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Students can Download History Chapter 7 Freedom Movement Questions and Answers, Notes Pdf, KSEEB SSLC Class 10 Social Science Solutions helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Social Science History Chapter 7 Freedom Movement

Class 10 Social Science Freedom Movement Textual Questions and Answers

I. Fill in the blanks with suitable answers:

Question 1.
The Indian National Congress was found in the year ………….
Answer:
1885.

Question 2.
The Drain Theory was forwarded by ………….
Answer:
Dadabai Navoroji.

Question 3.
Swarajya is my birth right was declared by ……………..
Answer:
Bal Gangadhar Tilak.

Question 4.
Bala Gangadhar Tilak published ……………… newspaper in Marathi.
Answer:
Kesari.

Question 5.
A secret organization by name ‘Abhinava Bharathi’ belonged ……………….
Answer:
Revolutionaries.

II. Choose the right option and fill in the blanks:

Question 1.
The founder of Indian National Congress is ………………
a) Mahatma Gandhiji
b) A.O. Hume
c) Balagandhar Tilak
d) Gopala Krishna Gokhale
Answer:
(b) A. O. Hume

Question 2.
‘Maratha’ paper was published by ………………
a) Jawahara Lai Nehru
b) Ras Bihari Bose
c) Balagandhara Tilak
d) V.D.Saavarkar
Answer:
(c) Bal Gangadhar Tilak

Question 3.
Muslim League was founded in ……………….
a)1924
b) 1922
c) 1929
d) 1906
Answer:
(d) 1906

Question 4.
The Viceroy who implemented the Bengal division was ………………….
a) Lord Cornwallis
b)Dalhousie
c) Lord Curzon
d)Robert Clive
Answer:
(c) Lord Curzon

III. Discuss in a group and answer the following:

Question 1.
Which were the organizations that were present before the founding of the Indian National Congress?
Answer:
The Hindu Mela, the East Indian Association, Poona Public Sabha, and the Indian Association.

Question 2.
What were the demands of Moderates placed in front of the British?
Answer:
The demands put forward before the British by the moderates were:

1. Development of Industries in India,
2. Reduction of military expenses in the British army.
3. Improvement in the educational standard.
4. Forcing the British Government to take up studies about poverty in the country.

Question 3.
Explain the Drain Theory.
Answer:
Moderates were the first to study the ill effects of British rule on India. They explained the drain of resources of India into England through scientific statistics and called it ‘Drain Theory’. By increasing the import and reducing the export, the British facilitated the draining out of precious Indian resources into India. Just like Dadabai Navoroji, R.C. Datta too published books explaining the draining of Indian resources into England.

Question 4.
Name the revolutionaries of the Indian Independence Movement.
Answer:
Aurobindo Gosh, V. D. Saavarakar, Ashwini Kumar Datta, Rajanarayana Bose, Rajguru, Chakikar brothers, Vishnu Shastri, Champukar, Shyamaji Krishnaverma, Ras Bihari Gosh, Madam Cama, Kudiram Bose, Ramprasad Bismil, Ashvakulla Khan, Bagath Singh, Chandrashekar Azad, Jatin Das are more prominent among the revolutionaries of Indian Independence Movement.

Question 5.
Discuss the role of Balagangadhar Tilak in the Indian Independence Movement.
Answer:
Bala Gangadhar Tilak was one of the members of the Radical group. The aim of the Radical group was an Independent India. Bala Gangadhar Tilak started preparing the common people for freedom struggle. He declared “Swaraj is My Birth Right and I will get it back”. Through religious functions like Shivaji Jayanti, Ganesh festival he started organizing people for the freedom movement. He published ‘Kesari’ in Marathi and ‘Maratha’ in English news paper and used them as weapons to criticize the British administration. He called the people for the active participation in the freedom struggle. Thus Tilak played an important role in the freedom struggle.

Question 6.
What were the reasons for the withdrawal of the Bengal Division?
Answer:
The division of Bengal in 1905 was opposed by the Indian National Congress. Still, the Bengali language could unite the Hindu and Muslim communities. Raksha Bandhan, a Cultural festival was held to bring in unity among Hindus and Muslims. The division of Bengal resulted in widespread protection across the country. The radicals took the issue to the doorsteps of common people. They called for boycotting of foreign goods and the institutions that encourage it. Indians were encouraged to use local goods. The British government withdrew the Bengal division order in 1911.

Class 10 Social Science Chapter 7 Freedom Movement Additional Questions and Answers

I. Fill In The Blanks With Suitable Answers:

Question 1.
Due to their differences in ideology, beliefs and execution styles they are identified as ……….. and ………..
Answer:
Moderates, Radicals.

Question 2.
The period between …………. is called as the Age of Moderates.
Answer:
CE 1885 and 1905.

Question 3.
Bengal had more concentration of ………….. and ………… people.
Answer:
Hindu, Muslim.

Question 4.
The British divided Bengal in ……………
Answer:
1905.

Question 5.
Bal Gangadhar Tilak published …………… paper in the English language.
Answer:
‘Maratha’.

II. Multiple Choice Questions:

Question 1.
The participation of Indians in the legislature process was
a. 1833
b. 1861
c. 1909
d. 1919
Answer:
b. 1861

Question 3.
Vernacular press Act was implemented by
a. Lord Lytton
b. A.O. Hume
c. Warren Mattings
d. Lord Rippon
Answer:
a. Lord Lytton

Question 5.
Moderates had faith in the rule of
a. British rule and Judiciary
b. British admisnistration
c. Self rule and Hema pulse
d. Own government
Answer:
a. British rule and Judiciary

Question 7.
‘Abhianava Bharatha’ and ‘Anushecla Sarniti’ were the Iwo important secret organizations of_______
a. Revolutionaries
b. Radicals
e. Moderates
d. Gandh Ian era
Answer:
Revuluthnuiries

Question 9.
“Swaraj is my Birth Right. I would definitely get it back”, declared by
a. Gandhiji
b. Nehru
c. Tilak
d. C.R.Das
Answer:
c. Tilak

Four Marks Questions:

Question 1.
Who founded INC? Which were its aims?
Answer:
A.O.Hume plays ed an important role in the formation of Indian national congress in 1885. Hume was a retired British civil servant and met political leaders in cities like Madras, Bombay, and Calcutta and discussed the various issues of public importance.

Aims of INC:

• The congress declared that achieving national unity as its primary aim during its first national convention.
• It thrived to achieve unity among the diverse cultural and social paths of India.
• The leaders of this period also had the commitment to achieve it.

Question 4.
What was role of Revolutionaries in freedom struggle?
Answer:

• Revolutionaries dreamed ofattaining complete freedom.
• They believed that they can drive away the British by employing violent methods. Aurobindo Gosh.
• D. Saavarkar, Ashwini Kumar Datta. Bagath Singh, Chandrashekar Azad are important revolutionaries.
• They established secret associations across the country
• Started collecting weapons and money for an armed struggle against the British.
• A secret organization named ‘Lotus and Dragger was founded in England.
• “Gadha’ in USA can be recalled here
• Abhinava Bharatha’ and ‘Anusheela Samiti’ were two important secret

We hope the KSEEB SSLC Class 10 History Solutions Chapter 7 Freedom Movement help you. If you have any query regarding Karnataka SSLC Class 10 History Solutions 7 Freedom Movement, drop a comment below and we will get back to you at the earliest.

Students can Download History Chapter 4 Opposition to British Rule in Karnataka Questions and Answers, Notes Pdf, KSEEB SSLC Class 10 Social Science Solutions helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Social Science History Chapter 4 Opposition to British Rule in Karnataka

Class 10 Social Science Opposition to British Rule in Karnataka Textual Questions and Answers

I. Fill in the blanks with suitable answers :

Question 1.
The First Anglo-Mysore war took place between ……….. and ………
Answer:
Hyder ali and British.

Question 2.
The Second Anglo-Mysore war ended with …….. treaty.
Answer:
Salbhai.

Question 3.
Kittur Chenamma adopted a boy named ………….
Answer:
Shivalingappa.

Question 4.
Rayanna of Kittur state belonged to ………. village.
Answer:
Sangolli.

Question 5.
Surapura’ is in the present district of ………..
Answer:
Yadagiri.

Question 6.
The Bedas of ……….. village of Belgaumdistrict rebelled against the British.
Answer:
Hulagali.

Question 7.
The Amara Sulya rebellion was basically a ………. rebellion.
Answer:
Farmers.

