NCERT Solutions for Class 10

NCERT Solutions for Class 10 Geography Chapter 7 – Lifelines of National Economy (Social Science), contains solutions to various questions in Exercise for Chapter 7.  At the end of the Solutions, all the keywords which are important to understand Chapter 7 Lifelines of National Economy Class 10 Geography, have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Geography Chapter 7 – Lifelines of National Economy

NCERT Solutions for Class 10 Geography Chapter 7 – Lifelines of National Economy Exercises includes Question/Answers which helps you to understand the topic covered in Lifelines of National Economy Class 10 Geography (Social Science), in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Geography Chapter 7 – Lifelines of National Economy – NCERT Exercises

Ques.1: Multiple choice questions.

(i) Which two of the following extreme locations are connected by the east-west corridor?

(a) Mumbai and Nagpur
(b) Silchar and Porbandar
(c) Mumbai and Kolkata
(d) Nagpur and Siligudi

(ii) Which mode of transportation reduces trans-shipment losses and delays?

(a) Railways
(b) Roadways
(c) Pipeline
(d) Waterways

(iii) Which one of the following states is not connected with the H.V.J. pipeline?

(a) Madhya Pradesh
(b) Maharashtra
(c) Gujarat
(d) Uttar Pradesh

(iv) Which one of the following ports is the deepest land-locked and well-protected port along the east coast?

(a) Chennai
(b) Paradwip
(c) Tuticorin
(d) Vishakhapatnam

(v) Which one of the following is the most important modes of transportation in India?

(a) Pipeline
(b) Railways
(c) Roadways
(d) Airways

(vi) Which one of the following terms is used to describe trade between two or more countries?

(a) Internal trade
(b) International trade
(c) External trade
(d) Local trade

Ans.1: Correct alternatives:

(i) (b) Silchar and Porbandar
(ii) (c) Pipelines
(iii) (b) Maharashtra
(iv) (d) Vishakhapatnam
(v) (b) Railways
(vi) (b) International trade

Ques.2: Answer the following questions in about 30 words.

(i) State any three merits of roadways.
(ii) Where and why is rail transport the most convenient means of transportation?
(iii) What is the significance of the border roads?
(iv) What is meant by trade? What is the difference between international and local trade?

Ans.2: Solutions:

(i) The merits of roadways are as follows:

• Road transport is economical for the transportation of people and goods over short distances.
• Road transportation provides door-to-door service, thereby reducing the loading and unloading cost.
• Roads can be constructed in much lower costs than the railways.

(ii) The railways are the most convenient means of transportation in the northern plains of India as these plains are flat stretches of land. The region houses a highly dense population with a large number of agricultural fields, thereby supporting transportation across the region.

(iii) The exchange of goods among people, states and countries are known as trade. The trade between countries is known as international trade, and the trade which takes place among the states, cities, towns, and villages of the same country is known as local trade.

Ques.3: Answer the following questions in about 120 words.

(i) Why are the means of transportation and communication called the lifelines of a nation and its economy?

(ii) Write a note on the changing nature of international trade in the last fifteen years.

Ans.3: Solutions:

(i) The means of transportation and communication are called lifelines of a nation for the following reasons:

• Transportation facilitates the movement of goods and people from one place to another.
• The demands for various goods, services and raw materials are fulfilled by transportations.
• Communication helps in the exchange of cultural diversity across regions and nations, connecting people across borders.
• The means of transportation and communication like railways, airways, newspapers, radio, cinema, etc. connect the world despite socio-cultural and language diversities.
• Transportation has helped increase communication. Hence, transport, communication and trade are complementary to each other.
• Most importantly, efficient transport and communication are prerequisites for the fast development of a nation.

(ii) The trade activities between two or more countries are termed as international trade. The nature of international trade has changed in a variety of ways in the last fifteen years. The trade has moved from being restricted to trade in goods to exchange of information, technology, and knowledge. The advent of globalisation has changed the picture of international trade. For example, India has emerged as a software giant and the field of information technology is growing increasingly in India. Also, in the last fifteen years, many countries have emerged in the global trade scenario, altering global trade relations.

NCERT Solutions for Class 10 Geography

• Chapter 1 – Resources and Development
• Chapter 4 – Agriculture
• Chapter 6 – Manufacturing Industries

NCERT Solutions for Class 10 Geography Chapter 6 – Manufacturing Industries (Social Science), contains solutions to various questions in Exercise for Chapter 6.  At the end of the Solutions, all the keywords which are important to understand Chapter 6 Manufacturing Industries Class 10 Geography, have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Geography Chapter 6 – Manufacturing Industries

NCERT Solutions for Class 10 Geography Chapter 6 – Manufacturing Industries Exercises includes Question/Answers which helps you to understand the topic covered in Manufacturing Industries Class 10 Geography (Social Science), in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Geography Chapter 6 – Manufacturing Industries – NCERT Exercises

Ques.1: Multiple choice questions.

(i) Which one of the following industries uses limestone as a raw material?

(a) Aluminium
(b) Cement
(c) Plastic
(d) Automobile

(ii) Which one of the following agencies markets steel for the public sector plants?

(a) HAIL
(b) SAIL
(c) TATA Steel
(d) MNCC

(iii) Which one of the following industries uses bauxite as a raw material?

(a) Aluminium Smelting
(b) Cement
(c) Paper
(d) Steel

(iv) Which one of the following industries manufactures telephones, computer, etc.

(a) Steel
(b) Electronic
(c) Aluminium Smelting
(d) Information Technology

Ans.1: Correct alternatives:

(i) (b) Cement
(ii) (b) SAIL
(iii) (a) Aluminium
(iv) (b) Electronics

Ques.2: Answer the following briefly in not more than 30 words.

(i) What is manufacturing?

(ii) Name any three physical factors for the location of the industry.

(iii) Name any three human factors for the location of an industry.

(iv) What are the basic industries? Give an example.

(v) Name the important raw materials used in the manufacturing of cement?

Ans.2:

(i) The processing of raw materials in industries to produce valuable goods in large quantities is known as manufacturing. For example, making furniture from wood.

(ii) The physical factors for the location of the industry are the availability of raw materials, availability of energy source and proximity to the market.

(iii) The human factors for the location of an industry are the availability of labour, availability of cheap labour and accessibility to the cities.

(iv) The industries whose produced goods are used as raw materials for other industries are known as basic industries. These industries provide the raw material for other industries. For example, iron and steel industry.

(v) The important raw materials used in the manufacturing of cement are limestone, silica and gypsum.

Ques.3: Write the answers to the following questions in 120 words.

(i) How are integrated steel plants different from mini steel plants? What problems does the industry face? What recent developments have led to a rise in the production capacity?

(ii) How do industries pollute the environment?

(iii) Discuss the steps to be taken to minimise environmental degradation by industry.

Ans.3:

(i) The integrated steel plants, as the name suggests, are large plants where all the simple and complex are carried out from the assembling of the raw material to its rolling, shaping, and steel making. Whereas mini steel plants are smaller. They use steel scrap and sponge iron. They use electric furnaces and re-rollers. Mild and alloy steel is produced by these plants. The problems faced by the industry are poor infrastructure, lower productivity of labour, high costs and limited availability of coking coal, and irregular supply of energy. The recent developments that have led to a rise in the production capacity of the iron and steel industry are liberalisation and foreign direct investment.

(ii) The industries pollute the environment in the following ways:

• The industries pollute the air by discharging the smoke into the atmosphere without filtering. Brick kilns and chemical factories emit smoke which consists of dust and other harmful particles and gases which are atmospheric pollutants. Burning of fossil fuels also adds toxins into the atmosphere. Major atmospheric pollutants are Sulphur dioxide and carbon monoxide.
• Water pollution is caused by the discharge of untreated industrial effluents into water bodies. This not only pollutes the water but is also harmful to aquatic life as it damages the aquatic ecosystems. Such industries include paper industries, petroleum refineries, fertilizer and pesticide industry, etc.
• Thermal pollution is another type of pollution whereby the hot water from the industries is discharged into lakes and other water bodies without cooling. This disturbs the natural temperature of the water bodies and results in the death of many plants and animals in the aquatic life, thereby damaging the aquatic ecosystem.
• Noise pollution caused by large industries mainly results in psychological and physiological impacts, it may cause feelings of irritation and anger and can cause impaired hearing, risk of high blood pressure, etc.

(iii) The steps to be taken to minimise the environmental degradation by industry can be:

• The water used in the industries could be recycled. This would minimise the use of water.
• Rainwater harvesting should be encouraged.
• The water should be treated before being discharged into the water bodies.
• Energy-efficient technology must be out to use.
• Filtering of smoke should be done before discharging into the atmosphere.
• Silencers could be used to minimise noise pollution.

NCERT Solutions for Class 10 Geography

• Chapter 1 – Resources and Development
• Chapter 4 – Agriculture
• Chapter 7 – Lifelines of National Economy

NCERT Solutions for Class 10 Geography Chapter 4 – Agriculture (Social Science), contains solutions to various questions in Exercise for Chapter 4.  At the end of the Solutions, all the keywords which are important to understand Chapter 4 Agriculture Class 10 Geography, have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Geography Chapter 4 – Agriculture

NCERT Solutions for Class 10 Geography Chapter 4 – Agriculture Exercises includes Question/Answers which helps you to understand the topic covered in Agriculture Class 10 Geography (Social Science), in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Geography Chapter 4 – Agriculture – NCERT Exercises

Ques.1: Multiple choice questions.

(i) Which one of the following describes a system of agriculture where a single crop is grown on a large area?

(a) Shifting Agriculture
(b) Plantation Agriculture
(c) Horticulture
(d) Intensive Agriculture

(ii) Which one of the following is a rabi crop?

(a) Rice
(b) Gram
(c) Millets
(d) Cotton

(iii) Which one of the following is a leguminous crop?

(a) Pulses
(b) Jawar
(c) Millets
(d) Sesamum

Ans.1: Correct alternatives:

(i) (b) Plantation Agriculture
(ii) (b) Gram
(iii) (a) Pulses

Ques.2: Answer the following questions in 30 words.

(i) Name one important beverage crop and specify the geographical conditions required for its growth.

(ii) Name one staple crop of India and the regions where it is produced.

(iii) Enlist the various institutional reform programmes introduced by the government in the interest of farmers.

(iv) The land under cultivation has got reduced day by day. Can you imagine its consequences?

Ans.2:

(i) An important beverage crop is tea. Tea was introduced in India by the British. Tea requires tropical or sub-tropical climate along with well-drained, fertile soils which are rich in humus and organic matter. The climate required for the cultivation of tea is warm and moist frost-free climate. Evenly distributed frequent showers ensure continuous growth of tea leaves. The major human factor is the availability of cheap and skilled labour. Major tea-producing states are Assam, West Bengal, Tamil Nadu, Kerala, Himachal Pradesh, Uttarakhand, Meghalaya, Andhra Pradesh and Tripura.

(ii) Rice is a major staple crop in India. It is cultivated in the plains of north and north-eastern India, coastal areas and the deltaic regions.

(iii) The various institutional reform programmes introduced by the government in the interest of farmers are:

• Minimum Support Price (MSP)
• Subsidy on fertilizers
• Crop insurance
• Grameen banks which provide low-interest loans

(iv) The reduction in land under cultivation can have many adverse consequences. Some of them can be a shortage of food crops. This would increase the price of the crops. Shortage of food crops and high demand will lead to an increase in the import of food grains. Lower production would also result in increased unemployment.

Ques.3: Answer the following questions in about 120 words.

(i) Suggest the initiative taken by the government to ensure the increase in agricultural production.

(ii) Describe the impact of globalisation on Indian agriculture.

(iii) Describe the geographical conditions required for the growth of rice.

Ans.3:

(i) The Government of India has taken several measures to ensure an increase in agricultural production. The major reforms were brought about in the 1960s and 1970s. the Green Revolution and White Revolution were introduced to boost Indian agriculture. In the 1980s and 1990s, a comprehensive plan for institutional and technical reforms was introduced. Grameen banks were set up to provide loans to farmers on low interest and crop insurance against flood, drought and other natural calamities were also provided. Schemes like Kisaan Credit Card (KCC) and Personal Accident Insurance Scheme (PAIS) were introduced. Agricultural and weather forecast programmes were introduced on television and radio to disseminate information to the farmers. The policy of Minimum Support Price (MSP) was introduced.

(ii) Globalisation has resulted in the world becoming a global village. And in this process, agriculture has also been majorly affected. Right from the time when India was ruled by the British, India is engaged in global trade. Back then, spices and other raw materials were exported from India to England to meet the demands of the growing population and increasing industries. It was because of the export of cotton from India that Manchester emerged as a major city in textile production. Globalisation has helped Indian agriculture to open up to new opportunities of trade and helped in its technological development. However, the advent of globalisation has also brought challenges to the Indian farmers to compete with the developed countries.

(iii) The geographical conditions required for the growth of rice are as follows:

• The cultivation of rice requires high temperatures, above 25⁰C.
• The rainfall levels need to be over 100 cm with high humidity.

Important Terms Relevant for NCERT Solutions for Class 10 Geography Chapter 4 – Agriculture

Rabi: This cropping season lasts from October to June. The crops are sown in October-December and harvested in April-June.

Kharif: These crops are sown with the onset of monsoon and harvested in September-October.

Zaid: This is a short cropping season between rabi and Kharif seasons.

Globalisation: The linking of the world with the advent of technology.

