Class 10

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SSLC Chemistry Chapter 6 Notes Questions and Answers Pdf Metals

SCERT Class 10 Chemistry Chapter 6 Metals Notes Pdf

SSLC Chemistry Chapter 6 Questions and Answers – Let Us Assess

Question 1.
Define the following and answer the given questions.
(1) Ore – What is the ore of aluminium?
(2) Roasting – Which type of ores are subjected to roasting?
(3) Reducing agent – What is the reducing agent used in the manufacture of aluminium?
(4) Flux – Which flux is used in the manufacture of copper? Why?
(5) Leaching – Which metal ore is leached with sodium cyanide?
Answer:
(1) Ore – Any mineral from which a metal can be extracted easily and economically is called an ore of that metal. Ore of Aluminium is Bauxite

(2) Roasting – Roasting refers to heating the concentrated ore at a temperature below the melting point in the presence of air. Sulphide ores are subjected to roasting,

(3) Reducing agent – The chemical species that helps reduction is the reducing agent. The reducing agent gets oxidised in a chemical reaction. The reducing agent used in the manufacture of aluminium is electricity.

(4) Flux – The ore may contain impurities that cannot be removed in the first stages of concentration methods. In order to remove these impurities, a substance is added during the metallurgical stage. This substance is known as Flux.
Flux used in the manufacture of Copper is Sand (SiO₂). The ore contains basic impurities such as iron oxide even after concentration. So acidic flux is used.

(5) Leaching – The powdered ore is mixed with a suitable solvent. The ore dissolves in it or undergoes a chemical reaction to become a solution. The gangue remains insoluble in it. It is then filtered out. This process is called leaching. Metal ore leached with sodium cyanide is gold and silver.

Question 2.
Find the relation and write the answer.
Zinc sulphide: Roasting
Calcium carbonate: …………(a)…………..
Magnetite: Magnetic separation
Bauxite: …………(b)…………..
Answer:
(a) Calcination
(b) Leaching

Question 3.
The chemical reaction that occurs when calcium carbonate is heated is given.
CaCO₃ CaO + CO₂
How is this chemical change of calcium carbonate utilised in the industrial preparation of iron?
Answer:
The chemical change of calcium carbonate (CaCO₃) is utilised in the industrial preparation of iron to remove impurities from the iron ore. The product of this reaction, calcium oxide (CaO), acts as a flux in the blast furnace.

When heated, calcium carbonate decomposes into calcium oxide and carbon dioxide. The calcium oxide then reacts with the silica (SiO₂) impurities present in the iron ore, forming a molten substance called slag (calcium silicate, CaSiO₂). This slag is less dense than the molten iron, allowing it to float on top and be easily separated, thus purifying the iron. The overall reaction for slag formation is:
CaO(s) + SiO₂(s) → CaSiO₂(l)

Question 4.
Following are two facts related to the manufacture of an industrially important metal.
• The ore is treated with hot NaOH solution.
• Electricity is used as the reducing agent to extract the metal.
(i) These facts are related to the production of which metal?
(ii) What is the reason for using electricity as the reducing agent?
(iii) Which substance is used as the electrolyte here?
(iv) Which gas is liberated at the anode?
Answer:
(i) Aluminium
(ii) Aluminium is a highly reactive and electropositive metal.
(iii) Molten cryolite (Na₃AlF₆)
(iv) Oxygen (O₂)

Question 5.
A portion of the periodic table is given.

(i) Which method given below is more possible for the production of gallium?
(Reduction using carbon, electrolysis)
(ii) If gallium chloride (GaCl₃) is electrolysed, what is the product formed at the cathode. Write the chemical equation.
(iii) Write the subshell electron configuration of the outermost shell of gallium.
Answer:
(i) Electrolysis. Gallium is a highly reactive metal.
(ii) Gallium metal.
Ga³⁺ + 3e^– → Ga
(iii) 4s² 4p¹ (Atomic number: 31)

Question 6.
Corrosion is a process by which iron is converted into its oxide. This is an oxidation reaction.
(i) What is oxidation?
(ii) Complete the following chemical equation.
….Fe + …O₂ → ..Fe₂O₃
(iii) Suggest two methods for preventing the corrosion of iron.
Answer:
(i) The process involving loss of electrons in a chemical reaction is called oxidation.
(ii) 4Fe + 3O₂ → 2Fe₂O₃
(iii)

1. Painting or coating
2. Galvanization

Question 7.
Alloys containing iron are given. Find a, b, c, d.

Alloys	Constituent elements	Uses
Alnico	(a)	(b)
(c)	Fe, Cr, Ni, C	Resist the corrosion of iron. Making of utensils
Silicon steel	Fe, Si, C	(d)

Answer:
(a) Fe, Al, Ni, CO
(b) For the manufacture of permanent magnets
(c) Stainless steel
(d) Also known as electrical steel. As it reduces electric loss, it is used in the core of electromagnetic instruments like motors, generators, transformers etc.

Chemistry Class 10 Chapter 6 Notes Kerala Syllabus Metals

Question 1.
Metals are used in many forms in different fields of our life. Try to find some of them and complete the table.

Answer:

Transport sector	Cars, cycles. Trains. Buses. Airplanes. Ships, Trucks
Agricultural sector	Agricultural tools, tillers. Tractors. Combine harvesters, ploughs. Watering cans
Construction sector	Buildings, bridges, pipes. Steel beams. Nails. Screws. Cranes
Technology sector	Phones, laptops, electrical wires. Microchips. Computer casings. Circuit boards
Consumer goods	Jewellery, household articles. Utensils, Appliances. Furniture. Coins

Question 2.
Which metals are. mostly used in these situations? List them.
Answer:
Iron, Gold, Steel, Aluminium, Copper, Silver, Platinum
Metals are used in various fields because of some of their common properties.

Question 3.
What properties of metals are utilised in various fields?
Answer:

• Sonority
• Ductility
• High melting point
• Thermal conductivity
• Malleability
• Metallic lustre
• Hardness

Question 4.
The nature of ores is given. Complete the table using appropriate concentration methods.

Answer:

Nature of ore	Concentration method
1. Ores are less dense than the impurities	Froth floatation
2. Ores have magnetic properties but impurities are non magnetic	Magnetic separation
3. A solvent which dissolves the ore is used.	Leaching
4. Ores are denser than the impurities.	Levigation or Hydraulic washing

Question 5.
How are metals arranged in the reactivity series?
Decreasing order of reactivity / increasing order of reactivity
Answer:
Decreasing order of reactivity
Here are some metals in the reactivity series.

Mercury can be easily separated from its ores by controlled heating.

Metals with comparatively moderate reactivity like, are generally found as oxides, sulphides or carbonates.
Eg. ZnCO₃, ZnS, Fe₂O₃, SnO₂, PbS
There are two stages for the production of metal from them.

1. Conversion of concentrated ore to oxide
Two methods are commonly used for this.

(a) Calcination
Metal carbonates or hydroxides are converted to their oxides by a process called calcination.

Calcination refers to heating the concentrated ore in limited quantities or absence of air at a temperature below the melting point.

Eg. CaCO₃(s) CaO(s) + CO₂(g)
ZnCO₃ (s) ZnO(s) + CO₂ (g)

(b) Roasting

Roasting refers to heating the concentrated ore at a temperature below the melting point in the presence of air.

Zinc sulphide and lead sulphide can be easily

c) converted into their oxides by this method
2ZnS(s) + 3₂ (g) 2ZnO(s) + 2S₂ (g)
2PbS (s) + 3₂ (g) 2PbO (s) + 2S₂ (g)

When the concentrated ore is subjected to roasting, the water content in them gets released as vapours. They become pure and moisture free oxide ores.

2. Reduction of oxide ore
The metal part of the ores is in the form of positive ions, and hence it can be extracted by a reduction process.

Question 6.
Note the process that separates zinc from zinc oxide.
ZnO + C → Zn + CO
Which is the reducing agent in this case?
Answer:
Carbon
Reducing agents such as coke (an allotrope of carbon) and carbon monoxide are used to extract moderately reactive metals such as iron, zinc and tin from their oxide ores. Carbon has a greater tendency to react with oxygen at high temperatures, so oxygen is easily removed from the ore.

Strong reducing agents are required to extract highly reactive metals such as sodium, potassium, calcium, magnesium and aluminium from their ores. Such metals are produced by electrolysis.

Question 7.
Complete the table regarding the manufacture of metals.

Answer:

Method of separating metals from ores	Metals
Electrolysis	Sodium, Potassium, Calcium, Magnesium and Aluminium
Reduction using coke	Iron, Zinc and Tin
Heating in controlled ways	Mercury

Question 8.
The order of reactivity of some metals is given.
Al > Zn > Fe > Au
a) Which metal forms the most stable compound?
Answer:
Aluminium

b) Which metal is produced by electrolysis?
Answer:
Aluminium

c) Which metal can displace zinc from compounds of zinc? Why?
Answer:
Aluminium can displace Zinc from its compounds. A metal can displace another from a compound if it is more reactive. Since A1 is higher than zinc in the reactivity series, it can displace Zinc.

d) Which metal is found in the free state in nature?
Answer:
Gold (Least reactive one)

e) Which metal is produced by reduction using carbon?
Answer:
Zinc and Iron

Question 9.
Complete the table regarding the manufacture of iron.

Answer:

Ores of iron and its chemical formula	Haematite (Fe2O3) Magnetite (Fe3O4)
Methods used for concentration	Washing in stream of water, roasting. Magnetic separation
Substance used to reduce the ore	Carbon (Coke)

Iron is produced in steel towers, known as blast furnaces, which are 25 to 30 meters high. Its interior is lined with bricks (Refractory bricks) capable of withstanding high temperatures

The concentrated ore mixed with limestone ( CaCO₃) and coke (C) is fed from the top of the furnace via cup and cone arrangement. This mixture is called charge.

A blast of hot air is blown into the furnace from its bottom. As the charge descends from the top of the furnace, various chemical changes take place inside it. The coke combines with the oxygen in the air, when the blast of hot air flows from the bottom to the top of the blast furnace.
C + O₂ → CO₂ + heat
This is an exothermic reaction. Coke reacts with carbon dioxide to produce carbon monoxide. This is an endothermic reaction.
CO₂ + C+ heat → 2CO
The main chemical reaction taking place in the blast furnace is the reduction of iron ore using carbon monoxide.
Fe₂O₃ + 3CO → 2Fe + 3CO₂
Even though the iron ore fed into the furnace is in the most refined form, it may contain acidic impurities such as sand (silica SiO₂) that cannot be removed in the first stages of concentration methods. For this, a substance known as flux is added during the metallurgical stage. As the impurities are acidic in nature, a basic substance is used as the flux. The flux combines with gangue to form molten slag.
Gangue + Flux → Slag
When heated at high temperature, the limestone decomposes to calcium oxide (Quick lime). Calcium oxide produced in this reaction acts as flux.
The chemical changes of limestone inside the blast furnace.
CaCO₃ CaO(s) + CO₂(g)
Calcium oxide reacts with sand (SiO₂) to form calcium silicate (slag), which melts easily.

If the gangue in the ore is basic in nature, acidic flux should be used.

Since the molten slag is less dense, it floats on the molten iron. These are removed at regular intervals through a separate tube/ tap. This slag is used in the production of cement and in the construction of roads.

The molten iron obtained from the blast furnace contains 4% carbon and small amounts of other impurities such as manganese, silicon, phosphorus and sulphur. This is called pig iron.

If the percentage of impurity in it is reduced, the iron obtained can be cast as desired. It is called cast iron. It contains more than 2% carbon.

Wrought iron is a relatively pure form of iron, containing 0.02% – 0.05% carbon and small amounts of phosphorus, silicon, manganese, sulphur etc.

Question 10.
Prepare a list of different types of steels, their methods of manufacture and use. ICT can be utilised.
Answer:

• Tool Steel: Contains elements like tungsten, molybdenum, and cobalt to increase hardness, heat resistance, and wear resistance. It is used for tools, dies, and cutting instruments.
• Structural Steel: Used in construction and contains manganese to improve strength. It is used in buildings, bridges, and other large structures.

Methods of Manufacture
The primary method for producing steel involves the following steps:

1. Blast Furnace: Iron ore is melted in a blast furnace to produce pig iron.
2. Steel Making: The pig iron is then processed in a Basic Oxygen Furnace (BOF) or an Electric Arc Furnace (EAF) to reduce carbon content and add alloying elements.
3. Casting: The molten steel is cast into various shapes, such as slabs, blooms, or billets.
4. Finishing: The cast steel is then rolled, shaped, and heat-treated to achieve the desired properties.

Alloy steel
The use of metals in their pure form is relatively less. Most of the metals are converted into alloys to be used for various purposes in daily life.

Question 11.
List the alloys familiar to you.
Answer:
Brass, Bronze, Solder, Amalgam, Duralumin
Alloys are mixtures of two or more metals. Nonmetals like carbon, nitrogen and phosphorus are also used for the production of alloys.

Question 12.
What are the constituents of steel?
Answer:
Iron, Chromium, Nickel, Carbon
Alloy steels are made by adding other metals to steel. Their properties are different from those of steel.
Some important alloy steels:

Alloy steels	Constituent elements	Properties and uses
Stainless steel	Fe, Cr, Ni, C	Resists corrosion. Manufacture of household appliances and blades.
Manganese steel	Fe, Mn, C	High hardness. Used for the manufacture of drills, safes, plates and railway tracks.
Silicon steel	Fe, Si, C	Also known as electrical steel. As it reduces electric loss, it is used in the core of electromagnetic instruments like motors, generators, transformers etc.

Question 13.
Expand the list by finding more alloys, their constituents and uses.
Answer:

Alloy steels	Constituent elements	Properties	Uses
Alnico	Fe, Al, Ni, Co	Magnetic nature	For the manufacture of permanent magnets
Nichrome	Fe, Ni, Cr, C	High resistance	For making heating coils

Question 14.
Complete the table regarding the production of zinc.

Answer:

Sulphide ore of zinc (Common name, chemical formula)	Zinc blende (ZnS)
Concentration method	Froth floatation
Carbonate ore of zinc (Common name, chemical formula)	Calamine (ZnCO3)

Question 15.
What are the processes used in the conversion of zinc ore to zinc oxide after the concentration of ore?
Answer:
Roasting. Zinc sulphide can be easily converted into their oxides by this method.
2ZnS (s) + 3O₂(g) 2ZnO(s) + 2SO₂ (g)

When the concentrated ore is subjected to roasting, the water content in them gets released as vapours. They become pure and moisture free oxide ores.

Coke is mixed with the zinc oxide obtained after these processes.

This mixture is put into a furnace similar to a blast furnace. At high temperature in the furnace, the coke reduces zinc oxide to zinc.

ZnO (s) + C (s)Zn(s) + CO (g)
Zinc has comparatively low boiling point (907°C) and hence it vapourises at high temperature and comes out. It is immediately cooled to liquid form.

Question 16.
You are familiar with the purification process of zinc. What is it known as?
Answer:
Distillation
Brass is an important alloy containing zinc.

Question 17.
Prepare a note on the production of zinc and its use. Record it in the science diary.
Answer:
Production and Use of Zinc
Zinc is primarily produced from the mineral sphalerite through a process that involves roasting, followed by either electrolysis or smelting. Its main use is for galvanizing to protect steel and iron from rust. It is also used in alloys like brass, in die-casting, and for batteries.

Question 18.
What method of concentration is suitable for these sulphide ores?
Answer:
Froth floatation

Question 19.
The ore contains basic impurities such as iron oxide even after concentration. What type of flux is suitable to remove it? (Acidic / Basic)
Answer:
Acidic flux
Sand (SiO₂) having acidic nature is added to the ore and heated to form cuprous sulphide (Cu₂S) of maximum purity. It is partially converted to cuprous oxide (Cu₂O) when heated under controlled flow of air. The cuprous oxide converts the remaining cuprous sulphide into copper.

Note that no reducing agent is added here. Molten copper is cooled in moulds. The copper lumps thus obtained are called blister copper. The sulphur dioxide gas produced during the production of copper is released when the metal condenses. As a result blisters are formed and they appear as such on the surface of copper lumps.

Question 20.
Brass and bronze are important alloys that contain copper. List the constituent metals in them.
Answer:
Brass is an alloy of copper and zinc. Bronze is traditionally an alloy of copper and tin.

Purification of Copper
The method of refining copper is electrolytic refining. Copper used for electrical purposes should be 99.99% pure. A high quality product is obtained through electrolytic refining. The diagram of refining of copper is given.

Question 21.
Observe the diagram and complete the table.

Answer:

Anode	Impure Copper
Cathode	Pure copper
Electrolyte	Copper Sulphate solution with H2SO4
Chemical equation of the reaction taking place at anode.	Cu → Cu2+ + 2e–
Chemical equation of the reaction taking place at cathode.	Cu2+ + 2e– → Cu

Question 22.
Prepare a note on the process involved in the purification of copper and record it in the science diary.
Answer:
Copper is purified using electrolytic refining. Impure copper acts as the anode (positive electrode) and pure copper as the cathode (negative electrode). Both are placed in a copper sulphate solution. When electricity is applied, copper from the anode dissolves and is deposited onto the cathode, leaving impurities behind.

Question 23.
What reaction takes place at the cathode?
Answer:
Al³⁺ + 3e^– → Al

Question 24.
What reaction takes place at the anode?
Answer:
2O^2- → O₂(g) + 2e^–

Question 25.
Which gas is liberated at the anode?
Answer:
Oxygen

Question 26.
In the electrolytic cell, the positively charged carbon rods are replaced periodically. What is the reason?
Answer:
The oxygen produced at the anode will cause the carbon rods to oxidise into carbon dioxide. As a result, the size of the anode rod decreases. So, Carbon rods need to be replaced periodically.

Question 27.
Complete the – Table related to the electrolysis of alumina.

Answer:

Anode	Carbon lining
Cathode	Molten cryolite and Alumina
Electrolyte	Oxygen gas
The product formed at anode	Aluminium
The product formed at cathode	Electricity
Reducing agent	Carbon lining

Duralumin and alnico are important alloys which contain aluminium. The constituents of alnico are Fe, Al, Ni, CO etc. Alnico is used for making permanent magnets.

Question 28.
What are the factors influencing the corrosion of iron?
Answer:

• Presence of Water and Oxygen
• Presence of Electrolytes
• pH of the Environment
• Temperature

The metals occupying higher positions in the reactivity series readily corrode upon contact with atmospheric air. The rate of corrosion decreases as we move down. Metals higher up the reactivity series are more electropositive, meaning they have a greater tendency to lose electrons and form positive ions. Corrosion, or oxidation, is the process where a metal loses electrons. Therefore, more reactive metals readily lose electrons to elements in the atmosphere, like oxygen and water, causing them to corrode faster than less reactive metals.

Even though aluminium is a metal at the top of the reactivity series, it resists corrosion to some extent. Aluminium’s corrosion resistance stems from its immediate reaction with atmospheric oxygen to form a thin, durable, and non-porous layer of aluminium oxide (Al₂O₃) on its surface. This protective layer acts as a barrier, preventing further oxidation and corrosion of the metal beneath. The oxide coatings on metals such as aluminium, zinc and tin are thin and nonporous and therefore protect the metal from further corrosion.

The corrosion of iron differ from this, because the hydrated iron oxide (rust) that forms on the surface of iron is porous and powdery. So this process of corrosion continues until the metal is destroyed.

The factors given below influence the corrosion of metals.?

• Nature of the metal
• Moisture in which gases or salts are dissolved
• Contact with other metals

Question 29.
Let us do an experiment.

Prepare a very dilute solution of sodium chloride in two large test tubes. Place iron nails, one tightly wrapped with copper wire and the other with magnesium wire. Observe the test tube for five to six days. Record the results of the observation in the science diary.

a) In which test tube, the iron nail undergoes rusting?
Answer:
The iron nail in contact with copper.

b) It is found that the iron nail in contact with copper corrodes easily while that in contact with magnesium is protected. Why does this happen?
Answer:
When a metal comes in contact with another metal in the presence of an electrolyte, it acts as a simple voltaic cell. This phenomenon is due to galvanic corrosion, an electrochemical process. When iron is in contact with copper, the more reactive iron acts as the anode and corrodes to protect the copper, the cathode. Conversely, when in contact with magnesium, the highly reactive magnesium acts as a sacrificial anode, corroding preferentially and preventing the less reactive iron from rusting.

c) When iron comes into contact with copper, which metal will release electrons? Why?
Answer:
When iron comes into contact with copper, iron will release electrons. This is because iron is more reactive than copper, meaning it has a greater tendency to lose electrons. As a result, in the presence of an electrolyte, iron becomes the anode and corrodes, while copper acts as the cathode and is protected from corrosion.

The iron atoms, that lose electrons, change into ions. The rate of corrosion of iron increases. Here iron acts as anode and copper acts as cathode.

d) Why is iron protected when it is in contact with magnesium?
Answer:
It is because the more reactive magnesium donates electrons. In the cell formed here, magnesium acts as anode and iron as cathode.

Question 30.
Which metal will corrode when two metals are in contact in the presence of moisture in the air?
The more reactive / The less reactive
Answer:
The more reactive
It is seen that the metal acting as anode is destroyed and the metal acting as cathode is protected. This is an important method to prevent corrosion of iron. This is known as cathodic protection.

Question 31.
Why is it not advisable to join aluminium wires and copper wires, and to join iron materials to copper piping
for electrical purposes?
Answer:
Joining aluminium and copper wires or iron and copper pipes is a bad idea due to galvanic corrosion and thermal expansion.

• Galvanic Corrosion: When in contact, the more reactive metal (aluminium or iron) corrodes faster to protect the less reactive one. This increases resistance and can cause fires.
• Thermal Expansion: Aluminium expands more than copper when heated. This difference can loosen the connection over time, increasing fire risk.

Question 32.
Why are zinc or magnesium blocks always attached to sea bridges and ship hulls?
Answer:
Zinc or magnesium blocks are attached to ships and bridges as sacrificial anodes. Being more reactive than steel, they corrode instead of the structure, providing cathodic protection.

Question 33.
Discuss various methods to prevent corrosion of metals and list them in the science diary.
Answer:
To prevent corrosion, you can:

• Coat the metal with paint or oil.
• Use a sacrificial coating, like galvanizing steel with zinc.
• Create alloys like stainless steel.
• Use cathodic protection with sacrificial anodes or electric current.
• Electroplate with a non-corrosive metal.

Metals play a crucial role in the human body also. We can learn more about it in the higher classes.

Std 10 Chemistry Chapter 6 Notes – Extended Activities

Question 1.
Thermite process

Mix equal amounts of anhydrous aluminium powder and iron oxide in a crucible. Immerse it in a tin pot filled with sand. Place a little of barium peroxide and magnesium powder mixture on top of the first mixture. Fix a 6 cm long polished magnesium ribbon in slanting position on its top. After lighting the ribbon, watch it from a distance.

After the reaction stops, examine the crucible and you will find a small globule of iron.
Fe₂O₃ + 2Al Al₂O₃ + 2Fe
a) Analyse the chemical changes that took place here based on the metallurgical process.
b) Find instances where thermite process is used in daily life.
c) Write the chemical equation for the production of chromium from chromic oxide (Cr₂O₃) in this way.
Answer:
a) The thermite process is a redox reaction. The aluminium powder acts as a strong reducing agent because it is more reactive than iron. It has a greater tendency to lose electrons and combine with oxygen. Aluminium reduces the iron oxide (Fe₂O₃) to molten iron (Fe), while the aluminium itself gets oxidised to aluminium oxide (Al₂O₃). The reaction is highly exothermic, releasing a large amount of heat that melts the iron. The barium peroxide and magnesium ribbon mixture acts as an igniter to initiate this high-temperature reaction.

b) The thermite process is utilised in daily life for its ability to produce extremely high temperatures and molten metal quickly. Common uses include:

Thermite Welding: It is widely used to weld railway tracks and other large metal parts in situ, as the process can generate molten iron to join the ends of the metal sections.

Demolition and Cutting: The intense heat from the reaction can be used to cut through thick metal plates.

c) The chemical equation for the production of chromium from chromic oxide (Cr₂O₃) using the thermite process is:
Cr₂O₃ + 2Al → Al₂O₃ + 2Cr
In this reaction, aluminium acts as the reducing agent, reducing the chromic oxide to pure chromium metal.

Question 2.
Prepare a note on how the rusted iron materials collected from the scrap iron shop are used in making steel. Present it in the class.
Answer:
Using Rusted Iron to Make Steel
Rusted iron from scrap yards is recycled into new steel through a process that primarily uses an electric arc furnace (EAF).

1. Melting: The scrap is placed in a high- temperature EAF, where the rust (iron oxide) and other impurities are melted.
2. Purification: Impurities like silicon and manganese are oxidised and then removed as slag.
3. Carbon and Alloying: Once purified, specific amounts of carbon and other elements (like chromium or nickel) are added to the molten iron to give it the desired properties of steel.
4. Casting: The new molten steel is then cast into various shapes for use in manufacturing.
   This process efficiently recycles metal, reducing the need for new iron ore mining and making steel production more sustainable.

Question 3.
Prepare a flow chart showing the various steps in the purification of bauxite.
Answer:

Metals Class 10 Notes

Metals Notes Pdf

• Minerals – The metallic compounds generally seen in the earth’s crust.
• Ores – A mineral from which a metal is economically, easily and quickly extracted.
• 3 Steps in metallurgy:
  1. Concentration of ores (removing impurities (gangue) from ore)
  2. Extraction of metals
  3. Refining of metals (Other metals, metal oxides, and small amounts of non-metals may be impurities in reduced metal. Metal refining removes impurities to make pure metal.)
• Concentration of ores
  1. Levigation (Gangue – Lighter, Ore – Heavier)
  2. Froth floatation (Gangue – Heavier, Ore – Lighter)
  3. Magnetic separation (Either ore or Gangue has magnetic nature)
  4. Leaching (when ore particles are soluble in suitable solution)
• Extraction of metals
  a) Conversion of the concentrated ore into its oxide.
  i) Calcination (Heating the ore at a temperature below its melting point in absence of air)
  ii) Roasting (Heating the ore at a temperature below its melting point in presence of air)
  b) Reduction of the oxide.
  Extraction of metal from the oxide.
• Refining of metals
  1. Liquation – Low melting metals
     Ex: Tin & Lead
     (Heating metals on an inclined surface)
  2. Distillation – Metals with low boiling points Zinc, Cadmium, Mercury
  3. Electrolytic Refining – Electrolysis (electrolyte – Solution of salt of metal, -ve electrode – small piece of pure metal, +ve electrode – impure metal)
• Industrial production of Iron
  Ore – Haematite –
  
  The chemical reactions taking place in blast furnace
  C + O₂ → CO₂ + Heat
  CO₂ + C + Heat → 2CO
  Fe₂O₃ + 3CO → 2Fe + 3CO₂
  CaCO₃ → CaO + CO₂
  CaO + SiO₂ → CaSiO₃
  Flux + Gangue → Slag
• Extraction of Zinc

Sulphide ore of zinc (Common name, Chemical formula)	Zinc blende (ZnS)
Concentration method	Froth floatation
Carbonate ore of zinc (Common name, Chemical formula	Calamine (ZnCO3)

The concentrated ore is subjected to roasting. Zinc sulphide converted to Zinc Oxide. Then Coke is mixed with Zinc oxide and put into furnace at high temperature, the coke reduces Zinc oxide to Zinc
Purification method of Zinc – Distillation

• Extraction of Copper – Copper is extracted from copper pyrites (CuFeS₂) through a process of concentration by froth flotation, followed by roasting and smelting with silica to remove iron as slag, which ultimately yields impure blister copper that is refined by electrolysis to obtain a high purity metal.
• Extraction of Aluminium – Aluminium is extracted from its ore, bauxite, in a two-stage process: first process purifies the bauxite to produce pure alumina (Al₂O₃), which is then dissolved in molten cryolite (Na₃ AlF₆) and reduced to pure aluminium metal by electrolysis in the Hall-Heroult process.
• Corrosion of metal is a process in which the metal reacts with a surrounding medium and undergoes chemical change.
• The factors given below influence the corrosion of metals move left.
• Nature of the metal
• Moisture in which gases or salts are dissolved Contact with other metals

INTRODUCTION

The use of metals accelerated the growth of people and the birth of an original culture. The discovery of metals and their role in human progress is amazing. The term ‘metal age’ refers to the transition period from the Stone Age to the use of metals. Our ancient period is marked by the names Copper Age, Bronze Age, Iron Age, etc. At present, metals are used in the making of a wide range of products, ranging from small paper screws to large-scale machinery. The unique features of each metal make them important in different fields.

Occurrence of metals in nature

• Naturally occurring metals or compounds are generally called minerals. These include metallic compounds and non metallic compounds.
• Any mineral from which a metal can be extracted easily and economically is called an ore of that metal.

Metallurgy

• The process of extraction of metal from ores involves three major stages. Such processes are generally referred to as metallurgy.
  1. Concentration of ores
  2. Extraction of metal from concentrated ore
  3. Refining of metal
• Concentration of ores is the process of increasing the metal content of the ore by removing impurities, known as gangue.
• Concentration methods include Levigation or Hydraulic washing, Froth floatation, Magnetic separation and Leaching. These methods are adopted depending on the nature of the ore and gangue.
• Extraction of metal from concentrated ore – 1. Conversion of concentrated ore to oxide, 2. Reduction of oxide ore
• Calcination refers to heating the concentrated ore in limited quantities or absence of air at a temperature below the melting point.
• Roasting refers to heating the concentrated ore at a temperature below the melting point in the presence of air.
• Refining of metals is the process of removing the impurities to produce pure metal. Refining methods include Liquation, Distillation and Electrolytic refining and these are adopted depending upon the metal to be purified and impurities contained in it.

