Students can Download Maths Chapter 3 Mathematics of Chance Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
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Students can Download Maths Chapter 11 Statistics Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Students can Download Physics Chapter 3 Electromagnetic Induction Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
You can Download Production of Metals Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.
→ All ores are minerals, but are all minerals ores?
Answer:
NO
→ Which metal’s ore is calamine?
Answer:
Zinc (Zn)
→ Which is the ore of aluminum?
Answer:
Bauxite (Al2O2 2H2O)
→ Which metals have sulfide ores?
Answer:
Copper, Zinc
Production Of Metals Class 10 Kerala Syllabus Text Book Page No: 65
→ Complete the table 4.2.
Answer:
→ Write suitable method of concentration for the ores given in the Table (4.3)
Answer:
Sslc Chemistry Chapter 4 Kerala Syllabus Text Book Page No: 67
→ Complete The table 4.4
Answer:
→ Haematite, magnetite, iron pyrites, etc. are the minerals of iron. Which are the ores of iron among these minerals?
Answer:
Haematite, magnetite.
Chemistry Class 10 Chapter 4 Kerala Syllabus Text Book Page No: 69
→ Complete the table 4.5.
Answer:
→ Which alloy steel is used for the production of heating coils? Explain the reason.
Answer:
Nichrome, because of high resistance, easily it becomes hot.
→ Even though nichrome and stainless steel contain the same components they possess different properties. Find out the reason.
Answer:
The ratio of the component elements are different.
→ Which alloy steel is used for making permanent magnets?
Answer:
Alnico
Sslc Chemistry Chapter 4 Notes Kerala Syllabus Text Book Page No: 70
→ Complete the table 4.7
Answer:
→ How ¡s alumina obtained from this aluminum hydroxide?
Answer:
The precipitate is separated, washed and then heated strongly to get alumina.
Production Of Metals Class 10 Notes Kerala Syllabus Text Book Page No: 71
→ Complete the flow diagram, related to concentration of bauxite, which is given below.
Answer:
→ Complete the chemical equation for the reaction taking place when Aluminium hydroxide is heated.
Ans. 2Al(OH)3 → Al2O3 + 3H2O
→ Which method can be used for separating aluminum from alumina?
Answer:
Electrolysis
→ Can we use carbon as the reducing agent? Why?
Answer:
Can’t use carbon as the reducing agent, because the reactivity of aluminum is very high hence they require very strong reducing agent.
Sslc Chemistry Chapter 4 Notes Pdf Kerala Syllabus Text Book Page No: 72
→ To which electrode does Al3+ move?
Answer:
Towards negative electrode (Cathode)
→ What about oxide ion?
Answer:
Towards positive electrode (Anode)
→ Complete the table related to the electrolysis of Alumina.
Answer:
Sslc Chemistry Chapter 4 Notes English Medium Question 1.
Which of the properties of metals is utilized in the following instances?
a. Aluminum utensils are used for cooking.
b. Copper is used for making vessels.
c. Gold wires are used in ornaments.
Answer:
a. Heat conductivity, lightweight, can be molded in any shape, low price, etc.
b. Malleability
c. Ductility
Sslc Chemistry 4th Chapter Kerala Syllabus Question 2.
What are the factors to be considered while selecting minerals for the extraction of metals?
Answer:
High availability
Extraction of metal should be easy
The percentage of metal content in the mineral should be comparatively high.
Cost of production should below.
Kerala Syllabus 10th Standard Chemistry Chapter 4 Question 3.
Write the different stages involved in metallurgy.
Answer:
a. Concentration of the ore methods are:
b. Extraction of metal from the concentrated ore has 2 stages.
c. Refining of metals. Different methods are:
Production Of Metals Class 10 Notes Pdf Kerala Syllabus Question 4.
What are the different methods for the refining of metals?
Answer:
Chemistry Chapter 4 Class 10 Kerala Syllabus Question 5.
How is iron extracted industrially?
Answer:
The ore is crushed into small lumps and then wash out the soluble impurities in running water. Then it undergoes roasting. Thus impurities like sulfur, arsenic, moisture, etc. get removed. The roasted ore is mixed with coke and limestone and the mixture is changed into a blast furnace.
Reaction in blast furnace:
At high temperature, CaCO3 (limestone) decomposes to form CaO.
CaCO2 (s) → CaO(s) + CO2 (g)
This CaO combines with acidic gangue, SiO2 forming slag.
CaO(s)+ SiO2(g) → CaSiO3 (s)
Coke reacts with oxygen and form CO2
C(s) + O2(g) → CO2(g)
CO2 combines with more carbon and produces CO.
CO2(g) + C2(s) + Heat → 2 CO (g)
This CO acts as reducing agent.
Fe2O3 is reduced to form Fe.
Fe2O3(s)+ 3 CO(g) → 2Fe(s)+ 3CO2(g)
Hss Live Guru 10th Chemistry Kerala Syllabus Question 6.
Write the uses of the following:
a. Nichrome
b. Stainless steel
c. Alnico
Answer:
a. Nichrome – For making heating coils
b. Stainless steel – For the manufacture of utensils, parts of vehicles
c. Alnico-To make permanent magnets
Kerala Syllabus 10th Standard Chemistry Guide Question 7.
Explain the process of producing alumina from bauxite.
Answer:
Bauxite is treated with hot concentrated NaOH. Aluminum oxide reacts with NaOH solution forming sodium aluminate solution (NaAlO2). The unreacted impurities are removed from the solution. To the remaining solution of sodium aluminate, add Al(OH)3 in small quantity and dilute with water. Then the whole aluminum in the solution gets precipitated as Al(OH)3. The Al(OH)3 precipitated is separated from the solution. Then it is washed and heated strongly. Then Al(OH)3 decomposes to give pure AL2O3 or alumina.
2Al(OH)3(s)→ Al2O3(s) + 3H2O(1).
Hsslive Chemistry Class 10 Kerala Syllabus Question 8.
Explain the method of obtaining pure aluminum from alumina by electrolysis. In this process, the carbon rods are replaced from time to time. Why?
Answer:
Pure Al2O3 mixed with cryolite undergo electrolysis.
At cathode :
Al2+ + 3e– → Al
At anode :
2O2 → O2 + 4e–
Carbon at anode reacts with O2and form CO2. Thus anode gets used up. So anode is replaced frequently.
Hss Live Guru Class 10 Chemistry Kerala Syllabus Question 1.
You know that metals can be separated from molten compounds of metals by electrolysis. Find out how metals like Na, Ca and Mg are extracted.
Answer:
Na
Produced by the electrolysis of molten NaCl. KCl is added to reduce the melting point of NaCl.
Chemical equation:
NaCl → Na++Cl-
At cathode : Na++ le– → Na
At anode : 2Cl– → Cl2 + 2e–
Ca:
Mixture of CaCl2 and 16% CaF2 are melted and undergo electrolysis.
Mg:
By the electrolysis of molten mixture of camalite (KCl.MgCl2.6H2O) and NaCl in equal amounts.
Hss Live Guru 10th Chemistry Malayalam Medium Question 1.
Complete the table
Answer:
Chemistry Solutions Class 10 Kerala Syllabus Question2.
Features of ore and impurity are given in the table. Write down the method used for the separation of the ore.
a. How can we convert ore into its oxide form? Explain with proper examples
b. ZnCO3/Cu2S in these two calcination is used for ………. and Roasting is used for ………..
c. Give Examples for reducing agents for reducing oxide ores.
d. Strongest reduction agent
e. Which reducing agent used fur reducing ZnO, Fe2O3, Al2O3 ?
Answer:
i. Levigation/Hydraulic washing
ii. Magnetic Separation
iii. Froth Floatation
iv. Leaching
a. Calcination
Calcination is the process of heating the concentrated ore at a temperature below its melting point to remove the volatile impurities. When subjected to calcination, impurities like water, organic matter, and other volatile impurities are expelled from the ore. Metal Carbonate and hydroxides decompose to form oxides,
eg: ZnCO is converted to ZnO by calcination
Roasting:
Roasting is the process of heating the concentrated ore at a temperature below its melting point in a current of air. During roasting the ore gets converted into its oxide. When the concentrated ore is subjected to roasting, the water present in it is removed as vapor. Other impurities like sulfur, phosphorus and organic matter are oxidized and expelled. The sulfide combines with oxygen to form oxide eg: Cu2S ore is converted to Cu2O by roasting;
b. Calcination – ZnCO3, Roasting – Cu2S
c. Carbon monoxide, Carbon, Electricity
d. Electricity
e. Carbon Monoxide is used as the reducing agent to extract iron from haematite. (Fe2O3)
ZnO :- Carbon is used as the reducing agent to extract Zinc from Zinc Oxide. Electricity is used to extract aluminum from Al2O3.
Kerala Syllabus Class 10 Chemistry Solutions Question 3.
Complete the table:
Answer:
a. Low Melting point.
b. Low Boiling points
c. Metals having high electropositivity
Question 4.
a. Complete the table.
b. Stainless steel and Nichrome are having same content (Fe, Ni, Cr, C). But nature of both alloys are different Why?
c. Bauxite and clay are minerals of aluminum. But bauxite is the only ore of Aluminium. Why?.
Answer:
i. Obtained from Blast Furnace Contains 4% Carbon and other impurities like manganese silicon, phosphorus, etc.
ii. Pig iron mixed with scrap iron and coke melted in a special furnace contains 3% carbon.
iii. Made by purifying cast Iron.
iv. Prepared by varying the amount of carbon from 0.1 to 1.5%.
b. Stainless steel and Nichrome are having the same content but the nature of both alloys are different because ratio of constituent elements are different.
c. A mineral from which a metal is economically, easily and quickly extracted is called the core of the metal. Among the minerals of aluminum, bauxite possesses these properties. Hence bauxite is the ore of Aluminium. Since clay does not possess these properties it is not an ore of Aluminium.
Question 5.
Given below are the equations for the reaction taking place inside the blast furnace.
C + O2→ CO2
CO2+ C → 2CO
CaCO3 + SiO2 → CaSiO3
Fe2O3+3CO → 2Fe + 3CO2
a. Name the ore of iron.
b. Which is the gangue in iron ore?
c. Name the flux used in blast furnace,
d. Gangue + flux → ………… Which product is formed in blast furnace?
e. Reducing agent used in blast furnace.
f. Subjects dropped in blast furnace are …………., …………….
Answer:
a. Haematite
b. SiO2
c. CaO
d. slag(CaSiO3 )
e. Carbon monoxide
f. A mixture of roasted haematite, coke and limestone.
Question 6.
a. Write down the names of Anode, Cathode, Electrolyte used in the Electrolyte cell for the manufacturing of copper,
b. Write down the equations for the reaction in anode and cathode.
Answer:
Anode – Copper to be refined
Cathode – Pure Copper
Electrolyte – Aqueous Copper Sulphate Solution mixed with H2SO4
b. Anode – Cu → Cu2 + 2e–
Cathode – Cu2+ +2e– → Cu
Question 7.
Manufacturing of iron.
a. Name the furnace used for producing iron.
b. Name the materials using for producing iron.
c. Write down the reaction occurring on coke when hot is blasted on it?
d. Why CaCO3 is dropping inside the furnace?
e. Write down the nature of gangue with iron ore.
f. Gangue + flux → ……….. Write down the uses of the product formed in blast furnace.
g. Reducing agent in blast furnace
h. Write down the reactions taking place inside the blast furnace.
i. Iron formed from the blast furnace is called ……………
j. How can we change iron into steel?
k. What are the different types of steel?
Answer:
a. Blast Furnace
b. Mixture of roasted haematite, coke and limestone.
c. At the bottom of the blast furnace, coke combines with oxygen in the hot current of air. The CO2 which rises up along with the hot air current is reduced by coke.
d. CaCO3 in the furnace decomposes to form CaO and CO2. CaO which is basic combines with SiO2 (acidic) to form slag. Molten slag which is lighter floats over the heavier molten iron. CaCO3 is added to the furnace for the production of slag.
e. Acidic
f. Slag. (CaSiO3) used for the production of cement and in the Construction of Road.
g. Carbon monoxide
h. CaCO3(s) → CaO + CO2
CaO+SiO2 → CaSiO3
C+O2 CO2 + Heat
CO2(g) + C(s) + Heat → 2CO (g)
Fe2O3+3CO → 2Fe + 3CO2
i. Pig iron
j. Different types of steel can be prepared by varying the amount of carbon from 0.1 to 1.5
k. Mild steel, Medium steel, High carbon steel
Mild steel:- If the carbon content in steel is from 0.05% to 0.2 % then it in called mild steel. I
Medium steel :- Medium steel contains carbon from 0.2 % to 0.6 % medium steel
High carbon steel: – If the content of carbon is from 0.61 % to 15 % then it is known as high carbon steel.
Question 8.
Flow chart of Manufacturing Aluminium.
a. Draw the Electrolyte cell and then write answers for the following questions.
b. Anode, Cathode in this cell are …………….
c. Write down the reactions taking place in Anode and Cathode.
d Why Carbon power dropped above the electrolyte?
e. Which gas is evolved out from graphite.
f. Uses of Cryolite.
Answer:
a.
b. Anode – Carbon Rod, Cathode – Carbon lining
c. Anode -2O2– → O2 + 4e-
Cathode – AI3+ +3e– → AI
d. To prevent the chemical reaction of carbon rods with the oxygen in atmosphere.
e. Oxygen
f Cryolite is added to alumina to reduce its melting point and increase its electrical conductivity.
Question 1.
Nature of some ores are given. Pick ore concentration from the bracket.
(Magnetic Separation, Froth Floatation, Levigation, Leaching)
i. Ores are lighter and impurities are heavier.
ii. Ore is magnetic. But impurities are non-magnetic.
iii. Uses a solution which dissolves the ore.
iv. Ore is heavier and impurities are lighter.
Answer:
i. Froth floatation
ii. Magnetic separation
iii. Leaching
iv. Levigation
Question 2.
Some metals and ores are given. Match them suitably
Answer:
Question 3.
Calcination is used to convert zinc carbonate into zinc oxide. But cuprous sulfate is converted into cuprous oxide by roasting.
a. What is the difference between calcination and roasting?
b. What happens to the ore when it is subjected to calcination?
Answer:
a. Calcination: It is the process in which the ore is heated in the absence of air at a temperature below its melting point so that the moisture content of ore, volatile impurities, bio substances, etc. can be removed. Along with this, metal carbonates and hydroxides are converted into its oxide.
Roasting: It is the process in which the ore is heated in the presence of air at a temperature below its melting point so that the moisture content of ore, sulfur, phosphorous, etc, are converted into its oxide and can be removed. Sulfide ores also get converted into oxides.
b. Zinc carbonate ore is converted to Zinc oxide.
Question 4.
a. Some metals and their methods of concentration are given. Match them suitably.
Mercury, Zinc, Tin, Copper, Lead
Liquation, Electrolytic refining, Distillation
b. Write the reason for selecting the methods for the concentration of mercury and tin.
Answer:
a. Liquation – Tin, Lead
Electrolytic refining – Copper Distillation – Mercury, Zinc
b. Boiling point of mercury is low. Melting point of Tin is low
Question 5.
The order the reactivity of some metals are given. Answer the following questions by analyzing it.
Al >Zn >Cu >Au
a Which metal is produced bytfie electrolysis of its molten salt ?
b. Metal occur in free state in nature,
c. Metal produced by the self oxidation reduction reaction.
d. Metal ore which is reduced by carbon.
Answer:
a. Al
b. Au
c. Cu
d. Zn
Question 6.
A reducing agent is required to extract the metal from its ore. Why ? Explain with ex ample.
Answer:
In ores metals are in positive oxidation state, reducing agent is needed (electron giving substance) to get the metal.
Eg: Carbon is used to extract zinc from zincoxide. Zno(s) + C(s) → Zn(s) + CO(g)
Question 7.
The equations of the production of iron in the blast furnace are given. Answer the following questions.
C + O2 → CO2
CO2 + C → 2CO
CaCO3 → CaO + CO2
CaO + SiO2 → CaSiO3
Fe2O3+3CO → 2Fe + 3CO2
a. Which substance reduces haematite in the metallurgy of iron? How this reducing agent is produced in the furnace?
b. Which is the main impurity found in haematite? Which substance is used to remove the gangue?
c. Write the chemical equation of the formation of slag in blast furnace.
Answer:
a. CO, Oxygen in the blast of hot air reacts with coke, to form CO2. This CO2 again reacts with coke to produce CO.
b. SiO2 – Impurity; CaO – remove gangue
c. CaO + SiO2 → CaSiO3
Question 8.
a. How pig iron is converted into cast iron?
b. Molten cast iron is poured into moulds to make different shapes. Which speciality if cast iron is based for it?
Answer:
a. Pig Iron mixed with scrap iron and coke, is melted in a special furnace to make cast iron.
b. Molten cast iron expands a little on solidifiation.
Question 9.
Alloys containing iron are given. Find out a, b, c and
Answer:
a. Fe, Ni, Al, Co
b. For the manufacture of permanent magnets
c. Stainless steel
d. For making heating coils
Question 10.
Aluminium is prepared industrially by Hall-Heroult process. Various steps in the concentration of ore are given below. Write them
in the correct order.
i. The precipitate formed is separated, washed and strongly heated to get alumina,
ii. Crushed bauxite is leached with hot sodium hydroxide solution.
iii. Impurities are removed from the sodium aluminate solution by filtration.
iv. Solution is diluted after adding a little aluminium hydroxide, to precipitate aluminum hydroxide.
Answer:
Order:
(ii),
(iii),
(iv),
(i)
Question 11.
a. Carbon monoxide cannot be used as reducing agent to extract aluminium from alumina. Why?
b. The electrolytic cell for alumina is given below.
i. Al2O3 dissolved in molten cryolite is used as the electrolyte. What is the purpose of adding cryolite to alumina?
ii. Anode is replaced from time to time while producing aluminium. Why?
iii. Write the chemical equation of the reaction at the cathode.
Answer:
a. Aluminium compounds are very stable,
b. i. The melting point of alumina is very high. Cryolite is added to alumina to reduce its melting point and increase its electrical conductivity, .
ii. Oxygen liberated at the anode reacts with carbon, forming CO
iii. Al3+ + 3e– → Al
Question 12.
a. Illustrate the arrangement of refining copper and label the anode, cathode and electrolyte.
b. Write the chemical equations at the anode and cathode and sustain it as a redox reaction.
Answer:
As oxidation and reduction takes place it is a redox reaction
Question 13.
Clay, cryolite and bauxite are the minerals of aluminium.
a. Which among them is the ore of aluminium? What is its chemical formula?
b. What are the features of an ore?
Answer:
a. Bauxite, Al2O3, 2H2O
b. 1. Easily available
2. Metal can be separated easily
3. High content of metal
Question 14.
The chemical reaction of calcium carbonate while heating is given. .
\(\mathrm{CaCO}_{2} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}\)
How the reaction is made use in the metallurgy of iron?
Answer:
CaO is framed by the decomposition of Ca- CO3. This CaO acts as a flux and combines with SiO2 (gangue) to form CaSiO3 (Slag).
Question 15.
Find the relation and answer the following.
a. Zinc Sulphate : Roasting,
Calcium Carbonate:…………
b. Haematite: Magnetic Separation;
Bauxite:…………..
Answer:
a. Calcination
b. Leaching
Question 16.
The flowchart of the process of concentrate of aluminium ore is given. Complete the flowchart.
Answer:
a. NaAlO2(Sodium aluminate)
b. Al(OH)3 /Aluminum hydroxide)
c. The precipitate is separated, washed well and strongly heated;
d. Al2O3
Very Short Answer Type Questions (Score 1)
Question 1.
What is the importance of adding cryolite in the electrolysis of alumina? ‘
Answer:
Alumina has a very high melting point. Cryolite lowers the melting point of alumina and improves the conductivity of alumina.
Question 2.
What are gangue, flux and slag?
Answer:
Gangue: Impurities in the metal ore 1
Flux: Chemical substances used to convert impurities which are not easily removable (gangue) into slag.
Slag: substances which are easily removable when gangue and flux are combined.
Short Answer Type Questions (Score 2)
Question 3.
a. Choose the ore from those given below which can be used in the above figure. Bauxite, Tin stone, Copper pyrites, Calamine
b. Give reason for your answer:
Answer:
a. Tin stone (SnO2)
b. This figure depicts magnetic separation. SnO2 is a magnetic ore. But the gangue, iron tungstate present in this also has magnetic property.
Question 4.
Haematite, the ore of iron undergo roasting.
a. Which impurity is not removed by this method?
b. How is it removed then? Explain.
Answer:
a. Silicon dioxide (SiO2)
b. It is removed during metallurgy. At high temperature, CaC03 (limestone) decomposes to form CaO, which is the flux. CaO combines with SiO2 and forms CaSiO3 (slag).
CaCO3(s) → CaO(s) + CO2(g)
CaO(s) + SiO2(s) → CaSiO3(s)
Question 5.
During the concentration process of bauxite,
a. Why is hot concentrated NaOH used?
b. Why isAl(OH)3 added in small quantity and diluted with water to sodium aluminate solution?
Answer:
a. Only bauxite is soluble in hot concentrated-. ted NaOH. As the impurities are insoluble, they can be easily filtered. This process is known is leaching.
b. When Al(OH)3 is added in small quantity and diluted with water to sodium aluminate solution, the whole aluminium in the solution gets precipitated as Al(OH)3. The Al(OH)3 precipitate is separated from the solution. Then it is washed and heated strongly to form pure Al2O3 or alumina.
Question 6.
Define:
a. Cast iron
b. Wrought iron
Answer:
a. Cast iron is formed by heating pig iron, scrap iron and coke. Cast iron contains about 3% of carbon. Cast iron expands on solidification. So these are used to make molds. Though it is strong, it is brittle.
b. When cast iron is purified, it becomes wrought iron which is comparatively a pure form of iron. Wrought iron contains about 0.2% to 0.5% of carbon. Small amounts of phosphorus, silicon, etc. are also present.
Short Answer Type Questions (Score 3)
Question 7.
Match columns A, B and C suitably.
Answer:
Question 8.
The concentration methods of certain ores are given below. Why these methods are used?
a. Bauxite – Leaching
b. Magnetite – Magnetic separation
c. Copper pyrites – Froth floatation
Answer:
a. Bauxite gets dissolved in hot concentrate NaOH. But the impurities does not dis-solve in this.
b. Magnetite, the ore of iron has magnetic properties but the impurities does not have.
c. Copper pyrites have a lesser density and impurities have higher density. Also, only the ore particles float in pine oil.
Question 9.
Explain the relationship between reactivity series of metals and metallurgy.
Answer:
Metals such as K,”Na, Ca, Mg and Al are placed above in the reactivity series. Strong reducing agent like electricity is used in the production of these metals by the electrolysis of their molten metallic compounds. Weak reducing agent such as carbon/carbon monoxide is used for the production of Zn, Fe, Ni, Sn and Pb. Copper is produced by the oxidation-reduction reactions of metal sulfide. Ag, Au, etc
Question 10.
Purification of copper is depicted here.
a Identify the anode, cathode and electrolyte.
b. Write the chemical equation during electrolysis.
c. What is seen below the positive electrode?
Answer:
a. Anode : Impure copper
Cathode : Pure copper
electrolyte : aqeous solution of CuSO4 to which H2SO4 is added.
b. At anode : Cu → Cu2+ + 2e–
At cathode : Cu2+ + 2e– → Cu
c. Anode mud – impurities in impure copper
Question 11.
Certain alloys are given below.
i. Nichrome
ii. Stainless steel
a. What are the constituent elements in them?
b. What is the reason for the difference in their properties?
c. Write one use of each.
Answer:
i. Nichrome: Fe, Ni, Cr, C
ii. Stainless steel: Fe, Cr, Ni, C
b. The constituent elements are same in both but their ratios are different. So their properties also differ.
c. Nichrome : for making heating coils Stainless steel: to make utensils
Long Answer Type Questions (Score 4)
Question 12.