II. Discuss in groups and answer later :

Question 1.
How did Hyder Ali come to power?
Answer:
After the death of Aurangzeb, Moghul empire was weakened. As a result the Moghul lost political control over South India. A lot of political struggles took place in Carnatic region. Before this, the death of Chikkadevaraj wodeyar in 1704 created the various political challenges in Mysuru state. His death created the problems of succession and administration.

All these developments Cloded the politics of Mysuru. Hyder Ali gained prominence in this scenario of uncertainty that clouded over the Mysuru and carnatic region. He joined Mysore Army as an ordinary soldier, but was known for his shrewd political moves. He came into prominence during the siege of Devanahalli and militaiy action against Nizam of Arcot.

Question 2.
What are the effects of Second Anglo-Mysore war?
Answer:
The Second Anglo-Mysore was started in 1780. Hyder Ali was defeated in a battle held in Port Novae by the British. This increased the confidence of the British and also changed the direction of the battle. But they suffered financial setbacks in Pulieat and Soliungur.

Meanwhile, by entering ‘Salbai Agreement’, the British were successful in winning over the Marathas and Nizam of Hyderabad to their side. Hyder Ali died due to illness during the war. The British tried to take advantage of Hyder Ali’s death by invading Mangalore and Bidanoor. They also tried to instigate the rulers of Calicut and Malabar regions against Tippu Sultan. Tippu Sultan defeated the British. The “Treaty of Mangalore” ended the Second Anglo-Mysore War in 1784.

Question 3.
What were the conditions of Srirangapatanam treaty?
Answer:
Following are the conditions of Srirangapatanam treaty:

1. Tippu was forced to. part with half of his Kingdom, was forced pay three crore rupees as war damage fee.
2. He had to pledge two of his children as a guarantee against the payment.
3. He was also forced to release the Prisoners of War.

Question 4.
The Fourth Anglo Mysore strengthened the position of British in Mysore. Discuss.
Answer:
Fourth Anglo-Mysore War took place between Tippu Sultan and British in 1799. The British were able to destroy the strong fort. Tippu died while fighting the British in 1799. With the death of Tippu Sultan, the British were happy as if the whole of India came under their rule. Most of the territories under Tippu’s rule were shared among the British, Marathas, and Hyderabad Nizam. A small territory was handed were to the royal representative of Mysore Wodeyars. This region came to be known as Mysore Princely State.

Question 5.
Explain the method of resisting British power by Dondiya Wagh.
Answer:
Dondiya started his career as a cavalry soldier in Hyder Ali’s army and grew to the position of military general. He built his own private army and fought along with Tippu Sultan. Due to differences with Tippu, he was imprisoned. The British released him from prison after the Fourth Anglo-Mysore War. He built a small army and started his operations. He organized the army with the unhappy soldiers of Tippu’s army and the feudatory rulers who had lost power. He captured Bidanoor and Shivamogga forts and made an unsuccessful attempt to capture Chitradurga fort. Lord Wellesley tried check this rebellion.

An attack was organized on Shivamogga, Honalli, Harihara and other places under the control of Dondiya.
Dondiya lost his base. After the capture of Shikaripura, Dondiya ran away towards Gutti, which was under the control of Nizam of Hyderabad. When Nizam’s army attacked Gutti, Dondiya had to run towards the regions of Maratha. The Maratha army attacked him and captured most of his horses, camels and arms. In spite of these, he continued his war fare. When he was caught in between Maratha Army and Nizam’s army, the British attacked him near Yelaparavi and killed him at Konagal.

Question 6.
Explain the method adopted by Rayanna to fight the British.
Answer:
Sangoli Rayanna developed a sense of nationalism and went on organizing an army. He went on organizing secret meetings at sensitive geographical locations. He aimed at looting the treasury and taluk offices of the British. He had an army of five hundred men. He became furious with the villagers who were assisting the British army. The British devised a cunning strategy to capture Rayanna. They encouraged Desais who yere opposing Rani Chenamma. An Amaldhar named Krishnaraya joined hand with them. Thus Rayanna was cunningly captured and brought down to Dharwad. He was declared as an offender and was hanged till death.

Question 7.
Explain the contribution of Puttabasappa of Kodagu in freedom struggle.
Answer:
Puttabasappa took over the leadership of the rebellion. The rebellion started in the hilly region. Puttabasappa organized the rebels and calmed down the people. He declared that tax on tobacco and saflt will be with drawn, if the rebel government assumes power. The rich farmers, land owners and local chieftains were assured of this move. Puttabasappa killed an Amaldhar who was known for his brutality further increased the popularity of Puttabasappa.

The rebels marched towards Mangalore to capture it. The British were engaged in fortifying their fort in Mangalore. The rebels marched towards Mangalore through Panimangalore and Bantwal. They looted the treasury and prison of Bantwal.

The British sought the army of Thalacheri, Kannur and Bombay to quell this uprising. On hearing this development, Puttabasappa and his associated fled towards Sulya. The British captured them with the help of people in Kodagu. Puttabasappa, Lakshmappa, Bangarasa, Kedambadi Ramayaih Gowda and Guddemane Appaih were hanged till death.

Question 8.
Discuss the Surapura rebellion in brief.
Answer:
Surapura is at fifty kilometers from the present day Yadgir. Venkatappa Nayaka came to throne at an early age. His ascendance to throne was opposed by Krishna Nayaka’s brother Peddanayaka. This resulted in internal struggles. The British interfered in the affairs of Surapura. In 1842, they appointed Medes Taylor as their political agent and gained proxy power over Surapura.

In 1857, it came to the notice of the government that the representatives of Nana Saheba were present in Surapura. The British appointed an officer named Campbell to report on the various activities of the King. The officer submitted a report to the resident of Hyderabad that the King is involved in misadministration. The British army captured Surapura in 1858. The war continued, there are confusions regarding Venkatappa Nayaka’s end.

Class 10 Social Science Opposition to British Rule in Karnataka Additional Questions and Answers

Question 1.
British entered an agreement with Hyder Ali through ……….. treaty.
Answer:
Madras.

Question 2.
The treaty of ……….. ended the second Anglo-Mysore War.
Answer:
Mangalore.

Question 3.
Tippu Sultan signed the Treaty of ……….. and with this Third Anglo-Mysore War came to an end.
Answer:
Srirangapatna.

Question 4.
The British attacked Kittur under the leadership of ………….
Answer:
Colonel Deak.

Question 5.
In the year ………. Rayanna was hanged till death.
Answer:
1831.

Question 6.
The rebellion of Hulagali is called ………… rebellion.
Answer:
Bedas.

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• Karnataka SSLC Maths Model Question Papers
• Karnataka SSLC Science Model Question Papers
• Karnataka SSLC Social Science Model Question Papers
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• Karnataka SSLC English Model Question Papers
• Karnataka SSLC Hindi Model Question Papers

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Tili Kannada Text Book Class 10 Answers Solutions Guide

Class 10 Tili Kannada Gadya Bhaga Karnataka State Board Solutions

• Chapter 1 Ona Marada Gili
• Chapter 2 Asi Masi Krishi
• Chapter 3 Ganayogi Pandita Puttaraja Gawai
• Chapter 4 Hakkigudugala Nigudha Jagattu
• Chapter 5 Kaphi Kappu
• Chapter 6 Tussauds Wax Museum
• Chapter 7 Nanna Gopala

Class 10 Tili Kannada Padya Bhaga Karnataka State Board Solutions

• Chapter 1 Eni
• Chapter 2 Bodhivrukshada Hadu
• Chapter 3 Savi Chaitra
• Chapter 4 Saddu Madadiru!
• Chapter 5 Guri
• Chapter 6 Moodal Kunigal Kere
• Chapter 7 Vachanagalu
• Chapter 8 Nittotadali Haydanu Bittamandeyali

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• Chapter 1 Kallara Guru
• Chapter 2 Honge Bevugala Hadu

Karnataka SSLC Class 10 Tili Kannada Grammar and Writing Skills

• Anvayika Vyakarana
• Gadegalu
• Patra Lekhana
• Prabandha Lekhana

KSEEB SSLC Class 10 Maths Solutions
KSEEB SSLC Class 10 Science Solutions

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Students can download Class 10 Science Chapter 8 How do Organisms Reproduce? 11 Human Eye and Colourful World Important Questions, KSEEB SSLC Class 10 Science Important Questions and Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka SSLC Class 10 Science Important Questions Chapter 8 How do Organisms Reproduce?

Question 1.
Define reproduction.
Answer:
The biological process by which organisms produce young ones, which resemble themselves, is known as reproduction.