Manufacturing methods of some industrially important metals

• Extraction of Iron – Iron is extracted from its iron oxide ores, such as haematite, in a blast furnace by reducing the iron oxide with carbon at high temperatures, which separates the iron from the oxygen.
• Extraction of Zinc – Zinc is extracted from its ores, primarily zinc sulphide (ZnS), by first converting the ore to zinc oxide (ZnO) through roasting, then reducing the oxide with carbon in a furnace at high temperatures, and finally purifying the resulting metal.
• Extraction of Copper – Copper is extracted from copper pyrites (CuFeS₂) through a process of concentration by froth flotation, followed by roasting and smelting with silica to remove iron as slag, which ultimately yields impure blister copper that is refined by electrolysis to obtain a high purity metal.
• Aluminium is extracted from its ore, bauxite, in a two-stage process: first process purifies the bauxite to produce pure alumina (Al₂O₃), which is then dissolved in molten cryolite (Na₃ AlF₆) and reduced to pure aluminium metal by electrolysis in the Hall-Heroult process.

Corrosion of metals

• Corrosion of metal is a process in which the metal reacts with a surrounding medium and undergoes chemical change.
• The factors given below influence the corrosion of metals.
  • Nature of the metal
  • Moisture in which gases or salts are dissolved
  • Contact with other metals

OCCURRENCE OF METALS IN NATURE
The earth’s crust is the major source of metals. Metals are generally found in combined state as they are highly reactive. Metal compounds like sodium chloride and magnesium chloride are dissolved in sea water also. A very few metals are seen in free state too. The most common examples include Gold (Au), Platinum (Pt), Silver (Ag).

Naturally occurring metals or compounds are generally called minerals. These include metallic compounds and non metallic compounds.

The same metal can be found in the form of various minerals in nature. Metals are separated from the most suitable ones among these. For example, clay (Al₂O₃.2SiO₂.2H₂O) is an abundant mineral of aluminium. Bauxite
(Al₂O₃.2H₂O) and cryolite (Na₃AlF₆) are also minerals of aluminium. But among these, bauxite is the mineral that is commonly used to produce aluminium.

The qualities of a mineral that is used to extract a metal

• Abundance
• Ease of extraction
• High metal content
• Low production cost.

Based on these characteristics, bauxite is the most suitable mineral for the production of aluminium.
The ores of some metals are listed in the Table

Any mineral from which a metal can be extracted easily and economically is called an ore of that metal.

METALLURGY
Highly reactive metals are found in nature in combined form.
The process of extraction of metal from ores involves three major stages. Such processes are generally referred to as metallurgy.
1. Concentration of ores
2. Extraction of metal from concentrated ore
3. Refining of metal

CONCENTRATION OF ORES
Ores are mined from the earth’s crust. A lot of earthy impurities like soil and sand may be mixed with this. Concentration of ores is the process of increasing the metal content of the ore by removing such impurities, known as gangue. Depending on the nature of the ore and gangue, various methods are adopted for this purpose.

a) Levigation or Hydraulic washing
This method is employed when the density of the ore is more than that of the gangue. The powdered ore is washed in a stream of water. The less dense gangue, which floats, is filtered out. The more dense ore remains at the bottom. Generally, oxide ores can be concentrated by this method.

b) Froth floatation
The froth floatation process is used when the ore is less dense than the gangue. The powdered ore is added to a mixture of water and pine oil and is stirred in a strong current of air. The ore particles stick to the froth formed, and float. It is removed and then the ore is separated and dried. Generally, sulphide ores are concentrated by this method.

c) Magnetic separation
When the powdered ore is passed through a magnetic roller, either the ore or the impurities having magnetic property is separated. In this way, pure ore is obtained by magnetic separation.

For example, magnetite (Fe₃O₄) ore has magnetic properties, but the gangue is non magnetic. Tinstone (SnO₂) ore is non magnetic, but iron tungstate, the gangue present in it is magnetic.

d) Leaching
The powdered ore is mixed with a suitable solvent. The ore dissolves in it or undergoes a chemical reaction to become a solution. The gangue remains insoluble in it. It is then filtered out. This process is called leaching.

Bauxite, the ore of aluminium, is concentrated in this way.

Metals such as gold and silver occur almost freely in nature. The ores of these metals are leached with a dilute solution of sodium cyanide (NaCN) or potassium cyanide (KCN) in the presence of air to extract the metals.

EXTRACTION OF METAL FROM CONCENTRATED ORE
You have learnt that the reactivity of metals differ. You also know about the reactivity series that is prepared based on this.

REFINING OF METALS
The metal extracted from concentrated ore may contain other elements or their compounds as impurities. Refining of metals is the process of removing these impurities to produce pure metal.

Various methods are adopted for this purpose depending upon the nature of the metal to be purified and the impurities contained in it. Some are given below.

(a) Liquation: When metals with low melting point contain impurities with high melting point, the metal is heated on the inclined surface of the furnace. The pure metal separates from the impurities, melts, flows down and it is collected. This is known as liquation. Metals such as tin (Sn) and lead (Pb) having low melting point can be purified by this method.

(b) Distillation: This method is used to purify metals with relatively low boiling points. When the metal containing impurities is heated under suitable conditions, only the pure metal vapourises. When this vapour condenses, pure metal is obtained. Metals like zinc (Zn), cadmium (Cd) and mercury (Hg) can be purified using this method.

(c) Electrolytic refining: In an electrolytic cell, an impure metal is used as the anode and a thin piece of pure metal is used as the cathode. A suitable salt solution of the same metal is used as the electrolyte. When electric current is passed, the pure metal alone is separated from the anode and deposited at the cathode.
Anode: M → Mⁿ⁺ + ne^–
Cathode : Mⁿ⁺ + ne^– → M
(Hint: M – metal, n – number of electrons involved in the reaction)
Copper (Cu) and zinc (Zn) can be purified in this way.

MANUFACTURING METHODS OF SOME INDUSTRIALLY IMPORTANT METALS
EXTRACTION OF IRON
Iron comprises about 5% of the earth’s crust and it is the second most abundant metal.

Steel
The pig iron obtained from the blast furnace is not suitable either for hammering into various shapes or for drawing into wires. The steel obtained from this is used for various industrial purposes. Steel is iron containing 0.05% to 1.5% carbon. There are different types of carbon steels depending on the amount of carbon. They differ in properties also.

Carbon steel	Amount of carbon (%)	Features
Mild steel	0.05 – 0.2	Easy to draw into wires and hammer into plates. Strong and hard. Used for making agricultural tools.
Medium steel	0.2 – 0.6	High hardness. Used for the construction of railway tracks, handrails, rafters etc.
High carbon steel	0.6 – 1.5	Very high elasticity and hardness. Used for the manufacture of surgical instruments, springs, knives, drills etc.

EXTRACTION OF ZINC
Zinc is another industrially important metal.

EXTRACTION OF COPPER
Principal ores of copper
Copper pyrites – CuFeS₂ Copper glance – Cu₂S

EXTRACTION OF ALUMINIUM
Aluminium is produced by Hall-Heroult process.
It has two stages.

1. Concentration of bauxite
Bauxite is concentrated by leaching. The powdered ore is treated with hot concentrated
sodium hydroxide (NaOH) solution. The ore alone dissolves in it to form sodium aluminate
(NaAlO₂) solution. Impurities are then filtered off.
Al₂O₃.2H₂O + 2NaOH → 2NaAlO₂ + 3H₂O

A little aluminium hydroxide (Al(OH))₃ is added to the solution of sodium aluminate and stirred vigorously. It is diluted by adding water. As a result, pure aluminium hydroxide separates out from the solution.
NaAlO₂ + 2H₂O → Al(OH)₃ + NaOH
If the aluminium hydroxide thus obtained is heated strongly, anhydrous alumina (Al₂2O₃) is obtained.
2Al(OH)₃ Al₂O₃ + 3H₂O

2. Electrolysis of aluminium oxide
Molten cryolite (Na₃AIF₆) is added to the alumina, obtained after concentration, to get a solution. This is the electrolyte used in the production of aluminium. The melting point of alumina is 2017°C. Cryolite is added to reduce the melting point and to increase the electrical conductivity.

Aluminium is separated at the cathode when a high voltage electric current is passed through the molten mixture. Examine the chemical equations of the reactions.
Al₂O₃ → 2Al³⁺ + 3O^2-

CORROSION OF METALS
Corrosion of metal is a process in which the metal reacts with a surrounding medium and undergoes chemical change.

Students rely on Class 10 Chemistry Notes Kerala Syllabus Chapter 5 Important Questions with Answers Electrochemistry to help self-study at home.

SSLC Chemistry Chapter 5 Important Questions Kerala Syllabus

Electrochemistry Class 10 Important Questions

Question 1.
Which substance is deposited at the cathode when aqueous Sodium Chloride is electrolysed? [Oxygen, Chlorine, Sodium, Hydrogen]
Answer:
Hydrogen

Question 2.
In the process of electroplating copper on an iron bangle, the bangle is connected to which terminal of the battery?
Answer:
Negative terminal

Question 3.
Statement 1: A Galvanic cell is an electrochemical cell that converts chemical energy into electrical energy.
Statement 2: A galvanic cell requires an external power source to function.
Which of the following is correct regarding the above statements?
A. Both statements are correct
B. Both statements are not correct
C. Only statement 1 is correct
D. Only statement 2 is correct
Answer:
C. Only statement 1 is correct
Galvanic cells generate electrical energy spontaneously and do not require an external power source

Question 4.
a) Which of the following metals reacts vigorously with diluted hydrochloric acid?
a) Gold
b) Magnesium
c) Lead
d) Iron
Answer:
b) Magnesium

Question 5.
Match the following.

Column A	Column B
1. Primary Cells	a) Can be recharged and used again.
2. Secondary Cells	b) Cannot be recharged.
3. Fuel Cells	c) Convert chemical energy into electrical energy by the combustion of fuel.

Choose the correct answer from the options given below.
a) 1 – b, 2 – a, 3 – c
b) 1 – c, 2 – a, 3 – b
c) 1 – a, 2 – c, 3 – b
d) 1 – c, 2 – a, 3 – b
Answer:
a) 1 – b, 2 – a, 3 – c

Column A	Column B
1. Primary Cells	b) Cannot be recharged.
2. Secondary Cells	a) Can be recharged and used again.
3. Fuel Cells	c) Convert chemical energy into electrical energy by the combustion of fuel.

Question 6.
Find out whether the following statements are true or false. If false, correct the statement.
I. Oxidation takes place at the cathode in the galvanic cell.
II. Electrical energy is converted into chemical energy in an electrolytic cell.
III. Electrical energy is converted into chemical energy in a galvanic cell.
Answer:
I) False – Oxidation takes place at the anode in a galvanic cell.
II) True
III) False – Chemical energy is converted into electrical energy.

Question 7.
a) In which test tube displacement reaction take place

b) Why doesn’t the displacement reaction take place in one of the test tubes?
Answer:
a) Test tube 1
b) In the second test tube, Ag is less reactive than copper. So, a displacement reaction cannot occur.

Question 8.
Copper is electroplated on an iron ring.
a) Which electrolyte is used here?
b) Which metal is connected to the positive terminal of the battery?
Answer:
a) Copper sulphate solution
b) Cu

Question 9.
Name the electrolytes used for the electrolysis of copper, silver and gold.
Answer:

Metals to be covered	Electrolyte
Copper	Copper sulphate
Silver	Silver nitrate (or) Sodium cyanide + Silver cyanide solution
Gold	Sodium cyanide + Gold Cyanide solution

Question 10.
A student draws the arrangement of a galvanic cell as shown below. Find out if there is any mistake? If so, draw the correct picture of the galvanic cell.

Answer:

Each metal should be placed in its own salt solution; that is, the Zn rod should be dipped in ZnSO₄, and the Cu rod should be dipped in CuSO₄.If the arrangement is correct, electron flow is from Zinc to copper.

Question 11.
a) Which is the anode and cathode in the Mg – Ag galvanic cell?
b) Write the reaction taking place at anode and cathode of this cell.
c) Write the equation of a redox reaction.
Answer:
a) Anode – Mg, Cathode – Ag

b) Anodic reaction: Mg → Mg²⁺ + 2e^–
Cathodic reaction: 2Ag⁺ + 2e^– → 2Ag

c) Redox reaction: Mg + 2Ag⁺ → Mg²⁺ + 2Ag

Question 12.
Observe the figure in which a copper plate is immersed in AgNO₃ (Silver Nitrate) solution. (Reactivity: Cu > Ag)

a) The reaction taking place here is a redox reaction. Why?
b) What change can be observed on the copper plate?
c) Write the equation of the oxidation reaction taking place here.
Answer:
a) Oxidation and reduction take place simultaneous!}’, so this reaction is a redox reaction.
b) A layer of Ag is deposited on the Copper plate, and the silver nitrate solution turns blue.
c) Cu → Cu²⁺ + 2e^–

Question 13.
A galvanic cell is constructed using Mg and Cu as electrodes.
a) What is the energy change taking place in a galvanic cell?
b) Which is the anode of the given galvanic cell?
c) Write the equation of the chemical reaction taking place at the cathode.
Answer:
a) Chemical energy is converted into electrical energy.
b) Mg
c) Cu²⁺ + 2e^– → Cu

Question 14.
The figure given below represents metals like Fe, Cu, and Mg kept in dilute HCl

a) Which among these test tubes didn’t show any reaction? Which metal was kept in that test tube?
b) Which metal is present in test tube 1? Give an equation for this reaction.
c) Arrange the given metals in the increasing order of their reactivity towards acids.
Answer:
a) 3^rd test tube, Cu

b) Mg. Because Mg reacts vigorously with HCl, more H₂ gas is produced.
Mg + 2HCl → MgCl₂ + H₂
c) Mg > Fe > Cu

Question 15.
Analyse the picture of an experiment.
a) What change took place on the surface of the iron nail?

b) Which among the following reactions are involved in the above change?
Fe²⁺ + 2 e^– → Fe
Ag⁺ + 1 e^– → Ag
Fe → Fe²⁺ + 2e^–
Ag → Ag⁺ + 1 e^–

c) Which is oxidised? Which is reduced?
Answer:
a) Ag is deposited on the nail.
b) Fe → Fe²⁺ + 2e^–
Ag⁺ + 1 e^– → Ag
c) Fe is oxidised, Ag⁺ is reduced.

Question 16.
Some materials are given below:

a) From the given materials, choose the appropriate ones to constmct a galvanic cell and draw the diagram of the cell. (Order of reactivity: Mg > Fe > Cu > Ag)
b) Which is the anode of this cell?
c) At which electrode does reduction take place?
d) Write the chemical equation of the reaction taking place at the cathode.
Answer:
a) Mg, MgSO₄, Cu, CuSO₄, Salt bridge, connecting wire with voltmeter connected.

b) Mg/Magnesium
c) Cathode (Cu rod)
d) Cu²⁺ + 2e^– → Cu

Question 17.
The zinc rod is dipped in CuSO₄ solution. Observe the changes taking place after some time.
a) Which metal is more reactive, Zn or Cu?
b) Which metal is displaced during this reaction?
c) Is this a redox reaction? Give reason.
Answer:
a) Zn

b) Cu

c) Yes. Here, Zn gets oxidised, and Cu is reduced. Thus, both oxidation and reduction take place simultaneously. So, this is a redox reaction.

Question 18.
Analyse the picture of an experiment.

a) What is the reason for the decrease in the green colour of the FeSO₄ solution?
b) The equation for this reaction is given below. Complete it.
Zn + FeSO₄ → ……… + ZnSO₄
c) If Pb is used instead of Zn, no reaction takes place. From this, what inference regarding the reactivity of lead can be obtained?
Answer:
a) Zn displaces iron from the FeSO₄ solution.
b) Zn + FeSO₄ → Fe + ZnSO₄
c) Pb is less reactive than Zn.

Question 19.
Observe the pictures given below. Based on the reactivity series, predict which of these undergoes a displacement reaction. Complete the table.

Metal	Solution	Displacement reaction
Mg	CuSO4	(a)
Ag	CuSO4	(b)
Mg	ZnSO4	Takes place
Mg	AgNO3	(c)
Cu	MgSO4	(d)

Answer:
a) Takes place
b) Does not take place
c) Takes place
d) Does not take place

Question 20.
a) Draw a diagram of the setup for electroplating gold onto a copper bangle and label its parts.
b) What are the advantages of electroplating?
c) A picture of the arrangement of copper plating on an iron bangle is given below. What are the mistakes in this?

Answer:
a)

b)

• It enhances the beauty of metal
• Prevents corrosion

c) Copper should be connected to the positive terminal of the battery.
The iron bangle should be connected to the negative terminal of the battery.
Copper sulfate solution should be used as the electrolyte.

Students rely on SCERT Kerala Syllabus 10th Standard Chemistry Textbook Solutions and Class 10 Chemistry Chapter 5 Electrochemistry Notes Questions and Answers English Medium to help self-study at home.

SSLC Chemistry Chapter 5 Notes Questions and Answers Pdf Electrochemistry

SCERT Class 10 Chemistry Chapter 5 Electrochemistry Notes Pdf

SSLC Chemistry Chapter 5 Questions and Answers – Let Us Assess

Question 1.
Examine the diagram of a galvanic cell and find out whether the given statements are true or false.

a) As the cell operates, the mass of the zinc rod, Zn(s) decreases.
b) The copper electrode is the anode.
c) Electrons flow from the zinc electrode to the copper electrode through the external circuit.
d) During cell reaction, reduction takes place at the copper electrode.
e) During cell reaction, the concentration of Cu2+ decreases.
Answer:
a) True
b) False: Copper is the cathode (site of reduction). Zinc is the anode.
c) True
d) True
e) True

Question 2.
The chemical equation of the reaction taking place in a galvanic cell is given.
Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)
a) Represent diagrammatically the above galvanic cell.
b) Which electrode has a negative charge?
c) What is the reaction taking place at each electrode?
Answer:
a)
b) The zinc electrode is negatively charged.
c) Anode: Zn(s) → Zn²⁺ + 2e^– [Oxidation]
Cathode: 2Ag⁺ + 2e^– → 2Ag(s) [Reduction]

Question 3.
A galvanic cell is made up of the following half cells:
• A magnesium electrode immersed in magnesium sulphate solution.
• A nickel electrode immersed in nickel sulphate solution.
The reactions taking place in the half-cells are given.
Mg (s) → Mg²⁺ (aq) + 2e^–
Ni²⁺(aq) + 2e^– → Ni(s)
Draw the cell diagram and label the anode, cathode and direction of flow of electrons.
Answer:

Question 4.
When an aqueous sodium chloride solution is electrolysed, the reaction that takes place at the anode is:
(i) Na⁺(aq) + e^– → Na(s)
(ii) 2H₂O → 4H⁺(aq) + O₂(g) + 4e^–
(iii) H⁺ (aq) + e^– → 1/2H₂(g)
(iv) Cl^– (aq) → 1/2 Cl₂ +e^–
Answer:
The reaction at the anode is:
Cl^– (aq) → Cl₂(g) + e^–

Question 5.
Write the products obtained when the following solutions undergo electrolysis.
(i) Aqueous solution of NaCl
(ii) Aqueous solution of CuSO₄ (using copper electrodes)
Answer:
(i) At Cathode (-): Water is reduced → Hydrogen gas (H₂) is released.
2H₂O + 2e^– → H₂ + 2OH^–
At Anode (+): Chloride ions are oxidised → Chlorine gas (Cl₂) is released.
2Cl → Cl₂ + 2e^–
Products: Hydrogen gas, Chlorine gas, and Sodium hydroxide (NaOH in solution).

(ii) At Cathode (-): Copper ions are reduced → Copper metal is deposited.
Cu²⁺ + 2e^– → Cu
At Anode (+): Copper electrode dissolves (oxidation).
Cu → Cu²⁺ + 2e^–
Products: Copper is deposited on the cathode and at anode copper dissolves as ions.

Question 6.
Brass is an alloy of zinc and copper. When brass comes into contact with saline water, corrosion of metal takes place and zinc gradually dissolves in the solution, leaving copper behind. Explain why zinc dissolves in comparison to copper.
Answer:
Zinc is more reactive than copper as it lies above copper in the reactivity series. When brass is exposed to saline water (which contains electrolytes), zinc atoms lose electrons more readily and go into solution as Zn²⁺ ions. Copper, being less reactive, does not dissolve as easily and remains behind.

Question 7.
List the cells we use in our daily life and classify them into primary and secondary cells.
Answer:
Primary cells (non-rechargeable): Dry cell (used in torches, clocks, toys), Mercury cell, Lithium cell (in calculators, watches).

Secondary cells (rechargeable): Lead-acid battery (car battery), Nickel-Cadmium cell, Lithium-ion battery (in mobiles, laptops).

Question 8.
Define fuel cells and write their advantages.
Answer:
A fuel cell is a device that converts the chemical energy produced by the combustion of fuel directly into electrical energy.

Advantages:

1. High efficiency.
2. Environment-friendly (produces water as a by-product in hydrogen fuel cells).
3. Continuous supply of electricity as long as fuel is provided.

Chemistry Class 10 Chapter 5 Notes Kerala Syllabus Electrochemistry

Question 1.
a) What did you observe?
Answer:
The bulb glows.

b) What is the reason for the glowing of the bulb?
Answer:
The bulb glows because chemical energy is converted into electrical energy in the electrochemical cell.
We can understand that there is a flow of electric current through the bulb.

Separate the copper wires from the bulb and connect them to a galvanometer as shown in the figure given below. When the copper rod is connected to the positive terminal of the galvanometer and the zinc rod to the negative terminal, it can be seen that the pointer of the galvanometer is deflecting towards the right (towards the positive terminal)

This proves that:
• Electrons flow from the zinc rod to → copper rod.
• Electric current flows in the opposite direction (copper → zinc) through the galvanometer.
Connect a voltmeter instead of the galvanometer as shown in the figure below.

SALT BRIDGE

• A salt bridge is a U-shaped glass tube filled with an inert electrolyte.
• Common electrolytes used:
  • Potassium chloride (KCl)
  • Ammonium chloride (NH₄Cl)
  • Potassium nitrate (KNO₃)
• The electrolyte is made into a gel using agar-agar.
• It connects the oxidation and reduction half-cells in a galvanic cell.
• It allows electrical contact between the two solutions without mixing them.
• Maintains electrical neutrality by allowing ion flow.
• Enables smooth operation of the galvanic cell.

Examine the metal rods and the solutions in which they are immersed in the figure below.

Question 2.
Find the voltage between the two electrodes?
Answer:
The voltmeter shows a reading of about 1.1 volts. This means the voltage (potential difference) between zinc and copper electrodes is approximately 1.1 V.
The reaction that occurs at each electrode is given below.
Zinc electrode: Zn(s) → Zn²⁺ (aq) + 2e^–

Question 3.
What type of chemical reaction takes place at the zinc electrode? (Oxidation / Reduction)
Answer:
Oxidation
Copper electrode: Cu²⁺(aq) + 2e^– → Cu(s)

Question 4.
What type of chemical reaction takes place at the copper electrode?
Answer:
Reduction
Anode: The electrode where oxidation occurs is called the anode
Cathode: The electrode where reduction occurs is called the cathode.
The processes that take place in the Daniel cell can be summarised as:
Anode: Zn(s) → Zn²⁺ (aq) + 2e^–
Copper electrode: Cu²⁺(aq) + 2e^– → Cu(s)
Cell reaction: Zn(s) + Cu²⁺ (aq) → Zn²⁺(aq) + Cu(s)
The electric current is produced as a result of a redox reaction taking place in the galvanic cell.

In the Daniel cell, replace the zinc rod in zinc sulphate with aluminium, silver and magnesium rods (plates) and immerse these metals in their own salt solution. Repeat the experiment by connecting each to the copper rod immersed in copper sulphate.

Question 5.
a) List out the direction of flow of electrons, the equation of reaction at each electrode, the cell voltage, and analyse them.

Answer:

b) Which has the higher value of voltage, Zn – Cu galvanic cell or Al – Cu galvanic cell?
Answer:
Al – Cu galvanic cell has the higher voltage.
The tendency of aluminium to get oxidised in redox reactions is greater than that of zinc. That is, aluminium is more reactive than zinc.
Zn – Cu cell voltage = 0.34 – (-0.76) = 1.10 V
Al – Cu cell voltage = 0.34 – (-1.66) = 2.00 V

c) Which has the higher value of voltage, Mg – Cu galvanic cell or Al-Cu galvanic cell?
Answer:
The voltage value for the Mg-Cu galvanic cell is higher.

d) What is the direction of flow of electrons in a galvanic cell made up of a pair of copper and silver electrodes?
Answer:
Cu acts as anode and Ag acts as cathode. So, Electrons flow from copper to silver.

Question 6.
Draw a diagram of the galvanic cell having silver and copper as electrodes and mark the direction of flow of electrons.
Answer:

CELL VOLTAGE
The cell voltage is the potential difference between the two electrodes of a cell when no current is flowing.

Cell Voltage (EMF) = Potential of Cathode – Potential of Anode

Question 7.
Prepare a list of the metals, familiarised through the experiment, in the order of their tendency to get oxidised in redox reactions.

Answer:

In this way, we can construct galvanic cells using different elements as electrodes and compare their reactivities.

REACTIVITY SERIES.
The series in which elements are arranged in decreasing order of their reactivity is known as the reactivity series.

APPLICATION OF THE REACTIVITY SERIES

1. To identify which metal is being displaced
Place a zinc rod in a beaker of copper sulphate solution. Observe for some time.
a) Did the colour of the CuS04 solution change?
Answer:
Yes, the blue colour of the CuS04 solution faded and eventually disappeared.

b) What change was observed in the zinc rod?
Answer:
The zinc rod became coated with a reddish- brown deposit.

c) What may be the reason for the fading of the colour of the solution?
Answer:
A decrease in the concentration of Cu2+ ions is the reason for the fading of the blue colour.

d) With the help of the reactivity series, find out which metal is more reactive, Zn or Cu?
Answer:
Zinc (Zn) is more reactive than copper (Cu).

e) What are the ions fonned when CuSO₄ dissolves in water?
Answer:
Cu²⁺ (copper ions) and SO₄^2- (sulphate ions).

f) What changes have occurred to the Cu²⁺ ions?
Answer:
The Cu²⁺ ions have gained two electrons and have been converted to solid copper atoms (Cu). This process is called reduction.

g) Write the chemical equation of the reaction.
Answer:
Zn + Cu²⁺SO₄^2- → Zn²⁺SO₄^2- + Cu

h) What changes have occurred to Zn?
Answer:
The zinc (Zn) atoms have lost two electrons and have been converted into zinc ions (Zn²⁺). This process is called oxidation.

i) Which metal is displaced here?
Answer:
In the reaction between zinc (Zn) and copper sulfate (CuSO₄), copper (Cu) is the metal being displaced. This is a single displacement reaction, where the more reactive metal, zinc, displaces the less reactive metal, copper, from its compound.

Take a little silver nitrate solution in a beaker and dip a copper rod in it. Observe for some time.
a) Is there any change in the colour of the solution?
Answer:
Yes, the solution gradually turns blue because of the formation of Cu²⁺ ions.

b) What change was observed in the copper rod?
Answer:
A greyish-white layer of silver metal gets deposited on the copper rod.

c) Write the equation of the chemical reaction taking place here.
Answer:
Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

d) Explain the reaction by comparing the positions of Ag and Cu in the reactivity series.
Answer:
In the reactivity series, copper (Cu) is above silver (Ag), which means copper is more reactive than silver.

e) Complete this chemical equation by assigning oxidation numbers.
Answer:

f) Which ion is responsible for the change in the colour of the solution?
Answer:
The Cu²⁺ ion is responsible (makes the solution blue).

g) Which metal is oxidised here?
Answer:
Copper (Cu) is oxidised.

h) Complete the equation for oxidation.
Answer:
Cu → Cu²⁺ + 2e^–

i) Which one is reduced here?
Answer:
Silver ion (Ag⁺) is reduced to metallic silver (Ag).

j) Write the equation for reduction.
Answer:
Ag⁺ + e^– → Ag

The more reactive metal displaces the less reactive metal from its salt solution.

Question 8.
Silver nitrate solution cannot be stored in a copper vessel. Why?
Answer:
Copper (Cu) is more reactive than silver (Ag) in the reactivity series. So, copper will displace silver from the silver nitrate solution:
Cu + 2AgNO₃ → Cu (NO₃)₂ + 2Ag
As a result, the copper vessel will gradually dissolve (forming blue copper nitrate solution), and silver will get deposited. This reaction would spoil both the vessel and the solution.

2. To identify the oxidising agent and the reducing agent
Question 9.
Consider the displacement reaction given below.

a) Which metal is oxidised here?
Answer:
Magnesium (Mg)

b) Write the equation for oxidation.
Answer:
Mg → Mg²⁺ + 2e^–

c) Which metal is reduced here?
Answer:
Copper (Cu)

d) Write the equation for reduction.
Answer:
Cu²⁺ + 2e^– → Cu

e) Which reaction occurs to the more reactive metal? (Oxidation/Reduction)
Answer:
Oxidation

f) Which reaction occurs to the less reactive metal?(Oxidation/Reduction)
Answer:
Reduction

The more reactive metal gets oxidised while the ion of the less reactive metal gets reduced.

g) What is the oxidising agent and the reducing agent in this reaction?
Answer:
Cu²⁺ is the oxidising agent and Mg is the reducing agent.

The substance that causes oxidation is called an oxidising agent and the substance that causes reduction is called a reducing agent.

h) What happens to the oxidising agent in this redox reaction?
Answer:
The oxidising agent gets reduced

i) What happens to the reducing agent?
Answer:
The reducing agent gets oxidised.

Question 10.
Identity the oxidising agent and reducing agent in the reaction Given below.
2AgNO₃ + Cu → Cu(NO₃)₂ + 2Ag
Answer:
Oxidising agent: AgNO₃ (because Ag+ is reduced) Reducing agent: Cu (because it is oxidised)

In redox reactions, the more reactive metal acts as reducing agent and the less reactive metal acts as an oxidising agent.