Describe the following:
a. Calcination
b. Roasting v
Answer:
Extraction of metals from the concentrated ore have 2 stages:-
a. Calcination: It is the process in which the ore is heated in the absence of air at a temperature below its melting point so that the moisture content of ore, volatile impurities, bio substances, etc. can be removed. Along with this, metal carbonates and hydroxides are converted into its oxide.
b. Roasting: It is the process in which the ore is heated in the presence of air at a temperature below its melting point so that the moisture content of ore, sulfur, phosphorous, etc, are converted into its oxide and can be removed. Sulfide ores also get converted into oxides.
Question 13.
Complete the table.
Answer:
a. Haematite
b. Levigation
c. Roasting
d. Impurities with less density are removed.
e. Moisture content is removed. Sulfur, Arsenic, phosphorous, etc. are converted into its oxides. As they are in gaseous state, these are also removed.
Question 14.
Minerals of certain metals are given below. Write down the refining method of each.
a. Tin
b. Copper
c. Zinc
d. Lead
e. Cadmium
f. Silver
g. Mercury
Answer:
a. Tin – Liquation
b. Copper – Electrolysis
c. Zinc – Distillation
d. Lead – Liquation
e. Cadmium – Distillation
f. Silver – Electrolysis
g. Mercury – Distillation
Question 15.
The figure depicts the electrolysis of mixture of alumina and cryolite. Label anode, cathode and electrolyte in the figure.
Answer:
You can Download Mathematics of Chance Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.
Textbook Page No. 71
Mathematics Of Chance Questions And Answers Kerala Syllabus 10th Standard Question 1.
A box contains 6 black and 4 white balls. If a ball is taken from it. What is the probability of it being black? And the probability of it being white?
Answer:
Total no. of balls = 10
No. of black balls = 6
No. of white balls = 4
Probability of it to be black = \(\frac { 6 }{ 10 }\) = \(\frac { 3 }{ 5 }\)
Probablitiy of to be white = \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)
Mathematics Of Chance Class 10 Kerala Syllabus Kerala Syllabus Question 2.
There are 3 red balls and 7 green balls in a bag, 8 red and 7 green balls in another.
i. What is the probability of getting a red ball from the first bag?
ii. From the second bag?
iii. If all the balls are put in a single bag, What is the probability of getting a red ball from it?
Answer:
i. Total no. of balls in the first bag = 10
No. of red balls = 3
P(red ball) = 3/10
ii. Total no. of balls in the second bag= 15 No. of red balls = 8
P(red ball) = 8/15
iii. Total no. of balls in both bags = 25
Total no. of red balls = 11
P(red ball) = 11/25
Mathematics Of Chance Questions And Answers Pdf Kerala Syllabus 10th Standard Question 3.
One is asked to say a two-digit number. What is the probability of it being a perfect square?
Answer:
Total no. of two-digit numbers = 90
Perfect squares from 10 to 99 are 16, 25, 36, 49, 64, 81.
There are 6 favourable numbers Probability = \(\frac { 6 }{ 90 }\) = \(\frac { 1 }{ 15 }\)
Mathematics Of Chance Extra Questions And Answers Kerala Syllabus 10th Standard Question 4.
Numbers from 1 to 50 are written on slips of paper and put in a box. A slip is to be drawn from it; but before doing so, one must make a guess about the number, either prime number or a multiple of five. Which is the better guess? Why?
Answer:
Probablity of getting prime numbers from 1 to 50 = \(\frac { 15 }{ 50 }\) = \(\frac { 3 }{ 10 }\)
Probability of getting a multiple of 5 from 1 to 50 = \(\frac { 10 }{ 50 }\) = \(\frac { 1 }{ 5 }\)
∴ Better guess would be prime numbers as probability of that is more.
Sslc Maths Mathematics Of Chance Kerala Syllabus 10th Standard Question 5.
A bag contains 3 red beads and 7 green beads. Another contains one red and one green more. The probability of getting a red from which bag is more?
Answer:
Total no. of beads in first bag = 10
No. of red beads in first bag = 3
P(red ball) in first bag = \(\frac { 3 }{ 10 }\)
Total no. of beads in second bag = 12
No. of red beads in second bag = 4
P(red ball) in second bag = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
The probability of getting a red bead is more from the second bag since, \(\frac{1}{3}>\frac{3}{10}\left(\frac{10}{30}>\frac{9}{30}\right)\)
Textbook Page No. 72
In each picture below, the explanation of the green part is given. Calculate in each, the probability of a dot put without looking to be within the green part.
Mathematics Of Chance Pdf Kerala Syllabus 10th Standard Question 1.
A square got by joining the mid points of a bigger square.
Answer:
∴ Probability = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
Class 10 Maths Chapter 3 Mathematics Of Chance Kerala Syllabus Question 2.
A square with all vertices on a circle
Answer:
Side of square = a = 2cm Radius
ie., The area of the square is 2/π the area of the circle.
Probability = \(\frac{a^{2}}{\frac{\pi}{2} a^{2}}=\frac{4}{2 \pi}=\frac{2}{\pi}\)
Probablity ofthe dot falling on the square is \(\frac { 2 }{ π }\) = 0.64
Sslc Maths Chapter 3 Questions And Answers Kerala Syllabus Question 3.
Circle exactly fitting inside a square.
Answer:
If radius of circle is r
Side of the square = 2r
Area of the square = 4r2
Area of the circle= πr2
Mathematics Of Chance Extra Questions Kerala Syllabus 10th Standard Question 4.
A triangle got by joining alternate vertices of a regular hexagon.
Answer:
The area of regular hexagon having side a = \(\frac{3 \sqrt{3}}{2} a^{2}\)
Side of the triangle got by joining alternate vertices = √3a
Area of triangle = \(\frac{\sqrt{3}}{4}(\sqrt{3} a)^{2}=\frac{3 \sqrt{3}}{4} a^{2}\)
∴ Probability of the dot falling on the tri-angle = \(\frac{3 \sqrt{3}}{4} a^{2} \times \frac{2}{3 \sqrt{3} a^{2}}=\frac{2}{4}=\frac{1}{2}\)
Sslc Maths Chapter 3 Solutions Kerala Syllabus 10th Standard Question 5.
A regular hexagon formed by two overlapping equilateral triangles.
Answer:
Area of two equilateral triangles
Area of regular hexagon = \(\frac{3 \sqrt{3}}{2} a^{2}\)
Area of regular hexagon is half of the area of equilateral triangle.
∴ Probability of dot falling on the hexagon
Textbook Page No. 75
Class 10 Maths Chapter 3 Kerala Syllabus Question 1.
Raj ani has three necklaces and free pairs of earrings, of green, blue and red stones. In what all different ways can she wear them? What is the probability of her wearing the necklace and earrings of the same color? Of different colors?
Answer:
Possible ways of wearing:
| Necklace | Earring |
| Green | Green |
| Green | Blue |
| Green | Red |
| Blue | Green |
| Blue | Blue |
| Blue | Red |
| Red | Green |
| Red | Blue |
| Red | Red |
Raj ani can wear the ornaments in 9 different ways.
Probability of wearing necklace and earring of same colour = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)
Probability of wearing necklace and earring of different colour = \(\frac { 6 }{ 9 }\) = \(\frac { 2 }{ 3 }\)
Sslc Maths Chapter 3 Notes Question 2.
A box contains four slips numbered 1, 2, 3, 4 and another box contains two slips numbered 1,2. If one slip is taken from each, what is the probability of the sum of numbers being odd? What is the probability of the sum being even?
Answer:
| First box | Second box | Sum |
| 1 | 1 | 2 |
| 1 | 2 | 3 |
| 2 | 1 | 3 |
| 2 | 2 | 4 |
| 3 | 1 | 4 |
| 3 | 2 | 5 |
| 4 | 1 | 5 |
| 4 | 2 | 6 |
Probability of sum being odd = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
Probability of sum being even = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)
Kerala Syllabus 10th Standard Maths Chapter 3 Question 3.
A box contains four slips numbers 1, 2, 3, 4 and another contains three slips numbered 1, 2, 3. If one slip is taken from each, what is the probability of the product being odd? The probability of the product being even?
Answer:
| First box | Second box | Product |
| 1 | 1 | 1 |
| 1 | 2 | 2 |
| 1 | 3 | 3 |
| 2 | 1 | 2 |
| 2 | 2 | 4 |
| 2 | 3 | 6 |
| 3 | 1 | 3 |
| 3 . | 2 | 6 |
| 3 | 3 | 9 |
| 4 | 1 | 4 |
| 4 | 2 | 8 |
| 4 | 3 | 12 |
Probability of sum being odd = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3 }\)
Probability of sum being even = \(\frac { 8 }{ 12 }\) = \(\frac { 2 }{ 3 }\)
10 Th Maths Text Book Questions And Answers Question 4.
From all two-digit numbers with either digit 1, 2 or 3 one number is chosen.
i. What is the probability of both digits being the same?
ii. What is the probability of the sum of the digits being 4?
Answer:
Two digit numbers
11, 12, 13, 21, 22, 23, 31, 32, 33.
i. P (both digits being same) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)
ii. P(sum of digits being 4) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)
10th Class Maths Textbook Answers Question 5.
A game for two players. First, each has to decide whether he wants odd number or even number. Then both raise some fingers of one hand. If the sum is odd, the one who chose odd at the beginning wins; if it is even, the one who chose even wins. In this game, which is the better choice at the beginning, odd or even?
Answer:
The results in the order when the first player raises one finger and second player raises one finger and so on.
Textbook Page No. 78
Maths Chapter 3 Class 10 Kerala Syllabus 10th Standard Question 1.
In class 10A, there are 30 boys and 20 girls. In 10B, there are 15 boys and 25 girls. One student is to be selected from each class.
i. What is the probability of both being girls?
ii. What is the probability of both being boys?
iii. What is the probability of one boy and one girl?
iv. What is the probability of at least one boy?
Answer:
i. Total no. of possible pairs = 50 × 40 = 2000
No. of pairs in which both are girls = 20 × 25 = 500
Probability of both being girls = \(\frac { 500 }{ 2000 }\) = \(\frac { 1 }{ 4 }\)
ii No. of pairs in which both are boys = 30 × 15 = 450
Probability of both being boys = \(\frac { 450 }{ 2000 }\) = \(\frac { 9 }{ 40 }\)
iii. No. of pairs in which one is boy and one is girl = 2000 – (500 + 450) = 2000 – 950 = 1050
Probability of one being boy and one girl = \(\frac { 1050 }{ 2000 }\) = \(\frac { 21 }{ 40 }\)
iv. No. of pairs in which atleast one is boy = 1050 + 450 = 1500
Probability in which atleast one is boy = \(\frac { 1050 }{ 2000 }\) = \(\frac { 3 }{ 4 }\)
Question 2.
One is asked to say a two-digit number.
i. What is the probability of both digits being the same?
ii. What is the probability of the first digit being larger?
iii. What is the probability of the first digit being smaller?
Answer:
i. Probability of two digits being same = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)
(11, 22, 33, 44, 55, 66, 77, 88, 99)
ii. Numbers in which first digit is greater than second digit are 10, 20, 21, 30, 31, 32, 40, 41, 42, 43, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 65, 70, 71, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 96, 97, 98,
45 outcomes, Required Probability = \(\frac { 45 }{ 90 }\) = \(\frac { 1 }{ 2 }\)
iii. Numbers in which first digit is smaller than second digit are
12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89,
36 total no.of favourable outcomes = 36
Probability =
Question 3.
Each two-digit number is written on a paper slip and these are all put in a box. What is the probability that the product of the digits of a number drawn is a prime number? What if three-digit numbers are used instead?
Answer:
Total two-digit numbers = 90
Product of the digits of a number drawn is a prime number 12, 13, 15, 17, 21, 31, 51, 71.
(1 is not a prime number)
Total number whose product of the digits drawn is a prime number = 8
Probability of product of the digits drawn is a prime number = \(\frac { 8 }{ 90 }\) = \(\frac { 4 }{ 45 }\)
Total three-digit numbers = 900
Product of the digits of a number drawn is a prime number 112, 113, 115, 117, 121, 131, 151, 171.
Total number whose product of the digits drawn is a prime number = 8
Probability of product of the digits drawn is a prime number = \(\frac { 8 }{ 900 }\) = \(\frac { 2 }{ 225 }\)
Question 4.
Two dice with faces numbered from 1 to 6 are rolled together. What are the possible sums? Which of these sums has the maximum probability?
Answer:
Each dice has the following numbers:
Worksheet 1
Question 1.
How many odd numbers are there below 25
Answer:
12 odd numbers
Question 2.
How many prime numbers are there below 30?
Answer:
10 prime numbers
Question 3.
Find the number of two-digit even numbers?
Answer:
90 two-digit even numbers
Question 4.
How many two digits perfect squares are there?
Answer:
6
Question 5.
Write all three-digit numbers that can be written using the digits 3, 6, 8 without repeating the digits.
Answer:
368, 386, 683, 638, 836, 863
Question 6.
How many multiples of 7 are there in between 100 and 300?
Answer:
First multiple of 7 = 105
Last multiple of 7 = 294
Number of multiples
= 
Question 7.
There are 50 children in a class. Thirty of them are girls. There are 40 children in another class. 25 of them are boys. One student is taken from each class at random. What is the number of outcomes? How many outcomes contain both boys. How many outcomes contain both girls. How many outcomes have one boy and one girl?
Answer:
Total pairs = 50 × 40 = 2000
No. of pairs in which both are boys = 20 × 25 = 500
No. of pairs in which both are girls = 30 × 15 = 450
No. of pairs in which one is a boy and the other a girl = 30 × 25 + 20 × 15 = 750 + 300 = 1050
Worksheet 2
Question 8.
A fine dot is placed into the picture with-out looking into it. What is the probability of falling the dot in the small semicircle? What is the probability of falling the dot outside the small semicircle but inside the big semicircle?
Answer:
Let the radius of circle be r, then the radius of
Question 9.
P, Q, R are the midpoints of the sides of triangle ABC. Another triangle is drawn by joining these points. A fine dot is placed into the figure without looking into the picture. What is the probability of falling the dot in triangle PQR?
What is the probability of falling the dot outside the triangle?
Answer:
Area of each small triangle is 1/4th of area of large triangle ABC.
Probability of the dot falling on triangle PQR is = 1/4
Probability of the dot falling on small triangle is = 1/4
Probability of the dot falling inside the triangle PQR is \(\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\) less than that of outside the triangle.
Question 10.
What is the probability of occurring 53 Sundays in a leap year
Answer:
Leap year have 366 days.
That is 52 weeks and 2 days.
There two days are
Sunday – Monday, Monday – Tuesday, Tuesday – Wednesday, Wednesday – Thursday, Thursday – Friday, Friday – Saturday, Saturday – Sunday.
∴ Probability for to occur 53 Sunday is 2/7.
Question 11.
You can see a triangle inside a square. ABCD is a square. P, Q are the midpoints of C D and C B. A fine dot is placed into the figure without looking into the figure. What is the probability of falling the dot in triangle APQ?
Answer:
Question 12.
The value of 21, 22, 23… 250 are written in small papers and put it in the box. A paper is taken at random. What is the probability of getting a number having 4 in ones place? What is the probability of falling 8 in the one’s place?
Answer:
The 13 numbers having 4 in one’s place are 22, 26, 210 …, 250.
∴ Probability = \(\frac { 13 }{ 50 }\)
The 12 numbers having 8 in one’s place are 23, 27, 211 …, 247.
∴ Probability = \(\frac { 12 }{ 50 }\)
Worksheet 3
Question 13.
Numbers from 1 to 10 are written in small papers and placed in a box. One number is taken from the box at random. What is the probability of getting a prime number?
Answer:
Probability of getting a prime number = \(\frac { 4 }{ 10 }\) = \(\frac { 2 }{ 5 }\)
Question 14.
Two boxes contain tokens on which numbers 1, 2, 3, 4 are written One token is taken from each box. What is the probability of getting sum of the face numbers a prime number
Answer:
Pairs of numbers in tokens are
(1.1) , (1,2), (1,3), (1,4)
(2.1), (2, 2), (2, 3), (2, 4)
(3, 1), (3,2), (3; 3), (3,4)
(4, 1), (4, 2), (4, 3), (4,4).
Pairs getting sum as prime numbers
(1.1) , (1,2), (1,4), (2, 1), (2, 3), (3, 2), (3, 4), (4, 1), (4, 3)
Probability of getting sum of the face numbers a prime number = \(\frac { 9 }{ 16 }\)
Question 15.
One box contains 8 black balls and 12 white balls. Another box contains 9 black and 6 white balls. One ball is taken from each box at random. What is probability of getting both black? What is the probability of getting both white? What is the probability of getting one black and one white?
Answer:
Total pairs = 20 x 15 = 300
Number of pairs getting both black= 8 x 9 = 72
Probability of getting both black = \(\frac { 72 }{ 300 }\) = \(\frac { 6 }{ 25 }\)
Probability of getting one black and one white
Question 16.
In the figure, a triangle is drawn by joining the alternate vertices of a regular hexagon. A fine dot is placed into the figure at random. What is the probability of falling the dot in the triangle?
Answer:
Probability of the dot falling on triangle
Question 17.
What is the probability of occurring four Wednesdays in 23 consecutive days in a month?
Answer:
23 days = 3 weeks + 2 days
Wednesday comes on Tuesday + Wednesday, Wednesday + Thursday when two days are taken.
Total probabilities (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
∴ Total probabilities = 7
∴ Probabilities = 2/7
Question 18.
One is asked to say a two digit number. What is the probability of being the number not a perfect square? [Score : 3, Time : 3 Minutes]
Answer:
Total number of two digit number : 90 (1)
Total number of two digit perfect squares: 6 (1)
Number of two digit numbers which are not perfeet squares : 90 – 6 = 84,
Probability = \(\frac { 84 }{ 90 }\) = \(\frac { 42 }{ 45 }\) (1)
Question 19.
A bag contains 10 blue balls and 12 yellow balls. Another contains 15 blue balls and 7 yellow balls.
a. What is the probability of getting a yellow ball from the first bag?
b. What is the probability of getting a yellow ball from the second bag?
C. If all the balls are put in a single bag, what is the probability of getting a yellow ball from it? [Score : 4, Time : 4 Minutes]
Answer:
a. Total number of balls in the first bag = 10 + 12 =22, Number of yellow balls = 12
Probability of getting a yellow ball = \(\frac { 12 }{ 22 }\) = \(\frac { 6 }{ 11 }\) (1)
b. Total number of balls in the second bag = 15 + 7 = 22, Number of a yellow ball = 7
Probability of getting a yellow ball = \(\frac { 7 }{ 22 }\) (1)
c. Total number of balls = 22 + 22 = 44
Number of yellow balls = 12 + 7 = 19 (1)
Probability of getting a yellow ball = \(\frac { 19 }{ 44 }\) (1)
Question 20.
A regular hexagon is drawn with its vertices on a circle. Without looking into the I picture, if one put dot in that picture, what is the probability of being the dot not in the regular hexagon? [Score: 4, Time: 3 Minutes] .
Answer:
Area of circle = πr2
Question 21.
In the figure, all the four shaded semicircles have same area. If we put a dot in the figure without looking into it, what is the probability of being the dot in the shaded semicircles? [Score : 3, Time: 5 Minutes]
Answer:
If the radius of the shaded semicircle is r,
Question 22.
What is the probability of getting 5 Sundays in December in a calendar year? [Score : 3, Time : 5 Minutes]
Answer:
There are 31 days in December. That means 4 full weeks and 3 days.
The probable three days are as shown below.
Question 23.
Two semicircles are drawn in a square as shown. If we put a dot in the figure, without looking into it, what is the probability of being the dot in the shaded region? [Score: 3, Time: 5 Minutes]
Answer:
If a is the side of a square,
Short Answer Type Questions (Score 2)
Question 24.
A bag contains 6 red balls, 8 green balls, and 8 white balls. One ball is drawn at random from the bag, find the probability of getting
i. A white or green ball
ii. Neither green ball nor a red ball.
Answer:
Red balls = 6
Green balls = 8
White balls = 8
Total number of balls = 6 + 8 + 8 = 22
a. The probability of getting a white or green ball = 16/20
b. The probability of getting neither green balls nor a red ball = 8/20
Question 25.
20 cards numbered 1, 2, 3, 4, ….19, 20 are put in a box. One boy draws a card from the box. Find the probability that the number on the card is:
i. Prime
ii. Divisible by 3
Answer:
Total number of outcomes = 20
a. Prime numbers from 1 to 17 are 2, 3, 5, 7, 11,13, 17, 19.
Number of outcomes = 8
The probability that the card drawn is prime number = 8/20
b. Numbers are divisible by 3 are 3, 6, 9, 12, 15, 18.
Number of outcomes = 6
The probability that the card drawn is divisible by 3 = 6/20
Question 26.
In a bag, there were 3 white balls and 5 black balls. From this one ball is taken, then
a. What is the probability of being blackball?
b. What is the probability of being white ball?
Answer:
Total no. of balls is 8 and in that 5 of then are black balls.
a. Probability of getting black balls = 5/8
b. 3 of the balls were white, so the probability of getting white balls = 3/8
Question 27.
a. How many two-digit natural numbers are there in all?
b. If we choose one number from the two-digit numbers, what is the probability that the sum of digits of that number will be 10?
Answer:
a. Number of two-digit natural numbers = 90
b. Numbers whose sum of digits will be 10 is 19, 28, 37, 46, 55, 64, 73, 82, 91
Probability that the sum of digits of that number will be 10 = \(\frac { 9 }{ 90 }\) = \(\frac { 1 }{ 10 }\)
Short Answer Type Questions (Score 3)
Question 28.
In selecting a two-digit number up to 50.
a. What is the probability of the digit in the ten’s place to be larger than the digit in the one’s place?
b. What is the probability of the digit in the tens place to be smaller than the digit in the one’s place?
Answer:
Total numbers of two-digit numbers up to 50 = 41
a. Number of numbers with the digit in the tens place to be larger than the digit in the units place = 11
Probability that the number with tens place digit is larger than the digit in the unit place = 11/41
b. Probability that the number with the tenth place digit is smaller than the digit in the unit place = 26/41
Question 29.
A black dice and a white dice are thrown at the same time,
a. Write all the possible outcomes.
b. What is the probability that the sum of the two numbers is to be 8.
c. What is the probability that being the same number to be on both dice?
Answer:
a. Total outcomes
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) ‘
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6) .
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4.1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5.1) , (5,2), (5,3), (5,4), (5,5), (5,6)
(6.1), (6,2), (6,3), (6,4), (6,5), (6,6)
Total = 36
b. Sum of two numbers is to be 8 = 5
(2,6), (3,5), (4,4), (5,3), (6,2),
Probability = 5/36
c. Probability of being same number =(1,1),(2,2), (3,3), (4,4), (5,5), (6,6),
Total = 6
Probability = \(\frac { 6 }{ 36 }\) = \(\frac { 1 }{ 6 }\)
Question 30.
A box contains 400 electronic toy cars. Among them, 12 are defective. One toy is taken out at random. What is the probability that
a. It is a defective toy.
b. It is a non-defective toy.
Answer:
Total number of cars = 400
Number of defective toys = 12
Number of non-defective toys = 400 – 12 = 388
a. Probability of getting a defective toy = \(\frac { 12 }{ 400 }\) = \(\frac { 3 }{ 100 }\)
b. Probability of getting a non-defective toy = \(\frac { 388 }{ 400 }\) = \(\frac { 97 }{ 100 }\)
Long Answer Type Questions (Score 4)
Question 31.
Natural numbers from 1 to 30 are written on paper slips and kept in a box. If one slip is taken from the box,
a. What is the probability of this number to be even?
b What is the probability of this number to be a multiple of 3?
c. What is the probability of this number to be a multiple of 3 and 5?
d. What is the probability that this number be a natural number?
Answer:
Question 32.