Question 2.
Explain the importance of reproduction in organisms.
Answer:
Reproduction is an important characteristic feature of living organisms. It is an essential life process, which not only helps in survival of the species but also helps in continuity of that race and group. Young ones replace the old and dying ones. These young ones feed, grow and reproduce again.

Reproduction also helps in increasing the population of the species. Reproduction acts as a vehicle of organic evolution by transmitting advantageous variations to the offspring. In the absence of reproduction, species would perish and life on earth would be wiped out of existence.

Question 3.
How are the modes for reproduction different in unicellular and multicellular organisms?
Answer:
Unicellular organisms generally reproduce asexually by fission, budding or spore formation. Their body has just one cell, which can easily multiply by simple cell division, budding or spore formation.

Such methods are difficult in multicellular organisms as they have a complex organization of specialized tissues. Therefore, multicellular organisms usually reproduce by sexual method.

Question 4.
How does reproduction help in providing stability to populations of species?
Answer:
Population of any species will remain stable when the birth rate is equal to the death rate. Reproduction produces new individuals, which will compensate for the deaths. Thus reproduction provides stability to population of a given species.

Question 5.
Is reproduction essential for the survival of the individual?
Answer:
No, reproduction is not essential for the survival of the individual. An organism incapable of reproduction will also survive and complete its life cycle. However, reproduction is most essential for the growth in population and survival of the species. In the absence of reproduction, species will be wiped out of existence.

Question 6.
Why do organisms of the same species look similar?
Answer:
Reproducing organisms create new individuals similar to themselves. The individuals of the same species are the products that emerge from a common but similar blueprint. The DNA of the parents decides the blueprint. During reproduction, similar copies of the blueprint are produced by a process called DNA copying or DNA replication.

Question 7.
What is variation? Why are hereditary variations important?
Answer:
The young ones of any species may resemble their parents broadly but differ from them in many ways. The differences in the characters among the offspring compared to their parents is called variation. Variation is necessary for organic evolution.

Question 8.
What is DNA? What is its structure called? Where is it located in the cell?
Answer:
DNA stands for Deoxyribo Nucleic Acid. It is a macromolecule found in chromosomes of living beings and carries the genetic information of the organism. The structure of DNA is called ‘double helix’ structure.

Every eukaryotic cell has a nucleus, which contains chromosomes. These chromosomes carry information for the inheritance of characteristics from the parents to their offspring in the form of DNA molecules.

Question 9.
What is meant by DNA copying? What is its importance?
Answer:
The process of producing two identical copies of one original DNA molecule is called DNA copying. It is also called DNA replication. DNA replication is important because replication is needed during cell division. Replication allows the cell to be duplicated so that it can continue the process of cell division.

DNA copying enables organisms to beget young ones similar to their own kind. The process of DNA copying brings some variation each time. The surviving cells are similar to but subtly different from each other. This inbuilt tendency for variation during reproduction brings variations among individuals of the same species. This is useful for ensuring survival of the species.

Question 10.
Why is DNA copying an essential part of the process of reproduction?

OR

What is the importance of DNA copying in reproduction?

Answer:
The process of reproduction produces offspring similar to the parents. The exact blueprint of the body is inherited by the offspring due to DNA copying in the parent cell. This is why DNA copying is an essential part of reproduction.

Question 11.
Briefly describe the process of DNA copying.
Answer:
The replication of DNA occurs in every cell during cell division. The process requires the uncoiling of the two strands of the DNA helix. The two open strands serve as templates for the assembly of daughter strands. The assembly of new strands takes place in a complimentary mode.

This process is accompanied by the creation of an additional cellular apparatus. Then, the two DNA copies separate out each having its own cellular apparatus. Effectively, a cell divides to give rise to two cells.

Question 12.
Why is variation beneficial to the species but not necessarily for the individual?
Answer:
Populations of organisms reside in well-defined places or niches in the ecosystem. However, habitats or niches can change because of reasons beyond the control of the organisms. Variations allow organisms to exist in diverse habitats or niches. In the absence of variations, a species may be restricted to a particular area.

If this area gets drastically altered, due to various natural or human-made causes, the species may be wiped out. However, if some variations were present in a few individuals, they could colonize other habitats, reproduce and could survive. But if variations were present in a single organism there would be very little chance for it to survive. Therefore, variation is beneficial to the species but not necessarily for the individual.

Question 13.
Mention the major modes of reproduction seen in animals.
Answer:
Reproduction in animals takes place principally by two modes. They are asexual reproduction and sexual reproduction.

Question 14.
What is asexual reproduction?
Answer:
A type of reproduction in which the young one arises from a single parent and is genetically identical to the parent is called asexual reproduction. This process does not involve the fusion of gametes.

Question 15.
Give examples of animals that reproduce by asexual method.
Answer:
Lower order organisms like amoeba, yeast and hydra reproduce by asexual method.

Question 16.
What is the method of reproduction in amoeba?
Answer:
Amoeba is a unicellular organism. It reproduces asexually by the method of binary fission.

Question 17.
What is binary fission in reproduction? Name an organism that reproduces by this method.
Answer:
A kind of asexual reproduction in which a unicellular parent cell divides itself into two daughter cells is known as binary fission. Amoeba reproduces by binary fission.

question 18.
Explain the process of reproduction in amoeba.
Answer:

Amoeba is a unicellular eukaryotic organism. Its body is made of a single cell. It reproduces asexually by the method of binary fission. The adult amoeba becomes ready for binary fission. First the nucleus divides into two. A constriction develops dividing the cytoplasm into two parts.

The parent cell breaks at the point of constriction resulting in two daughter amoebae, which are identical to that of the parent. This is a form of asexual reproduction called binary fission.

Question 19.
What is multiple fission in reproduction? When does amoeba reproduce by multiple fission?
Answer:
A type of asexual reproduction in which a unicellular organism divides itself into three or more daughter cells that are identical to the parent organism is called multiple fission. Amoeba reproduces by the method of multiple fission under unfavourable or drought conditions to increase the chances of survival of daughter cells.

Question 20.
How does binary fission differ from multiple fission?
Answer:
Binary fission is a mode of asexual reproduction in which a unicellular parent organism divides itself into two daughter organisms through the process of cell division. This method produces two young ones. This happens under favourable conditions in the absence of any stress.

Multiple fission is a mode of asexual reproduction in which a unicellular parent organism undergoes cell division to produce more than two (many) young ones. This type of reproduction occurs under unfavourable or stressful conditions.

Question 21.
Explain binary fission in Leishmania.
Answer:
In Leishmania, binary fission occurs in a definite orientation in relation to flagella. Splitting of parent cell during fission takes place in a definite plane (longitudinally) with respect to flagellum at its ends.

Question 22.
Name the disease caused-by Leishmania.
Answer:
The disease kala-azar is caused by Leishmania.

Question 23.
What is the difference between binary fission in amoeba and leishmania?
Answer:
In amoeba, the splitting of the two cells during division can take place in any plane. Binary fission in Leishmania happens in a longitudinal plane.

Question 24.
What is the mode of reproduction in yeast? Explain briefly.
Answer:
Yeast reproduces asexually by a process known as budding. The parent yeast cell forms a small bud on its own body, which eventually develops into a new individual. On maturity, the bud breaks away from the parent body and gets independent existence.

Question 25.
Describe a simple activity to show the reproduction in yeast cells.
Answer:
Yeasts reproduce asexually by budding. This can be observed under a powerful microscope. Dissolve about 10 g of sugar in 100 mL of water. Take 20 mL of this solution in a test tube and add a pinch of yeast granules to it. Put a cotton plug on the mouth of the test tube and keep it in a warm place.

After 1 or 2 hours, put a small drop of yeast culture from the test tube on a slide and cover it with a cover slip. Observe the slide under a microscope. Now we see many yeast cells and some of them show budding on the outside wall of their cell wall.

Question 26.
How do fungi reproduce?
Answer:
Fungi reproduce asexually by the method of spore formation. Spore formation is a method of asexual reproduction in which the parent plant produces hundreds of tiny spores, which can grow into new organisms under favourable conditions.

Question 27.
Describe how growth of fungus on a piece of moist slice of bread takes place.
Answer:
Take a slice of moistened bread. Keep it on a plate in a cool, moist and dark place. Observe the surface of the slice with a magnifying glass for over a week. A white cottony mass appears on the surface of the bread which gradually turns black. This is due to the growth of bread mould (a fungus). The spores of fungus will always be hanging in air.