3. To identify the displacement of hydrogen from the acid
Take equal amounts of dilute HCl in different test tubes as shown in Figure. Treat equal masses of polished Fe, Mg, Cu, Pb, Zn with dilute hydrochloric acid.

a) In which of the test tubes is hydrogen gas produced?
Answer:
Hydrogen gas is produced in the test tubes containing:

• Fe (Iron)
• Mg (Magnesium)
• Pb (Lead)
• Zn(Zinc)

Not produced in the Cu (Copper) test tube, because copper is below hydrogen in the reactivity series and cannot displace hydrogen from dilute HCl.

b) Is the rate of formation of hydrogen gas the same in all the test tubes?
Answer:
No, the rate of formation of hydrogen gas is not the same in all test tubes.
It depends on the reactivity of the metal with dilute HCl.
Mg > Zn > Fe > Pb > Cu (no reaction)

c) Examine the position of hydrogen and the metals used here in the reactivity series and record your observations.
Answer:
Metals above hydrogen in the reactivity series
(Mg, Zn, Fe, Pb) can displace hydrogen from dilute acids and produce hydrogen gas.
Copper (Cu) is below hydrogen in the reactivity series, so it cannot displace hydrogen from dilute HCl. Hence, no reaction is observed.

The metals placed above hydrogen in the reactivity series can displace hydrogen from dilute acids.

Question 11.
Which of the following metals can displace hydrogen from hydrochloric acid?
[Sodium, gold, silver, aluminium]
Answer:
Sodium (Na) and Aluminium (Al) can displace hydrogen from hydrochloric acid because they are above hydrogen in the reactivity series.

Gold (Au) and Silver (Ag) are below hydrogen in the reactivity series, so they cannot displace hydrogen from hydrochloric acid.

Question 12.
a) Bring a burning candle close to the mouth of the test tubes and record the observations.
Answer:
When you bring a burning candle to the test tube near the cathode, a “pop” sound will be heard, and the gas will ignite. This indicates the presence of hydrogen gas.

When you bring the burning candle to the test tube near the anode, the flame will bum more brightly. This indicates the presence of oxygen, as it supports combustion.

b) Which gas is produced in each test tube?
Answer:
At the cathode (negative electrode): Hydrogen gas (H₂) is produced. This is the test tube with the larger volume of gas.

At the anode (positive electrode): Oxygen gas (O₂) is produced. This is the test tube with the smaller volume of gas.

c) Write the chemical equation of the reaction.
Answer:
The overall chemical equation for the electrolysis of water is:

Here, acidified water decomposes into hydrogen and oxygen. This is because of the flow of electricity from the external electrical source into the solution through the electrodes.

Observe the given figure

Carry out this electrolysis with the help of the teacher.
Question 13.
Record your observations.
a) Which metal is connected to the positive terminal of the battery?
Answer:
Copper

b) Which metal is connected to the negative terminal of the battery?
Answer:
Iron

c) Which solution is used as the electrolyte?
Answer:
Copper sulphate solution

d) What are the ions present in the electrolyte?
Answer:
The ions present are copper cations (Cu²⁺) and sulphate anions (SO₄^2-).

e) Which electrode is connected to the positive terminal of the battery? Anode/Cathode
Answer:
Anode

f) What is the reaction taking place at the anode?
Answer:
Oxidation

g) Complete the equation of the oxidation reaction taking place at the copper plate.
Answer:
Cu → Cu²⁺ + 2e^–

h) Which electrode is connected to the negative terminal of the battery?
Anode/Cathode
Answer:
Cathode

i) What is the reaction taking place at the cathode?
Answer:
The reaction taking place at the cathode is reduction.

j) Which ions will be attracted to the cathode, the iron ring, from the solution?
Answer:
Copper ions (Cu²⁺) will be attracted to the cathode (the iron ring) from the solution.

k) Complete the equation of the reduction reaction taking place here.
Answer:
Cu²⁺ + 2e^– → Cu

The reactions taking place at the anode and cathode:
Anode: Cu → Cu²⁺ + 2e^–
Cathode: Cu²⁺ + 2e^– → Cu
That is, at the anode, copper undergoes oxidation and enters into the solution as Cu²⁺ ions. At the same time, Cu²⁺ ions from the solution are reduced on the surface of the iron ring, which is the cathode, forming a thin coating of copper on it.

Electroplating is the process of coating a layer of one metal onto another metal through electrolysis.

Question 14.
What are the benefits of electroplating?
Answer:
It enhances the beauty of the metal and prevents its corrosion.

Question 15.
Explain the process of electroplating.
Answer:

• The metal that is to be plated should be connected to the positive terminal of the battery.
• The object on which the plating is to be done should be connected to the negative terminal of the battery.
• A salt solution of the metal to be plated is used as the electrolyte

The metals to be coated	Electrolyte
Silver	Silver nitrate solution/Sodium cyanide + Silver Cyanide solution
Gold	Sodium cyanide + Gold cyanide solution

Std 10 Chemistry Chapter 5 Notes – Extended Activities

Question 1.
The following are certain observations about metals A, B, C and D.
(a) When a plate of metal A is placed in a solution containing B²⁺ ions, no reaction is observed.
(b) When the plate of A is placed in a solution containing C⁺ ions, no change occurs.
(c) When a plate of metal D is placed in a solution containing C⁺ ions, a black precipitate of C is formed on the surface of D, and the presence of D²⁺ ions can be detected in the solution.
(d) When a plate of B is placed in a solution of D²⁺ ions, D appears on the surface of B and B²⁺ ions appear in the solution.
List A⁺, B²⁺, C⁺ and D²⁺ in the ascending order of their ability to attract electrons.
Answer:
The tendency to lose an electron varies in the following order.
C < D < B
Therefore, the ascending order of their ability to attract electrons is.
B²⁺ < D²⁺ < C⁺ < A⁺

Question 2.
As shown in the galvanic cell of the given figure, place a silver electrode in a solution of silver nitrate and a lead electrode in a solution of lead nitrate. Connect the two electrodes using a copper wire. Also connect the two breakers using a salt bridge. Next, find the answers to the questions given below.

(a) What is the anode
(b) What is the cathode?
(c) Where does oxidation occur?
(d) Where does reduction occur?
(e) In which direction do electrons flow through the copper wire?
(f) What will be the cell voltage?
(g) Will the cell voltage vary if the two solutions are diluted alternately?
Answer:
a) Anode is where oxidation occurs.
Lead (Pb) electrode

b) Cathode is where reductions occur.
Silver (Ag) electrode

c) At the lead (Pb) electrode

d) At the silver (Ag) electrode

e) From the lead electrode to the silver electrode

f) Standard electrode potentials:

• Pb²⁺/Pb = -0.13 V
• Ag⁺ /Ag = +0.80 V
  Cell Voltage =Potential of Cathode – Potential of Anode = 0.80 V – (-0.13 V) = 0.93 V

g) Yes, the cell voltage will vary if the solutions are diluted. The cell voltage depends on the concentration of the ions. Diluting the solutions changes the ion concentrations, which affects the electrode potential and thus alters the overall cell voltage.

Electrochemistry Class 10 Notes

Electrochemistry Notes Pdf

• Electrochemistry: Study of electricity-related chemical reactions.
• Electrochemical cells: Devices that convert chemical energy to electrical energy or vice versa.
• Types of Electrochemical Cells
  • Galvanic cells: Convert chemical energy to electrical energy.
  • Electrolytic cells: Convert electrical energy to chemical energy.
• Galvanic Cell (Daniell Cell)
  • Oxidation takes place at the anode (Negative electrode).
  • Reduction takes place at the cathode (Positive electrode).
• Electrolytic cell
  • Oxidation takes place at the anode (Positive electrode).
  • Reduction takes place at the cathode (Negative electrode).
• Reactivity Series: Arrangement of elements in decreasing order of reactivity is called the reactivity series.
• Displacement Reactions: More reactive metals displace less reactive metals from salt solutions.
  Zn + CuSO₄ → ZnSO₄ + Cu
• Oxidising & Reducing Agents
  • Oxidising agent: Gains electrons (gets reduced).
  • Reducing agent: Loses electrons (gets oxidised).
• Hydrogen Displacement
  • Metals above hydrogen in the reactivity series displace H₂ from acids.
  • Reactive metals (Mg, Zn, Fe, Pb) react with HCl to release H₂ gas.
  • Unreactive metals (Cu, Ag, Au) do not.
• Electrolysis of molten sodium chloride
  • Ions: Na⁺, Cl^–
  • At Cathode: Na⁺ + e^– → Na (molten)
  • At Anode: 2Cl^– → Cl₂ + 2e^–
  • Products: Na (metal), Cl₂ (gas)
• Electrolysis of aqueous sodium chloride
  • At Cathode: 2H₂O + 2e^– → H₂ + 2OH^–
  • At Anode: 2Cl^– → Cl₂ + 2e^–
  • Products: H₂ gas, Cl₂ gas, NaOH solution
• Electroplating
  • Anode (positive): Metal to be plated (e.g., Cu)
  • Cathode (negative): Object to be plated (e.g., Iron ring)
  • Electrolyte: Salt solution of plating metal (e.g., CuSO₄)
• Benefits
  • Enhances appearance
  • Prevents corrosion
• Applications of electrolysis:
  • Production of metals: Na, K, Ca, Al
  • Production of non-metals: H₂, O₂, Cl₂
  • Manufacture of compounds: NaOH, KOH
  • Electroplating: Jewellery, utensils
  • Purification of metals: Cu, Au
• Different types of cells
  • Primary Cells: Non-rechargeable [Examples: Dry cell, Button cell]
  • Secondary Cells: Rechargeable [Examples: Lead-acid battery, Li-ion battery, Ni-Cd cell]
    Fuel Cells: Fuel cells are galvanic cells that use fuels like H,, CH₄, or CH₃OH to continuously and efficiently convert chemical energy directly into electrical energy in an environmentally friendly manner

INDRODUCTION

In our daily lives, we use many devices, such as vehicles, mobile phones, laptops, and medical equipment. All these works with the help of stored electrical energy from cells or batteries. Chemical reactions inside them produce this energy. Electrochemistry is the branch of chemistry that studies how chemical reactions can produce electricity and how electricity can be used to carry out chemical reactions. The devices that make this possible are called electrochemical cells.

In this unit, we will learn about electrochemical cells, the reactivity series and its applications, electrolytic cells, electrolysis of molten sodium chloride, electrolysis of aqueous sodium chloride solution, electroplating and its uses, and different types of cells

Electrochemical Cells

• Devices that convert chemical energy into electrical energy (or vice versa).
• Made of two electrodes (anode and cathode) and an electrolyte.
• Work on redox reactions where oxidation occurs at the anode and reduction at the cathode.
• Examples: Galvanic (Voltaic) cells and Electrolytic cells.

Electrolysis of Molten Sodium Chloride

• Process of breaking down molten NaCl using electricity.
• At the cathode: Na⁺ ions are reduced to sodium metal.
• At the anode: Cl^– ions are oxidised to chlorine gas.
• Overall: Production of sodium metal and chlorine gas.

Electrolysis of Aqueous Sodium Chloride Solution

• Involves electrolysis of aqueous sodium chloride solution.
• At the cathode: Water is reduced to hydrogen gas.
• At the anode: Cl^– is oxidised to chlorine gas.
• Solution left behind is sodium hydroxide (NaOH).

Electroplating

• Coating one metal with another using electrolysis.
• Cathode: Object to be plated.
• Anode: Metal to be deposited.
• Electrolyte: Salt solution of the coating metal.
• Used for decoration, corrosion protection, and improving durability.

Applications of Electrolysis

• Extraction of metals (e.g., aluminium from alumina, sodium from NaCl).
• Purification of metals (e.g., copper).
• Electroplating (jewellery, utensils, machine parts).
• Production of metals and non-metals.

Different Types of Cells

• Primary Cells: Non-rechargeable [Examples: Dry cell, Button cell]
• Secondary Cells: Rechargeable [Examples: Lead-acid battery, Li-ion battery, Ni-Cd cell]
• Fuel Cells: Generate electricity from a continuous supply of fuel and oxygen (e.g., hydrogen fuel cell).

ELECTROCHEMICAL CELLS
Electrochemistry is the branch of chemistry that deals with the study of the processes that produce electricity through chemical reactions and use electricity to bring about chemical reactions. Electrochemical cells are the devices that make such changes possible. Electrochemical cells can be divided into two types.

1. Galvanic cells
2. Electrolytic cells

GALVANIC CELL
Galvanic cells are devices that convert chemical energy into electrical energy. Daniel cell is an example of a galvanic cell.

METHOD OF CONSTRUCTING A DANIEL CELL
Materials Required:

• Beaker
• Zinc rod
• Copper rod
• Zinc sulphate solution
• Copper sulphate solution
• Salt bridge
• NH₄Cl/KCl/KNO₃
• Copper wires
• Galvanometer
• Voltmeter
• LED bulb

Experimental procedure

• Take two beakers.
  • In one beaker, pour zinc sulphate solution and dip a zinc rod (Half cell 1).
  • In the other beaker, pour copper sulphate solution and dip a copper rod (Half cell 2).
• Make a salt bridge:
  • Roll a piece of filter paper.
  • Add some KCl (Potassium chloride) / NH₄Cl (Ammonium chloride) / KNO₃ (Potassium nitrate) crystals.
  • Sprinkle water to moisten it.
  • Bend it into a ‘U’ shape.
• Place the salt bridge so that it connects the two beakers.
• Connect the zinc rod and copper rod to a bulb using copper wires.

ELECTROLYTIC CELLS
The cells that utilise electrical energy to bring about chemical changes are called electrolytic cells.

EXPERIMENT: ELECTROLYSIS OF produced?
ACIDIFIED WATER
Procedure of the experiment:
Take a plastic cup. Make two holes at the bottom and attach rubber stoppers as shown in the Figure. Insert carbon electrodes (graphite rods from old torch cells can be used for this purpose) into these rubber stoppers. Connect the electrodes to a 6V battery.

Fill the cup with water so that the electrodes are immersed. Add 2 – 3 mL of dilute sulphuric acid. Dilute sulphuric acid is added to increase the ionisation of water and thereby increase its electrical conductivity. Take two test tubes filled with water and invert them over the carbon electrodes. Turn on the switch and leave the experimental setup undisturbed for some time. You will observe bubbles forming at both the electrodes. After some time, remove the test tubes carefully.

ELECTROLYTES
Electrolytes are substances that undergo chemical changes when electricity passes through them.

Electrolytes are substances that conduct electricity in molten state or in an aqueous solution and undergo chemical changes.

ELECTROLYSIS
Electrolysis is the process by which an electrolyte undergoes chemical changes when electricity is passed through it.
Examine the figure that represents the electrolytic process.

Ions are free to move in the liquid state or in an aqueous solution of an electrolyte. These ions are responsible for the electrical conductivity in the electrolyte.

• Acids (like HCl, H₂SO₄, HNO₃, CH₃COOH) contain free ions in aqueous solutions (when mixed with water).
• Alkalies (like KOH, NaOH) also contain free ions in aqueous solutions.
• Salts (NaCl, KCl,.) contain free ions in their aqueous solution and molten state.
  For example, sodium chloride dissociates into Na⁺ and Cf ions in the molten state or in the aqueous solution.

ELECTRODES

• Electrodes are conductors through which electricity enters or leaves the electrolyte.
• During electrolysis:
  • Anode is the electrode connected to the positive terminal of the battery.
  • Cathode is the electrode connected to the negative terminal of the battery.
• Reduction happens at the cathode.
• Oxidation happens at the anode.

ELECTROLYTIC CELL	GALVANIC CELL
Oxidation take place at anode (positive electrode). Reduction take place at cathode (negative electrode).	Oxidation take place at anode (negative electrode). Reduction take place at cathode (positive electrode).

ELECTROLYSIS OF MOLTEN SODIUM CHLORIDE
In the solid state, sodium chloride does not conduct electricity. The reason is that, though the ions can vibrate about their fixed positions, they are not free to move. However, molten NaCl is a good conductor because its ions move freely.

Examine the figure given below.

Two graphite electrodes are connected to a direct current (DC) source through wires. They are dipped in a vessel containing molten sodium chloride. When electricity is passed through them, the following observations can be made.
1. Chlorine gas (Cl₂), having a pale green colour, is liberated at the anode.
2. Silvery molten sodium metal (Na) is formed at the cathode.
a) What are the ions in molten sodium chloride?
Answer:
The ions present are sodium cations (Na⁺) and chloride anions (Cl^–). When sodium chloride (NaCl) is in a molten (liquid) state, its ionic bonds are broken, allowing these ions to move freely.

b) What is the chemical reaction taking place at the anode?[Oxidation/Reduction|
Answer:
Oxidation occurs at the anode (the positive electrode). The chloride anions (Cl^–) lose an electron to form chlorine gas (Cl₂).
Write the equation of the reaction. 2Cl^– → Cl₂ + 2e^–

c) What is the chemical reaction taking place at the cathode? [Oxidation/Reduction]
Answer:
Reduction occurs at the cathode (the negative electrode). The sodium cations (Na⁺) gain an electron to form molten sodium metal (Na).
Write the equation of the reaction. Na⁺ + e^– → Na

d) What are the products obtained as a result of the electrolysis of molten sodium chloride?
Answer:
The products are chlorine gas (Cl₂) at the anode and molten sodium metal (Na) at the cathode.

ELECTROLYSIS OF AQUEOUS SODIUM CHLORIDE
The figure below representing the electrolysis of aqueous sodium chloride solution.

When a suitable voltage is applied across the electrodes of the cell, the following observations can be made.

1. H₂ gas is liberated at the cathode. If phenolphthalein indicator is present in the solution, the solution around the cathode turns pink. That is, the solution becomes basic in nature.

2. Cl₂ gas is liberated at the anode.
At the anode, chloride ions are getting oxidised to Cl₂ gas.
a) Write the equation of the reaction.?
Answer:
2Cl^– → Cl₂ + 2e^–

b) Based on the reactivity series, which has greater oxidising tendency-sodium or hydrogen?
Answer:
Hydrogen has a greater oxidising tendency. In the reactivity series of metals, hydrogen is below sodium, which means it is less reactive.

c) If so, which of the following is getting reduced at the cathode, Na⁺ or H₂O?
Answer:
At the cathode, H₂O molecules are reduced to produce H₂ gas and OH^– ions.
2H₂O + 2e^– → H₂ + 2OH^–
As a result of the processes taking place in the cell gases such as H₂, Cl₂ and an aqueous solution of NaOH are formed. This aqueous solution can
basic in nature. be evaporated to produce solid NaOH.

Electrodes	Chemical change	Product
Anode	2Cl– → Cl2 + 2e–	Chlorine gas
Cathode	2H2O + 2– → H2 + 2OH–	Hydrogen gas

APPLICATION OF ELECTROLYSIS

1. Production of metals
   Metals like sodium, potassium, calcium and aluminium are produced by the electrolysis of some of their compounds.
2. Production of nonmetals
   Electrolysis is used in the industrial production of nonmetals like hydrogen, oxygen and chlorine.
3. Manufacture of compounds
   Electrolysis is used in the production of sodium hydroxide and potassium hydroxide.
4. Electroplating
   Gold plated jewellery, silver plated utensils, chromium plated iron objects etc are produced through electroplating.
5. Purification of metals
   Electrolysis is useful in the purification of metals such as copper, gold etc.

ELECTROCHEMICAL CELLS AS ENERGY SOURCES
A cell is a device that helps to convert the energy released in a chemical reaction directly into electricity. Cells perform two main functions.

1. As portable sources of electrical energy:
   Examples range from the button cells, used in electronic watches, to the lead-acid cells used in vehicles.
2. As storage devices of electrical energy provided by an external source:
   Such cells can be used for powering electric vehicles, emergency power distribution and storing solar energy.

DIFFERENT TYPES OF CELLS
There are two main types of cells – Primary cells and secondary cells.
1. PRIMARY CELLS

• In primary cells, the cell becomes dead, when the electrical energy produced by chemical reaction is used up.
• These cells cannot be recharged and reused.
• Examples of primary cells are dry cells and button cells. [Dry cells are commonly used in clocks and button cells are used in watches]

2. SECONDARY CELLS

• A secondary cell can be recharged and used again.
• The most important type of secondary cell is the lead acid cell used in vehicles.
• The nickel-cadmium cell used in flashlights is also a secondary cell.

Uses of Lithium-ion batteries
Lithium-ion batteries are secondary cells that have become an integral part of our daily life.

• They are used as power sources for devices ranging from smartphones and laptops to electric vehicles.
• Lithium-ion batteries are used in portable electronic devices due to their high energy density, long lifespan and low self-discharge rate.
• Lithium-ion batteries are also used as satellite batteries, a crucial component in the energy systems of spacecrafts.

FUEL CELLS

• Fuel cells are a type of galvanic cell.
• They convert chemical energy produced by the combustion of fuel into electrical energy
• Fuels used include: – Hydrogen, Methane, Methanol
• They operate continuously as long as fuel is supplied.
• Main advantage: They produce electricity with high efficiency.

The comprehensive approach in SCERT Class 10 Physics Solutions Chapter 7 Mechanical Advantage in Action Important Questions with Answers ensure conceptual clarity.

SSLC Physics Chapter 7 Important Questions Kerala Syllabus

Mechanical Advantage in Action Class 10 Important Questions

Question 1.
If the toothed wheel attached to the axle is connected to the wheel attached to the engine as in the given figure, the speed of the vehicle …………………………….. .

(increases/deceases/ does not change)
Answer:
speed decreases

Question 2.
Which of the following have mechanical advantage equal to one.
a) Fixed pulley
b) Lemon squeezer
c) Common balance
d) forceps
Answer:
a) Fixed pulley and
c) Common balance

Question 3.
Some statements are given.
i) Mechanical advantage of first order lever is equal to one or greater than one or lesser than one.
ii) Mechanical advantage of second order lever is always less than one.
iii) Mechanical advantage of third order lever is always less than one.
iv) Mechanical advantage of third order lever is always greater than one.
Choose the correct option,
a) i & ii
b) i & iv
c) i & iii
d) none of these
Answer:
c) i & iii

Question 4.
If the toothed wheel attached to the axle is connected to the wheel attached to the engine as in the given figure, the efficiency of the vehicle to rotate ……………………………. .

(increases/deceases/does not change)
Answer:
increases

Question 5.
What is an example of a first-order lever in the human body?
Answer:
The neck is an example, where the skull (load) pivots on the atlas vertebra (fulcrum), and neck muscles (effort) control head movement.

Question 6.
Select the odd one.
(Scissors, Nut cracker, Common balance, seesaw)
Answer:
Nut cracker

Question 7.
Find the relation between the first word pair and complete the second pair.
Lemon Squeezer : Second order lever
Fishing pole : ……………………
Answer:
Third order lever

Question 8.
Give an example of a second-order lever used in everyday life?
Answer:
A nutcracker is a second-order lever, where the fulcrum is at one end, the load is in the middle, and the effort is applied at the handles.

Question 9.
What is an example of a third-order lever in the human body?
Answer:
The elbow joint is a third-order lever, where the fulcrum is the elbow, the effort is applied by the biceps muscle, and the load is in the hand or forearm.

Question 10.
Statement: The mechanical advantage of a second-order lever is always greater than one.
Reason: The effort arm and the load arm are equal. Which of the following is correct?
a) Both the statement and the reason are correct.
b) The statement is correct, but the reason does not justify the statement.
c) Both the statement and the reason are incorrect.
d) The statement is incorrect, but the reason is correct.
Answer:
b) The statement is correct, but the reason does not justify the statement.

Question 11.
What is the difference between a wedge and a screw?
Answer:
Both are simple machines. A wedge is a double inclined plane with a sharp edge . A screw can be considered as an inclined plane wrapped around a cylinder.

Question 12.
Why do second-order levers always have a mechanical advantage greater than 1?
Answer:
Because the effort arm (distance from fulcrum to effort) is always longer than the load arm (distance
from fulcrum to load), less effort is needed to lift the load, resulting in a mechanical advantage greater than 1.

Question 13.
A force of 30 N was applied on a nail puller to pull a nail. If the mechanical advantage of the nail puller was two, what would be the load applied by the nail?
Answer:
Mechanical advantage = Load /Effort
Mechanical advantage = 2
Effort = 30 N
Load = ?
Load = Mechanical advantage Effort
= 2 × 30 N
= 60 N

Question 14.
Say whether the following statements are true or false. Correct them if wrong.
a) If the effort comes in between the load and the fulcrum, it is a second order lever.
b) There is no gain in work by using simple machine.
Answer:
a) False
If the effort comes in between the load and the fulcrum, it is a third order lever.

b) True

Question 15.
What is a third-order lever, and how does it differ from other levers?
Answer:
In a third-order lever, the effort is applied between the fulcrum and the load. The load arm is longer than the effort arm, resulting in a mechanical advantage less than 1, meaning more effort is required.
Example: A pair of tweezers, fishing rod

Question 16.
What is a second-order lever, and how does it work?
Answer:
In a second-order lever, the load is located between the fulcrum and the effort. The effort arm is always longer than the load arm, providing a mechanical advantage greater than 1, making it easier to lift heavy loads.
Example: Wheelbarrow, nut cracker, lemon squeezer

Question 17.
Classify the following into those with a mechanical advantage greater than 1 and those less than 1: Nutcracker, fishing rod, lemon squeezer, bottle opener, wheelbarrow, forceps, tongs.
Answer:

Mechanical advantage greater than 1	Mechanical advantage less than 1
Nutcracker lemon squeezer bottle opener wheelbarrow	fishing rod forceps tongs

Question 18.
Figure shows a lever.

a) Identify the type of lever shown in figure above? Give an another example for the same type of lever.
b) Explain about the type of lever identified.
c) How does it work?
Answer:
a) First-order lever
Example: Scissors

b) First-order lever has the fulcrum positioned between the effort and the load.

c) It works by pivoting around the fulcrum, allowing the effort to move the load. The mechanical advantage depends on the relative distances from the fulcrum to the effort and load.

Question 19.
How does the position of the fulcrum affect a lever’s function?
Answer:
The fulcrum’s position determines the lever’s mechanical advantage and function. In first-order levers, moving the fulcrum closer to the load increases mechanical advantage. In second-order levers, the fulcrum at one end maximizes force. In third-order levers, the fulcrum’s position enhances speed and range of motion at the cost of force.

Question 20.
What are the differences between first, second, and third order levers?
Answer:
First order lever: In a first order lever, fulcrum comes in between effort and load. The load can be equal to, lesser or greater than the effort. The mechanical advantage of a first order lever is equal to one or greater than one or lesser than one.

Second order lever: In a second order lever, the load is always more than the effort. The load comes in between effort and fulcrum. Mechanical advantage of a second order lever is always more than one.

Third order lever: In a third order lever, the load is always less than the effort. The effort comes in between fulcrum and load. Its mechanical advantage is always less than one.

Question 21.
Classify the following levers into first, second, and third order lever.
Nut cracker, Fishing pole, Scissors, Wheel barrow, Lemon Squeezer, Seesaw, Forceps, Common balance, Fixed pulley. Elbow, Ice tongs
Answer:
First order lever

• Scissors
• Common balance
• Fixed pulley

Second order lever

• Nut cracker
• Wheelbarrow
• Lemon Squeezer

Third order lever

• Fishing pole
• Forceps
• Elbow
• Ice tongs

Question 22.
Simple machines are basic mechanical devices that mechanical devices that make work easier.
a) What do you mean by simple machines?
b) There are mainly six types of simple machines. Which are they?
c) What is mechanical advantage (MA)?
d) Mechanical advantage of a wheel and axle system is ……………………………..
Answer:
a) Simple machines are devices that change the magnitude of the effect of force or the direction of the force or both. They are devices that make exertion easier.

b) Lever, Pulley, Wheel and axle, Inclined plane, Screw, Wedge

c) Mechanical advantage (MA) is the ratio of the load to the effort. It is a number indicating how many times of the load is the effort. Mechanical advantage is only a ratio. It has no unit.
Mechanical advantage, MA = \(\frac{\text { Load }}{\text { Effort }}\)

d) \(\frac{R(\text { Radius of wheel })}{r(\text { Radius of axle })}\)

The comprehensive approach in SCERT Kerala Syllabus 10th Standard Physics Textbook Solutions and Class 10 Physics Chapter 7 Mechanical Advantage in Action Notes Questions and Answers English Medium ensure conceptual clarity.

SSLC Physics Chapter 7 Notes Questions and Answers Pdf Mechanical Advantage in Action

SCERT Class 10 Physics Chapter 7 Mechanical Advantage in Action Notes Pdf

SSLC Physics Chapter 7 Questions and Answers – Let’s Assess

Question 1.
Write down an example of a lever where the load arm and effort arm are equal. What is its mechanical advantage?
Answer:
Common balance
Mechanical advantage = 1

Question 2.
In which order of lever is the load arm longer than the effort arm? Write down an example. Which may be the possible mechanical advantage? Choose from the brackets.
(less than one/one/more than one)
Answer:
Third order of lever
Example: ice tongs
less than one

Question 3.
As shown in the figure, an object is suspended from one end of a meter scale. When 400 gwt is suspended from the midpoint of the other side, the scale is balanced. What is the mass of the object? What is the weight of the object?

Answer:
Load = x
Load arm = 50 cm
Effort = 400 gwt
Effort arm = 25 cm
\(\frac{\text { Load }}{\text { Effort }}\) = \(\frac{\text { Effort arm }}{\text { Load arm }}\)
\(\frac{x}{400}\) = \(\frac{25}{50}\)
x = 200 gwt
Mass of the object = 200 g
Weight of the object = 200 gwt

Question 4.
In which type of lever is the mechanical advantage always greater than one?
Answer:
Second order lever

Question 5.
Draw a schematic diagram of a second order lever. Mark the fulcrum, load, and effort. Explain why the mechanical advantage of this is neither one nor less than one.
Answer:

Mechanical advantage = \(\frac{\text { Effort arm }}{\text { Load arm }}\)
As the load comes in between the fulcrum and effort, the effort arm will always be greater than the load arm. So the mechanical advantage of a second order lever is neither one nor less than one.

