0

Question 16.
12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Solution:
We have number of good pens = 132 and number of defective pens = 12
∴ Total number of pens = 132 + 12 = 144 = Total possible outcomes
There are 132 good pens.
∴ Number of favourable outcomes = 132

Question 17.
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:
Since, there are 20 bulbs in the lot.
Total number of possible outcomes = 20
(i) ∵ Number of defective bulbs = 4
∴ Favourable outcomes = 4

(ii) ∵ The bulb drawn above is not included in the lot.
∴ Number of remaining bulbs = 20 – 1 = 19.
⇒ Total number of possible outcomes = 19.
∵ Number of bulbs which are not defective = 19 – 4 = 15
⇒ Number of favourable outcomes = 15

Question 18.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Solution:
We have total number of discs = 90
∴ Total number of possible outcomes = 90
(i) Since the two-digit numbers are 10, 11, 12, ………, 90.
∴ Number of two-digit numbers = 90 – 9 = 81
∴ Number of favourable outcomes = 81

(ii) Perfect square from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64 and 81.
∴ Number of perfect squares = 9
∴ Number of favourable outcomes = 9

(iii) Numbers divisible by 5 from 1 to 90 are 5, 10,15, 20, 25, 30, 35,40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
i. e., There are 18 numbers from (1 to 90) which are divisible by 5.
∴ Numbers of favourable outcomes = 18

Question 19.
A child has a die whose six faces show the letters as given below:

The die is thrown once. What is the probability of getting (i) A? (ii) D?
Solution:
Since there are six faces of the given die and these faces are marked with letters

∴ Total number of letters = 6
∴ Total number of possible outcomes = 6
(i) ∵ Number of faces having the letter A = 2
∴ Number of favourable outcomes = 2

(ii) ∵ Number of faces having the letter D = 1
∴ Number of favourable outcomes = 1

Question 20.
20. Suppose you drop a die at random on the rectangular region shown in figure. What is the probability that it will land inside the circle with diameter 1 m?

Solution:
Here, area of the rectangle = 3m × 2m = 6 m²
And, the area of the circle = πr²

Question 21.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it? (ii) She will not buy it?
Solution:
Total number of ball pens = 144
⇒ Total number of possible outcomes = 144
(i) Since there are 20 defective pens.
∴ Number of good pens = 144 – 20 = 124
⇒ Number of favourable outcomes = 124

(ii) Probability that Nuri will not buy it = 1 – [Probability that she will buy it]
= \(1-\frac{31}{36}=\frac{36-31}{36}=\frac{5}{36}\)

Question 22.
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. An event is defined as the sum of the two numbers appearing on the top of the dice.
(i) Complete the following table

(ii) A student argues that’there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\) Do you agree with this argument? Justify your answer.
Solution:
∵ The two dice are thrown together.
∴ Following are the possible outcomes :
(1, 1) ; (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1) ; (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1); (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1) ; (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6,.2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ Total number of possible outcomes is 6 × 6 = 36
(i) (a) The sum on two dice is 3 for (1, 2) and (2, 1)
∴ Number of favourable outcomes = 2
⇒ P(3) = \(\frac{2}{36}\)

(b) The sum on two dice is 4 for (1, 3), (2, 2) and (3, 1).
∴ Number of favourable outcomes = 3
⇒ P(4) = \(\frac{3}{36}\)

(c) The sum on two dice is 5 for (1, 4), (2, 3), (3, 2) and (4,1)
∴ Number of favourable outcomes = 4
⇒ P(5) = \(\frac{5}{36}\)

(d) The sum on two dice is 6 for (1, 5), (2, 4), (3, 3), (4, 2) and (5,1)
∴ Number of favourable outcomes = 5
⇒ P(6) = \(\frac{5}{36}\)

(e) The sum on two dice is 7 for (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6,1)
∴ Number of favourable outcomes = 6
⇒ P(7) = \(\frac{62}{36}\)