They settle on the moist bread which provides favourable conditions for their germination. The cottony growth on the bread is a mass of vegetative filaments of the fungus. These filaments develop asexual sporangia which are black in colour. This is why the cottony mass turns black after a couple of days.

Question 28.
How will an organism be benefited if it reproduces through spores?
Answer:
The following are some benefits for the organism that reproduces through spore formation:

• Reproduction through spores is simple as well as faster.
• The spores are light in weight, thus keep floating in the air. This helps in their dispersal.
• The spores are covered with a thick layer, which enables them to remain dormant in unfavourable conditions.

Question 29.
What is fragmentation? Name an organism that reproduces by this method.
Answer:
Fragmentation is a form of asexual reproduction wherein a parent organism breaks into fragments and each fragment is capable of growing independently into a new organism. Spirogyra is a filamentous alga. It reproduces by the method of fragmentation.

Question 30.
Why do most of the multicellular organisms need more complex methods of reproduction?
Answer:
Multicellular organisms cannot simply divide cell-by-cell or regenerate from its fragments. This is because, many multicellular organisms are not simply a random collection of cells. They have many different specialised cells, which are organised into tissues, and tissues are organised into organs designed to carry out varied but different activities.

These organs are placed at definite positions in their body. In such a carefully organised situation, cell-by-cell division or regeneration of fragments would be impractical. Therefore, multicellular organisms need to use more complex ways of reproduction.

Question 31.
What is regeneration? Name an organism that reproduces by this method.
Answer:
Regeneration is a type of asexual reproduction in which the organism has the ability to give rise to new individual organisms from their body parts. This method of asexual reproduction is found in animals like planaria, earthworm, starfish etc.

Question 32.
With the help of a suitable diagram explain asexual reproduction in planaria.
Answer:

Planaria reproduces asexually by the method of regeneration. In this mode of reproduction, planaria simply constricts its body until it actually separates into two parts. One part is the anterior end and the other is the posterior end. Each of the parts then regenerates the missing portion and thus two complete individuals arise.

Even if planaria is cut into a number of pieces, each piece grows into a complete organism by regeneration. Specialised cells carry out regeneration. These cells proliferate and make large numbers of cells. From this mass of cells, different cells undergo changes to become various cell types and tissues. These changes take place in an organised sequence.

Question 33.
Differentiate between regeneration and fragmentation.
Answer:
Regeneration:

1. This type of reproduction is seen in fully differentiated organisms.
2. It is carried out by specialized cells.
3. Specialized cells proliferate to form a mass of cells from which various cell types and tissues emerge.

Fragmentation:

1. This type of reproduction is seen in organisms with simple body organization.
2. No specialized cells are involved in this process.
3. The parent body breaks into segments and each segment develops into a new organism

Question 34.
With the help of a suitable diagram explain the method of reproduction in hydra.

Answer:
Hydra is a multicellular organism. It reproduces asexually by the method of budding. First, a small outgrowth called ‘bud’ is formed on the side of its body by the repeated mitotic divisions of its regenerative cells.

The bud grows gradually to form a small hydra. The young hydra develops a mouth and tentacles. Finally the tiny hydra detaches itself from the body of parent and lives as a separate organism. This method of asexual reproduction is called budding.

Question 35.
Can you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?
Answer:
Multicellular organisms are not mass of random collection of similar cells. They consist of many different types of specialised cells, which are organized into tissues and tissues are organised into organs. These organs with different cell structures are located in specific positions of the body.

Regeneration happens through mitosis and a particular type of tissue can give rise to its own kind only. Hence, regenerating a different kind of tissue from another kind is not possible. Therefore, complex organisms cannot give rise to new individuals through regeneration.

Question 36.
What is the reproductive method in Leishmania, plasmodium and yeast?
Answer:
Leishmania, plasmodium and yeast reproduce by asexual methods. Leishmania reproduces by the method of binary fission. Plasmodium reproduces by multiple fission. Yeast reproduces by the method of budding.

Question 37.
What is vegetative propagation? Name some plants that are multiplied by this method.
Answer:
A form of asexual reproduction in plants, in which new plants grow from the vegetative parts, such as leaves, stem or roots, is called vegetative propagation. New plants of sugarcane, roses, grapes, banana, jasmine, rose etc., are grown by vegetative propagation.

Question 38.
Why is vegetative propagation considered as asexual reproduction?
Answer:
In vegetative propagation only one parent is involved and the offspring is genetically identical to the parent. This mode of reproduction does not involve the fusion of gametes. Therefore, vegetative propagation is considered as asexual reproduction.

Question 39.
Why is vegetative propagation practiced for growing some types of plants?
Answer:
Some plants do not produce seeds or they produce seeds from which a new plant cannot be grown. Vegetative propagation is practiced for growing such type of plants.

Question 40.
Give examples of three plants that have lost the ability to bear seeds.
Answer:
Examples of plants that have lost the ability to bear seeds include plantain, rose and jasmine.

Question 41.
List the advantages of vegetative propagation.
Answer:
The following are the advantages of vegetative propagation

1. The new plant is genetically a replica of the parent plant. Therefore, purity of the variety is maintained in vegetative propagation.
2. In this method, it is possible to propagate those plants that are not able to produce seeds.
3. New plants can be grown quickly by vegetative propagation. Plants raised by this method can bear flowers and fruits earlier than those produced by seeds.

Question 42.
How do you show vegetative propagation in potato?
Answer:
Potato is modified stem and therefore it is a vegetative part of the plant. Take a potato and cut it into small pieces such that some pieces contain a notch or bud and some do not. Spread some cotton on a tray and wet it. Place the potato pieces that contain notch and other pieces separately on the wet cotton.

Ensure that the cotton is kept moist all the time. Observe the changes taking place in the potato pieces over the next few days. It is observed that the potato pieces with notch will develop fresh green shoots and roots from the sites of notches.

Pieces of potato that have no notches do not develop shoots and roots. This shows that a new potato plant can be grown from a potato piece that has notches. This demonstrates vegetative propagation in potato.

Question 43.
How does a bryophyllum plant reproduce?
Answer:

Bryophyllum reproduces by vegetative propagation. There are notches along the leaf margin of bryophyllum. When a leaf of this plant falls into the soil, buds emerge from these notches and develop into new plants.

Question 44.
How does a money plant propagate itself?
Answer:
A money plant propagates vegetatively from its stem.

Question 45.
How does rhizopus reproduce? Explain.
Answer:
Rhizopus is a fungus. It reproduces asexually by spore formation. Rhizopus is a filamentous alga. The thread-like body structure of this organism is called hyphae. Each hyphae has a tiny blob called sporangia at its free end. These blobs contain ‘seeds’ of reproduction called spores.

Sporangia break and release spores to the surrounding medium. The spores are covered by thick walls that protect them until they get favourable conditions. When these spores come into contact with a moist surface, they begin to grow new filaments.

Question 46.
What are the reproductive structures in rhizopus called?
Answer:
The reproductive structures in rhizopus are called sporangia. These are blob-like structures that contain spores.

Question 47.
What is sexual reproduction?
Answer:
A type of reproduction, which involves the fusion of male and female gametes, is known as sexual reproduction.

Question 48.
What are the advantages of sexual reproduction?
Answer:
Sexual reproduction promotes genetic variability, favours the development of adaptive features, and speeds up evolution.

Question 49.
What is meiosis? What is its need?
Answer:
A type of cell division that results in daughter cells with half the number of chromosomes of the parent cell is called meiosis. Meiosis is required for producing reproductive cells such as gametes. Sexual reproduction allows for the production of unique offspring of the species with variable traits.

This becomes possible because the genetic material from two parents gets mixed and produces a unique combination. This requires that the daughter cells produced by cell division must contain only half the chromosomes present in the parent cell.

When this happens, the zygote can again become diploid and have the same number of chromosomes as the parent cell. Therefore, meiosis is necessary for sexual reproduction.

Question 50.
What are the advantaees of sexual reproduction over asexual reproduction?
Answer:
Sexual reproduction involves the fusion of male and female gametes and usually involves two parents. Since the characteristics are inherited from both the mother and the father, the offspring shows variation in characteristics. This is useful in adaptation and survival. The offspring in asexual reproduction are genetically identical to the parent and hence no variation is possible.