Question 6.
A log weighing 5000 kgwt is lifted through a height of 2 m onto a lorry using an inclined plane. A force of 2000 N was applied. What is the length of the inclined plane used to lift the log? What is the mechanical advantage in this situation? (Consider g = 10 m/s²)
Answer:
Mechanical advantage of an inclined plane, MA
= (L/E) = (l\h)
L = Load = 5000 kgwt = 5000 ×
10 N = 50000 N
E = Effort = 2000 N
Mechanical advantage = \(\frac{L}{E}\) = \(\frac{50000}{2000}\) = 25
l/h = 25
h = 2 m
Length of the inclined plane, = 2 × 25 = 50 m

Question 7.
How will you calculate the mechanical advantage of a screw? Describe an experiment to prove that a screw is an inclined plane.
Answer:
Mechanical advantage of a screw
=length of one thread /pitch Cut a paper in the shape of an inclined plane. Colour its slanting edge as in Fig.7.35(a)
Wrap this paper around a cylindrical pencil as in Fig.7.35(b).
We can see the coloured lines that appear as rings on the pencil look,like a screw.

Question 8.
A load of 1600 kgwt is suspended from an axle with a diameter of 6 cm. What should be the force applied in newton, to turn a 3 m long lever on one side of the axle, to lift this load? (Consider g = 10 m/s²)
Answer:
Mechanical advantage of axle and wheel = \(\frac{R}{r}\) = \(\frac{L}{E}\)
Diameter of the axle = 6 cm
Radius of the axle, r = 3 cm
Load = L = 1600 kgwt
= 1600 × 10 N
= 16000 N
\(\frac{L}{E}\) = \(\frac{R}{r}\)
\(\frac{16000}{E}\) = \(\frac{1}{2}\)
Effort, E = 160 N

Question 9.
Explain the necessity of gears in vehicles.
Answer:
While going uphill, large wheel is connected to the toothed wheel attached to the engine. When connected to a large toothed wheel, the speed of the vehicle decreases, but its efficiency to rotate increases.

To increase the speed of the vehicle, the toothed wheel attached to the engine must be connected to the small toothed wheel that helps to turn the tyre of the vehicle.

Question 10.
Match appropriately.

Lever	A	B	C
First order	Load is always more than the effort	Fulcrum comes in between effort and load	Mechanical advantage is always less than one
Second order	Load can be equal to, lesser or greater than the effort	Effort comes in between fulcrum and load	Mechanical advantage is always more than one
Third order	Load is always less than the effort	Load comes in between effort and fulcrum	Mechanical advantage is always equal to one or greater than one or lesser than one.

Answer:

Lever	A	B	C
First order	Load can be equal to, lesser or greater than the effort	Fulcrum comes in between effort and load	Mechanical advantage is equal to one or greater than one or lesser than one.
Second order	Load is always more than the effort	Load comes in between effort and fulcrum	Mechanical advantage is always more than one
Third order	Load is always less than the effort	Effort comes in between fulcrum and load	Mechanical advantage is always less than one

Physics Class 10 Chapter 7 Notes Kerala Syllabus Mechanical Advantage in Action

Question 1.
Observe the pictures given below. Name the simple machines shown in the pictures.


Answer:
Fig.7.1 (a) – Ramp as inclined plane
Fig.7.1 (b) – Spoon
Fig.7.1 (c) – Nail cutter
Fig.7.1 (d) – Bottle opener
Fig.7.1 (e) – Lever in the pump

Question 2.
How did these devices alleviate exertion?
Answer:
These devices reduce exertion by using simple machines (such as pumps, levers, and wedges) to change force or change its direction and thus helps to complete the task easily.

Question 3.
Observe the pictures. Analyse the situations in which a nail is pulled out.

a) In which situation [Fig. 7.2 (a), Fig. 7.2 (b)] was more force applied?
Answer:
In Fig. 7.2 (a)

b) Which device helped to increase the effect of the force we applied many folds?
Answer:
Nail puller

Question 4.
Observe the two ways of drawing water from a well.

a) In both the situations [Fig. 7.3(a), Fig. 7.3 (b)], is the force applied in the same direction?
Answer:
No, the force applied is not in the same direction in both situations.

b) In which direction is it more convenient to apply force?
Answer:
It more convenient to apply force in the downward direction as in situations in Fig. 7.3(b)

c) What is the benefit of using a pulley?
Answer:
Pulley helps to change the applied force in a more convenient direction, that is in the downward direction.

Question 5.
Analyse the situations in Fig. 7.2(a), Fig. 7.2(b) and Fig. 7.3(a), Fig. 7.3(b). Answer the questions given below.


a) In which situation did the effect of the force we applied increase many fold?
Answer:
In Fig. 7.2(b) and Fig. 7.3(b)

b) Which device increased the effect of the applied force many fold?
Answer:
Nail puller

c) Which device helped to change the direction of the applied force?
Answer:
Pulley

Advantages of using simple machines.

• Changes the magnitude of the effect of the force,
• Changes the direction of the applied force.

Simple machines are devices that change the magnitude of the effect offorce or the direction of the force or both.

Question 6.
How many types of simple machines are there? What are they?
Answer:
There are mainly six types of simple machines.
They are:

1. Lever
2. Pulley
3. Wheel and axle
4. Inclined plane
5. Screw
6. Wedge

Question 7.
When we squeeze a lemon using lemon squeezer,

a) What is the force we apply on the lemon squeezer?
(F₁/F₂)
Answer:
F₁

b) What is the force the lemon applies against the force we apply? (F₁/ F₂)
Answer:
F₂

When a lemon squeezer is used as a simple machine, the force we apply to squeeze the lemon is the effort. The force the lemon applies is the load.

The force we apply to a simple machine is the effort (E). The force the simple machine has to overcome is the load (L).

Question 8.
Identify the load and the effort while removing a nail using a nail puller [Fig. 7.2 (b)]. Note them down in your science diary.

Answer:
The force we apply to the nail puller is the effort (E).
The force the nail applies to the nail puller is the load (L).

Question 9.
Observe the picture of lifting a stone weighing 400 N, using a crowbar.

a) What is the weight lifted using the crowbar?
Ans:
400 N

b) What is the magnitude of force applied?
Answer:
100 N

c) How many times did the effect of the force applied increase?
Answer:
Four times

d) What is the advantage of using a crowbar?
Answer:
The effect of the force applied increased four times. That is the effect of the effort is increased by four times.

Question 10.
If a simple machine can lift a load four times the effort,
a) What is the ratio between the load and the effort?
Answer:
The ratio between the load and the effort is four.

b) How much is the mechanical advantage in the above case?
Answer:
mechanical advantage in the above case is four.

c) How many times did the effect of the force applied increase?
Answer:
Here the effect of the effort is increased by four times.

Mechanical advantage (MA) is the ratio of the load to the effort. It is a number indicating how many times of the load is the effort. Mechanical advantage is only a ratio. It has no unit Mechanical advantage, MA = \(\frac{\text { Load }}{\text { Effort }}\)

Question 11.
A force of 40 N was applied on a nail puller to pull a nail. If the mechanical advantage of the nail puller was three, what would be the load applied by the nail?
Answer:
Mechanical advantage = \(\frac{\text { Load }}{\text { Effort }}\)
Mechanical advantage = 3
Effort = 40 N
Load = ?
Load = Mechanical advantage × Effort
= 3 × 40 N
= 120 N

LEVER
Question 12.
Observe figure.

a) Name the position that supports the rod used as the simple machine.
(effort, load, fulcrum)
Answer:
fulcrum

b) Where is the fulcrum in common balance, seesaw, etc.?

Answer:
In the middle

Question 13.
A crowbar, beam of a common balance, seesaw, etc., are rigid rods. In all these, where is the load and effort?
Answer:
In crowbar, beam of a common balance, seesaw, etc., load is at one end and effort at the other end.

A lever is a rigid rod that can rotate around a fixed point called fulcrum.

Question 14.
What is the load arm here?
Answer:
The load arm here is 25 cm.

Question 15.
What is the perpendicular distance from the effort to the fulcrum known as?
Answer:
The perpendicular distance from the effort to the fulcrum known as the effort arm.

Question 16.
How much is the effort arm here?
Answer:
The effort arm here is 25 cm.

Question 17.
Change the loads and their positions and use appropriate efforts to bring the meter scale to equilibrium. Complete the table with the data obtained in each case.

Answer:

Question 18.
What inference can you arrive at from these activities?
Answer:
The inference that can be arrived at is that when a lever is in equilibrium, Load × Load arm = Effort × Effort arm.

When a lever is in equilibrium, Load × Load arm = Effort × Effort arm. This is the principle of a lever.

Question 19.
How can we calculate the mechanical advantage of a lever?
Answer:
Mechanical advantage = \(\frac{\text { Load }}{\text { Effort }}\)
According to the principle of the lever,
Load × Load arm = Effort × Effort arm
\(\frac{\text { Load }}{\text { Effort }}\) = \(\frac{\text { Effort arm }}{\text { Load arm }}\)
Mechanical advantage of levers = \(\frac{\text { Effort arm }}{\text { Load arm }}\)

Question 20.
You are given a meter scale, weights, a stand, a thread, and a mango. Find out and present how to
determine the mass of the mango.
Answer:
Suspend a meter scale at the centre of gravity. Suspend the mango at a fixed distance from that point. On the opposite side, at the same distance from the pivot, suspend the required weight to balance it. That will be the weight of the mango.

Question 21.
Is the fulcrum in a lever always between the load and the effort?
Answer:
No

Levers can be classified into three types based on the relative positions of the fulcrum, effort, and load. * First Order Lever * Second Order Lever * Third Order Lever

First Order Lever
Question 22.
Observe figure 7.11(a). What is the position of the fulcrum?

Answer:
The position of the fulcrum is in between the load and the effort.
First order lever

If the fulcrum is in between the load and the effort, it is a first order lever.Examples: common balance, scissors, seesaw.

Question 23.
In a common balance, the effort arm and the load arm are(equal/not equal).
Answer:
equal
Hence the load and effort will be equal.

Question 24.
What about while using scissors?
Answer:
In scissors the effort arm and load arm can vary depending on the use of the scissors.

Question 25.
While using a crowbar to move a stone, as shown in figure 7.11 (a), should you increase the effort arm or the load arm to enhance the mechanical advantage?
Answer:
While using a crowbar to move a stone, we should increase the effort arm to enhance the mechanical advantage.

Question 26.
What is the mechanical advantage in this case? (greater than one, one, less than one)
Answer:
greater than one

Question 27.
If the length of the load arm is increased more than the length of the effort arm, what will the mechanical advantage be?
(greater than one, one, less than one)
Answer:
less than one

Question 28.
Complete table 7.2 with the appropriate terms from the brackets.
(load is more than the effort, load and effort are equal, load is less than the effort)

Answer:

If the mechanical advantage is less than one	load is more than the effort
If the mechanical advantage is one	load and effort are equal
If the mechanical advantage is greater than one.	load is less than the effort

Question 29.
Write down examples for first order levers from daily life situations.
Answer:
Fixed pulley, common balance, scissors, seesaw, pliers, human elbow

Second Order Lever
Question 30.
Observe figure 7.13

a) Mark the position of the load, fulcrum, and effort in the figure.
Answer:

b) Where is the position of the load in the figure?
Answer:
The load comes in between the effort and the fulcrum

If the load comes in between the effort and the fulcrum, it is a second order lever. Examples: Lemon squeezer, nut cracker, bottle opener, staplers, wheelbarrows

Question 31.
Mark the load, fulcrum, and effort in the pictures given below.

Answer:

Question 32.
Where will the fulcrum in a second order lever be?
Answer:
At one end

Question 33.
What about the position of the load?
Answer:
The load comes in between the effort and the fulcrum.

Question 34.
Which of the following is correct in a second order lever?
(effort arm and load arm are equal, effort arm is longer than the load arm, effort arm is shorter than the load arm)
Answer:
Effort arm is longer than the load arm

Question 35.
The mechanical advantage of second order levers will always be
(less than one, one, greater than one)
Answer:
greater than one

Question 36.
Write down more examples for second order lever from daily life situations.
Answer:
Lemon squeezer, nut cracker, bottle opener, staplers, wheelbarrows

Third Order Lever
Question 37.
Observe figure 7.16 and mark the load, effort, and fulcrum.

Answer:

If the effort comes in between the load and the fulcrum, it is a third order lever. Examples: Tools used to pick sweets in bakeries, forceps, Tongs

Question 38.
In third order levers, which arm is longer?
(load arm, effort arm)
Answer:
load arm

Question 39.
What will be the mechanical advantage of a third order lever?
(less than one, one, greater than one)
Answer:
less than one

Question 40.
Write down more examples for third order levers from daily life situations.
Answer:

1. Fishing pole
2. Broom
3. ’ Tweezers
4. ’ Tools used to pick sweets in bakeries
5. ’ Forceps
6. ’ Tongs

Question 41.
The effort is greater than the load in a third order lever. If so, what is the advantage of using a third order lever?
Answer:
The advantage is that it helps to handle objects safely with ease.

Question 42.
Mark the fulcrum, load and effort in the case of a forceps.
Answer:

Question 43.
Complete the table

Answer:

Order of a lever	Relative position of fulcrum, load and effort	Mechanical advantage
First order	Fulcrum is in between the load and the effort	Greater than one /equal to one/less than one
Second order	Load is in between the fulcrum and the effort	Alwavs greater than one
Third order	Effort is in between the load and the fulcrum	Alwavs less than one

Fixed Pulley
Question 44.
What are fixed pulleys?
Answer:

Fixed pulleys are pulleys that rotate around a stationary axle. A fixed pulley can be imagined as many spokes rotating around a central pivot named fulcrum. In these spokes, the position where the load is experienced is marked as L and the position where the effort is applied is marked as E. In this, the effort arm and the load arm are the radii of the pulley.

Question 45.
What type of lever a fixed pulley is?
Answer:
A fixed pulley is similar to a first order lever.

Question 46.
What will be the relation between the load arm and the effort arm of the pulley?
(load arm is longer, effort arm is longer, effort arm and load arm are equal)
Ans:
effort arm and load arm are equal

Question 47.
When we consider a fixed pulley without friction, we have to apply …………………………
(more effort than the load, less effort than the load, an effort equal to load)
Answer:
an effort equal to load

Question 48.
What will be the mechanical advantage of a fixed pulley?
(less than one, one, greater than one)
Answer:
one

Question 49.
What is the advantage of using friction less fixed pulley to lift objects? Choose the correct option from those given below.
(can reduce the magnitude of the applied force / can change the direction of the applied force)
Answer:
can change the direction of the applied force

Question 50.
In a movable pulley the effort and the fulcrum are at either ends and the load is at the middle. Which is the order of this lever?
Answer:
Second order lever

Question 51.
Which is the load arm and effort arm?
Answer:
Load arm = Radius of the pulley
Effort arm = Diameter of the pulley

Question 52.
What is the mechanical advantage of a movable pulley?
Answer:
Mechanical advantage
= \(\frac{\text { Effort arm }}{\text { Load arm }}\) = \(\frac{\text { Diameter of the pulley }(2 r)}{\text { Radius of the pulley }(r)}[latex]
= 2

Question 53.
The mechanical advantage of a single movable pulley is 2. If so, how many times of the load to be lifted should the applied effort be?
Answer:
Half

Question 54.
The rope is pulled 2 m by applying effort. How much will the load rise?
Answer:
The load rises only one metre.

Question 55.
What should be the displacement of the rope, when the effort applied raises the load by 1 m?
(1 m, 2 m, 0 m)
Answer:
2 m

Question 56.
Is there a gain in work when the applied effort is halved? Prove.
Answer:
There is no gain in work when the applied effort is halved
Work, W = Fs
When the effort is reduced by half, the displacement produced by it is doubled.
That is, W = (F/2) × 2s = Fs. That is there is no change in the quantity of work done. No gain in work is achieved while using a movable pulley.

Question 57.
A movable pulley is used to lift 600 N load. When the rope is pulled 8 m by applying a force of 300 N then the load raises by 4 m. Prove that there is no gain in the work even though there is a gain in the effort applied.
Answer:
The displacement of the object, s = 4 m
Weight of the object, F = 600 N
Work done on the object
= F × s
= 600 × 4 = 2400 J
The displacement of the rope on applying 300 N force,
s₁ = 8 m
F₁ = 300N

The work done by the force we applied = F₁ s₁
= 30o × 8 = 2400 J
The work done on the object and the work done by the effort are equal.

Question 58.
Is there a gain in work by using simple machines?
Answer:
There is no gain in work by using simple machines.

Question 59.
The radius of the wheel is R and the radius of the axle is r. When the wheel is rotated once, What will be the distance moved by a point on the wheel’s circumference?
Answer:
When the wheel is rotated once, the distance moved by a point on the wheel’s circumference be 2πR.

Question 60.
What will be the distance moved by the object tied to the rope attached to the axle? (2πR / 2πr)
Answer:
2πr

Question 61.
What is the distance moved by the point on the wheel when rotated once due to the effort applied (E)?
Answer:
The distance moved by the point on the wheel when rotated once due to the effort applied (E) is 2πR.

Question 62.
What is the distance moved by the load?
Answer:
The distance moved by the load is 2πr.
The work done by the effort and the work done by the load are equal here.
Work done by the effort = Work done by the load
Work done = Fs
E × 2πR = L × 2πr
= L/E = R/r
Since the radius of the wheel is larger and the radius of the axle is smaller, the mechanical advantage of wheel and axle will be greater than one.
As the radius of the wheel increases, the mechanical advantage also increases.

• The train bogies were slowly lifted by turning the wheel using long levers.

Gear

Question 63.
What are gears? What are the uses of these gears?
Answer:
A gear is a mechanical device used in machines.
They are used to transfer motion or force from one part of the machine to another.

Question 64.
What is the main part of a gear system?
Answer:
The main part of a gear system is the system of two interlocking toothed wheels.
In gears, energy is transferred from a large toothed wheel to a small toothed wheel or from a small toothed wheel to a large toothed wheel.

Question 65.
When the large toothed wheel rotates once, how many times will the small toothed wheel rotate? (once, more than once, less than once)
Answer:
more than once

Question 66.
What changes will this cause in the speed of the second toothed wheel?
Answer:
The speed of the second toothed wrheel increases.

Question 67.
A toothed wheel is attached to the engine of most of the vehicles. What is the function of this toothed wheel?
Answer:
This toothed wheel is connected to a system of wheels of different sizes attached to the axle which transfers motion to the tyres.

Question 68.
Which wheel is connected to the toothed wheel attached to the engine to increase the speed of the vehicle?

To increase the speed of the vehicle, the toothed wheel attached to the engine must be connected to
the small toothed wheel that helps to turn the tyre of the vehicle.

Question 69.
Which wheel is connected to the toothed wheel attached to the engine while going uphill?
(small/large)

Answer:
large

Question 70.
What is the advantage of connecting the large wheel to the toothed wheel attached to the engine while going uphill?
Answer:
The speed of the vehicle decreases, but its efficiency to rotate increases.

Question 71.
What happens to the speed of the large toothed wheel when energy is transferred from the small toothed wheel to the large toothed wheel? (Increases/decreases)
Answer:
decreases

Question 72.
Try lifting the same object to the same height using inclined planes of different lengths. When did it feel easier?
Answer:
It felt easier when using inclined plane of more length.

Question 73.
An object is to be lifted to a height h metre by pushing along an inclined plane of length l metre. Here, what is the work done by the effort?
Answer:
If the effort is E and the load is L, then the work done by the effort E to move it over a distance,
l = (F_s) = El

Question 74.
An object is to be lifted to a height h metre by pushing along an inclined plane of length 1 metre.
Here, what is the work done by the load?
Answer:
Work done by the load L to lift to a height h is Work = Force × Displacement
i.e., Weight × height = Lh

Question 75.
An object is to be lifted to a height h metre by pushing along an inclined plane of length 1 metre. Here, what is the mechanical advantage?
Answer:

Question 76.
Does using a longer inclined plane or a shorter inclined plane provide more mechanical advantage?
Answer:
Mechanical advantage = [latex]\frac{\text { Length of inclined plane }}{\text { Height of inclined plane }}\)
So, using a longer inclined plane provides more mechanical advantage.

Question 77.
Inclined planes are often used in situations where objects are to be lifted.
a) What is the work done to lift a 600 N load by 3 m?
b) The force applied was 200 N along an inclined plane to lift this load. Displacement of the load was 9 m. What is the work done?
c) Is there a gain in the work done?
Answer:
a) Force, F = 600 N
Displacement, s = 3 m
Work done, W = Fs = 600 × 3 = 1800 Nm= 1800 J

b) Force, F = 200 N
Displacement, s = 9 m
Work done, W = Fs = 200 × 9 = 1800 Nm= 1800 J

c) There is no gain in work done.

Question 78.
What are the ways to increase the mechanical advantage of an inclined plane?
Answer:
We can increase the length of an inclined plane to increase the mechanical advantage.

Question 79.
Why are long inclined roads built in hairpin roads on ghats and other such places?


Answer:
The concept of mechanical advantage of inclined plane is utilised here. As we increase the length of an inclined plane the mechanical advantage increases. So the inclined roads are built with maximum length.
• It is understood that large heavy stones were taken up to build the pyramids using inclined roads.

Question 80.
Observe the wedges of different length and thickness. Which of these is easier to use? Why?

Answer:
Fig. 7.32 (c),

The wedge in figure 7.32 (c) has less thickness and more length. So, its mechanical advantage is greater than that of others. If a wedge is made longer and thinner, it will be easier to use.

Question 81.
Write down the use of a screw jack.
Answer:
A screw jack is used to lift vehicles to change a tyre or for minor repairs.

Question 82.
A 600 kgwt load is lifted by 4 m along an inclined plane of length 8 m. Calculate the mechanical advantage. What force must be applied along the inclined plane?
Answer:
Mechanical advantage of an inclined plane

Question 83.
Label the name of the simple machines given below.

Answer:

The bicycles and sewing machines we use today are often said to be the simplest machines. But they are a combination of many types of the simple machines mentioned above.

Std 10 Physics Chapter 7 Notes – Extended Activities

Question 1.
Plan a project to find the density of a meter scale.
Answer:
Find the centre of gravity of the meter scale. Hang the scale at a distance of 10 cm, 20 cm, 30 cm from the centre of gravity and determine the position required to balance it in each case.

Assuming the mass of the part that is 1 cm as m. Using the equation load × load arm = effort × effort arm, the mass of the scale and the density of it can be found out. When balancing it at 10 cm, the load = 60 × m
load arm = 30 cm
Effort = 40 × m + suspended mass (y)
load × load arm = 60 × m × 30
Effort × Effort arm = 40m × 20 + y × x (distance from the suspended mass ‘y’ to the fulcrum). Find the value of x, for a value of y (say 50 g or 100 g). Repeat the experiment by hanging the scale at different distances (as mentioned above 20 cm, 30 cm from the centre of gravity).
Solve for the mass from the euqation.
Find 100 × m = M, to find the total mass of the meter scale.

Measure the Length, breadth (b) and thickness (t) (dimensions) of the meter scale.
Calculate volume V= Length × Breadth × Thickness and calculate density using the formula
Density = \(\frac{M(\text { mass })}{V(\text { Volume })}\)

Question 2.
Observe a bicycle and list the simple machines that forms a part of it.
Answer:

1. Wheel and axle
2. Lever
3. Pulley
4. screw
5. Inclined plane

Mechanical Advantage in Action Class 10 Notes

Mechanical Advantage in Action Notes Pdf

• Devices that make exertion easier are called simple machines.
• Simple machines are devices that change the magnitude of the effect of force or the direction of the force or both.
• There are mainly six types of simple machines. They are:
  Lever, Pulley, Wheel and axle, Inclined plane, Screw and Wedge
• Mechanical advantage (MA) is the ratio of the load to the effort. It is a number indicating how many times of the load is the effort. Mechanical advantage is only a ratio. It has no unit.
  Mechanical advantage, MA = \(\frac{\text { Load }}{\text { Effort }}\)
• Centre of gravity is the point at which the entire weight of an object is considered to act.
• When a lever is in equilibrium, Load × Load arm = Effort × Effort arm. This is the principle of a lever.
• Levers can be classified into three types based on the relative positions of the fulcrum, effort, and load.
  • First Order Lever
  • Second Order Lever
  • Third Order Lever
• First order lever: In a first order lever, fulcrum comes in between effort and load. The mechanical advantage of a first order lever is equal to one or greater than one or lesser than one.
• Second order lever: The load comes in between effort . and fulcrum. Mechanical advantage of a second order . lever is always more than one.
• Third order lever: The effort comes in between fulcrum and load. Its mechanical advantage is always less than one.
• A Pulley is a simple machine. It is of two types:
  1. Fixed pulley
  2. Movable pulley
• In wheel and axle systems, the wheel is rotated by applying force (Effort) to it. When the wheel is rotated once, the axle also rotates once.
• Since the radius of the wheel is larger and the radius of the axle is smaller, the mechanical advantage of wheel and axle will be greater than one.
• A gear is a mechanical device used in machines. They are used to transfer motion or force from one part of the machine to another. The main part of a gear system is the system of two interlocking toothed wheels.
• A wedge is a double inclined plane.
• A screw jack is used to lift vehicles, to change a tyre or for minor repairs.

INTRODUCTION

Simple machines are basic mechanical devices that make work easier by changing the magnitude or direction of a force. Even the bicycles and sewing machines we use today are often said to be the simplest machines. Simple machines are the building blocks of more complex machines and have been used for centuries to perform tasks more efficiently. There are six classic simple machines. They are lever, pulley, wheel and axle, inclined plane screw and wedge. These machines operate on the principle of mechanical advantage, allowing a smaller force to move a larger load or perform tasks with less effort. They are fundamental in everyday tools and complex machinery, simplifying tasks like lifting, cutting, or moving objects. This unit deals with the basic ideas about simple machines.

Simple machines

• Devices that make exertion easier are called simple machines.
• Advantage of using simple machines.
  • Changes the magnitude of the effect of the force.
  • Changes the direction of the applied force.
• Simple machines are devices that change the magnitude of the effect of force or the direction of
  the force or both.
• There are mainly six types of simple machines. They are lever, pulley, wheel and axle, inclined plane, screw and wedge.
• The force we apply to a simple machine is the effort (E). The force the simple machine has to overcome is the load (L).
• Mechanical advantage (MA) is the ratio of the load to the effort. It is a number indicating how many times of the load is the effort. Mechanical advantage is only a ratio. It has no unit.

Mechanical advantage, MA = \(\frac{\text { Load }}{\text { Effort }}\)

Lever

• A lever is a rigid rod that can rotate around a fixed point called fulcrum.
• Centre of gravity is the point at which the entire weight of an object is considered to act.
• Levers can be classified into three types based on the relative positions of the fulcrum, effort, and load.
  • First Order Lever
  • Second Order Lever
  • Third Order Lever
• When a lever is in equilibrium,
  Load × Load arm = Effort × Effort arm. This is the principle of a lever.
  Mechanical advantage of levers = \(\frac{\text { Load }}{\text { Effort }}\) = \(\frac{\text { effort arm }}{\text { load arm }}\)
• If the fulcrum is in between the load and the effort, it is a first order lever.
  Examples: common balance, scissors, seesaw.
• The mechanical advantage of first order lever can be greater than one /equal to one/less than one.
• If the load comes in between the effort and the fulcrum, it is a second order lever. Examples: Lemon squeezer, nut cracker, bottle opener, wheelbarrows
• The mechanical advantage of second order levers will always be greater than one.
• If the effort comes in between the load and the fulcrum, it is a third order lever. Examples: Tools used to pick sweets in bakeries, forceps, Tongs
• The mechanical advantage of third order lever will be always less than order one.

Pulley

• Pulley is a simple machine. It is of two types. Fixed pulley and movable pulley.
• Fixed pulleys are pulleys that rotate around a stationary axle. A fixed pulley can be imagined as many spokes rotating around a central pivot named fulcrum. In these spokes, the position where the load is experienced is marked as L and the position where the effort is applied is marked as E. In this, the effort arm and the load arm are the radii of the pulley.
• Mechanical advantage of a fixed pulley is one
• When effort is applied on a movable pulley, the pulley and the load will be lifted upwards along the other side of the rope. This lifting is caused by the rotation of the pulley along the rope.
• Mechanical advantage of a movable pulley is 2 Mechanical advantage = \(\frac{\text { Diameter of the pulley (2r) }}{\text { Radius of the pulley (r) }}\)

Wheel and axle

• In wheel and axle systems, the wheel is rotated by applying force (Effort) to it. When the wheel is rotated once, the axle also rotates once. If the radius of the wheel is R and the radius of the axle is r, and when the wheel is rotated once, the distance moved by a point on the wheel’s circumference would be 2πR.
• Since the radius of the wheel is larger and the radius of the axle is smaller, the mechanical advantage of wheel and axle will be greater than one.
  Mechanical Advantage = \(\frac{\text { L(load) }}{\text { E (effort) }}\) = \(\frac{\text { Radius of wheel (R) }}{\text { Radius of axle (r) }}\)
• As the radius of the wheel increases, the mechanical advantage also increases
• A gear is a mechanical device used in machines. They are used to transfer motion or force from one part of the machine to another. The main part of a gear system is the system of two interlocking toothed wheels.

Inclined plane

• Mechanical advantage = \(\frac{\text { Length of inclined plane (l) }}{\text { Height of inclined plane(h) }}\)
• Using a longer inclined plane provides more mechanical advantage.

Wedge

• A wedge is a double inclined plane. Household tools like knife, axe, and work tools like chisel are variants of wedges.
• Mechanical advantage of a wedge \(\frac{\text { length of the inclined plane of the wedge }(l)}{\text { thickness of the wedge }(\mathrm{h})}\)

Screw

• A screw can be considered as an inclined plane. The distance between two consecutive threads is the pitch of a screw.
• The mechanical advantage of a screw = length of the inclined plane (l) / height of the inclined plane (h) = length of one thread (l) / pitch (h)

You may have seen loading a lorry using an inclined plane.
• When it is difficult to climb stairs, we use ramps. From this, it is understood that inclined planes are used to reduce exertion.
Devices that make exertion easier are called simple machines.

SIMPLE MACHINES
Let’s examine the ways in which simple machines make exertion easier.

TYPES OF SIMPLE MACHINES
LEVER
Activity to understand the principles related to levers
Suspend a wooden meter scale on a stand. Balance it as shown in figure 7.8.

Identify the point at which it is to be balanced.

Entire weight of the meter scale act through this point. This point is the centre of gravity of the meter scale. The meter scale can be suspended or pivoted along a perpendicular line passing through the centre of gravity as shown in the figure.