(f) The sum on two dice is 9 for (3, 6), (4, 5), (5, 4) and (6, 3)
∴ Number of favourable outcomes = 4
⇒ P(9) = \(\frac{4}{36}\)

(g) The sum on two dice is 10 for (4, 6), (5, 5), (6,4)
∴ Number of favourable outcomes = 3
⇒ P(10) = \(\frac{3}{36}\)

(h) The sum on two dice is 11 for (5, 6) and (6,5)
∴ Number of favourable outcomes = 2
⇒ P(11) = \(\frac{2}{36}\)

Thus, the complete table is as follows:

(ii) No. The number of all possible outcomes is 36 not 11.
∴ The argument is not correct.

Question 23.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.
Solution:
Let T denotes the tail and H denotes the head.
∴ All the possible outcomes are:
{H H H, H H T, H T T, T T T, T T H, T H T, T H H, H T H)
∴ Number of all possible outcomes = 8
Let the event that Hanif will lose the game denoted by E.
∴ Favourable events are: {HHT, HTH, THH, THT, TTH, HTT}
⇒ Number of favourable outcomes = 6
∴ P(E) = \(\frac{6}{8}=\frac{3}{4}\)

Question 24.
A die is thrown twice. What is the probability that
(i) 5 will not come up either time?
(ii) 5 will come up at least once?
[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]
Solution:
Since, throwing a die twice or throwing two dice simultaneously is the same.
∴ All possible outcomes are:
(1, 1); (1, 2); (1, 3); (1, 4); (1, 5); (1, 6).
(2, 1); (2, 2); (2, 3); (2, 4); (2, 5); (2, 6).
(3, 1) ; (3, 2); (3, 3); (3, 4); (3, 5); (3, 6).
(4, 1); (4, 2); (4, 3); (4, 4); (4, 5); (4, 6).
(5, 1) ; (5, 2); (5, 3); (5, 4); (5, 5); (5, 6).
(6, 1) ; (6, 2); (6, 3); (6, 4); (6, 5); (6, 6).
∴ All possible outcomes = 36
(i) Let E be the event that 5 does not come up either time.
∴ Numebr of favourable outcomes = [36 – (5 + 6)] = 25
∴ P(E) = \(\frac{25}{36}\)
(ii) Let N be the event that 5 will come up at least once, then number of favourable outcomes = 5 + 6 = 11
∴ P(N) = \(\frac{11}{36}\)

Question 25.
Which of the following arguments are correct and which are not correct? Give reasons for your answer.
(i) If two coins are tossed simultaneously there are three possible outcomes – two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\).
(ii) If a die is thrown, there are two possible outcomes – an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{3}\)
Solution:
(i) Given argument is not correct. Because, if two coins are tossed simultaneously then four outcomes are possible (HH, HT, TH, TT). So total outcomes is 4.
∴ The required probability = \(\frac{1}{4}\).
(ii) Given argument is correct.
Since, total numebr of possible outcomes = 6
Odd numbers = 3 and even numbers = 3
So, favourable outcomes = 3 (in both the cases even or odd).
∴ Probability = \(\frac{3}{6}=\frac{1}{2}\)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.4

Use π = \(\frac{22}{7}\), unless stated otherwise.

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
We have,

d₁ = 4 cm
∴ r₁ = \(\frac{d_{1}}{2}\) = 2 cm
and d₂ = 2 cm
r₂ = \(\frac{d_{2}}{2}\) = 1 cm
and h = 14 cm
Volume of the glass

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
We have,
Slant height (l) = 4 cm
Circumference of one end = 2πr₁ = 18 cm
and Circumference of other end = 2πr₂ = 6 cm

⇒ πr₁ = \(\frac{18}{2}\) = 9 cm
and πr₂ = \(\frac{6}{2}\) = 3 cm
∴ Curved surface area of the frustum of the cone
= π(r₁ + r₂) l = (πr₁ + πr₂) l = (9 + 3 ) × 4 cm²
= 12 × 4 cm² = 48 cm².

