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:

Ethanol	Ethanoic acid
This is in liquid form at room temperature. Its melting point is 156° K.	Its melting point is 290K and hence it often freezes during winter in cold climates.
Difference in chemical properties
Ethanol will not react with metallic carbo­nates.	Ethanoic acid reacts with carbo­nates and Hydrogen carbonate and forms salts, carbon dioxide and water.

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

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 14 Statistics Ex 14.1 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 14 Statistics Ex 14.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?
Solution:
We can calculate the mean as follows :

∴ Mean = \(\overline{x}=\frac{\sum f_{i} x_{i}}{N}=\frac{162}{20}=8.1\)
Thus, mean number of plants per house is 8.1 Since, values of x_i and f_i are small, so we have used the direct method.

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Let the assumed mean, a = 150
∵ Class size, h = 20
∴ \(u_{i}=\frac{x_{i}-a}{h}=\frac{x_{i}-150}{20}\)
∴ We have the following table:

Now, \(\overline{x}\) = a + h × {\(\frac{1}{N}\) Σf_iu_i}
= 150 + 20 × \(\left(\frac{-12}{50}\right)\) = 150 – \(\frac{24}{5}\)
= 150 – 4.8 = 145.20

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f.

Solution:
Let the assumed mean, a = 18
∵ Class size, h = 2
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-18}{2}\)
Now, we have the following table:

⇒ [f + 44] (0) = 2[f – 20]
⇒ 2[f – 20] = 0 ⇒ f = 20
Thus, missing frequency is 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Solution:
Let the assumed mean, a = 75.5
∵ Class size, h = 3
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-75.5}{3}\)
Now, we have the following table:

Thus, the mean heart beats per minute is 75.9.

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained a varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Let the assumed mean, a = 57
∴ d_i = x_i – 57
Now, we have the following table:

Thus, the average number of mangoes per box = 57.19. We choose assumed mean method.

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.
Solution:
Let the assumed mean, a = 225
∵ Class size, (h) = 50
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-225}{50}\)
Now, we have the following table:

Thus, the mean daily expenditure on food is ₹ 211.

Question 7.
To find out the concentration of S0₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of S0₂ in the air.
Solution:
Let the assumed mean, a = 0.14
∵ Class size, (h) = 0.04
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-0.14}{0.04}\)
Now, we have the following table:

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:
Using the direct method, we have the following table:

Thus, mean number of days a student remained absent = 12.48.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:
Let the assumed mean, a = 70
∵ Class size, (h) = 10
∴ u_i = \(\frac{x_{i}-a}{h}=\frac{x_{i}-70}{10}\)
Now, we have the following table:

Thus, the mean literacy rate is 69.43%

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 11 Constructions Ex 11.2 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 11 Constructions Ex 11.2

In each of the following, give also the justification of the construction:

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of Construction :
I. Draw a circle of radius 6 cm. Let its centre be O.
II. Take a point P such that OP = 10 cm. Join OP.

III. Bisect OP and let M be its midpoint.
IV. Taking M as centre and MP or MO as radius draw a circle.
Let the new circle intersects the given circle at A and B. Join PA and PB.
Thus, PA and PB are the required tangents. By measurement, we have : PA = PB = 8 cm.
Justification:
Join OA and OB
Since PO is a diameter.
∴ ∠OAP = 90° = ∠OBP [Angles in a semicircle]
Also, OA and OB are radii of the same circle.
⇒ PA and PB are tangents to the circle.

Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Solution:
Steps of Construction :
I. Draw two concentric circles with centre O and radii 4 cm and 6 cm.
II. Take any point P on outer circle.
III. Join PO and bisect it and let the midpoint of PO is represented by M.
IV. Taking M as centre and OM or MP as radius, draw a circle such that this circle intersects the circle (of radius 4 cm) at A and B.
V. Join AP.
Thus, PA is the required tangent.
By measurement, we have PA = 4.5 cm

Justification:
Join OA. As PO is diameter
∴ ∠PAO = 90° [Angle in a semi-circle]
⇒ PA⊥OA
∵ OA is a radius of the inner circle.
∴ PA has to be a tangent to the inner circle.
Verification:
In right ∆PAO, PO = 6 cm, OA = 4 cm.

Hence both lengths are approximately equal.

Question 3.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of Construction :
I. Draw a circle of radius 3 cm with centre O and draw a diameter.
II. Extend its diameter on both sides and cut OP = OQ = 7 cm
III. Bisect PO such that M be its mid-point.
IV. Taking M as centre and MO as radius, draw a circle. Let it intersect the given circle at A and B.
V. Join PA and PB.
Thus, PA and PB are the two required tangents from P.
VI. Now bisect OQ such that N is its mid point.
VII. Taking N as centre and NO as radius, draw a circle. Let it intersect the given circle at C and D.
VIII. Join QC and QD.
Thus, QC and QD are the required tangents from Question

Justification:
Join OA to get ∠OAP = 90°
[Angle in a semi-circle]
⇒ PA⊥OA
⇒ PA is a tangent.
Similarly, PB⊥OB
⇒ PB is a tangent.
Now, join OC to get ∠QCO = 90°
[Angle in a semi-circle]
⇒ QC⊥OC ⇒ QC is a tangent.
Similarly, QD⊥OD
⇒ QD is a tangent.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Solution:
Steps of Construction :
I. With centre O and radius = 5 cm, draw a circle.
II. Taking a point A on the circle draw ∠AOB = 120°.

III. Draw a perpendicular on OA at A.
IV. Draw another perpendicular on OB at B.
V. Let the two perpendiculars meet at C.
Thus CA and CB are the two required tangents to the given circle which are inclined to each other at 60°.
Justification:
In a quadrilateral OACB, using angle sum
property, we have
120° + 90° + 90° + ∠ACB = 360°
⇒ 300° + ∠ACB = 360°
⇒ ∠ACB = 360° – 300° = 60°.

Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:
Steps of Construction:
I. Draw a line segment AB = 8 cm
II. Draw a circle with centre A and radius 4 cm, draw another circle with centre B and radius 3 cm.
III. Bisect the line segment AB. Let its mid point be M.
IV. With centre as M and MA (or MB) as radius, draw a circle such that it intersects the two circles at points P, Q, R and S.
V. Join BP and BQ.
Thus, BP and BQ are the required two tangents from B to the circle with centre A.
VI. Join RA and SA.
Thus, RA and SA are the required two tangents from A to the circle with centre B.

Justification:
Let us join A and P.
∵ ∠APB = 90° [Angle in a semi-circle]
∴ BP⊥AP
But AP is radius of the circle with centre A.
⇒ BP has to be a tangent to the circle with centre A.
Similarly, BQ has to be tangent to the circle with centre A.
Also AR and AS are tangents to the circle with centre B.

Question 6.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Steps of construction :
I. Draw ∆ABC such that AB = 6 cm, BC = 8 cm and ∠B = 90°.
II. Draw BD⊥AC. Now bisect BC and let its midpoint be O.
So O is centre of the circle passing through B, C and D.

III. Join AO
IV. Bisect AO. Let M be the mid-point of AO.
V. Taking M as centre and MA as radius, draw a circle intersecting the given circle at B and E.
VI. Join AB and AE. Thus, AB and AE are the required two tangents to the given circle from A.
Justification:
Join OE, then ∠AEO = 90° [Angle in a semi circle]
∴ AE⊥OE.
But OE is a radius of the given circle.
⇒ AE has to be a tangent to the circle.
Similarly, AB is also a tangent to the given circle.

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:
Steps of Construction:
I. Draw the given circle using a bangle.
II. Take two non parallel chords PQ and RS of this circle.

III. Draw the perpendicular bisectors of PQ and RS such that they intersect at O. Therefore, O is the centre of the given circle.
IV. Take a point N outside this circle.
V. Join ON and bisect it. Let M be the mid-point of ON.
VI. Taking M as centre and OM as radius, draw a circle. Let it intersect the given circle at A and B.
VII. Join NA and NB. Thus, NA and NB are the required two tangents.
Justification:
Join OA and OB.
Since ∠OAN = 90° [Angle in a semi circle]
∴ NA⊥OA.
Also NA is a radius.
∴ NA has to be a tangent to the given circle.
Similarly, NB is also a tangent to the given circle.

MP Board Class 10th Science Solutions Chapter 11 Human Eye and Colourful World

MP Board Class 10th Science Chapter 11 Intext Questions

Class 10th Science Chapter 11 Intext Questions Page No. 190

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye to focus the distant objects as well as the nearby objects on the retina by changing focal length or converging power of its lens is called accommodation. The normal eye has a power of accommodation which enables the object as close as 25cm & as far as infinity to be focused on its retina.

Question 2.
A person with a myopic eye can not see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?
Answer:
The person is able to see nearby objects clearly, but he is unable to see objects beyond 1.2m. This happens because the image of an object beyond 1.2 m is formed in front of the retina and not at the retina, as shown in the figure.

To correct this defect of vision, he must use a concave lens. The concave lens will bring the image back to the retina is shown in the given figure.

Fig. 11.1 Myopic eyes and their correction.

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
Far point of the human eye with normal vision is near than infinity and near point is 1.2 m.

Question 4.
A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?
Answer:
A student has difficulty reading the black board while sitting in the last row means he is suffering from myopia. This defect can be corrected using concave lens of suitable power.

MP Board Class 10th Science Chapter 11 NCERT Textbook Exercises

Question 1.
The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to:
(a) presbyopia
(b) accommodation
(c) near-sightednes
(d) far-sightedness
Answer:
(b) Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens.