Question 51.
Distinguish between sexual reproduction and asexual reproduction.
Answer:
Sexual reproduction:

1. Involves the fusion of gametes.
2. Usually two parents are involved.
3. The offspring is unique.
4. The offspring is genetically different from the parents.
5. Produces genetic variations in characteristics.
6. One or more offspring are produced

Asexual reproduction:

1. Does not involve the fusion of gametes.
2. Only one parent is involved.
3. The offspring is a copy of the parent.
4. The offspring is genetically identical to the parent.
5. Does not produce any genetic variations in characteristics.
6. Two or more offspring are produced.

Question 52.
What are gametes? What are its types?
Answer:
Specialized cells, which are directly involved in the process of reproduction in higher order plants and animals, are called gametes. They are als called sex cells or germ cells. There are two types of gametes: Male gamete and female gamete. The male gamete is called sperm and the female gamete is called ovum (also called egg).

Question 53.
What is fertilization ?
Answer:
The fusion of a male sex cell and a female sex cell resulting in a single cell, which develops into a new organism, is called fertilization.

Question 54.
Give the meaning of the following terms: Gamete, gonads, sperm, ovum, fertilization, and zygote.
Answer:

• Gamete: The cells involved in sexual reproduction are called gametes.
• Gonads: The reproductive glands that produce sex cells are called gonads.
• Sperm: The male sex cell (male gamete) is called sperm.
• Ovum: The female sex cell (female gamete) is called ovum or egg.
• Fertilization: The fusion of sperm and ovum resulting in a single cell is called fertilization.
• Zygote: A fertilized ovum is called zygote.

Question 55.
Name the sex cell produced in female reproductive system and the sex cell produced in male reproductive system.
Answer:
The sex cell produced in female reproductive system is called egg or ovum. The sex cell produced in male reproductive system is called sperm.

Question 56.
Distinguish between male gamete and female gamete.
Answer:
Male gamete:

1. Male gamete is produced in the male reproductive system.
2. The male gamete in animals is called sperm.
3. Male gamete is usually motile and small in size.
4. Generally male gamete does not store food.

Female gamete:

1. Female gamete is produced in the female reproductive system.
2. Female gamete is called egg or ovum.
3. Female gamete is non-motile and usually bigger in size.
4. Female gamete generally stores food.

Question 57.
In which group of green plants do we find sexual mode of reproduction?
Answer:
We find sexual mode of reproduction in flowering plants, which are also known as angiosperms.

Question 58.
What are angiosperms? Name the organ of reproduction in angiosperms.
Answer:
Plants that bear seeds that are enclosed in fruit are called angiosperms. These are also known as flowering plants. The organ of reproduction in angiosperms is flower.

Question 59.
Draw a labelled diagram of the loneitudinal section of a flower.
Answer:

• Pistil – Female reproductive part
• Stigma – Terminal part of pistil; the part that receives pollen
• Style – Middle elongated part of pistil; the connective tissue between stigma and ovary
• Ovary – Bottom part of pistil; the part that contains ovules
• Stamen – Male reproductive part; the part that produces pollen grains
• Anther – Pollen-producing organs
• Filament – Stalk supporting anthers .
• Petals – Collectively called corolla
• Sepals – Collectively called calyx; protective leaf-like enclosures for the flower buds
• Stamen (Male) – Anther, Filament
• Pistil or carpel (Female) – Stigma, Style, Ovary

Question 60.
Which are the four prominent parts (whorls) present in a typical flower?
Answer:
The four prominent parts in a typical flower are sepals, petals, stamens and carpels.

Question 61.
What is a sepal? What is its function?
Answer:
The outermost whorl of a flower consisting of green coloured units is called sepal. The sepals protect the flower, especially in the bud condition.

Question 62.
What are petals? What is its function?
Answer:
As we move from outside towards inside of a flower, the second whorl consists of bright coloured segments called petals. Petals are modified leaves that surround the reproductive parts of flowers. Together, all of the petals of a flower are called a corolla. Petals protect the reproductive parts that lie inside the flower. Their function is to attract insects for pollination.

Question 63.
What are stamens ? What is their function?
Answer:
TheThe male reproductive structures in a flower are called stamens. The chief function of stamens is to produce pollen grains, which produce male sex cells.

Question 64.
What are carpels? What is each carpel made of? What is their function?
Answer:
The female reproductive structures in a flower are called carpels. Each carpel consists of an ovary, which contains one or more ovules, a style and the stigma. A carpel in a flower consists of ovary, style and stigma.

The ovary of a carpel produces ovules. The ovule produces female sex cells. The ovary protects the ovule. After fertilisation, the ovule will become the fruit. The style of a carpel is a tubular structure between the stigma and the ovary.

The style helps the movement of pollens from the stigma to the ovary. Stigma is the receptacle for pollen. It has a sticky surface so that pollens stick. The stigma also transports the pollen to the ovum for fertilization.

Question 65.
Describe briefly the structure of a carpel.
Answer:
Each carpel consists of an ovary, which contains one or more ovules, a style and the stigma. The ovary is at the base of the flower. A tubular structure that extends from the ovary is called the style. The top of the style is a surface called stigma, which receives pollens.

Question 66.
Name the structure that gives rise to and contains the female reproductive cells in a flower. Where is it located?
Answer:
The structure that gives rise to and contains the female reproductive cells in a flower is called ovule. It is located inside the ovary.

Question 67.
Which are the male and female reproductive parts in a flower?
Answer:
Stamens and carpels are the reproductive parts in a flower. Stamen is the male reproductive part and carpel is the female reproductive part in a flower.

Question 68.
What are unisexual flowers? Give two examples.
Answer:
Flowers, which contain either stamen or carpel, are called unisexual flowers. Such flowers will have to undergo only cross-pollination. For example, flowers of papaya and watermelon are unisexual.

Question 69.
What are bisexual flowers? Give two examples.
Answer:
Flowers that contain both stamen and carpel are called bisexual flowers. Such flowers can undergo both self-pollination and cross-pollination. Example: Mustard, hibiscus, etc.

Question 70.
Which are the two major steps involved in reproduction in angiosperms?
Answer:
Angiosperms reproduce by sexual method. Reproduction in angiosperms involves two steps namely pollination and fertilization.

Question 71.
What are pollen grains? What do pollen grains contain?
Answer:
A fine powdery substance, typically consisting of microscopic yellow grains discharged from the stamen of a flower is called pollen grain. Each grain of pollen contains a male gamete that can fertilize the egg.

Question 72.
What is pollination ? Mention its importance in reproduction.
Answer:
The transfer of pollen from the anther of a flower to the stigma of the same flower or of another flower is called pollination. Pollination is a prerequisite for fertilization. Seeds cannot form without pollination.

Question 73.
Name the two types of pollination found in plants.
Answer:
The two types of pollination in plants are

• self-pollination
• cross-pollination.

Question 74.
Distinguish between self-pollination and cross-pollination.
Answer:
The transfer of pollen from the anther of a flower to the stigma of the same flower is called self-pollination. The transfer of pollen from the anther of a flower to the stigma of another flower of the same plant or a different plant of the same species is called cross-pollination.

Question 75.
What is fertilization in plants? When and where does it occur?
Answer:
The fusion of male and female gametes resulting in the formation of a zygote in a flowering plant is known as fertilization. Fertilization occurs after pollination. Fertilization occurs in the ovaries.

Question 76.
How do pollens reach the ovaries of a flower? Explain.
Answer:
Pollen somehow gets deposited on the stigma of the carpel. For fertilization to occur, these pollens must reach the ovule through the style. The pollen grains on the stigma of a flower will develop a pollen tube that extends up to the ovule through the style. Pollens, which carry the male sex cells, will now move through the pollen tube.

Once the pollen tube reaches an ovule, it delivers sperm cells with a burst, which goes to the embryo sac. The sperm cell fertilizes the egg cell to form an embryo, which will develop into a seed. The seed is a potential future plant.

Question 77.
Draw a neat, labelled diagram showing the germination of pollen on stigma.
Answer:

Question 78.
After fertilization in a flower, which part develops into fruit, seed coat and seed?
Answer:
After fertilization in a flower, the ovary grows rapidly and ripens to form a fruit. The ovule becomes a seed. The covering of the ovule develops into seed coat.

Question 79.
What changes occur in the flower after pollination and fertilization?
Answer:
After pollination and fertilization get over, the following changes occur in the flower

1. The embryo develops into a seed, which later can grow into a new plant.
2. A nutritive tissue develops around the embryo, which stores nutrients.
3. The covering of the ovule forms the seed coat.
4. Ovary part of the carpel gets transformed into a fruit.
5. Sepals, petals and other parts of the flower will fall off.