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Solution:
Here, the radius of the open side (r₁) = 10 cm
The radius of the upper base (r₂) = 4 cm
Slant height (l) = 15 cm
∴ Area of the material required = [Curved surface area of the frustum] + [Area of the top end]
= π(r₁ + r₂)l + πr₂²

Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 8 per 100 cm². (Take π = 3.14)
Solution:
We have, r₁ = 20 cm, r₂ = 8 cm and h = 16 cm



Area of the bottom = πr₂²
= (\(\frac{314}{100}\) × 8 × 8) cm² = 200.96 cm²
∴ Total area of metal required
= 1758.4 cm² + 200.96 cm² = 1959.36 cm²
Cost of metal required for 100 cm² = ₹ 8
∴ Cost of metal required for 1959.36 cm²
= ₹ \(\frac{8}{100}\) × 1959.36 = ₹ 156.75

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{1}{16}\) find the length of the wire
Solution:
Let us consider the frustum DECB of the metallic cone ABC


Thus, the required length of the wire = 7964.44 m

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 10 Circles Ex 10.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 10 Circles Ex 10.2

In questions 1 to 3, choose the correct option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution:
(A): ∵ QT is a tangent to the circle at T and OT is radius

∴ OT⊥QT
Also, OQ = 25 cm and QT = 24 cm
∴ Using Pythagoras theorem, we get
OQ² = QT² + OT²
⇒ OT² = OQ² – QT² = 25² – 24² = 49
⇒ OT = 7
Thus, the required radius is 7 cm.

Question 2.
In figure, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ =110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:
(B): TQ and TP are tangents to a circle with centre O and ∠POQ = 110°
∴ OP⊥PT and OQ⊥QT
⇒ ∠OPT = 90° and ∠OQT = 90°
Now, in the quadrilateral TPOQ, we get
∠PTQ + 90° + 110° + 90° = 360° [Angle sum property of a quadrilateral]
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 360° – 290° = 70°

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
(A) : Since, O is the centre of the circle and two tangents from P to the circle are PA and PB.
∴ OA⊥AP and OB⊥BP
⇒ ∠OAP = ∠OBP = 90°

Now, in quadrilateral PAOB, we have
∠BPA + ∠PAO + ∠AOB + ∠OBP = 360°
⇒ 80° + 90° + ∠AOB + 90° = 360°
⇒ 260° + ∠AOB = 360°
⇒ ∠AOB = 360° – 260° ⇒ ∠AOB = 100°
In right ∆OAP and right ∆OBP, we have
OP = OP [Common]
∠OAP = ∠OBP [Each 90°]
OA = OB [Radii of the same circle]
∴ ∆OAP ≅ ∆OBP [By RHS congruency]
⇒ ∠POA = ∠POB [By CPCT]
∴ ∠POA = \(\frac{1}{2}\) ∠AOB = \(\frac{1}{2}\) × 100° = 50°

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
In the figure, PQ is diameter of the given circle and O is its centre.
Let tangents AB and CD be drawn at the end points of the diameter PQ.
Since, the tangents at a point to a circle is perpendicular to the radius through the point.

∴ PQ⊥AB
⇒ ∠APQ = 90°
And PQ⊥CD
⇒ ∠PQD = 90° ⇒ ∠APQ = ∠PQD
But they form a pair of alternate angles.
∴ AB || CD
Hence, the two tangents are parallel.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
In the figure, the centre of the circle is O and tangent AB touches the circle at P. If possible, let PQ be perpendicular to AB such that it is not passing through O.
Join OP.
Since, tangent at a point to a circle is perpendicular to the radius through that point.