There is one spot at one side of the cube, two on another side, three on the third side and so on. There are spots on all the six faces; in this order. Another cube which is marked in the same way is taken.
a. If both the cubes are thrown, what is the probability that the total number of spots on the upper faces is 6?
b. What is the probability that the sum of the spots on the upper faces is 9?
c. What is the probability that the sum of the spots be one?
d. What is the probability that the sum of the spots be a prime number?
Answer:
Question 33.
In a pack of 52 cards, half are red and the rest are black. There are 4 Suits of 13 cards each and having the signs ‘hearts’, ‘spade’, ‘clubs’ and ‘diamond’. If a card is picked from this pack.
a. What is the probability of it being black?
b. Whatistheprobabilityofitbeingaspade?
c. What is the probability of it being a spade or a diamond?
Answer:
a. Total number of cards = 52
Number of black cards = 26
Probability of the picked card being black = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)
b. No. of spade cards = 13
Probability of a picked card being spade = \(\frac { 12 }{ 52 }\) = \(\frac { 1 }{ 4 }\)
c. No. of cards spade or diamond =13 + 3 = 26
Probability of a picked card being spade or diamond = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)
Long Answer Type Questions (Score 5)
Question 34.
A man is asked to say a 3 digit number,
a. What is the probability that the first and last digits be equal?
b. What is the probability that the last two digits be ‘O’?
c. What is the probability that the last digits being greater than the first?
Answer:
Total 3 digit numbers = 900
a. Numbers with first and last digits are equal
101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393
There are 90 such numbers.
∴ Probability = \(\frac { 90 }{ 900 }\) = \(\frac { 1 }{ 10 }\)
b. The numbers with last two digits zero are 100, 200, 300,……… 900 total numbers 9.
∴ Probability = \(\frac { 9 }{ 900 }\) = \(\frac { 1 }{ 100 }\)
c. The numbers with the last digit greater than the first digit is 36
∴ probability = \(\frac { 36 }{ 90 }\) = \(\frac { 2 }{ 5 }\)
Question 35.
There are 10 black pearls and 5 white pearls in the box A. There are 8 black pearls and 7 white pearls in box B.
a. Which box has more probability of be ing the pearls black, when a pearl from each of the boxes is taken?
b. What is the probability to get a white pearl from the box A?
c. What is the probability to get a black pearl from box B?
d. If all the pearls in the box B is dropped in the box A, then what is the probability to get a black pearl from it?
Answer:
a. Box A because it has more black pearl
Question 36.
Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number of the card is
a. an even number
b. a number less than 16
c. a number which is a perfect square
d. a prime number less than 25
Answer:
When probabilities are explained in terms of numbers it is the ratio of number of favorable outcomes to the total number of outcomes.
The least probability will be 0 and the highest will be 1. The probability will be a number between 0 and 1.
If an event can be completed in ‘m’ ways and another event can be completed in ‘n’ ways then both the events can be completed one after the other in m x n ways.
Students can Download Chemistry Chapter 5 Compounds of Non-Metals Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Chemistry Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
You can Download Nomenclature of Organic Compounds and Isomerism Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemistry Solutions Chapter 6 help you to revise complete Syllabus and score more marks in your examinations.
Text Book Page No: 97
Nomenclature Of Organic Compounds And Isomerism Class 10 Question 1.
The given structures indicate the valency of carbon. Imagine that hydrogen atoms are added to these structures. Complete the given structures.
Answer:
Text Book Page No: 98
Sslc Chemistry Chapter 6 Kerala Syllabus Question 2.
Complete the table (6.2) given below.
Answer:
Sslc Chemistry Chapter 6 Solutions Kerala Syllabus Question 3.
What is the relationship between the number of atoms of carbon and hydrogen in alkane?
Answer:
Number of hydrogen atoms are equal to adding 2 with twice of number of carbon atoms.
Chemistry Chapter 6 Class 10 Kerala Syllabus Question 4.
If an alkane contains ‘n’ carbon at-oms, how many hydrogen atoms will be there?
Answer:
2n + 2
Chemistry Class 10 Chapter 6 Kerala Syllabus Question 5.
If so, can you deduce a general formula for alkanes?
Answer:
CnH2n+2
Text Book Page No: 99
Nomenclature Of Organic Compounds And Isomerism Kerala Syllabus Class 10 Question 6.
What is the difference between the number of carbon atoms and hydrogen atoms in CH4 and C2H6 ?
Answer:
One carbon and two hydrogen atoms (CH2)
Sslc Chemistry Chapter 6 Notes Kerala Syllabus Question 7.
Is the difference same in the case of C2H6 and C3H8.
Answer:
Yes
Nomenclature Of Organic Compounds Class 10 State Syllabus Question 8.
What is the difference between the molecular formulae of any two successive alkanes?
Answer:
CH2
Class 10 Chemistry Chapter 6 Question 9.
Complete the table given below (Table 6.3).
Answer:
Text Book Page No: 100
Class 10 Chemistry Chapter 6 Kerala Syllabus Question 10.
Analyze Table 6.3 and find the number of hydrogen atoms in an alkene with ‘n’ carbon atoms.
Answer:
2n
Sslc Chemistry Chapter 6 Questions Kerala Syllabus Question 11.
If so, can a general formula of alkenes be deduced? Try to write it.
Answer:
CnH2n
Sslc Chemistry Chapter 6 Notes Pdf Kerala Syllabus Question 12.
Complete the table 6.4.
Answer:
Chemistry Class 10 Chapter 6 Kerala Syllabus Question 13.
Analyse Table 6.4 and find out how many hydrogen atoms would be present in an alkyne within carbon atoms.
Answer:
2n – 2
Sslc Organic Chemistry Kerala Syllabus Question 14.
If so, can a general formula of alkynes be deduced? Try to write the general formula of alkynes?
Answer:
CnH2n-2
Sslc Chemistry Nomenclature Of Organic Compounds Kerala Syllabus Question 15.
Check whether the alkynes given in the above table are members of a homologous series.
Answer:
Yes
Text Book Page No: 102
Chemistry Chapter 6 Class 10 Kerala Syllabus Question 16.
Write the IUPAC names of all the alkanes in Table 6.2.
Answer:
CH4 — Methane
C2H6 — Ethane
C3H8 — Propane
C4H10 — Butane
C5H12 — Pentane
Text Book Page No: 103
Class 10 Chemistry Chapter 6 Notes Kerala Syllabus Question 17.
number for the carbon atom carrying the branch?
b. Number of carbon atoms in the main chain.
c. Word root
d. Suffix
e. Name of the alkyl radical coming as branch.
f. Position of the branch.
Answer:
a. 2
b. 4
c. But
d. -ane
e. methyl
f. 2
Kerala Syllabus 10th Standard Chemistry Chapter 6 Questions 18.
Complete the table 6.3.
Answer:
Text Book Page No: 104
Hsslive Guru 10th Chemistry Kerala Syllabus Question 19.
Some structural formula are given below. Name them.
Answer:
Text Book Page No: 105
10th Class Chemistry Chapter 6 Malayalam Kerala Syllabus Question 20.
Number the carbon atoms in the main chain of the above compound given above. Put an ✓ against the correct position numbers of the branches.
Answer:
a. What is the IUPAC name?
Answer:
2,4 – Dimethyl hexane.
b. Which is the second branch?
Answer:
CH3
c. When does this branch get the lowest number? Put a ✓ mark against the correct one.
d. While numbering from left to right
e. While numbering from right to left
Answer:
While numbering from right to left ✓
Class 10 Organic Chemistry Questions Kerala Syllabus Question 21.
Write the IUPAC name of the compound given below.
Answer:
2, 3, 6 – trimethylheptane
Hsslive Guru Chemistry Kerala Syllabus Question 22.
See the given compound
Answer:
No. of branches in this compound: 2
Names of the branches: methyl
Position numbers of branches: 2, 2
IUPAC name: 2, 2-dimethylbutane
Text Book Page No: 106
Hss Live Guru 10th Chemistry Kerala Syllabus Question 23.
How can the structure of 2, 3- dimethylbutane be written?
Answer:
a. How many carbon atoms are present in its main chain?
Answer:
4
b. Let us represent the main chain.
Answer:
C – C – C- C
c. Which are the branches?
Answer:
Methyl groups – 2 in number
d. What are the positions?
Answer:
2, 3
Question 24.
Complete the table given below. (Table 6.4).
Answer:
Text Book Page No: 107
Question 25.
Complete the table 6.5.
Answer:
Question 26.
Can you write the structural formula of the compound C2H4 ?
Answer:
CH2 = CH2
Question 27.
If so, what will be structural formula of But – 2- ene?
Answer:
CH3 – CH = CH – CH3
Text Book Page No: 108
Question 28.
Which is IUPAC name of the com-pound CH3 – CH2 – CH = CH – CH3? Tick (✓) the right one.
Pent – 3 – ene
Pent – 2 – ene
Answer:
Pent – 2 – ene ✓
Question 29.
CH3 – C = C – CH3 But – 2 – yne
How many hydrocarbons can be written by changing the position of triple bond in this compound? Try to write their IUPAC names also.
Answer:
CH3 = C – C – CH3 But – 1 – yne
CH3 – C = C – CH3 But – 2 – yne
CH3 – C – C = CH3 But – 1 – yne
Text Book Page No: 109
Question 30.
Try to write down the molecular formula of Benzene.
Answer:
C6H6
Text Book Page No: 110
Question 31.
See the compound given below.
CH3 – CH2 – CH2 – OH
Write its molecular formula.
Answer:
C3H8O
Question 32.
Write the molecular formula.
Answer:
C3H8O
Question 33.
Then try to write the IUPAC name of the second compound.
Answer:
Propan – 2 – ol
Text Book Page No: 112
Question 34.
Complete the table 6.6.
Answer:
Question 35.
See the two compounds given below.
a. What are the similarities between these two compounds?
Answer:
No change in the molecular formula and functional group
b. Molecular formula : C3HgO
c. Functional group: OH
d. What is the difference between them?
Answer:
Position is different
Question 36.
Examine the two compounds given below.
a. Try to write the molecular formula. Can’t you write the IUPAC names of these compounds?
Answer:
Molecular formula: C4H10
IUPAC Name: Butane, 2 – Methyl Propane.
b. What is the difference between them?
Answer:
The structure of carbon chains differs from each other.
Text Book Page No: 113
Question 37.
What are the functional groups in CH3 – CH2 – OH and CH3 – O – CH3 ?
Answer:
-OH, CH3-O-
Question 38.
Try to write down their molecular formula
Answer:
C2H6O
Question 39.
Are they isomers?
Answer:
Yes. They show functional isomerism
Text Book Page No: 114
Organic Chemistry Chapter 2 Problem 15s Question 40. Try to write down all position isomers of the compounds CH3 -CH2 – CH2 – Cl.
Answer:
Question 41.
Examine the compounds given below and find out the isomeric pairs. To which type do they belong?
Answer:
Pairs of isomers :
1. CH3 – CH2 -CH2 -CH2 -CH3,
: Chain isomerism.
2. CH3 -CH2 -CH2 -CH2 -OH,
: Position isomerism.
5. CH3 -CH2 -CH2 -OH
6. CH3 -CH2 -O -CH3
: Functional Isomerism
Organic Chemistry Chapter 2 Problem 5s Question 42. How many position isomers are possible for CH3 -CH2 – CH2 – CH2 – CH2 – OH.
Answer:
3
a. Write the structure and IUPAC names of its functional isomers?
Answer:
Question 43.
How many position isomers are possible for CH3 – CH2 – CH2 – CH2- CH2 – CH3 ? Write them down.
Answer:
5
i. CH3 -CH2 -CH2 -CH2 -CH2 -CH3
Organic Chemistry Chapter 1 Problem 6s Question 44.
The structural formula of various compounds are given. Tabulate them into different pairs of isomers. Write down their IUPAC names also.
Answer:
Chain isomers
1. CH3 -CH2 -CH2 -CH2 -CH2 -CH3
Hexane
Functional group isomers
2. CH3 – CH2 – O – CH3
Methoxy ethane
4. CH3 – CH2 – CH2 – OH
Propan -1 -ol
Let Us Assess
SSLC Chemistry Chapter 6 Question 1. Mark the main chain in the compounds given below.
Answer:
Question 2.
See how the carbon chains are numbered. Correct the wrong ones.
Answer:
Question 3.
Write down the IUPAC names of the given compounds.
Answer:
Question 4.
Write down the structural formulae of compounds given below.
a.2, 2 – Dimethylhexane
b. But – 2 – ene.
Answer:
b. CH3 – CH = CH – CH3
Question 5.
Write down the structural formula of compound C2H10. Write down the structural formula of one of its isomers which is an alicyclic compound.
Answer:
Extended Activities
Question 1.
Given below are certain hints about a hydrocarbon.
1. The molecular formula is C5H10.
2. Has a methyl radical as branch.
a. Write the structural formula of any two possible isomers of their compound.
b. Write their IUPAC names.
Answer:
b. i. 3-Methyl but – 1 – ene
ii. 2 – Methyl but – 2 – ene
iii. 2 – Methyl but -1 – ene
Organic Chemistry Chapter 2 Problem 14s Question 2. Write down the IUPAC names of the compounds given below.
d. CH3 – CH2 – CH2 – C = CH
e. CH3 – CH2 – CH2 – OH
f. CH3 – CH2 – O – CH3
g. CH3 – CH2 – CH2 – CH2 – CH2 – COOH
Answer:
a. 4-Methylhept-l-ene
b. 3, 5 – diethylheptane
c. 2, 4-dimethyl hexane
d. Pent – l – yne
e. Propane – 1 – ol
f. Methoxyethane
g. Hexanoic acid
Question 3.
Write the structural formulae of all possible isomers of the compound with molecular formula C4H10. Identify the different isomer pairs from them and find the type of isomerism to which each pair belongs.
Answer:
a. CH3 – CH2 – CH2 – CH2 – OH
c. CH3 – O – CH2 – CH2 – CH3
d. CH3 – CH2 – O – CH2 – CH3
e. CH3 – CH – CH2 – OH
Question 4.
Find three pairs of isomers from the compounds given below. Identify the type of isomerism to which each pair belongs.
a. Propan -1 – ol
b. 2, 2, 3, 3 – Tetra methyl butane
c. Octane
d. Propan – 2 – ol
e. Methoxyethane
Answer:
1. Propan – l – ol
Propan – 2 – ol (Position isomerism)
2. Octane
2, 2, 3, 3 -Tetramethyl butane (Chain Isomerism)
3. Propan – l – ol Methoxyethane (Functional Isomerism)
Question 5.
The structural formulae of two organic compounds are given.
i. CH3 – O – CH2 – CH3
ii. CH3 – CH2 – CH2 – OH
a. What are the IUPAC name of these compounds?
b. Write one similarity and one difference between these two compounds,
c. What is this phenomenon known as?
Answer:
a. i. Methoxy ethane ii. Propan – l – ol
b. Similarity: Molecular formula is same
Difference: Functional group is different
c. Functional group isomerism.
Question 6.
Write down the structural formulae of the following compounds.
a. Cyclopentane
b. Cyclobutene
Answer:
Question 1.
Write the structure, Root word, Suffix and IUPAC name of
CH4, C3H6, C3H8 ,C4H10, C5H12, C6H14, C2H4, C3H6, C4H8, C5H10, C6H12, C2H2, C3H4, C4H6, C5H8, C6H10.
Question 2.
Write down the IUPAC name of the compound
CH3
CH3 – CH2 – CH – CH3
a. How many Carbon sare there in the main Chain?
b. Position number of the branch,
c. Name of the branch,
d. IUPAC Name.
Answer:
a. Four
b. Two
c. Methyl
d. 2 – Methyl Butane
Question 3.
Write down the IUPAC name of the following structures by finding the branch no, Branch name and no.of carbons in the main chain.
Answer:
Question 4.
Draw the structure of the following compounds.
a 2, 2 – Dimethyl pentane
b. 2, 4 – Dimethyl octane
Answer:
a. 2, 2 – Dimethyl pentane
b. 2, 4 – Dimethyl octane
Question 5.
Complete the table.
Answer:
Question 6.
Find out the pairs exhibiting same type of Isomerism from those given below.
Answer:
Evaluation Questions
Question 7.
Write down the structure of C4H10 then write possible isomeric forms of the same.
Answer:
C4H10 → CH3 – CH2 – CH2 – CH3 Butane Chain Isomerism
Question 8.
Write down all the position isomers of CH3 – CH2 – CH2 – CH2 – CH2 – OH
Answer:
CH3 – CH2 – CH2 – CH2 – CH2 – OH – pent – l – ol
Question 9.
Write down the IUPAC name of the compounds given above and then write the name of the isomerism exhibited by it.
Answer:
Functional Isomerism
CH3 – CH2 – CH2 – O – CH2 – CH2 – CH3 – Propoxy propane
Question 10.
Write down all the possible structures of C5H10. Name them, what type of isomerism are they exhibiting?
Answer:
Position Isomerism
Question 11.
Write down all the position isomers and functional group isomers of the compound. Name all of the compounds.
CH3 – CH2 – CH2 – O – CH2 – CH2 – CH3
Answer:
Question 12.
Complete the table.
Answer:
a. C7H16
b. CH3 – CH2 – CH2 – CH2 – CH2 – CH2 – CH3
c. Hexane
d. CH3 – CH2 – CH2 – CH2 – CH2 – CH3
Question 13.
The structure of a hydrocarbon having 5 carbon atoms is given below.
a. Complete the structure by adding hydrogen atoms.
b. Write the molecular formula of the compound
c. Write a possible chain isomer of the compound
d. Write its IUPAC name.
Answer:
b. C5H12
d. Pentane or 2, 2 – dimethyl propane
Question 14.
The parts of the structure of an organic compound are given below.
a. Write a completed structure of an organic compound by connecting all the groups given above.
b. Write the IUPAC name of the compound
c. Write the structure of a position isomer of the compound.
Answer:
Question 15.
A hydrocarbon chain with molecular C7H16 is numbered in four different ways.
a. Which of the above is numbered correctly?
b. What is the name of alkyl radical found as the branch?
c. Write the IUPAC name of the compound.
Answer:
a. Cb. Methyl
c. 3 – methyl hexane
Question 16.
Analyze the following structural formula and answer the questions.
a. How many carbon atoms are there in the longest chain?
b. What are the positions of the branches?
c. Write the IUPAC name of the compound.
Answer:
a. 6
b. 2, 4
c. 2,4 dimethyl hexane
Question 17.
The features of an organic compound are given.
1. It’s an alkane.
2. There are 7 carbon atoms in the longest chain.
3. There is a methyl radical on the 3rd carbon and ethyl radical on the 4th carbon.
a. Write the structural formula of the compound.
b. Write the IUPAC name of the compound
Answer:
b. 4- ethyl – 3 – methyl heptane
Question 18.
The functional groups of two compounds with same molecular formula are given. Analyze it and complete the boxes.
ii. What is the name of this
Answer:
i. (a) CH3 – CH2 – CO – CH3 or CH3 – CO – CH2 – CH3
(b) Butan – 2 – one
(c) CH3-CH2-CH2CHO
ii. Functional isomerism
Question 19.
Analyse the given organic compounds and answer the following questions.
i. CH3 – CH2 – CH2 – CH3
iii. CH3 – CO – CH3
iv. CH3 – CH2 – CH2 – CH2 – OH
v. CH3 – CH2 – CH2 – CHO
a. Identify the isomer pairs. Write the type of isomerism observed in them.
b. Write the structure of the isomer of compound (iii). Write the IUPAC name.
Answer:
& CH3 – CH2 – CH2 – CH2 – OH
Position isomerism
b. CH3 – CH2 – CHO, Propanal
Question 20.
Complete the table.
Answer:
a. Hydroxyl
b. Amino
c. Butane – l – amino
d. CH3 – CH2 – O – CH2 – CH2 – CH3
Question 21.
a, b, c, are the different isomers of C4H10O.
i. identify a, b, c
ii. identify a pair of functional isomers among them.
Answer:
Question 22.
a. Write the IUPAC name of this compound.
b. Write the molecular formula.
c. Write the structural formula of its isomer
d. Identify the type of isomerism in the above.
Answer:
a. 2 – Methyl propane
b. C4 – H10
c. CH3 – CH2 – CH2 – CH3
d. Chain isomerism
Question 23.
The structural formula of two organic compounds are given below.
i. CH3 – CH2 – CH2 – OH
ii. CH3 – O – CH2 – CH3
a. What is the similarity between these two? What is the phenomenon known as?
b. Is their chemical properties the same? What is the reason?
c. Write the functional groups of these two compounds.
Answer:
a. Same molecular formula, Isomerism.
b. No, In these functional groups are different
c. Hydroxyl, Alkoxy
Question 24.
Analyze the IUPAC names given to the following organic compounds and correct them, if incorrect.
iii. CH3 – CH2 – CH = CH – CH3 : Pent – 3 -ene
Answer:
i. 3 – Methylhexane
ii. Pent – 2 – ene
Question 25.
An organic compound is given below. CH3 – CH2 – CH2 – CH = CH – CH3
Pick out suitable statements for the given compound from below.
a. It’s a saturated compound b The general formula is CnH2n
c. It’s an alkene
d. IUPAC name is hex – 4 – ene
e. Similar with the molecular formula of cyclohexane.
f. IUPAC name is hex – 2 – ene.
Answer:
b.The general formula is CnH2n
c. It’s an alkene
e. Similar with the molecular formula of cyclohexane.
f. IUPAC name hex – 2 ene.
Question 26.
CH3 – CH2 – C = C – CH2 – CH3
a. Write the IUPAC name of this organic compound.
b. Write the structure of any two isomers of this compound.
Answer:
a. Hex-3-yne
b. CH3 – C = C – CH2 – CH2 – CH3
CH3 – CH2 – C = C – CH2 – CH3
Question 27.
Some details about the structure of an organic compound are given below.
i. There are 5 carbon atoms in the main chain.
ii. There is a double bond between 1st and 2nd carbon atoms.
iii. There is a methyl radical on the 3rd carbon as a branch.
a. Write the structural formula of this compound.
b. Identify the category of organic compounds.
Answer:
a. CH2 = CH – CH – CH2 – CH3
CH3
b. Alkene
Question 28.
Match the following.
Answer:
Question 29.
CH3 – CH2 – CH2 – CH2 – OH
a. Write the structure of a chain isomer of this compound.
b. Write the IUPAC name of a position isomer of the given compound.
c. Write the structure of the functional isomer of the given compound. What is the name of the functional group in the isomer.
Answer:
b. Butan – 2 – ol
c. CH3 – CH2 – O – CH2 – CH3
CH3 – O – CH2 – CH2 – CH3
Alkoxy group
Question 30.
CH3 – CH2 – CH = CH2
a. Write the IUPAC name of this organic compound.
b. Give the structure of the alicyclic compound having the same molecular formula. Write its IUPAC name.
Answer:
a. But – 2- ene
b. Cyclobutane
Very Short Answer Type Questions (Score 1)
Question 31.
Examine the organic compound given below
a. How many carbons atoms are there in the longest carbon chain?
b. Which are the substituents?
c. How is the longest carbon chain numbered in this? (Left to right or right to left)
d. Give the IUPAC name of the compound.
Answer:
a. 8
b. Methyl, Ethyl
c. Right to left
d. 3 – Ethyle – 6 – Methyloctane
Question 32.
The names of some organic compounds are given below. Identify the wrong ones and correct them.
a Butan – 3 – ol
b. Hexanoic acid
c. 3 – ethyl – 2 – methyl pentane
d. 2, 2, 3 – methyl hexane
e. Methanol
Answer:
a. In this Butane – 3 – ol is incorrect. Butane – 2 – 0l is correct name.
d. 2, 2, 3 – Trimethylhexane
Question 33.
The name of an organic compound was written as 5- Methyl hexane.
a. Is this name correct?
b. If not draw its structural formula and give reason.
c. Give the correct IUPAC name.
Answer:
a. Name of the compound is incorrect.