NCERT Solutions for Class 10 Geography

• Chapter 1 – Resources and Development
• Chapter 6 – Manufacturing Industries
• Chapter 7 – Lifelines of National Economy

NCERT Solutions for Class 10 Geography Chapter 1 – Resources and Development (Social Science), contains solutions to various questions in Exercise for Chapter 1.  At the end of the Solutions, all the keywords which are important to understand Chapter 1 Resources and Development Class 10 Geography, have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Geography Chapter 1 – Resources and Development

NCERT Solutions for Class 10 Geography Chapter 1 – Resources and Development Exercises includes Question/Answers which helps you to understand the topic covered in Resources and Development Class 10 Geography (Social Science), in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Geography Chapter 1 – Resources and Development – NCERT Exercises 

Ques.1: Multiple choice questions.

(i) Which one of the following types of a resource is iron ore?

(a) Renewable
(b) Biotic
(c) Flow
(d) Non-renewable

(ii) Under which of the following type of resource tidal energy cannot be put?

(a) Replenishable
(b) Human-made
(c) Abiotic
(d) Non-recyclable

(iii) Which one of the following is the main cause of land degradation in Punjab?

(a) Intensive cultivation
(b) Deforestation
(c) Over irrigation
(d) Overgrazing

(iv) In which one of the following states is terrace cultivation practised?

(a) Punjab
(b) Plains of Uttar Pradesh
(c) Haryana
(d) Uttarakhand

(v) In which of the following states black soil is predominantly found?

(a) Jammu and Kashmir
(b) Maharashtra
(c) Rajasthan
(d) Jharkhand

Ans.1: Correct alternatives:

(i) (d) Non-renewable
(ii) (a) Replenishable
(iii) (c) Over-irrigation
(iv) (d) Uttarakhand
(v) (b) Maharashtra

Ques.2: Answer the following questions in about 30 words.

(i) Name three states having black soil and the crop which is mainly grown in it.

(ii) What type of soil is found in the river deltas of the eastern coast? Give three main features of this type of soil.

(iii) What steps can be taken to control soil erosion in the hilly areas?

(iv) What are the biotic and abiotic resources? Give some examples.

Ans.2:

(i) Three states having black soil are Gujarat, Maharashtra and Madhya Pradesh. The crop mainly cultivated in black soil is cotton.

(ii) Alluvial soil is found in the river deltas of the eastern coast. This is the most fertile type of soil and is very suitable for the cultivation of crops. It also consists of varying amounts of sand, silt and clay. Also, the alluvial soils are rich in lime, phosphoric acid and potash which makes it suitable for the cultivation of sugarcane and paddy.

(iii) To prevent soil erosion in India, the following steps can be taken:

• Contour farming
• Strip cropping
• Terrace farming

(iv) The resources which are obtained from the environment and have life are known as biotic resources. For example, humans, crops, fish, livestock, etc.
The non-living resources are known as abiotic resources. For example, rocks, metals, etc. 

Ques.3: Answer the following questions in about 120 words.

(i) Explain the land use pattern in India and why has the land under forest not increased much since 1960-61?

(ii) How have technical and economic development led to more consumption of resources?

Ans.3:

(i) The land use pattern is majorly determined by physical and human factors. The physical factors include topography, climate, soil, and human factors include population density, technology and society and culture. The land use pattern in India is very diverse. Land under permanent pasture has decreased over time. The net sown area also varies from one state to another in India. In Punjab and Haryana, the net sown area is over 80% and in the states of Arunachal Pradesh, Mizoram, Manipur and Andaman and the Nicobar Islands. A part of the land in Indi ais also termed as wasteland which includes rocky land and deserted areas. The land put to non-agricultural uses has also increased over time. However, land under forest cover has not shown any significant growth since 1960-61. It is because the land under agriculture and non-agricultural uses has increased to meet the demands of the growing population.

(ii) The technical and economic development has led to increased consumption of resources in the following ways:

• As production was increased to fulfil the demands of the growing population, more and more resources are exploited.
• Improvement in technology has increased the human capacity to harness natural resources.

Important Terms Relevant for NCERT Solutions for Class 10 Geography Chapter 1 – Resources and Development

Resource: The things available in our environment that can be used to satisfy our needs is a resource. A resource is technologically accessible, economically feasible and culturally acceptable.

Biotic: The resources which are obtained from the environment and have life are known as biotic resources.

Abiotic: The non-living resources are known as abiotic resources

Renewable: These resources can be renewed by a physical, chemical or mechanical process.

Non-renewable: These resources take a long time to replenish themselves. These take nearly a million years to rejuvenate.

Soil erosion: The removal of topsoil is known as soil erosion.

NCERT Solutions for Class 10 Geography

• Chapter 4 – Agriculture
• Chapter 6 – Manufacturing Industries
• Chapter 7 – Lifelines of National Economy

NCERT Solutions for Class 10 Science Chapter 2 – Acids, Bases and Salts, contains solutions to various questions in Exercises for Chapter 2. Acids, Bases and Salts Class 10 NCERT Solutions have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Science Chapter 2 – Acids, Bases and Salts

NCERT Solutions for Class 10 Science Chapter 2 – Acids, Bases and Salts – NCERT Exercises

NCERT Solutions for Class 10 Science Chapter 2 – Acids, Bases and Salts, Exercises includes all in text and exercise solved questions which helps you to understand the topic covered in Chapter 2 Acids, Bases and Salts Class 10, in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Science Chapter 2 – Intext Questions – Page 18

1. You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?

Answer:

We can identify the contents of each test tube by using only litmus paper because it changes color when it comes in contact with base or distilled water. If we dip the red litmus in a basic solution, the color changes to blue color.

There is no change in the color if it is dipped in acid, and changes to purple color when dipped in distilled water.

Intext Questions – Page 22

1. Why should curd and sour substances not be kept in brass and copper vessels?

Answer:

Acids react with metals. Curd and sour food substances are acidic in nature. If kept in metal vessels, they would react with the metal and form compounds that are poisonous for humans.

2.Which gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas?

Answer:

Acid and metals react to form salt and Hydrogen. Hence, hydrogen gas is released.

We can test the presence of hydrogen by inserting a lighted splint into the test tube. The lighted splint extinguishes with a “pop” sound if hydrogen gas is present.

3. Metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.

Answer:

The metal compound A is Calcium Carbonate and the gas produced with effervescence is carbon dioxide. The reaction between calcium carbonate and HCl is:

Intext Questions – Page 25

1. Why do HCl, HNO₃ , etc. show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?

Answer:

When added in water, the hydrogen ions dissociate from acid molecules and show acidic nature. In compounds like alcohol and glucose, the hydrogen atoms remain attached to them because of stronger covalent bonds.

2. Why does an aqueous solution of an acid conduct electricity?

Answer:

An aqueous solution of acid contains free charged particles called ions. These ions move freely and carry the electricity making aqueous solution of an acid conduct electricity.

3. Why does dry HCl gas not change the colour of the dry litmus paper?

Answer:

There are no free hydrogen ions in dry HCl. Hence, dry HCl does not show acidic nature and does not change the color of the dry litmus.

4. While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?

Answer:

Water and concentrated acids react violently and release huge amount of heat which can result in explosion. To avoid any mishap, it is recommended to add acid to water and not water to the acid.

5. How is the concentration of hydronium ions (H₃O⁺) affected when a solution of an acid is diluted?

Answer:

The concentration of hydronium ions decreases when diluted with water.

6. How is the concentration of hydroxide ions (OH ^–) affected when excess base is dissolved in a solution of sodium hydroxide?

Answer:

Hydroxide ions increase initially when excess base is dissolved in a solution of sodium hydroxide. However, after a point, the saturation limit is achieved and the concentration of hydroxide ions is not affected any further.

Intext Questions – Page 28

1. You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?

Answer:

The pH of any substance is inversely proportional to the hydrogen ion concentration. The more the hydrogen present in the solution, the more acidic the solution is. Since solution A has less pH, it contains more hydrogen ions and is acidic in nature, and solution B with pH of 8 is basic.

2. What effect does the concentration of H⁺ (aq) ions have on the nature of the solution?

Answer:

The concentration of H⁺ (aq) ions determines the nature of the solution. If the concentration of H⁺ (aq) ions is more, the solution is said to be more acidic and  if the concentration is low, its either less acidic or basic.

3. Do basic solutions also have H⁺ (aq) ions? If yes, then why are these basic?

Answer:

Yes, basic solutions also have H⁺ ions. These solutions are basic because the concentration of OH^–  (Hydroxide ) ions is more than that of H⁺ ions.

4. Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?

Answer:

If the soil is acidic in nature and has a pH below 7, it should be treated with basic compounds like quick lime, slaked lime or chalk.

NCERT Solutions for Class 10 Science Chapter 2 – Intext Questions – Page 33

1. What is the common name of the compound CaOCl₂?

Answer:

Bleaching Powder

2. Name the substance which on treatment with chlorine yields bleaching powder.

Answer:

Calcium Hydroxide [Ca(OH)₂]

3. Name the sodium compound which is used for softening hard water.

Answer:

Sodium Carbonate

4. What will happen if a solution of sodium hydrocarbonate is heated? Give the equation of the reaction involved.

Answer:

If we heat a solution of sodium hydrocarbonate, sodium carbonate will be released and carbon dioxide gas will be liberated.

5.Write an equation to show the reaction between Plaster of Paris and water.

Answer:

NCERT Solutions for Class 10 Science Chapter 2 – NCERT Exercise – Page 34

1. A solution turns red litmus blue, its pH is likely to be

(a) 1
(b) 4
(c) 5
(d) 10

Answer:

(d) 10 (pH>7 are bases)

2. A solution reacts with crushed egg-shells to give a gas that turns lime-water milky. The solution contains

(a) NaCl
(b) HCl
(c) LiCl
(d) KCl

Answer:

(b) HCl ( CaCO₃ + 2HCl ⇒ CaCl₂ + H₂O + CO₂ )

3. 10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HCl. If we take 20 mL of the same solution of NaOH, the amount HCl solution (the same solution as before) required to neutralise it will be

(a) 4 mL
(b) 8 mL
(c) 12 mL
(d) 16 mL 

Answer:

(c) 16mL (8×2=16mL)

4. Which one of the following types of medicines is used for treating indigestion?

(a) Antibiotic
(b) Analgesic
(c) Antacid
(d) Antiseptic

Answer:

(c). Antacid    (Indigestion is due to excess production of acid in stomach.)

5. Write word equations and then balanced equations for the reaction taking place when –

(a) dilute sulphuric acid reacts with zinc granules.

(b) dilute hydrochloric acid reacts with magnesium ribbon.

(c) dilute sulphuric acid reacts with aluminium powder.

 (d) dilute hydrochloric acid reacts with iron filings.

Answer:

(a) dilute sulphuric acid + zinc → Zinc Sulphate + Hydrogen Gas

H₂SO₄(aq) + Zn → ZnSO₄(aq) + H₂(g)

(b) dilute Hydrochloric + Magnesium → Magnesium Chloride + Hydrogen Gas

2HCl(aq) + Mg → MgCl₂(aq) + H₂(g)

(c) dilute Sulphuric Acid + Aluminium → Aluminium Sulphate + Hydrogen Gas

3H₂SO₄(aq) + 2Al(s) → Al₂(SO₄)₃(aq) + 3H₂(g)

(d) dilute Hydrochloric Acid + Iron → Ferrous Chloride + Hydrogen Gas

6HCl(aq) + 3Fe(s) → 3FeCl₂(aq) + 3H₂(g)

6. Compounds such as alcohols and glucose also contain hydrogen but are not categorised as acids. Describe an activity to prove it.

Answer:

In compounds such as alcohols and glucose even though have hydrogen, the hydrogen in these compounds are not free and do not from ions as they are attached to the compounds with strong covalent bonds. Since acidic nature is determined by number free hydrogen ions, these compounds are not considered acidic. If we mix glucose powder in water and put blue litmus paper, it will not turn red. Hence, it will be proved that glucose is not acidic.

7. Why does distilled water not conduct electricity, whereas rain water does?

Answer:

Distilled water does not have any free ions. Rain water contains many ions as various gases mix with it to from acidic or basic solutions. For example, carbon dioxide mixes with water to from carbolic acid. This solution has ions of hydrogen and carbonate that conduct electricity easily.

8. Why do acids not show acidic behaviour in the absence of water?

Answer:

To produce hydrogen ions, the presence of water is important. It is these hydrogen ions that make any substance acidic. Hence, acids do not show acidic behaviour in the absence of water.

9. Five solutions A,B,C,D and E when tested with universal indicator showed pH as 4,1,11,7 and 9, respectively. Which solution is

(a) neutral?

(b) strongly alkaline?

(c) strongly acidic?

(d) weakly acidic?

(e) weakly alkaline?

Arrange the pH in increasing order of hydrogen-ion concentration.

Answer:

In increasing order of hydrogen ion concentration:

pH 11(b) < pH 9(e) < pH 7(a) < pH 4(d) < pH 1 (c)

PH11 – Strongly alkaline

pH9 – weakly alkaline

PH7 – Neutral

pH4 – Weakly acidic

pH1 – Strongly acidic

10. Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A, while acetic acid (CH₃COOH) is added to test tube B. Amount and concentration taken for both the acids are same. In which test tube will the fizzing occur more vigorously and why?

Answer:

Fizzing occurs because of the production of the hydrogen gas due to reaction between an acid and the magnesium ribbon. Since HCl is a very strong acid and acetic acid is weak, there is a lot of liberation of hydrogen gas from test tube A. Therefore, more fizzing takes place in test tube A.

11. Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain your answer.

Answer:

Curd contains lactic acid. Since acid has less pH, the pH of milk will be reduced as it changes to curd.

12. A milkman adds a very small amount of baking soda to fresh milk.

(a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline?
(b) Why does this milk take a long time to set as curd?

Answer:

(a) To avoid/delay the souring of milk, the milkman adds a very small amount of baking soda to make it slightly alkaline.

(b) This milk takes long time to set into curd because the lactic acid produced in the process of making curd first neutralises the pH and then the pH is reduced to turn milk to curd.