∴ OP⊥AB
⇒ ∠OPB = 90° ……….. (1)
But by construction, PQ⊥AB
⇒ ∠QPB = 90° ………….. (2)
From (1) and (2),
∠QPB = ∠OPB
which is possible only when O and Q coincide. Thus, the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
∵ The tangent to a circle is perpendicular to the radius through the point of contact.
∴ ∠OTA = 90°
Now, in the right ∆OTA, we have
OA² = OT² + AT² [Pythagoras theorem]

⇒ OT² = 5² – 4²
⇒ OT² = (5 – 4)(5 + 4)
⇒ OT² = 1 × 9 = 9 = 3²
⇒ OT = 3
Thus, the radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
In the figure, O is the common centre, of the given concentric circles.
AB is a chord of the bigger circle such that it is a tangent to the smaller circle at P.
Since, OP is the radius of the smaller circle.
∴ OP⊥AB ⇒ ∠APO = 90°
Also, radius perpendicular to a chord bisects the chord.

∴ OP bisects AB
⇒ AP = \(\frac{1}{2}\) AB
Now, in right ∆APO,
OA² = AP² + OP²
⇒ 5² = AP² + 3² ⇒ AP² = 5² – 3²
⇒ AP² = 4² ⇒ AP = 4 cm
⇒ \(\frac{1}{2}\) AB = 4 ⇒ AB = 2 × 4 = 8 cm
Hence, the required length of the chord AB is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure).

Prove that AB + CD = AD + BC
Solution:
Since, the sides of quadrilateral ABCD, i.e., AB, BC, CD and DA touches the circle at P, Q, R and S respectively, and the lengths of two tangents to a circle from an external point are equal.
∴ AP = AS, BP = BQ,
DR = DS and CR = CQ
Adding them, we get
(AP + BP) + (CR + RD) = (BQ + QQ) + (DS + SA)
⇒ AB + CD = BC + DA

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B.

Prove that ∠AOB = 90°.
Solution:
∵ The tangents drawn to a circle from an external point are equal.
∴ AP = AC ……… (1)
Join OC.
In ∆PAO and ∆CAO, we have
AO = AO [Common]
OP = OC [Radii of the same circle]
AP = AC [From (1)]
⇒ ∆PAO ≅ ∆CAO [SSS congruency]
∴ ∠PAO = ∠CAO
⇒ ∠PAC = 2∠CAO …………. (2)
Similarly, ∠CBQ = 2∠CBO ……………… (3)

Again, we know that sum of internal angles on the same side of a transversal is 180°.
∴ ∠PAC + ∠CBQ = 180°
2∠CAO + 2∠CBO = 180° [From (2) and (3)]
⇒ ∠CAO + ∠CBO = \(\frac{180^{\circ}}{2}\) = 90° …………… (4)
Also, in ∆AOB,
∠BAO + ∠OBA + ∠AOB = 180° [Sum of angles of a triangle]
⇒ ∠CAO + ∠CBO + ∠AOB = 180°
⇒ 90° + ∠AOB = 180° [From (4)]
⇒ ∠AOB = 180° – 90°
⇒ ∠AOB = 90°

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Solution:
Let PA and PB be two tangents drawn from an external point P to a circle with centre O.

Now, in right ∆OAP and right ∆OBP, we have
PA = PB [Tangents to circle from an external point]
OA = OB [Radii of the same circle]
OP = OP [Common]
⇒ ∆OAP ≅ ∆OBP [By SSS congruency]
∴ ∠OPA = ∠OPB [By CPCT]
and ∠AOP = ∠BOP
⇒ ∠APB = 2∠OPA and ∠AOB = 2∠AOP
In right ∆OAP,
∠AOP + ∠OPA + ∠PAO = 180°
⇒ ∠AOP = 180° – 90° – ∠OPA
⇒ ∠AOP = 90° – ∠OPA
⇒ 2∠AOP = 180° – 2∠OPA
⇒ ∠AOB = 180° – ∠APB
⇒ ∠AOB + ∠APB = 180°