You can balance a book on the tin of your finger

A book can be balanced when it is pivoted on the perpendicular line passing through its centre of gravity.
Centre of gravity is the point at which the entire weight of an object is considered to act

Activity to understand the principles related to levers

Suspend a wooden meter scale along a perpendicular line passing through the centre of gravity as shown in the figure on a stand. Balance it as shown in figure. Suspend a mass of 50 g at a distance of 25 cm from the balancing point on one side of the meter scale. Suspend another mass of 50 g on the other side of the meter scale to balance it.

The weight suspended first is considered as the load. The perpendicular distance from the load to the fulcrum is the load arm. The weight suspended or the force applied to produce equilibrium in the rod is the effort.

PULLEY
A Pulley is a simple machine. It is of two types:
1. Fixed pulley
2. Movable pulley

Movable Pulley

When effort is applied on a movable pulley, the pulley and the load will be lifted upwards along the other side of the rope. This lifting is caused by the rotation of the pulley along the rope.

WHEEL AND AXLE
There were instances when modern machines failed to lift train bogies that had fallen into a lake. But they were brought ashore using a wheel and axle.

In wheel and axle systems, the wheel is rotated by applying force (Effort) to it. When the wheel is rotated once, the axle also rotates once.

INCLINED PLANE
You may have seen an inclined plane being used to load large logs, machine parts, etc., on to a lorry.

WEDGE
A wedge is a double inclined plane. Household tools like knife, axe, and work tools like chisel are variants of wedges.

SCREW JACK
A screw can be considered as an inclined plane. The distance between two consecutive threads is the pitch of a screw.

Activity to explain pitch of a screw.
Cut a paper in the shape of an inclined plane. Colour its slanting edge. (Fig 7.35(a))
Wrap this paper around a cylindrical pencil. (Fig.7.35(b))

The vertical distance from the beginning of the slanting edge of the paper (P) to where one full turn is completed (Q) is the pitch (h). That is, the distance between two consecutive threads is the pitch.

Activity to explain the length of one thread of a screw
Cut a paper in the shape of an inclined plane. Colour its slanting edge. (Fig.7.35(a))

Wrap this paper around a cylindrical pencil.(Fig.7.35(b)) Unwrap the paper and straighten it. Measure the distance from P to Q (Fig.7.35(c)). This Distance is the length of the thread PQ (l). That is the length of one thread of a screw.

The mechanical advantage of a screw = length of the inclined plane (l) / height of the inclined plane (h) = length of one thread (l) / pitch (h)

The comprehensive approach in SCERT Class 10 Physics Solutions Chapter 6 Electromagnetic Induction in Daily Life Important Questions with Answers ensure conceptual clarity.

SSLC Physics Chapter 6 Important Questions Kerala Syllabus

Electromagnetic Induction in Daily Life Class 10 Important Questions

Question 1.
When there is a change in magnetic flux linked with a conductor, an emf is induced in it. Name the phenomenon?
Answer:
Electromagnetic Induction

Question 2.
The secondary coil of a transformer has double turns than its primary coil. If the voltage applied in the primary coil is 25 V, what will the voltage be in the secondary?
(25 V, 50 V, 2 V, 12.5 V)
Answer:
50 V

Question 3.
What is the energy conversion taking place in a windmill?
Answer:
Wind Energy → Mechanical Energy → Electrical Energy

Question 4.
Statement: A current is induced in a coil when a magnet is moved towards or away from it
Reason: The/elative motion between the coil and the magnet changes the magnetic flux through the coil
(a) Both statement and reason are true; the reason explains the statement.
(b) Both statement and reason are true; but the reason does not explain the statement.
(c) Both the statement and the reason are incorrect.
(d) Statement is incorrect; reason is correct.
Answer:
(a) Both statement and reason are true; the reason explains the statement.

Question 5.
Some characteristics of stepup transformers is given below. Select the correct statements suitable for stepup transformers.
(a) Secondary voltage is greater than primary voltage.
(b) Thick wires are used in the secondary.
(c) Primary voltage is greater than secondary voltage.
(d) The current in the primary coil is greater than that in the secondary coil.
(i) (a), (b) and (d)only
(ii) (a) and (d) only
(iii) (a) and (c) only
(iv) (b) only
Answer:
(ii) (a) and (d) only

Question 6.
Write any two precautions to be taken to avoid electric shock.
Answer:

• Never handle electric equipment or operate switches when the hands are wet.
• Do not fly kites near electric lines.

Question 7.
The primary and secondary currents of a transformer are 2 A and 4 A respectively. The output voltage of this transformer is 100 V.
(a) Identify the transformer.
(b) Calculate the primary voltage of the transformer.
Answer:
(a) Stepdown Transformer

(b) V_P × Ip = Vs × Is
∴ V_P = (Vs × Is) / Ip
= (100 × 4) / 2
= 200 V

Question 8.
The current in the secondary coil of a transformer with no power loss is 5 A and that is the primary is 0.5 A.
(a) What type of transformer is this?
(b) If the input voltage of this transformer is 240 V calculate the output voltage.
Answer:
(a) Step down transformer

(b) I_S = 5 A
I_P = 0.5 A
V_P = 240 V
V_P = ?
I_S × V_S = I_P × V_P
V_S = \(\frac{V_P \times I_P}{I_S}\) = \(\frac{0.5 \times 240}{5}\) = 24 V

Question 9.
Fuse, MCB, ELCB, and RCCB are used to ensure safety in household electric circuits.
(a) What is the difference between the functioning of Fuse and MCB?
(b) What is the advantage of MCB over Fuse?
(c) What is the function of ELCB and RCCB in electric circuits?
Answer:
(a) Fuse – In fuse, fuse wire melts due to excess current. It works based on the heating effect of electric current.
MCB – It has an internal switch that gets tripped when there is excess current. It works based on the magnetic effect and the heating effect of electrical energy.

(b) In MCB, after solving the problem, the switch can be turned on. But in the fuse, the melted fuse wire is to be replaced by another suitable wire. Moreover, MCB is more sensitive than fuse. MCB is simple to resume the supply.

(c) ELCB helps to break the circuit automatically when there is leakage of current due to insulation failure or short circuit, it protects the user from electric shock. RCCB also breaks the circuit, so that there’is no harm due to electric shock. RCCB ensures more safety.

Question 10.
A transformer has 100 turns in the primary and 1000 turns in the secondary.
(a) Which coil of this transformer is made using thick wire? Give reason.
(b) Explain/how electrical energy is transferred from the primary to the secondary of the transformer
Answer:
(a) Primary coil is made using thick wire.
(Hint: Number of turns is less in the Primary here. So it is a step-up transformer. So thick wire is used in the primary coil)
When the thick wire is used, the resistance of the coil can be minimized. There are two benefits to this.
(i) Overheating of the coil can be prevented.
(ii) Energy loss can be minimized.

(b) It is through mutual induction power is transferred from the primary coil to the secondary coil.
When the strength or direction of the current in one of the two adjacent coils changes, the magnetic flux around it changes. As a result, an emf is induced in the secondary coil. This phenomenon is mutual induction.

Question 11.
AC generators are used in power stations in our Country.
(a) What is the voltage produced by the generators in our power stations?
(b) What do you mean by transmission loss?
(c) Explain how it is minimized.
Answer:
(a) 11000 V (11 kV)

(b) When electrical energy is transmitted through conducting wires, energy is wasted in the form of heat.

(c) As H = I²Rt, for reducing H, I and R are to be decreased.
As P = VI, For decreasing I without affecting the power, voltage V is to be increased before transmission.

Question 12.
A schematic diagram of a generator is given:

(a) Which type of generator is this? (AC/DC)
(b) Name the parts of this generator marked as
(c) State the working principle of this device.
1, 2, 3, 4,
1. ________________
2. ________________
3. ________________
4. ________________
(c) State the working principle of this device.
Answer:
(a) AC Generator

(b) 1. Field magnet
2. Armature
3. Slip rings
4. Brush

(c) Electromagnetic induction: Whenever there is a change in the magnetic flux linked with a coil, an emf is induced in the coil. This phenomenon is electromagnetic induction.

Question 13.
Observe the circuit of household electrification.

(a) Which device is used to measure the electrical energy consumed in household circuits?
(b) Write any two advantage of connecting the devices in parallel in a household circuit.
(c) Write the function of ELCB.
Answer:
(a) Watt-hour meter.
(b) Devices work according to the marked power.
Devices can be controlled using switches as needed.

(c) ELCB helps to break the circuit automatically whenever there is a current leak due to insulation failure or other reasons.

Question 14.
A transformer without power loss works at an input of 250 V. A current of 1 A flows through the secondary coil when an electrical device of power 50 W is connected to the secondary.
(a) Which type of transformer is used here?
(b) What is the working principle of a transformer?
(c) Calculate the current through the primary.
Answer:
(a) Ps = Vs × Is
50 = Vs × 1
Vs = 50 V
Since Vs is less than Vp, it is a step-down transformer.

(b) Mutual Induction

(c) Vp × Ip = Vs × Is
Vp = 250 V, Vs = 50 V
Is = 1 A
Ip = (Vs × Is) / Vp = (50 × 1) / 250 = 0.2 A

Question 15.
Analyze the following schematic diagram and answer the questions.

(a) Name the device.
(b) Which are the main parts of this device?
(c) Illustrate the graphic representation of the output when the field magnet is rotated by keeping the armature stationary.
Answer:
(a) DC Generator.

(b) Armature, field magnet, split ring commutator, and brushes.

(c)

The comprehensive approach in SCERT Class 10 Physics Solutions Chapter 5 Electric Energy: Consumption and Conservation Important Questions with Answers ensure conceptual clarity.

SSLC Physics Chapter 5 Important Questions Kerala Syllabus

Electric Energy: Consumption and Conservation Class 10 Important Questions

Question 1.
If intensity of electric current in a circuit is doubled the heat energy developed in this circuit increases _______________ times.
(2,12,4,14)
Answer:
4
Hint: Heat produced is directly proportional to the square of current. If current is doubled, heat energy will be increased to four times.

Question 2.
Which among the following is the energy transformation taking place when a storage battery is being charged?
a) chemical energy to electrical energy.
b) electrical energy to heat energy.
c) electrical energy to chemical energy.
d) magnetic energy to electrical energy.
Answer:
c) electrical energy to chemical energy

Question 3.
Some statements related to activities to be done to reduce carbon footprint is given
i. Increase domestic energy consumption.
ii Avoid wasting food.
iii. Use public transport.
iv. Reduce the use of reusable products.

Which are the correct statements related to it?
A. i and ii only
B. iv only
C. All are correct
D. ii and iii only
Answer:
D. ii and iii only

Question 4.
Some statements related to electrical power is given.
i. Electric power is measured in joules.
ii 1 kilowatt-hour is equal to 1000 joules.
iii Power is the quantity of work done by an electrical appliance per unit time.
iv. Power, P = \(\frac{V}{I}\)
Which are the correct statements related to it.
a) i and ii only
b) iii and iv only
c) iv only
d) iii only
Answer:
d) iii only

Question 5.
Statement: A heater A operating at 230 V draws 2 A current, its power obtained is 460 W.
Reason: The power can be calculated as the, product of voltage and current.
a) Statement and reason are correct; the reason explains the statement.
b) Statement and reason are correct; however, the reason does not explain the statement.
c) both the statement and the reason are incorrect.
d) Statement is incorrect; reason is correct
Answer:
a) Statement and reason are correct; the reason explains the statement.

Question 6.
The heat generated in a current carrying conductor can be explained by a famous law.
a) Write the name of this law.
b) Write the mathematical equation for this law, explain each letters used in the equation.
Answer:
a) Joule’s law

b) H = I²Rt or H ∝ I² Rt
H – Heat generated in a conductor
I – Intensity of electric current
R – Resistance of the conductor
t – time for which current flows

Question 7.
Resistance of a 230 V heating device is 460 Ω. Calculate the heat energy produced by it in 10 minutes.
Answer:
H = \(\frac{V^2 t}{R}\)t, t = 10 × 60 s, R = 460 Ω
= \(\frac{230^2}{460}\) × 600 = 69000 J
H = 69000J

Question 8.
a) Name the part of a heating equipment in which the electric energy is converted into heat energy.
b) Name the substance used to make this part.
Answer:
a) Heating coil
b) Nichrome

Question 9.
In a house, 5 CF lamps each of 20 W work for 4 hours, and 4 fans each of 60 W work for 5 hours in a day. What will be the daily consumption of energy shown by the watt-hour meter?
Answer:
Electric energy (in kilowatt hour) = \(\frac{\text { Power in watt × time in hour }}{1000}\)
Energy consumed by 5 CF lamps = \(\frac{5 \times 20 \times 4}{1000}\) = 0.4 kWh
Energy consumed by 4 fans = \(\frac{4 \times 60 \times 5}{1000}\) = 1.2 kWh
Daily consumption of energy
= 0.4 + 1.2 = 1.6 kWh
= 1.6 units

Question 10.
How many units of energy will be consumed by a 100 W bulb in 30 days, if it works 10 hours everyday?
Answer:
Electric energy (in kilowatt hour) = \(\frac{\text { Power in watt × time in hour }}{1000}\)
Energy consumed by bulb in a day
= \(\frac{1 \times 100 \times 10}{1000}\)
= 1 unit
Energy consumed by bulb in a month
= 1 × 30 = 30 units

Question 11.
The government is promoting the installation of solar panels by giving incentives to boost electricity generation. Explain its significance in the context of the energy crisis.
Answer:
The over use of non-renewable fossil fules like coal, oil, and natural gas which contribute to pollution and global warming has resulted in an energy crisis for the entire world.

Because it is abundant, renewable, environmentally friendly, and accessible even in remote locations, solar energy is an excellent substitute. Solar thermal power plants use the sun’s heat to produce electricity, whereas solar cells directly turn sunlight into electrical power. In everyday life, gadgets like solar water heaters and stoves can help cut down on the amount of power and fossil fuels-used. We can lessen the effects of the energy crisis, conserve resources, and safeguard the environment by utilising more solar energy.

Question 12.
100W bulb works on 230V supply.
a) If voltage is halved, how much will be the power?
b) If voltage decreases to 1/4 times, how much will be the power?
c) If the voltage is doubled what happen to the bulb?
Answer:
a) R = \(\frac{\mathrm{V}^2}{\mathrm{P}}\) = \(\frac{230^2}{100}\) = 529
V = 115 V (voltage is halved)
P = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) = \(\frac{115^2}{100}\) = 25 W
(When voltage is halved, power becomes 1/4 times the original power)
P = 25 W

b) V = 57.5 V (Voltage decreases to 1/4 times)
P = \(\frac{57.5^2}{529}\)
P = 6.25 W

c) Device will be damaged due to high voltage.

Question 13.
Fill in the blanks of the table suitably.

Device	Energy change	Effect of electricity
Electric bulb	(a)	Lighting effect
(b)	Electrical energy is converted to heat energy	(c)
Mixie	(d)	Mechanical effect
Electric kettle	(e)	(f)

Answer:
(a) – Electrical energy,
(b) – Electric Iron,
(c) – Heating effect,
(d) – Electric energy to mechanical energy,
(e) – electrical energy to heat energy,
(f) – heating effect.

Question 14.
A heating coil with 60 Ω resistance is connected to a 240V supply.
a) Calculate the power of the appliance.
b) Calculate the amount of heat generated by this heating coil in 5 minutes.
c) If this appliance continuously work for 10 hours, calculate the energy consumed in commercial units.
Answer:
a) Power of the appliance
P = \(\frac{\mathrm{V}^2}{\mathrm{R}}\) = \(\frac{240^2}{60}\) = 960 W

b) t = 5 min = 5 × 60 = 300 s
Heat generated = \(\frac{\mathrm{V}^2}{\mathrm{t}}\)
H = 960 × 300 = 288000 J

c) Electric energy (in kilowatt hour)
= \(\frac{\text { Power in watt } \mathrm{x} \text { time in hour }}{1000}\)
Electric energy = \(\frac{1 \times 960 \times 10}{1000}\) = 9.6 units

Question 15.
Energy conservation is very important for a better tomorrow.
a) What do you mean by energy crisis?
b) How do people contribute to energy crisis (any two)
c) Write two actions to reduce your daily electrical energy consumption, (any two)
Answer:
a) Energy crisis is the increase in demand for energy and the decrease in its availability.

b)

• Careless use of electricity at homes, transport, and industries
• Over dependence on fossil fuels like coal, oil, and natural gas
• Not making proper use of natural energy resources.
• Slow action and lack of awareness about saving energy

c)

• Make a transition to energy-efficient appliances.
• Use LED Bulbs-LEDs use less power and last longer than normal bulbs.
• Turn off electric appliances when not in use- Always switch off fans, lights, and appliances when you leave a room.
• Use natural light during the day

Question 16.
An electric heater has a resistance of 690 Ω and is designed to operate at 230 V.
(a) What are the characteristics of a good heating coil?
(b) Calculate the heat energy produced by this heater when it works for 30 minutes.
Answer:
Resistance, R = 690 Ω,
time, t = 30 minute = 30 × 60 = 1800 s
Voltage, V = 230 V

(a) Characteristics of a good heating coil are

• High oxidation resistance
• Ability to remain in red hot condition for a long time.
  • High melting point
  • High resistivity

(b) Heat generated,
H = \(\frac{V^2 t}{R}\) = \(\frac{230^2 \times 1800}{690}\) = 138000 J

Question 17.
An electric heater of 1000 Ω works on a 230 V supply
a) Write down the energy change taking place in the electric heater.
b) State the law to find the quantity of heat generated in an electric heater?
c) Calculate the electrical energy consumed when the heater works for three hours.
Answer:
a) Electrical energy is converted to heat energy

b) The quantity of heat produced in a current carrying conductor is directly proportional to the square of the current (I²), the resistance of the conductor (R) and the time (t) for which the current flows. This is Joule’s law.

c) V = 230V, R = 1000 Ω, t = 3 hour = 3 × 60 × 60 = 10800 s
H = \(\frac{V^2 t}{R}\) = \(\frac{230^2 \times 10800}{1000}\) = 571320J

Question 18.
Observe the figure.

a) Write down the name of the device.
b) What is it used for?
c) In a house 5 LED lamps each of 20 W works for 5 hours and one laptop of 50 W works for 2 hours daily.. Calculate electrical energy consumption in one month in commercial units.
Answer:
a) Watt-hour meter

b) The quantity of electric energy consumed in houses can be measured directly by a watt hour meter.

c) Electric energy (in kilowatt hour)
= \(\frac{\text { Power in watt × time in hour }}{1000}\)
Electric energy consumed by lamps
= \(\frac{5 \times 20 \times 5}{1000}\) = 0.5 units
Electric energy consumed by laptop
= \(\frac{1 \times 50 \times 2}{1000}\) = 0.1 units
Total energy consumption in a day
= 0.5 + 0.1 = 0.6 units
Total energy consumption in a month
= 0.6 × 30 = 18 units

Question 19.
The figure shows two heating coils connected in an electric circuit.

a) Find the current through each coil.
b) If current flows through the circuit for 5 minutes, which coil gets heated more? Find the heat
developed in that coil.
Answer:
a) V = 230 V, R = 115 Ω
Current through 115 Ω coil
I = \(\frac{V}{R}\) = \(\frac{230}{115}\) = 2 A
V = 230 V, R = 46 Ω
Current through 46 Ω coil
I = \(\frac{V}{R}\) = \(\frac{230}{46}\) = 5 A

b) The coil having 46 Ω. resistance gets heated up more (Even though V (voltage), t (time of current flow) is same in both coils, 46 Ω coil has less resistance, So more current flows through it and more heat energy is produced)
Heat energy produced in 46 Ω coil.

Question 20.
2 A current is drawn by a heating coil when 230 V potential difference is applied.
a) Which material is commonly used to make a heating coil?
b) What is the quantity of charge that flows through this coil in 5 minutes?
c) What is the resistance of the coil?
a) Find the current through each coil.
Answer:
a) Nichrome is used to the bating coil.

b) Charge (Q) = I × t = 2 × 300 = 600 C

c) Resistance (R) = \(\frac{V}{R}\) = \(\frac{230}{2}\) = 115 Ω

The comprehensive approach in SCERT Kerala Syllabus 10th Standard Physics Textbook Solutions and Class 10 Physics Chapter 6 Electromagnetic Induction in Daily Life Notes Questions and Answers English Medium ensure conceptual clarity.

SSLC Physics Chapter 6 Notes Questions and Answers Pdf Electromagnetic Induction in Daily Life

SCERT Class 10 Physics Chapter 6 Electromagnetic Induction in Daily Life Notes Pdf

SSLC Physics Chapter 6 Questions and Answers – Let’s Assess

Question 1.
Choose the correct answer from the brackets.
a) What is the working principle of a generator?
(motor principle, mutual induction, electro-magnetic induction, all of these)
b) What type of electricity is generated in the armature of a DC generator?
(AC, DC, current at constant voltage, none of these)
c) At what voltage is electricity generated in power stations in India?
(11 kV, 11 V, 110 V, 230 V),
d) What is the voltage of electricity supplied for household use in our state?
(230 V, 230 kV, 11 kV, 11 V)
Answer:
a) Working principle of a generator: Electromagnetic induction.
b) Type of electricity in DC generator armature: AC
c) Voltage generated in power stations in India: 11 kV
d) Voltage supplied for household use: 230 V

Question 2.
Observe the figure.

a) If the magnet is rapidly moved into the coil, what will you observe in the galvanometer?
b) What will be the nature of the current obtained if the magnet is rapidly moved up and down inside the coil?
c) What are the ways to increase the current induced in the coil?
d) By what name is this phenomenon known as?
Answer:
a) The galvanometer needle will deflect.

b) The current will be alternating (changes direction).

c)

• Moving the magnet faster
• Using a stronger magnet
• Using more turns of wire in the coil

d) This phenomenon is called electromagnetic 6. induction.

Question 3.
Classify the following that operate on AC and that on DC.
Torch, grinder, microwave oven, emergency lamp, calculator
Answer:

Operates on AC	Operates on DC
Grinder	Torch
Microwave oven	Emergency lamp
	Calculator

Question 4.
Observe the figure.

a) Which device is shown in the figure?
b) Name the parts marked X, Y, and Z.
c) Which type of current is obtained in the external circuit?
d) What is its working principle?
Answer:
a) AC Generator
b) X-Field magnet Y – Armature
Z -Slip ring

c) Alternating current (AC)

d) The working principle of AC Generator is Electromagnetic induction.

Question 5.
What are the problems faced while transmitting electricity over long distances using conducting wires? How can these problems be solved?
Answer:
Problems:

• Energy loss as heat due to the large current in wires
• Energy loss as heat due to the resistance in wires.
• Voltage drop over a long distances.

Solutions:

• Use high voltage to reduce the current.
• Use thick wires with low resistance.
• Use transformers to step up voltage for transmission and step down for distribution purpose.

Question 6.
Observe the figure.

a) Which line is the red coloured wire?
b) To which part of the appliance is the green coloured wire connected ?
c) What is the advantage of connecting a three pin plug to appliances?
Answer:
a) Phase

b) Earth (to ground) connection of the appliances.

c) Advantages of connecting a three-pin plug to appliances:

1. Prevents electric shock by using the earth wire.
2. Protects appliances from damage due to extra current.
3. Ensures proper current flow through live and neutral wires.
4. Works with MCB or fuse to stop faults safely.

Question 7.
Write down any 4 precautions to avoid electric shock.
Answer:

1. Do not handle electrical appliances or operate switches with wet hands.
2. Plug into or unplug from a socket only after turning off the switch.
3. Do not operate high power appliances in a normal socket.
4. Do not touch the inside of a cable TV adapter. Ensure that the adapter has an insulator cover.
5. Do not fly kites near power lines.

Question 8.
A transformer has 600 turns in its primary and 1800 turns in its secondary. 450 V is obtained across its secondary. Then,
a) Which type of transformer is this?
b) What will be the voltage supplied across the primary?
Answer:
a) Step-up transformer
b) Step-up transformer
\(\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{~V}_{\mathrm{P}}}\) = \(\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{~N}_{\mathrm{p}}}\), so
V_P = \(\frac{\mathrm{V}_{\mathrm{S}} \times \mathrm{N}_{\mathrm{P}}}{\mathrm{~N}_{\mathrm{S}}}\) = \(\frac{450 \times 600}{1800}\) = 150 V
Voltage across primary = 150 V

Question 9.
When 0.1 A current is supplied to the primary of a transformer, 1 A current is obtained in the secondary.
a) Which type of transformer is this?
b) If 1000 V is supplied across the primary of this transformer, what will be its power?
c) What will be the power in the secondary?
d) What will be the voltage induced across the secondary?
Answer:
a) Type of transformer: Step-down transformer (secondary current > primary current)
b) Power in primary: P = V × I = 1000 × 0.1 = 100 W

c) Power in secondary: Approximately 100 W (In a transformer with no power loss (Ideal transformer), the power in the primary is equal to the power in the secondary.)

d) Voltage across secondary:
V_S = \(\frac{\mathrm{P}}{\mathrm{I}_{\mathrm{s}}}\) = \(\frac{1}{2}\)= 100 V

Question 10.
a) In a household electric circuit, to which device does the line coming from the electric pole reach first?
b) What is the key feature of the main switch?
c) What is the necessity of ELCB in a domestic electric circuit?
Answer:
a) The line from the electric pole reaches the watt- hour meter first.

b) Key feature of the main switch: It can cut off electricity to the whole house.

c) Necessity of ELCB: It protects people from electric shocks by detecting leakage current and disconnecting the supply.

Physics Class 10 Chapter 6 Notes Kerala Syllabus Electromagnetic Induction in Daily Life

Question 1.
What happens when a conductor is moved in a magnectic field?
Answer:
An electric current is produced.

Question 2.
What happens when a magnetic field is moved near a stationary conductor?
Answer:
An electric current is produced.

Question 3.
Based on the table answer the questions below.
a. What happens to the magnetic field lines associated with the conductor when the conductor and the magnet are in relative motion?
(a change in flux / no change in flux)
Answer:
Undergo a change in flux

b. Which are the situations where galvanometer needle deflected?
(when the magnetic flux associated with the conductor is changed / when there is no change in the magnetic flux associated with the conductor)
Answer:
The galvanometer needle deflected when the magnetic flux associated with the conductor is changed.

c. Why did the galvanometer needle deflect?
Answer:
Galvanometer needle deflects due to flow of electricity.

d. When did the direction of deflection of the galvanometer needle change?
Answer:
The direction of deflection of the galvanometer needle changed when the direction of motion of the magnet or solenoid was reversed, i.e., when the magnetic flux changed in the opposite way.

e. Why did the direction of deflection of the galvanometer needle change?
Answer:
The direction of deflection of the galvanometer needle changed because whenever the magnet or solenoid is moved the other way, the magnetic field through the coil changes in the opposite direction, so the needle moves the other way too.

Whenever the magnetic flux linked with a closed circuit changes, an emf is induced in the circuit. This phenomenon is electromagnetic induction.

The phenomenon of inducing an emf across a conductor due to a change in the magnetic flux linked with the conductor is electromagnetic induction.

• It is evident that when the direction of the change in magnetic flux is reversed, the direction of the induced emf also reverses.
• The emf (electromotive force) developed due to electromagnetic induction is the induced emf and the current thus produced is the induced current.
• It was Michael Faraday who discovered experimentally that electricity can be produced using a conductor and a magnetic field.

Repeating the previous experiment by making the following changes:

Experiment	Deflection of the galvanometer needle	Induced emf
	Increase / Decrease	Increase / Decrease
Using strong magnets	Increases	Increases
Using weak magnets	Decrease	Decrease
Number of turns of solenoid increased	Increases	Increases
Number of turns of solenoid decreased	Decrease	Decrease
Magnet/ Solenoid moves with greater speed	Increases	Increases
Magnet/ Solenoid moves with lesser speed	Decrease	Decrease

The induced current resulting from the induced emf caused the deflection of the galvanometer needle.

Activity	Deflection of the galvanometer needle (with respect to the previous experiment)
When the number of turns per unit length of a coil is increased	More
When the strength of the magnet is increased	More
When the speed is increased	More

Question 4.
Analysing table 6.2. write down methods to increase the emf and current?
Answer:
To increase the emf and current we should,

• Increase the number of turns per unit length of a coil
• Increase the strength of the magnet
• Increase the speed of movement of the magnet or the solenoid

Question 5.
Is there any difference between the current induced as a result of electromagnetic induction and the current obtained from a battery?
Answer:
Yes

Question 6.
The deflection of the galvanometer
(deflects to both sides / deflects in only one direction)
Answer:
The galvanometer needle deflects only in one direction. The direction of the flow of current does not change here.

Current that flows only in one direction is Direct Current (DC).

Question 7.
What change do you observe in the deflection of the galvanometer needle?
Answer:
The galvanometer needle deflects in both direction.

Question 8.
Is the induced current in the same direction?
Answer:
No, the induced current is not in the same direction, it changes direction.

Current that continuously changes direction at regular intervals of time is Alternating Current (AC).

Devices that operate on AC and DC currents are:

Operates on DC	Operates on AC
• Mobile phone • TV Remote • Laptop • Torch (flashlight)	• Mixie • Ceiling fan • Refrigerator • Air conditioner

Question 9.
Is it in the same manner that electricity is generated on a large scale? What arrangements are required to generate electricity on a large scale?
Answer:

• Powerful magnet
• Mechanism for movement
• Large coils of wire
• Energy source
• Transmission system
• Generator

Observe the figure 6.4.

Question 10.
What is its use?
Answer:
It is a device that produces electricity continuously by the movement of a magnet or coiled conductor, within a magnetic field.

Question 11.
What is the energy transformation that occurs here?

Answer:
Mechanical energy → Electrical energy

A generator is a device that converts mechanical energy into electric energy based on the principle of electromagnetic induction. Generators are of two types: • AC generator • DC generator

When a magnet is move towards an insulated coil or bringing an insulated coil towards a magnet, the work done which is the mechanical energy, which is converted into electrical energy.