Question 80.
What are the advantages of seed formation for a plant?
Answer:
Seed formation provides the following advantages to a plant

1. Seeds have reserve food sources which provide nourishment to the developing embryo.
2. They protect the embryo from harsh environmental conditions and enable the embryo to survive for long periods.
3. The seeds are easily dispersed by natural agents. The dispersal of seeds to far-off places prevents competition among the members of the same species, thus preventing their extinction.
4. Seeds can survive without water. If there is a drought, the plants can survive.
5. Seeds are produced in large numbers, which facilitates plants to multiply.

Question 81.
How is the process of pollination different from fertilisation?
Answer:
Transfer of pollen grains from anther to stigma is called pollination. On the other hand, fusion of male and female gametes is called fertilization. Pollination is a physical process in which movement of pollen grains is achieved by certain physical factors. Fertilization is a biological process.

Ovary is the site of fertilization and embryo is the product of fertilization. Pollination involves only the male gamete (pollen grain). Fertilisation involves both male and female gametes.

Question 82.
What is germination of a seed? Explain.
Answer:
The development of a plant from a seed is called seed germination. The development of a seed into a plantlet after it has been planted in soil and remained dormant for a certain period of time is called germination. When seeds get into the soil, they remain inactive until conditions are suitable for germination. When they get favourable conditions, they germinate and a new young plant emerges.

Question 83.
Describe a simple activity to show the different parts of a seed.
Answer:

Soak a.few seeds of Bengal gram and keep them overnight. Drain the excess water. Cover the soaked seeds with a wet cloth and leave them for a day. Make sure that the seeds do not become dry. Cut open the seeds carefully and observe the different parts.

We observe the brownish seed coat that encloses the two cotyledons. The cotyledons are laden with stored food material which help the seedling in the early stage of its development. We can also see a small embryo in between the two cotyledons.

Question 84.
What type of reproduction and development happens in humans?
Answer:
Fluman beings reproduce by sexual method. The development is direct. This means a newly born baby resembles the adult in appearance, body parts and their functioning.

Question 85.
What is fertilization? Where does it occur in humans?
Answer:
The process of formation of a single cell by the union of male sex cell (sperm) and female sex cell (ovum) is called fertilization. In humans, fertilization is internal. It occurs inside the female body.

Question 86.
What is a zygote? Compare the number of chromosomes present in a sex cell with those present in the zygote.
Answer:
A single cell formed by the union of male gamete and female gamete is called zygote. Zygote contains twice the number of chromosomes present in a sex cell that caused its formation.

Question 87.
Why are sex cells said to be haploid?
Answer:
Both the male sex cell (sperm) and the female sex cell (ovum) contain half the number of chromosomes present in other cells of the organism. For example, the cells in our body contain 46 chromosomes. However, the male and female sex cells of a human being will have only 23 chromosomes each. This is why sex cells are said to be haploid.

Question 88.
A flowering plant has 24 chromosomes in its male gamete,

1. How many chromosomes are present in its female gamete?
2. What is the number of chromosomes in its zygote?

Answer:

1. 24 chromosomes (both gametes should contain same number of chromosomes).
2. 48 chromosomes are present in its zygote. The union of male and female gametes forms zygote and hence the chromosomes add up.

Question 89.
What is puberty?
Answer:
The period in life when a boy or girl becomes sexually mature and capable of sexual reproduction is called puberty. It is marked by several marked changes in the body. It is associated with the development of secondary sex characteristics and rapid growth. It is a process that usually happens between ages 10 and 14 in girls and ages 12 and 16 in boys.

Question 90.
What are the changes seen in eirls at the time of puberty?
Answer:
The following changes are seen in girls at the time of puberty

1. The size of the breasts begins to increase.
2. Hairs grow in the arm pits and around the genitals.
3. The nipples at the tip of the breasts become relatively dark.
4. Menstruation cycle begins at around this time and continues periodically.
5. The voice becomes shriller and more feminine.
6. The pelvic region becomes broader.
7. The reproductive organs mature and become functional.

Question 91.
What are the changes seen in boys at the time of puberty?
Answer:
The following changes are seen in boys at the time of puberty

1. Hairs appear in armpits, around the genitals, face (beard and moustach) and all over the body.
2. The voice breaks and becomes hoarse.
3. Gain height and weight. Hairs may appear on the chest.
4. The body becomes muscular and shoulders broaden.
5. Reproductive organs develop and become functional.

Question 92.
What are some of the common changes observed in boys and girls at puberty?
Answer:
The following are some of the common changes in boys and girls during puberty:

1. Hairs appear in both boys and girls in armpits and genital area between the thighs.
2. The genital area becomes darker in both boys and girls.
3. Hairs on arms and legs appear and become darker.
4. The skin becomes oily and pimples may appear on the face in both boys and girls.
5. Both boys and girls become conscious of their own body as well as those of others.

Question 93.
Name the parts of the male reproductive system in humans.
Answer:
The male reproductive system in humans consists of

• two testicles (testes)
• the duct system that includes epididymis and vas deferens
• the accessory glands which include seminal vesicles and prostate gland, and
• penis.

Question 94.
Draw a diagram of the male reproductive system in humans and label the parts.
Answer:

Question 95.
What are testes? Where are they located in the human body?
Answer:
Oval shaped reproductive gland in the body of a human male is called testis. There are two testes in the human body. Testes are located within the loose pouch of skin called scrotum that hangs outside the body behind the penis.

Question 96.
What is scrotum? What is their function?
Answer:
A small muscular sac, which forms a part of the external male genitalia located behind and underneath the penis and houses the testicles is called scrotum. The scrotum holds and protects the testicles and keeps them at the right temperature.

The production of sperms requires relatively lower temperature and therefore the temperature in the scrotum is lesser than the temperature of the rest of the body. Under cold conditions, scrotum pulls the testicles closer to the body and when the conditions are warmer, testicles hang away from the body.

Question 97.
What are the functions performed by the testis in human beings?
Answer:
Testis in human beings has twin functions. It produces male gametes namely the sperms. It also produces male sex hormone called testosterone, which is responsible for the development of secondary sexual characteristics in males.

Question 98.
What is a sperm? Describe the structure of a sperm cell.
Answer:
The male reproductive cell is called sperm. A human sperm cell consists of an oval shaped head, a mid-piece and a long tail. The tail helps to make the sperm move easily in a fluid medium.

Question 99.
What is the function of the tail in a sperm?
Answer:
The long tail of a sperm has a wavy motion with the help of which the sperm cell moves easily to the site of fertilization in the body of a female.

Question 100.
What is sperm duct? What is its other name? What is its function?
Answer:
The duct that conveys sperms in a male from the testicle to the urethra is called sperm duct. It is also called vas deferens. It is a coiled duct that provides a pathway for the transport of sperms to the penis.

Question 101.
What is penis? What is its function?
Answer:
The external muscular genital structure in males that carries the duct for the transfer of sperms to the female body is called penis. Biologically, penis has two main functions. It is an organ for urination, and passing on the sperms to the reproductive system of the female body for procreation.

Question 102.
Name any two reproductive accessory glands in the body of humans.
Answer:
The accessory glands associated with the reproductive system in humans include seminal vesicles and prostate gland.

Question 103.
What is the role of the seminal vesicles and the prostate eland?
Answer:
Seminal vesicles and prostate gland form an important part of the male reproductive system. The seminal vesicles and prostate gland add their secretions to sperms to make up the semen. The secretions of seminal vesicles and prostate gland provide nutrition to the sperms and also provide a suitable medium for their easy transport.

Question 104.
How are the sperms produced in testes delivered to the female body for fertilization?
Answer:
Sperms are produced in the testes and are stored there temporarily. They are delivered through the vas deferens, which unites with a tube coming from the urinary bladder. The urethra thus forms a common passage for both the sperms and urine. Along the path of the vas deferens, glands like the prostate and the seminal vesicles add their secretions to the sperms to form semen.

The sperms are now in a fluid medium, which makes their transport easier. The substances in the fluid will also provide nutrition to the sperms. The semen containing sperms is delivered by the penis into the vaginal passage of the female during intercourse. The sperms further travel upwards and reach the oviduct where one of the sperms may fertilize the egg.

Question 105.
Mention the three major functions of the female reproductive system.
Answer:
The three major functions of the female reproductive system are

• production of eggs (ovum),
• protection and nourishment of the fertilized egg until it fully develops into a baby, and
• giving birth to the baby.

Question 106.
Mention the parts of the female reproductive system in humans.
Answer:
The female reproductive system in humans consists of

• A pair of ovaries
• A pair of fallopian tubes
• Womb, and
• Vagina.