Question 2.
The human eye forms the image of an object at its:
(a) cornea
(b) iris
(c) pupil
(d) retina
Answer:
(d) The human eye forms the image of an object at its retina.

Question 3.
The least distance of distinct vision for a young adult with normal vision is about
(a) 25 m.
(b) 2.5 cm
(c) 25 cm
(d) 2.5 m
Answer:
(c) 25 cm

Question 4.
The change in focal length of an eye lens is caused by the action of the
(a) pupil.
(b) retina
(c) ciliary muscles
(d) iris
Answer:
(c) ciliary muscles

Question 5.
A person needs a lens of power – 5.5 dioptres for correcting his distant vision. For correcting his distant vision. For correcting his near vision he needs a lens of power + 1.5 dioptre. What is the focal length of the lens required for correcting (i) distinct vision, and (ii) near vision?
Answer:
i) Lens required for correcting
distant vision = – 5.5
Focal length of lens F \(=\frac{1}{P}\)
\(\mathrm{F}=\frac{1}{-5.5}=0.181 \mathrm{m}\)
Lens required for correcting this defect =-0.181 M
ii) Lens required for correcting near vision = + 1.5 D
Focal length of lens F \(=\frac{1}{P}\)
\(F=\frac{1}{1.5}=0.667 \mathrm{m}\)

Question 6.
The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Answer:
The person is suffering from an eye defect called myopia. In this defect, the image is formed in front of the retina. Hence, a concave lens is used to correct this defect of vision.
Object-distance, u = infinity
Image-distance, v = -80 cm
Focal length = f
According to the lens formula,

A concave lens of Power – 1.25 D is required by the person to correct his defect.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Answer:
A person suffering from hypermetropia can see distinct objects clearly but faces difficulty in seeing nearby objects clearly. It happens because . the eye lens focuses the incoming divergent rays beyond the retina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following figure.

Fig. 11.2: Correction for hypermetropic eye.

The convex lens actually creates a virtual image of a nearby object (N’ in the figure) at the near point of vision (N) of the person suffering from hypermetropia. The given person will be able to clearly see the object kept at 25 cm (Near point of the normal eye), if the image of the object is formed at his near point, which is given as 1 m.

Object-distance, u = -25 cm
Image-distance, v = -1m = -100m
Focal length, f = ?

Using the lens formula,
\(\frac { 1 }{ v } \) – \(\frac { 1 }{ u } \) = \(\frac { 1 }{ f } \)
\(\frac { 1 }{ -100 } \) – \(\frac { 1 }{ -25 } \) = \(\frac { 1 }{ f } \)
\(\frac { 1 }{ f } \) = \(\frac { 1 }{ 25 } \) = \(\frac { 1 }{ f } \)
\(\frac { 1 }{ f } \) = \(\frac { 1 }{ 25 } \) – \(\frac { 1 }{ 100 } \)
\(\frac { 1 }{ f } \) = \(\frac { 4-1 }{ 100 } \)
f = \(\frac { 100 }{ 3 } \) = 33.3 cm = 0.33cm

A convex lens of power + 3.0 D is required to correct the defect.

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
The maximum accommodation of a normal eye is reached when the object is at a distance of 25 cm from the eye. The focal length of the eye lens cannot be decreased below this minimum limit. Thus an object placed closer than 25cm [or very close to eye] cannot be seen clearly by a normal eye.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
The distance eye lens and retina is the image distance inside the eye. The image distance is fixed. It cannot be changed at all. Therefore, when we increase the distance of an object from the eye, there is no change in the image distance inside the eye.

Question 10.
Why do stars twinkle?
Answer:
Stars emit their own light and they twinkle due to the atmospheric refraction of light. Stars are very far away from the earth. Hence, they are considered as point sources of light. When the light coming from stars enters the earth’s atmosphere, it gets refracted at different levels because of the variation in the air density at different levels of the atmosphere. When the star’s light refracted by the atmosphere comes more towards us, it appears brighter than when it comes less towards us. Therefore, it appears as if the stars are twinkling at night.

Question 11.
Explain why the planets do not twinkle.
Answer:
Planets do not twinkle because they appear larger in size than the stars as they are relatively closer to earth. Planets can be considered as a collection of a large number of point-size sources of light. The different parts of these planets produce either brighter or dimmer effect in such a way that the average of brighter and dimmer effect is zero. Hence, the twinkling effects of the planets are nullified and they do not twinkle.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:
Light from the Sun near the horizon passes through thicker layers of air and larger distance in the earth’s atmosphere before reaching our eyes. However, light from the Sun overhead would travel relatively shorter distance. At noon, the Sun appears white as only a little of the blue and violet colours are scattered. Near the horizon, most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelengths. This gives rise to the reddish appearance of the Sun.

Question 13.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelength more strongly than red. The scattered blue light enters our eyes. If the earth had no atmosphere, there would not have been any scattering. Then the sky would have looked dark. The sky appears dark to astronaut flying at very high attitude, as scattering is not prominent at such heights.

MP Board Class 10th Science Chapter 11 Additional Important Questions

MP Board Class 10th Science Chapter 11 Multiple Choice Questions

Question 1.
When our eye see a near object, its lens?
(a) Expand
(b) Remain same
(c) Contract
(d) None
Answer:
(a) Expand

Question 2.
What transfer information from eyes to brain?
(a) Neuro trans
(b) Iris
(c) Optical nerve
(d) Retina
Answer:
(c) Optical nerve

Question 3.
What controls the size of pupil?
(a) Iris
(b) Cornea
(c) Lens
(d) Aqueous humour
Answer:
(a) Iris

Question 4.
Near point of human eye is approximately:
(a) 50 cm
(b) 1 m
(c) 25 cm
(d) 25 m
Answer:
(c) 25 cm

Question 5.
Far point of human eye is approximately:
(a) 10 km
(b) infinity
(c) 5 km
(d) 25 cm
Answer:
(a) 10 km

Question 6.
Cataract is caused mostly to
(a) Farm workers
(b) Factory workers
(c) Kids watching TV and Mobiles
(d) Old age people
Answer:
(d) Old age people

Question 7.
In which kind of defect, image of a distant object is formed in front of retina?
(a) Myopia
(b) Cataract
(c) Hypermetropia
(d) Presbyopia
Answer:
(a) Myopia

Question 8.
A concave lens with correct power can help a person with
(a) Hypermetropia
(b) Myopia
(c) Presbyopia
(d) None of these
Answer:
(b) Myopia

Question 9.
Adjusting “cording to object’s distance from eye is called
(a) Vision power
(b) Motivation
(c) Accommodation power
(d) Presbyopia
Answer:
(c) Accommodation power

Question 10.
For an eye donation, eyes must be removed within ……… after death.
(a) 8 hours
(b) 25 hours
(c) 4-6 hours
(d) None.
Answer:
(c) 4-6 hours

Question 11.
What type of lens are required to correct a vision of hypermetropic eyes?
(a) Concave
(b) Convex
(c) both
(d) Plane mirror is required ;
Answer:
(b) Convex

Question 12.
When we pass a light through a prism, it is
(a) Refracted
(b) Scattered
(c) Reflected
(d) Inverted
Answer:
(a) Refracted

Question 13.
Which colour lies in centre of a rainbow?
(a) Red
(b) Violet
(c) Yellow
(d) Green
Answer:
(d) Green

Question 14.
If two prism are placed inverted to each other and white light is passed from one prism, what will we get refracted by other prism?
(a) VIBGYO
(b) white light
(c) All light will be absorbed
(d) Black spot
Answer:
(b) white light

Question 15.
We see twinkling of stars due to:
(a) Reflection of light.
(b) Refraction of light
(c) Scattering of light
(d) Polarization of light
Answer:
(b) Refraction of light

Question 16.
Particle looks moving in smoke filled place, when a light beam coming from a whole passes through it, because:
(a) Tyndall effect
(b) atmospheric refraction
(c) White light spectrum
(d) reflection
Answer:
(a) Tyndall effect

MP Board Class 10th Science Chapter 11 Very Short Answer Type Questions

Question 1.
Where image is formed in our eyes?
Answer:
Retina.

Question 2.
Which part of eye helps in adjustment of lens according to distance?
Answer:
Lens.

Question 3.
Which nerve transfer information regarding image to brain?
Answer:
Optic nerve.

Question 4.
What is power of accommodation?
Answer:
The ability of the eye to focus both near and distant objects, by adjusting its focal length is called power of accommodation of eyes.

Question 5.
How many points of accommodation are set to observe clarity of a person’s vision?
Answer:
Two-far point and near point.

Question 6.
What is the maximum distance till which a person with normal eye can see?
Answer:
No limit, infinite distance.

Question 7.
Name common refractive defects of eyes.
Answer:
Myopia, Hypermetropia and Presbyopia.

Question 8.
What kind of lens is used to correct a myopic defect in a person?
Answer:
Concave.

Question 9.
What kind of treatment is available to correct a cataract?
Answer:
Surgery.

Question 10.
What is bifocal lens?
Answer:
Lens made by both concave and convex lens is called bifocal lens.

Question 11.
What happens to the light ray entering a prism?
Answer:
The light ray get refracted.

Question 12.
What happens if white light is passed through a prism?
Answer:
It Scatters to seven colours spectral pattern.

Question 13.
What is an emergent ray?
Answer:
Ray which comes out after being refracted from a prism is called emergent ray.

Question 14.
Why sky looks red at sunrise and sunset?
Answer:
Scattering of white light.