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
We have ABCD, a parallelogram which circumscribes a circle (i.e., its sides touch the circle) with centre O.
Since, tangents to a circle from an external point are equal in length

∴ AP = AS
BP = BQ
CR = CQ
DR = DS
On adding, we get
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
But AB = CD [Opposite sides of parallelogram]
and BC = AD
∴ AB + CD = AD + BC ⇒ 2AB = 2BC
⇒ AB = BC
Similarly, AB = DA and DA = CD
Thus, AB = BC = CD = DA
Hence, ABCD is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Solution:
Here ∆ABC circumscribes the circle with centre O. Also, radius = 4 cm
Let AC and AB touches the circle at E and F, respectively and join OE and OF.
∵ The sides BC, CA and AB touches the circle at D, E and F respectively.
∴ BF = BD = 8 cm
[ ∵ Tangents to a circle from an external point are equal]
CD = CE = 6 cm
AF = AE = x cm (say)

∴ The sides of the ∆ABC are 14 cm, (x + 6) cm and (x + 8) cm
Perimeter of ∆ABC
= [14 + (x + 6) + (x + 8)] cm
= [14 + 6 + 8 + 2x] cm
= (28 + 2x) cm
⇒ Semi perimeter of ∆ABC,
s = \(\frac{1}{2}\) [28 + 2x] cm = (14 + x) cm
∴ s – a = (14 + x) – (8 + x) = 6
s – b = (14 + x) – (14) = x
s – c = (14 + x) – (6 + x) = 8
where, a = AB, b = BC, c = AC

Squaring both sides, we get
(14 + x)² = (14 + x)3x
⇒ 196 + x² + 28x = 42x + 3x²
⇒ 2x² + 14x – 196 = 0
⇒ x² + 7x – 98 = 0
⇒ (x – 7)(x + 14) = 0
⇒ x – 7 = 0 or x + 14 = 0
⇒ x = 7 or x = -14
But x = -14 is rejected.
∴ x = 7
Thus, AB = 8 + 7 = 15 cm, BC = 8 + 6 = 14 cm and CA = 6 + 7 = 13 cm

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
We have a circle with centre O. A quadrilateral ABCD is such that the sides AB, BC, CD and DA touches the circle at P, Q, R and S respectively.
Join OP, OQ, OR and OS.
We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠2
∠3 = ∠4
∠5 = ∠6 and ∠7 = ∠8
Also, the sum of all the angles around a point is 360°.
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
∴ 2(∠1 + ∠8 + ∠5 + ∠4) = 360°
⇒ (∠1 + ∠8 + ∠5 + ∠4) = 180° …………. (1)
and 2(∠2 + ∠3 + ∠6 + ∠7) = 360°
⇒ (∠2 + ∠3 + ∠6 + ∠7) = 180° ……………. (2)
Since, ∠2 + ∠3 = ∠AOB, ∠6 + ∠7 = ∠COD, ∠1 + ∠8 = ∠AOD and ∠4 + ∠5 = ∠BOC
∴ From (1) and (2), we have
∠AOD + ∠BOC = 180°
and ∠AOB + ∠COD = 180°

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 15 Probability Ex 15.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 15 Probability Ex 15.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Solution:
Here, the number of all the possible outcomes = 5 × 5 = 25
(i) For both customers visiting on same day:
Favourbale outcomes are (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)
∴ Number of favourable outcomes = 5
∴ Required probability = \(\frac{5}{25}=\frac{1}{5}\)

(ii) For both the customers visiting on consecutive days:
Favourable outcomes are (Tue., Wed.), (Wed., Thu.), (Thu., Fri.), (Fri., Sat.), (Sat., Fri.), (Wed., Tue.), (Thu., Wed.), (Fri., Thu.)
∴ Number of favourable outcomes = 8
∴ Required probability = \(\frac{8}{25}\)

(iii) For both the customers visiting on different days:
We have probability for both visiting on same day = \(\frac{1}{5}\)
∴ Probability for both visiting on different days = 1 – [Probability for both visiting on the same day]
= \(1-\left[\frac{1}{5}\right]=\frac{5-1}{5}=\frac{4}{5}\)
∴ The required probability = \(\frac{4}{5}\).