13. Plaster of Paris should be stored in a moisture-proof container. Explain why?

Answer:

Plaster of Paris should be stored in moisture-proof container because moisture can affect Plaster of Paris by slowing down the setting of the plaster as a result of hydration and could not be used.

14. What is a neutralisation reaction? Give two examples.

Answer:

When an acid reacts with base it gives out water and salt. Such a reaction is called as neutralisation reaction.

Examples: NaOH + HCl → NaCl + H₂O

Mg(OH)₂ + H₂CO₃ → MgCO₃ + 2H₂O

15. Give two important uses of washing soda and baking soda.

Answer:

Uses of Washing Soda:

1. Water softener for laundry purposes.
2. Electrolyte

Uses of Baking Soda:

1. Can be used to test acidity of the soil.
2. Washing car and other machine parts.

Topics Covered in Acids, Bases and Salts Class 10 Science

• Chemical Properties of Acids and Bases
  • Acids and Bases in the Laboratory
  • How do Acids and Bases React with Metals?
  • How do Metal Carbonates and Metal Hydrogencarbonates React with Acids?
  • How do Acids and Bases React with each other?
  • Reaction of Metallic Oxides with Acids
  • Reaction of a Non-metallic Oxide with Base
• Common Properties in Acids and bases
• Acid or Base Solutions
• Salts

NCERT Solutions for Class 10 Science Chapter 1 – Chemical Reactions and Equations, contains solutions to various questions in Exercises for Chapter 1. Chemical Reactions and Equations Class 10 NCERT Solutions have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Science Chapter 1 – Chemical Reactions and Equations

NCERT Solutions for Class 10 Science Chapter 1 – Chemical Reactions and Equations – NCERT Exercises

NCERT Solutions for Class 10 Science Chapter 1 – Chemical Reactions and Equations, Exercises includes all in text and exercise solved questions which helps you to understand the topic covered in Chapter 1 Chemical Reactions and Equations Class 10, in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Science Chapter 1 – Intext Questions – Page 6

1. Why should a magnesium ribbon be cleaned before burning in air?

Answer:

The upper layer of magnesium reacts with air to form magnesium oxide. In order to get the required reaction, this layer must be removed. Hence, to remove the layer of magnesium oxide, magnesium ribbon is cleaned before burning.

2. Write the balanced equation for the following chemical reactions.

(i) Hydrogen + Chlorine → Hydrogen chloride

(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride

(iii) Sodium + Water → Sodium hydroxide + Hydrogen

Answer:

(i) H₂ + Cl₂ → 2HCl

(ii) 3BaCl₂ + Al₂(SO₄)₃ → 2AlCl₃ + 3BaSO₄   

(iii) 2Na + 2H₂O → 2NaOH + H₂               

3. Write a balanced chemical equation with state symbols for the following reactions.

(i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.

(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

Answer:

Intext Questions – Page 10

1. A solution of a substance ‘X’ is used for whitewashing.

(i) Name the substance ‘X’ and write its formula.

(ii) Write the reaction of the substance ‘X’ named in (i) above with water.

Answer:

(i) Quick lime or Calcium oxide is used in whitewashing. Hence, the substance ‘X’ is CaO.

(ii) The chemical reaction involved is:

CaO + H₂O ⇒ Ca(OH)₂

2. Why is the amount of gas collected in one of the test tubes in Activity 1.7 double of the amount collected in the other? Name this gas.

Answer:

In Activity 1.7, the process of hydrolysis takes place and breaks water molecules into H₂ and O₂ gases. Since two molecules of hydrogen are released and only one molecule of Oxygen is released, the test tube with Hydrogen has double the amount than Oxygen.

Intext Questions – Page 13

1. Why does the colour of copper sulphate solution change when an iron nail is dipped in it?

Answer:

Iron is more reactive than copper. When iron nail is dipped in copper sulphate, it displaces copper and forms ferrous sulphate. Hence, the colour of the solution changes. We can see it in the given reaction:

Fe + CuSO₄  ⇒ FeSO₄ + Cu

2. Give an example of a double displacement reaction other than the one given in Activity 1.10.

Answer:

Another example of double displacement reaction is the reaction between silver nitrate and sodium chloride.

AgNO₃ + NaCl ⇒ AgCl + NaNO₃

3. Identify the substances that are oxidised and the substances that are reduced in the following reactions.

(i) 4Na(s) + O₂ (g) → 2Na₂O(s)

(ii) CuO(s) + H₂ (g) → Cu(s) + H₂O(l)

Answer:

(i) Sodium (Na) is oxidised because Oxygen (O₂) is added.

(ii) Copper (Cu) is reduced because Hydrogen (H₂) is added.

NCERT Solutions for Class 10 Science Chapter 1 – NCERT Exercise – Page 14

1. Which of the statements about the reaction below are incorrect?

2PbO(s) + C(s) ⇒ 2Pb(s) + CO₂ (g)

(a) Lead is getting reduced.
(b) Carbon dioxide is getting oxidised.
(c) Carbon is getting oxidised.
(d) Lead oxide is getting reduced.

(i) (a) and (b)
(ii) (a) and (c )
(iii) (a), (b) and (c)
(iv) All

Answer:

(i) (a) and (b)

2. Fe₂O₃ + 2Al → Al₂O₃ + 2Fe

The above reaction is an example of a

(a) combination reaction.
(b) double displacement reaction.
(c) decomposition reaction.
(d) displacement reaction.

Answer:

(d) Displacement reaction

3. What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer.

(a) Hydrogen gas and iron chloride are produced.

(b) Chlorine gas and iron hydroxide are produced.

(c) No reaction takes place.

(d) Iron salt and water are produced.

Answer:

(a) Hydrogen gas and iron chloride are produced

4. What is a balanced chemical equation? Why should chemical equations be balanced?

Answer:

The chemical equation in which there are equal number of atoms on both side of the equation is called a balanced chemical equation.

The equations need to balanced in order to justify the laws of conservation of mass and energy.

5. Translate the following statements into chemical equations and then balance them.

(a) Hydrogen gas combines with nitrogen to form ammonia.

(b) Hydrogen sulphide gas burns in air to give water and sulpur dioxide.

(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.

(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

Answer:

(a) Unbalanced: H₂ + N₂ → NH₃

Balanced: 3H₂ + N₂ → 2NH₃

(b) Unbalanced: H₂S + O₂ → H₂O + SO₂

Balanced: 2H₂S + 3O₂ → 2H₂O + 2SO₂

(c)  Unbalanced:

BaCl₂ + Al₂(SO₄)₃ → AlCl₃ + BaSO₄

Balanced: 3BaCl₂ + Al₂(SO₄)₃ → 2AlCl₃ + 3BaSO₄

(d) Unbalanced: K + H₂O → KOH + H₂

Balanced:  2K + 2H₂O → 2KOH + H₂

6. Balance the following chemical equations.

(a) HNO₃ +Ca(OH)₂ → Ca(NO₃ ) ₂ + H₂O
(b) NaOH + H₂SO₄ → Na2SO4 + H2O
(c) NaCl + AgNO₃ → AgCl + NaNO₃
(d) BaCl₂ + H₂ SO₄ → BaSO₄ + HCl

Answer:

(a) 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O
(b) 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
(c) NaCl + AgNO₃ → AgCl + NaNO₃
(d) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

7. Write the balanced chemical equations for the following reactions.

(a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
(b) Zinc + Silver nitrate → Zinc nitrate + Silver
(c) Aluminium + Copper chloride → Aluminium chloride + Copper
(d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride

Answer:

(a) 2Ca(OH)₂ + 2CO₂ → 2CaCO₃ + 2H₂O
(b) Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag
(c) 2Al + 3CuCl₃ → 2AlCl₃ + 3Cu
(d) BaCl₂ + K₂SO₄ → BaSO₄ + 2KCl

8. Write the balanced chemical equation for the following and identify the type of reaction in each case.

(a) Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)
(b) Zinc carbonate(s) → Zinc oxide(s) + Carbon dioxide(g)
(c) Hydrogen(g) + Chlorine(g) → Hydrogen chloride(g)
(d) Magnesium(s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen(g)

Answer:

(a) 2KBr + BaI₂ → 2KI + BaBr₂ (Double Displacement Reaction)
(b) ZnCO₃ → ZnO + CO₂ (Decomposition Reaction)
(c) H₂ + Cl₂ → 2HCl (Combination Reaction)
(d) Mg + 2HCl → MgCl₂ + H₂ (Displacement Reaction)

9. What does one mean by exothermic and endothermic reactions? Give examples.

Answer:

An exothermic reaction is one in which system releases its energy in the surroundings. An explosion is the example of exothermic reaction.

Endothermic reactions absorb energy from their surroundings. Melting of ice is an example of endothermic reaction.

10. Why is respiration considered an exothermic reaction? Explain.

Answer:

To get energy we need food. Glucose is formed by breaking down food during digestion. Glucose combines with oxygen to give us energy. The oxygen is taken from our cells by the process of respiration. When energy is released, it is called a exothermic reaction. Hence, we can say that respiration is an exothermic reaction.

Reaction: C₆H₁₂O₆ + 6O₂   →   6CO₂ + 6H₂O + Energy

11. Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.

Answer:

Combination reactions form a larger molecule from two or more smaller molecules. In decomposition reaction, the larger molecule is split into two or more smaller molecules. Since the opposite actions are occurring, we can say that combination and decomposition reactions are opposite.

Decomposition Reaction:

CaCO₃ + Energy ⇒ CaO + CO₂

Combination Reaction:

NH₂ (g) + HCl (g) ⇒ NH₄Cl (g)

12. Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.

Answer:

Decomposition by Heat (Thermolysis):

2KClO₃ + Heat  ⇒ 2KCl + 3O₂

Decomposition by Light (Photolysis):

Decomposition by Electricity (Electrolysis):

13. What is the difference between displacement and double displacement reactions?

Write equations for these reactions.

Answer:

When one or more reactive substance displaces a less reactive substance from its salt solution, it is called displacement reaction.

When two such displacements take place in the same reaction, it is called double displacement reactions.

Example of displacement reaction:

Mg + 2HCl ⇒ MgCl₂ + H₂

Example of double displacement reaction:

2KBr + BaI₂ ⇒ 2KI + BaBr₂

14.In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.

Answer:

The reaction involved is:

15. What do you mean by a precipitation reaction? Explain by giving examples.

Answer:

Two solutions containing soluble salts are combined resulting in a double displacement reaction. When one of the compounds formed is in solid form (insoluble in aqua), it settles down at the bottom of the container. This solid is known as the precipitate and the reaction is called the precipitation reaction. Examples of a precipitation reaction are:

16. Explain the following in terms of gain or loss of oxygen with two examples each.

(a) Oxidation
(b) Reduction

Answer:

(a) When Oxygen is added to an element to from its oxide, the element is said to be oxidised and is called oxidation.

H₂S + O₂ ⇒ H₂O + SO₂

(b) When Oxygen is removed from a compound, it is said to be reduced and the process is called reduction.

2HgO ⇒ 2Hg + O₂

17. A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.

Answer:

The brown coloured element ‘X’ is Copper (Cu). On heating, it reacts with oxygen in the atmosphere and makes black Copper Oxide.

Reaction:

Cu(s) + O₂(g) ⇒ 2CuO(s)

18.Why do we apply paint on iron articles?

Answer:

Iron reacts with atmospheric oxygen and moisture to form rust. To avoid iron articles from rusting, they are painted.

19.Oil and fat containing food items are flushed with nitrogen. Why?

Answer:

When oil and fats react with oxygen they undergo rancidity making them taste and smell unpleasant. To avoid this condition, nitrogen is flushed into food packets.

20. Explain the following terms with one example each. (a) Corrosion (b) Rancidity

Answer:

(a) Corrosion: When refined metals oxidise to form more stable compounds, such as oxides, it is called corrosion. The metals lose its properties and gradually turn into their oxides. Rusting of iron is a good example of corrosion. It can be prevented by painting the metals or by the process of galvanization.

(b) Rancidity: When oil and fats react with oxygen they undergo rancidity making them taste and smell unpleasant. To reduce this, we can keep food in refrigerators or add nitrogen in food packets.

Topics Covered in Chemical Reactions and Equations Class 10 Science

• Chemical Equations
  • Writing a Chemical Equation
  • Balanced Chemical Equations
• Types of Chemical Reactions
  • Combination Reaction
  • Decomposition Reaction
  • Displacement Reaction
  • Double Displacement Reaction
  • Oxidation and Reduction
  • Oxidation Reactions

NCERT Solutions for Class 10 Science Chapter 3 Metals and Non metals, contains solutions to various questions in Exercises for Chapter 3. Metals and Non metals Class 10 NCERT Solutions have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Science Chapter 3 Metals and Non metals

NCERT Solutions for Class 10 Science Chapter 3 – Metals and Non metals – NCERT Exercises

NCERT Solutions for Class 10 Science Chapter 3 – Metals and Non metals, Exercises includes all in text and exercise solved questions which helps you to understand the topic covered in Chapter 3 Metals and Non metals Class 10, in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Science Chapter 3 – Intext Questions – Page 40

1. Give an example of a metal which

(i) is a liquid at room temperature.

(ii) can be easily cut with a knife.

(iii) is the best conductor of heat.

(iv) is a poor conductor of heat.

Answer:

(i) Mercury

(ii) Sodium

(iii) Silver

(iv) Mercury

2. Explain the meanings of malleable and ductile.

Answer:

Malleable: If a metal holds the property of being able to beaten into sheets, it is called malleable.

Ductile: If a metal holds the property of being able to drawn into sheets, it is called ductile.

Intext Questions – Page 46

1. Why is sodium kept immersed in kerosene oil?

Answer:

Sodium reacts highly with oxygen at room temperature making it a volatile metal. Hence, it is kept immersed in kerosene.