In this compound, the longest carbon chain can numbered from right to left. Therefore 2 is the position value of branched carbon.
c. 2 – Methylhexane
Question 34.
Ether is a functional isomer of alcohol.
a. Which is the alcohol that has no functional isomer?
b. Write its IUPAC name.
Answer:
a. CH3 – OH
b. Methanol
Question 35.
From the below organic compounds, identify the pairs of isomers.
a. CH3 – CH2 – CH = CH2
c. CH3 – C = CH
Answer:
(a), (d) are isomers,
(b), (c) are isomers.
Short Answer Type Questions (Score 2)
Question 36.
Some hints are given below about an organic compound:
1. It contains 4 carbons.
2. It contains 1 oxygen atoms and 10 hydrogen atoms.
a. Write the structural formulas of the possible compounds using the above hints.
b. Write the IUPAC names of each.
Answer:
Question 37.
The names of some organic compounds are given below. Identify their functional group.
a. 2 – Methoxybutane
b. Heptane – 2 – ol
c. Propan – l – amine d Propanone
e. Pentanal
Answer:
a. Methoxy (CH3 – O -): Ether
b. Hydroxine (-OH): Alcohol
c. Amino (-NH2): Amines
d. Keto (-CO-): Ketones
e. Aldehyde (-CHO): Aldehydes
i. 1 – Methoxypropane
ii. Ethoxyethane
Question 38.
Look at the compounds given in the bracket [ Methanoic acid, Propane, Chlorobutane, Butyne, Methoxy-methane]
a. Which is the hydrocarbon that doesn’t have a chain isomer?
b. Which one has position isomer?
c. Which one has only one carbon atom?
d. Which belongs to the family of ether?
Answer:
a. Propane.
b. Chloroquine.
c. Mehtanoic acid
d. Methoxymethane
Short Answer Type Questions (Score 3)
Question 39.
a. Write all possible organic compounds with formula C3H8O. Give their IUPAC names.
b. Find out the isomers from this. Which type of isomerism do each exhibit?
Answer:
Methoxyethane
b. (i), (ii) are isomers.
Position isomerism
(i) , (iii) are isomers.
Functional isomerism
(ii), (iii) are isomers.
Functional isomerism
Question 40.
Give numbers in the correct order to the carbon chains given below:
Answer:
Answer:
Question 41.
Write the names of the following groups.
a. CH2 – CH3
b. -CO
c. -O – CH2 – CH3
d. -Cl
e. -COOH
f. -OH
Answer:
a. Ethane
b. Keto
c. Ethoxy
d. Chloro
e. Carboxyl
f. Hydroxyl
Long Answer Type Questions (Score 4)
Question 42.
Write the IUPAC names of the following organic compounds:
b. CH3 – CH2 – CH2 – CH2 – CH2 – OH
c. CH3 – COOH
Question 43.
Draw the structural formula of the following compounds.
a. Butanone
b. Methoxyethane
c. 2-Chloropropane
d. 2, 2, 3, 3- Tetramethylbutane
e. Ethanal
f. Ethyne
Answer:
Question 44.
Complete the following table.
Answer:
a. CH3 – CH2 – CH2 – CH2 – CH3 b.
d. Pentane
e. 2 – Methylbutane
f. 2, 2 – Dimethylpropane
g. CH3 – CH2 – CHO
h. CH3 – CO – CH3
i. Propanol
j.propanone
Question 45.
Match the molecular formula with the IUPAC names of the following compounds.
Pentanal, 1 – Methoxypentane, Methanoic acid, 3 – Ethyl – 3 – Propyl Pentane, 1 – Methoxybutane, 3, 3-Diethylhexane, Hex – 3 -yne.
Answer:
You can Download Magnetic Effect of Electric Current Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Physics Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.
Textbook Page No. 33
Observe the depiction of magnetic fields of two types of magnets.
Sslc Physics Chapter 2 Notes Kerala Syllabus Question 1.
The magnetic field of which magnets are depicted?
Answer:
Kerala Syllabus 10th Standard Physics Chapter 2 Question 2.
How can you identify the direction of the magnetic fields?
Answer:
The presence of the magnetic field and the polarity can be understood using a magnetic compass. The direction of the magnetic fields can also determine using the Right hand thumb rule of James Clark Maxwell. Magnetic field lines are continuous, forming closed loops without beginning or end. They go from the north pole to the south pole.
Sslc Physics Chapter 2 Questions And Answers Kerala Syllabus Question 3.
How can you find out the polarity of these magnets using a magnetic compass?
Answer:
If the North end of the compass needle is j pointing toward your magnet, if it attract you have found the South pole of your j magnet. Rotate the other side of your magnet toward the compass; the South end of the compass needle will now be pointing directly to the North pole of your magnet.
Sslc Physics Chapter 2 Questions And Answers Pdf Kerala Syllabus Question 4.
What are the main differences between the magnets in the picture?
Answer:
The first figure indicates the magnetic field lines of the bar magnet, second figure indites the magnetic field lines of a electromagnet.The magnetic strength magnetism of an electromagnet is temporary while magnetic strength of a bar magnet is permanent.
Textbook Page No. 35
Class 10 Physics Chapter 2 Questions And Answers Kerala Syllabus Question 5.
Arrange a circuit above a pivoted magnetic needle in such a way that the part AB of the conductor is parallel and close to the magnetic needle, as shown in Fig 2.3 (a).
Switch on the circuit. Observe the direction in which the North Pole(N) of the magnetic needle deflects and complete the Table 2.1.
→ When the direction of electric current is from A to B, what will be the direction of the electron flow through it?
Answer:
From B to A
→ Repeat the experiment after reversing the current and record your observations in the table.
Answer:
→ Repeat the experiment keeping the conductor below the magnetic needle and record the observations in the table.
Answer:
Sslc Physics Chapter 2 Questions Kerala Syllabus Question 6.
Find out the answer for the following based on the experiment.
→ What might be the reason for the deflection of the magnetic needle?
Answer:
A magnetic field is created which repels the magnetic needle. It is due to the current flow through the conductor. A magnetic field is created around a conductor when current flows through it.
→ Does the deflection depend on the direction of current?
Answer:
Yes
Textbook Page No. 36
Sslc Physics Chapter 2 Notes Pdf Kerala Syllabus Question 7.
Is the direction of current in the circuit between A and B from A to B or from B to A?
Answer:
From A to B
→ Examine whether the direction of magnetic field lines around X are in the . clockwise or anticlockwise direction by observing the North Pole of the magnetic compass.
Answer:
Anticlockwise direction
Physics Chapter 2 Class 10 Kerala Syllabus Question 8.
Compare the directions of the fingers of the right hand encircling the conductor and the magnetic field lines.
Answer:
According to Right Hand Thumb Rule of James Clark Maxwell holding a current carrying conductor with the right hand in such a way, that the thumb points in the direction of the current. The direction in which the other fingers encircle the conductor gives the direction of the magnetic field.
Textbook Page No. 37
Class 10 Physics Chapter 2 Kerala Syllabus Question 9.
Are the magnetic field lines inside the coil seen in the same direction?
Answer:
No
Class 10 Physics Chapter 2 Solutions Kerala Syllabus Question 10.
What is the difference observed in the direction of magnetic field lines on reversing the current through the solenoid?
Answer:
Direction of magnetic lines reversed.
Physics Class 10 Chapter 2 Kerala Syllabus Question 11.
When the coil is viewed in such a way that the current is in the clockwise direction, I how are the magnetic fields marked? (Into the coil/ out from the coil)
Answer:
Into the coil
Physics 10 Class Chapter 2 Kerala Syllabus Question 12.
How will the magnetic field lines appear when the coil is viewed in such a way that the current is in the anticlockwise direction?
Answer:
out from the coil
Textbook Page No. 38
Physics Class 10 Chapter 2 Notes Kerala Syllabus Question 13.
Record in the science diary the various factors affecting the magnetic effect of electricity.
Answer:
The strength of the magnetic field in-creases when the number of turns of the coil or current is increased.
10 Physics Chapter 2 Kerala Syllabus Question 14.
Take an insulated copper wire of length not less than 1 m and make a solenoid (preferably a wire of gauge number 26)
→ It will act as a magnet when current from a cell is passed through it after inserting a soft iron core. What is this device known as?
Answer:
Electromagnet
→ With the help of a magnetic compass check the specialty of the magnetism at either ends of the solenoid.
Answer:
Magnetism at either ends of the solenoid are different. The needle of the magnetic compass will be attracted by the south pole and repelled by the north pole.
→ What is the change observed in the movement of the magnetic needle when the experiment is repeated after removing the soft iron core?
Answer:
The strength of the magnetic field will decrease.
Physics Class 10 Chapter 2 Notes Kerala Syllabus Question 9.
From the movements of the magnetic needle in the magnetic compass, find out the polarity of the solenoid and mark them.
Answer:
When current flows through the solenoid, it behaves like a bar magnet.
→ Hold a current carrying solenoid with one end facing you. Note the direction of current at that end. Is it clockwise or anticlockwise?
Answer:
When a solenoid is hold in the right hand and if the four fingers represent the direction of current flow, then the thumb represents the direction of the North pole.
→ Find out the relationship between the direction of current and the polarity.
Answer:
The end of the solenoid through which current flows in the clock wise direction is the south pole and the end through which current flows in the anti clockwise direction is north pole.
Textbook Page No. 39
Chapter 2 Physics Class 10 Kerala Syllabus Question 15.
Based on the above activities, tabulate the factors affecting the strength of the magnetic field of a solenoid carrying current.
Answer:
10th Class Physics Chapter 2 Kerala Syllabus Question 16.
Analyse and compare the ability of solenoid and bar magnet to bring changes in permanency of the magnetism, polarity and the strength of the magnetism.
Answer:
| Bar magnet | Solenoid |
| 1. The magnetism is permanent | 1. The magnetism is temporary |
| 2. Permanent magnet | 2. Electromagnet It act as a magnet when current passed through it |
| 3. Weak magnetic field | 3. Strong magnetic field |
| 4. Strength cannot be changed. | 4. Strength can be changed by changing the current through it or changed the number of turns. |
| 5. Polarity is fixed and cannot be easily reversed. | 5. The polarity can be reversed by changing the direction of current through it |
Textbook Page No. 40
Physics Class 10 Chapter 2 Kerala Syllabus Question 17.
The figure shows a copper wire suspended between the pole pieces of a U shaped magnet, using thin conductors in such a way that the wire is perpendicular to the magnetic field and it is free to oscillate in the magnetic field.
→ Does the conductor move when the circuit is switched on? Observe in which direction it is moving.
Answer:
The copper wire deflects. Conductor will move perpendicular to direction of current.
→ Repeat the experiment by changing the direction of current.
Answer:
The direction of deflection of the copper wire also changes to the opposite direction.
→ Repeat the experiment by interchanging the position of the magnetic poles
Answer:
When the polarity of the magnetic field is changed the deflection is in the opposite direction.
→ Aren’t the direction of the magnetic field and the direction of current mutually perpendicular in this arrangement?
Answer:
Yes
Textbook Page No. 41
Physics 2nd Chapter Class 10 Kerala Syllabus Question 18.
Armature is the metallic coil wound round a soft iron core so that it is free to rotate. It is fixed firmly on the axis XY. In the figure, are the forces acting on sides AB and CD in the same direction? Find out on the basis of Fleming’s Left Hand Rule and write it down.
Answer:
No, AB moves forward and CD moves back wards.
→ What are the effects produced by these forces on the armature?
Answer:
Forces produced are in the opposite directions. They are experiences on the different positions of same object. So it rotates.
Physics 2 Chapter Class 10 Kerala Syllabus Question 19.
Observe the structure of a loud speaker.
→ Where is the voice coil situated?
Answer:
In the magnetic field
→ To which the diaphragm is connected?
Answer:
It is connected with the voice coil.
→ From where current reaches to the voice coil
Answer:
Current reaches from the amplifier.
→ What happens when current reaches through the voice coil
Answer:
It vibrates.
10th Physics Chapter 2 Kerala Syllabus Question 1.
Current is passed from South to North through a conductor placed below a freely pivoted magnetic needle.
a. To which direction will the North Pole of the magnetic needle turn?
b.Which is the rule used to arrive at this inference?
c. State the rule.
d. If the current flows in the conductor in the East West direction, what do you guess about the deflection of the magnetic needle? Explain
Answer:
a. Towards east
b. Ampere’s swimming rule
c. Ampere’s swimming rule
Suppose a man swims in the direction j of current flow in a conductor by looking towards a magnetic needle, the north pole deflects towards the direction of the left hand.
d. A freely suspended magnetic needle remains in the north south direction. When the current flows in the east west direction, the magnetic field produced will be in the north south direction. Magnetic needle does not deflect.
Hss Live Guru 10th Physics Kerala Syllabus Question 2.
How can we find the polarity when current flows through a solenoid? Write the methods to increase the strength of magnetic field around a current carding solenoid?
Answer:
If the direction of current flow in the end of the solenoid is in the clock wise direction. South pole is formed there . If the direction of current flow is in the anti clockwise direction, the pole formed there is North. methods to increase the magnetic strength of solenoid.
Question 3.
The figure shows an insulated copper wire AB made into a coil. Suppose current flows from A to B through this.
a. What will be the direction of electron flow through it?
b. Can you find out the direction of the j magnetic field around the conductor AB? State the rule that substantiates this.
c. Explain how you can find out the direction of the magnetic field inside the coil,
Answer:
a. From B to A
b. The magnetic field across the conductor AB can be formed out. The direction of the magnetic field will be below the table, The rule which helps to find out this is the right hand thumb rule.
According to Right Hand Thumb Rule of James Clark Maxwell holding a current carrying conductor with the right hand in such a way, that the thumb points in the direction of the current, i The direction in which the other fingers j encircle the conductor gives the direction of the magnetic field.
c. The direction of current flow in the coil will 1 be from B to A. That is when viewed from top in the clockwise direction so the direction of magnetic field lines will be from outside to inside the coil.
Question 4.
The magnetic field around the current carrying conductor AB is depicted
Based on the Maxwell’s Right Hand Cork Screw Rule find out the direction of cur¬rent and record it.
Answer:
If a right hand screw is rotated in such a wave that its tip advances along the direction of the current in the conductor, then the direction of rotation of the screw gives the direction of the magnetic field around the conductor. In figure the current flows from B to A.
Question 5.
Electricity flows through a very long solenoid. Some statements are given below related to the magnitude of the magnetic field developed. Find out the correct ones and write them down.
a. It is zero
b. It will be the same at all points
c. It gradually decreases towards the ends,
d. It gradually increases towards the ends.
Answer:
It will be the same at all points
Question 6.
The direction of movement of electrons through a magnetic field is depicted. “The force felt by the electrons due to the influence of the magnetic field is into the plane of the paper”. Is this statement correct? Explain based on the Fleming’s Left Hand Rule.
Answer:
Yes. Current flows on the opposite direction of electrons. According to Fleming’s Left hand rule, When the thumb, point finger and middle finger of the left hand are kept mutually perpendicular and if the point finger represents the direction of the magnetic field middle finger the direction of current then the thumb represents the direction of motion experienced on the conductor.
Question 7.
In an experiment to know the intensity of magnetic field around a current carrying coil, why is the coil kept in the North South Direction.
Answer:
When the coil is kept in south north direction the magnetic field becomes free. That is when kept in south north direction the geomagnetic does not influence the experiment.
Question 8.
In the split ring commutator of a DC motor, semi circular rings are used. What is the need for this?
Answer:
Tn the motor the split rings rotate, according to the armature rotation. When the position of the semiconductor rings in the split ring changes the direction of the current in the armature also change. In this way the continuous rotation of the DC motor is possible.
Question 9.
A current carrying solenoid is stretched to increase the distance between the coils. What change will occur in its magnetic field? Describe.
Answer:
The magnetic intensity will decrease. The magnetic intensity decreases as the number of magnetic lines decreases through as area of 1 unit.
Question 10.
State the Motor Rule. If the directions of current in the conductor and the magnetic field are the same, in which way will the conductor move?
Answer:
Principle of motor:
A freely suspended current carrying conductor when kept in a magnetic field moves when current flows through it. If the direction of current in the coil and the direction of the magnetic field are same, the conductor doesn’t move.
Very Short Answer Type Questions (Score 1)
Question 1.
Observe the figure. If current flows from P to Q in the conductor PQ, then find the direction of the magnetic field in ABCD?
Answer:
Clockwise
Question 2.
Find the odd one in the group and write the reason.
[Voice coil, field magnet, slip rings, armature]
Answer:
Voice coil. Others are parts of microphone.
Question 3.
Find out the relation and fill in the blanks.
Slip ring : AC Generator
…………. : DC Generator
Answer:
Split ring commutator
Question 4.
Choose the incorrect statement related with electromagnets.
a. The magnetism is permanent
b. Strength can be changed by changing
c. The polarity can be reversed by changing the direction of current through it.
Answer:
The magnetism is permanent
Question 5.
When current passes through a conductor a magnetic field is produced around it What rule helps to find the direction of magnetic field.
Answer:
Right hand thumb rule
Question 6.
is a coiled conductor wound up in the shape of a spring
Answer:
Solenoid
Question 7.
Which effect of electricity is made used of in a solenoid?
Answer:
Magnetic effect of electricity
Very Short Answer Type Questions (Score 2)
Question 8.
a. Which are the components of an electric motor
b. Explain an armature?
Answer:
a. Magnetic poles, Armature, Split rings, graphite brushes and axis
b. An armature is the will wound upon a soft iron core which is suspended such that it can rotate freely.
Question 9.
Draw the direction of magnetic flux lines around a solenoid when current flows through it. Show the direction of current flow and that of magnetic field?
Answer:
Question 10.
a. Find out the polarity in A&B.
b. Which are the way to increase the magnetic strength ?
Answer:
a. A – South pole
B – North pole
b. 1. Increase the number of turns
2. Increase the strength of current flow.
3. Use soft iron as the core.
4. Increase the area of cross section of the solenoid.
Question 11.
List some devices which use electromagnets.
Answer:
Electric bell, MCB, ELCB, Generator & Crane
Question 12.
a The direction of current flow at one end of a solenoid is given above. Which pole of the solenoid is this,
b. What is the relationship between the direction of current flow and magnetic polarity?
Answer:
a. South pole
b.The end of the solenoid through which current flows in the clock wise direction is the south pole and the end through which current flows in the anti clockwise direction is north pole.
Short Answer Type Questions (Score 3)
Question 13.
Observe the U magnet and the direction of current flow in the figure,
a Which is the direction of rotate an of the wheel?
b. Name and state the rule which is the base of this experiment?
Answer:
a. Clockwise direction,
b. Fleming’s left hand rule.
When the thumb point finger and middle finger of the left hand are kept mutually perpendicular and if the point finger represents the direction of the magnetic field middle finger, the direction of current then the thumb represents the direction of motion experienced on the conductor
14. When current is passed through a conductor a magnetic field is produced. The direction of the magnetic field can be found out using the right hand thumb rule,
a State this rule.
b. Name and state another rule used for this purpose.
Answer:
a. Right hand thumb rule:
Imagine you are holding a current carrying conductor with the right hand in such a way, that the thumb points in the direction of the current. The direction in which the other fingers encircle the conductor gives the direction of the magnetic field.
b. Right Hand Screw Rule:
If a right hand screw is rotated in such a way that its tip advances along the direction of the current in the conductor, then the direction of rotation of the screw gives the direction of the magnetic field around the conductor.
Question 15.
Write down the working of a moving coil loud speaker by rearranging the following in correct sequence.
a. strengthened electrical pulses are sent through the voice coil of a loudspeaker.
b. The voice coil, which is placed in the magnetic field, moves to and fro rap¬idly, in accordance with the electrical pulses.
c. The electrical pulses from a micro-phone.
d. Make the diaphragm vibrate, thereby reproducing sound
e. Electrical pulses are strengthened using an amplifier
Answer:
Answer:
c. The electrical pulses from a microphone.
e. Electrical pulses are strengthened using an amplifier
a. strengthened electrical pulses are sent through the voice coil of a loudspeaker
b.The voice coil, which is placed in the magnetic field, moves to and fro rapidly, in accordance with the electrical pulses.
d.Make the diaphragm vibrate, thereby reproducing sound
Short Answer Type Questions (Score 4)
Question 16.
a. Current flows towards west in a straight electric line. Find out the direction of the 1 magnetic field below and above the electrie line.
b. When the switch of the below given circuit is ON. What will be the direction of the north pole of the needle in the compass box ?
Answer:
a. According to Right Hand Thumb rule, the j direction of the magnetic field above the electric line will be from south to north whereas the direction of magnetic field below the current line will be from north to douth.
Since the direction of current flows is in clockwise direction. The compass needle will be directed towards the south pole.
Question 17.
The structure of a loud speaker is given.
a. What A and B represent?
b. Explain the working of this device?
Answer:
a. A – Diaphragm
B – Soft iron core
b. A voice coil which is kept in a magnetic field deflects, when current reaches on it. Sound is produced when the dia-phragm which is connected to the coil vibrated.
Students can Download Maths Chapter 7 Tangents Questions and Answers, Notes PDF, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Maths Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
Students can Download Physics Chapter 2 Magnetic Effect of Electric Current Questions and Answers, Notes Pdf, Activity in Malayalam Medium, Kerala Syllabus 10th Standard Physics Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
You can Download Second Degree Equations Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 4 help you to revise complete Syllabus and score more marks in your examinations.
Textbook Page No. 81
Second Degree Equation Class 10 Kerala Syllabus Question 1.
When each side of a square was reduced by 2 metres, the area became 49 square metres. What was the length of a side of the original square?
Answer:
Let the length of each side of the original square be x, then
The length of each side of the original square = 9 cm
Second Degree Equation Class 10 Questions And Answers Kerala Syllabus Question 2.
A square ground has a 2 metre wide path all around it. The total area of the ground and the path is 1225 square metres. What is the area of the ground alone?
Answer:
’Each side of the ground = x
Each side of ground+path = x + 4
(x + 4)2 = 1225 = 352;
x + 4 = 35
x= 35 – 4 = 31
Area of the ground alone = 312 = 961 m2
Second Degree Equation Class 10 Extra Questions Kerala Syllabus Question 3.
The square of a term in the arithmetic sequence 2, 5, 8, ……., is 2500. What is its position?
Answer:
Let 2500 is the square of the nth term of the arithmetic sequence 2, 5, 8, …………
It is the 17th term.
To solve a system of linear equations with steps, use the system of linear equations calculator.
Sslc Maths Second Degree Equations Kerala Syllabus Question 4.
2000 rupees was deposited in a scheme in which interest is compounded annually. After two years the amount in the account was 2205 rupees. What is the rate of interest?
Answer:
Amount (P) = 2000
Compound interest (r %)
Year (n)
Textbook Page No. 86
Sslc Second Degree Equation Questions Kerala Syllabus Question 1.
1 added to the product of two consecutive even numbers gives 289. What are the numbers?
Answer:
Let the two consecutive even numbers be x, x+2
x, x+2
x(x+2) + 1 = 289
x2 + 2x = 288
x2+ 2x – 288 = 0
x2+ 2x = 288
(x+1)2 = 288 + 1
(x+1)2 = 289
x+1 = ± 17
x = 16, – 18
The numbers are 16, 18
Second Degree Equation Questions And Answers Kerala Syllabus Question 2.
9 added to the product of two consecutive multiples of 6 gives 729. What are the numbers?
Answer:
Let the two consecutive multiples of 6 be x, x+6
x, x+6
x(x + 6)+ 9 = 729
x2 + 6 x + 9 = 729
x2 + 6x = 720
x2+ 6x – 720 = O
x2 +6x = 720
(x+3)2= 720+9
(x+3)2 = 729
x+3 = ± 27
x = 24, – 30
The numbers are 24, – 30
10th Maths Second Degree Equation Kerala Syllabus Question 3.
How many terms of the arithmetic sequence 5, 7, 9, …, must be added to get 140?