Question 15.
Which colour light bend most after coming out of a triangular prism?
Answer:
Violet.

Question 16.
Why we never see a shadow without light?
Answer:
When light passes through an object, it get partially deviated or absorbed so, disappear from that very way which forms a shadow.

Question 17.
Why light ray in a dusty room look sparkling and moving?
Answer:
Due to tyndall effect.

Question 18.
Why we can observe a star twinkling?
Answer:
Due to atmospheric refraction.

Question 19.
What is the time difference between actual and observed sunrise and sunset?
Answer:
2 minutes.

Question 20.
Why sky looks blue to all of us?
Answer:
Due to scattering of light.

MP Board Class 10th Science Chapter 11 Short Answer Type Questions

Question 1.
Why does the Sun appear reddish early in the morning?
Answer:
Early in the morning, the sun is near the horzion. Sunlight reaches us after covering a large thickness of the atmosphere. So, shorter waves of blue region are almost completely scattered away by the air molecules. Red waves of longer wavelength are least scattered and reach our eyes. The sun appears red.

Question 2.
Why does the sky appear dark instead of blue to an astronaut?
Answer:
The atmosphere is quite thin at very high altitudes. There is almost no scattering of sunlight. So, the sky appears dark to an astronaut.

Question 3.
Why is a concave lens used to correct myopia or short-sightedness?
Answer:
A concave lens of suitable focal length diverges the parallel rays from the distant object as if they are coming from the far point F of the myopic eye. This helps the eye lens to form a clear image at the retina.

Question 4.
Dispersion is caused by refraction not by reflection. Why?
Answer:
The reason is that for a given angle of incidence, the angle of reflection is same for all the wavelengths of white light while the angle of refraction is different for different wavelengths.

Question 5.
A hypermetropic person prefers to remove his spectacles, while driving. Give reason.
Answer:
When a hypermetropic person wearing the spectacles looks at a distant object, the parallel rays from the distant object get converged in front of the retina. The image appears blurred. In order to avoid this, the person prefers to remove his spectacles.

Question 6.
Give reasons for the following:

1. The stars appear to twinkie.
2. The planets do not twinkle.

Answer:

1. Stars appear to twinkle due to atmospheric refraction of starlight and physical conditions of the earth’s atmosphere are not being stationary.
2. Planets are much closer to the earth and are seen as extended sources. The fluctuations caused in the amount of light due to atmospheric refraction are negligible. Hence, planets do not twinkle.

MP Board Class 10th Science Chapter 11 Long Answer Type Questions

Question 1.
Why does the Sun seem to rise two minutes before the actual sunrise and set two minutes after the actual sunset? Explain with the help of labelled diagram.
Answer:
Advance sunrise and delayed sunset. Apparent shift in the position of sun at sunrise and sunset:
The sun is visible before actual sunrise and after actual sunset, because of atmospheric refraction. With altitude, the density and hence, refractive index of air-layer decreases. As shown in Fig. given the light rays starting from the sun travel from rarer to denser layers. They bend more and more towards the normal. To an observer on the earth, light rays appear to come from position S. The sun which is actually in position S below the horizon appears in position S above the horizon.

Fig. 11.6: Atmospheric refraction effect at sun rise.

Thus, the sun appears to rise early by about two minutes and for the same reason, it appears to set late by about two minutes. This increases the length of the day by about four minutes.

Question 2.
“Stars appear higher than they actually are.” Give reason.
Or
Is the position of a star as seen by us its true position? Justify your answer.
Answer:
Since, the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. The stars appear slightly higher (above) than their actual position when viewed near the horizon.

Question 3.
Why do we observe the apparent random wavering or flickering of objects when seen through a turbulent stream of hot air rising above fire, a stove or radiator?
Answer:
This is due to atmospheric refraction i.e., refraction of light by the earth’s atmosphere. The air just above the fire becomes hotter than air further up. Hotter air is lighter (less denser) than the cooler air (denser) above it. This causes refraction of light due to decrease of refractive index with decreasing density or increasing temperature. Since, the physical conditions of the refracting medium (air) are not stationary, the apparent positions of the objects, as seen through the hot air, fluctuate. Consequently, the objects seen through such air show a wavering effect.

Question 4.
Why do we see stars appear twinkling whereas planets do not twinkle?
Or
A star sometimes appear brighter and some other times fainter. What is this defect called, state reason for this effect.
Answer:
Twinkling of stars: Differential the apparent position of a star is slightly different from the actual position due to refraction of starlight by the atmosphere. Further, this apparent position is not stationary but keeps on changing due to the change in atmospheric conditions like density, temperature etc. The path of the rays of light coming from the star goes on varying slightly. The amount of light entering our eyes from a particular star increases or decreases randomly with time.

Fig. 11.7: Apparent star position due to atmospheric refraction.

Sometimes, the star appears bright and other times, it appears fainter. This gives rise to the twinkling effect of the star.

The planets do not show twinkling effect. As the planets are much closer to the earth, the amount of light received from them is much greater and the fluctuations caused in the amount of light due to atmospheric refraction are negligible as compared to the amount of light received from them.

Question 5.
What is tyndall effect? What is its cause? Name two phenomena observed in daily life which are based on Tyndall effect.
Or
A beam of light is allowed to pass through two beakers A and B containing a true solution and colloidal solution respectively. What do you observe? Name the phenomenon responsible for your observation.
Answer:
Tyndall effect: When a beam of light is passed through a colloidal solution, placed in a dark room, the path of beam becomes illuminated (or visible), when observed through a microscope placed perpendicular to the path of light. This effect is called Tyndall effect.

On the other hand, the path of a beam of light is not visible through a true solution, as shown in figure 11.8.

Fig. 11.8: Tyndall effect.

Cause of Tyndall effect: The size of the colloidal particle is relatively larger than the solute particle of a true solution. The colloidal particles first absorb energy from the incident light and then scatter a part of this energy from their surfaces. Thus, Tyndall effect is due to scattering of light by the colloidal particles and the colloidal particles are seen as points of light moving against a dark background.

Some daily life phenomena based on Tyndall effect are as follows:

1. When a fine beam of sunlight enters a smoke filled room through a small hole, the smoke particles become visible due to the scattering of light.
2. When sunlight passes through a canopy of a dense forest, the tiny water droplets in the mist scatter light and become visible.

MP Board Class 10th Science Chapter 11 NCERT Textbook Activities

Class 10 Science Activity 11.1 Page No. 192

• Fix a sheet cf white paper on a drawing board using drawing pins.
• Fife a glass prism on it in such a way that it rests on its triangular fee. Trace the outline of the prism using a pencil.
• Draw a straight line PE inclined to one of the refracting surfaces, say AB, of the prism.
• Fix two pins, say at points P and Q, on the line PE as shown in Fig. 11.3.
• Look for the images of the pins, fixed at P and Q, through the other face AC.
• Fix two more pins, at points R and S, such that the pins at R and S and the images of the pins at P and Q lie on the same straight line.
• Remove the pins and the glass prism.
• The line PE meets the boundary of the prism at point E (see Fig. 11.3). Similarly, join and produce the points R and S. Let these lines meet the boundary of the prism at E and F, respectively. Join E and F.
• Draw perpendiculars to the refracting surfaces AB and AC of the prism at points E and F, respectively.
• Mark the angle of incidence (∠i), the angle of refraction (∠r) and the angle of emergence (∠e) as shown in Fig. 11.3.

PE – Incident ray – ∠i – Angle of incidence
EF – Refracted ray ∠r – Angle of refraction
FS – Emergent ray ∠e – Angle of emergence
∠A – Angle of the prism – ∠D – Angle of deviation

Fig. 11.3 : Refraction of tight through a triangular glass prism

Observations:

• The light rays enter the prism and emerge out as the emergent ray. The light ray bent towards the normal upon refraction. At the second surface, the light ray enters from glass to air. The emergent ray appears to be coming along the rays FRS, the angle between the incident ray produced forward and the emergent ray produced backward, is called the angle of deviation.

Class 10 Science Activity 11.2 Page No. 193

• Take a thick sheet of cardboard and make a small hole or narrow slit in its middle.
• Allow sunlight to fall on the narrow slit. This gives a narrow beam of white light.
• Now, take a glass prism and allow the light from the slit to fall on one of its faces as shown in Fig. 11.4.
  
  Fig. 11.4: Dispersion of white light by the glass prism.
• Turn the prism slowly until the light that comes out of it appears on a nearby screen.
• What do you observe? You will find a beautiful band of colours. Why does this happen?

Observations:

• The white light gets dispersed into seven colour components by a prism. The band of the coloured components of a light beam is called spectrum. The splitting of light into its component colours is called dispersion. The colours bends at different angles as they pass through prism and are observed separately as VIBGYOR.

Class 10 Science Activity 11.3 Page No. 196

• Place a strong source (S) of white light at the focus of a converging lens (L). This lens provides a parallel beam of light.
• Allow the light beam to pass through a transparent glass tank containing clear water.
• Allow the beam of light to pass through a circular hole (e) made in a cardboard. Obtain a shaip image of the circulai hole on a screen (MN) using a second converging lens (L₂), as shown in Fig. 11.5
  
  Fig. 11.5: An arrangement for observing scattering of light in colloidal solution.
• Dissolve about 200 g of sodium thiosulphate (hypo) in about 2 L of clean water taken in the tank. Add about 1 to 2 mL of concentrated sulphuric acid to the water. What do you observe?

Observations:

• The microscopic particles of sulphur precipitate, in about 2 to 3 minutes. As they precipitate, the blue light from the three sides of the glass tank is seen. This happens due to scattering of short wavelengths by minute colloidal sulphur particles.