2. Write equations for the reactions of

(i) iron with steam

 (ii) calcium and potassium with water

Answer:

(i) 3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

(ii) Ca(s) + 2H₂O(I) → Ca(OH)₂(aq) + H₂(g)

2K(s) + 2H₂O(I) → 2KOH(aq) + 2H₂(g)

3. Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows.

Use the Table above to answer the following questions about metals A, B, C and D.

(i) Which is the most reactive metal?

(ii) What would you observe if B is added to a solution of Copper(II) sulphate?

(iii) Arrange the metals A, B, C and D in the order of decreasing reactivity.

Answer:

(i) Metal B

(ii) When metal B is added to copper (II) sulphate solution, a displacement reaction will take place and we observe that the blue colour of copper (II) sulphate solution will fade and a red-brown deposit of copper will be formed on metal B.

(iii) The correct order is:

B > A > C > D

Metal B is the most reactive because it displaces iron from its salt solution.

Metal A is less reactive because it displaces copper from its salt solution.

Metal C is still less reactive because it can displace only silver from its salt solution.

Metal D is the least reactive because it did not displace any metal from its salt solution.

4. Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H₂SO₄ .

Answer:

Hydrogen gas is liberated when dilute HCl is added to a reactive metal. The reaction can be given as:

Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g)

5. What would you observe when zinc is added to a solution of iron(II) sulphate? Write the chemical reaction that takes place.

Answer:

Since zinc is more reactive than iron, we observe that zinc displaces iron from its salt solution and converts the pale green solution into a colorless solution.

FeSO₄ + Zn → ZnSO₄ + Fe(s)

NCERT Solutions for Class 10 Science Chapter 3 – Intext Questions – Page 49

1. (i) Write the electron-dot structures for sodium, oxygen and magnesium.

(ii) Show the formation of Na₂O and MgO by the transfer of electrons.

(iii) What are the ions present in these compounds?

Answer:

(i) Sodium:

Oxygen:

Magnesium:

(ii) Formation of Na₂O

Oxygen requires two atoms to complete its octet. Two sodium atoms give their outermost single electron and sodium oxide is formed.

Formation of MgO

Magnesium has two electrons in the outermost orbit and they share it with Oxygen to form magnesium oxide.

(iii) Sodium oxide has two Na⁺ ions and one O^2-

Magnesium oxide has one Mg²⁺ and one O^2- ions.

2.Why do ionic compounds have high melting points?

Answer:

Heat is required to overcome the forces of attraction. Since ionic compounds have both positive and negative ions, they form a very strong bond of attraction which is difficult to overcome. Hence, ionic compounds have high melting points.

Intext Questions – Page 53

1. Define the following terms.

(i) Mineral
(ii) Ore
(iii) Gangue

Answer:

(i) Mineral: Compounds found naturally in earth’s crust are called minerals. Example: Alums.

(ii) Ore: Ores contain metals. Ores are purified and metals are extracted from them. For example, Magnetite is an iron ore, and Bauxite is an aluminium ore.

(iii) Gangue: The impurities and contaminates present in the ores like sand, etc are called gangue.

 2. Name two metals which are found in nature in the free state.

Answer:

Two metals found in nature in the free state are (i) gold (ii) silver.

3. What chemical process is used for obtaining a metal from its oxide?

Answer:

Reduction is the process of removing oxygen. Reduction is the chemical process used to obtain a metal from its oxide. For example, Lead oxide is reduced to lead by heating it with carbon:

PbO + C ⇒ Pb + CO

NCERT Solutions for Class 10 Science Chapter 3 – Intext Questions – Page 55

1. Metallic oxides of zinc, magnesium and copper were heated with the following metals.

In which cases will you find displacement reactions taking place?

Answer:

If a metal is more reactive than other, it displaces it in its oxide. Reactivity in increasing order is: Copper < Zinc < Magnesium. Following the reactivity series, we can fill the table.

2. Which metals do not corrode easily?

Answer:

Gold and Platinum do not corrode easily.

3. What are alloys?

Answer:

Alloys are homogeneous mixtures of two or more metals, or a metal and a non-metal. For example, steel is an alloy of iron, carbon, and small amounts of other elements like silicon, copper, etc.

NCERT Solutions for Class 10 Science Chapter 3 – NCERT Exercise – Page 56

1. Which of the following pairs will give displacement reactions?

(a) NaCl solution and copper metal

(b) MgCl₂ solution and aluminium metal

(c) FeSO₄ solution and silver metal

(d) AgNO₃ solution and copper metal.

Answer:

(d). AgNO₃ solution and copper metal.

Copper displaces silver.

2. Which of the following methods is suitable for preventing an iron frying pan from rusting?

(a) Applying grease

(b) Applying paint

(c) Applying a coating of zinc

(d) All of the above

Answer:

(c) Applying a coating of Zinc.

It is not appropriate to coat frying pans with grease and paint, so zinc is used to coat them to avoid rusting.

3. An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be

(a) calcium

(b) carbon

(c) silicon

(d) iron

Answer:

(a) Calcium

Calcium reacts with oxygen to give calcium oxide which has a high melting point and is soluble in water.

4. Food cans are coated with tin and not with zinc because

(a) zinc is costlier than tin.

(b) zinc has a higher melting point than tin.

(c) zinc is more reactive than tin.

(d) zinc is less reactive than tin.

Answer:

(c ) zinc is more reactive than tin.

5. You are given a hammer, a battery, a bulb, wires and a switch.

(a) How could you use them to distinguish between samples of metals and non-metals?

(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.

Answer:

(a) We know metals are good conductor of electricity, are malleable and ductile, where as non-mental do not conduct electricity and are break when hit with force. We can use the given things to distinguish metals on the basis of these properties. If we connect the metals with battery and bulb using the wires to form a circuit, the bulb will glow. Similarly, if we beat a metal with a hammer, it will be easy to beat it into sheets whereas a non-metal will break from the force.

(b) These tests can be quite useful in distinguishing the metals and non-metals on the basis of their physical properties.

6. What are amphoteric oxides? Give two examples of amphoteric oxides.

Answer:

Oxides that can react with both acids and bases to give salt and water are called amphoteric oxides. Aluminium oxide (Al₂O₃) is an amphoteric oxide.

7. Name two metals which will displace hydrogen from dilute acids, and two metals which will not.

Answer:

Two metals which will displace hydrogen from dilute acids are: 1. Magnesium 2. Zinc. (Very reactive in nature.)

Two metals that will not displace hydrogen from dilute acids are: 1. Gold 2. Silver. (Less reactive, noble metals.)

8. In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?

Answer:

In electrolytic refining of metal M, the electrolyte is a solution of suitable salt of metal M, the anode is the block of impure metal M, and the cathode is a thin strip of pure metal M.

9. Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it, as shown in figure below.

(a) What will be the action of gas on
(i) dry litmus paper?
(ii) moist litmus paper?
(b) Write a balanced chemical equation for the reaction taking place.

Answer:

(a) The heating of sulphur powder results in formation of sulphur dioxide gas which is collected in test tube.
(i) It will have no effect on dry litmus paper as no acid will be formed.
(ii) After coming in contact with moisture, sulphur dioxide turns into sulfurous acid and turns blue litmus into green.

(b) The reactions involved are:

S(s) + O₂(g) → SO₂(g)

SO₂(g) + H₂O(l) →H₂SO₃(aq)

10. State two ways to prevent the rusting of iron.

Answer:

Two ways to prevent rusting of iron are:

(i) Coating things made from iron with paint and grease.
(ii) Galvanization can be done to apply a zinc coating on the iron.

11. What type of oxides are formed when non-metals combine with oxygen?

Answer:

Either acidic or neutral oxides are formed when non-metals combine with oxygen. For example:

Nitrogen combines with oxygen to form acidic N₂O₅
Carbon combines with oxygen to from neutral CO.

12. Give reasons

(a) Platinum, gold and silver are used to make jewellery.

(b) Sodium, potassium and lithium are stored under oil.

(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.

(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.

Answer:

(a) Since gold and silver are not very reactive, they do not react with the air and water present in the atmosphere and do not lose their shine easily. They last longer. Hence, they are used to make jewellery.

(b) Sodium, potassium and lithium are highly reactive metals, at room temperature, they combine with water present in atmosphere and burn violently. To store them safely, they are kept in oil (usually kerosene).

(c) The surface of aluminium reacts with oxygen in present in atmosphere to form a thin layer of aluminium oxide. This layer of aluminium oxide is very stable and prevents aluminium from reacting any further. Because of this, we can use aluminium to make vessels as it is cheap and a good conductor of electricity.

(d) It is easier to extract metals from their oxides by the methods or roasting and calcination than extracting metals from their carbonates and sulphides. Hence, we convert metals into oxides during the process of extraction.

13. You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.

Answer:

Copper reacts with carbon dioxide and moisture present in air to form a greenish layer of copper carbonate (CuCO₃.Cu(OH)₂). The acids present in these sour substances dissolve this layer, restoring the color and shine of the metal.

14. Differentiate between metal and non-metal on the basis of their chemical properties.

Answer:

The difference between metals and non-metals on the basis of their chemical properties are as given:

15. A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?

Answer:

The solution used is acidic in nature. Popularly known as Aqua Regia (royal water in Latin), it is mixture of concentrated HCl and concentrated HNO₃ in the ration 3:1. It can easily react with noble metals like gold and dissolve their upper layer which results in lesser weight.

16. Give reasons why copper is used to make hot water tanks and not steel (an alloy of iron).

Answer:

Copper does not react with water or steam whereas steel, which contains iron easily reacts with water to form rust and corrodes. So, we use copper to make hot water tanks.

Topics Covered in Metals and Non metals Class 10 Science

• Physical properties of metals
• Physical properties of non-metals
• Chemical Properties of Metals
• Reactions of Metals and Non-Metals
• Occurrence of Metals
• Corrosion

NCERT Solutions for Class 10 Science Chapter 4 – Carbon and its Compounds, contains solutions to various questions in Exercises for Chapter 4. Carbon and its Compounds Class 10 NCERT Solutions have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Science Chapter 4 – Carbon and its Compounds

NCERT Solutions for Class 10 Science Chapter 4 – Carbon and its Compounds – NCERT Exercises

NCERT Solutions for Class 10 Science Chapter 4 – Carbon and its Compounds, Exercises includes all in text and exercise solved questions which helps you to understand the topic covered in Chapter 4 Carbon and its Compounds Class 10, in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Science Chapter 4 – Intext Questions – Page 61

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

Answer:

Two oxygen atoms share their two outermost electrons with carbon to complete their octet.

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

Answer:

Intext Questions – Page 68

1. How many structural isomers can you draw for pentane?

Answer:

There are three structural isomers of pentane.

1. n-pentane

2. 2-methylbutane

3. 2,2 – dimethylpropane

2. What are the two properties of carbon which lead to the huge number of carbon compounds we see around us? 

Answer:

The two properties of carbon which lead to the huge number of carbon compounds we see around us are:

(i) Carbon has high valency of 4. It requires 4 more electrons to complete its octet. So it can share electrons with a large number of other elements.

(ii) It is easy to from strong covalent bonds with carbon because of its small size.

3.What will be the formula and electron dot structure of cyclopentane?

Answer:

A cyclopentane has 5 carbon atoms and 10 hydrogen atoms. Each carbon atom shares 2 of its electrons with adjacent carbons and the rest two with 2 hydrogen atoms.

Formula: C₅H₁₀

Electron-dot structure:

4. Draw the structures for the following compounds.

(i) Ethanoic acid
(ii) Bromopentane*
(iii) Butanone
(iv) Hexanal

*Are structural isomers possible for bromopentane?

Answer:

(i) Ethanoic Acid

(ii) Bromopentane

Yes, there are two other isomers of Bromopentane.

Names: 2-Bromopentane , 3-Bromopentane

(iii) Butanone

(iv) Hexanal

5. How would you name the following compounds?

(i) CH3—CH2—Br

 (iii )

Answer:

(i) Bromoethane
(ii) Methanal or Formaldehyde
(iii) Hexyne

Intext Questions – Page 71

1. Why is the conversion of ethanol to ethanoic acid an oxidation reaction?

Answer:

Conversion of ethanol to ethanoic acid is an oxidation reaction because it involves the removal of hydrogen and the addition of oxygen.

Reactions Involved:

2. A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?

Answer:

When ethyne is burnt in oxygen, it burns completely and produces much more heat with a clean blue flame. If burnt in air, it will not combust completely leaving some residue and would not reach the required high temperature for welding.

Intext Questions – Page 74

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

Answer:

When we add sodium carbonate to carboxylic acid, we get a milky white color due to production of carbon dioxide gas. We don’t observe any such reaction in the case of alcohols.

The reaction involved is:

2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂

2. What are oxidising agents?

Answer:

Agents that add oxygen and remove hydrogen from any compound are known as oxidising agents. Nitric acid, halogens, and potassium nitrate are examples of oxidising agents.

Intext Questions – Page 76

1. Would you be able to check if water is hard by using a detergent?

Answer:

No, we cannot check if water is hard by using a detergent. Detergents are salts of ammonium or sulphonates of long chain carboxylic acids. They can not react with magnesium and calcium present in hard water to form scum unlike soaps.   

2.People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes?

Answer:

People beat the clothes after adding soap to aid the micelles present in soap to trap and remove dirt, oil, and grease easily. On beating, the dirt particles dissociate the clothes and go away with water leaving the clothes clean.