Answer:
First term f= 5,
Common difference = 2
10 terms should be added to get 140
Second Degree Equation Class 10 Model Question Paper Kerala Syllabus Question 4.
16 added to the sum of the first few terms of the arithmetic sequence 9, 11, 13, gave 256. How many terms were added?
Answer:
9, 11, 13,…………
First term f= 9,
Common difference d = 2
Sum of first n terms
Second Degree Equation Class 10 Notes Kerala Syllabus Question 5.
An isosceles triangle has to be made like this
The height should be 2 meters less than the base. The area of the triangle should be 12 square meters. What should be the length of its sides?
Answer:
Base = AB = x
Height= CD = x – 2
Sslc Maths Chapter 4 Questions And Answers Kerala Syllabus Question 6.
A 2.6-meter long rod leans against a wall, its foot 1 meter from the wall. When the foot is moved a little away from the wall, its upper end slides the same length down. How much farther is the foot moved?
Answer:
Textbook Page No. 91
Sslc Maths Chapter 4 Solutions Kerala Syllabus Question 1.
The product of a number and 2 more than that is 168, what are the numbers?
Answer:
Let the number be x
x(x+2) = 168
x2 + 2x = I68
x2 + 2x + 1 = I68 + 1
(x+1)2 = 169
x + 1 = ± 13
x + 1 = 13
x + 1= – 13
x = 13 – 1 = 12
x = -13 – 1 = – 14
The number is 12 and 14 or – 12 and – 14
Maths Chapter 4 Class 10 Kerala Syllabus Question 2.
Find two numbers with sum 4 and product 2.
Answer:
Sum of the numbers = 4
If, First number = x
Then second number = 4 – x
Class 10 Maths Second Degree Equation Kerala Syllabus Question 3.
How many terms of the arithmetic sequence 99, 97, 95, … must be added to get 900?
Answer:
The spectrally extended signal produced by nonlinear function calculator 520 is likely to have a pronounced dropoff in amplitude as frequency increases.
Class 10 Maths Chapter 4 Kerala Syllabus Kerala Syllabus Question 4.
A rod 28 centimeters long is to be bent to make a rectangle.
i. Can a rectangle of diagonal 8 centimeters be made?
ii. Can a rectangle of diagonal 10 centimeters be made?
iii. How about a rectangle of diagonal 14 centimeters?
Calculate the lengths of the sides of the rectangles that can be made.
Answer:
∴ Cannot make a rectangle of diagonal 14 centimeters.
Textbook Page No. 97
Maths Second Degree Equation Kerala Syllabus Question 1.
The perimeter of a rectangle is 42 meters and its diagonal is 15 meters. What are the lengths of its sides?
Answer:
Sslc Second Degree Equation Questions And Answers Kerala Syllabus Question 2.
How many consecutive natural numbers starting from 1 should be added to get 300?
Answer:
Sum of natural numbers starting from 1:
Second Degree Equation Class 10 Kerala Syllabus Kerala Syllabus Question 3.
What number added to 1 gives its own square?
Answer:
Let the number be x
Second Degree Equation Kerala Syllabus Question 4.
In writing the equation to construct a rectangle of specified perimeter and area, the perimeter was wrongly written as 24 instead of 42. The length of a side was then computed as 10 meters. What is the area in the problem? What are the lengths of the sides of the rectangle in the correct problem?
Answer:
If length is 10 and perimter is 24
2 (l + b) = 24
2(10 + b) = 24
10 + 5 = 12
b = 2
Width= 2
Area= lb =10 x 2 = 20 sq.cm
Area in the correct problem =20
lb = 20
2(l + b) = 42
l + b = 21
l + \(\frac { 20 }{ l }\) = 21
Kerala Syllabus 10th Standard Maths Chapter 4 Kerala Syllabus Question 5.
In copying a second-degree equation to solve it, the term without x was written as 24 instead of – 24. The answers found were 4 and 6. What are the answers of the correct problem?
Answer:
Worksheet 1
Second Degree Equations Exercises Kerala Syllabus Question 1.
Write two numbers whose square is 25?
Answer:
Let x be the number
x2 = 25
x= √25 = ±5
numbers are –5, +5
Second Degree Equation Class 10 Formulas Kerala Syllabus Question 2.
When the square of a number is added to the number we get 30. What are the numbers?
Answer:
Let x be the number ,
x2 + x = 30
x2 + x – 30 = 0, (x + 6) (x – 5) = 0
x = – 6, 5
numbers are – 6, +5
Second Degree Equations Class 10 Kerala Syllabus Question 3.
Find the side of the square whose area and perimeter are numerically equal.
Answer:
Let x be the length of side.
Perimeter = Area
x2 = 4x
x = 4
Length of side = 4
Question 4.
How many odd numbers from 1 makes the sum 961?
Answer:
Sum of continuous n odd numbers = n2
n2 = 961
n = √961 = ± 31
Sum of continuous 31 odd numbers
Question 5.
A man’s age after 15 years will be the square of his age 15 years ago. What is his present age?
Answer:
Worksheet 2
Form the equation
Question 6.
The sum of a number and its square is ten times that number
Answer:
Let x be the number
x2 + x = 10 x
x2 – 9x = 0
Question 7.
The sum of a number and its square root is 6.
Answer:
Let x be the number
x + √x = 6
x – 6 = √x
(x – 6)2 =(–√x)2
x2 – 12x + 36 = x
x2 – 13 x + 36 = 0
Question 8.
The sum of first n natural numbers is 210.
Answer:
\(\frac { n(n+1) }{ 2 }\) = 210
n2 + n = 420
n2 + n – 420 = 0
Question 9.
The area of a rectangle whose length is 5 more than its width
Answer:
Let x be the length of side.
other side = x + 5
x (x + 5) = 150
x2 + 5x – 150 = 0
Question 10.
The sum of a number and its reciprocal is \(\frac { 5 }{ 2 }\)
Answer:
Let x be the number
Question 11.
The sum of even numbers from 2 in an order is 240
Answer:
2 + 4 + 6 + + 2n = 420
2(1 + 2 + 3 + ….+ n) = 420
Question 12.
A man’s age after 15 years will be the square of his age 15 years ago.
Answer:
Let x be the present age
(x – 15)2 = x + 15
x2 – 30 x + 225 = x +15
x2 – 31x + 210 = 0
Worksheet 3
Question 13.
When 8 times a number is added to its square we get 8. Find the number by making the equation properly.
Answer:
x2 + 2 x = 8
x2 + 2x + 1 = 8 + 1
(x + 1)2 =9
x + 1 =3
x = 2
Number = 2
Question 14.
Which term in the sequence 2, 5, 8 …….. gives its square 2500?
Answer:
nth term = 3n + (2 – 3) = 3n – 1
(3n – 1)2 = 2500,
3n – 1 = 50,
n = 17
square of 17th term is 2500
Question 15..
A man’s age after 15 years will be the square of his age 15 years ago. Find the age
Answer:
Let x be the age
x + 15 = (x – 15)2
x + 15 = x2 – 30 x + 225
x2 – 31 x = – 210
Question 16.
The length of a rectangle is 2 more than its width. The area of the rectangle is 80. Find length and breadth.
Answer:
Let x be the width, then length = x + 2
x (x + 2) = 80
x2 + 2x = 80
x2 + 2x + 1 = 80 + 1
(x + 1)2 = 81
x + 1 =9
x = 8
length = 10
width = 8
Question 17.
The sum of a number and its reciprocal is \(\frac { 5 }{ 2 }\) Find the number.
Answer:
Let x be the number
Question 18.
The sum of some even numbers starting from 2 is 420. Find the number of even numbers added.
Answer:
Worksheet 4
Question 19.
Sum of the squares of three consecutive natural numbers is 110.
Answer:
If the numbers are x, x + 1, x + 2, then
x2 + ( x + 1)2 + (x + 2)2 = 110
x2 + x2 + 2x + 1 + x2 + 4x + 4 = 110
3x2 + 6x – 105 = 0
Question 20.
The product of the digits of a two-digit number is 12.When 36 is added to the number we get a two-digit number in which the digits are reversed. Find the two-digit number .
Answer:
Let the two digit number be 10 x + y, then
x y = 12
y = \(\frac { 12 }{ x }\)
If we change the positions of digits 10 y + x
10 y + x = 36+ 10 x + y
Let’s take x be the digits in the position of 10.
Let’s take \(\frac { 12 }{ x }\) as the digits in the position of unit.
Number which digits are reversed \(\frac { 120 }{ x }\) +x
Question 21.
Serena and Johan had 45 diamond stones. They sold 5 stones. The product of the remaining stones is 124. Find the number of stones each had
Answer:
Let x be the stones Serena has and 45 – x be the stone Johan has. After selling 5
(x – 5)(45 – x – 5) = 124
(x – 5)(40 – x) = 124
x2 – 45x + 200 = – 124
x2 – 45x + 324 = 0
Number of diamond stones Serena has = 36
Number of diamond stones Johan has = 9
Question 22.
The sum of a number and its reciprocal is \(1 \frac{1}{2}\) . Find the number.
Answer:
Let x be the number
Question 23.
The sum of two numbers is 15. Sum of its reciprocals is \(\frac { 3 }{ 10 }\). Find the numbers.
Answer:
Let x, 15 – x be the numbers
Question 24.
A two-digit number is four times sum of its digits. The number is three times product of the digits. Find the number
Answer:
Let x, y be the numbers
10 x + y = 4 (x + y)
6 x = 3 y
Worksheet 5
Question 25.
A train travels a distance of 300km constant speed. If the speed of the train is increased by 5 km, the journey would have taken 2 hours less. Find the original speed of the train
Answer:
Let x km/hr be the speed.
Let the time required to travel in the same speed be \(\frac { 300 }{ x }\) hour. If the speed is increased by 5 km/ hr the time will decreased by 2 hours.
Question 26.
An express train takes 3 hours less than a passenger train for a journey of 600 km. If the speed of the passenger train is 10 less than the speed of the express train find the speeds of both trains (Use Pythagoras theorem in distance, not in speeds)
Answer:
Let x km/hr be the speed of the express.
Speed of the express 50 km/hr
Speed of the passenger 50 km/hr
Worksheet 6
Question 27.
One year ago a man’s age is eight times the age of his son. At present man’s age is the square of son’s age. Find the present age.
Answer:
Let x be the present age of son
Question 28.
A man’s age after 15 years will be the square of his age 15 years ago. Find the present age by forming a second degree equation
Answer:
Let x be the present age .
x + 15 = (x – 15)2
x + 15 = x2 – 30x + 225
x2 – 31x + 210 = 0
Present age = 21
Question 29.
The product of Layas’s age before 5 years and after 8 years is 30. Find the present age.
Answer:
Let x be the present age of Laya
(x – 5)(x + 8) = 30
x2 – 3x – 40 = 30,
x2 – 3x – 70 = 0
Present age of Laya = 7
Worksheet 7
Question 30.
Sravani teacher asked the students to construct a rectangle having area 5 square unit and perimeter 8. Jeevan, a wise student of the class, after making some calculations told that it is not possible to construct such a rectangle. Can you agree with him. Justify reasonably
Answer:
Let x be the length, then the area will be 5 cm2, So the width will be \(\frac { 5 }{ x }\) cm. When the perimeter is 8 cm, 4 – x will be the width. The width which was obtained first will be \(\frac { 5 }{ x }\) and the width obtained now i.e., 4 – x will be equal.
– 4 < 0, x will not be obtained. Hence it is not possible to draw a rectangle with the given values.
Question 31.
The perimeter of a rectangle is 34 cm, area 60 square centimeter. Find the sides
Answer:
Let x be the length
Perimeter 34 cm, therefore width will be 17-x cm
Perimeter = 60 cm, therefore width be \(\frac { 60 }{ x }\) cm
x = 5 cm
length = 12 cm therefore width = 17 – 12 = 5 cm
length = 12 cm ,
width = 5 cm
Question 32.
The length of the rectangle is 4 more than its breadth. Area of the rectangle is 140 square centimeter. Calculate length and breadth
Answer:
Let x be the width of rectangle
length = x + 4 cm
x(x + 4) = 140
x2 – 4 x – 140 = 0
width =10 cm
length = 14 cm
Question 33.
When the sides of a square are increased by 4, area become 256. Find the length of the first square.
Answer:
Let x be the side of square
x will not -20
The side of the first square = 12 cm
Question 34.
The area of a right-angled triangle is 60 square unit. The one of the perpendicular sides is 10 more than other. Find the sides of the triangle.
Answer:
Let x cm, x + 10 cm be the perpendicular sides
Question 35.
The area of an isosceles triangle is 60 square meters. One of the equal sides is is 13 cm. Find the third side. Take base x then h = \(\sqrt{13^{2}-x^{2}}\)
Answer:
Question 36.
When the sides of a square are increased by 8 cm each, its area becomes 1225 sq. cm Frame an equation using the above data by taking the side of the smaller square as x cm. Find the sides of both the squares. [Score: 3, Time: 5 Minutes]
Answer:
One side of smaller square = x
One side of bigger square = x + 8
Area = (x + 8)2 = 1225 (1)
x + 8 = 35
x = 35 – 8 = 27 (1)
Side of smaller square = 27 cm
Side of bigger square = 35 cm (1)
Question 37.
The difference of two positive numbers is 6. Their product is 216. Find the numbers. [Score: 3, Time: 4 Minutes]
Answer:
Let the numbers be x, x + 6
Products : x (x+6) = 216 (1)
x2 + 6x = 216
x2 + 6x + 9 = 216 + 9 = 225
(x + 3)2 = 225 (1)
x + 3 = 15
x = 15 – 3 = 12
Numbers : 12, 18 (1)
Question 38.
In a right triangle one of the perpendicular sides is one less than 2 times the smaller side. Hypotenuse is one more than 2 times the same smaller side. By taking the smaller side as x cm, write the algebraic expression for the other two sides. Compute all the 3 sides of the right triangle. [Score: 4, Time: 5 Minutes]
Answer:
Smaller side = x
Perpendicular side = 2x – 1
Hypotenus = 2x + 1 (1)
x2 + (2x – 1)2 = (2x + 1)2 (1)
x2 + 4x2 – 4x + 1 = 4x2 + 4x + 1
x2 – 8x = 0 (1)
x (x – 8) = 0
x = 0 or x = 8
Side=8 cm, 15 cm, 17 cm (1)
Question 39.
Find the sides of a rectangle whose perimeter is 100 metres and area 600 sq. metres. [Score : 4, Time : 5 Minutes]
Answer:
Perimeter = 100 m .
Length+Breadth = 50 m.
Length = 25 + x.
Breadth = 25 – x
Area = (25 + x)(25 – x) = 600 (1)
252 – x2 = 600 (1)
x2 = 625 – 600 = 25, x = 5 (1)
Sides, 25 + 5 = 30 m
25 – 5 = 20 m (1)
Question 40.
The one’s place of a two-digit number is 4. The product of the number and digit sum is 238.
a. If ten’s place digit is taken as x, Write the number.
b. Frame a second-degree equation and find the number. [Score : 4, Time : 6 Minutes]
Answer:
Digit in the ten’s place = x
Number = 10 x + 4 (1)
The product of the number and digit sum
= (x + 4) (10 x + 4) = 238 (1)
Question 41.
How many consecutive natural numbers from 1 should be added to get 465? [Score : 4, Time : 5 Minutes]
Answer:
Sum of first n natural numbers
Question 42.
The product of the digits of a two-digit number is 12. When 36 is added to this number, got a new number with digits reversed. Find the number. [Score : 5, Time : 7 Minutes]
Answer:
Question 43.
For a two-digit number, one’s place is 3 more than its ten’s place. The product of this number and its digit sum is square of double the digit sum. What is the number? [Score : 4, Time : 6 Minutes]
Answer:
Question 44.
A pavement of Width 2 metres is built around a square shaped garden. The area of the pavement alone is 116 square metres. Find one side of the garden.
[Score : 3, Time: 4 Minutes]
Answer:
Side of garden = x
Area of the pavement Area of the pavement Area of the pavement Side of garden including pavement = x + 4 (1)
Area of the pavement = (x + 4)2 – x2 = 116
x2 + 8x + 16 – x2 = 116 (1)
8x + 16 = 116,
\(x=\frac{116-16}{8}=12.5 \mathrm{m}\)
Side of garden = 12.5m (1)
Question 45.
A rectangle of Width 8 centimetres is cut off from a square sheet along its side. The remaining rectangular portion has an area 84 sq. metres. Calculate the side of the square. [Score : 4, Time : 6 Minutes]
Answer:
Let the side of square = x
Sides of rectangle = x, x – 8 (1)
Area = x(x – 8) = 84
x2 – 8x = 84 (1)
x2 – 8x + 16 = 84 + 16
(x – 4)2= 100, (1)
x – 4= 10, x= 14
Side of square = 14 centimetre
Question 46.
The lengh of a rectangle is 3 metre more than 3 times its breadth. Its diagonal is 1 metre more than the length. Find the lengh and breadth of the rectangle. [Score: 4, Time: 7 Minutes]
Answer:
Sidc of rectangle = x
Length =3x +3, Diagonal = 3x + 4 (1)
(3x + 4)2 = x2 + (3x + 3)2 (1)
9x2 + 24 x + 16 = x2 + 9x2 + 18x + 9
x2 – 6x = 7, x2 – 6x = 7 (1)
x2 – 6x + 9 = 16, (x – 3)2 = 16
x – 3 = 4, x = 4 + 3 = 7 (1)
length of rectangle = 3 × 7 + 3 = 24 metre breadth of rectangle = 7 metre
Question 47.
In the figure AB is the diameter of the circle. CD = 10 cm. BC is 15 cm less than Ac. Find AB?
Answer:
Question 48.
The chords AB and CD of a circle intersect at. If MA = 6 cm, MB = 8 cm and CD = 16 cm. Find MC and MD.
Answer:
CD = 16 centimetre MC = 8 – x, MD = 8 + x
then, MA x MB = MC x MD (1)
6 x 8 = (8 – x)(8 + x)
48 = 64 – x2, x2 = 16, x = 4 (1)
MC = 8 – 4 = 4 centimetre (1)
MD = 8 + 4 = 12 centimetre (1)
Question 49.
In the figure, AD is drawn perpendicular to the side opposite to the right angled vertex A. BC-13 cm and AD=6 cm.
a. Take BD = x and express DC in terms of x.
b. Frame a second-degree equation and find the lengths BD and DC. [Score : 4, Time: 5 Minutes]
Answer:
Question 50.
Can the sum of a number and it’s reciprocal be 2/3? [Score: 4, Time: 5 Minutes]
Answer:
Number = x , then its reciprocal = 1/x
Since the discriminant is negative. We do not get a solution.
∴ The sum of a numbrer and it’s reciprocal never gives 2/3. (1)
Question 51.
The sum of a number and its reciprocal is \(\frac { 13 }{ 6 }\) What is the number ? [Score: 4, Time: 5 Minutes]
Answer:
Let number = x, then its reciprocal = 1/x
Question 52.
The sum of a number is 12 and sum of its reciprocal is \(\frac { 3 }{ 8 }\) Find the numbers. [Score: 4, Time: 5 Minutes]
Answer:
If we take numbers as 6 + x, 6 – x (1)
Question 53.
Consider an arithmetic sequence with common difference 20. If the sum of reciprocals of two consecutive terms of this sequence is 1/24, find the first term of the arithmetic sequence. [Score: 4,Time: 6 Minutes]
Answer:
’Terms x – 10, x + 10 (1)
Question 54.
Prove that the difference of a number and its reciprocal will be always positive. [Score: 3, Time : 5 Minutes]
Answer:
Number = x, its reciprocal = 1/x (1)
then, \(x-\frac{1}{x}=k\) (1)
Since k being a positive number, k2 + 4 also negative. (1)
Question 55.
In copying a second-degree equation, the number without x was written as -30 in¬stead of 30. The answers found were 15 and -2. What are the answers of the correct problem? [Score: 5, Time: 8 Minutes]
Answer:
Short Answer Type Questions (Score 2)
Question 56.
Find two numbers whose sum is 6 and product is 9.
Answer:
’First number = x
Second number = 6 – x
x ( 6 – x ) = 9
6x – x2 = 9
0 = 9 – 6x + x2
x2 – 6x + 9 = 0
(x – 3)2 = 0
(x – 3 ) (x – 3) = 0
x = 3
Numbers are 3, 3.
Question 57.
Ammu is 7 years younger than Divya. If 2 is added to the product of their ages, we get 200. Find their ages.
Answer:
Age of Ammu = x
Age of Divya = x – 7
x(x – 7)+2 = 200
x2 – 7x + 2 = 200
x2 – 7x – 198 = 0 (x – 18) (x + 11) = 0
x= 18, x = – 11
(age will not be a negative number)
Age of Ammu = 18,
Age of Divya =18 – 7 = 11
Question 58.
In order to solve a quadratic equation, Arun did the following steps. Find the values of x by completing each step.
Answer:
Question 59.
The perimeter of a rectangle is 42 cm and its diagonal is 15 cm. Find the dimensions of the rectangle.
Answer:
Let breadth = x; perimeter = 2 (length + breadth) length + breadth = 42/2 = 21
∴ length = 21 – x
x2 + (21 – x) = 152
x2 + 441 – 42x + x2 = 225
2x2 + 216 – 42x =0
x2 – 21x + 108 = 0
(x – 12) (x – 9) = 0
x = 12 or x = 9
breadth = 9 unit
length = 12 unit
Question 60.
If the equation 4x2 – 5 x + k = 0 has two equal solutions. Find the value of k.
Answer:
If there is one solution, the discriminant = 0
Short Answer Type Questions (Score 3)
Question 61.
The product of a number and the number 8 more than it is 105.
a. What is the least number to be added to make the product a perfect square?
b. What are the numbers in the problem?
Answer:
a. Let the number be x, then second number is x+ 8
x(x + 8) = 105
x2 + 8x = 105
The number to be added to get a perfect square is 16
Question 62.
On Arts day, 180 sweets were distributed equally among the students. Tasting one sweet Deepa said, “It is a pity that 9 of our friends are absent today.”
Hearing this Deepu said, “Because of that we got one sweet more.”
a. Find out the total number of students,
b. How many students were present on that day?
Answer:
a. Let the number of students = n
Number of sweets given to one student = \(\frac { 180 }{ n }\)
Since 9 students were absent = \(\frac { 180 }{ n-9 }\)
b. Total present = 45 – 9 = 36
Question 63.
One of the perpendicular side of a right angled triangle has length 4 cm more than twice of the other side. It has surface area 80 cm2. Find out the lengths of the perpendicular sides.
Answer:
Let one of the perpendicular side = x
Second side = 2x + 4, Area = 80
1/2 × (2x + 4) = 80
2x2 + 4x = 160
x2 + 2x = 80
x2 + 2x + 12 = 80 +12
(x + 1)2 = 81
x +1 = ± 9
x + 1 = 9
x = 8
x = – 10 cannot be accepted as – ve sign cannot exist for the length of the side
∴ One side = 8cm
Second side = 2 × 8 + 4 = 20 cm
Long Answer Type Questions (Score 4)
Question 64.
Reji’s father bought several note books of the same price. Total price Rs. 360. If the price of each book were less by 2 rupees, he would get 2 books more.
a. Find the number of books if the price of one book is x.
b. Find the number of books if the price of one books is less by 2 rupees.
c. Form a quadratic equation.
d. Find the number of books bought.
Answer:
(a) Price of each book be x rupees,
No. of Notebooks = \(\frac { 360 }{ x }\)
(b) If the cost of each book is Rs (x – 2)
No. of books brought for Rs. 360 = \(\frac { 360 }{ x – 2 }\)
Question 65.
1235 Sacks of rice is to be brought from Trivandrum to Kottayam. For this a minitruck takes 6 trips more than that of an ordinary truck. An ordinary truck can carry 30 sacks more than that of a mini truck. Find the capacity of each truck.