MP Board Class 10th Science Solutions

In this article, we will share MP Board Class 10th Sanskrit Solutions व्याकरण प्रत्यय-प्रकरण Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Sanskrit व्याकरण प्रत्यय-प्रकरण

प्रत्यय की परिभाषा-यः शब्दः धातोः प्रातिपदिकस्य वा अन्ते संयुज्य विशेषार्थस्य बोधं कारयति सः प्रत्ययः।

अथवा

धातूनां प्रातिपदिकानां च अन्ते ये प्रयुज्यन्ते ते प्रत्ययाः।

अर्थात् किसी धातु अथवा शब्द के बाद में किसी विशेष अर्थ को प्रकट करने के लिए जो शब्दांश जोड़ा जाता है, उसे प्रत्यय कहते हैं।

जैसे-
बालकः गच्छति।
बालकः पठति।

उपरोक्त दो वाक्यों को प्रत्यय के प्रयोग से एक वाक्य बना सकते हैं।

बालकः गत्वा पठति।

प्रत्यय के भेद-प्रत्ययों से बनने वाले शब्दों को निम्नलिखित छ: वर्गों में विभक्त किया जाता है

१. सुबन्त
सुप् आदि २१ प्रत्ययों के योग से संज्ञा, सर्वनाम, विशेषण आदि के शब्द रूप बनते हैं। जैसे-रामः, सीता आदि।

२. तिङन्त
तिङ आदि १८ प्रत्ययों के योग से क्रिया पदों (धातु रूपों) का निर्माण होता है। जैसे-पठति, गच्छति आदि।

३. समासान्त –
समास करने के बाद समस्त पद के अन्त में कप, टच आदि प्रत्यय जुड़ते हैं। समास के अन्त में जुड़ने के कारण इन प्रत्ययों को समासान्त कहते हैं।

जैसे-
महाराजः,,सपत्नीकः आदि।

४. कृदन्त
जो प्रत्यय संज्ञा, सर्वनाम, विशेषण आदि बनाने के लिए धातुओं के साथ जोड़े जाते हैं, वे कृत् प्रत्यय कहे जाते हैं और कृत् प्रत्यय से बने हुए शब्दों को कृदन्त कहते हैं।
जैसे-
क्त, क्तवतु, तव्यत्, अनीयर्, क्त्वा, ल्यप्, तुमुन्, शतृ, शानच्, क्तिन् इत्यादि।

❖ क्त

• पठ् + क्तः = पठितः – (पढ़ा हुआ)
• गम् + क्तः = गतः – (गया हुआ)
• हस् + क्तः = हसितः – (हँसा हुआ)
• दृश + क्तः = दृष्टः – (देखा हुआ)

❖ क्तवतु

• पठ् + क्तवतु = पठितवान् – (पढ़ा)
• गम् + क्तवतु = गतवान् – (गया)
• कृ + क्तवतु = कृतवान् – (करा)
• क्रीड् + क्तवतु = क्रीडितवान् – (खेला)

❖ तव्यत्

• पठ् + तव्यत् = पठितव्यम् – (पढ़ने योग्य)
• खाद् + तव्यत् = खादितव्यम् – (खाने योग्य)
• गम् + तव्यत् = गन्तव्यम् – (जाने योग्य)
• लिख् + तव्यत् = लेखितव्यम् – (लिखने योग्य)
• अनीयर । पठ् + अनीयर् = पठनीयम् – (पढ़ने योग्य)
• चल् + अनीयर् = चलनीयम् – (चलने योग्य)
• कृ + अनीयर् = करणीयम् – (करने योग्य)
• क्रीड् + अनीयर् = क्रीडनीयम् – (खेलने योग्य)
• क्त्वा पठ् + क्त्वा = पठित्वा – (पढ़कर)
• क्रीड् + क्त्वा = क्रीडित्वा

❖ (खेलकर)

• गम् + क्त्वा = गत्वा – (जाकर)
• पा + क्त्वा = पीत्वा – (पीकर)
• ल्यप् प्र + दा + ल्यप् = प्रदाय – (देकर)
• आ + गम् + ल्यप् = आगम्य/आगत्य – (आकर)
• प्र + नम् + ल्यप् = प्रणम्य – (प्रणाम करके)
• वि + हस् + ल्यप् = विहस्य – (हँसकर)

❖ तुमुन्

• पठ् + तुमुन् = पठितुम् – (पढ़ने के लिए)
• गम् + तुमुन् = गन्तुम् – (जाने के लिए)
• श्रु + तुमुन् = श्रोतुम् – (सुनेन के लिए)
• हस् + तुमुन् = हसितुम् – (हँसने के लिए)

❖ शतृ

• पठ् + शतृ = पठन् – (पढ़ते हुए)
• गम् + शतृ = गच्छन् – (जाते हुए)
• कृ + शतृ = कुर्वन् – (करते हुए)
• क्रीड् + शतृ = क्रीडन् – (खेलते हुए)

❖ शानच

• सेव् + शानच् = सेवमानः – (सेवा करता हुआ)
• लभ् + शानच् = लभमानः – (प्राप्त करता हुआ)
• वृध् + शानच् = वर्धमानः – (बढ़ता हुआ)
• कम्प, + शानच् = कम्पमानः – (काँपता हुआ)

❖ क्तिन्

• भू + क्तिन् = भूतिः – (ऐश्वर्य)
• गम् + क्तिन् = गतिः – (गति)
• स्तु + क्तिन् = स्तुतिः – (स्तुति)

५. तद्धितान्त
जो प्रत्यय संज्ञा, सर्वनाम और विशेषण शब्दों में जोड़े जाते हैं, उन्हें तद्धित प्रत्यय कहते हैं और तद्धित प्रत्ययों के बने हुए शब्दों को तद्धितान्त कहते हैं। जैसे-मतुप्, इनि इत्यादि।

❖ मतुप

• विद्या + मतुप् = विद्यावान् – (विद्या वाला)
• शक्ति + मतुप् = शक्तिवान् – (शक्ति वाला)
• गुण + मतुप् = गुणवान् – (गुण वाला)
• धन + मतुप् = धनवान् – (धन वाला)

❖ इनि (इन्)

• मान + इनि = मानी (मानिन्) – (मान वाला)
• धन + ‘इनि = धनी (धनिन्) – (धन वाला)
• ज्ञान + इनि = ज्ञानी (ज्ञानिन्) – (ज्ञान वाला)
• गुण + इनि = गुणी (गुणिन्) – (गुण वाला)

❖ ठक (इक)

• समाज + ठक् = सामाजिकः – (समाज से सम्बन्धित)
• इतिहास + ठक् = ऐतिहासिक: – (इतिहास से सम्बन्धित)
• धर्म + ठक् = धार्मिकः – (धर्म से सम्बन्धित)

❖ त्व

• मनुष्य + त्व = मनुष्यत्वम् – (मुनष्यपन या मनुष्य का कर्म)
• विद्वस् + त्व = विद्वत्त्वम् – (विद्वान का भाव या कर्म)
• लघु + त्व = लघुत्वम् – (छोटापन)
• गुरु + त्व = गुरुत्वम् – (बड़प्पन)

❖ त्रल

• बहु + ल् = बहुत्र – (बहुत जगह)
• सर्व + त्रल् = सर्वत्र – (बहुत जगह)
• यद् + त्रल् = यत्र – (जहाँ)
• तद् + त्रल् = तत्र – (वहाँ)

६. स्त्री-प्रत्ययान्त
शब्दों को स्त्रीलिंग वाची बनाने के लिए टाप, ङीष्, ङीप्, आदि प्रत्ययों का प्रयोग होता है। इन प्रत्ययों से बनने वाले शब्दों E को स्त्री-प्रत्ययान्त कहते हैं। जैसे-रमा, उमा, अजा आदि।

❖ टाप

• बालक + टाप् = बालिका – (लड़की)
• अज + टाप् = अजा – (बकरी)
• कोकिल+ टाप् = कोकिला – (कोयल)
• अश्व + टाप् = अश्वा – (घोड़ी)

❖ ङीप्

• स्वामिन्+ ङीप् = स्वामिनी – (मालकिन)
• देव + ङीप् = देवी – (देवी)
• कुमार + ङीप् = कुमारी – (कुमारी)
• किशोर + ङीप् = किशोरी – (किशोरी)
• ङीष गौर + ङीष् = गौरी – (गौरी)
• सुन्दर + ङीष् = सुन्दरी – (सुन्दरी)
• नर्तक + ङीष् = नर्तकी – (नर्तकी)

❖ डीन्

• नर + ङीन् = नारी – (नारी)
• ब्राह्मण + ङीन् = ब्राह्मणी – (ब्राह्मणी)

वस्तुनिष्ठ प्रश्न

बहु-विकल्पीय

प्रश्न १.
पठ् + तव्यत्: ” ……………………… ” होता है।
(अ) पठनीयम्,
(ब) पठिव्य,
(स) पठित्वा,
(द) पठितव्यम्।
उत्तर-
(द) पठितव्यम्।

२. हस + शत = ………………………” होगा।
(अ) हसन्,
(ब) हसमान्,
(स) हसथ,
(द) हसामि।
उत्तर-
(अ) हसन्,

३. ‘कथितः’ को पृथक करने पर होगा
(अ) कथ् + तव्यत्,
(ब) कथ् + क्त,
(स) कथ् + क्त्वा,
(द) कथ् + शतृ।
उत्तर-
(ब) कथ् + क्त,