NCERT Solutions for Class 10 Science Chapter 4 – NCERT Exercise – Page 77

1. Ethane, with the molecular formula C₂H₆ has

(a) 6 covalent bonds

(b) 7 covalent bonds

(c) 8 covalent bonds

(d) 9 covalent bonds

Answer:

(b) 7 covalent bonds

2. Butanone is a four-carbon compound with the functional group

(a) carboxylic acid

(b) aldehyde

(c) ketone

(d) Alcohol

Answer:

(c ) Ketone

3. While cooking, if the bottom of the vessel is getting blackened on the outside, it means that

(a) the food is not cooked completely.

(b) the fuel is not burning completely.

(c) the fuel is wet.

(d) the fuel is burning completely.

Answer:

(b) the fuel is not burning completely

4. Explain the nature of the covalent bond using the bond formation in CH₃Cl.

Answer:

Carbon can only form covalent bonds. It requires four more electrons to complete its octet. Three hydrogen atoms share their electrons and become stable and one chlorine atom shares its electrons to complete the octet of both carbon and chlorine.

5.Draw the electron dot structures for

(a) ethanoic acid

(b) H₂ S

(c) propanone

(d) F₂ 

Answer:

(a) ethanoic acid

(b) H₂S

(c) propanone

(d) F₂

6. What is a homologous series? Explain with an example.

Answer:

Homologous series are a series of compounds which have the same functional group. They have similar general formula and chemical properties.

Physical properties of such compounds differ because of different molecular masses. For example, ethene, propene, etc. are all part of the alkene homologous series. Similarly, Methanol, ethanol, propanol, etc are the part of homologous series of alcohols.

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

Answer:

Ethanol (C₂H₅OH) and ethanoic acid (CH₃COOH) can be differentiated as follows:

8. Why does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?

Answer:

Soaps are sodium or potassium salts of long chain fatty acids. These salts have a polar and a non-polar end. The polar end of the salt is hydrophilic in nature and the non-polar end is hydrophobic. When added to water, these molecules orient themselves such that the hydrophilic ionic end stays towards the water and the hydrophobic end made of hydrocarbons protrudes out of water. In this arrangement, hydrophobic tails are in interior of the cluster and the ionic ends are on the surface of the cluster. Clusters of such molecules trap the dirt and result in micelle formation.

No, micelle is not formed in solvents like ethanol because these salts get easily dissolved in ethanol.

9. Why are carbon and its compounds used as fuels for most applications?

Answer:

Due to high calorific value, carbon compounds that contain carbon and hydrogen atoms in majority, give optimum ignition temperature and release high amount of energy. It is also easy to control their rate of combustion. Hence, carbon and its compounds are used as fuels for most applications.

10. Explain the formation of scum when hard water is treated with soap.

Answer:

Hard water contains calcium and magnesium ions. A soap contains salts of sodium and potassium. When we add soap to hard water, the soap molecules react with the ions present in water and give calcium and magnesium salts. These are sticky and insoluble, and are known as scum.

11. What change will you observe if you test soap with litmus paper (red and blue)?

Answer:

Soaps are alkaline in nature. Since they are salts of metals, they will be basic. If we dip a red litmus paper in a soap solution, it turns blue. If we dip a blue litmus paper, it doesn’t change its color.

12. What is hydrogenation? What is its industrial application?

Answer:

Hydrogenation is the process by which hydrogen is added to a compound. It is a chemical reaction between hydrogen and another compound that needs to be reduced in the presence of a catalyst like platinum, palladium or nickel. Unsaturated hydrocarbons are converted into saturated hydrocarbons by this method. Its industrial application is to convert liquid vegetable oils and other compounds into solid or semi-solid fats.

13. Which of the following hydrocarbons undergo addition reactions: C₂H₆ , C₃H₈ , C₃H₆ , C₂H₂ and CH₄ .

Answer:

Hydrocarbons that are not saturated and contain multiple bonds between two carbon atoms undergo addition reactions. Hence, C₃H₆ and C₂H₂ will undergo addition reaction.

14. Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.

Answer:

Bromine water test can be used to differentiate between saturated and unsaturated hydrocarbons. It is an addition reaction and hence, only unsaturated compounds will undergo a change.

Bromine water is a reddish-brown colored solution made by dissolving bromine in water. When this solution is added to an unsaturated hydrocarbon ( such as alkenes and alkynes), its brown color disappears and the water is decolorised. No such change is observed when added to alkanes.

15. Explain the mechanism of the cleaning action of soaps.

Answer:

The soap molecules have different properties at different ends. One is hydrophilic end which dissolves in the water and is attracted towards the water and the second is the hydrophobic end which is dissolved in the hydrocarbons and is repulsive to water. The hydrophobic tail aligns itself along the surface of the water and is attracted to the hydrocarbons present in grease on dirty clothes. The cluster of such formation is called micelle. The oily dirt is collected in the middle of the micelle. The micelles stay in solution as colloid and form an emulsion. These are easy to wash away and leave clothes neat.

Topics Covered in Carbon and its Compounds Class 10 Science

• Bonding in Carbon-The Covalent Bond
• Versatile Nature of Carbon
  • Saturated and Unsaturated Carbon Compounds
  • Chains, Branches and Rings
  • Homologous Series
  • Nomenclature of Carbon Compounds
• Chemical Properties of Carbon Compounds
  • Combustion
  • Oxidation
  • Addition Reaction
  • Substitution Reaction
• Some Important Carbon Compounds-Ethanol and Ethanoic acid
• Soaps and Detergents

NCERT Solutions for Class 10 Science Chapter 10 – Light – Reflection and Refraction, contains solutions to various questions in Exercises for Chapter 10. Light – Reflection and Refraction Class 10 NCERT Solutions have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Science Chapter 10 – Light – Reflection and Refraction

NCERT Solutions for Class 10 Science Chapter 10 – Light – Reflection and Refraction – NCERT Exercises

NCERT Solutions for Class 10 Science Chapter 10 – Light – Reflection and Refraction, Exercises includes all in text and exercise solved questions which helps you to understand the topic covered in Chapter 10 Light – Reflection and Refraction Class 10, in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Science Chapter 10 – Intext Questions – Page 168

1. Define the principal focus of a concave mirror.

Answer:

Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is called the principal focus of the concave mirror.

2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:

Given: Radius of curvature ( R ) = 20 cm

We know, Radius of curvature of a spherical mirror = 2 x Focal length

So, Focal length = R/2

= 20/2

= 10 cm.

Hence, the focal length of the given spherical mirror is 10 cm.

3. Name a mirror that can give an erect and enlarged image of an object.

Answer:

Concave mirror gives an erect and enlarged image of an object when the object is placed between the pole and the principle focus.

4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer:

Convex mirror forms a virtual, erect, and diminished image of the object placed in front of it. We prefer a convex mirror as a rear-view mirror in vehicles to have a wider view of the road. It helps in assessing the traffic better.

Intext Questions – Page 171

1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer:

Given: Radius of curvature (R) = 32 cm

We know, Radius of curvature = 2 × Focal length (f)

R = 2f

f= R/2

= 32/2 = 16 cm

Hence, the focal length of the given convex mirror is 16 cm.

2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Answer:

The relation between magnification, height of images, and image distances can be given by:

m = height of the image/ height of the object = – image distance/object distance

m = h₁/h₀  = – v / u

Let height of the object h₀ = h

Then the height of the image h₁ will be = – 3h (real image is formed)

On putting in equation and comparing:

– 3h / h = – v / u

v / u = 3

Image distance v = 3 X (- 10) = – 30 cm.

An inverted image is formed in front of the given concave mirror at a distance of 30 cm.

Intext Questions – Page 176

1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

Answer:

When a light ray enters from an optically rare medium (with lower refractive index) to an optically denser medium (with a higher refractive index), its speed slows down and bends towards the normal. Since water is denser than air, the ray will bend towards the normal.

2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10⁸ m s^–1 .

Answer:

Refractive index of a medium (n_m) = Speed of light in vacuum/Speed of light in the medium

We know, Speed of light in vacuum (c) = 3 × 10⁸ m/s

Given: Refractive index of glass (n_g) = 1.50

So, Speed of light in the glass (v) = Speed of light in vacuum/ Refractive index of glass

= c/n_g

=3 x 10⁸/1.50

= 2x 10⁸ ms^-1

Hence, speed of light in glass is 2x 10⁸ ms^-1.

3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

 Answer:

A medium with the highest refractive index will have the highest optical density and vice-versa.

On the given basis:

Lowest Optical Density: Air with lowest refractive index of 1. 0003.

Highest Optical Density: Diamond with refractive index of 2.42.

4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.

Answer:

Light will travel fastest in water as compared to kerosene & turpentine.

The speed of light is inversely proportional to the refractive index. Since according to the given table, the refractive index of water is lower than that of kerosene and turpentine, light will travel fastest in water.

5. The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:

We know,

Refractive index of a medium (n_m) = Speed of light in vacuum/Speed of light in the medium.

Diamond has a refractive index of 2.42.

Using the relation, Speed of light in medium = Speed of light in vacuum / 2.42

This shows that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in the vacuum.

NCERT Solutions for Class 10 Science Chapter 10 – Intext Questions – Page 184

1. Define 1 dioptre of power of a lens.

Answer:

Power of a lens is the reciprocal of its focal length. Dioptre is the SI unit of power of the lens.

P = 1 / f (in m)

Hence, 1 dioptre can be defined as the power of a lens with focal length 1 metre.

2. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

Answer:

For the image to be equal in the size of the object, the position of the image should be 2F.

Given: Object distance (u) = – 50 cm.

So, image distance (v) = 50 cm.

Let focal length = f

Focal length formula =

1 / f = 1 / 50 – 1 / (-50)

= 1 / 50 + 1/ 50

=1 / 25

So, f = 25 cm = 0.25 m

Power, P = 1 / f

P = 1 / 0.25

P = +4D

Hence, power of lens is +4D.

3. Find the power of a concave lens of focal length 2 m.

Answer:

Focal length of concave lens (f) = 2 m

Power of lens (P) = 1/f

= 1/ (-2)

= -0.5D.

Hence, the power of the given lens is -0.5D.

NCERT Solutions for Class 10 Science Chapter 10 – NCERT Exercise – Page 185

1. Which one of the following materials cannot be used to make a lens? (a) Water (b) Glass (c) Plastic (d) Clay

Answer:

(d) Clay; as it is opaque and does not allow the light to pass through it.

2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

(a) Between the principal focus and the centre of curvature

(b) At the centre of curvature

(c) Beyond the centre of curvature

(d) Between the pole of the mirror and its principal focus.

Answer:

(d) Between the pole of the mirror and its principal focus.

3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?

(a) At the principal focus of the lens

(b) At twice the focal length

(c) At infinity

(d) Between the optical centre of the lens and its principal focus.

Answer:

(b) At twice the focal length

4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be

(a) both concave.

(b) both convex.

(c) the mirror is concave and the lens is convex.

(d) the mirror is convex, but the lens is concave.

Answer:

(a) Both concave

5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

(a) only plane

(b) only concave

(c) only convex

(d) either plane or convex

Answer:

(d) Either plane or convex

6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

(a) A convex lens of focal length 50 cm.

(b) A concave lens of focal length 50 cm.

(c) A convex lens of focal length 5 cm.

(d) A concave lens of focal length 5 cm.

Answer:

(c ) A convex lens of focal length 5 cm; When object is placed between radius of curvature and focal length.

7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Answer:

In a concave mirror, to obtain an erect image, we need to put the object between the pole and the focus. So, the range of the object from the mirror should be between 0 cm to 15 cm.

The image will be erect, virtual, and magnified.

8. Name the type of mirror used in the following situations.

(a) Headlights of a car.

(b) Side/rear-view mirror of a vehicle.

(c) Solar furnace

Support your answer with reason.

Answer:

(a) Concave Mirror is used in the headlights of a car. Concave mirrors can produce powerful parallel beams of light when light source is placed on their principal focus.

(b) Convex mirror is used in side/rear-view of a vehicle because it makes a virtual, erect, and diminished image. This helps in getting a wider view of the traffic behind the vehicle.

(c) Concave Mirror is used in solar furnace because it concentrates the parallel rays of sun at principal focus making it possible to concentrate heat a point.

9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer:

Yes, this lens will produce a complete image of the object even if one-half of the lens is covered with black paper. However, the intensity of the image will be low and the image will not be bright enough. This can be verified by observing the image of any distant object like a tree on some screen.

10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer:

Given, Height of the Object, h₀ = 5 cm

Distance of the object from the converging lens, u = -25 cm

Focal length of converging lens, f = 10 cm

Using lens formula,

v = 250 / 15

= 16.66 cm

For a converging lens, h_i /h_O= v / u

h_i = ( v / u ) x h_O

= 50 x 5 / 3 x (-25)

= 10 / -3 = -3.3 cm.

The image is inverted and formed at a distance of 16.7 cm behind the lens and size of the image is 3.3 cm.

11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer:

Given, Focal length of concave lens (OF₁), f = – 15 cm

Image distance, v = – 10 cm

Using the lens formula,

1/u = 1/v – 1/f                                               

= – 1/10 – 1/(-15)

= – 1/10 + 1/15

u = – 5 / 150 = – 30 cm.

Hence, we conclude that the object is placed 30 cm in front of the lens.

12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Answer:

Given: Focal length of convex mirror (f) = +15 cm

Object distance (u) = – 10 cm

=> 1/v = 1/f – 1/u

= 1/15 – 1/(-10)

= 2+3 / 30

=> v = 1/5/30 = 6 cm.

Magnification = -v/u = -6 / -10

= 0.6

The image is located at a distance of 6 cm from the mirror behind the mirror.

The positive and a value less than 1 of magnification indicates that the image formed is virtual, erect, and diminished.

13. The magnification produced by a plane mirror is +1. What does this mean?

Answer:

We know, magnification (m) = height of the image/ height of the object

The positive sign signifies that the image formed by a plane mirror is virtual and erect. The magnification 1 means that the size of the image is equal to the size of the object.