Question 107.
Dr aw a neat diagram of the female reproductive system in humans and label the parts.
Answer:

Question 108.
What are ovaries? What is their main function?
Answer:
A pair of oval-shaped glands of the female reproductive system that are located slightly above the uterus on its either side and involved in the production of female sex cells is called ovaries. The main function of the ovaries is to produce ova or eggs. They also produce female hormones.

Question 109.
What are fallopian tubes? What is their function?
Answer:
Either of the two tubes that connects ovary to the uterus is called fallopian tube. They are also called oviducts. Oviducts help the eggs produced in ovaries to move towards or into the uterus.

Question 110.
What is uterus? Where is it located?
Answer:
A sac-like elastic organ in the female reproductive system, shaped in the form of an inverted triangle, into which the oviducts open is called uterus. It is also called womb. Uterus is located in the lower abdomen of a woman, between rectum and urinary bladder.

Question 111.
What are the functions of uterus ?
Answer:
The uterus has several important functions. It includes receiving the developing zygote, implantation, gestation, labour, delivery and menstruation. Most importantly, the fertilized egg develops into a human baby inside the uterus.

Question 112.
What is vagina? What are its functions?
Answer:
An elastic, muscular canal with a soft, flexible lining that is connected to the uterus is called vagina. The vagina receives the semen and helps to pass the sperms to the fallopian tube through the uterus. It also serves as an outlet for menstrual flow from the uterus. During childbirth, the vagina expands to facilitate the passage of the baby during childbirth.

Question 113.
In which part of the female reproductive system does fertilization occur?
Answer:
Fertilization occurs in the fallopian tube (oviduct).

Question 114.
What is zygote? Where does it get formed? What happens to it after its formation?
Answer:
The single cell formed by the union of a sperm and ovum is called zygote. It is nothing but a fertilized ovum. Zygote is formed in the fallopian tube when a sperm succeeds in entering into the ovum. Fertilization results in the formation of the zygote.

The zygote undergoes cell division for about 5-6 days in the fallopian tube itself to form a small spherical mass. This mass enters the uterus and gets implanted on its wall. The embryo develops there until the baby is born.

Question 115.
Distinguish between a gamete and zygote. Explain their roles in sexual reproduction.
Answer:

Gamete	Zygote
1. The cells involved in sexual reproduction are called gametes. E.g.: sperm (male) and ova (female)	The fusion of male gamete and female gamete forms zygote during sexual reproduction.
2. Gametes are unfertilised reproductive cells.	Zygote is fertilised egg or fertilised ovum.
3. The fusion of sperm and egg forms a fertilised ovum or zygote.	Zygote undergoes development and forms a new organism.

Question 116.
What is implantation? What is its significance to a woman?
Answer:
The attachment of the fertilized egg or a ball of mass formed from it, to the wall of the uterus is called implantation. Implantation signifies that the woman has become pregnant. It marks the beginning of pregnancy.

Question 117.
What is a human embryo?
Answer:
The collection of human cells developed from the fertilized egg, especially from implantation on the uterine wall and the eighth week, is called embryo. This is the early stage of development before differentiation of the body parts occurs.

Question 118.
What is placenta? What are its functions?
Answer:
A disc-shaped special tissue embedded in the uterus wall with the help of which human embryo gets nutrition from mother’s blood is called placenta. Placenta provides large surface area for the passage of nutrients and oxygen from the mother’s blood to the embryo. The metabolic waste generated by the embryo is removed through the placenta into mother’s blood.

Question 119.
What is the role of placenta during pregnancy?

OR

Placenta is extremely essential for foetal development. Give reason.
Answer:
Placenta in the womb serves as the connecting link between mother’s blood and the growing embryo during pregnancy. It has ftnger-like projections called villi. The villi increase the surface area for fixation and absorption.

They serve as the passage for nutrients and oxygen from the blood of the mother to the embryo. They also serve as the passage for waste materials to move out of the embryo into the mother’s blood.

Question 120.
‘The structure and function of placenta plays an important role at the time of gestation period’. Explain.
Answer:
Placenta has a disc shaped structure that is embedded in the wall of the uterus. It has finger-like projections called villi. This provides the placenta a large surface area for the passage of nutrients and oxygen from the blood of the mother to the embryo. It also serves as the passage for waste materials to move out of the embryo into the mother’s blood.

Question 121.
What is menstruation? Explain.
Answer:
The discharging of blood and other materials from the lining of the uterus at intervals of about one month from puberty until menopause is called menstruation. Before ovulation, the uterus prepares itself to receive the developing zygote. The walls of the uterus become thick and spongy with plenty of blood supply.

If the egg is not fertilized, the soft tissues of the lining of the uterus break down. The broken tissues along with blood and the unfertilized egg are discharged from the uterus through the vagina. This flow of materials through the vagina is known as menstruation. A woman will generally experience menstrual flow from 2 to 8 days.

Question 122.
What is menstrual cycle? What is its period? How long does a woman discharge blood and other materials through the vagina during menstruation?
Answer:
The monthly cycle of changes in the ovaries and the lining of the uterus starting with the preparation for ovulation is called menstrual cycle. The period of the menstrual cycle is about 28 days.

Question 123.
What happens when fertilization fails to occur in a woman?

OR

Why does menstruation occur?
Answer:
The menstrual cycle begins when a girl reaches the age of puberty. It is the reproductive cycle that produces egg for fertilisation. During the menstrual cycle the uterus prepares itself every month for implantation and nurturing of the embryo. The lining thickens and is richly supplied with blood to nourish the growing embryo.

If fertilization and implantation do not occur, the uterus lining is shed from the body and moves out through the vaginal opening. This is how menstruation occurs when fertilization and implantation do not take place.

Question 124.
How does the embryo in the mother’s womb develop into a baby?

OR

How does the embryo get nourishment inside the mother’s body?
Answer:
The fertilized egg (zygote) turns into a mass of cells in the fallopian tube and gets implanted in the uterus. The embryo gets nutrition from the mother’s blood with the help of a special tissue called placenta. This is a disc that is embedded in the uterine wall. It contains villi on the embryo’s side of the tissue.

On the mother’s side are blood spaces, which surround the villi. This provides a large surface area for glucose and oxygen to pass from the mother to the embryo. The developing embryo will also generate waste substances. These are transferred to the mother’s blood through the placenta.

The development of the child inside the mother’s body takes approximately nine months. The child is born as a result of rhythmic contractions of the muscles in the uterus.

Question 125.
What are sexually transmitted diseases? Give examples.
Answer:
Those infections that can be transmitted through sexual contact with an infected individual are called sexually transmitted diseases. They are also called sexually transmitted infections. Sometimes, they are also called venereal diseases. eg: Syphilis, Gonorrhoea, Genital herpes, AIDS etc.

Question 126.
How are sexually transmitted infections passed from one person to another?
Answer:
Sexually transmitted infections are usually transmitted from one person to another when individuals engage in intimate sexual activity with someone who is infected.

Question 127.
Define pregnancy.
Answer:
The period between conception and birth during which a foetus develops into an infant baby inside a woman’s womb or uterus is called pregnancy.

Question 128.
What is contraception? What are the various types of contraception available to people?
Answer:
Any technique, procedure or device used to prevent pregnancy as a consequence of sexual intercourse is called contraception.
The following are some of the types of contraception employed by people to prevent possible pregnancy
1. Mechanical barrier method:
This technique employs a barrier that prevents the sperm from meeting the egg. eg: Use of condoms (males), cervical caps (females), and diaphragms.

2. Hormonal method:
This technique causes changes in hormonal balance, which prevents the release of egg. eg: Use of oral pills.

3. Use of IUD:
This technique employs placement of a device that either releases copper or hormones, which prevents pregnancy. eg: Copper-T, loop etc.

4. Surgical method:
This technique aims at preventing the passage of sperms in males and ova in females. eg: Tubectomy for females and vasectomy for males.

Question 129.
Write the expandedforms of the following: HIV, AIDS, IUD, and DNA.
Answer:

• HIV: Human Immunodeficiency Virus.
• AIDS: Acquired Immunodeficiency Syndrome.
• IUD: Intra-Uterine Device.
• DNA: Deoxyribonucleic Acid.

Question 130.
What are the different methods of contraception?
Answer:
The various methods of contraception include the use of condoms (both men and women), copper-T, oral pills, spermicidal pills, tubectomy and vasectomy.

Question 131.
What could be the reasons for adoptins contraceptive methods?
Answer:
The main reason for adopting contraceptive methods is to prevent unwanted pregnancy. Some contraceptive methods will also give protection against sexually transmitted diseases.