४. ‘कृ + तुमुन्’ को मिलाने पर होगा
(अ) कर्तुम्,
(ब) कर्तम्,
(स) कुर्तुम्,
(द) कतुम्।
उत्तर-
(अ) कर्तुम्,

५. ‘आगम्य’ में प्रकृति-प्रत्यय हैं
(अ) आ + गम्य + ल्यपु,
(ब) आ + गम् + ल्यप्,
(स) आ + गय + ल्यप्,
(द) आ + ल्यप् + गम्।
उत्तर-
(ब) आ + गम् + ल्यप्,

रिक्त स्थान पूर्ति-

१. बाल + टाप् = ………………………………
२. जि + तुमुन् = ………………………………।
३. वध् + शानच् = ………………………………।
४. गम् + क्त = ………………………………।
५. दृश् + अनीयर् = ………………………………
उत्तर-
१. बालिका,
२. जेतुम्,
३. वर्धमानः,
४. गतः,
५. दर्शनीयम्।

सत्य/असत्य-

१. पठितव्यम्’ में शानच् प्रत्यय है।
२. ‘गत्वा’ में क्त्वा प्रत्यय है।
३. विद्यवान्’ में शानच् प्रत्यय है।
४. ‘श्रीमान्’ में मतुप् प्रत्यय है।
५. ‘सर्वत्र’ में त्रल् प्रत्यय है।
उत्तर-
१. असत्य,
२. सत्य,
३. असत्य,
४. सत्य,
५. सत्य।

जोड़ी मिलाइए-

उत्तर-
१. → (iv)
२. → (i)
३. → (v)
४. → (ii)
५. → (iii)

In this article, we will share MP Board Class 10th Social Science Book Solutions Chapter 5 मानचित्र पठन एवं अंकन Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Social Science Solutions Chapter 5 मानचित्र पठन एवं अंकन

MP Board Class 10th Social Science Chapter 5 पाठान्त अभ्यास

MP Board Class 10th Social Science Chapter 5 वस्तुनिष्ठ प्रश्न

सही विकल्प चुनकर लिखिए

प्रश्न 1.
भारत में मौसम मानचित्रों का प्रकाशन प्रारम्भ हुआ
(i) 1853 में
(ii) 1947 में
(iii) 1950 में
(iv) 1875 में।
उत्तर:
(iv) 1875 में।

प्रश्न 2.
भारत में मौसम मानचित्रों का प्रकाशन होता है
(i) कोलकाता से
(ii) दिल्ली से
(iii) पुणे से
(iv) हैदराबाद से।
उत्तर:
(iii) पुणे से

प्रश्न 3.
भारतवर्ष में मौसम विभाग विभाजित है
(i) 6 क्षेत्रों में
(ii) 4 क्षेत्रों में
(iii) 5 क्षेत्रों में
(iv) 8 क्षेत्रों में।
उत्तर:
(iii) 5 क्षेत्रों में

रिक्त स्थानों की पूर्ति कीजिए –

1. अन्तर्राष्ट्रीय मौसम संकेतों को 1935 में मौसम विज्ञान संघ द्वारा ………….. में मान्यता दी गयी।
2. ब्यूफोर्ट ………….. की जल सेना से सम्बन्धित थे।
3. वायुवेग मापने का नियोजन सर्वप्रथम ………….. किया गया था।

उत्तर:

1. वारसा (इटली)
2. ब्रिटिश
3. 1805

सही जोड़ी मिलाइए

उत्तर:

1. → (ख)
2. → (क)
3. → (घ)
4. → (ग)

MP Board Class 10th Social Science Chapter 5 अति लघु उत्तरीय प्रश्न

प्रश्न 1.
नॉट क्या है ?
उत्तर:
नॉट वायु वेग नापने की इकाई, एक नॉट 1.85 किमी. के बराबर होता है। इसका अर्थ है कि वायु की गति 1.85 किमी. प्रति घण्टा है या 1 नॉट = एक समुद्री मील के बराबर होता है।

प्रश्न 2.
भारत में भूकम्पमापी केन्द्र कितने हैं ?
उत्तर:
भारत में 22 भूकम्पमापी केन्द्र हैं।

MP Board Class 10th Social Science Chapter 5 लघु उत्तरीय प्रश्न

प्रश्न 1.
मौसम संकेतों से क्या आशय है ?
उत्तर:
मौसम संकेतों का आशय-प्रेक्षण शालाओं से प्राप्त मौसम तत्वों को मानचित्र पर अंकों, चिह्नों या प्रतीकों द्वारा प्रदर्शित किया जाता है। ये अंक, चिह्न या प्रतीक अन्तर्राष्ट्रीय मान्यता प्राप्त होते हैं। इन्हें अन्तर्राष्ट्रीय मौसम संकेत कहा जाता है। इन संकेतों को 1935 में वारसा (इटली) में आयोजित अन्तर्राष्ट्रीय मौसम विज्ञान संघ द्वारा मान्यता प्रदान की गयी थी। प्रमुख मौसम संकेत निम्न प्रकार हैं

1. वायुमापन संकेत
2. वर्षा मापनी संकेत
3. मेघाच्छादन संकेत
4. समुद्री तरंग संकेत।

प्रश्न 2.
मौसम मानचित्र तैयार करने हेतु मौसम सूचनाएँ कैसे एकत्रित की जाती हैं ?
उत्तर:
मौसम मानचित्र तैयार करने हेतु मौसम सूचनाएँ निम्न प्रकार एकत्रित की जाती हैं –

1. मौसम मानचित्रों को तैयार करने हेतु वेधशालाओं, वायुयानों के पायलटों, गुब्बारों तथा जलयानों से मौसम सूचनाएँ प्राप्त की जाती हैं।
2. वेधशालाओं में निम्न मौसमी तत्वों की जानकारी एकत्रित की जाती है-तापमान, वर्षा, वायु की गति एवं दिशा आपेक्षिक आर्द्रता, सूर्य प्रकाश की अवधि, समुद्र की दशा, वर्तमान एवं पूर्व मौसम।
3. मौसम मानचित्रों में मौसम के तत्वों का चिह्नों द्वारा अंकन किया जाता है।
4. मौसम मानचित्र में सूचनाओं को प्रेषित करने हेतु कूट संख्याओं का प्रयोग किया जाता है, जिनका विशिष्ट अभिप्राय होता है।
5. मौसम मानचित्र तैयार करने हेतु मौसमी दशाओं/तत्वों का निरीक्षण एवं अभिलेखन वेधशालाओं में प्रातः 8:30 व सायंकाल 5.30 बजे होता है।

प्रश्न 3.
मौसम मानचित्र में मौसमी दशाओं को कैसे व्यक्त किया जाता है ?
उत्तर:
प्रेक्षण शालाओं से प्राप्त मौसम तत्वों को मानचित्र पर अंकों, चिह्नों या प्रतीकों द्वारा प्रदर्शित किया जाता है। ये अंक, चिह्न या प्रतीक अन्तर्राष्ट्रीय मान्यता प्राप्त होते हैं। इन्हें अन्तर्राष्ट्रीय मौसम संकेत कहा जाता है।

प्रश्न 4.
दिए गए वायु मापन में संकेत चिह्नों को पहचानिए व उनका नाम व वेग लिखिए –

उत्तर:

1. शान्त
2. धीर समीर
3. प्रबल समीर
4. झंझा
5. झंझावात।

प्रश्न 5.
दिए गए मेघाच्छादन संकेत चिह्नों को पहचानिए व उनकी मात्रा व स्तर को लिखिए –

उत्तर:

1. मात्रा 1/8; निम्नस्तर
2. 1/8; उच्च स्तर
3. मात्रा 3/8; निम्न स्तर
4. मात्रा 1/4; निम्न स्तर
5. मात्रा 7/8; निम्न स्तर
6. मात्रा 8/8; निम्न स्तर
7. मात्रा 1/2; उच्च स्तर
8. मात्रा 5/8; उच्च स्तर
9. सूर्य प्रकाश; उच्च स्तर

प्रश्न 6.
दिए गए समताप रेखाओं द्वारा निर्मित वायुमण्डलीय दशाओं को पहचानकर लिखिए –

उत्तर:

1. चक्रवात, एवं
2. प्रतिचक्रवात।

MP Board Class 10th Social Science Chapter 5 दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
मौसम मानचित्र से प्राप्त पूर्वानुमान कहाँ अत्यधिक उपयोगी है ? मौसम मानचित्रों का महत्त्व लिखिए।
अथवा
मौसम मानचित्र की विशेषताएँ लिखिए। (2012)
उत्तर:
मौसम मानचित्रों से प्राप्त पूर्वानुमानों की उपयोगिता-मौसम मानचित्र से प्राप्त पूर्वानुमान नौ संचालन, वायुयान की सुरक्षित उड़ान, प्राकृतिक आपदाओं के दुष्प्रभावों का निरीक्षण करने में, कृषि की उचित देखभाल तथा समुद्रतट पर रहने वाले मछुआरों को समुद्र की दिशा चक्रवात (समुद्री तूफान) से सावधान करने में यह अत्यन्त उपयोगी हैं।