Answer:
Let the capacity of mini truck be x sacks and ordinary truck be x + 30 sacks
Long Answer Type Questions (Score 5)
Question 66.
a. The sum of a number and it’s reciprocal is \(\frac { 25 }{ 12 }\) What is the number?
b. Prove that the sum of a positive number and it’s reciprocal is always greater than or equal to 2.
Answer:
Free Limit using Substitution Calculator – Find limits using the substitution method step-by-step.
Question 67.
Sum of the area of two squares is 500 m2. If the difference of their perimeters is 40m, find the sides of the two squares.
Answer:
Let the side of the squares be x and y meters.
According to the condition,
x2 + y2 = 500 (1)
4x – 4y = 40
(x – y) = 10
y = x – 10
Substituting the value of y in (1), we get
x2 + (x – 10)2 – 500
2x2 – 20x – 400 = 0
x2 – 10x – 200 = 0
x = 20 or x = – 10
As the side cannot be negative, x = 20
Hence, side of the first square, x = 20 m
Side of the second square, y = 20 – 10 = 10 m
Question 68.
a By increasing the speed of a bus by 10 km/hr, it takes one and half hours less to cover a journey of 450km. Find the original speed of the bus.
b. 250 Rupees is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.
Answer:
Let speed of the bus be x km/hr
The value of x in the second-degree equations can be found out by mainly 3 ways.
1. Factorization
2. Completing the square
3. By using equation
You can Download Gas Laws Mole Concept Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 10th Standard Chemsitry Solutions Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.
Gas Laws And Mole Concept Questions And Answers Kerala Syllabus 10th Text Book Page No: 33
→ Complete the table 2.1
Answer:
→ If a gas which is kept in a cylinder having a volume of 1 liter, is completely transferred to another 5-liter cylinder then what will be the volume of the gas?
Answer:
5 liter
→ Press the piston after closing the nozzle of the syringe. What will happen to the volume of air inside the syringe?
Answer:
Volume decreases
→ Explain it on the basis of the distance between the molecules of gas and their freedom of movement?
Answer:
Gases molecules are separated from each other by a large distance. As a result, there will be a lot of vacant spaces. So as the piston is pressed, the molecUles come closer and this volume decreases
Use the Molarity Calculator Chemistry to calculate the mass, volume or concentration required to prepare a solution of compound of known molecular weight.
Sslc Chemistry Chapter 2 Kerala Syllabus Text Book Page No: 34
→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions
→ What assumption can be made regarding the possibility of collision between gas molecules?
Answer:
The molecules collide each other.
→ Which energy gained due to the movement of molecules? Potential energy/Kinetic, energy.
Answer:
Kinetic energy
→ When a gas is heated, temperature is increased. What happens to the movement of molecules if the temperature of the gas is increased?
Answer:
Speed of motion increased
→ Asa result, what happens to the energy of the molecules?
Answer:
Energy of molecules increase
→ Volume
Answer:
The space needed for a substance to occupy is its volume. The volume of solids and liquids are definite. But the volume of a gas is the volume of its container in which it is present.
→ Pressure
Answer:
The force exerted at unit area is pressure, Therefore, force at unit area/ pressure
\(=\frac{\text {Force exerted at the surface}}{\text {Area of the surface}}\)
→ Temperature
Answer:
The average Kinetic energy of all the molecules in a substance is its temperature
Sslc Chemistry Chapter 2 Questions And Answers Text Book Page No: 35
→ Is there any change in the number of molecules?
Answer:
No
→ What happens to the pressure when the volume is decreased?
Answer:
Pressure increases
→ What is the specialty of the movement of the molecules?
Answer:
Molecules move in all possible directions
Mass to Moles Calculator — The quantity of substance n in moles is equal to the mass m in grams divided by the molar mass M in g/mol.
→ What changes can you observe in the volume of the gas inside the syringe?
Answer:
Volume is formed to be decreasing.
→ What about decreasing the pressure?
Answer:
Volume increasing
→ What relation do you arrive at between pressure and volume of the gas?
Answer:
As pressure increases volume decreases when pressure is reduced volume increased.
Gas Laws And Mole Concept Kerala Syllabus 10th Text Book Page No: 36
→ The size of the air bubbles rising from the bottom of an aquarium increases. Can you explain the reason?
Answer:
As the bubbles move upward, the pressure on them decreases. This causes increase in volume. So as the bubbles move upward, their size increases.
→ What do you observe?
Answer:
Ink rises through the tube
→ What is the reason for the rising of the ink upwards?
Answer:
When the bottle is placed in hot water, air inside the bottle becomes hot. This causes expansion of air. This pushes ink in the tube. So ink rises through the tube.
→ What did you observe on cooling the bottle after taking it out? Why?
Answer:
Ink comes down. Because as air becomes cool, Its volume decreases.
→ What can you infer about the relation between the volume and temperature of a gas?
Answer:
When temperature is increased volume of gas increases. Similarly, when temperature is decreased, The volume decreases.
→ Complete the table 2.2
Answer:
Sslc Chemistry Chapter 2 Notes Kerala Syllabus Text Book Page No: 37
→ In which unit is the temperature stated?
Answer:
Kelvin (K)
→ What happens to the volume when the temperature is increased?
Answer:
Volume increases.
→ If an inflated ballon is kept in sunlight, it will burst. What may be the reason for this?
Answer:
When the ballon placed in sun light, temperature increases. so volume of air inside the ballon increases. Thus ballon expands and finally bursts.
→ What happens to the volume of the gas when its pressure is decreased or temperature is increased. volume increased/decreased.
Answer:
volume increased
→ If the temperature and pressure are kept constant how can we increase the volume?
Answer:
Fill some more gas.
→ Fill the cylinder with a little more gas. Does the number of molecules increase or decrease now?
Answer:
Then number of molecules increases
→ What is the relation between the volume and number of molecules?
Answer:
When the number of molecules increases, volume increases.
→ According to Avagadro’s law when the temperature and pressure remain constant on which factor does the volume of gas depend?
Answer:
Depends on the number of molecules.
Kerala Syllabus 10th Standard Chemistry Chapter 2 Text Book Page No: 38
→ If the mass of a coin 5g, then what will be the mass of thousand coins?
Answer:
5 x 1000 = 5000g
→ If the mass of coins in a bag is 50,000 g, then how many coins will be there?
Answer:
\(\frac { 50000 }{ 5 }\) = 10000
→ Like this we can calculate the number of coins on the basis of mass. Can‘t we?
Answer:
Yes, it becomes easy.
→ Is their any relation between the mass and the number, if the particles are of the same mass.
Answer:
Yes
→ What may be the method of stating the mass of atoms?
Answer:
The mass of an atom is compared to the mass, of another atom and expressed as a number which shows how many times it is heavier than the other atom. The atomic mass of elements are expressed by considering 1/12 mass of an atom of carbon-12 as one unit.
→ What do you understand from the statement that the atomic mass of Helium is 4?
Answer:
Atomic mass of Helium is 4. That is mass of one atom of Helium is 4 times of 1/12th mass of carbon atom.
Enter the formula and press “calculate” to work out the molar mass calculator with steps, the number of moles in 1 g and the percentage by mass of each element.
Chemistry Class 10 Chapter 2 Kerala Syllabus Text Book Page No: 39
→ How many oxygen atoms combine with one carbon atom?
Answer:
2 Oxygen atoms
→ How many oxygen atoms combine with 1000. carbon atoms?
Answer:
2000 oxygen atoms.
→ How many atoms are present in 12g carbon?
Answer:
6.022 x 10B carb Answer:
2 × 6.022 × 1023 oxygen atoms
→ What will be the mass of these tabs ?
Answer:
2 × 16 = 32g
Gas Laws And Mole Concept Extra Questions 10th Text Book Page No: 40
→ Complete the table 2.5
Answer:
→ 1GAM sodium means 23g sodium. This contains 6.022 × 1023 atoms. If so, how many GAM is present in 69 g sodium? How many atoms are present in it?
Answer:
\(\frac { 69 }{ 23 }\) = 3 GAM,
3 × 6.22 × 1023 sodium atoms.
Gas Laws And Mole Concept Notes Pdf Kerala Syllabus 10th Text Book Page No: 41
→ How many GAMs are present in each the samples given below? Calculate the, number of atoms present in each of die sample ? (Atomic mass N= 14, O= 16)
1. 42g Nitrogen,
2. 80g Oxygen
Answer:
1. 42g Nitrogen:
No.of GAM = \(\frac { 42 }{ 14 }\) = 3GAM
No.of atoms = 3 × 6.022 × 1023
2. 80g Oxygen:
No.of GAM = \(\frac { 80 }{ 16 }\) = 5GAM
No.of atoms = 5 × 6.022 × 1023
→ Complete the table 2.6
Answer:
→ Calculate the molecular mass of glucose (C6 H12 O6) and sulphuric acid (H2 SO4)
Answer:
Molecular mass of glucose
= 6 × 12 + 12 × 1 + 6 × 16 = 72 + 12 + 96 = 180 g
Molecular mass of sulphuric acid
2 × 1+ 1 × 32 + 4 × 16 = 2 + 32 + 64 = 98 g
Gas Laws And Mole Concept Questions And Answers Pdf 10th Text Book Page No: 42
→ Complete the table 2.7
Answer:
→ What is the molecular mass of oxygen?
Answer:
32 g
→ How many GMM is present in 32g oxygen?
Answer:
1 GMM
→ How many molecules are present in it?
Answer:
6.022 × 1023 oxygen molecules
→ How many GMM is present in 28 gm nitrogen?
Answer:
1 GMM
→ How many molecules are present in N2?
Answer:
6.022 × 1023 N2, molecules
→ How many GMM is present in 18 gm water?
Answer:
1 GMM
→ How many H2O molecules arf present in it?
Answer:
6.022 × 1023 H2O Molecules
→ Calculate the number of GMM present in 96g oxygen?
Answer:
\(\frac { 96 }{ 32 }\) = 3GMM
Sslc Chemistry Chapter 2 Notes Pdf Kerala Syllabus Text Book Page No: 43
How many GMM are present in each of the given samples? Calculate the number of molecules present in each sample ?
→ 360 g glucose (Molecular mass = 180)
Answer:
No.of GMM = \(\frac { 360 }{ 180 }\) = 2 GMM
No. of molecules = 2 × 6.022 × 1023
→ 90g water (Molecular mass = 18)
Answer:
No.of GMM = \(\frac { 90}{ 18 }\) = 5 GMM
No. of molecules = 5 × 6.022 × 1023
→ How many molecules of water are present . in one mole of water ?
Answer:
6.022 × 1023 water molecules
→ What is its mass?
Answer:
18 g
→How many GMM is present in it?
Answer:
1 GMM
Sslc Chemistry 2nd Chapter Notes Kerala Syllabus Text Book Page No: 44
22.4 L of a gas at STP = 1 mole
44.8 Lofa gas at STP = = 2 mole
224 L of a gas at STP = \(\frac { 224 }{ 22.4 }\) = 10 mole
Gas Law And Mole Concept Kerala Syllabus 10th Text Book Page No: 45
Complete the flow chart given below, related to one mole of substance.
Answer:
Gas Laws And Mole Concept Notes Kerala Syllabus 10th Question 1.
Examine the date given in the table (Temperature and number of molecules of the gas are kept constant).
a. Calculate P × V
b. Which is the gas law related to this?
Answer:
a. 8L atm
b. Boyle’s law
Gas Laws And Mole Concept Pdf Kerala Syllabus 10th Question 2.
Analyse the situations given below and explain the gas law associated with it.
a. When an inflated balloon is immersed in water, its size decreases.
b. A balloon is being inflated
Answer:
a. Avogadro’s law
b. Boyle’s law
Gas Laws And Mole Concept Class 10 Kerala Syllabus Question 3.
Certain data regarding various gases kept under the same conditions of temperature and pressure are given below.
a. Complete the table?
b. Which gas law is applicable here?
Answer:
a.
b. Avogadro’s law
Hss Live Guru 10th Chemistry Kerala Syllabus Question 4.
a. Calculate the mass of 112 L CO2 gas kept at STP (molecular mass = 44)
b. How many molecules of CO2 are present in it?
Answer:
Sslc Chemistry Chapter 2 Gas Laws And Mole Concept Question 5.
Calculate the volume of 170g of ammonia at STP ? (Molecular mass 17)
Answer:
Number of moles = \(\frac { Given mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10 moles
Volume at STP = mole × 22.4 L = 10 × 22.4 L = 224 L
Hsslive Chemistry 10th Kerala Syllabus Question 6.
Find out the number of moles of molecules present in the samples given below (GMM-N2=28g, H2O= 18g)
a. 56g N2
b. 90g H2O
Answer:
a. Number of mol molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 56 }{ 28 }\) = 2
b. Number of mol molecules = \(\frac { 90 }{ 18 }\) = 5
10th Chemistry 2nd Chapter Kerala Syllabus Question 7.
The molecular mass of ammonia is 17.
a. How much is the GMM of ammonia?
b. Find out the number of moles of molecules present in 170g of ammonia.
c. Calculate the number of ammonia molecules present in the above sample of ammonia?
Answer:
a. GMM of ammonia (NH3) = 14 + 3 × 1 = 17g = 1GMM
b. Number of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 170 }{ 17 }\) = 10
c. Number Of molecules = Mole × 6.022 × 1023
= 10 × 6.022 × 1023
Class 10 Chemistry Chapter 2 Kerala Syllabus Question 8.
The molecule’s mass of oxygen is 32.
a. What is the GMM of O2
b. How many moles of molecules are there in 64g of oxygen? How many molecules are there in it?
c. Calculate the number of oxygen atoms present in 64g of oxygen?
Answer:
a. GMM of O2 =2 × 16 = 32 g
b. No. of mole molecules = \(\frac { Mass }{ GMM }\) = \(\frac { 64 }{ 32 }\) = 2
Number of molecules = mole × 6.022 × 1023
= 2 × 6.022 × 1023
c. c. Number of atoms = Number of molecules × number of atoms in one molecules Number of atoms in one molecules of oxygen (O2) =2
∴ total number of atoms = 2 × 6.022 × 1023 × 2
=4 × 6.022 × 1023
Hss Live Guru Chemistry 10 Kerala Syllabus Question 1.
How many grams of carbon and oxygen are required to get the same number of atoms as in one gram of Helium?
Answer:
GAM of Helium = 4 g
Number of mole atoms in 4 g of Helium = 6.022 × 1023
Number of atoms in 1 gofHelium = \(\frac { 1 }{ 4 }\) × 6.022 × 1023
GAM ofCarbon = 12 g
∴ Number of atoms in 12gofCarbon=6.022 x 1023
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 1023
Carbon atoms = \(3 \mathrm{g}\left(12 \times \frac{1}{4}\right)\)
GAM of Oxygen = 16 g.
∴ Number of atoms in 16 g of Oxygen = 6.022 × 1023
∴ Mass required for \(\frac { 1 }{ 4 }\) × 6.022 × 1023
Oxygen atoms = \(4 \mathrm{g}\left(16 \times \frac{1}{4}\right)\)
Kerala Syllabus 10th Standard Chemistry Guide Question 2.
Examine the samples given:
a. 20 g of He
b. 44.8 L of NH3 at STP
c. 67.2 L of N2 at STP
d. 1 mol of H2SO4
e. 180 g of water.
i. Arrange the samples in increasing order of the number of molecules in each.
ii. What will be the ascending order of the total number of atoms?
iii. What will be the masses of samples b, c, and d?
Answer:
Mole Concept Class 11 Question 3.
In 90 grams of water.
a. How many molecules are present?
b.What will be the total number of atoms?
c. What will be the total number of electrons in this sample?
Answer:
a. GMM= 18 g
∴ Number of molecules
\(\frac { 90 }{ 18 }\) × 6.022 × 1023 = 5 × 6.022 × 1023 18
b. ∴ Number of atoms = 3 × 5 × 6.022 × 1023 = 15 × 6.022 × 1023
c. Electrons in 1 H atom = 1
Electrons in 1 O atom = 8
Total electrons in H2O molecule? 10
Total electrons in 90 g H2O = 10 × 5 × 6.022 × 1023 = 50 × 6.022 × 1023
Question 1.
a. Complete the table based on the data given in the box.
b. Express atomic weight and molecular weight in grams. How many moles is this? Find out the number of Atoms or molecules in it?
Answer:
a.
b.
Question 2.
Complete the table based on the molecules given in the first column and then answer the question given below.
a. 10 Mole of water = …………. g …………… Molecules
5 mole of CaO = ……….. g ………… Molecules
2 Mole of H2SO4 = …………. g ………… Molecules
1/2 Mole of Al2O3= …………. g ………….. Molecules
Answer:
a. 10 Mole of water = 10 × 18 = 180g, 10 × 6.022 × 1023 Molecules
5 mole of CaO = 280g, 5 × 6.022 × 1023 Mol-ecules
2 Mole of H2SO4 = 196g, 2 × 6.022 × 1023 Mol-ecules
2- Mole of AlO = 100g, 1/2 × 6.022 × 1023
Question 3.
Complete the table.
Answer:
Question 4.
Complete the data.
Answer:
Question 5.
Based on the reaction given below, write the answers for the questions.
N2 + 3H2 → 2NH3 ;
a. Write the ratio of reactant molecules and product molecules.
b. How many moles of Ammonia forms when we take 2 moles of Nitrogen and six moles of Hydrogen?
c. Two moles of Nitrogen and three moles of hydrogen are taken in jar? Will they react together?
d. How many moles of Nitrogen and Hydrogen is needed for rearing 20 moles of Ammonia?
Answer:
a. 1:3:2
b. 2 Mole
c. No
d. 10 Mole Nitrogen 30 Mole Hydrogen
Question 6.
Balance the given equation and then write down the answers for the questions given below.
CH4+O2 → CO2 + H2O
a. How many moles of C02 formed when 20 moles of Methane burns in air?
2C2H6 + 7O2 → 4CO2 + 6H2O
b. Based on the equation above, How many moles of CO2 is formed when 10 moles of Ethane is burned in air ?
Answer:
a. When 1 mole of methane bum in air 1 mole of CO2 gas is formed.
When 20 moles methane burn 20
moles of CO2 are formed.
The molecular weight of 20 moles = 20 × 44 = 880 g
b. When 2 moles of ethane is burned 4 moles of CO2 is formed.
The number of moles of CO2 when 1 mole of ethane is burned = 4/2 = 2 mole.
The number of moles when 10 moles of ethane bums = 2 x 10 = 20 mole.
Weight of 20 moles = 20 × 44 = 880 g
Question 7.
Based on the given equation write down the answers.
2H2+ O2 → H2O
a. How much Oxygen and Hydrogen is needed for making 1800g of water vapor?
b. How many moles of Oxygen is needed for the reaction with one mole of Hydrogen?
Answer:
a. 36 g of water vapour can be made using 4 g hydrogen.
The amount of hydrogen required to make 1 g water vapour = 4/36
The amount of hydrogen required to produce 1800 g watervapour= 4/26 × 1800 = 200 g
Mass of oxygen = 1800 – 200 = 1600 g
b. 0.5 Mole
Question 1.
Find out the number of moles of hydrogen and Oxygen atoms present in 10 moles of HCI.
Answer:
One mole HC1 contains 1 mole of hydrogen and 1 mole of chlorine.
Hydrogen contained in 10 moles of HC1 = 1 × 10 =10 mol,
Clatom = 10 × 1 = 10 mol
Question 2.
Find out the mass of Hydrogen atom and chlorine atom in 10 moles of HCI.
Answer:
Mass of 10 mole hydrogen atom = 10 × 1 = 10g
Mass of 10 mole chlorine atom =10 × 35.5 = 355 g.
Question 3.
a Find out the mass of one mole of CaCO3. How many moles of calcium present in 1000g CaCO3?
b. How many moles of Oxygen present in 1000gms of CaCO3?
Answer:
a. Mass of 1 mole of CaCO3 = 40 + 12 + 48 = 100 gram.
No of moles ofCa in 1000 g CaCO3 = \(\frac { 1000 }{ 100 }\) = 10 mol
b. No of moles of Ca in 10 moles CaCO3 =10 × 1 = 10 mol
No of moles of oxygen in 10 moles of CaCO3 = 10 × 3 = 30mol.
This Combined Gas Law Calculator can help you estimate either the pressure, temperature or the volume of gas.
Question 4.
Find out number of moles of water formed when 4gms of Hydrogen and 32 gms of Oxygen combined together. What is the result when 5 gms of Hydrogen and 32 gms of Oxygen combined together?
Answer:
Ans. When 4 g of hydrogen and 32g of oxygen are combined 36 g of water \(\frac { 37 }{ 18 }\) = 2 mol
5g H + 32 gO → 37g H2O
No of moles in 3 7 g of water = \(\frac { 37 }{ 18 }\) = 2.055 mol
Question 5.
a How much grams of NaCl is needed for making 2 molar solution (NaCl – 58.5). Wh-at is the amount of water needed for this?
b. How will you change a two molar solution of Sodium Chloride into 5 major?
Answer:
a. Mass of 2 moles of NaCl = 2 × 58.5 = 117 g
1 liter water is required for this.
b. When 2 moles of NaCl is dissolved in 4 liter of water 5 molar solution is obtained.
Finally, you encounter how to find molar concentration step-by-step manually, and if your preference indulges with instant calculations.
Question 6.
How many moles of Cl2 present in 11.2 L of same in STP? Find out the mass of this?
Answer:
No.of moles present in 11.2 litre of chlorine = \(\frac { 11.2 }{ 22.4 }\) = 0.5 mol
Mass of 0.5 moles of chlorine = 0.5 × 35.5 = 17.759 g.
Question 7.
Find out the mass of Oxygen atom in 44.8L of CO2 in STP.
Answer:
No of moles in 44.8 litre of CO2 \(\frac { 44.8 }{ 22.4 }\) 2 mol,
1 Mole of CO2 contains 1 mole of C and 1 mole of O2
∴ 2 Mole of CO2 contains 2 mole of O2 or 4 mole of oxygen atom.
Man of oxygen atom 4 × 16 = 64 g
Question 8.
Find out the amount of CO2 formed when the burning of one mole of Ethane.
Answer:
2 moles of CO2 is formed when 1 mole of ethane burns.
Mass of 2 moles of CO2 = 2 × 44 = 88g
Question 9.
Why are atomic mass of some elements are infractions ?
Answer:
The atomic masses of some elements are infractions because they exist as a mixture of isotopes of different masses. The fractional atomic masses arise because of this mixture.
Average mass = \(\frac { Total mass of all atoms }{ numbers of atoms }\)
Question 1.
One GAM substance contains Avogadro number of particles in it.
a. How many particles are there in Avogadro number ?
b. Write the number of atoms present in each of the following.
i. 32g Sulphur
ii. 32g Oxygen
iii. 32g Carbon
(Atomic mass S = 32, O = 16, C = 12)
Answer:
a. 6.022 × 1023
b. i.6.022 × 1023
ii. 2 x 6.022 × 1023
iii \(\frac { 32 }{ 12 }\) × 6.022 × 1023
Question 2.
a. Group the following into pairs having same number of atoms.
A. 2g Hydrogen
B. 16g Oxygen
C. 14g Nitrogen
D. 8g Helium (Atomic mass H=1, O= 16, N =14, He=4)
b. How many atoms are present in each pair?
Answer:
a. A, D2g Hydrogen, 8g Helium
B, C 16g Oxygen, 14g Nitrogen
b. A, D – 2 × 6.022 × 1023
B,C – 6.022 × 1023
Question 3.
N2 + 3H2 → 2NH3
a. What is the ratio between the reactant molecules in the above reaction?
b. Complete the following table.
Answer:
a. 1:3
b. a – 2 NH3,
b – l H2,
c – 12H2,
d – 4NH3
Question 4.
2H2 + O2 → 2H2O
a. What is the ratio between the reactant molecules in the above reaction?
b How many O2 molecules are required to react 100 H2 molecules completely?
c. How many water molecules are formed when 1000 H2 molecules are reacted completely
Answer:
a. 2:1
b. 50 O2 molecules
c. 1000 H2O molecules
Question 5.