Parts	AC Generator	DC Generator
NS	Field magnet	Field magnet
ABCD	Armature (coil)	Armature (coil)
B1, B2	Brushes	Brushes
R1, R2	Slip rings	Split ring

Question 12.
What happens to the magnetic field lines linked with the armature coil, when the armature of the generator rotates?
Answer:
When the armature rotates, the magnetic field lines linked with the coil changes continuously.

Question 13.
Will current be induced in the armature when there is change in flux?
Answer:
Yes, current will be induced in the armature when there is a change in magnetic flux.

Question 14.
Will there be a change in the direction of the current induced in the armature?
Answer:
Yes, the direction of the induced current in the armature changes after every half rotation.

Question 15.
Will the current induced in the armature be AC or DC?
Answer:
The current induced in the armature will be AC (Alternating current).

Question 16.
Which mechanism transfers the current induced in the armature to the external circuit?
AC generator: Slip rings, Brushes
DC generator: __________’ ____________
Answer:
Split rings, Brushes

Question 17.
What is the function of split rings in a DC generator?
Answer:
The function of split rings in a DC generator is to reverse the connection of the coil with the external circuit every half rotation(that is after every half rotation direction of current in armature coil reverses), so that the current in the external circuit flows in one direction only (making it DC).

Question 18.
Write down the similarities and differences between AC and DC generators.
Answer:
Similarities

• Works on the principle of electromagnetic induction
• Has a field magnet and an armature coil
• Uses brush to transfer current
• AC is induced in the armature of both generators.

Differences

AC Generator	DC Generator
• Current in the external circuit is alternating current • Uses slip rings	• Current in the external circuit is direct current • Uses split rings

Question 19.
What type of electricity will be obtained in the external circuit if the armature of a DC generator
is kept stationary and the field magnet is rotated? Why?
Answer:
If the field magnet is rotated and the armature is stationary, AC will be obtained, because the magnetic flux linked with the armature changes continuously.

Graphical Representation

Graphical representation of current obtained from AC generator, DC generator and cell

Question 20.
Is the electricity that reaches our homes produced by the same type of generators used in shops and other establishments?
Answer:
Yes, both use AC generators based on electromagnetic induction.

It is not practical to use small generators to produce the electricity needed for homes and big buildings. So, special centres called power stations are setup to generate electricity on a large scale. Power stations are named according to the type of energy used to run their generators.

The majority of the world’s electricity needs are met by the three types of power stations listed below.

Question 21.
By what name are the centres that produce electricity on a large scale using large generators known as?
Answer:
They are called Power stations.

Question 22.
Write down the energy transformation that takes place in each power station.

Answer:

Thermal power station	Heat energy → Electric energy
Nuclear power station	Nuclear energy → Electric energy
Hydroelectric power station	Potential energy (water) → Electric energy

There are power stations that generate electricity using wave energy, wind energy, solar energy, geothermal energy and tidal energy.

Question 23.
Find more information about such power stations, including their advantages and limitations, and present a seminar in class.
Answer:
An example is given below
Topic: Power Stations-advantages and limitations

Main points
There are different types of power stations used to generate electricity. Thermal power plants bum coal and produce a lot of energy but also cause air pollution. Hydro power plants use water from dams and are clean, but they need large rivers and big dams to work. Nuclear power plants create a large amount of electricity, but their waste is very dangerous. Solar power plants use sunlight, which is clean, but they only work during the day. Wind power plants use the wind to make electricity and are also clean, but only work when there is wind. Each type of power station has its own benefits and drawbacks.

Conclusion
Clean energy sources like solar, wind, and hydro are better for the environment. They do not pollute the air or produce harmful waste. To protect the Earth, we should use more renewable energy in the future.

Question 24.
Write down the similarities and differences between nuclear power station and thermal power station.
Answer:
Similarities

• Both produce electricity on a large scale.
• Both use steam to rotate turbines which drive generators.
• Both convert heat energy into electrical energy.
  Differences
• Thermal power station:
  • Heat is produced by burning coal, oil, or gas.
  • Causes more air pollution due to smoke and carbon dioxide.
  • Cheaper to build but costlier to run.
• Nuclear power station:
  • Heat is produced by nuclear fission of uranium or other fuels.
  • Produces radioactive waste instead of smoke.
  • Costly to build but cheaper to operate once set up.

Question 25.
Write down in order the energy transformations that occur in a hydroelectric power station.
Answer:
Potential energy of water → Kinetic energy → Mechanical energy (Turbine) → Electrical energy (Generator)

Question 26.
How does the electricity generated in power stations reach our houses?
Answer:
Electricity from power stations is sent through high- voltage transmission lines to substations, where the voltage is reduced. From there it travels through distribution lines and finally enters our homes at a safe voltage for use.

Observe the figures

The AC generated in power stations reach our homes through conducting wires.

Question 27.
Is there a possibility of energy loss when electricity is transmitted over long distances from power stations through conducting wires? Answer based on Joule’s Law.
Answer:
Yes, according to Joule’s Law, when current flows through wires, some electrical energy is lost as heat (H = I² Rt). Energy loss occurs during long distance transmission.

Question 28.
The energy loss primarily occurs in the form of heat. How can this be minimised?
Answer:
The energy loss can be minimised by:

1. Increasing the voltage of transmission (reduces current, I, so I² R loss decreases).
2. Using thicker wires or materials with low resistance.
3. Keeping transmission lines short wherever possible.

Question 29.
What are the three factors that influence the quantity of heat produced when current flows through a conductor?
Answer:

• Current (I)
• Time (t)
• Resistance of the conductor (R)

If the values of these factors are reduced, the quantity of heat produced can be reduced and thus energy loss can be minimised.

• It is not practical to reduce time.
• To reduce resistance, use materials with low resistivity.

Question 30.
Examine table 6.8 and answer the questions below.

Metal	Resistivity (Ωm)
Silver	1.59 × 10-8
Copper	1.68 × 10-8
Gold	2.44 × 10-8
Aluminium	2.65 × 10-8
Tungsten	5.60 × 10-8
Iron	9.71 × 10-8

a. Identify the metals with low resistivity?
Answer:

• Silver → 1.59 × 10^-8 Ωm
• Copper → 1.68 × 10^-8 Ωm
• Gold → 2.44 × 10^-8 Ωm
• Aluminium → 2.65 × 10^-8 Ωm

b. From these, find out the one with lowest cost and write down the most suitable metal?
Answer:
Among these lowest cost metal is Copper.

Reducing resistance beyond a certain limit is not practical. Current (I) is the third factor that needs to be reduced to minimise energy loss.

Question 31.
Can current be reduced?
Answer:
Yes.
We can reduce current by using higher voltage for the same power, higher voltage means less current, which lowers energy loss in the wires.

Question 32.
According to the formula P = VI, if the current (I) is reduced, what change occurs in the power (P)? (increases / decreases)
Answer:
Power (P) decreases

Question 33.
How can we reduce current without decreasing power?
(increase the voltage/ decrease the voltage)
Answer:
Increase the voltage

Question 34.
Write down the methods to minimise energy loss while transmitting electricity over long distances through conducting wires.
Answer:

1. Use suitable metal wires with low resistivity.
2. Increase voltage and decrease current without changing power.
3. Use thicker wires
4. Keep transmission lines short and direct where possible
5. Use efficient transformers
6. Regular maintenance of wires and connections

Question 35.
Which is the device that helps to increase voltage without change in power?
Answer:
Transformer

Question 36.
In which situations does the galvanometer needle deflect?
Answer:
The galvanometer needle deflects only when the switch is being turned on or turned off.

Question 37.
Is there a magnetic field around the coil when the switch is in the off position?
Answer:
No, a magnetic field is only there when electricity is flowing through the coil. When the switch is off, no electricity flows, so there is no magnetic field.

Question 38.
What about while turning on the switch?
Answer:
Yes, when you turn the switch on, a magnetic field is being created around the coil.

Question 39.
While turning on the switch, does the magnetic field linked with the second coil change?
Answer:
Yes, when the switch is turned on, the magnetic field linked with the second coil changes.

Question 40.
If so, will current be induced in the second coil due to electromagnetic induction?
Answer:
Yes, a current will be induced in the second coil due to electromagnetic induction

Question 41.
What methods can be adopted to induce current continuously in the second coil?
Answer:

1. Move a magnet or the second coil
2. Use AC current
3. Rotate the second coil

Conclusions:
The galvanometer deflects (momentarily) for the time when switch is turned On or off because there will be a change in magnetic field around the coil. There is no deflection when the switch remains in ON or off position. By moving a magnet or the second coil or by using AC current we can induce current continuously in the second coil.

Question 42.
To which coil of the transformer is AC supplied?
Answer:
Primary coil

Question 43.
In which coil is the AC induced?
Answer:
Secondary coil

Question 44.
What are the structural differences between stepup transformer and stepdown transformer?

Answer:

Stepup transformer	Stepdown transformer
• Thick wires are used in the primary	• Thin wires are used in the primary
• The primary has lesser number of turns than the secondary	• The secondary has less number of turns than the primary

Question 45.
How does a transformer change voltage?
Answer:
A transformer changes voltage by mutual induction.
In an ideal transformer, the induced emf per turn is equal in both coils.

Number of turns of coil in the primary (NP)	Voltage applied across primary coil (VP)	Voltage induced in one turn (e)	Number of turns of coils in the secondary (NS)	Induced voltage in the secondary (VS = NS × e )
100	100 V	1 V	100	100 × 1 V = 100 V
100	100 V	1 V	200	200 × 1 V = 200 V
200	100 V	0.5V	400	200 V
200	400 V	2 V	100	200 V
200	400 V	2 V	200	400 V

Question 46.
Analysing table 6.11, what conclusions can you arrive at?
Answer:
The ratio of the number of turns in the secondary to the primary in the transformer (\(\frac{N_s}{N_p}\)) will be the same as the ratio of the voltages across the secondary to the primary (\(\frac{V_s}{V_P}\)).
That is = \(\frac{V_s}{V_P}\) = \(\frac{N_s}{N_p}\)

Question 47.
A transformer with no power loss operating at 240 V AC supplies 12 V to an electric bell connected to it. Calculate the number of turns in the secondary, if the primary coil of the transformer has 4000 turns.
Answer:

Question 48.
A transformer with no power loss operating at an input voltage of 230 V has 120 turns in the secondary and 1200 turns in the primary. What is the output voltage of this transformer?
Answer:

In a transformer with no power loss (Ideal transformer), the power in the primary is equal to the power in the secondary.

Question 49.
When a transformer is used to change the AC voltage, does the current change?
Answer:
Yes. When a transformer changes the AC voltage, the current changes in the opposite way—if voltage increases current decreases, and if voltage decreases current increases.

An appliance connected to a stepdown transformer operating at 200 V consumes 1000 W of power. Suppose 100 V is obtained across its secondary.

Primary Power = Secondary Power
PP = 1000W	PS = 1000 W
VP × IP = 200 V × 5 A = 1000 W	VS × IS = 100 V × 10 A = 1000 W

Question 50.
Observe table 6.12 and write down the change that occurs in the current when the voltage decreases.
Answer:
When the voltage decreases, current increases.

Question 51.
What happens to the current, when the voltage increases?
Answer:
When the voltage increases, current decreases.

V_P I_P = V_S × I_S or = \(\frac{\mathbf{I}_{\mathbf{P}}}{\mathbf{I}_{\mathbf{S}}}\) = \(\frac{V_s}{V_p}\)
The secondary voltage of a stepup transformer will be higher than its primary voltage and the secondary current will be lower than its primary. The secondary voltage of a stepdown transformer will be lower than its primary voltage and secondary current will be higher.

Question 52.
In a transformer with no power loss (Ideal transformer), the primary has 3000 turns and the secondary has 150 turns. The primary voltage is 120 V and the current is 0.1 A. Calculate the secondary voltage and current.
Answer

Secondary voltage V_S = 6 V
Secondary current I_S = 2 A

Question 53.
Question 53.
Observe figure 6.12 and answer the questions below.
a) Discuss how electricity generated in a power station reaches houses and other establishments?
Answer:
Electricity is produced in a power station at about 11 kV.

A stepup transformer raises it to very high voltage (like 220 kV) for long-distance transmission through big power lines.

A stepdown transformer reduces it to 110 kV at substation and to 11 kV at secondary substation.

Finally smaller transformers reduce it further to 230 V and send it through local lines to houses and other buildings.

b. At what voltage is electricity generated in a power station?
Answer:
Electricity is generated at about 11 kV in a power station.

c. Which type of transformer is used in a power station?
Answer:
A stepup transformer is used in a power station to increase the generated voltage for long-distance transmission.

d. 11 kV electricity is generated in the power station.
To what voltage is it stepped up? (220 kV/ 110kV)
Answer:
It is stepped up to about 220 kV for long-distance transmission.

e. Where does the electricity reach after travelling through high voltage transmission lines?
Answer:
After travelling through high-voltage transmission lines, the electricity reaches substation .The voltage is reduced using step-down transformers in the secondary substation (1 lkV). This voltage is then reduced to the voltage required for distribution (230 V) using a distribution transformer.

f. Where are stepdown transformers used?
Answer:
The voltage from the transmission lines needs to be changed to a safe and usable voltage (230 V). To reduce high voltage (such as 11 kV, 110 kV or 220kV), stepdown transformers are used in substations (secondary substations, distribution substations) near homes and other establishments.

g. To what voltage is 11 kV AC reduced in a distribution transformer?
Answer:
230 V

h. What is the voltage of electricity supplied to houses?
Answer:
230 V

Question 54.
Observe figure 6.12, analyse the answers to the questions and prepare a short note on power transmission and distribution.
Answer:
1. Power Transmission:

• It is the process of transporting the electric power from power stations to substations near populated areas.
• Transmission uses the high voltage (e.g., 110 kV, 220 kV) to reduce current and minimize energy loss over long distances.
• Stepup transformers are used at the power stations to increase the voltage before transmission.

2. Power Distribution:

• It is the process of delivering electric power from substations to consumer’s premises (homes, offices, factories).
• Stepdown transformers reduce high voltage to safe, usable levels (230 V single-phase, 400 V three-phase).
• Ensures that the electricity has reached consumers efficiently and safely.

Question 55.
Explain the role of transformers in power transmission.
Answer:
1. Stepup Transformers (at Power Station):

• Increase the voltage from 11 kV (generated in the power station) to a high transmission voltages
  (e.g., 110 kV, 220 kV).
• Purpose: Higher voltage → lower current → reduces energy loss (heat) in transmission lines (P = I²R).

2. Step-down Transformers (at Substations):

• Reduce high transmission voltage to lower voltages that is suitable for consumer’s use (e.g., 230 V single-phase, 400 V three-phase).
• Purpose: Makes electricity safe and usable in house, offices, and industries.

The role of transformers in enabling efficient long-distance power transmission is very big. They can prevent excessive energy loss. They will make electricity safe for users.

Question 56.
What is the necessity of increasing the voltage of electricity generated at 11 kV in a power station?
Answer:
Electricity generated at 11 kV is stepped up to high voltage to reduce current and minimize energy loss during long-distance transmission.

Question 57.
What is the voltage used for domestic distribution?
Answer:
The voltage used for domestic distribution is 230 V AC (single-phase).

Question 58.
How does electricity reach the electrical appliances we use in our homes?
Answer:
Electricity reaches our home appliances through distribution wires after being stepped down to 230 V by a distribution transformer.

Question 59.
Observe the diagram of a household electric circuit and answer the questions below.

a. Where is the watt hour meter connected?
Answer:
It is placed at the beginning of the household electric circuit. The watt-hour meter is connected in series with the live (phase) wire and also to the neutral of the household supply.

b. How many wires reach the watt hour meter?
Answer:
Two wires

c. Which device is connected between the watt hour meter and the main switch?
Answer:
Main fuse

d. How is it connected? (in series / in parallel)
Answer:
In series

e. Where is the main switch placed?
Answer:
Main switch is placed right after the main fuse.

Main Switch
The main switch is a device used connect or disconnect the phase (live wire) and neutral line reaching the house hold circuit from the electric pole. The position of the main switch is right after the main fuse. Main switch is functioning as a double switch.

In addition, an ELCB (Earth Leakage Circuit Breaker) is connected to ensure more safety for the circuit. From there, the lines reach the MCB (Miniature Circuit Breaker) board and are distributed as branches to each part of the house.

• Appliances are connected across these lines in parallel.
• An appliance and the switch that controls it is connected in series to the phase line.
• When high power appliances are included in the circuit, they should be connected to the earth wire for better safety.
• For high power appliances, separate branch lines should be used with thick wires.
• Power plugs must be used to connect high power appliances.
• Red coloured wires are used commonly for the phase line, black for the neutral line and green for the earth line.

Question 60.
What are the situations that can cause excess current in a circuit?
Answer:
A short circuit occurs when the positive and negative terminals of a battery, or two wires in an AC main, come into contact with negligible resistance. This results in excessive current flow, which can lead to various hazards like fire.

Excess current and subsequent dangers can also arise in a circuit when a device with power more than the permissible limit, or many devices that consume excessive power, are connected. This type of excessive current flow in a circuit is overloading.

Question 61.
Why is it that a 2000 W induction cooker is not usually connected to a normal plug?
Answer:
A 2000 W induction cooker needs high current which can cause overload in a normal plug and cause heating that might damage the induction cooker and cause fire too.
A normal plug and its connected wires are designed to withstand a current of 5 A.

Question 62.
Will the current in a 2000 W induction cooker operating at 230 V be more or less than 5 A?
Answer:
According to the equation P = V × I
P = 2000 W
V = 230 V
Amperage = \(\frac{\text { Wattage }}{\text { Voltage }}\) = \(\frac{2000 \mathrm{~W}}{230 \mathrm{~V}}\) = 8.7 A

This means that more current than permissible limit will flow through the circuit. Connecting an appliance with higher power than permitted in a circuit in this manner is overloading. Therefore, connecting a high power appliance from a normal plug or using an extension cord to connect multiple devices, or using a multipin plug to connect more devices, will cause overloading.

Question 63.
Draw a circuit diagram for constructing a branch circuit with two bulbs, one three pin socket, one two pin socket, one fuse or MCB and necessary switches.
Answer:

Question 64.
What are the measures to be taken to protect household electrical appliances?
Answer:
Simple measures to protect household electrical appliances:

1. Use proper fuses or circuit breakers to prevent overloading.
2. Avoid wet hands or water contact while operating appliances.
3. Do not overload sockets or extension boards.
4. Switch off appliances when not in use.
5. Use voltage stabilizers for sensitive appliances.
6. Regularly check wiring and plugs for damage.
7. Keep appliances clean and dust-free.

Measures to ensure safety in domestic electricity distribution:
Safety fuse, ELCB (Earth Leakage Circuit Breaker), MCB (Miniature Circuit Breaker), three pin plug and earthing are commonly used safety measures in household electrical circuits.

1. Safety Fuse
Overloading, short circuits, lightning, etc., can cause excessive current flow through a circuit. A safety fuse is a device to protect living beings and equipment from the dangers caused by this.
It works based on the heating effect of electricity. The important part of a safety fuse is the fuse wire. Generally, alloys (eg: an alloy of tin and lead) are used to make fuse wire. Fuse wire has a relatively low melting point.

For each circuit, use a fuse wire that is appropriate for the current flowing through the circuit.

Question 65.
Which are the situations that could lead to excessive current causing the fuse wire to melt?
Answer:
Some situations that could lead to excessive current causing the fuse wire to melt are

• Connecting too many appliances at once
• Wires touching by mistake (short circuit)
• Bad wiring or damaged appliance
• Sudden high voltage

Question 66.
How is the fuse wire connected in the circuit?
(series / parallel)
Answer:
Series

2. MCB (Miniature Circuit Breaker)

MCB is a device used in branch circuits instead of safety fuse. When there is excessive current in a circuit due to short circuit or overload, the MCB automatically operates and disconnects the circuit ie., it trips. After resolving the circuit problem, the MCB switch can be turned on to restore the circuitto its original state. MCB works by utilising the magnetic effect and heating effect of electricity.

3. ELCB (Earth Leakage Circuit Breaker)

ELCB helps to disconnect the circuit automatically if there is current leak due to insulation failure or other reasons. This prevents electric shock to those who come in contact with the electric circuit or device. In household electric circuits, branch circuits start after the ELCB. Usually, one ELCB is sufficient for a household electric circuit. Subsequently, each branch starts with MCB.

4. Three Pin Plug and Earthing
The three pin plug is another device to ensure greater safety in household electric circuits.

Question 67.
Observe the figure and answer the questions below
a. Which line does the letter E indicate?
Answer:
Earth line

b. Where should this line be connected to the device?
Answer:
The earth line should be connected to the metal body of the device.

c. Where will the current flow, if a short circuit occurs in the device?
Answer:
The current will go safely to the earth through the earth wire.

d. Will this cause excess current flow?
Answer:
Yes. Excess current will flow to the earth if there is a fault.

e. If so, what will happen to the MCB/ safety fuse?
Answer:
The MCB or safety fuse will trip to stop the excess current and protect the device.

f. Observe the figure and analyse the answers. Then prepare a note on how the three pin plug ensures safety.
Answer:
In a three pin plug ,one of the three pins are connected to live, one to neutral and one to earth wire.

1. Live Wire: Carries current to the device.
2. Neutral Wire: Returns current to the supply.
3. Earth Wire: Connected to the metal body of the device. If a fault occurs, excess current flows through the earth wire.

The earth pin in a three-pin plug is longer and thicker, making it the first to make contact and the last to break when inserted into or pulled out of a socket. Its middle slit ensures a tight fit. The earth pin E makes contact with the earth line. The appliance’s body is attached to this pin. In the instance when the body makes any kind of electrical contact, the earth wire conducts electricity to the earth. Current is increased as it travels to the ground through a low resistance circuit. The circuit breaks as a result of the fuse wire’s increased generation of heat. This ensures both the instrument’s and the user’s safety.

RCCB (Residual Current Circuit Breaker)

Instead of ELCB, RCCB (Residual Current Circuit Breaker) is now used to ensure greater safety. It identifies current leaks by detecting, the difference between phase current and neutral current and then disconnects the circuit.

Question 68.
What will we do if we get an electric shock due to failure in safety system or carelessness?
Answer:
If someone gets an electric shock, switch off the main power immediately and take the person to a safe place. Then call for medical help.

Question 69.
What are the circumstances in which electric shock occur?
Answer:
Electric shock can occur when:

1. We touch a live wire.
2. There is faulty or damaged wiring.
3. Appliances are not earthed properly.
4. Water is near electricity.
5. Safety devices fail.

Question 70.
What dangers can people face due to electric shock?
Answer:
People can face these dangers due to electric shock:

1. Burns on the skin or inside the body.
2. Heart problems or irregular heartbeat.
3. Muscle injury or paralysis.
4. Death in severe cases.
5. Falls or accidents if shocked while standing on a height or near machines.

Safety must be ensured while working with electric circuits.

• If there is an electric shock, turn off the main switch immediately. Separate the person who gets an electric shock from the electric contact using an insulator.
• Under any circumstances do not touch with bare hands a person who received an electric shock.

Let’s look at the precautions to be taken to avoid electric shock. Precautions

• Do not handle electrical appliances or operate switches with wet hands.
• Plug into or unplug from a socket only after turning off the switch.
• Do not operate high power appliances in a normal socket.
• Do not touch the inside of a cable TV adapter. Ensure that the adapter has an insulator cover.
• Do not fly kites near power lines.
• Do not use iron/aluminium ladders, poles, etc., near electric lines.
• While carrying out repairs on household electric circuits, ensure that the main switch is turned off.
• During lightning, do not perform activities that involve contact with electric circuits (there is a possibility of excessive current in the circuit).
• Unplug appliances from sockets if there is a possibility of lightning.
• During rain and wind, transmission lines may touch the ground, creating the risk of accidents. If water enters houses (due to floods or other reasons), disconnect the power supply. After the water recedes, restore power only after the switch boards and the main switch are completely dry.

First Aid for Electric Shock
Provide first aid only after disconnecting the person who gets the shock from the electric wire.

• Rub the body to increase blood circulation and raise body temperature.
• Administer artificial respiration.
• Rub the muscles to restore them to their normal state.
• Start first aid to restart the heart perform chest compressions rhythmically and forcefully (Cardio Pulmonary Resuscitation). Take the person to the nearest hospital as soon as possible.

Std 10 Physics Chapter 6 Notes – Extended Activities

Question 1.
Open a three pin .top, understand how the wires are connected, and then connect wires in another three pin top in the same manner.
Answer:
Aim: To learn how wires are connected in a three-pin plug and safely replicate the connections.
Materials Required:
Three-pin plug (one for observing, one for connecting)
Screwdriver
Wires of an appliance

Procedure:

1. Open a three-pin plug carefully.
2. Observe how the live (red), neutral (black), and earth (green) wires are connected.
3. Note the positions of each wire: live to L, neutral to N, earth to E.
4. Connect the wires in the second plug in the same way.
5. Tighten screws and close the plug safely.

Conclusion: The experiment helps understand safe wiring of a three-pin plug and the role of each wire in protecting users and appliances.

Question 2.
Which are the types of power stations used to generate electricity in countries worldwide, including India?
Calculate the percentage of electricity generated by each category in the total power production and prepare a table in the descending order of their percentage of production. ,
Answer:
An example is given below

Types of Power Stations

1. Thermal Power Stations
   Coal-based
   Gas-based
   Oil-based
2. Hydroelectric Power Stations
3. Nuclear Power Stations
4. Solar Power Plants
5. Windmills
6. Biomass Power Plants

Global Electricity Generation (mention a particular year for which study is conducted)

Source	Percentage (%)
Coal	34%
Natural Gas	22%
Hydropower	16%
Wind & Solar	13%
Nuclear	9%
Other Renewables	3%
Oil	2%
Total	100%

India’s Electricity Generation (mention a particular year for which study is conducted)

Source	Percentage (%)
Coal	73%
Hydropower	9%
Solar	8%
Wind	5%
Nuclear	3%
Biomass	2%
Gas & Oil	1%
Total	100%

(The table shows indicative values, original values may vary)
Coal remains the dominant source of electricity generation globally and in India.Renewable sources (wind, solar, hydro) are growing rapidly but still contribute a smaller share compared to fossil fuels.

Question 3.
Construct and demonstrate a branch circuit including two electric bulbs, a three pin socket, a two pin socket, a fuse or MCB, electric bulbs operated by a two way switch and switches required for that. (This activity should be done only under adult supervision).
Answer:
Hint
Aim: To construct and understand a household branch circuit with bulbs, sockets, switches, and protection devices.

Materials Required: 2 electric bulbs with holders, 3-pin socket, 2-pin socket, Fuse or MCB, Two-way switch, Single-pole switches, Connecting wires, Screwdriver, insulating tape

Procedure:

1. Connect the main supply to a fuse or MCB for protection.
2. From the MCB, connect wires to the three-pin socket and two-pin socket.
3. Connect the two bulbs in the circuit so that they can be operated by two-way and single-pole switches.
4. Use proper live (red), neutral (black), and earth (green) connections.
5. Test the circuit carefully with adult supervision. Conclusion: The activity demonstrates how a branch circuit works in homes, how switches control bulbs, how sockets are connected, and how a fuse or MCB protects against overcurrent.

This setup helps understand safe wiring and control of electricity in a household.

Electromagnetic Induction in Daily Life Class 10 Notes

Electromagnetic Induction in Daily Life Notes Pdf

• Whenever the magnetic flux linked with a closed circuit changes, an emf is induced in the circuit. This phenomenon is electromagnetic induction.
• The phenomenon of inducing an emf across a conductor due to a change in the magnetic flux linked with the conductor is electromagnetic induction.
• The emf (electromotive force) developed due to electromagnetic induction is the induced emf and the current thus produced is the induced current.
• To increase the emf and current we should,
  • Increase the number of turns per unit length of a coil
  • Increase the strength of the magnet
  • Increase the speed of movement of the magnet or the solenoid
• Current that flows only in one direction is Direct Current (DC).
• Current that continuously changes direction at regular intervals of time is Alternating Current (AC).
• A generator is a device that converts mechanical energy into electric energy based on the principle of electromagnetic induction. Generators are of two types:
  • AC generator
  • DC generator
• Steps to minimise energy loss while transmitting electricity over long distances through conducting wires.
  • Use suitable metal wires with low resistivity.
  • Increase voltage and decrease current without changing power.
  • Use thicker wires
  • Keep transmission lines short and direct where possible
  • Use efficient transformers
  • Regular maintenance of wires and connections
• Transformer is a device that works on the principle of mutual induction. Transformers change AC voltage without change in power.
  Transformers are of two types.
  1. Stepup transformer: transformers that increases the AC voltage
  2. Stepdown transformer: transformers that decreases the AC voltage
• The ratio of the number of turns in the secondary to the primary in the transformer (\(\frac{N_S}{N_P}\)) will be the same as the ratio of the voltages across the secondary to the primary (\(\frac{V_S}{V_P}\))
  That is \(\frac{V_S}{V_P}\) = \(\frac{N_S}{N_P}\)
• In a transformer with no power loss (Ideal transformer), the power in the primary is equal to the power in the secondary.
• V_P × I_P =V_S × I_S or \(\frac{I_P}{I_S}\) = \(\frac{V_S}{V_P}\)
• The secondary voltage of a stepup transformer will be higher than its primary voltage and the secondary current will be lower than its primary. The secondary voltage of a stepdown transformer will be lower than its primary voltage and secondary current will be higher.
• Measures to ensure safety in domestic electricity distribution: Safety fuse, ELCB (Earth Leakage Circuit Breaker), MCB (Miniature Circuit Breaker), three pin plug and earthing are commonly used safety measures in household electrical circuits.
• Electric shock is the impact caused by the flow of current through the body.

INTRODUCTION

One basic energy source that is necessary for modern living is electricity. Our communication systems, schools, hospitals, businesses, and houses are all powered by it. Many resources, including coal, water, wind, sunlight, and nuclear energy, can be used to produce electricity. Michael Faraday’s discovery of electromagnetic induction was one of the most significant in the history of electricity. Generators in power plants use this idea to create energy. Additionally, it is utilized in gadgets like wireless chargers, transformers, and induction cookers. Conductors, such as copper wires, carry electricity, which can be controlled with switches and circuits. Electricity must be handled and used properly to prevent accidents and shocks and also to conserve energy. This chapter explores topics like electromagnetic induction, alternating and direct current generator, transformer, power transmission and distribution, household electrification and electric shock.