मौसम मानचित्रों का महत्त्व-मौसम मानचित्रों के प्रमुख महत्त्व निम्नलिखित हैं –

1. मौसम मानचित्रों की सहायता से प्राकृतिक आपदाओं; जैसे–बाढ़, भूकम्प, सूखा आदि के जनजीवन पर पड़ने वाले प्रभावों का पूर्वानुमान लगता है।
2. ये मानचित्र नाविकों तथा वैज्ञानिकों के लिए बहुत महत्त्वपूर्ण होते हैं।
3. वायुयान चालकों के लिए यह मानचित्र महत्त्वपूर्ण होते हैं।
4. इन मानचित्रों की सहायता से मौसम का पूर्वानुमान लगाया जाता है। इसे समाचार-पत्रों एवं दूरदर्शन के माध्यम से प्रसारित कर अतिवृष्टि, भूकम्प, ओलावृष्टि, तूफान एवं हिमपात जैसी प्राकृतिक आपदाओं से जन-सामान्य को सुरक्षा प्रदान करने का प्रयास किया जाता है।

प्रश्न 2.
निम्न मौसमी दशाओं को स्पष्ट करने हेतु संकेत बनाइए

1. कुहरा (2009, 10, 11, 14)
2. ओला (2009, 11, 13, 14, 16, 17)
3. सम्पूर्ण मेघाच्छादन (2018)
4. हिम (2009, 10, 11, 14, 15)
5. वर्षा (2009, 10, 14, 15, 17)
6. कुहासा (2013, 16)
7. धुन्ध (2013, 14, 18)
8. तड़ित झंझा (2015, 17)
9. फुहार (2016, 18)

उत्तर:

प्रश्न 3.
दिए गए मौसम मानचित्र की व्याख्या निम्नलिखित बिन्दुओं पर कीजिए –

1. चक्रवात व गौण चक्रवात का क्षेत्र
2. वायुफान का क्षेत्र
3. प्रतिचक्रवात का क्षेत्र।

उत्तर:
(1) चक्रवात व गौण चक्रवात का क्षेत्र-चक्रवात की समदाब रेखाएँ मन्द होती हैं और इसके भीतर अल्पतम दाब होता है। इसीलिए इसको अल्पदाब अवस्था भी कहते हैं। अल्पतम दाब केन्द्र गर्त रेखाओं का प्रतिच्छेदन बिन्दु होता है, इसीलिए बाहर से हवाएँ भीतर की ओर जाती हैं। उत्तरी गोलार्द्ध चक्रवात की वायु वामावर्त दिशा में और दक्षिणी गोलार्द्ध में दक्षिणावर्त दिशा में चलती हैं। ये चक्रवात स्थायी वायुदाब के प्रवाह की निश्चित दिशा में आगे बढ़ते हैं। उपर्युक्त मानचित्र में चक्रवात व गौण चक्रवात की स्थिति को 990 मिलीबार व 992 मिलीबार की समदाब रेखाओं द्वारा प्रदर्शित किया गया है जिससे स्पष्ट होता है कि गहरे अवदाब में केन्द्र का वायुदाब बहुत कम होता है और छिछले अवदाब में केन्द्र का वायुदाब थोड़ा ही कम होता है। गहरा अवदाब एक से अधिक समदाब रेखाओं से घिरा होता है और छिछला अवदाब केवल एक समदाब रेखा से घिरा होता है और अन्य समदाब रेखाओं से अंशतः घिरा होता है। छिछले अवदाब में समदाब रेखाएँ दूर-दूर और गहरे अवदाब में निकट-निकट अंकित रहती हैं। इसमें वायुराशि एकत्रित होती है, ऊपर उठती है और ठण्डी होकर बादल तथा वर्षा का रूप ग्रहण करती है।

(2) वायुफान का क्षेत्र-यह एक त्रिभुजाकार उच्चदाब का क्षेत्र होता है। मानचित्र (5.5) में यह 994 मिलीबार की समदाब की रेखा द्वारा प्रदर्शित किया गया है जिससे स्पष्ट होता है इसकी समदाब रेखाएँ वी-आकार की होती हैं, जिनका शीर्ष गोल होता है और अल्पदाब के क्षेत्र की ओर इंगित करता है। इसके मध्य में सबसे अधिक वायुदाब रहता है और शीर्ष तथा किनारे की दाब क्रमशः कम होती जाती है। प्रधान चक्रवात के साथ इसका बढ़ाव आगे होता है। सर्वोच्च दाब बिन्दु और शीर्ष बिन्दु को मिलाने वाली रेखा शिखर रेखा कहलाती है।

(3) प्रतिचक्रवात का क्षेत्र-चक्रवात के विपरीत प्रतिचक्रवात होते हैं। इनके केन्द्र में उच्च दाब का स्थान होता है। इसको उच्चदाब अवस्था भी कहते हैं। मानचित्र (5.5) में इसे 1000 मिलीबार की समदाब रेखा द्वारा प्रदर्शित किया गया है जिससे स्पष्ट होता है कि इसकी समदाब रेखाएँ प्रायः वृत्ताकार होती हैं और हवाओं की दिशा उत्तरी गोलार्द्ध में दक्षिणावर्त तथा दक्षिणी गोलार्द्ध में वामावर्त होती हैं। इसमें केन्द्र से बाहर की ओर वायु चलती है। इसमें दाब प्रवणता कम होती है। प्रतिचक्रवात शक्तिहीन होते हैं और एक ही स्थान पर देर तक रुके रहते हैं।

MP Board Class 10th Social Science Chapter 5 अन्य परीक्षोपयोगी प्रश्न

MP Board Class 10th Social Science Chapter 5 वस्तुनिष्ठ प्रश्न

बहु-विकल्पीय

प्रश्न 1.
अन्तर्राष्ट्रीय मौसम संकेत को कब और कहाँ मान्यता प्रदान की गई ?
(i) 1945 में दिल्ली में
(ii) 1948 में फ्रांस में
(iii) 1935 में वारसा (इटली) में
(iv) 1955 में जेनेवा में।
उत्तर:
(iii) 1935 में वारसा (इटली) में

प्रश्न 2.
पश्चिमी तट पर वायु वेग है
(i) 5 नॉट के लगभग
(ii) 7 नॉट के लगभग
(iii) 15 नॉट के लगभग
(iv) 10 नॉट के लगभग।
उत्तर:
(iv) 10 नॉट के लगभग।

रिक्त स्थानों की पूर्ति कीजिए

1. मौसम मानचित्र में वायुदाब ………….. में दर्शायी जाती है।
2. मौसम मानचित्र तैयार करने हेतु मौसम दशाओं एवं तत्वों के निरीक्षण एवं अभिलेखन का केन्द्र ………….. कहलाता है।

उत्तर:

1. मिलीबार
2. वेधशाला।

सत्य/असत्य

प्रश्न 1.
भारत में मौसम विभाग की स्थापना 1880 ई. में हुई थी।
उत्तर:
असत्य

प्रश्न 2.
मेघ की छाया वृत्तों द्वारा प्रदर्शित की जाती है।
उत्तर:
सत्य

प्रश्न 3.
मौसम मानचित्र में वायुदाब मिलीबार में दर्शाया जाता है।
उत्तर:
सत्य

प्रश्न 4.
वायु वेग नापने की इकाई, एक नॉट 1.85 किमी. के बराबर होता है।
उत्तर:
सत्य।

जोड़ी मिलाइए

उत्तर:

1. → (ख)
2. → (ग)
3. → (क)

एक शब्द/वाक्य में उत्तर

प्रश्न 1.
भारत में मौसम विभाग की स्थापना किस सोसाइटी के सुझाव पर हुई ?
उत्तर:
एशियाटिक सोसायटी

प्रश्न 2.
मौसम विभाग का.प्रधान कार्यालय कहाँ है ?
उत्तर:
दिल्ली

प्रश्न 3.
भारतीय मानचित्रों में किस ऊँचाई के मेघ दिखाये जाते हैं ?
उत्तर:
निम्न व मध्यम

प्रश्न 4.
वायु वेग नापने की इकाई क्या है ?
उत्तर:
नॉट

प्रश्न 5.
वायु के आकस्मिक प्रचण्ड झोंके जो स्वत: कुछ समय बाद समाप्त हो जाते हैं, को क्या कहते हैं ?
उत्तर:
झंझा।

MP Board Class 10th Social Science Chapter 5 अति लघु उत्तरीय प्रश्न

प्रश्न 1.
प्रमुख वर्षा संकेत स्पष्ट कीजिए। (2009)
उत्तर:
प्रमुख वर्षा संकेत –

प्रश्न 2.
झंझा का क्या प्रभाव पड़ता है ?
उत्तर:
पेड़ की टहनियाँ टूटने लगती हैं और चलना-फिरना कठिन हो जाता है।

MP Board Class 10th Social Science Chapter 5 लघु उत्तरीय प्रश्न

प्रश्न 1.
मौसम मानचित्र से क्या तात्पर्य है ?
उत्तर:
मौसम मानचित्र-इस प्रकार के मानचित्र जिसमें पृथ्वी के किसी भाग में घटित मौसम सम्बन्धी दशाओं (जैसे-तापमान की स्थिति, वायुदाब, वर्षा, पवन की दिशा एवं वेग, मेघाच्छादन) को अन्तर्राष्ट्रीय स्तर पर निर्धारित संकेतों के माध्यम से व्यक्त किया जाता है, मौसम मानचित्र कहलाते हैं।

प्रश्न 2.
मौसम मानचित्र की व्याख्या के प्रमुख बिन्दु कौन-कौनसे हैं ?
उत्तर:
मौसम मानचित्र की व्याख्या के प्रमुख बिन्दु