14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer:

Given, Object distance (u) = – 20 cm

Object length (h) = 5 cm

Radius of curvature (R) = 30 cm

We know,

Radius of curvature = 2 × Focal length

R = 2f

=> f = 15 cm

=> 1/v = 1/f – 1/u

= 1/15 + 1/20

= 7/60

=> v = 8.57 cm

Magnification, m = – image distance/object distance

= -8.57 / – 20

= 0.428

The image is formed behind the mirror and is virtual in nature.

Also, height of image/height of object = m;

=> h_i = m x height of object

= 0.428 x 5 = 2.14 cm.

The image formed is erect and smaller in size.

15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.

Answer:

Give, Object distance (u) = – 27 cm

Object height (h) = 7 cm

Focal length (f) = – 18 cm

=> 1/v = 1/f – 1/u

= – 1/18 + 1/27

= -1/54

=> v = – 54 cm

Hence, the screen should be placed 54 cm in front of the mirror.

Magnification, m = – image distance/object distance

= -54 / 27

=  – 2

The image is real in nature.

Also, height of image/height of object = m;

=> h_i = m x height of object

= 7 x – 2 = -14 cm.

The image formed is inverted and larger in size.

16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Answer:

Given, P = -2D

We know,

Power of lens (P) = 1/f

=> f = -1/2 = -0.5 m

Since it has a negative focal length, it is a concave lens.

17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer:

Given, P = 1.5D

We know,

Power of lens (P) = 1/f

=> f = 1/1.5 = 10/15 = 0.66 m

Since it has a positive focal length, we can say that it is a convex lens which is converging in nature.

Topics Covered in Light – Reflection and Refraction Class 10 Science

• Reflection
• Spherical Mirrors
  • Image Formation by Spherical Mirrors
  • Representation of Images Formed by Spherical Mirrors Using Ray Diagrams
  • Sign Convention for Reflection by Spherical Mirrors
  • Mirror Formula and Magnification
• Refraction of Light
  • Refraction through a Rectangular Glass Slab
  • The Refractive Index
  • Refraction by Spherical Lenses
  • Image Formation by Lenses
  • Sign Convention for Spherical Lenses
  • Lens Formula and Magnification
  • Power of a Lens

NCERT Solutions for Class 10 Science Chapter 12 – Electricity, contains solutions to various questions in Exercises for Chapter 12. Electricity Class 10 NCERT Solutions have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

Download NCERT Solutions for Class 10 Science Chapter 12 – Electricity

NCERT Solutions for Class 10 Science Chapter 12 – Electricity – NCERT Exercises

NCERT Solutions for Class 10 Science Chapter 12 – Electricity, Exercises includes all in text and exercise solved questions which helps you to understand the topic covered in Chapter 12 Electricity Class 10, in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Science Chapter 12 – Intext Exercise – Page 200

1. What does an electric circuit mean?

Answer:

A continuous closed path made of electric components such as conductors, cells, switches, wires, etc, through which the current flows is called as electric circuit.

2. Define the unit of current.

Answer:

The flow of one coulomb of charge through wire in one second is known as one ampere. This is the unit of current.

3. Calculate the number of electrons constituting one coulomb of charge.

Answer:

Charge is a quantized entity, such that charge Q = ne;

n = number of electrons and e = charge on electrons.

We know, e = 1.6 × 10^-19 C

=> 1 C = n x 1.6 × 10^-19 C

=> n = 1 / 1.6 × 10^-19

=> n = 6.25 x 10¹⁸

Hence, we can say that 1 coulomb of charge has 6.25 x 10¹⁸ electrons.

Intext Exercise – Page 202

1. Name a device that helps to maintain a potential difference across a conductor.

Answer:

Battery or electric cells help to maintain a potential difference across a conductor.

2. What is meant by saying that the potential difference between two points is 1 V?

Answer:

When an object does 1 J of work to move a charge of 1 C from one point to another, it is said that the potential difference between two points is 1 V.

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer:

We know, V = W/Q ;

W = work done in moving the charge from one point to another

Q = charge

On rearranging the above equation, we get:

W = V x Q

W = 6 x 1 = 6 J

Hence, 6 J of energy is given to each coulomb.

Intext Exercise – Page 209

1. On what factors does the resistance of a conductor depend?

Answer:

Resistance R is given by = ρ l/A

From this equation we can say that resistance is directly proportional to the length and inversely proportional to the area of the conductor.

Other than this, resistance also depends on temperature of the conductor and the nature of its material.

2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?

Answer:

Resistance  is given by, R = ρ l/A

where,

ρ is the resistivity of the material of the wire,

l is the length of the wire

A is the area of the cross-section of the wire.

From this equation we can say that resistance is inversely proportional to the area of the conductor.

Hence, current will flow more easily in a thick wire as it offers low resistance.

3. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?

Answer:

Ohm’s Law determines the change in the current flowing through any electrical component.

It states that:

I = V / R ;

V = potential difference, R = resistance, and I = current

Given: Potential difference is reduced by half keeping resistance constant.

So, let new potential difference be V’.

=> V’ = V/2

Let new resistance = R’ = R and new current = I’

According to Ohm’s law:

I’ = V’ / R’

I’ = V/2 / R

I’ = 1/2 ( V/R)

I’ = I / 2

Hence, we can see that the current flowing through the component is reduced by half.

4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

Answer:

Alloys have higher melting points than that of metals. They have high resistance and do not melt easily. Since these items gain very high temperatures, coils of electric toasters and electric irons made of an alloy rather than a pure metal.

5. Use the data in Table 12.2 to answer the following –

(a) Which among iron and mercury is a better conductor?

(b) Which material is the best conductor?

Answer:

(a) According to the table, we see that iron has lesser resistivity as compared to mercury, so we can say that mercury is a better conductor.

(b) In the given materials, silver has the lowest resistivity. So, silver is the best conductor.

QUESTIONS (PAGE 213):

1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.

Answer:

The given diagram represents the above circuit.

2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer:

We can obtain the reading of the ammeter and the voltmeter by applying Ohm’s law.

The total resistance of the circuit = 5 Ω + 8 Ω +12 Ω = 25 Ω.

Given potential difference of the circuit = 6 V,

=>Current flowing through the circuit or the resistors can be calculated as follows:

I = V/R = 6/25 = 0.24A

Let the potential difference across the 12 Ω resistor be V₁.

=>V1 = 0.24A × 12 Ω = 2.88 V

Hence, the ammeter will read 0.24 A and the voltmeter reading will be 2.88 V.

NCERT Solutions for Class 10 Science Chapter 12 – Intext Exercise – Page 216

1. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10⁶ Ω, (b) 1 Ω and 10³ Ω, and 10⁶ Ω.

Answer:

When resistance are connected in parallel, their equivalent is given by:

(a) Using the formula:

1 / R_Total = 1/1 + 1/ 10⁶

R_Total = 10⁶/ 10⁶ + 1

Since 1 is very small as compared to 10^6,

R_Total = 1 Ω

(b) Using the formula:

1 / R_Total = 1/1 + 1/ 10³ + 1/ 10⁶

R_Total  = 10⁶ / 10⁶ + 10³ + 1

R_Total  = 1000000/ 1000001

R_Total  = 0.999 Ω

2. An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer:

The above circuit is shown in given diagram:

=> 1 / R_Total = 1/100 + 1/50 + 1/500

=>1 / R_Total = 5+10+1 / 500

=>R_Total = 500 / 16 Ω

Using Ohm’s Law:

I = V/R

I = 220 / (500/16)

I = 220 x 16 / 500

I = 7.04 A

Since all the devices are connected in parallel, all of them draw a current of 7.04 A.

Using the same formula we can get the resistance of iron connected to the same source:

R = V / I

R = 220 / 7.04

R = 31.25 Ω

3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer:

When electrical devices are connected in parallel, the total effective resistance is reduced. The potential difference across all devices remains same. Hence, there is no division of voltage across appliances.

4. How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Answer:

(a) We connect the 3 Ω and 6 Ω resistors in parallel to obtain an equivalent of 2 Ω.

1/R = 1/3 + 1/6

= 2+1 / 6

= 3/6

R = 2 Ω

Then we add the given 2 Ω resistor in series to get a total of 4 Ω resistance. We can see the arrangement in the following diagram.

(b) We connect all the resistances in parallel to get a total resistance of 1 Ω.

1/R = 1/2 + 1/3 + 1/6

= 3 + 2 + 1 / 6

= 6 / 6

R = 1 Ω.

We can see the arrangement in the following diagram.

5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer:

(a) We can obtain the highest resistance by connecting all the coils in series.

The total resistance will be: 4 + 8 + 12 + 24 = 48 Ω.

(b) We can obtain the lowest total resistance by connecting all the coils in parallel.

The total resistance in this case will be:

1/R = 1/4 + 1/8 + 1/12 + 1/24

R = 24 / 6 + 3 + 2 + 1

R = 24 / 12

R = 2 Ω.

NCERT Solutions for Class 10 Science Chapter 12 – Intext Exercise – Page 218

1. Why does the cord of an electric heater not glow while the heating element does?

Answer:

The cord of an electric heater is made of pure metals like copper and aluminium. Such elements have low resistance and do not dissipate a lot of heat. The heating element of an electric heater is made of an alloy which has a high resistance. When the current flows through this heating element, due to high resistance, the heating element becomes hot and glows red.

2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Answer:

According to Joule’s Law:

H = VIt

where,

V = voltage, Given: V = 50 V

I = current

t = time in seconds, => 1 hour = 3600 seconds

Amount of current = Amount of charge / Time taken in flow of charge

Amount of current = 96000 / 3600

=> I = 26.66 A

The amount of heat generated = 50 x 26.66 x 3600 = 4.8 x 10⁶ J.

Hence the amount of heat generated is 4.8 x 10⁶ J.

3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Answer:

The amount of heat generated can be calculated using the Joule’s law.

H = VIt

Where:

V = voltage

I = current

t = time in seconds

Voltage = Current x Resistance

V = 5 x 20 = 100 V.

Substituting the given values in the above equation, we get,

H = 100 × 5 × 30 = 1.5 × 10⁴ J

Hence, the amount of heat developed by the electric iron in 30 s is 1.5 × 10⁴ J.

Intext Exercise – Page 220

1. What determines the rate at which energy is delivered by a current?

Answer:

The rate of consumption of electrical energy by electric appliances is known as electric power. The rate at which energy is delivered by a current is the power of the appliance.

2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Answer:

The power of the motor can be calculated by the equation:

P = VI

P = 220 V × 5 A = 1100 W

The energy consumed by the motor is given by:

E = P × T

P = power = 1100 W and T = 2 h = 2 x 60 x 60 = 7200 s.

Now putting these values in energy equation,

E = 1100 W × 7200 = 7.92 × 10⁶ J.

Hence, the power of the motor is 1100 W and the energy consumed by the motor in 2 hours is 7.92 × 10⁶ J.

NCERT Solutions for Class 10 Science Chapter 12 – NCERT Exercise – Page 221

1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is –

(a) 1/25
(b) 1/5
(c) 5
(d) 25

Answer:

(d) 25

As, 1/R’ = 5/R + 5/R + 5/R + 5/R + 5/R

R’ = R/25

=> R/ R’ = 25.

2. Which of the following terms does not represent electrical power in a circuit?

(a) I ²R
(b) IR²
(c) VI
(d) V ²/R

Answer:

(a) IR²

3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W

(d) Answer:25 W

P = VI = V²/R

R = V² / P

R = (220)²/100 = 484 Ω

The power consumed when it is operated on 110 V:

P = V²/R

=> P = (110)² V/484 Ω = 25 W

Hence, the power consumed when the electric bulb operates at 110 V is 25 W.

4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be –

(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1

Answer:

( c) 1:4

Let R_s and R_p be the equivalent resistance of the wires when connected in series and parallel respectively.

For the same potential difference V, the ratio of the heat produced in the circuit is given by

H_s / H_p = R_p / R_s

R_s = 2R

R_p = R/2

On substituting:

=> H_p / H_s = 1/4

5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Answer:

The voltmeter should be connected parallel between two points in order to measure the potential difference.

6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer:

Resistance R is given by = ρ l/A

The area of cross-section of the wire can be calculated by:

=> l = R x A / ρ

=> l = (10 x 3.14 x 25) / (4 x 1.6)

On solving we get, l = 122.72 m.

To find the resistance when the diameter is doubled:

We use, R = ρ l/A

R = 1.6 × 10^–8x 122.72 / 3.14 x (0.001/2)²

R = 250.2 x 10^-2 Ω = 2.5 Ω

7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

I (amperes) 0.5 1.0 2.0 3.0 4.0

V (volts) 1.6 3.4 6.7 10.2 13.2

Plot a graph between V and I and calculate the resistance of that resistor.

Answer:

We plot the current on Y axis and the voltage on X axis. This graph is called as IV characteristic graph. The slope of this graph gives the value of resistance.

Slope = 1/R = BC/AC = 2/6.8

To calculate R,

R = 6.8/2 = 3.4 Ω

The resistance of the resistor is 3.4 Ω.

8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer:

Value of the resistance is calculated by Ohm’s law.

R = V / I

R = 12 / 2.5 x 10^-8 = 4.8 x 10³ Ω

R = 4.8 kΩ

Hence, the value of the resistance of the resistor.

9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω , 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?

Answer:

When resistors are connected in series, the current flowing through them remains the same as there is no distribution of current.

The total resistance when the given resistances are connected in series is:

R = 0.2Ω + 0.3Ω + 0.4Ω + 0.5Ω + 12Ω = 13.4Ω

Now, we can use Ohm’s law to calculate the current flowing through the resistors:

I = V / R

I = 9 / 13.4

I = 0.671 A

Hence, we can say that a current of 0.671A flows through the 12Ω resistor.