Question 2.
A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score in the two throws:

What ist the probability that the total score is
(i) even?
(ii) 6?
(iii) at least 6?
Solution:
The complete table as follows

∴ Number of all possible outcomes = 36
(i) For total score being even:
Favourable outcomes = 18
[∵ The even outcomes are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8,12]
∴ The required probability = \(\frac{18}{36}=\frac{1}{2}\)

(ii) For total score being 6 :
In list of scores, we have four 6’s.
∴ Favourable outcomes = 4
∴ Required probability = \(\frac{4}{36}=\frac{1}{9}\)

(iii) For toal score being at least 6:
The favourable scores are : 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12
∴ Number of favourable outcomes = 15
∴ Required probability = \(\frac{15}{36}=\frac{5}{12}\)

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Solution:
Let the number of blue balls in the bag be x.
Total number of balls = x + 5 Number of possible outcomes = (x + 5).
For a blue ball, favourable outcomes = x
Probability of drawing a blue ball = \(\frac{x}{x+5}\)
Similarly, probability of drawing a red ball = \(\frac{5}{x+5}\)
Now, we have \(\frac{x}{x+5}=2\left[\frac{5}{x+5}\right]\)
⇒ \(\frac{x}{x+5}=\frac{10}{x+5}\) ⇒ x = 10
Thus the required number of blue balls = 10.

Question 4.
A box contains 12 balls out of whichxare black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
∵ The total number of balls in the box = 12
∴ Number of possible outcomes = 12
Case – I: For drawing a black ball
Number of favourable outcomes = x
∴ Probability of getting a black ball = \(\frac{x}{12}\)

Case – II: When 6 more black balls are added
Then, the total number of balls = 12 + 6 = 18
⇒ Number of possible outcomes = 18
Now, the number of black balls = (x + 6)
∴ Number of favourable outcomes = (x + 6)
∴ Required probability = \(\frac{x+6}{18}\)
According to the given condition,
\(\frac{x+6}{18}=2\left(\frac{x}{12}\right)\)
⇒ 12 (x + 6) = 36x ⇒ 12x + 72 = 36x
⇒ 36x – 12x = 72 ⇒ 24x = 72
⇒ x = \(\frac{72}{24}\) = 3
Thus, the required value of x is 3.

Question 5.
Ajar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\) Find the number of blue marbles in the jar.
Solution:
There are 24 marbles in the jar.
∴ Number of possible outcomes = 24.
Let there are x blue marbles in the jar.
∴ Number of green marbles = 24 – x
⇒ Favourable outcomes = (24 – x)
∴ Required probability for drawing a green marbles \(\frac{24-x}{24}\)
Now, according to the given condition,
\(\frac{24-x}{24}=\frac{2}{3}\)
⇒ 3(24 – x) = 2 × 24 ⇒ 72 – 3x = 48
⇒ 3x = 72 – 48 ⇒ 3x = 24 ⇒ x = \(\frac{24}{3}\) = 8
Thus, the required number of blue marbles is 8.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 10 Circles Ex 10.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 10 Circles Ex 10.1

Question 1.
How many tangents can a circle have?
Solution:
A circle can have an infinite number of tangents.