1. प्रस्तावना – दिन, तिथि तथा समय।
2. तापमान का विचलन – शीत लहर और उष्ण लहर ज्ञात करने में सहायक।
3. वायुदाब का विचलन –
   • वायुदाब का सामान्य वितरण उच्च व न्यूनदाब क्षेत्र।
   • समदाब रेखाओं की उपनति, वायुदाब की प्रवणता, चक्रवातों की गति।
   • सामान्य वायु दाब से प्रमाण।
4. वायु दिशा एवं वायु वेग।
5. वृष्टि-सामान्य से अधिक या कम वर्षा के क्षेत्र, वर्षा की मात्रा।
6. मेघ-बादल के प्रकार (उच्च व निम्न) तथा मेघाच्छादन या मेघ की प्रकृति तथा मात्रा।
7. वायुमण्डलीय घटनाएँ-धुन्ध, कोहरा, विद्युत, धूल-आँधी, वज्रघोष, शिलावृष्टि।
8. समुद्र की दिशा।
9. भविष्यवाणी-मानचित्रों के आधार पर मौसम की भविष्यवाणी की जाती है।

MP Board Class 10th Social Science Chapter 5 दीर्घ उत्तरीय प्रश्न

प्रश्न 1.
प्रमुख समुद्री तरंग-संकेतक स्पष्ट कीजिए।
उत्तर:
समुद्री तरंग-संकेतक समुद्री तरंग की दिशा अक्षरों के रूप में प्रदर्शित की जाती है। प्रमुख समुद्री तरंग संकेतक निम्न प्रकार दर्शाये गये हैं –

प्रश्न 2.
प्रमुख वायु मापन संकेतों को स्पष्ट करते हुए उनके प्रभावों को स्पष्ट कीजिए।
उत्तर:
वायु मापन संकेत

प्रश्न 3.
भारत के मानचित्र में निम्नलिखित को दर्शाइए
1.

1. कराकोरम पवर्तमाला
2. सतपुड़ा
3. अरावली
4. नीलगिरि।

2. भारत में हिमालय का सर्वोच्च शिखर (माउण्ट एवरेस्ट)।
3. छोटा नागपुर का पठार। (2009, 10, 16, 18)

4.

1. गंगा (2013, 16)
2. ब्रह्मपुत्र
3. कृष्णा
4. कावेरी
5. नर्मदा। (2010, 14, 16, 17)

5. सर्वाधिक वर्षा का क्षेत्र (चेरापूँजी/मौसिनराम)। (2018)
6. बंगाल की खाड़ी। (2009, 13)
7. कन्याकुमारी। (2009)
उत्तर:

प्रश्न 4.
भारत के मानचित्र पर निम्नलिखित को दर्शाइए –

1. जूट उत्पादक क्षेत्र
2. गन्ना उत्पादक क्षेत्र
3. काली मिट्टी का क्षेत्र या कपास उत्पादक क्षेत्र
4. चाय उत्पादक क्षेत्र (2009)
5. रबर उत्पादक क्षेत्र (2012)
6. चावल उत्पादक क्षेत्र।

उत्तर:

प्रश्न 5.
भारत के दिए गये मानचित्र में निम्नलिखित को दर्शाइए
1. कोयला उत्पादक क्षेत्र

2.

• प्रमुख खनिज तेल उत्पादक क्षेत्र (2009)
• डिग्बोई (असम) (2012)

3. यूरेनियम उत्पादक क्षेत्र

• राजस्थान में यूरेनियम (2009)

4. प्राकृतिक गैस उत्पादक क्षेत्र
5. लक्षद्वीप (2010)
6.

• नाभिकीय ऊर्जा केन्द्र (2013)
• तारापुर अणु विद्युत गृह। (2016)

उत्तर:

प्रश्न 6.
भारत के मानचित्र में निम्नलिखित को दर्शाइए
(1)

1. दिल्ली (2010, 15)
2. विशाखापट्टनम (2018)
3. भिलाई
4. भोपाल
5. मुम्बई (2009, 11, 15)
6. चेन्नई (2009, 12, 15)
7. कोलकाता (2010)

(2) दिल्ली-कोलकाता रेलवे लाइन (2014)

(3)

1. सान्ताक्रुज
2. पालम व
3. दमदम हवाई अड्डा (2009)

(4) अरब सागर (2011, 12, 15)
(5) आगरा से मुम्बई राष्ट्रीय राजमार्ग (2012)
(6) थार का मरुस्थल (2015)
(7) चिल्का झील (2011, 16)
(8) कोई दो प्रमख आंतरिक जल परिवहन मार्ग
(9) सीमा सड़क विकास बोर्ड का क्षेत्र।
उत्तर:

प्रश्न 7.
भारत के मानचित्र में निम्नांकित को दर्शाइए –

1. शीतकालीन वर्षा का क्षेत्र
2. काजीरंगा राष्ट्रीय उद्यान (2017)
3. भाखड़ा-नंगल बाँध
4. कच्छ का रन (2013, 18)
5. कर्क रेखा (2009, 14, 17)
6. हजीरा-जगदीशपुर गैस पाइप लाइन (2009, 14, 18)
7. आनन्द-अहमदाबाद दुग्ध पाइप लाइन। (2009)

उत्तर:

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.4 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² – 3x + 5 = 0
(ii) 3x² – 4\(\sqrt{3} x\) + 4 = 0
(iii) 2x² – 6x + 3 = 0
Solution:
(i) Comparing the given quadratic equation with ax² + bx + c = 0, we get a = 2, b = -3, c = 5
b² – 4 ac = (-3)² – 4(2)(5) = 9 – 40 = -31 < 0
Since b² – 4ac is negative.
∴ The given quadratic equation has no real roots.

(ii) Comparing the given quadratic equation with ax² + bx + c = 0, we get

Since, b² – 4ac is zero.
∴ The given quadratic equation has two real roots which are equal. Hence, the roots are

(iii) Comparing the given quadratic equation with ax² + bx + c = 0, we get a = 2,b = -6,c = 3
∴ b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24 = 12 > 0
Since, b² – 4ac is positive.
∴ The given quadratic equation has two real and distinct roots, which are given by

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx(k – 2) + 6 = 0
Solution:
(i) 2x² + kx + 3 = 0
Here, a = 2, b = k, c = 3
b² – 4ac = (k)² – 4(2) (3)
= k² – 24
∵ For a quadratic equation to have equal root, b² – 4ac = 0

(ii) kx(k – 2) + 6 = 0
Comparing kx² – 2kx + 6 = 0 with ax² + bx + c, we get
a = k, b = -2k, c = 6
∴  b² – 4ac = (-2k)² – 4(k)(6) = 4k² – 24k
Since, the roots are equal
∴  b² – 4ac ⇒ 4k² – 24k = 0.
⇒ 4k(k – 6) = 0
⇒ 4k = 0 or k – 6 = 0
⇒ k = 0 or k = 6
But k cannot be 0, otherwise, the given equation is not quadratic. Thus, the required value of k is 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m² ? If so, find its length and breadth.
Solution:
Let the breadth be x m.
∴ Length = 2x m
Now, Area = Length × Breadth = 2x × x m² = 2x² m²
According to the given condition,
2x² = 800

Therefore, x = 20 or x = -20
But x = -20 is not possible
[ ∵ Breadth cannot be negative]
∴ x = 20 ⇒ 2x = 2 × 20 = 40
Thus, length = 40 m and breadth = 20 m

Question 4.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the age of one of the friends be ‘x’.
Age of another friend is (20 – x).
4 years back age of 1st friend is (x – 4)
4 years back age of 2nd friend = (20 – x – 4) = (16 – x)
Product of their ages is 48.
∴ (x – 4) (16 – x) = 48
16x – x² – 64 + 4x = 48
-x² + 20x – 64 = 48
-x² + 20x – 64 – 48 = 0
-x² + 20x – 112 = 0
x² – 20x + 112 = 0
Here, a = 1, b = -20, c = 112
b² – 4ac = (-20)² – 4( 1)( 112) = 400 – 448 = – 48
Here, b² – 4ac = -48 < 0.
∴ It has no real roots.
∴ This situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Solution:
Let the breadth of the rectangle be x m.
Since, the perimeter of the rectangle = 80 m
∴ 2(Length + breadth) = 80
2(Length + x) = 80

⇒ Length = (40 – x) m
∴ Area = (40 – x) × x m² = (40x – x²) m²
According to the given condition,
Area of the rectangle = 400 m²
⇒ 40x – x² = 400
⇒ – x² + 40x – 400 = 0
⇒ x² – 40x +400 = 0 …(1)
Comparing equation (1) with ax² + bx + c = 0, we get
a = 1, b = -40, c = 400

Since, Length = Breadth
⇒ This rectangle is a square.
Thus, it is not possible to design a rectangular park of given perimeter and area.

In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Ex 3.7 Pdf, These solutions are solved subject experts from the latest edition books.

MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let the age of Ani = x years
and the age of Biju’s = y years
Case I:
y > x
According to 1^st condition : y – x = 3 …. (1)
∵ [Age of Ani’s father] = 2[Age of Ani] = 2x years


Substituting the value of x in equation (1),
we get y – 21 = 3 ⇒ y = 3 + 21 = 24
∴ Age of Ani = 21 years
Age of Biju = 24 years

Substituting the value of y in equation (1),
we get 19 – y = 3 ⇒ y = 16
∴ Age of Ani = 19 years
Age of Biju = 16 years

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
[Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)]
Solution:
Let the capital of 1^st friend = ₹ x,
and the capital of 2^nd friend = ₹ y
According to the condition,
x + 100 = 2(y -100)
⇒ x + 100 – 2y + 200 = 0
⇒ x – 2y + 300 = 0 … (1)
Also, 6(x – 10) = y + 10 ⇒ 6x – y – 70 = 0 …. (2)
From (1), x = -300 + 2y (3)
Substituting the value of x in equation (2), we get
6[-300 + 2y] – y – 70 = 0
⇒ -1870 + 11y = 0

Now, Substituting the value of y in equation (3), we get, x = – 300 + 2y
= – 300 + 2(170) = – 300 + 340 = 40
Thus, 1^st friend has ₹ 40 and the 2^nd friend has ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the actual speed of the train = x km/hr
and the actual time taken = y hours
Distance = speed × time
According to 1^st condition: (x +10) × (y – 2) = xy
⇒ xy – 2x + 10y – 20 = xy
⇒ 2x – 10y + 20 = 0 … (1)
According to 2^nd condition: (x -10) × (y + 3) = xy
⇒ xy + 1ox -10y – 30 = xy
⇒ 3x – 10y – 30 = 0 ….(2)
Using cross multiplication for solving (1) and (2), we get

Thus, the distance covered by the train = 50 × 12 km = 600 km.

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of students = x
and the number of rows = y
∴ Number of students in each row

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.
Solution:
Sum of angles of a triangle = 180°
∴ ∠A + ∠B + ∠C = 180° … (1)
∵ ∠C = 3∠B = 2(∠A + ∠B) … (2)
From (1) and (2), we have ∠A + ∠B + 2 (∠A + ∠B) = 180°
⇒ ∠A + ∠B + 2∠A + 2∠B = 180°
⇒ ∠A + ∠B = 60° …. (3)
Also, ∠A + ∠B + 3∠B = 180°
⇒ ∠A + 4∠B = 180° ….(4)
Subtracting (3) from (4), we get
∠A + 4∠B – ∠A – ∠B = 180°- 60°

Substituting ∠B = 40° in (4) we get,
∠A + 4(40°) = 180°
⇒ ∠A = 180° – 160° = 20°
∴ ∠C = 3∠B = 3 × 40° = 120°
Thus, ∠A = 20°, ∠B = 40° and ∠C = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and they-axis.
Solution:
To draw the graph of 5x – y = 5, we get

x	1	2	0
y	0	5	-5

and for equation 3x – y = 3, we get

x	2	3	0
y	3	6	-3

Plotting the points (1, 0), (2, 5) and (0, -5), we get a straight line l₁. Plotting the points (2, 3), (3, 6) and (0, -3), we get a straight line l₂.

From the figure, obviously, the vertices of the triangle formed are A(l, 0), B(0, -5) and C(0,-3).

Question 7.
Solve the following pair of linear equations:
(i) px + qy = p – q, qx – py = p + q
(ii) ax + by = c, bx + ay = 1 + c
(iii) \(\frac{x}{a}-\frac{y}{b}\)= 0, ax + by = a² + b²
(iv) (a – b)x + (a + b)y = a² – 2ab – b²,
(a + b)(x + y) = a² + b²
(v) 152x – 378y = -74, – 378x + 152y = -604
Solution:

Question 8.
ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Solution:
ABCD is a cyclic quadrilateral.
∴ ∠A + ∠C = 180° and
∠B + ∠D = 180°
⇒ [4y + 20] + [- 4x] = 180°
⇒ 4y – 4x + 20° -180° = 0
⇒ 4y – 4x – 160° = 0
⇒ y – x – 40° = 0 … (1)

In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 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 13.1

Unless stated otherwise, take π = \(\frac{22}{7}\)

Question 1.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of each cube = 64 cm³

Let the edge of each cube = x
∴ x³ = 64 cm³
⇒ x = 4 cm
Now, Length of the resulting cuboid (l) = 2x cm = 8 cm
Breadth of the resulting cuboid (b) = x cm = 4 cm
Height of the resulting cuboid (h) = x cm = 4 cm
∴ Surface area of the cuboid = 2 (lb + bh + hl)
= 2 [(8 × 4) + (4 × 4) + (4 × 8)] cm²
= 2 [32 + 16 + 32] cm² = 2 [80] cm²
= 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
For hemispherical part,
radius (r)= \(\frac{14}{2}\) = 7cm
∴ Curved surface area = 2πr²
= 2 × \(\frac{22}{7}\) × 7 × 7cm²
= 308cm² 7
Total height of vessel = 13 cm

∴ Height of cylinder = (13 – 7)cm = 6 cm and radius(r) = 7 cm
∴ Curved surface area of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 6cm² = 264cm² 7
∴ Inner surface area of vessel = (308 + 264)cm² = 572 cm²

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Let h be the height of cone and r be the radius of cone and hemisphere.

∴ h = [height of toy – radius of hemi sphere]
= (15.5 – 3.5) cm = 12 cm
Also l² = h² + r² = 12² + (3.5)² = 156.25 cm²
∴ l = 12.5 cm
Curved surface area of the conical part = πrl
Curved surface area of the hemispherical part = 2πr²
∴ Total surface area of the toy = πrl + 2πr²
= πr(l + 2 r)
= \(\frac{22}{7} \times \frac{35}{10}\) (12.5 + 2 × 3.5) cm²
= 11 × (12.5 + 7) cm² = 11 × 19.5 cm²
= 214.5 cm²

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
Let side of the block, l = 7 cm

∴ The greatest diameter of the hemisphere = 7 cm
Surface area of the solid = [Total surface area of the cubical block] + [C.S.A. of the hemisphere] – [Base area of the hemisphere]
= 6 × l² + 2πr² – πr²
[where l = 7 cm and r = \(\frac{7}{2}\) cm]

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let l be the side of the cube.

∴ Diameter of the hemisphere = l
⇒ Radius of the hemisphere (r) = \(\frac{l}{2}\)
Curved surface area of hemisphere = 2πr²
= 2 × π × \(\frac{l}{2} \times \frac{l}{2}=\frac{\pi l^{2}}{2}\)
Base area of the hemisphere = πr²
= \(\pi\left(\frac{l}{2}\right)^{2}=\frac{\pi l^{2}}{4}\)
Surface area of the cube = 6 × l² = 6l²
∴ Surface area of the remaining solid = [Total surface area of cube + C.S.A. of hemisphere – base area of hemisphere]

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:
Radius of the hemispherical part

Curved surface area of one hemispherical part = 2πr²
∴ Surface area of both hemispherical parts
= 2(2πr²) = 4πr² = [4 × \(\frac{22}{7} \times\left(\frac{25}{10}\right)^{2}\)] mm²
= \(\left(4 \times \frac{22}{7} \times \frac{25}{10} \times \frac{25}{10}\right)\) mm²
Entire length of capsule = 14 mm
∴ Length of cylindrical part = [Length of capsule – Radius of two hemispherical part]
= (14 – 2 × 2.5)mm = 9mm Area of cylindrical part = 2πrh
= (2 × \(\frac{22}{7}\) × 2.5 × 9 ]mm² = (2 × \(\frac{22}{7} \times \frac{25}{10}\) × 9) mm²
Total surface area
= [Surface area of cylindrical part + Surface area of both hemispherical parts]

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)
Solution:

For cylindrical part:
Radius (r) = \(\frac{4}{2}\) m = 2m and height (h) = 2.1 m
∴ Curved surface area = 2πrh = (2 × \(\frac{22}{7}\) × 2 × \(\frac{21}{10}\))m²
For conical part:
Slant height (l) = 2.8 m
and base radius (r) = 2 m
∴ Curved surface area
= πrl = (\(\frac{22}{7}\) × 2 × \(\frac{28}{10}\)) m²
∴ Total surface area = [Curved surface area of the cylindrical part] + [Curved surface area of conical part]

Cost of the canvas used :
Cost of 1 m² of canvas = ₹ 500
∴ Cost of 44 m² of canvas = ₹ (500 × 44)
= ₹ 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Solution:

For cylindrical part :
Height (h) = 2.4 cm and diameter = 1.4 cm
⇒ Radius (r) = 0.7 cm
∴ Total surface area of the cylindrical part
= 2πrh + 2πr² = 2πr [h + r]
= 2 × \(\frac{22}{7} \times \frac{7}{10}\) [2.4 + 0.7]
= \(\frac{44}{10}\) × 3.1 = \(\frac{44 \times 31}{100}\) = \(\frac{1364}{100}\) cm²
For conical part:
Base radius (r) = 0.7 cm and height (h) = 2.4 cm

Base area of the conical part
= \(\pi r^{2}=\frac{22}{7} \times\left(\frac{7}{10}\right)^{2}=\frac{22 \times 7}{100} \mathrm{cm}^{2}=\frac{154}{100} \mathrm{cm}^{2}\)
Total surface area of the remaining solid = [(Total surface area of cylindrical part) + (Curved surface area of conical part) – (Base area of the conical part)]
= \(\left[\frac{1364}{100}+\frac{550}{100}-\frac{154}{100}\right] \mathrm{cm}^{2}=\frac{1760}{100} \mathrm{cm}^{2}\)
Hence, total surface area to the nearest cm² is 18cm².

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:
Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm
∴ Curved surface area = 2πrh
= 2 × \(\frac{22}{7} \times \frac{35}{10}\) × 10cm² = 220cm²
Curved surface area of a hemisphere = 2πr²
∴ Curved surface area of both hemispheres
= 2 × 2πr² = 4πr² = 4 × \(\frac{22}{7} \times \frac{35}{10} \times \frac{35}{10}\) cm²
= 154 cm²
Total surface area of the remaining solid = (220 + 154) cm² = 374 cm².