10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer:

Let the number of required resistor be ‘n’.

We know the total resistance when resistors are connected in parallel is:

On putting 176Ω in the above equation and solving, we get:

=> R_Total = 176 / n = R

To calculate the ‘n’, we will use Ohm’s law:

R = V / I

176 / n = 220 / 5

n = 176 x 5 / 220

n = 4

Hence, we require four 176Ω resistors in parallel to carry 5A on a 220 V line.

11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer:

(i)

The two parallel resistors give : 1/R = 1/6 + 1/6

R = 3 Ω

Combining with series we get: 3 Ω + 6 Ω = 9 Ω.

(ii)

Resistors in series give: 6 Ω + 6 Ω = 12 Ω

Adding another 6 Ω in parallel we get: 1/R = 1/6 + 1/12

R = 12 / 3 = 4 Ω

12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Answer:

Given: Power (P) = 10 W.

To find the resistance of one bulb we can use the formula:

P = V² / R

=> R = V² / P

=> R = (220)² / 10

=> R = 4840 Ω

We can find the total resistance to be provided by ‘n’ number of bulbs by applying Ohm’s law:

R_total = V / I

R_total = 220 / 5

R_total = 44 Ω

Finally to find the number of bulbs with resistance 4840Ω required in parallel:

1/R_total = n x 1/4840

n = 4840 / 44 = 110

Hence, we require 110 bulbs of 4840Ω resistance connected in parallel.

13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?

Answer:

When the coils are used separately:

Since both have same resistance, using Ohm’s law we get:

I = V / R

I = 220 / 24

I = 9.166 A

When the coils are connected in series:

In series, total resistance = 24Ω + 24Ω = 48 Ω

Using Ohm’s law to find the current:

I = V / R

I = 220 / 48

I = 4.58 A

When the coils are connected in series:

When connected in parallel = 1/24 + 1/24

R = 576/48 = 12 Ω

Using Ohm’s law to find the current:

I = V / R

I = 220 / 12

I = 18.33 A

14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer:

We define power (P) = I²R = V² / R

(i) Since the given resistors are connected in series, the total resistance will be: 1Ω + 2Ω = 3Ω.

We can use Ohm’s law to find the current flowing through the resistors:

I = V / R

I = 6 / 3

I = 2 A

Now, using the obtained data to get the power used in 2Ω resistor:

P = (2)²x2

P = 8 W

(ii) Since the given resistors are connected in parallel, the potential difference across them remains the same:

P = V² / R

P = 4² / 2

P = 8 W.

15.Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?

Answer:

Since the two lamps are connected in parallel, the voltage across them will remain equal.

To find the current drawn by the first lamp, we use the equation: P = V x I

=> I = V / P

=> I = 100 / 220 A

Similarly, for the second lamp:

I = 60 / 220 A

Hence, the total current drawn

I _total = (100 / 220) + (60 / 220)

I _total = 160 / 220 A.

I _total = 0.727 A

Hence, 0.727 A of current is drawn from the line if the supply voltage is 220 V.

16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?

Answer:

We know; H = P x t

Where P = power and t = time in seconds.

Energy consumed by a 250 W TV set in 1 hr (3600 s) :

H = 250 W x 3600 s = 9 x 10⁵ J

Energy consumed by a 1200 W toaster in 10 minutes (600s) :

H = 1200 W x 600 s = 7.2 x 10⁵ J

We see that watching one hour of TV consumes more energy than using a toaster for 10 minutes.

17. An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Answer:

We know power is energy consumed per second.

Power P is given by : P = I²R

Given: R = 8 Ω and I = 15 A

P = (15)² x 8

P = 1800 J/s

Hence, the rate at which heat is developed in the heater is 1800 J/s.

18. Explain the following.

(a) Why is the tungsten used almost exclusively for filament of electric lamps?

(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

(c) Why is the series arrangement not used for domestic circuits?

(d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer:

(a) Electric lamps operate at high temperatures. Since tungsten has a very high resistance and melting point, it does not melt at the temperatures on which the the lamps operate. Hence, tungsten used almost exclusively for filament of electric lamps.

(b) Since the resistance given by an alloy is much higher than that of pure metals, they are able to dissipate more heat. They also have higher melting points as compared to metal. Because of these two properties, the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal.

(c) In series circuit, the voltage is divided across the resistors. As a result, each component in the circuit receives a small voltage because of which the amount of current decreases and the device generates a lot of heat and does not work properly or can be damaged. This is the reason why series circuits are not used in domestic circuits.

(d) Resistance is inversely proportional to the area of cross section.

R = ρ l/A

where,

ρ is the resistivity of the material of the wire,

l is the length of the wire

A is the area of the cross-section of the wire.

As the area of the cross section increases, the resistance decreases and vice-verse.

(e ) Copper and aluminium are good conductors of electricity and have low resistivity. These properties reduce the loss of energy in the form of heat.  Hence, they are usually employed for electricity transmission.

Topics Covered in Electricity Class 10 Science

• Electric Current and Circuit
• Electric Potential and Potential Difference
• Circuit Diagram
• OHM’S LAW
• Resistance of conductors
• System Of Resistors
• Heating Effect
• Electric Power

NCERT Solutions for Class 10 Science Chapter 5 – Periodic Classification of Elements, contains solutions to various questions in Exercises for Chapter 5. Periodic Classification of Elements Class 10 NCERT Solutions have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

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

NCERT Solutions for Class 10 Science Chapter 5 – Periodic Classification of Elements, Exercises includes all in text and exercise solved questions which helps you to understand the topic covered in Chapter 5 Periodic Classification of Elements Class 10, in a better manner to help you to score good marks in your examinations.

NCERT Solutions for Class 10 Science Chapter 5 – Intext Questions – Page 81

1. Did Döbereiner’s triads also exist in the columns of Newlands’ Octaves? Compare and find out.

Answer:

Yes, Döbereiner’s triads did exist in Newlands’ Octaves. The elements Lithium (Li), Sodium (Na), and Potassium (K) make a Döbereiner’s triad but are also present in the second column of Newlands’ octaves.

2. What were the limitations of Döbereiner’s classification?

Answer:

The Döbereiner’s classification had limitations because:

1. They did not apply for elements with very low mass or very high mass.
2. The validity of these triads began to decrease with emerging methods of calculation of mass.
3. Döbereiner’s triads could not accommodate all the elements present and the new elements that were being discovered.

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

Answer:

Newland assumed that there are only 56 elements in nature and did not consider discovery of new elements. The law is such that it is applicable only for elements till Calcium. It fits two elements in same slot and many elements with differing properties in same columns. For example, Iron (Fe) was placed far away from elements like Nickel and Cobalt, that share its properties.

Intext Questions – Page 85

1. Use Mendeléev’s Periodic Table to predict the formulae for the oxides of the following elements:

K, C, AI, Si, Ba.

Answer:

In Mendeléev’s Periodic Table, Oxygen is a member of group VI A making its valency two. According to the table, the formula for oxides of:

K is K₂O, as K is the member of group I A with valency 1.

C is CO₂ or C₂O₄, as C is a member of group IV A with valency 4.

Al is Al₂O₃, as Al is a member of group III A with valency 3.

Si is SiO₂ or Si₂O₄, as Si is a member of group IV A with valency 4.

Ba is BaOor Ba₂O₂, as C is a member of group II A with valency 2.

2. Besides gallium, which other elements have since been discovered that were left by Mendeléev in his Periodic Table? (any two)

Answer:

Besides gallium, Germanium and Scandium were also the elements that were later discovered and were left by Mendeléev in his Periodic Table.

3. What were the criteria used by Mendeléev in creating his Periodic Table?

Answer:

Mendeléev used the following criteria to sort elements in his Periodic Table:

1. Elements were segregated on the basis of various compounds formed by them with Hydrogen and Oxygen.
2. He also observed the relationship between atomic masses of the elements to distribute them across the table in various columns.

4. Why do you think the noble gases are placed in a separate group?

Answer:

Noble gases are placed in group zero. Due to their inert nature, they don’t react with other elements easily. Their concentration in our atmosphere is also low. In order to distinguish them and maintain the order of periodic table, they are placed in a separate group.

Intext Questions – Page 90

1. How could the Modern Periodic Table remove various anomalies of Mendeléev’s Periodic Table?

Answer:

The Modern Periodic Table removes the anomalies of Mendeléev’s Periodic Table by arranging the elements on the basis of atomic numbers. Since Mendeléev arranged his periodic table on the basis of atomic mass, the anomaly due to elements with same atomic mass are removed. The isotopes of same elements with different atomic masses are assigned with a fixed position in the modern periodic table. Also, atomic number of cobalt is 27 and nickel is 28, cobalt will come before nickel even though its atomic mass is greater.

2. Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?

Answer:

Calcium and Beryllium will have chemical reactions similar to magnesium. Both Calcium and Beryllium have two valence electrons and belong to the same group to which magnesium belongs to.

3. Name

(a) three elements that have a single electron in their outermost shells.

(b) two elements that have two electrons in their outermost shells.

(c) three elements with filled outermost shells.

Answer:

(a) Lithium (Li), Sodium (Na), Potassium (K).

(b) Magnesium (Mg), Calcium (Ca).

(c) Neon (Ne), Argon (Ar), Xenon (Xe).

4. (a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?

(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?

Answer:

(a) Lithium, sodium and potassium are all alkali metals. They have only one electron in their outermost shell. This makes them highly unstable. They readily lose this electron and react with water to liberate hydrogen gas.

(b) Both helium and neon are noble gases. They have completely filled outermost electrons making them very stable and unreactive.

5. In the Modern Periodic Table, which are the metals among the first ten elements?

Answer:

Lithium (Li) and Beryllium (Be) are the metals among the first ten elements in modern periodic table.

6. By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristic?

Ga Ge As Se Be

Answer:

In the given elements, Gallium and Beryllium have the maximum metallic characteristics. However, Gallium (Ga) has a larger size than beryllium so it loses its outermost electrons more easily making it the most metallic among the given compounds.

NCERT Solutions for Class 10 Science Chapter 5 – NCERT Exercise – Page 91

1. Which of the following statements is not a correct statement about the trends when going from left to right across the periods of periodic Table.

(a) The elements become less metallic in nature.

(b) The number of valence electrons increases.

(c) The atoms lose their electrons more easily.

(d) The oxides become more acidic.

Answer:

(c ) The atoms lose their electrons more easily.

We can say that this statement is incorrect because on moving left to right, the non-metallic character increases in elements. It becomes harder to lose electrons.

2. Element X forms a chloride with the formula XCl2 , which is a solid with a high melting point. X would most likely be in the same group of the Periodic Table as

(a) Na (b) Mg (c) AI (d) Si

Answer:

(b) Mg (Mg has a valency of 2 and Cl has a valency of 1, so they form MgCl₂).

3. Which element has

(a) two shells, both of which are completely filled with electrons?
(b) the electronic configuration 2, 8, 2?
(c) a total of three shells, with four electrons in its valence shell?
(d) a total of two shells, with three electrons in its valence shell?
(e) twice as many electrons in its second shell as in its first shell?

Answer:

(a) Neon

(b) Magnesium (atomic number 12)

(c) Silicon (atomic number 14)

(d) Boron (atomic number 5)

(e) Carbon (2,4)

4. (a) What property do all elements in the same column of the Periodic Table as boron have in common?

(b)What property do all elements in the same column of the Periodic Table as fluorine have in common?

Answer:

(a) All the elements in the same column of the Periodic table as boron have the same valency of 3.

(b) All the elements in the same column of the Periodic table as fluorine have the same valency of 1. They gain one electron to complete their octet.

5. An atom has electronic configuration 2, 8, 7.

(a) What is the atomic number of this element?

(b) To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.)

N(7) F(9) P(15) Ar(18)

Answer:

(a) 2+8+7 = 17.

Hence, the atomic number of this element is 17.

(b) Electronic configuration of:

N(7) = 2,5

F(9) = 2,7

P(15) = 2,8,5

Ar(18) = 2,8,8

Since Fluorine has the same number of valence electrons in its outermost shell as the given element, it would be chemically similar to it.

6. The position of three elements A, B and C in the Periodic Table are shown below –

(a) State whether A is a metal or non-metal.

(b) State whether C is more reactive or less reactive than A.

(c) Will C be larger or smaller in size than B?

(d) Which type of ion, cation or anion, will be formed by element A?

Answer:

Keeping the trends of the Periodic Table in mind, we can answer these questions.

(a) Element A is a non-metal.

(b) Element C is less reactive than A.

(c) C is smaller in size than B.

(d) A will form an anion. ( negative ion)

7. Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?

Answer:

Electronic Configuration of:

Nitrogen: 2,5

Phosphorus: 2,8,5

As we move down the group in a periodic table the electronegativity of the elements decreases because of increasing distance between nucleus and outermost shell. So, Nitrogen will be more electronegative than phosphorus.

8. How does the electronic configuration of an atom relate to its position in the Modern Periodic Table?

Answer:

The electronic configuration of an atom determines the number of valence electrons. Atoms with the same number of electrons in its valence shell are kept in the same group as they share similar chemical properties. Hence, electronic configuration helps decide the position of an element on the basis of its valence electrons.

9. In the Modern Periodic Table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium?

Answer:

The outermost shell of calcium has two electrons. Hence, the element with atomic number 12 with same number of valence electrons will have properties resembling to calcium.