Complete the following table. (All the elements given are diatomic. Atomic mass O=16, N=14, CI=35.5)
Answer:
a. 6.022 × 1023
b. 6.022 × 1023
c. 71 g
d. 14 g
Question 6.
A sample of substances are given.
Hint: Molecular mass NH3 = 17, N2 = 28, H2SO4 = 98, O2=32
a. Which of these samples have same number of molecules?
b. Which of these samples has least number of molecules?
Answer:
a. 68g NH3, 128gO3
b. 49 g H2SO4
Question7.
Pick out the correct statements from the following. Also, correct the incorrect statements.
a The number of molecules present in 1 mol hydrogen and 1 mol oxygen are same,
b. 2 mol chlorine contains 4 x 6.022 x 1023 chlorine molecules.
c. The mass of 1/2 mol nitrogen gas is 14 g.
d. 0.5 mol water has the mass 9g. There are 6.022 × 1023 H20 molecules in it. (Atomic mass H = 1, O = 16, CI= 35.5, N= 14)
Answer:
Correct statements – a, c
No. of molecules in 2 mol chlorine is 2 × 6.022 × 1023
Mass of 0.5 mol water is 9g. So it contains 0.5 × 6.022 × 1023 H2O molecules.
Question 8.
Complete the following.
Answer:
a. 2 × 6.022 × 1023
b. 1GMM
c. 6.022 × 1023
Question 9.
67.2 L of Carbon dioxide gas is filled in a cylinder at STP.
a. Calculate the mass of CO2 present in it. (Atomic mass- C = 12, O = 16)
b. Calculate the number of molecules present in the cylinder.
Answer:
a. Molecular mass of CO2=12 × 1+16 × 2 = 12 + 32 = 44
No. of moles in 67.2L CO2 at STP = \(\frac { 67.2L }{ 22.4L }\) = 3
b. 3 × 6.022 × 1023
Question 10.
Answer:
a. 2
b. 2 × 6.022 × 1023
c. 17 g
d. 51 g
e. 3
f. 3 × 22.4 L
Question 11.
CH4 + 2O2 → CO2 + 2H2O
The equation describes the combustion of methane in air.
a. How many moles of oxygen is required for the complete combustion of 16g CH4?
b. Calculate the amount of CO2 formed when 100g of CH4 is completely burnt?
Answer:
a. 2 mol
b. Amount of CO2 produced by the combustion of 16g CH4 = 44g
Amount of CO2 produced by the combustion of 1 gm CH4 = \(\frac { 44 }{ 16 }\) g
Amount of CO2 produced by the combustion of 100g CH4 = \(\frac { 44 }{ 16 }\) × 100g
Question 12.
45 g glucose is taken in a beaker and made into 1 L (MM = 180).
a. Calculate the molarity of the solution,
b. Above solution is made up to 2 L by adding more water. What will be the molarity of the resultant solution?
c. How will you prepare IM solution of glucose with the same quantity (45 g) of glucose?
Answer:
a. 0.25
b. M = \(\frac { n }{ v }\) = \(\frac { 0.25 }{ 2 }\)
c. Add 250 ml water in 45 g glucose
Question 13.
Two gases occupy equal volume at STP are shown below.
(Atomic mass S = 32, O =16, N = 14
a. Find the mass of the gas in B.
b. Calculate the number of molecules present in B.
Answer:
a. No. of moles in 320g SO2 = \(\frac { 320 }{ 64 }\) = 5
Mass of 5 mol NO2 = 5 × 46 = 230 g
b. No.of molecules present in B = 5 × 6.022 × 1023
Question 14.
The balanced chemical equation of a reaction (at STP) is given below.
2H2(g) + O2(g) → 2H2O(g)
a. Calculate the volume of oxygen required to combine completely with 224 L of the hydrogen at STP.
b. Calculate the mass of water formed as a result of the reaction (a).
Answer:
a. 112 L
b. Volume of water obtained when 224 L hydrogen completely reacts with oxygen = 224 L
No. of moles in 224 L water = \(\frac { 224 }{ 22.4 }\) = 10
Massof 10 mol water = 10 × 18 = 180 g
Question 15.
Complete the table.
Hint: (MM – CO2 = 44, CH4 = 16, SO2 = 64)
Answer:
a. 67.2 l
b. 132 g
c. 1/4.
d. 4g
e. 11.2 L
f. 1/2.
Question 16.
Analyse the following equation
2NO(g) + O2(g) → 2NO2(g)
a. Calculate the number of the moles of NO required to combine completely with 112 L of Oxygen at STP.
b. Calculate the mass of NO2 formed when 112L of oxygen reacts completely?
Answer:
a. 10 mol
b. 2NO(g) + O2(g) → 2NO2(g) (2 : 1: 2)
No. of moles in 112L O2 = 5 mol
According to equation no. of moles of NO2 obtained by reacting oxygen completely with nitric oxide = 2
No. of moles of NO2 obtained by reacting 5 mol oxygen completely = 10
Massof 10 mol NO2 = 10 × 46 = 460g
Question 17.
The chemical equation of the decomposition of calcium carbonate is given below.
CaCO3→ CaO + CO2
(HintMM: CaCO3 – 100, CaO – 56, CO2 – 44)
a. Calculate the mass of CaCO3 required to get 224 g of CaO?
b. Calculate the number of CO2 molecules fronted when 224g of CaO is obtained?
Answer:
a. CaCO3 → CaO + CO2
100g 56g 44g
I I : I
Amount of CaCO3 required to get 56g of CaO = 100g
Amount of CaCO3 required to get 1 g of CaO = \(\frac { 100 }{ 56 }\)
Amount of CaCO3 required to get 224 g of CaO = \(\frac { 100 }{ 56 }\) × 224 = 400g
b. 4 × 6.022 × 1023
Question 18.
You are requested to make 20 moles of NaCl into packets of 100g each. (Hint^Molecular mass of NaCl is 58.5)
a. How many packets of NaCl can be prepared?
b. Is there any NaCl remaining? If so, how much?
Answer:
a. Mass of 20 mol NaCl = 20 × 58.5 = 1170g
1170 g NaCl can be made into 11 packets with 100g each.
b. Remaining NaCl = 1170 – 1100 = 70g
Very Short Answer Type Questions (Score 1)
Question 1.
GAM of Hydrogen is 1 g.
a. How many number of atoms are there in 1 g of Hydrogen?
b. Find the mass of 1 atom of hydrogen.
Answer:
a. 1g hydrogen = 6.022 × 1023 atoms
b. Mass of 1 atom of hydrogen
= \(\frac{1 g}{6.022 \times 10^{23}}\) = 1.66 × 10-24 g
Question 2.
Number of molecules of substance is 3.011 × 1024.
a. What is the number of molecules of 1 mole of any substance?
b. Find the number of moles of 3.011 × 1024 molecules.
Answer:
a. 6.022 ×1023
b. Number of moles of molecules = \(\frac { Number of molecules }{ NA }\)
= \(\frac{3.011 \times 10^{24}}{6.022 \times 10^{23}}=5\)
Short Answer Type Questions (Score 2)
Question 3.
Fill the patterns.
Answer:
a. GMM
b. 22.4
c. Number of molecules
d. Number of molecules
e. Volume in litres
f. Mass
Question 4.
Identify the incorrect statements from those given with respect to the arrangements of molecules in gases.
a. The minute molecules are present without any freedom of movement
b. Collision take place between the molecules,
c. Increasing the number of molecules at constant volume causes the decrease in number of collisions.
d. The energy of molecules are comparatively high.
Answer:
a. The statements (a) and (c) are not correct
Question 5.
The pressure of 20L of a gas kept at 300 K is found to be 2 atoms. If the pressure is increased to 3 atom at the same temperature, what will be the new volume?
Answer:
According to Boyles law, PV = a constant
Therefore, P1 V1 = P2 V2
Here, P1 = 2atm V1 = 20L P2 = 3atm V2=?
∴ 2 × 20 = 3 × V2
Thus, V2 = \(\frac { 2 × 20 }{ 3 }\) = 13.3 L
Charles Law Calculator is a free online tool that displays the volume of gas that tends to expand when heated.
Question 6.
If the temperature of 5L of a gas at atmospheric pressure is changed from 200K to 50 K, what will be the volume?
Answer:
According to Charles law, \(\frac { V }{ T }\) = a constant
Question 7.
What will be mass of 89.6 L of ammonia (NH3) gas at STP?
Answer:
89.6 L of NH3 gas at STP = \(\frac { 89.6 }{ 22.4 }\) = 4 mol
GMM of NH3 = 14 + 3 = 17g
Mass = mole × GMM = 4 × 17 = 68g
Question 8.
8. a. What is molar volume?
b. What is the molar volueofa gas at STP?
Answer:
a. The volume of one mole of a gas is called molar volume,
b. 22.4 L
Question 9.
Look at the balanced equation given.
2NaOH + CO2 → Na2 CO3 + H20
a. Find out the mass of NaOH needed for 264 g CO2 to react completely.
b. Find out the total number of moles of water molecules when CO2 reacts.
Answer:
a. GMM of CO2 = 44 g
∴Number of moles in the molecule of 264 g CO2 = \(\frac { 264 }{ 44 }\) = 6
According to the equation NaOH needed for the reaction of 1 mole CO2= 2 moles
∴ NaOH needed for the reaction of 6 moles CO2 = 2 × 6 = 12 moles
GMM of NaOH = 23 + 16 + 1 = 40 g
Total mass of NaOH = 12 × 40 = 480 g
b. H2O formed when lmole CO2 reacts = 1 mole
∴ Total number of moles of water molecules when 6 mole CO2 reacts = 6 moles
Question 10.
2C4H10 + 13O2 → 8CO2 + 10H2O
This is the equation of ignition of cooking gas butane.
Calculate the volume of CO2 in STP during the complete ignition of 14 kg of cooking gas.
Answer:
Mass of Butane (C4H10) = 14 kg = 1400g
GMM of C4H10 = 4 × 12 + 10 × 1 = 58 g
∴ Number of moles in molecules = \(\frac { 1400 }{ 58 }\) = 241.38
Amount of CO2when 2 moles of C4 H10 ignites = 8 moles of C4H10 ignites = \(\frac { 8 }{ 2 }\) × 965.52 moles
∴ Volume of CO2 formed in STP
= 965.52 × 22.4 L = 21627.65 L
Question 11.
Write down the preparation of 100 ml NaOH solution of 0.1 M.
Answer:
GMM of NaOH = 40 g
Molarity = \(\frac{\text { Number of moles of solute }}{\text { Volume of solution in litres }}=\frac{\mathbf{n}}{\mathbf{v}}\)
M = 0.1 V=100ml = 0.1 L
0.1 = n/0.1
∴ n = 0.1 × 0.1 = 0.01
Mass needed to prepare 100 ml NaOH in 0.1M = 0.01 × 40 = 0.4g
Take 0.4 g NaOH in a beaker. Dissolve it Hilly by adding a little amount of water. Then, again add water to make it 100 ml.
Question 12.
The molarity of250 ml of Na2CO3 solution is 0.5 M. Find the mass of Na2CO3.
Answer:
Question 13.
63 g HNO3 is in the dilute solution of 200 ml HNO3 (Nitric acid). Find the molarity.
Answer:
a.
Short Answer Type Questions (Score 3)
Question 14.
Some equations related to gas laws are given below.
i. V α P
ii. \(\frac { V }{ T }\) = a constant
iii. V α n
iv. Pv = a constant
a. Which of these are correct?
b. Write the gas law to which it is related for the correct equations.
Answer:
a. Equations (ii), (iii) and (iv) are correct
b. (ii) Charles law
(iii) Avogadro’s law
(iv) Boyles law
Question 15.
In 100 g of CaCO3
a. Find out the number of moles of each element and atom.
b. Find out the total number of atoms of each element.
Answer:
Question 16.
Fill the blanks in the given table.
Answer:
Long Answer Type Questions (Spore 4)
Question 17.
See CO2 gas is taken in a cylinder provided with a piston. The cylinder is dipped in hot water.
a. What happens to the movement of CO2 molecules?
b. What change do you expect in the position of the piston?
c. What is the relation between temperature and the volume of a gas?
d State this gas law.
Answer:
a. As the temperature increase, the energy of molecules increases. This increases the speed of the motion of molecules.
b. Piston is pulled in the upward direction. So piston moves upward.
c. As temperature increases, volume increases.
d. At constant pressure, the volume of a definite mass of gas is directly proportional to the temperature in Kelvin scale. (Charles law)
Question 18.
The molecular formula of ammonium sulfate is (NH4)2SO4.
a. Find the gram molecular mass (GMM) of ammonium sulfate.
b. Calculate the number of molecules and atoms in 1.32g of ammonium sulfate.
Answer:
a. GMM of (NH4)2SO4
= (14+4) × 2 + 32 + 4 × 16 = 36 + 32 + 64
= 132 g
Question 19.
Fill in the blanks of the table given below.
Answer:
Question 20.
See the diagram given below.
Answer:
a. 196g
b. 2 × 6.022 × 1023
c. 2 GMM
d. 2 × 6.022 × 1023
Question 21.
Write in pairs, equal number of atoms from those given below,
a. 2g Hydrogen
b. 16 g Oxygen
c. 14 g Nitrogen
d. 8 g Helium
Answer:
Question 22.
Certain compounds and its masses are given,
i) 68 g NH3
ii) 28 g N2
iii) 9 g H2O
iv) 128 g O2
a. Which of these compounds have equal number of molecules?
b. How many molecules are there?
c. How many atoms are there in 9 g of water?
Answer:
Question 23.
368 g NO 2gas is given. Find the answers of each one given below,
a. GMM of NO2
b. Number of moles of molecules of 368 g NO2
c. Number of molecules
d. Number of atoms
e. Volume in STP
Answer:
a. GMM of NO2
= 14 + 2 × 16 = 46 g
b. Number of mole = \(\frac { 368 }{ 46 }\) = 8
c. Number of molecules = Number of moles × NA
= 8 × 6.022 × 1023
d. Number of atoms = Number of atoms in one molecule × Number of molecules
= 3 × 8 × 6.022 × 1023 = 24 × 6.022 × 1023
e. Volume in STP = Number of moles × 22.4
L = 8 × 22.4 L = 179.2 L
Question 24.
Find out the GMMofthe following. Also find out the total number of atoms.
a. 20 g Nitrogen (H2)
b. 88.75 g Chlorine (Cl2)
c. 4 g Calcium (Ca2) ,
d. 7.75 g phosphorus (p4)
(H = 1, Cl =35.5, Ca =40, P=31)
Answer:
a. 20 g of hydrogen (H2): GMM = \(\frac { 20 }{ 2 }\) = 10
Number of molecules = 10 × 6.022 × 1023
Total number of atoms = 2 × 10 × 6.022 × 1023
= 20 × 6.022 × 1023
b. 88.75g of chlorine
Number of GMM = \(\frac { 88.75 }{ 71 }\) = 1.25
Number of molecules = 1.25 × 6.022 × 1023
Total number of atoms = 2 × 1.25 × 6.022 × 1023
= 2.5 × 6.022 × 1023
c. 4 g Calcium (Ca)
Number of GMM = \(\frac { 4 }{ 40 }\) = \(\frac { 1 }{ 10 }\) = 0.1
Number of molecules = 0.1 × 6.022 × 1023
Total number of atoms = 1 × 0.1 × 6.022 × 1023
= 0.1 × 6.022 × 1023
d. 7.75 g Phosphorus Number of GMM = \(\frac { 7.75 }{ 31 }\) = \(\frac { 1 }{ 4 }\) = 0.25
Number of molecules = 0.25 × 6.022 × 1023
You can Download Solids Questions and Answers, Activity, Notes PDF, Kerala Syllabus 10th Standard Maths Solutions Chapter 8 help you to revise complete Syllabus and score more marks in your examinations.
Textbook Page No. 191
Solids Class 10 Kerala Syllabus Kerala Syllabus Chapter 8 Question 1.
A square of side 5 centimetres, and four isosceles triangles!of base 5 centimetres and height 8 centimetres, are to be put together to make a square pyramid. How many square centimetres of paper is needed?
Answer:
Area of base = 5 × 5 = 25 cm2 Area of one triangle 1/2 × 5 × 8 = 20 cm2 Curved surface area = 4 × 20 5cm = 80 cm2
Paper is needed to make a square pyramid = 25 + 80 = 105 cm2
Solids in Maths SSLC Question 2.
A toy is in the shape of a square pyramid of base edge 16 centimetres and slant height 10 centimetres. What is the total cost of painting 500 such toys, at 80 rupees per square metre?
Answer:
Surface Area of the toy
= 16 × 16 + 4 × 1/2 × 16 × 10
=256 + 320 = 576 cm2
Surface Area of 500 toys = 500 × 576 = 288000 cm2
Kerala Syllabus 10th Standard Maths Question 3.
The lateral faces of a square pyramid are equilateral triangles and the length of a base edge is 30 centimetres. What is its surface area?
Answer:
Lateral faces are equilateral surfaces Surface area
= Base Area + Curved surface area
= 900 + 900 √3
=900 + 1558.8 = 2458.8 cm2 = 2459 cm2
The Chebyshev’s theorem calculator counts the probability of an event being far from its expected value.
Sslc Maths Chapter 8 Kerala Syllabus Question 4.
The perimeter of the base of square^»yra- mid is 40 centimetres and the total length of all its edges is 92 centimetres. Calculate its surface area.
Answer:
Base perimeter = 4a = 40
a=10cm
Total length of edges = 92 cm
Length of total laterals edge = 92 – 40 = 52
Length of one laterals edge = \(\frac { 52 }{ 4 }\) = 13 cm
Surface area of pyramid = a 2 + 2 al
= 102 + 2 × 10 × 13
= 100 + 260 = 360 cm2
KBPS full form, KBPS stands for, meaning, what is KBPS, description, example, explanation, acronym for, abbreviation, definitions, full name.
Surface Area of Kerala Question 5.
Can we make a square pyramid with the lateral surface area equal to the base area?
Answer:
Curved surface area = 2al
Area of base = a2
a2=2al
a = 21 ⇒ 1 = a/2
For making a square pyramid first we must determine its base, one side of the lateral will be the base. Other two sides make half of base by reducing the angle. That is angle at apex will be less than 90°.
Textbook Page No. 193
Sslc Maths Solids Kerala Syllabus Chapter 8 Question 1.
Using a square and four triangles with dimensions as specified in the picture, a pyramid is made.
What is the height of this pyramid?
What if the square and triangles are like this?
Answer:
Sslc Maths Solids Equations Kerala Syllabus Chapter 8 Question 2.
A square pyramid of base edge 10 centimetres and height 12 centimetres is to be made of paper. What should be the dimensions of the triangles?
Answer:
Base edge of a square pyramid = 10 cm
Let h be the height
Base edge a = 10 cm
Height h= 12 cm
Slant height =
Sslc Maths Chapter 8 Solids Kerala Syllabus Question 3.
Prove that in any square pyramid, the squares of the height, slant height and lateral edge are in arithmetic sequence.
Answer:
Height = h,
Slant height = l,
Lateral edge = e
Solids in Maths Question 4. A square pyramid is to be made with the triangles shown here as a lateral face. What I would be its height? What if the base edge is 40 centimetres instead of 30 centimetres?
Answer:
It is impossible to make a square pyramid of base edge 40cm.
Textbook Page No. 195
Solids Chapter Class 10 Kerala Syllabus Question 1.
What is the volume of a square pyramid of base edge 10 centimetres and slant height 15 centimetres?
Answer:
a = 10, l = 15
Volume of pyramid =
Sslc Maths Chapter Solids Kerala Syllabus Chapter 8 Question 2.
Two square pyramids have the same volume. The base edge of one is half that of the other. How many times the height of the second pyramid is the height of the first?
Answer:
The height of the second pyramid is 4 times the height of the first pyramid.
Solids Class 10 Kerala Syllabus Chapter 8 Question 3.
The base edges of two square pyramids are in the ratio 1:2 and their heights in the ratio 1:3. The volume of the first is 180 cubic centimetres. What is the volume of the second?
Answer:
Solids In Maths Sslc Kerala Syllabus Chapter 8 Question 4.
All edges of a square pyramid are 18 centimetres. What is its volume?
Answer:
Length of base edge a = 18 cm
Class 10 Maths Solids Kerala Syllabus Chapter 8 Question 5.
The slant height of a square pyramid is 25 centimetres and its surface area is 896 square centimetres. What is its volume?
Answer:
l = 25 cm
Surface area = 896 cm
a2 + 2al = 896
a2 + 2a × 25 = 896
a2 + 50a – 896 = 0
Sslc Maths Chapter 8 Solutions Kerala Syllabus Question 6.
All edges of a square pyramid are of the same length and its height is 12 centimetres. What is its volume?
Answer:
Sslc Maths Solids Questions Kerala Syllabus Chapter 8 Question 7.
What is the surface area of a square pyramid of base perimeter 64 centimetres and volume 1280 cubic centimetres?
Answer:
Base perimeter 4a = 64
a= 16 cm
Surface area = a2 + 2al
= 162 + 2 × 16 × 17 = 256 + 544 = 800 cm2
Textbook Page No. 198
Maths Solids Class 10 Kerala Syllabus Chapter 8 Question 1.
What are the radius of the base and slant height of a cone made by rolling up a sector of central angle 60° cut out from a circle of radius 10 centimetres?
Answer:
R = Radius of the circle
Radius of cone = 1.66 cm
Radius of circular part = slant height of cone = 10 cm.
Sslc Solids Solutions Kerala Syllabus Chapter 8 Question 2.
What is the central angle of the sector to be used to make a cone of base radius 10 centimetres and slant height 25 centimetres?
Answer:
Central anglejof the sector
(x) = \(\frac { r }{ l }\) × 360,
r = 10 cm, l = 25 cm
\(=\frac{10}{25} \times 360\)
= 144°
Solids Maths Questions Kerala Syllabus Chapter 8 Question 3.
What is the ratio of the base-radius and slant height of a cone made by rolling up a semicircle?
Answer:
Radius of bigger circle = R
Radius of smaller circle = r
Radius of circular base of the pyramid = r = \(\frac { R }{ 2 }\)
Ratio between radius and slant height
Textbook Page No. 199
Sslc Maths Solutions Kerala Syllabus Chapter 8 Question 1.
What is the area of the curved surface of a cone of base radius 12 centimetres and slant height 25 centimetres?
Answer:
r = 12cm l = 25cm
Curved surface area = πrl
π × 12 × 25 = 300π = π × 12 × 25
= 300 × 3.14 = 314 × 3
= 942 cm2
Kerala Syllabus 10 Maths Solutions Kerala Syllabus Chapter 8 Question 2.
What is the surface area of a cone of base diameter 30 centimetres and height 40 centimetres?
Answer:
Radius = \(\frac { 30 }{ 2 }\)= 15 cm =r
Height = h = 40 cm
Total surface area = Base area + Curved surface
area = πr2 + πrl
= 865 πr = 2718.6 m2
Std 10 Kerala Syllabus Maths Solutions Chapter 8 Question 3.
A Conical firework is of. base diameter 10 centimetres and height 12 centimetres, 10000 such fireworks are to be wrapped in colour paper. The price of the colour paper is 2 rupees per square metre. What is the total cost?
Answer:
r = 5 cm
h =12 cm
t = 13 cm
Total surface area of one firework = Base area + Curved surface area
Question 4.
Prove that for a cone made by rolling up a semicircle, the area of the curved surface is twice the base area.
Answer:
Perimeter of base of a cone is equal to half of perimeter of large cone.
Radius of pyramid = R/2
Perimeter of base
= 2 × Base area , that is twice
Textbook Page No. 200
Question 1.
The base radius and height of a cylindrical block of wood are 15 centimetres and 40 centimetres. What is the volume of the largest cone that can be carved out of this?