Question 2.
Fill in the blanks :
(i) A tangent to a circle intersects it in _________ point(s).
(ii) A line intersecting a circle in two points is called a ______.
(iii) A circle can have ______ parallel tangents at the most.
(iv) The common point of a tangent to a circle and the circle is called _______.
Solution:
(i) Exactly one
(ii) Secant
(iii) Two
(iv) Point of contact

Question 3.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) \(\sqrt{119}\) cm
Solution:
In right ∆QPO,
OQ² = OP² + PQ²

Question 4.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:
We have the required figure, as shown

Here, l is the given line and a circle with centre O is drawn.
Line n is drawn which is parallel to l and tangent to the circle. Also, m is drawn parallel to line 1 and is a secant to the circle.

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.3 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 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:
(i) Let the vertices of the triangle be A(2, 3), B(-1, 0) and C(2, – 4)
Here, x₁ = 2, y₁ = 3
x₂ = -1, y₂ = 0
x₃ = 2, y₃ = -4
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [2{0 – (-4)} + (-1){-4 – (3)} + 2{3 – 0}]
= \(\frac{1}{2}\) [2(0 + 4) + (-1)(-4 – 3) + 2(3)]
= \(\frac{1}{2}\) [8 + 7 + 6] = \(\frac{1}{2}\) [21] = \(\frac{21}{2}\) sq. units

(ii) Let the vertices of the triangle be A(-5, -1), B(3, -5) and C(5, 2)
Here, x₁ = -5, y₁ = -1
x₂ = 3, y₂ = -5
x₃ = 5, y₃ = 2
∵ Area of a triangle
= \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
∴ Area of the ∆ABC
= \(\frac{1}{2}\) [-5{-5 – 2} + 3{2 – (-1)} + 5{-1 – (-5)}]
= \(\frac{1}{2}\) [-5{-7} + 3{2 + 1} + 5{-1 + 5}]
= \(\frac{1}{2}\) [35 + 3(3) + 5(4)]
= \(\frac{1}{2}\) [35 + 9 + 20] = \(\frac{1}{2}\) × 64 = 32 sq. units

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:
The given three points will be collinear if the triangle formed by them has zero area.
(i) Let A(7, -2), B(5, 1) and C(3, k) be the vertices of a triangle.
∴ The given points will be collinear, if ar (∆ABC) = 0
or \(\frac{1}{2}\) [7(1 – k) + 5(k + 2) + 3(-2 – 1)] = 0
⇒ 7 – 7k + 5k + 10 + (-6) – 3 = 0
⇒ 17 – 9 + 5k – 7k = 0
⇒ 8 – 2k = 0 ⇒ 2k = 8 ⇒ k = \(\frac{8}{2}\) = 4
The required value of k = 4.
(ii) \(\frac{1}{2}\) [8(- 4 + 5) + k(- 5 -1) + 2(1 + 4)] = 0
⇒ 8 – 6k + 10 = 0 ⇒ 6k = 18 ⇒ k = 3.

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:
Let the vertices of the triangle be A(0, -1), B(2, 1) and C(0, 3).
Let D, E and F be the mid-points of the sides BC, CA and AB respectively.
∴ Coordinates of D are


ar(∆DEF) : ar(∆ABC) = 1 : 4.

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Solution:
Let A(- 4, – 2), B(- 3, – 5), C(3, – 2) and D(2, 3) be the vertices of the quadrilateral.
Let us join diagonal BD.

Now, ar(∆ABD)

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(A, -6), B(3, -2) and C(5, 2).
Solution:
Here, the vertices of the triangle are A(4, -6), B(3, -2) and C(5, 2).
Let D be the midpoint of BC.
∴ The coordinates of the point D are
\(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\) or ( 4, 0)
AD divides the triangle ABC into two parts i.e., ∆ABD and ∆ACD.

ar(∆ADC) = \(\frac{1}{2}\) [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)]
= \(\frac{1}{2}\) [-8 + 32 – 30] = \(\frac{1}{2}\) [-6] = -3
= 3 sq. units (numerically) ………… (2)
From (1) and (2),
ar(∆ABD) = ar(∆ADC)
Thus, a median divides the triangle into two triangles of equal areas.