Electromagnetic Induction
• Electromagnetic induction is the process of producing electricity from motion. An electric current is created in the wire when a magnet moves near a coil of wire, or when the coil moves near a magnet. This happens because of the movement of magnet changes the magnetic field around the coil, which makes the electrons in the wire start to flow.

• This idea was discovered by Michael Faraday and is one of the most important principles in science. It is used in generators to produce electricity in power stations, in transformers to change voltage, in bicycle dynamos, induction stoves, and even in wireless chargers. Without electromagnetic induction, we would not have the easy and powerful supply of electricity that we enjoy today.

Alternating current (AC), Direct Current (DC)
• Electricity flows in two main ways: Alternating Current (AC) and Direct Current (DC).

• Direct Current (DC):
In DC, the electric current flows in one constant direction. It is steady and does not change with time. Batteries, solar cells, and mobile phone chargers give DC.
DC is used in gadgets and electronics

• Alternating Current (AC):
In AC, the electric current changes direction back and forth many times every second. This type of electricity is used in homes and industries because it is easier to send over long distances. The electricity we get from power stations is AC.
AC powers our homes, schools, and cities.

Generator
• A generator is a device that produces electricity. It works on the principle of electromagnetic induction, discovered by Michael Faraday. When a coil of wire is made to rotate inside a magnetic field (or when a magnet rotates near a coil), it creates an electric current.

• Generators are used in power stations to produce the electricity that we are using at our home, in schools, and in factories. They are also found in bicycles (dynamos), cars, and backup power systems. In simple words, a generator changes the mechanical energy (movement) into electrical energy, helping to light up our world and run our machines we use in our day to day life.

Transformer
• A transformer is a device that is used to increase or decrease the voltage of electricity without changing the amount of power. It works on the principle of mutual induction. A transformer has two sets of coils which are primary and secondary coils wrapped around an iron core. When alternating current (AC) flows through the primary coil, it creates a magnetic field that induces a Voltage in secondary coil.

• Transformers have very importance with electricity. In the power stations, stepup transformers used to increase the voltage so that electricity can travel long distances with less energy loss. Near our home, stepdown transformers are used to reduce the voltage to a safe level for household use. In simple words, a transformer helps electricity to move safely and efficiently from the power plants to our homes and workplaces.

Power transmission and distribution
• Power transmission and distribution is the process of carrying the electricity from the power stations to our home, schools, and factories. After electricity is generated in a power plant, it travels to a long distances to reach people safely and efficiently.

• At first stepup transformers will increase the voltage so that the electricity can move through high-voltage transmission lines with less energy loss. Next step is that when it reaches cities or towns, stepdown transformers which lowers the voltage to safer levels. Finally the electricity can pass through distribution lines and enters our homes at the right voltage for everyday use.

Household wiring

• Household wiring is the system of electrical wires and connections that brings the electricity safely into our homes. It carries the electric current from the main supply to lights, fans, sockets, and other appliances we use in our day to day life.
• In a house wiring system, three main wires we used are:
  • Live (Phase) wire – which carries the current from the supply.
  • Neutral wire which completes the circuit back to the supply.
  • Earth wire – which provides safety by directing excess current into the ground in case of a fault.
• Important safety devices like fuses, MCBs (Miniature Circuit Breakers), and switches are included so that to protect against overload and short circuits. So we can say that household wiring is like the road network of electricity which brings power to every room safely and efficiently.

Electric Shock

• An electric shock happens when the electricity is passed through the human body. Our body can conduct the electricity because it contains water and salts. When someone touches a live wire or faulty appliance, the electric current flows through their body and then to the ground. This can cause pain, burns, muscle cramps, or even serious injury and death to the affected person.
• Electric shocks can be prevented by:
  • Using insulated wires and proper household wiring.
  • Installing earthing and fuses/MCBs for safety.
  • Avoiding wet hands while handling electrical devices.
• In simple words, an electric shock is the harmful effect of electricity on our body. By the careful use of electrical appliances we can keep us safe and protected.

There are various types of electromagnets are used for different purposes. Some of them are given below.

• Electric crane
• Electric motors
• Electric bell
• Electric rails
• Electric fan
• Loud speakers

In all these, electric energy is converted into magnetic energy. A magnet can be made using electricity by making an electromagnet by winding insulated copper wire over a soft iron core and passing electricity through the wire. . From this idea, Michael Faraday developed the principle of electromagnetic induction, which later led to the invention of the generator.

ELECTROMAGNETIC INDUCTION
A current carrying conductor placed in a magnetic field experiences a force and develops a tendency to move.

ACTIVITY
Arrange a magnet, a conductor in the shape of a solenoid and a galvanometer as shown in the figure.

Observation

a) ACTIVITY 1
Connect a 1.5 V cell, a 470 Ω resistor, a switch and a galvanometer in series. Turn on the switch.

b) ACTIVITY 2
Repeat the activity with the magnet and coil. Move the solenoid connected to the galvanometer rapidly towards and away from the magnet. Observe the galvanometer needle.

TRANSFORMER
These are different types of transformers.

ACTIVITY 1
Aim: To understand how transformers work.
Materials required: • PVC pipe -1 [12 cm long with 4 cm (1.5″) diameter]

• PVC pipe – 1 [12 cm long with a 2.5 cm (1″) diameter)
• Insulated copper wire-28 gauge (250 g)
• 9 V DC and 9 V AC source
• Galvanometer
• Soft iron

Procedure:
Wind approximately 600 turns of insulated copper wire around each PVC pipe.

Connect a 9 V DC source and a switch to the ends of the first coil. Connect a galvanometer between the ends of the second coil. Arrange them as close as possible, without touching each other.

Observation:

Activity	Galvanometer needle Deflects / does not deflects
The switch is turned on	Deflects
The switch is in the on position	Does not deflects
The switch is turned off	Deflects
The switch is in the off position	Does not deflects

ACTIVITY 2
Observation :

Activity	Galvanometer needle Deflects / does not deflects
The switch is turned on	Deflects
The switch is in the on position	Deflects
The switch is turned off	Deflects
The switch is in the off position	Does not deflect

Conclusion: The galvanometer deflects whenever the circuit is complete (i.e., when the switch is on). This is because the AC current is constantly changing, which creates a constantly changing magnetic field that induces a current in the second coil. When the switch remains in off position no current flows, so there is no magnetic field and thus no induction. (There is a momentary deflection for the time when switch is turned off).

Consider two coils kept close to each other. When the intensity or direction of current in one of them changes, the magnetic field around it changes. As a result, an emf is induced in the second coil. This phenomenon is mutual induction.

ACTIVITY 3
Repeat the activity by inserting the smaller coil inside the larger coil.

Observations: The galvanometer needle will deflect more strongly.

Conclusion: By inserting the smaller coil inside the larger one, the magnetic field from the larger coil is more concentrated within the smaller coil. This results in a greater change in magnetic flux, which in turn induces a stronger current in the second coil.

ACTIVITY 4
Repeat the activity by inserting a soft iron core inside the smaller coil.

Observation: The galvanometer needle will deflect much more strongly.

Conclusion: A soft iron core acts as a concentrator for the magnetic field lines. When you place it inside the coil, it greatly increases the strength of the magnetic field. This stronger magnetic field leads to a larger change in magnetic flux, which induces a much greater current in the second coil. This is why a soft iron core is used in devices like transformers to maximize the induced current.
Transformer is a device that works on the principle of mutual induction. Transformers change AC voltage without change in power. Transformers are of two types.
1. Stepup transformer
2. Stepdown transformer
i) Stepup transformer: transformers that increases the AC voltage
ii) Stepdown transformer: transformers that decreases the AC voltage

ELECTRIC SHOK
Electric shock is the impact caused by the flow of current through the body.

CARDIO PULMONARY RESUSCITATION (CPR)

The letters C-A-B are used to remember the sequence for performing the steps of CPR.
C-Compressions
A-Airway
B-Breathing

Compressions: Restore Blood Flow (Chest Compressions)
Compression refers to pressing firmly and quickly on the person’s chest in a specific rhythm using your hands. Compressions are the most crucial step in CPR. To perform CPR compressions, follow the steps below:

1. Lay the person flat on a firm surface.
2. Place the palm of one of your hands at the centre of the person’s chest.
3. Place your other hand on top of your first hand. Your elbows should be straight and your shoulders should be directly above your hands.
4. Press down on the chest by at least 2 inches (under no circumstances, should it exceed 2.4 inches). While compressing the chest, use not just your hands, but your body weight as well.
5. Press hard and fast at the centre of the chest. Try to perform 30 compressions in 15-20 seconds. Allow the chest to fully return to its original position after each compression.
6. If you are not trained in CPR, continue chest compressions until there are signs of movement or emergency medical help arrives. If you are trained in CPR, begin rescue breaths.

Open the Airway
After 30 chest compressions, perform the following to open the person’s airway. This action is called the Head – Tilt, Chin – Lift.

1. Place the palm of your hand on the person’s forehead.
2. Gently tilt the head backward.
3. With your other hand, gently lift the chin to open the airway.

Rescue Breathing
After opening the airway using the Head-Tilt, Chin-Lift method, do the following.

1. For mouth to mouth breathing, close the person’s nose and cover their mouth with your mouth (you can place a handkerchief with a hole in the middle, between the mouths).
2. Give the first rescue breath. This should last one second-observe if the chest rises.
3. If the chest rises, give another breath. If the chest does not rise, perform the Head- Tilt, Chin-Lift again and give a breath.

After thirty chest compressions, give two breaths. This action is considered one cycle. Repeat this cycle until there are signs of movement or emergency medical help arrives.

The comprehensive approach in SCERT Kerala Syllabus 10th Standard Physics Textbook Solutions and Class 10 Physics Chapter 5 Electric Energy: Consumption and Conservation Notes Questions and Answers English Medium ensure conceptual clarity.

SSLC Physics Chapter 5 Notes Questions and Answers Pdf Electric Energy: Consumption and Conservation

SCERT Class 10 Physics Chapter 5 Electric Energy: Consumption and Conservation Notes Pdf

SSLC Physics Chapter 5 Questions and Answers – Let’s Assess

Question 1.
An electric heater operating at 230 V draws 2 A current.
a) What is the resistance of the heater?
b) Calculate the heat produced when this heater operates for 10 minutes.
c) What is the power of this appliance?
Answer:
V = 230 V, I = 2A, t = 10 × 60 s
a) R = \(\frac{V}{I}\) = \(\frac{230}{2}\) = 115 Ω

b) H = VIt
H = 230 × 2 × 600
= 276000 J

c) H = Pt
p = \(\frac{276000}{600}\) = 460 W

Question 2.
a) A heating appliance operating at 230 V supply consumes 2 A current. What is the quantity of heat produced in five minutes?
b) What is the energy consumed by this appliance in five minutes while operating at 115 V?
Answer:
a) V = 230 V, I = 2 A, t = 5 × 60 = 300 s
H = 230 × 2 × 300
= 138000J

b) V = 1115 V
R = \(\frac{230}{2}\) = 115 Ω
I = \(\frac{V}{R}\) = \(\frac{115}{115}\) = 1A
H = 115 × 1 × 300
= 34500 J

Question 3.
In a house, a 500 W electric iron operates for one hour, two 40 W fans for 8 hours, and five 12 W LED bulbs for 10 hours. Calculate in unit the energy consumption per day in that house.
The tariff of electricity for domestic consumers in Kerala is given below (subject to change).

Answer:
Electric energy (in kilo watt hour)
= \(\frac{\text { Power in watt } x \text { time in hour }}{1000}\)
Energy consumed by electric iron (in kWh)
= \(\frac{500 \times 1}{1000}\) = 0.5 kWh = 0.5 unit
Energy consumed by fan (in kWh)
= \(\frac{2 \times 40 \times 8}{1000}\) = 0.64 kWh = 0.64 unit
Energy consumed by LED bulb (in kWh)
= \(\frac{5 \times 12 \times 10}{1000}\) = 0.6 kWh = 0.6 unit
Total energy consumption in a day = 0.5 + 0.64 + 0.6 = 1.74 kWh
Total energy consumption in a day in units =1.74 unit

Question 4.
In a house, a 600 W grinder operates for one hour, three 60 W fans for six hours, and ten 10 W LED bulbs for 10 hours daily.
a) What will be the electric energy consumed in units per month?
b) If the appliances operate in the same way, calculate the electricity bill for two months in that house.
(Use the tariff given in table 5.8)
The tariff of electricity for domestic consumers in Kerala is given below (subject to change).

Answer:
a) Energy consumed by grinder
= \(\frac{600 \times 1}{1000}\) = 0.6 kWh = 0.6 unit
Energy consumed by fans
= \(\frac{3 \times 60 \times 6}{1000}\) = 1.08 kWh = 1.08 unit
Energy consumed by bulbs
= \(\frac{10 \times 10 \times 10}{1000}\) = 1 kWh = 1 unit
Total energy consumption in a day
= 0.6 + 1.08 + 1 = 2.68 kWh
= 2.68 unit
Total energy consumption in a month (30 days) = 80.4 units

b) Total energy consumption in for two months (60 days) = 80.4 × 2 = 160.8 units
For up to 250 units of energy consumption it is known as Telescopic and above that it is non telescopic.
For 2 months the energy consumed is 160.8 units. Here the energy consumption comes under the telescopic section.

Check for the energy consumption charge for each range of unit.
For first 50 units of the total consumption ,the charge = unit × energy charge per unit = 50 × 3.35 = 167.5
For the second 50 units of the total consumption,
the charge = 50 × 4.25 = 212.5
For the third 50 units of the total consumption,
the charge = 50 × 5.35 = 267.5
For the remaining 10.8 .units the total
consumption charge = 10.8 × 7.2 = 77.76

So the total consumption charge = Rs. 725.26 The charge for the single phase (usually for household connections) for a month lies in the electricity consumption range of 50-100, is also added up with this to give the total electricity bill.

Total electricity bill = 725.26 + (85 × 2) = 725.26 + 170 = Rs. 895.26 (other charges may also add up to the bill you receive. In the given bill only energy charge per unit and single phase fixed charge is added up based on the tariff | given in table 5.8)

Question 5.
Two heaters A and B operate at 230 V. Heater A draws 2 A current and heater B draws 2.5 A current.

a) Calculate the power of the heaters in figure 5.12 (a) and (b).
b) Which heater has higher resistance?
c) If both heaters operate for 5 minutes each, which heater will produce more heat?
Answer:
a) Power, P = VI
For heater A
V = 230 V, I = 2A
P = 230 × 2 = 460 W
For heater B
V = 230 V, I = 2.5 A
P = 230 × 2.5 = 575 W

b) Resistance of heater A, R = \(\frac{V}{I}\) = \(\frac{230}{2}\) = 115 Ω
Resistance of heater B, R = \(\frac{V}{I}\) = \(\frac{230}{2.5}\) = 92 Ω
Heater A has more resistance

c) H =VIt
t = 5 × 60 = 300 s
For heater A, H = 230 × 2 × 300 = 138000 J
For heater B, H = 230 × 2.5 × 300 = 172500 J
Heater B produces more heat.
When the time of operation is same, the heater with more power produces more heat.

Question 6.
Which of the given statements is correct with regard to the heating element?

(a) Low melting point
(b) High resistivity
(c) Ability to remain in red hot state
(d) Low oxidation resistance
(e) High melting point
(f) High oxidation resistance
i. abfc
ii. abfe
iii. bdfe
iv. beef
Answer:
iv. bcef

Question 7.
Two bulbs, 230 V, 40 W and 230 V, 60 W are arranged as shown in the figures.

If 230 V is applied, which bulb will be brighter in each circuit?
Answer:
For 40 W bulb, R = \(\frac{V^2}{P}\) = \(\frac{230^2}{40}\) = 1322.5 Ω
For 60 W bulb, R = \(\frac{V^2}{P}\) = \(\frac{230^2}{60}\) = 881.7 Ω
40 W has higher resistance than 60 W.

In the parallel circuit (5.13(a)), the voltage across both bulbs is the same (equal to the source voltage, 230 V). The power dissipated by each bulb is given by P = \(\frac{V^2}{R}\). Since the voltage is the same, the bulb with lower resistance will dissipate more power and thus be brighter. Since resistance of 60 W bulb is lesser than resistance of 40 W bulb, the 60 W bulb will be brighter in the parallel circuit.

In the series circuit (5.13(b)), the current flowing through both bulbs is the same. The power dissipated by each bulb is given by P = VI = (IR)I = I²R. Since the current I is the same, the bulb with higher resistance will dissipate more power and thus be brighter. Since resistance of 40 W bulb is higher than resistance of 60 W bulb, 40 W bulb will be brighter in the series circuit.

In general, in a parallel connection, device of higher power will have lower resistance ,it will draw more current, produce more heat and glow brighter and in a series connection, device of lower power will have higher resistance, it will share more voltage and glow brighter.

Question 8.
To reduce electricity consumption in our houses, we can maximise the use of sunlight dining the day. Do you agree with this statement? Explain how.
Answer:
Yes, I agree with the statement.
I do believe that utilising sunlight during the day lowers the amount of electricity used. One free and natural light source that we may use in our home is sunlight. By opening the windows and drapes ,rooms can be made brighter without the use of artificial lighting. Our electricity costs are reduced as a result. To maximize daylight, we should choose locations for work or study next to windows. Mirrors and light-colored walls can improve the room’s ability to reflect sunlight. When we rely upon natural light, it reduces the reliability on electricity from power plants, thus contributing to pollution reduction. Using sunlight to save electricity at home is easy and environmentally beneficial.

Question 9.
In the context of the energy crisis, write down any two suggestions that can be implemented to reduce energy consumption in newly constructed houses.
Answer:

• New houses should have large windows to use sunlight and fresh air instead of lights and fans.
• Solar panels can be installed to produce electricity and lower electricity bills.
• Energy-saving appliances like LED lights and 5-star rated appliances should be used.
• Good insulation in walls and roofs can keep the house cool or warm without using much electricity.

Question 10.
What is the necessity for reducing carbon footprint?
Answer:
Both direct and indirect emissions of greenhouse gases come from people, families, businesses, events, services, and goods. The amount of greenhouse gases that are released, transformed, expressed as equivalent to the measure of carbon dioxide, is carbon footprint.

Reducing the carbon footprint is important because it helps slow down climate change by lowering greenhouse gas emissions. In turn, this reduces the occurrence of severe weather phenomena like droughts and floods. Preserving natural habitats for plants and animals and limiting pollution are two further ways in which it protects the environment. Decreased emissions lead to cleaner air and water, which enhances public health by lowering respiratory and other illnesses. By conserving limited natural resources like fossil fuels, more efficient energy use promotes sustainability. We contribute to a safer and better Earth for coming generations by reducing carbon emissions.

Physics Class 10 Chapter 5 Notes Kerala Syllabus Electric Energy: Consumption and Conservation

Question 1.
Which are the appliances in houses that operate using electricity?

Answer:

• Tube light
• Electric kettle
• Fan
• Electric iron
• Induction cooker
• Electric drill machine
• Refrigerator

Different electrical appliances are used for different purposes. These appliances convert electric energy into different forms of energy.

Question 2.
What forms of energy are produced when a mixie operates? Write them down.
Answer:

• Mechanical energy
• Sound energy
• Heat energy

The electric energy we supplied is transformed into various forms when the mixie is operated. Effect of electric current-The useful form of energy into which an appliance mainly converts electric energy is considered the effect of electric current in that appliance.

Question 3.
Observe the appliances in figure 5.1 and the appliances already listed. Based on their function, identify the effect of electric current utilised in each and complete table 5.1 given below.

Answer:

Name of the appliance	Function	Energy conversion	Effect of electric current
Electric kettle	Heats water	Electric energy → heat energy	Heating effect
Mixie	Grind or mix ingredients	Electrical energy → mechanical energy	Mechanical effect
Electric iron	To iron clothes	Electric energy → heat energy	Heating effect
Tube light	Provides light	Electric energy → Light energy	Lighting effect

Question 4.
From the table, you may have understood that electric current can produce various effects. Which of these appliances produce heat?
Answer:

• Electric iron
• Electric kettle

Question 5.
Connect a 5 cm long nichrome wire to a 9 V battery as shown in Figure 5.2.

Switch on and observe the nichrome wire.
What change do you notice?
Answer:
When current passes through the nichrome wire, heat energy is produced.

The process of production of heat, when electricity flows through a conductor is Joule heating or Ohmic heating. The part used to produce heat energy in heating appliances is the heating element.

Observe the pictures of the given electric heating appliances.

You may have noticed that the part common to all these appliances is the heating coil (heating element).
The heating coil is made of an alloy, nichrome.
(There are also heating appliances which does not make use of heated coils. Microwave ovens and
induction cookers are a few examples.)

Question 6.
What are the characteristics of nichrome that make it suitable for a heating element?
Answer:

• High oxidation resistance (oxidation resistance is the ability of a material to resist corrosion due to contact with oxygen or other oxidisers at high temperature).
• Ability to provide heat energy for a long time in a red hot state.
• High resistivity (due to this property even a nichrome wire of short length can provide sufficient resistance)
• High melting point(due to this property, nichrome is able to tolerate the high temperatures needed
  for efficient heating without melting or deforming.)

We know that all materials do not conduct electricity in the same way.

Resistivity is one of the intrinsic properties of a material that mainly determines the resistance which is the characteristic of a conductor. Resistivity is different for each material.

Resistivity and conductivity.
The resistance of a conductor with an area of cross section of one square metre and a length of one metre is the resistivity of the material of the conductor. It is denoted by the Greek letter ρ(rho). Its unit is ohmmetre (Ωm). Conductivity is the ability of a material to conduct electricity. This is the reciprocal of resistivity.

The factors which influence the resistance of a conductor are length of the conductor
(l), area of cross section (A) and the nature of the material (ρ). The relation of resistance (R) with these factors is given by the equation R = \(\frac{\rho l}{A}\)

Activity to find out the factors that influence the heat produced in a conductor carrying current.
• Construct a circuit which consists of a 5 cm long nichrome wire, 6 V battery eliminator, etc., as shown in figure 5.4. Take 20 mL of water in a boiling tube. Immerse the nichrome wire and a thermometer into it. Adjust the rheostat to vary the current. Record the temperature in table 5.2 every two minutes. Repeat the experiment by replacing the water in the boiling tube each time.

Observation
Temperature of water before starting the experiment = 31°C

Current (A)	Temperature (°C)
0.5	33
1	35
1.5	37

Analysis
On analysing table 5.2 given above we can understand that the temperature of water increases with increase in current. In this experiment a fixed quantity of water (20 mL) is heated by passing current. To increase the temperature of a fixed quantity of water more heat is required. Hence as the quantity of heat received by the water increases, the temperature also increases. Therefore, it is understood that when the current increases heat produced also increases.

• Adjust the rheostat to make the current 1 A. In table 5.3 given below record the temperature of water, every 2 minutes.
Observation

Time (in minutes)	Temperature (°C)
0	31
2	33
4	35
6	32

Analysis
Temperature increases because the current flowed for a longer time. We can understand that the heat produced increases as the time of flow of current increases.

Activity
Take nichrome wire and aluminium wire of equal length (5 cm) and equal area of cross section. Using them make a circuit as shown in the figure.

Pass current through the circuit for about 2 minutes using a 9 V, 50 W eliminator. Place a thin paper on both aluminium and nichrome wire.

Question 7.
What do you observe?
Answer:
Because the nichrome wire generates a lot more heat than the aluminium wire, we can see that the thin paper on the nichrome wire is likely to bum or char while the paper on the aluminium wire would either stay the same or only warm slightly.

Question 8.
Which conductor becomes hotter?
Answer:
Nichrome wire becomes hotter.

Question 9.
The magnitude of current in the nichrome wire and aluminium wire is (the same / different).
Answer:
Same

Question 10.
The resistance of nichrome wire and aluminium wire is (the same/different).
Answer:
Different
The nichrome wire become hotter because its resistance is greater than that of the aluminium wire. Resistance is another factor that influences the quantity of heat produced in a conductor carrying current.

Question 11.
From the experiments conducted, note down the three factors that influence the quantity of heat produced in a current carrying conductor:
Answer:

• Current (I)
• Resistance of the conductor(R)
• Time for which current flows(t)

The scientist James Prescott Joule discovered the relation between the heat produced in an electric conductor and the factors influencing the quantity of heat.

Joule’s law
The quantity of heat produced in a current carrying conductor is directly proportional to the square of the current (I²), the resistance of the conductor (R) and the time (t) for which the current flows. This is Joule’s law. If a current I flows through a conductor having resistance R for a time t, according to Joule’s law, the quantity of heat produced will be H = I²Rt. Its SI unit is joule (J).

Question 12.
A current of 0.5 A flows through a conductor with 100 Ω resistance for 5 minutes.
a) What will be the quantity of heat produced?
b) If a current of 1A flows through this conductor for 5 minutes, what will be the quantity of heat produced?
Ans:
a) R = 100 Ω
I = 0.5 A
t = 5 × 60s
H = ?
H = I²Rt
= 0.5 × 0.5 × 100 × 5 × 60 J
= 7500 J

b) I = 1A, H = I²Rt
H = 1 × 1 × 100 × 5 × 60 J
= 30000J

Question 13.
When the current was doubled, the heat produced in the conductor became 4 times. If so, what happens to the heat produced if the current is halved?
(halved/ quartered)
Answer:
quartered
The electrical appliances commonly used in houses operating at 230 V.

Other forms of Joule’s Law
We can formulate other equations to calculate the quantity of heat.
According to ohm’s law
V = IR
Hence I = \(\frac{V}{R}\)
On substituting the value of I in the equation
H = I²Rt
H = (\(\frac{V}{R}\))² Rt
So, we get H = \(\frac{V^2 t}{R}\)
Similarly, if we substitute R = \(\frac{V}{I}\) in the equation
H = I²Rt
We get H = VIt
H = I²Rt = \(\frac{V^2 t}{R}\) = VIt

Question 14.
If the resistance of a heating appliance operating at 230 V is 920 Ω , what is the quantity of heat produced in 10 minutes?
Answer:
V = 230 V, R = 920 Ω, t = 10 × 60 s
H = \(\frac{V^2 t}{R}\)
= \(\frac{230 \times 230}{920}\) × 10 × 60 = 34500 J

Question 15.
If 7500 J of heat is produced in 30 s in a conductor having 2500 resistance, what is the current through the conductor?
Answer:
H = 7500 J,
t = 30 s,
R = 250 Ω
H = I²Rt
\(\frac{H}{R t}\) = I²
I = \(\sqrt{\frac{H}{R t}}\) = \(\sqrt{\frac{7500}{250 \times 30}}\)
I = √1 = 1A

Question 16.
Calculate the quantity of heat produced if 2 A current flows for 10 minutes through a heating coil of an electric kettle having 100 Ω resistance?
Answer:
I = 2A,
t = 10 × 60 = 600 s,
R = 100 Ω
H = I²Rt
H = 2² × 100 × 600 = 240000 J

Question 17.
A potential difference of 230 V is applied across a circuit for 5 minutes. If the resistance in the circuit is as given below, calculate the current and heat in each case.
(a) 115 Ω
(b) 230 Ω
Answer:
V= 230 V, t = 5 × 60 = 300 s
(a) R = 115 Ω
Current, I = \(\frac{V}{R}\) = \(\frac{230}{115}\) = 2A
Heat, H = VIt = 230 × 2 × 300
H = 138000 J

(b) R = 230 Ω
Current, I = \(\frac{V}{R}\) = \(\frac{230}{230}\) = 1A
Heat, H = VIt = 230 × 1 × 300
H = 69000 J

Question 18.
Observe the circuits given below. Calculate the quantity of heat produced in each, in 5 minutes.

Answer:

Heat produced in circuit 5.6(a)	Heat produced in circuit 5.6(b)	Heat produced in circuit 5.6(c)
H = I2Rt = 0.25 × 0.25 × 48 × 300 = 900 J	H = I2Rt = 0.5 × 0.5 × 24 × 300 = 1800 J	H = I2Rt = 1 × 1 × 1 × 300 = 3600 J

Question 19.
Analysing the completed table, answer the following questions.
a) What is the difference in the quantity of heat in circuits 5.6 (a) and 5.6 (b)?
Answer:
Heat in circuit 5.6 (a) = 900 J
Heat in circuit 5.6 (b) = 1800 J
Difference in quantity of heat = 1800 – 900
= 900 J

b) Why is there a difference in the quantity of heat produced in these circuits?
Answer:
Even though the voltage and time are the same, different combinations of current and resistance cause a variation in the amount of heat generated. In particular, even with a lower resistance, the greater current in circuit 5.6 (b) -0.5 A when compared to 0.25 A in 5.6 (a) results in a considerably greater I² term, which generates more heat.

c) What is the resistance of the circuit in figure 5.6 (c)?
Answer:
12 Ω

d) What is the total heat produced in the circuit in figure 5.6 (c)?
Answer:
H = 3600 J

e) Less heat is produced in the resistor of higher resistance in the circuit in figure 5.6 (a). Why?
Answer:
When the current is constant, the heat generated is directly proportional to the resistance, as stated by Joule’s law (H = I²Rt). However, a larger resistance results in a smaller current (I = V/R) if the voltage is constant, for all these three circuits. The effect of square of current mainly contributes to heat generated because heat is dependent on I²R. Therefore, less heat is produced in the resistor of higher resistance in the circuit in figure 5.6 (a).

f) Among these circuits, in which resistor is the maximum quantity of heat produced?
Answer:
In the 12 resistor in circuit 5.6(c)

g) What are your inferences? Record them in the science diary.
Answer:
If voltage is constant, when the resistance in the circuit is decreased, the current increases. Hence the quantity of heat produced increases.