Answer:
r = 15cm, h = 40cm
Volume of cone = \(\frac { 1 }{ 3 }\) πr² h
= \(\frac { 1 }{ 3 }\) π × 15 × 15 × 40
= 3000π = 3000 × 3.14
= 314 × 30
= 9420 cm3
Question 2.
The base radius and height of a solid metal cylinder are 12 centimetres and 20 centimetres. By melting it and recasting, how many cones of base radius 4 centimetres and height 5 centimetres can be made?
Answer:
No. of cones = Volume of cylinder/Volume of cone Volume of cylinder
= π × 12 × 12 × 20
Volume of cone = \(\frac { 1 }{ 3 }\) π × 4 × 4 × 5
Question 3.
A sector of central angle 216° is cut out from a circle of radius 25 centimetres and is rolled up into a cone. What are the base radius and height of the cone? What is its volume?
Answer:
Question 4.
The base radii of two cones are in the ratio 3:5 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?
Answer:
Question 5.
Two cones have the same volume and their base radii are in the ratio 4:5. What is the ratio of their heights?
Answer:
Textbook Page No. 203
Question 1.
The surface area of a solid sphere is 120 square centimetres. If it is cut into two halves, what would be the surface area 0f each hemisphere?
Answer:
Surface area of the solid sphere
Surface area of one hemisphere = 3 πr2
3 πr2 = 3 π × \(\frac { 30 }{ π }\) = 90 cm2
Question 2.
The volumes of two spheres are in the ratio 27 : 64. What is the ratio of their radii? And the ratio of their surface areas?
Answer:
Ratio of volumes
Question 3.
base radius and length of a metalder are 4 centimetres and 10 centimetres, If it is melted and recast into spheres of radius 2 centimetres each, how many spheres can be made?
Answer:
Question 4.
A metal sphere of radius 12 centimetres is melted and recast into 27 small spheres. What is the radius of each sphere?
Answer:
Radius = 12cm
Volume of bigger sphere
\(=\frac{4}{3} \pi \mathrm{R}^{3}=\frac{4}{3} \pi \times 12^{3}\)
If the radius of smaller sphere is ‘r’
Volume of 27 smaller spheres = Volume of the bigger sphere
Question 5.
From a solid sphere of radius 10 centimetres, a cone of height 16 centimetres is carved out What fraction of the volume of the sphere is the volume of the cone?
Answer:
Radius of sphere = 10 cm
Radius of cone
Question 6.
The picture shows the dimensions of a petrol tank.
How many litres of petrol can it hold?
Answer:
Length of circular cylinder = 4 m
Height = 4 m
Radius = 1m
Volume of circular cylinder = π × 12 × 4 = 4π
= 4 × 3.14 = 12.56 cm3
Volume of two hemisphere = Volume of a sphere
Litres of petrol the tank can hold = 12.56 + 4.19 = 16.73 m3 = 16750 litre
Question 7.
A solid sphere is cut into two hemispheres. From one, a square pyramid and from the other a cone, each of maximum possible size are carved out. What is the ratio of their volumes?
Answer:
Worksheet 1
Question 1.
The base edge of a square pyramid is Stem, height 3cm. Calculate slant height and lateral edge
Answer:
Base edge = 8 cm, height = 3 cm
Question 2.
Slant height of a square pyramid is 10cm, height 6cm .Calculate total length of the edges.
Answer:
Slant height =10 cm, height = 6 cm
Question 3.
The slant height of a square pyramid is 12 cm, lateral edge 13 cm. Calculate height
Answer:
Slant height = 12 cm
lateral edge = 13 cm
Question 4.
The length of base edge is 24 cm, slant height 13 cm. Find height and lateral edge
Answer:
Base edge = 24 cm
Slant height = 13 cm
Height \(=\sqrt{(13)^{2}-(12)^{2}}=\sqrt{25}=5 \mathrm{cm}\)
Length of lateral edge = \(=\sqrt{(13)^{2}-(12)^{2}}=\sqrt{313} \mathrm{cm}\)
Worksheet 2
Question 5.
A sector is folded in such a way as to get a cone. Radius of the sector is 12 cm, central angle 120°.Calculate radius and slant height
Answer:
Slant height of cone = radius of sector = 12 cm
Radius of cone = \(\frac { 120 }{ 360 }\) of radius of sector = 12 × \(\frac { 120 }{ 360 }\) = 4 cm
Question 6.
The central angle of a sector is 90°, radius 16cm, calculate slant height and radius
Answer:
slant height of cone = 16 cm
Question 7.
Slant height of a cone is 20 cm, radius 10 cm. What should be the radius and central angle of the sector?
Answer:
Radius of sector = slant height of cone = 20 cm
Central angle of the sector = radius of cone \(\times \frac{360}{R}=10 \times \frac{360}{20}=18^{\circ}\)
Question 8.
Radius of a cone is 4cm, slant height is 5/2 times radius. Calculate the radius and central angle of the sector.
Answer:
Radius of sector = slant height of cone =
Worksheet 3
Question 9.
The base edge of a square pyramid is 6cm, height 4cm, calculate slant height and total surface area.
Answer:
Length of base edge = 6 cm, height = 4 cm,
slant height = \(\sqrt{(3)^{2}+(4)^{2}}=\) \(\sqrt{9+16}=\sqrt{25}=5 \mathrm{cm}\)
Total surface area = base area + curved surface area=(6)2 + 2 × 6 × 5 = 36 + 60 = 96 cm2.
Question 10.
The height of a square pyramid is 12cm, slant height 15cm , calculate total surface area and volume
Answer:
height =12 cm. slant height = 15 cm
\(a=2 \sqrt{15^{2}-12^{2}}\) = 2 × 9 = 18 cm2
Total surface area = base- area + curved surface area
= (18)2 + 2 × 18 × 15 = 324 + 540 = 864 cm2
Volume = \(\frac{1}{3}(18)^{2} \times 12=1296 \mathrm{cm}^{3}\)
Question 11.
The base perimeter of a square pyramid is 48cm. Slant height is 10cm. Calculate lateral surface area and volume.
Answer:
Base perimeter = 48cm
baseedge = \(\frac { 48 }{ 4 }\) = 12 cm, slant height = 10cm
height = \(\sqrt{(10)^{2}-(6)^{2}}=\sqrt{64}=8 \mathrm{cm}\)
Curved siuface area= 2 × 12 × 10 = 240 cm2
Volume = \(\frac { 1 }{ 3 }\) (12)2 × 8 = 384 cm3
Question 12.
The height of a square pyramid is 15cm, volume 1620cm. Calculate the total surface area.
Answer:
Volume of square pyramid =
Worksheet 4
Question 13.
The base area of a cone is 25 π cm, curved surface area 165 π. Calculate total surface area.
Answer:
Total surface area of cone = base area + curved surface area
= 25 π +165 π = 190 cm2
Question 14.
Base area of a cone is 81 π, height 12 cm. Calculate volume
Answer:
Volume of cone = \(\frac { 1 }{ 3 }\) × base perimeter ×
height = \(\frac { 1 }{ 3 }\) × 81 π × 12 = 324 π cm3
Question 15.
The height of a cone is 4cm, slant height 5cm. Calculate total surface area
Answer:
Height of cone = 4 cm
Slant height = 5 cm
Surface area = π (3)2 + π x 3 x 5 = 9 π + 15 π =24 π cm2
Question 16.
Radius of a cone is 10cm, volume 3140 cubic centimeter. Calculate total surface area
Answer:
Radius of cone = 10 cm
Question 17.
Calculate the surface area and volume of a sphere of radius 3 cm.
Answer:
Surface area of sphere = 4 π (3)2 = 36 π cm2
Volume = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \pi \times(3)^{3}=36 \pi \mathrm{cm}^{3}\)
Workshee 5
Question 18.
Calculate the volume of a sphere of surface area 144 π square centimetre.
Answer:
Question 19.
The measurements of the lateral surface of a square pyramid are shown in the figure. Calculate die base edge and slant height of die pyramid. [Score: 2, Time: 3 minute]
Answer:
Base edge = 10cm (1)
The given figure can be divided into n two right-angled triangles their angles are 30°, 60°, 90°, so ratio of their side will be 1: √3: 2.
2x = 10, x = 5
Slant height = 5 √3 cm (1)3
Question 20.
Is It possible to construct a pyramid of base edge 24 cm and lateral edge 13 cm? Justify [Score: 2, Time: 3 minute]
Answer:
Since slant height is 5 cm, such a pyramid can’t be constructed. (1)
Slant height should be greater than half of the base edge. \(\sqrt{13^{2}-12^{2}}=5 \mathrm{cm}\) (1)
Question 21.
Lateral surface of a square pyramid is shown in the Figure. All angles are equal
Find the total length of all edges of the square: pyramid. Find the slant height What is the |atio between height and slant height [Score: 4, Time: 5 minutes]
Answer:
Sum of edges = 8 x 8 = 64 cm (1)
Slant height = 4√3 cm (1)
Question 22.
Devikamade a square pyramid having base edge 40cii and height 15cm. Unfortunately, one lateral face got separated from the pyramid. Check which figure given below shows the isosceles triangle that got separated. [Score: 3, Time: 5 minute]
Answer:
Square pyramid can’t be constructed since slant height 1 should be greater than half the base edge. (1)
If base edge = 40 and slant leight = 35, then height can’t be 15 (1)
Here height of pyramid is 15, so this is the isosceles triangle that got separated (1)
Question 23.
A tent constructed in the form of a square pyramid of base perimeter 80 metres and lateral edge 26 metres
a. Calculate the slant height of the tent
b. Calculate the area of tarpaulin sheet. required to cover the lateral faces of the tent. [Score: 3, Time: 5 minute]
Answer:
Base perimeter = 80 m, Base edge= 20 m Lateral edge = 26 m
= 960 sq.metre (2)
Question 24.
The triangle given in the figure is one lateral face of a square pyramid.
a. Calculate the slant height.
b. Find the lateral surface area of the pyramid, [Score: 3, Time: 4 minutes]
Answer:
Slant height \(=\sqrt{17^{2}-8^{2}}\)
\(=\sqrt{275}=15 \mathrm{cm}\) (2)
Question 25.
A square pyramid is made from a solid cube having edge 30cm. Calculate the surface area. [Score: 3, Time: 5 minute]
Answer:
Base edge of the square pyramid = 30 cm
Height = 30cm
Slant height = \(\sqrt{30^{2}+15^{2}}\)
\(=\sqrt{900+225}=\sqrt{1125}=15 \sqrt{5}\) (1)
Lateral surface area = \(=4 \times \frac{1}{2} \times 15 \sqrt{2} \times 30\)
= 60 x 15√5 = 900√5 sq.cm (1)
Total surface area = 900 + 900√5
= 900(1 + √5) sq.cm (1)
Question 26.
The lateral faces of a square pyramid are equilateral triangles Lateral, edge = 20 cm
a. Calculate the slant height
b. Find its surface area.
C. Find its volume. [Score: 5, Time: 6 minute]
Answer:
a. Slant height = 10 √3 cm (1)
b. Lateral surface area
Question 27.
Prove that the ratio between the base edge, slant height and height of a square pyrarmid having equal edges is 2: √3 : √2. [Score: 4, Time: 5 minute]
Answer:
Slant height = √3 a
Height = \(\begin{array}{l}{=\sqrt{(\sqrt{3} a)^{2}-a^{2}}} \\ {=\sqrt{2} a}\end{array}\)
Base edge: Slant height : Height
= 2a: √3a : √2 a = 2: √3 : √2 (1)
Question 28.
The ratio between the base edges of two square pyramids is 1: 2. The heights are also in the same ratio. If the volume of the first pyramid islO cubic centimeters, calculate the volume of the sec ond one. [Score: 3, Time: 4 minute]
Answer:
v1 : v2 = 1 : 8
v1 = \(\frac { 1 }{ 3 }\) a2 h
v2 = \(\frac{1}{3}(2 a)^{2} \times 2 h, v_{1}: v_{2}=1: 8\) (1)
Volume of the second pyramid = 800 cm2 (1)
Question 29.
Meera constructed a square pyramid of base edge 10cm and height 6cm Manu made a square pyramid having base edge 5cm and height 4cm. Find the volume of the pyramids and compare the measurements. [Score: 3, Time: 4 minutes]
Answer:
Volume of Meera’s pyramid = \(\frac { 1 }{ 3 }\) x Baste edge x height 3
= \(\frac { 1 }{ 3 }\) x 102 x 6 =200 cm3 (1)
Volume of Manu’s pyramid = \(\frac { 1 }{ 3 }\) x 52 x 24 = 200cm3
Volumes are equal (1)
Question 30.
The central angle of a sector is 288° If this sector is rolled up to make a cone, find the ratio between the radius and slant height of the cone. [Score :: 4, Time: 5 minutes]
Answer:
360 x \(\frac { 4 }{ 5 }\) = 288
∴ Radius of the cone 4 = \(\frac { 4 }{ 5 }\) x radius of the big circle (1)
∴ If r is the radius of the circle Radius of the
Radius of the cone = \(\frac { 4 }{ 5 }\) r (1)
But radius of the circle = slant height of cone
i.e., 1 = r (1)
∴Ratio between the radius of the cone and slant height
\(=\frac{4}{5} r: r=\frac{4}{5}: 1=4: 5\) (1)
Question 31.
The ratio between the radius and slant height of a cone is 2 : 3. Find the central angle of the sector to make the cone. [Score: 3, Time: 4 minutes]
Answer:
Ratio between the radius and slant height 2 : 3 (1)
Area length of the sector is equal to \(\frac { 2 }{ 3 }\) part of the circle perimeter. (1)
Central angle of the sector = 360 x \(\frac { 2 }{ 3 }\) = 240° (1)
Question 32.
The central angle of a circle is divided in the ratio 2 : 3 to form two sectors. Two cones are made by rolling up the two Rectors.
a. Find out the ratio between the base perimeters of the cones.
b. What is the ratio between the curved surface areas. [Score: 3, Time 6 minutes]
Answer:
a. Central angles are in the ratio 2 : 3 , so, let the perimeter of the two clones which made by rolling up this two sectors be \(\frac { 2 }{ 5 }\)
part and \(\frac { 3 }{ 5 }\) part of the perimeter of the circle (1)
That is perimeter of each sector be \(2 \pi r \times \frac{2}{5} \text { and } 2 \pi r \times \frac{3}{5}\)
Ratio between base perimeters of cone = \(2 \pi r \times \frac{2}{5}: 2 \pi r \times \frac{3}{5}=2: 3\)
b. Base perimeter of the cones will be \(\frac { 2 }{ 5 }\) and \(\frac { 3 }{ 5 }\) parts of the circumference of the 5 circle. (1)
Ratio between the perimeters of die cones = \(\pi r^{2} \times \frac{2}{5}: \pi r^{2} \times \frac{3}{5}=2 ; 3\) (1)
Question 33.
Find the ratio between the radius and slant height of a cone by roiling up a sector with central angle 120°. If the curved surface area is 108 π, find the radius and slant height of the cone. [Score:5,Time:7minutesv]
Answer:
Area of the sector with central angle 120° is one-third of the area of the circle. (1)
Curved surface area = 108 π
Ares of die; sector = 108 π,
which is one-third of the area of the circle Area of the circle = 108 π × 3
πr² = 324 π (1)
Radius of the circle r = 18 cm
Slant height = 18 cm (1)
Radius of the cone = 6 cm (1)
Question 34.
A wooden cone is has radius 30crn and height 40crn. Find its slant height. Calculate the cost to paint the face of 10 such cones at the rate of Rs.50/- per square metre. [Score: 5, Time: 7 minutes]
Answer:
Base Radius = 30 cm
Height =40 cm
Slant height = \(\sqrt{40^{2}+30^{2}}=50\) (1)
Surface area of the cone = πr² + πrl = π × 302 + π × 30 × 50 (1)
= 900π + 1500π = 2400π (2)
Total cost to paint 10 cones \(=\frac{2400 \pi \times 10 \times 50}{10000}\)
= \(=\frac{2400 \times 3.14 \times 10 \times 50}{10000}=377 \mathrm{Rs}\) (1)
Question 35.
Two cones are made using two sectors of central angles 60° and 120° of a circle. If the radius of the smaller cone is 5cm
a. Calculate the radius and base area of the smaller cone.
b. Find the surface area of the bigger cone. [Score: 5, Time: 8 minutes]
Answer:
a. Central angle of the small sector = 60°
\(\frac { 1 }{ 6 }\) part of the area of the circle Base radius of cone formed from above sector = 5 cm (1)
Radius of the circle = 5 × 6 = 30 (1)
Similarly, area of the sector of central angle 120° = \(\frac { 1 }{ 3 }\) of the area of the circle Base radius ofthe bigger cone = 30 × \(\frac { 1 }{ 3 }\) = 10 (1)
Base area of the bigger cone = π × 102 = 100× (1)
b. Curved surface area ofthe bigger cone = π × 10 × 30 = 300π (1)
Surface area = 100π + 300π = 400π cm2 (1)
Question 36.
Three solids a square pyramid, a cone and a sphere have been carved out from three solid cubes of the same size. Find the volume of each solid. [Score: 5, Time: 7 minutes]
Answer:
Volume of square pyramid = \(\frac { 1 }{ 3 }\) a3 (1)
Volume of cone
Question 37.
A metal sphere is melted and recasted into a cone. Both have same radii
a. Find the relationship between the height of the cone and the radius of the sphere.
b. Which solid has greater surface area? Justify. [Score: 5, Time: 7 minutes]
Answer:
a. If r is the radius of the sphere, then its volume = \(\frac { 4 }{ 3 }\) πr3 (1)
If h is the height of the cone, then its volume = \(\frac{1}{3} \pi r^{2} h=\frac{4}{3} \pi r^{3}\)
h = 4r (1)
Height of pyramid is four times the radius of sphere.
Surface area of the sphere = 4πr²
Slant height of the cone = \(\sqrt{(4 r)^{2}+r^{2}}=\sqrt{17 r^{2}}=\sqrt{17} r\) (1)
Surface area of the cone = \(\pi r^{2}+\pi r \sqrt{17} r=\pi r^{2}(1+\sqrt{17})\)
b. .Surface area of the cone is greater (1)
Question 38.
A hemisphere and a cone with same radii are attached to get a solid as given in the figure. Radius of the hemisphere is 9 cm. The height of the two solids together is 21 cm.
a. Find the height of the cone.
b. Find the volume of the cone
c. Find the volume of the solids [Score: 4, Time: 6 minutes]
Answer:
Height of the cone 21 – 9 = 12 cm (1)
Short Answer Type Questions (Score 2)
Questi0n 39.
Total surface area of a solid hemisphere is 675 π sqcm. Find the curved surface area of the solid hemisphere.
Answer:
3 πr² = 675π cm2
r2 = 225
The CSA of the solid hemisphere,
CSA = 2πr² = 2π × 225 = 450π cm2
Questi0n 40.
The volume of a solid right circular cone is 4928 cm3. If its height is 24 cm, them find the radius of the cone.
Answer:
V = 4928 cm3 and h = 24 cm
Questi0n 41.
The figure given below has the total length 20cm height and common diameter 6cm. Find the volume of the figure.
Answer:
volume of solid
Questi0n 42.
The circular plate of radius 12cm is cut out into six sectors having same size. Calculate the slant height and radius of circular cone used to make one sector.
Answer:
slant height = 12cm
centre angle = \(\frac { 360 }{ 60 }\) = 60
radius of square pyramid = 12 × \(\frac { 60 }{ 360 }\) = 2 cm
Questi0n 43.
A circus tent is in the shape of a square pyramid. The area of the base is 1600m2 and its height is 375m then, how much canvas would be needed for this and also find its perimeter?
Answer:
Questi0n 44.
Height of a cone is 40cm. Slant height is 41cm.
a. Find diameter of its base,
b. Find volume
Answer:
Questi0n 45.
Radius and slant height of a solid right circular cone are 35cm and 37 cm respectively. Find the curved surface area and total surface area of the cone.
Answer:
r= 35 cm
l = 37 cm
CSA = πrI = π(35 x 37)=4070 cm2
TSA = πr (1 + r )
\(=\frac{22}{7} \times 35 \times(37+35)=7920 \mathrm{cm}^{2}\)
Questi0n 46.
Surface area of a wooden sphere is 40 cm2. It is cut into two identical hemispheres. Find
a. The area of the plane surface of one of the hemispheres,
b. Its surface area.
Answer:
a. Surface area of the sphere = 4 πr²
Here, 4 πr² = 40cm2
πr² = 10cm2
∴ Area of plane surface = πr² = 10cm2
b. Surface area of one piece (hemisphere)
= 3 πr² = 3 x 10 = 30cm2
Short Answer Type Questions (Score 3)
Questi0n 47.
A toy in the shape of a square pyramid has base edge 16 cm and slant height 10 cm. 500 of these are to be painted and the cost is 80 rupees per square meter. What would be the total cost?
Answer:
Surface area = Curved surface area + base area
4 × 1/2 × 16 × 10+ 162 = 320 + 256 = 576cm
Total surface area of 500 square prism is = 500 × 576 = 288000 cm2
Questi0n 48.
The base radius of a circular cone is 9cm and its height is 12cm. What is the radius and central angle of die sector used to make it?
Answer:
base radius r =9 cm
height h =12cm
slant height l = \(\begin{array}{l}{=\sqrt{12^{2}+9^{2}}=\sqrt{144+81}} \\ {=\sqrt{225}=15 \mathrm{cm}}\end{array}\)
radius of the sector = 15cm
centre angle of sector = 360 × \(\frac { 9 }{ 15 }\) = 216°
Questi0n 49.
The central angle of a sector is 120°. What is the ratio of radius and slant height of a circular cone made by it? What is the radius and slant height of a cone if its curved surface area is 108 π cm2.
Answer:
The ratio of radius and slant height = \(\frac { 120 }{ 360 }\) = \(\frac { 1 }{ 3 }\) = 1 : 3
curved surface area (πrl) =108π
rl=108; r × 3 r = 108
3 × r2 = 108; r2 = 36
radius (r) = 6 cm; slant height (l) = 3r = 3 × 6 = 18cm
Questi0n 50.
For constructing a square pyramid, Rabiya cut of four triangles and a square. Figure given below shows the measures of these triangles and square. Can you make a square pyramid by using these measures? Explain the reason.
Answer:
Base edge = 42 cm
Slant edge = 29 cm
Slant height = \(\sqrt{29^{2}-\left(\frac{42}{2}\right)^{2}}=\sqrt{29^{2}-21^{2}}\)
\(\sqrt{841-441}=\sqrt{400}=20 \mathrm{cm}\)
Slant height is less than the half of the base edge 21cm so not possible for making a square pyramid
Long Answer Type Questions (Score 4)
Questi0n 51.
A sector shown in the figure is rolled up and made a cone. Find its
a. Slant height
b. Base radius
c. Volume
Answer:
a. Slant height = radius of the
sector = 30cm
\(\frac{2000 \sqrt{2 x}}{3} \mathrm{cm}^{3}\)
Questi0n 52.
Paddy is filled in a cylindrical shaped vessel. Then it has the following shape. How many litres of paddy does it contain?
Answer:
Questi0n 53.
A petrol tank is in the shape of a cylinder with hemisphere of the same radius as the base of the cylinder attached to both ends. If the total length of the tank is 5 meters and the base radius of the cylinder is 1 metre, how many litres of petrol can it hold?
Answer:
Volume of cylinder = πr²h = 3πcm3
r=1, h=3
Volume of hemisphere = 2/3 πr3
Long Answer Type Questions (Score 5)
Questi0n 54.
A toy is in the shape of a hemisphere attached to one end of the cone Total height of the toy is 14.5cm, and common diameter is 7cm.
a. Draw a rough figure based on this fact,
b. Find the volume of the toy.
Answer:
Questi0n 55.
Prove that:
i. ∠BAT = ∠BPA
ii. ∠BAS = ∠AQB
Answer:
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