10. Compare and contrast the arrangement of elements in Mendeléev’s Periodic Table and the Modern Periodic Table.

Answer:

The Mendeléev’s Periodic Table and the Modern Periodic Table are different in the following ways:

Topics Covered in Periodic Classification of Elements Class 10 Science

• Classification of Elements
  • Döbereiner’s Triads
  • Newlands’ Law of Octaves
• MENDELÉEV’S PERIODIC TABLE
  • Achievements of Mendeléev’s Periodic Table
  • Limitations of Mendeléev’s Classification
• Modern Periodic Table
  • Position of Elements in the Modern Periodic Table
  • Trends in the Modern Periodic Table
  • Trends in the Modern Periodic Table-Metallic and Non-metallic Properties

NCERT Solutions for Class 10 Science Chapter 6 – Life Processes, contains solutions to various questions in Exercises for Chapter 6. Life Processes Class 10 NCERT Solutions have been explained in a simple and easy to understand manner. We are providing NCERT Solutions for Class 10 all subjects which can be accessed by clicking here.

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NCERT Solutions for Class 10 Science Chapter 6 – Intext Questions – Page 95

1. Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans?

Answer:

Diffusion is insufficient to meet the oxygen requirements of multicellular organisms like humans because it is a slow process and multicellular organisms require rapid diffusion of oxygen throughout their large bodies.

2. What criteria do we use to decide whether something is alive?

Answer:

There are several movements and processes like walking, digestion, breathing, requirement of food and water, etc that can be used as a criteria to decide if something is alive. Many times, a lot of processes are not visible to the naked eyes. In such cases, presence of life is considered as a criteria to decide whether something is alive or not.

3. What are outside raw materials used for by an organism?

Answer:

Outside raw materials used by an organism are food, water and oxygen. However, the requirements of these raw materials also varies with the complexity and environment of an organism.

4. What processes would you consider essential for maintaining life?

Answer:

Life processes like respiration, circulation, transportation, digestion, and excretion are essential for maintaining life. These processes help the body of the organism to generate energy, remove waste products and ensure a healthy life.

Intext Questions – Page 101

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

Answer:

In autotrophic nutrition, the organism prepares its own food with the help of water, sunlight and CO₂ in the presence of chlorophyll. This process is performed by plants and green bacteria and is also called photosynthesis.

Heterotrophic nutrition occurs when an organism is dependent on other organisms for food. All animals and most organisms have a heterotrophic mode of nutrition.

2. Where do plants get each of the raw materials required for photosynthesis?

Answer:

Plants make food by the process of photosynthesis. The materials required for photosynthesis are carbon dioxide, sunlight, water and chlorophyll.

Carbon dioxide is absorbed from the atmosphere through stomata. Water from ground is transported to the leaves through root, chlorophyll is present in the leaves of the plants and sunlight in daytime gets trapped in this pigment to aid the process of photosynthesis.

3. What is the role of the acid in our stomach?

Answer:

The acid present in our stomachs is HCl. It has the following roles:

• It creates an acidic environment in the stomach. Due to this, the protein digesting enzyme pepsinogen is activated and converted into pepsin.
• It plays a major role in digestion by breaking down large and complex molecules.
• It also kills of disease causing harmful pathogens o keep us healthy.

4. What is the function of digestive enzymes?

Answer:

Digestive enzyme aid the process of digestion and make digestion simpler and easier by breaking large food molecules into simpler form. These simpler molecules are easily absorbed by our body giving us energy to function.

5. How is the small intestine designed to absorb digested food?

Answer:

The surface of the small intestine has small bristles called micro villi. These protrusions increase the surface area of the intestine to facilitate the diffusion of food into blood vessels. The food is absorbed in blood vessels present in the intestine and is carried to the whole body by blood.

Intext Questions – Page 105

1. What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?

Answer:

Oxygen is more abundant in atmosphere than in water. Terrestrial organisms can easily obtain this oxygen by the process of respiration whereas aquatic organisms need to breathe faster to get the required amount of oxygen to survive.

2. What are the different ways in which glucose is oxidised to provide energy in various organisms?

Answer:

In a cell, the glucose first breaks into three carbon compounds called as called pyruvate. These compounds further breaks down by either of the two processes given below, depending on the organism.

(i) Aerobic respiration: The pyruvate is broken into water and carbon dioxide, and energy is released. It occurs in the presence of oxygen. This usually occurs in mitochondria.

(ii) Anaerobic Respiration in Muscles: When the oxygen supply is not sufficient in body while running or exercising, the pyruvate is broken into lactic acid.

(iii) Anaerobic Respiration in yeast: Pyruvate is converted into ethanol and carbon dioxide. There is no oxygen present in this process. This is also called fermentation.

 3. How is oxygen and carbon dioxide transported in human beings?

Answer:

Oxygen and Carbon dioxide are transported in human beings are transported by blood. Cells require oxygen and need to get rid of the carbon dioxide released. Oxygen is absorbed by the blood capillaries from the alveoli present in lung by diffusion and is transported to the cells. The carbon-dioxide brought from cells is absorbed by the alveoli from blood capillaries by diffusion.

4. How are the lungs designed in human beings to maximise the area for exchange of gases?

Answer:

When we breathe in, we lift our ribs, flatten our diaphragm and chest cavity becomes larger. This action leads in inhalation of air and fills the lungs with air. The passage inside the lungs keeps on subdividing into smaller tubes which terminate in a balloon like part called alveoli. This alveoli expands and is filled with air and is the place where exchange of gases takes place. It contains an extensive network of blood vessels where oxygen and carbon dioxide are exchanged.

NCERT Solutions for Class 10 Science Chapter 6 – Intext Questions – Page 110

1. What are the components of the transport system in human beings? What are the functions of these components?

Answer:

The transport system in human beings consists of the heart, the blood, and the blood vessels. All of these play important roles in our body.

The heart pumps oxygenated blood to be carried throughout the body. It receives deoxygenated blood from body and sends it to lungs for oxygenation. Then it receives oxygenated blood from lungs and pumps it out to be transported to the whole body.

Blood transports oxygen, nutrients, CO2, and nitrogenous wastes to specific site of the body, and blood vessels are the organs that carry the blood to and from the different parts of the body.

2. Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?

Answer:

Mammals and birds are warm blooded animals who have a constant body temperature irrespective of the environment. To maintain this temperature, a lot of oxygenated blood is needed to produce the required amount of energy. To get the maximum out of the  circulatory system and make them efficient to produce sufficient energy, it is necessary to separate oxygenated blood and deoxygenated blood.

3. What are the components of the transport system in highly organised plants?

Answer:

Highly organised plants have the following two main components:

(i) Xylem: Tissues that conduct water and minerals from roots to rest of the plant parts.

(ii) Phloem: Tissues that transport food materials from leaf to other parts of the plant

4. How are water and minerals transported in plants?

Answer:

Water and minerals in plants are transported by xylem. Xylem are conducting tissues that are interconnected with all the parts of a plant like roots, stem, and leaves to form a continuous system of water-conducting channel. When plant loses water through transpiration, a suction pressure is created that forces water from ground and through roots into the xylem tissues. These tissues carry minerals and water throughout the plants.

5. How is food transported in plants?

Answer:

Phloem is responsible for transporting food in plants. An osmotic pressure is created when ATP (energy) is spent. Due to this pressure, the food stored in leaves in the form of carbohydrates starts flowing in the phloem and reaches other parts of the plants as per requirement.

Intext Questions – Page 112

1. Describe the structure and functioning of nephrons.

Answer:

Nephrons are tubular structures made of glomerulus, Bowman’s capsule, and a long renal tube. They are basic filtration units present in kidneys.

The blood initially enters the renal artery which branches into capillaries associated with glomerulus. At Bowman’s capsule, the water and solute are transferred to the nephron. Some substances like glucose, amino acids, salts and a major amount of water, are selectively re-absorbed as the urine flows along the proximal tubule and unwanted substances are added to the urine. This filtrate then moves down into the loop of Henle, where more water is absorbed. After this, the remaining filtrate moves upwards into the distal tubule and gets collected in the collecting duct where urine from many nephrons is collected. This collected urine moves into the urinary bladder via ureter, where it is stored until excreted.

Diagram of nephron:

2. What are the methods used by plants to get rid of excretory products?

Answer:

Plants get rid of excretory products by many different methods. They get rid of dead cells and leaves by losing them. Some of these leaves may also contain waste products stored in cellular vacuoles. Other waste products are stored as resins and gums, especially in old xylem. Some waste is lost in the nearby soil.

They lose excess water by transpiration.

3. How is the amount of urine produced regulated?

Answer:

The amount of urine produced depends on the amount of water consumed and the amount of waste produced in the body. Various other factors like hormones, functioning of human, health, and environment.

NCERT Solutions for Class 10 Science Chapter 6 – NCERT Exercise – Page 113

1. The kidneys in human beings are a part of the system for

(a) nutrition
(b) respiration
(c) excretion
(d) transportation

Answer:

(c ) Excretion

2. The xylem in plants are responsible for

(a) transport of water
(b) transport of food
(c) transport of amino acids
(d) transport of oxygen

Answer:

(a) transport of water

3. The autotrophic mode of nutrition requires

(a) carbon dioxide and water
(b) chlorophyll
(c) sunlight
(d) all of the above

Answer:

(d ) all the above. (by the process of photosynthesis)

4. The breakdown of pyruvate to give carbon dioxide, water and energy takes place in

(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) nucleus

Answer:

(b ) mitochondria

5. How are fats digested in our bodies? Where does this process take place?

Answer:

Food is completely digested in our small intestines. Food components like fats, carbohydrates and proteins are broken down and digested here. To facilitate the process of digestion, the small intestine receives secretions from liver and the pancreas.

First, the acidified food coming from stomach is made alkaline in nature with the help of bile from liver. This allows the pancreatic enzymes to work on it.

The food contains large globules of fats which are difficult to digest. Bile juice breaks down these large globules into smaller units. Then pancreatic juice, which contains enzyme called lipase for breaking down emulsified fats, works on it.

Finally, the walls on the intestine contain glands that secrete intestinal juice. This juice contains enzymes that convert fats into fatty acids and glycerol which can be easily absorbed by the body.

6. What is the role of saliva in the digestion of food?

Answer:

Saliva contains an enzyme called salivary amylase to break the large starch molecule into smaller sugar molecules. It makes it easier to be absorbed by the alimentary canal, facilitating digestion.

7. What are the necessary conditions for autotrophic nutrition and what are its by-products?

Answer:

The necessary conditions for autotrophic nutrition include carbon dioxide, sunlight, chlorophyll, water, and ability to store food.

Plants make their own food by the process of photosynthesis. They take carbon dioxide, sunlight present in atmosphere and convert them into starch.

Starch are carbohydrates that give them energy. These are not necessarily used immediately and are sometimes stored for future use.

The by-products obtained in this process are oxygen and carbohydrates.

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

Answer:

Difference between aerobic an anaerobic respiration are:

Some organisms that use anaerobic mode of respiration are yeast, some kinds of bacteria and some parasites.

9. How are the alveoli designed to maximise the exchange of gases?

Answer:

The passage inside the lungs keeps sub-dividing in smaller tubes, which terminate in balloon-like structures, called as alveoli. Alveoli contains a dense network of blood vessels, where the exchange of gases takes place.

As we breathe, we lift our ribs, flatten our diaphragm and our chest cavity becomes larger and air is flushed into the lungs which fills the expanded alveoli.  Oxygen is absorbed by the blood capillaries from the alveoli present in lung by diffusion and is transported to the cells. The carbon-dioxide brought from cells is absorbed by the alveoli from blood capillaries by diffusion.

The lungs always contain a residual volume of air between the process of inhalation and exhalation so that there is sufficient time for oxygen to be absorbed and carbon dioxide to be released.

10. What would be the consequences of a deficiency of haemoglobin in our bodies?

Answer:

Haemoglobin binds oxygen to the blood cells. It is a protein, responsible for transportation of oxygen to the body cells for cellular respiration. In the case of deficiency of haemoglobin, oxygen will not be properly circulated throughout the body, leading to lack of oxygen and hampering some important processes.

The deficiency of haemoglobin is also called anaemia and occurs due to lack of iron.

11. Describe double circulation of blood in human beings. Why is it necessary?

Answer:

In double circulation of blood, the oxygenated and deoxygenated blood are separated.

In a single cycle, blood goes into heart twice. The double circulation includes pulmonary circulation and systematic circulation.

1. Pulmonary Circulation: In this step, the right ventricle pumps deoxygenated blood into the lungs where it is oxygenated. The oxygenated blood is brought back to the left atrium, then it is pumped into the left ventricle and finally blood goes into the aorta for systemic circulation.
2. Systematic Circulation: The oxygenated blood then, is pumped to various parts of the body from the left ventricle. The deoxygenated blood from different parts of the body passes through vena cava to reach right atrium. The right atrium transfers the blood into right ventricle. This completes the double circulation.

It is necessary to separate oxygenated and deoxygenated blood for efficient functioning of the circulatory system.

12. What are the differences between the transport of materials in xylem and phloem?

Answer:

Xylem transports water and minerals from roots to all plant parts. Phloem transports food from leaves to all other parts of plant body.

13. Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning.

Answer:

Alveoli:

1. Structure: Alveoli are balloon like structures, are one cell thick, and contain extensive network f blood capillaries.
2. Function: Site of exchange of gases in lungs. Oxygen and carbon dioxide are exchanged through the blood capillaries present in and around alveoli.

Nephrons:

1. Structure: Nephrons are tubular structures made of glomerulus, Bowman’s capsule, and a long renal tube.
2. Function: They are basic filtration units present in kidneys. Blood enters the kidney through renal artery and the nitrogenous waste in the form are collected by collecting duct in the form of urine.

Topics Covered in Life Processes Class 10 Science

• What are Life processes?
• Nutrition
  • Autotrophic Nutrition
  • Heterotrophic Nutrition
  • How do Organisms obtain their Nutrition?
  • Nutrition in Human Beings
• Respiration
• Transportation
  • Transportation in Human Beings
  • Transportation in Plants
• Excretion
  • Excretion in Human Beings
  • Excretion in Plants