Question 20.
If the voltage remains constant, what change will occur in the quantity of heat produced when the resistance varies? Explain based on Joule’s law.
Answer:
Joule’s law is given by H = I²Rt. However, when the voltage (V) is constant, expressing the current using Ohm’s law and substituting in joule’s law, we get H = \(\frac{V^2 t}{R}\). This equation shows that the heat generated (H) is inversely proportional to the resistance (R) when V and t are constant. This indicates that as resistance increases, heat generated decreases and when resistance decreases heat generated increases.

Question 21.
Calculate the quantity of heat produced if 2 A current flows through an electric heater operating at 230 V for 15 minutes. What is the resistance of the heater?
Answer:
I = 2A, t = 15 × 60 = 900s, V = 230 V
Heat produced, H =VIt
H = 230 × 2 × 900 =414000 J
Resistance of the heater = R = \(\frac{V}{I}\)
R = \(\frac{230}{2}\) = 115

Question 22.
An electric iron operating at a potential difference of 230 V has a heating coil of resistance 100 £2. If this electric iron operates for half an hour, how much electric energy will be converted to heat energy? What is the current through the electric iron?
Answer:
V = 230 V, R = 100 Ω, t = 30 × 60 = 1800 s
H = \(\frac{V^2 t}{R}\)
H = \(\frac{230^2 \times 1800}{100}\)
Current through the electric iron, I = \(\frac{V}{I}\) = \(\frac{230}{100}\) = 2.3 A

Question 23.
A heating appliance having a resistance of 92 Ω operates at 230 V. Using various equations, calculate the heat produced by the appliance in 14 minutes and write them down in the table below. What resistance is needed to double the heat energy?

Answer:

H = I2Rt	H = \(\frac{V^2 t}{R}\)	H = VIt
V = 230 V R = 92 Ω t = 14 minute = 14 × 60 = 840s I = \(\frac{V}{R}\) = \(\frac{230 \mathrm{~V}}{92 \Omega}\) = 2.5 A H = I2Rt = (2.5)2 × 92 × 840 J = 483000 J	V = 230 V R = 92 Ω t = 14 minute = 14 × 60 = 840s H = \(\frac{V^2 t}{R}\) H = \(\frac{230^2 \times 840}{92}\) = 483000 J	V = 230 V I = \(\frac{V}{R}\) = \(\frac{230 \mathrm{~V}}{92 \Omega}\) = 2.5 A t = 14 minute = 14 × 60 = 840 s H = VIt H = 230 × 2.5 × 840 = 483000 J

As the voltage and time is remaining constant, we take H = \(\frac{V^2 t}{R}\)
For H to become 2H(to double the heat energy),
we need to find new resistance value-R_new
2H = \(\frac{\mathrm{V}^2 \mathrm{t}}{\mathrm{R}_{\text {new }}}\)
We have 2\(\frac{V^2 t}{R}\) = \(\frac{\mathrm{V}^2 \mathrm{t}}{\mathrm{R}_{\text {new }}}\), \(\frac{2}{R}\) = \(\frac{1}{\mathrm{R}_{\text {new }}}\)
R_new = \(\frac{R}{2}\) = \(\frac{92}{2}\) = 46 Ω

Question 24.
Observe the values given on the label of the appliance in figure 5.7. What does marking 750 W indicate?

Answer:
750 W indicates the power at which the appliance work.
Energy conversion occurs when an appliance operates, is the work done. The quantity of work done per unit time is power. Power is the rate of work based on time.

Question 25.
The power of each electric appliance is marked on it. Complete the table given below.

Answer:

Device	Power (P)W	Energy consume in 1 second (Pt = p × 1)
LED Bulb	9	= 9 × 1 = 9 J
Projector	100	= 100 × 1 = 100 J
Fan	60	= 60 × 1 = 60 J
Laptop	40	= 40 × 1 = 40 J

While labelling the power of an appliance, the voltage at which it is to be operated is also indicated. In our country, the power obtained while operating at 230 V is labelled on electrical appliances. If the appliance operates above or below 230 V, the power will vary accordingly. That is, if the voltage increases, the power increases and if the voltage decreases, the power also decreases. This may lead to the damage of appliances. (There are also some appliances that operate at 400 V).

Countries design electrical appliances and power according to their power supply voltage. In all electrical appliances, the voltage required for it and its power will be marked on it. The indicated power is available from an appliance only when it operates at the specified voltage.

If the voltage is constant, according to P = \(\frac{V^2}{R}\), appliances with higher resistance will have lower power, and those with lower resistance will have higher power. While solving numerical problems related to household appliances, it is more desirable to use the formula P = \(\frac{V^2}{R}\). In some foreign countries, the supply voltage is 110 V.

Question 26.
At what applied voltage will an electric heating appliance marked 800 W, 240 V produce a power of 200 W?
Answer:
P₁ = 800W, V₁ = 240 V
R = \(\frac{\mathrm{V}_1^2}{\mathrm{P}}\) = \(\frac{240^2}{800}\)
R = 72 Ω
P₂ = 200 W, V₂ = ?
V₂² = P₂ × R
V₂² = 200 × 72 = 14400
V₂ = \(\sqrt{14400}\) = 120 V
Whenever voltage becomes half, power becomes 1/4 times the original power.

Question 27.
Calculate the quantity of heat produced when a heating appliance of power 500 W operates for five minutes.
Answer:
P = 500 W
t = 5 min = 5 × 60 = 300 s
H = Pt = 500 × 300 J = 150000 J

Question 28.
How much heat energy is produced when an electric heater with 600 W power operates for 20 minutes?
Answer:
P = 600 W, t = 20 × 60 = 1200 s
H = Pt = 600 × 1200 = 720000 J

Question 29.
How much heat energy will be produced if a heating appliance with 250 W power operates for 10 minutes?
Ans:
P = 250 W, t = 10 × 6 = 600 s
H = Pt = 250 × 600 = 150000 J

Question 30.
An electric heater produced 450000 J of heat energy when operated for 5 minutes. What would be the power of this heater?
Answer:
H = 450000 J, t = 5 × 60 = 300 s
H = Pt, P = \(\frac{H}{t}\)
P = \(\frac{450000}{300}\) = 1500 W

Question 31.
On a heating appliance it is marked 529 W, 230 V.
a) What do each of these mean?
b) What would be the power of this appliance if 100 V is supplied?
Answer:
a) 529 W, 230 V indicates that the appliance operating at voltage of 230 V consumes 529 W power.

b) R = \(\frac{\mathrm{V}^2}{\mathrm{P}}[latex] = [latex]\frac{230^2}{529}[latex] = 100 Ω
P = [latex]\frac{\mathrm{V}^2}{\mathrm{R}}[latex]
P = [latex]\frac{100^2}{100}[latex] = 100 W

Question 32.
Calculate the energy consumed by a 1 kW power appliance in one hour.
Answer:
E = Pt = 1kW × 1h
= 1kWh

One kilowatt hour or one unit of electric energy is the energy consumed by an appliance with 1000 W (1 kW) power in one hour.

One unit of electric energy in joules

1000W = 1kW

Energy consumed,
E = H = Pt
P = 1000W, t = 1 h = 60 × 60s
El = 1000 × 60 × 60 J
= 3600000 J = 3.6 × 10⁶ J

1kWh = one unit of electric energy = 3.6 × 106 J

Question 33.
How much heat energy will be produced when a heating appliance of 60 W operates for one hour?
Answer:
H = Pt
= 60 × 1 × 60 × 60 J
= 216000 J

Question 34.
The power and operating time of some appliances are given in the table. Complete the table.

Answer:

Power of device (P)W	Time (t)h	Electric energy spent (Pt) J	Energy utilized in unit (kWh) = [latex]\frac{\text { Power in watt }}{1000}\) × hour
1000	1 hour	1000 × 60 × 60 = 3600000 J	1k Wh = 1 unit
2000	1 hour	2000 × 60 × 60 = 7200000 J	2k Wh = 2 unit
500	1 hour	500 × 60 × 60 = 1800000 J	0.5 kWh = 0.5 unit
500	2 hour	500 × 2 × 60 × 60 = 3600000 J	Ik Wh = 1 unit

Question 35.
Calculate the electric energy consumed if a 500 W grinder and a 600 W electric iron each operate for 2 hours.
Answer:
Total power in watt = 500 W + 600 W
Time = 2 h
Electric energy (in kilo watt hour)
= \(\frac{\text { Power in watt × time in hour }}{1000}\)
= \(\frac{1100 \mathrm{~W} \times 2 \mathrm{~h}}{1000}\)
= 2.2 kWh
= 2.2 unit
Using this formula, we can calculate the monthly consumption of electricity.

Question 36.
Considering the total power of the appliances used and their operating time, calculate the daily electricity consumption in your house. Similarly, calculate the monthly consumption of electricity and present in class the main findings and suggestions based on the project report.
Answer:
An example is given below
• Prepare a table listing the given information

Appliance	Power (Watt)	Hours used per day	Number of appliances
Fan	75 W	10 hours	3
Tube light	40 W	6 hours	4
Television	100 W	4 hours	1
Refrigerator	150 W	24 hours	1
Washing Machine	500 W	1 hour	1
Steamer	1000 W	0.5 hours	1

• Calculate energy consumption in a day
  Electric energy (in kilowatt hour)

= 8.21 kWh = 8.21 units

• Calculate energy consumption in a month (30 days) = 8.21 × 30 = 246.3 units
• Analyze the findings. Identify which were among the high consuming appliances.
• Compare with energy consumption pattern in the previous month.
• Take action if the energy consumption is increased very much.
• Some actions that can be taken are
  • Turn off appliances when not in use to save energy.
  • Use energy-efficient appliances (like LED lights).
  • Avoid using high-power devices unnecessarily (like steamer).
  • Use natural light during the day to reduce lighting load.
  • Consider installing solar panels for long-term savings.
• Stay alert of your energy consumption by tracking it,which helps you to take appropriate action whenever necessary.

All houses have the different electricity tariff. Electricity charges are levied at a higher rate from consumers who use more electricity and at a lower rate from consumers who use less electricity. (Table 5.8)

Now we can understand that one of the reasons for the increased consumption of electricity is the unwanted and careless consumption of electricity. The amount in the electricity bill can be reduced by lowering electricity consumption.

Question 37.
What can we do to reduce electricity consumption at home?
Answer:

• Use energy efficient electrical appliances.
• Turn off switches immediately after use.
• Use LED tubes and bulbs.
• Choose the size of the fan according to the size of the room.
• Use BLDC fans. [BLDC – Brush Less Direct Current]
• Electronic appliances should be unplugged while not in use.
• Switch to energy star rated appliances which helps in reducing electricity bills, have less environmental impact and government benefits are applicable to them. (Like offer rebates or incentives to purchase them)

Further activities to reduce energy consumption can be implemented under the leadership of the School Energy Club in collaboration with the Energy Management Centre (EMC).

Question 38.
Do large scale electricity generating power plants cause environmental impacts? Discuss.
Answer:
Large scale electricity generating power plants are important for providing power to cities, industries, and homes. They produce a large amount of electricity and offer a steady and reliable supply. However, many of these plants, especially those using coal or gas, cause serious environmental problems like air pollution, global warming, and water pollution. They often use large amounts of water and land, which can harm wildlife and displace communities. Large scale hydropower projects can cause environmental problems. They often lead to the destruction of forests, loss of wildlife habitats, and the shifting of people due to dam construction. These projects can also disrupt river ecosystems. But compared to thermal and nuclear power plants hydro power is a reliable alternative. There are some demerits in all, but ways are to be identified to tackle the disadvantages. Switching on to other renewable energy sources which causes less pollution can become a smart and sustainable solution to the energy and environmental challenges caused by other large scale power plants.

Observe figure 5.9.
There are ways to produce electricity without causing pollution.

You know that the sun is the basis of all energy on the earth.

Question 39.
List the devices that utilise solar energy.
Answer:
• Solar water heater
• Solar cooker
• Solar furnace

Solar cells, panels and power plants
Solar cells are devices that convert solar energy into electric energy. Only negligible voltage and current can be obtained from a single cell. Therefore, a solar panel is constructed by arranging many solar cells suitably to provide electricity as per the requirement. The electricity obtained from such panels can be stored in a storage battery and utilised when needed. Or it can be directly supplied to the electricity distribution agencies. Solar panels are used for energy requirements in artificial satellites. In addition to small scale requirements, large scale solar power plants are also in operation.

• Cochin International Airport Limited (CIAL) in Nedumbassery, Kerala, produces electricity required for its entire operations from such a solar power plant.
• Cochin International Airport is the world’s first airport to operate entirely on solar power.

Question 40.
List the benefits of installing solar panels.
Answer:

• Can produce electricity required for homes.
• Can reduce environmental pollution.
• Transmission and distribution losses are minimised as electricity is generated at the place of consumption.

It is understood that atmospheric pollution is reduced when solar panels are used. We know that the quantity of greenhouse gases in the atmosphere is increasing daily. An increase in such gases cause global warming, resulting in climate change. Each of our interventions in this matter is very important. Global warming can be reduced by reducing the production of major greenhouse gases like carbon dioxide, methane etc. We need activities that can reduce the quantity of carbon and carbon compounds emitted by individuals, organisations, and products.

Question 41.
What can be done to reduce carbon footprint?
Answer:

• Reduce domestic energy consumption.
• Avoid wasting food.
• Use public transport.
• Reduce waste by utilising reusable products.
• Educate society about reducing carbon footprint.
• Perform energy consuming activities in an energy saving manner.

Question 42.
Prepare a seminar paper on how to reduce individual carbon footprint and present it in class.
Answer:
An example is given below.
Introduction
One of the most significant issues our world is facing now is climate change. It is mostly brought on by gases like carbon dioxide that are emitted into the atmosphere as a result of things like using electricity, driving a car, and creating garbage. These gases raise Earth’s temperature by trapping heat in the atmosphere, which can result in hazardous weather and environmental changes. Every individual’s, everyday activities add to this issue, and their carbon footprint is the overall quantity of greenhouse gases they create. Lowering carbon footprint helps in slowing down climate change and thus helps in preserving the earth for future generations, so lowering it is cmcial. We may take action to lessen our carbon footprint by being aware of its origins.

Methods for Lowering Carbon Footprint

• Conserve energy at home-When not in use, turn off fans, lights, and electrical equipment. Reduce the amount of electricity you use by using energy-efficient appliances and bulbs.
• Choose for greener transportation-Whenever feasible, take public transportation, walk, or cycling instead of driving a car. Carpooling with friends or travel mates reduces emissions as well. It is even better to use electric or fuel- efficient vehicles.
• Reducing, reusing, and recycling-Reduce the use of plastic bags and bottles. Use reusable bags and bottles instead of single-use plastics. Recycling glass, plastic, and paper contributes to a decrease in pollution and waste.
• Conserve water-As you brush your teeth, turn off the faucets and promptly address any leaks. Conserving water also reduces the energy required for water treatment.
• Encourage environmentally friendly products- Purchase goods that are manufactured sustainably and with less packaging. More green activities are encouraged when businesses that care about the environment are supported.
• Raise awareness-Discuss with loved ones the significance of lowering carbon footprints. To have a greater influence, join or form school-based environmental organizations.

Conclusion
It is not very easy to reduce carbon footprint, yet when many people make minor changes in their daily lives, the impact may be significant. Each of us may contribute to protecting the planet by minimizing trash, eating sensibly, adopting more environmentally friendly transportation, conserving energy, and raising awareness. Together we can make the planet a better and cleaner place to live.

‘Reducing atmospheric pollution and conserving energy is everyone’s duty. Energy conservation is equivalent to energy production. ’

‘Energy is precious, don’t waste it! ’

Std 10 Physics Chapter 5 Notes – Extended Activities

Question 1.
Observe the electricity bill of your house and find out the monthly electricity consumption. Present activities that can be implemented in your house to reduce electricity consumption in a seminar in the Energy Club. Also, prepare necessary posters for this.
Answer:
By observing the electricity bill one can find out the monthly electricity consumption. Total units consumed and amount to be paid can be identified and it can be compared with previous month to understand the pattern of energy consumption.

Activities that can be implemented to reduce electricity consumption

• Make a transition to energy-efficient appliances.
• Maximize the performance of your air conditioning and heating systems.
• Use LED bulbs-LEDs use less power and last longer than normal bulbs.
• Turn off electrical appliances when not in use-Always switch off fans, lights, and appliances when you leave a room.
• Unplug devices-Chargers, TVs, and microwaves still use power when plugged in. So electronics should be unplugged while not in use.
• Use natural light during the day-Open curtains and windows instead of turning on lights.
• Use washing machine for full loads only-This saves both electricity and water.
• Limit geyser use-Use it only for a short time and turn it off immediately after use.
• Check your meter weekly-Keep a track of electricity consumption which will make you aware of the
  energy usage.

Switch Off to Switch On a Better Future

Save Energy, Save Earth

Question 2.
List the activities that the School Science Club intends to undertake to reduce the school’s carbon footprint.
Answer:

• Plan tree-planting activities and create green spaces on the school’s property.
• To reduce waste, set up recycling facilities and encourage the use of reusable products.
• When not in use, remind employees and students to turn off lights and electronics.
• Encourage water-saving practices including repairing leaks and properly shutting off taps.
• Increase public knowledge of climate change and sustainability through seminars, discussions, and initiatives.
• Encourage environmentally sustainable modes of transportation, such as walking, bicycling, or ridesharing, to get to school.
• Initiate the process , of composting organic waste for school gardens.
• Examine the school’s energy usage on a regular basis and make recommendations for ways to reduce , electricity use.
• Join local or national green projects by working with environmental organizations.
• Reduce the amount of plastic used in the cafeteria and promote eco-friendly options for school supplies.

Electric Energy: Consumption and Conservation Class 10 Notes

Electric Energy: Consumption and Conservation Notes Pdf

• The conversion of electric energy mainly into heat energy is the heating effect of electric current. Appliances that utilise this effect are electric heating appliances. –
• The process of production of heat, when electricity flows through a conductor is Joule heating or Ohmic heating. The part used to produce heat energy in heating appliances is the heating element.
• The resistance of a conductor with an area of cross section of one square metre and a length of one metre is the resistivity of the material of the conductor. It is denoted by the Greek letter ρ(rho). Its unit is ohm metre (Ωm).
• Conductivity is the ability of a material to conduct electricity. This is the reciprocal of resistivity.
• The factors which influence the resistance of a conductor are length of the conductor (l), area of cross section (A) and the nature of the material (ρ). The relation of resistance (R) with these factors is given by the equation R = \(\frac{\rho l}{A}\)
• Factors that influence the quantity of heat produced in a current carrying conductor are
  • Current (I)
  • Resistance of the conductor (R)
  • Time for which current flows (t)
• The quantity of heat produced in a current carrying conductor is directly proportional to the square of the current (I²) the resistance of the conductor (R) and the time (t) for which the current flows. This is Joule’s law.
• If a current I flows through a conductor having resistance R for a time t, according to Joule’s law, the quantity of heat produced will be H = I²Rt. Its SI unit is joule (J).
• Power is the quantity of work done by an electrical appliance per unit time. The work done by an electrical appliance is the conversion of electric energy into another form of energy.
• The quantity of electric energy consumed in houses can be measured directly by a watt-hour meter connected to our household electric circuit. In this, electric energy is measured in kilowatt hour (kWh) units. This is the commercial unit of electric energy.
• One kilowatt hour or one unit of electric energy is the energy consumed by an appliance with 1000 W (1 kW) power in one hour.
• Energy crisis is the increase in demand for energy and the decrease in availability.
• Solar cells are devices that convert solar energy into electric energy.
• Greenhouse gases are directly and indirectly emitted by individuals, families, organizations, events, services, products, etc. The quantity of such emitted greenhouse gases, converted and expressed as equivalent to the measure of carbon dioxide, is carbon footprint.

INTRODUCTION

Electricity plays a crucial role in our everyday activities and is used in many forms. This chapter explores how electrical energy can be transformed into other types of energy, such as heat, light, and mechanical energy. A key topic is Joule heating, which explains how electric current produces heat in wires and devices. We also learn about electric power, which tells us the rate at which electrical energy is used, and how it is measured using a watt-hour meter. The electricity used in homes is measured in kilowatt-hours (kWh), which is the standard commercial unit. Excessive use of energy, especially from non-renewable sources, contributes to the energy crisis and increases our carbon footprint, damaging the environment. To solve these problems, switching to solar energy and other renewable sources is a cleaner and more sustainable option.

Heating effect of electric current

• The useful form of energy into which an appliance mainly converts electric energy is considered the effect of electric current in that appliance.
• The conversion of electric energy mainly into heat energy is the heating effect of electric current. Appliances that utilise this effect are electric heating appliances.
• The process of production of heat, when electricity flows through a conductor is Joule heating or Ohmic heating. The part used to produce heat energy in heating appliances is the heating element. The heating coil is made of an alloy, nichrome.

Resistivity and conductivity

• The resistance of a conductor with an area of cross section of one square metre and a length of one metre is the resistivity of the material of the conductor. It is denoted by the Greek letter ρ(rho). Its unit is ohmmetre (Ωm).
• Conductivity is the ability of a material to conduct electricity. This is the reciprocal of resistivity.
• The factors which influence the resistance of a conductor are length of the conductor (l), area of cross section (A) and the nature of the material (ρ). The relation of resistance (R) with these factors is given by the equation R = \(\frac{\rho l}{A}\)
• Factors that influence the quantity of heat produced in a current carrying conductor are
  • Current (I)
  • Resistance of the conductor(R)
  • Time for which current flows(t)

Joule’s law of heating

• The quantity of heat produced in a current carrying conductor is directly proportional to the square of the current (I²), the resistance of the conductor (R) and the time (t) for which the current flows. This is Joule’s law.
• If a current I flows through a conductor having resistance R for a time t, according to Joule’s law, the quantity of heat produced will be H = I²Rt. Its SI unit is joule (J).

Electric power

• Power is the quantity of work done by an electrical appliance per unit time. The work done by an electrical appliance is the conversion of electric energy into another form of energy.
• The quantity of electric energy consumed in houses can be measured directly by a watt-hour meter connected to our household electric circuit. In this, electric energy is measured in kilo watt hour (kWh) units. This is the commercial unit of electric energy.
• One kilowatt hour or one unit of electric energy is the energy consumed by an appliance with 1000 W (1 kW) power in one hour.

Energy crisis
• Though the demand for energy has increased many folds, production has not sufficiently increased. This situation is energy crisis. Simply, we say Energy crisis is the increase in demand for energy and the decrease in availability.

Solar cells, panels and power plants

• Solar cells convert sunlight into electrical energy, but a single cell produces only a small amount of power.
• To meet higher energy needs, many cells are combined to form solar panels, which can store electricity in batteries or supply it to the power grid.
• Solar panels are used in satellites, homes, and large-scale solar power plants for energy production with less pollution.

Carbon footprint

• Greenhouse gases are directly and indirectly emitted by individuals, families, organizations, events, services, products, etc. The quantity of such emitted greenhouse gases, converted and expressed as equivalent to the measure of carbon dioxide, is carbon footprint.
• Each individual should strive to reduce his/her personal carbon footprint. This can be achieved by paying attention to daily activities.

HEATING EFFECT OF ELECTRIC CURRENT
The conversion of electric energy mainly into heat energy is the heating effect of electric current. Appliances that utilise this effect are electric heating appliances.

ELECTRIC POWER
Power is the quantity of work done by an electrical appliance per unit time. The work done by an electrical appliance is the conversion of electric energy into another form of energy.

The function of electric heating appliances is to produce heat. You have learnt that the quantity of heat produced
H = I²Rt
Work = quantity of heat produced
W = H = I²Rt
Equation to find electric power
Power = \(\frac{Work}{time}\)
P = \(\frac{W}{t}\) = \(\frac{H}{t}\)
P = \(\frac{I^2 R t}{t}\)
P = I²R
According to Ohm’s law, I = \(\frac{V}{R}\)
Then P = \(\frac{V^2}{R}\)
We know that R = \(\frac{V}{I}\)
Hence P = VI
We will get the equations = P = \(\frac{V^2}{R}\), P = VI
P = I²R = \(\frac{V^2}{R}\) = VI
If we substitute the value of P in the equations used to calculate heat, H = I²Rt.We get
H = Pt
It is understood that the quantity of heat produced in a given time can be found if the power of an appliance is known.

Watt-hour meter

The quantity of electric energy consumed in houses can be measured directly by a watt hour meter connected to our household electric circuit. In this, electric energy is measured in kilowatt hour (kWh) units. This is the commercial unit of electric energy. Power is expressed in watt, time in second, and energy in joule.

The energy consumed if a 9 W bulb operates for one hour daily for 30 days is E = Pt = 9 × 30 × 3600J = 972000J (nine lakh seventy two thousand joules). So, if we consider the monthly energy consumption of all the appliances in a house, the total would be very large, making it inconvenient to record. Therefore, energy is calculated in kilowatt hour by measuring power in kilowatt and time in hour.

ENERGY CRISIS
Despite having many small scale electricity projects in addition to large power stations, the electricity they produce is not sufficient for our needs. Though the demand for energy has increased many folds, production has not sufficiently increased. This situation is energy crisis. This is why sometimes power cuts and load shedding are to be implemented.

‘Energy crisis is the increase in demand for energy and the decrease in availability.’

There are hydroelectric power plants, thermal power plants, nuclear power plants, etc., to produce electricity on a large scale. Yet, it is not easy to increase the electricity production.

Rooftop power project
Solar panels can be installed on rooftops of houses and other places receiving sunlight. The Rooftop Power Project is a scheme jointly initiated by the Central Government and KSEB for this purpose.

Carbon footprint
Greenhouse gases are directly and indirectly emitted by individuals, families, organizations, events, services, products, etc. The quantity of such emitted greenhouse gases, converted and expressed as equivalent to the measure of carbon dioxide, is carbon footprint.

Each individual should strive to reduce his/her personal carbon footprint. This can be achieved by paying attention to daily activities. How one travels, what food one eats, what clothes one uses, and how much waste is generated are all important.

Expert Teachers at HSSLive.Guru has created Kerala Syllabus SSLC 10th Standard Physics Solutions Guide Pdf Free Download of Chapter wise Questions and Answers, Notes are part of Kerala Syllabus 10th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT Kerala State Board Syllabus 10th Standard Physics Textbooks Solutions Pdf of Kerala SSLC Class 10 Part 1 and 2 HSS Live physics. Students can also read Kerala SSLC Physics Model Question Papers 2019-2020.

Kerala SCERT Class 10 Physics Solutions

Kerala Syllabus 10th Standard Physics Notes Textbook Solutions Pdf English Medium

SCERT Class 10 Physics Textbook Solutions Part 1

• Class 10 Physics Chapter 1 Notes Kerala Syllabus – Sound Waves
• Class 10 Physics Chapter 2 Notes Kerala Syllabus – Lenses
• Class 10 Physics Chapter 3 Notes Kerala Syllabus – The World of Colours and Vision
• Class 10 Physics Chapter 4 Notes Kerala Syllabus – Magnetic Effect of Electric Current

Class 10 Physics Notes Solutions Kerala Syllabus Part 2

• Class 10 Physics Chapter 5 Notes Kerala Syllabus – Electric Energy: Consumption and Conservation
• Class 10 Physics Chapter 6 Notes Kerala Syllabus – Electromagnetic Induction in Daily Life
• Class 10 Physics Chapter 7 Notes Kerala Syllabus – Mechanical Advantage in Action

10th Physics Important Questions and Answers Kerala Syllabus

SSLC Physics Important Questions and Answers Pdf

• Chapter 1 Sound Waves Class 10 Important Questions
• Chapter 2 Lenses Class 10 Important Questions
• Chapter 3 The World of Colours and Vision Class 10 Important Questions
• Chapter 4 Magnetic Effect of Electric Current Class 10 Important Questions
• Chapter 5 Electric Energy: Consumption and Conservation Class 10 Important Questions
• Chapter 6 Electromagnetic Induction in Daily Life Class 10 Important Questions
• Chapter 7 Mechanical Advantage in Action Class 10 Important Questions

Kerala Syllabus 10th Standard Physics Notes Textbook Solutions Pdf Malayalam Medium

10th Class Physics Notes Pdf Malayalam Medium Part 1

• Class 10 Physics Chapter 1 Malayalam Medium – ശബ്ദതരംഗങ്ങൾ
• Class 10 Physics Chapter 2 Malayalam Medium – ലെൻസുകൾ
• Class 10 Physics Chapter 3 Malayalam Medium – കാഴ്ചയും വർണ്ണങ്ങളുടെ ലോകവും
• Class 10 Physics Chapter 4 Malayalam Medium – വൈദ്യുതിയുടെ കാന്തികഫലം

SCERT Class 10 Physics Solutions Malayalam Medium Part 2

• Class 10 Physics Chapter 5 Malayalam Medium – വൈദ്യുതോർജം : ഉപഭോഗവും സംരക്ഷണവും
• Class 10 Physics Chapter 6 Malayalam Medium – വൈദ്യുതകാന്തികപ്രേരണം നിത്യജീവിതത്തിൽ
• Class 10 Physics Chapter 7 Malayalam Medium – യാന്ത്രികലാഭം പ്രയോഗത്തിൽ

SSLC Physics Important Questions and Answers Pdf Malayalam Medium

Physics Class 10 Kerala Syllabus Important Questions

1. ശബ്ദതരംഗങ്ങൾ Important Questions
2. ലെൻസുകൾ Important Questions
3. കാഴ്ചയും വർണ്ണങ്ങളുടെ ലോകവും Important Questions
4. വൈദ്യുതിയുടെ കാന്തികഫലം Important Questions
5. വൈദ്യുതോർജം : ഉപഭോഗവും സംരക്ഷണവും Important Questions
6. വൈദ്യുതകാന്തികപ്രേരണം നിത്യജീവിതത്തിൽ Important Questions
7. യാന്ത്രികലാഭം പ്രയോഗത്തിൽ Important Questions

We hope the given Kerala Syllabus 10th Standard Physics Solutions Guide Pdf Free Download in English Medium and Malayalam Medium of Chapter wise Questions and Answers, Notes will help you. If you have any queries regarding SCERT Kerala State Board Syllabus 10th Standard Physics Textbooks Answers Guide Pdf of Kerala SSLC Class 10 Part 1 and 2, drop a comment below and we will get back to you at the earliest.