Question 132.
A woman is using copper-T to prevent pregnancy. Will this device give protection against sexually transmitted diseases?
Answer:
No. The use of copper-T will not give protection from sexually transmitted diseases.

Question 133.
What could be the reasons for adopting condom as a contraceptive method?
Answer:
Use of condoms helps to prevent unwanted pregnancy and childbirth as a consequence of sexual intercourse. The use of condoms will give protection against sexually transmitted diseases.

Question 134.
How will vasectomy and tubectomy prevent pregnancy? Explain.

OR

How can pregnancy be prevented surgically?
Answer:
Tubectomy and vasectomy are surgical techniques of contraception. Tubectomy is performed on women while vasectomy is performed on men. In women, the fallopian tubes are cut and then the cut ends are tied, burned, or clipped.

Any of these methods prevents the eggs from travelling from the ovaries to the uterus. In men, the vas deferens is cut and then the cut ends are tied so that sperms eannot mix with semen.

Question 135.
What is sex ratio? Why is it falling in our country?
Answer:
The number of females for every 1000 males in a population is called sex ratio. The sex ratio in our country is falling due to the preference of parents for male children. This bias is reflected in reckless female foeticides, killing of female babies, neglect of girl children and discrimination against women. As a result of all these, the sex ratio in our country is falling.

Fill In The Blanks:

1. Amoeba usually reproduces asexually by binary fission
2. The male reproductive cell is called sperm
3. A single cell, which emerges as the product of fertilization, is called zygote
4. The part of the flower that develops into a fruit is ovary
5. Pollen grain in a flower contains a male gamete
6. Surgical method used on females to prevent pregnancy is tubectomy
7. The mode of reproduction that does not involve the fusion of gametes is called asexual reproduction
8. The primary genital organ in human males is called testis
9. The primary genital organ in human females is called ovary
10. The process by which the embryo in a seed develops into a seedling is called germination
11. A single cell formed by the fusion of an egg and a sperm is called zygote
12. Somatic cells : Diploid :: Germ cells : Haploid

Multiple Choice Questions:

Question 1.
sexual reproduction takes place through budding in
(A) amoeba.
(B) yeast.
(C) plasmodium.
(D) leishmania.
Answer:
(B) yeast.

Question 2.
Which of the following is not a part of the female reproductive system in human beings?
(A) Ovary
(B) Uterus
(C) Vas deferens
(D) Fallopian tube
Answer:
(C) Vas deferens

Question 3.
The anther contains
(A) sepals.
(B) ovules.
(C) carpel.
(D) pollen grains.
Answer:
(D) pollen grains.

Question 4.
Vegetative propagation refers to the formation of new plants from
(A) stem, roots, and flowers.
(B) stem, roots, and leaves
(C) stem, flowers, and fruits
(D) stem, leaves, and flowers.
Answer:
(B) stem, roots, and leaves

Question 5.
A feature of reproduction that is common to amoeba, spirogyra and yeast is that
(A) they reproduce asexually
(B) they are all unicellular
(C) they reproduce only sexually
(D) they are all multicellular.
Answer:
(A) they reproduce asexually

Question 6.
The correct sequence of reproductive stages seen in flowering plants is
(A) gametes → zygote → embryo → seedling
(B) zygote → gametes → embryo → seedling
(C) seedling → embryo → zygote → gametes
(D) gametes → embryo → zygote → seedling
Answer:
(A) gametes → zygote → embryo → seedling

Question 7.
Characteristics that are transmitted from the parents to their offspring are present in
(A) cytoplasm
(B) golgi bodies
(C) ribosomes
(D) genes
Answer:
(D) genes

Question 8.
The number of chromosomes in parents and offspring of a particular species remains constant due to
(A) doubling of chromosomes after gamete formation
(B) doubling of chromosomes after zygote formation
(C) halving of chromosomes after gamete formation
(D) halving of chromosomes during gamete formation
Answer:
(D) halving of chromosomes during gamete formation

Question 9.
Offspring formed as a result of sexual reproduction exhibit more variations because
(A) sexual reproduction is a lengthy process
(B) genetic material comes from two parents of the same species
(C) genetic material comes from two parents of different species
(D) genetic material comes from many parents
Answer:
(B) genetic material comes from two parents of the same species

Question 10.
The sequential route followed by sperms in the female reproductive system is
(A) vagina, fallopian tube, uterus
(B) uterus, vagina, fallopian tube
(C) vagina, uterus, fallopian tube
(D) fallopian tube, uterus, vagina
Answer:
(C) vagina, uterus, fallopian tube

Question 11.
The process of the union of an egg and a sperm is known as
(A) Asexual reproduction
(B) Multiple fission
(C) Zygote
(D) Fertilization
Answer:
(D) Fertilization

Question 12.
The event that marks the beginning of the reproductive phase in human females is
(A) changes in voice pattern
(B) menstruation (menarche)
(C) changes in hair pattern
(D) increase in height and weight
Answer:
(B) menstruation (menarche)

Question 13.
Hydra reproduces asexualiy by
(A) budding
(B) binary fission
(C) multiple fission
(D) vegetative propagation
Answer:
(A) budding

Question 14.
In a flower, the parts that produce male and female gametes (germ cells) are respectively
(A) stamen and anther
(B) filament and stigma
(C) anther and ovary
(D) stamen and style
Answer:
(C) anther and ovary

Question 15.
The mode of asexual reproduction in which the parent cell divides into many daughter cells is
(A) Binary fission
(B) Budding
(C) Fission
(D) Multiple fission
Answer:
(D) Multiple fission

Question 16.
During adolescence, several changes occur in the human body. A change that represents the development of secondary sexual characteristic in boys is
(A) loss of milk teeth
(B) increase in height
(C) cracking of voice
(D) weight gain
Answer:
(C) cracking of voice

Question 17.
Characteristics that are transmitted from parents to offspring during reproduction show
(A) only similarities with parents
(B) only variations with parents
(C) both similarities and variations with parents
(D) neither similarities nor variations with parents
Answer:
(C) both similarities and variations with parents

Question 18.
The part of a potato plant that can be propagated vegetatively is
(A) stem
(B) root
(C) leaves
(D) flowers
Answer:
(A) stem

Question 19.
Reproduction is essential for living organisms in order to
(A) keep the individual organism alive
(B) meet the energy requirement of the organism
(C) maintain growth
(D) ensure the continuation of species generation after generation
Answer:
(D) ensure the continuation of species generation after generation

Question 20.
Advantages of artificial methods of vegetative propagation include
(A) crops with better quality
(B) crops with more yield
(C) ability to produce plants with desirable qualities
(D) all of the above
Answer:
(D) all of the above

Question 21.
Which of the following reproduces by spore formation?
(A) Planaria
(B) Rhizopus
(C) Spirogyra
(D) Yeast
Answer:
(B) Rhizopus

Question 22.
Which one of the following is not a part of the male reproductive system in human beings?
(A) Fallopian tube
(B) Penis
(C) Testis
(D) Vas deferens
Answer:
(A) Fallopian tube

Question 23.
The correct sequence of organs in the male reproductive system for the transport of sperms is
(A) testis → vas deferens → urethra
(B) testis → ureter → urethra
(C) testis → urethra → ureter
(D) testis → vas deferens → ureter
Answer:
(A) testis → vas deferens → urethra

Question 24.
A disadvantage of natural vegetative propagation is that it
(A) lacks dispersal mechanisms for propagation
(B) does not involve production of gametes and fertilization
(C) quickly spreads and colonizes the entire area with similar plants
(D) produces daughter plants quickly which start yielding products
Answer:
(A) lacks dispersal mechanisms for propagation

Question 25.
The reason for increase in human population is
(A) improvement in health care services.
(B) the increase in birth rate and decrease in death rate.
(C) the increase in agricultural production.
(D) All of the above
Answer:
(D) All of the above

Question 26.
Offsprings of plants receive/possess varied characteristics in
(A) cross-pollination
(B) self-pollination
(C) wind pollination
(D) insect pollination
Answer:
(A) cross-pollination

Question 27.
The correct order of binary fission in Leishmania is

(A) II, III, IV, I
(B) I, III, IV, II
(C) IV, I, III, II
(D) III, I, II, IV
Answer:
(B) I, III, IV, II

Match The Following:

Column A	Column B
1. Ovulation	(a) Termination of human pregnancy
3.  Fertilization	(b) Attained during puberty
5.  Implantation	(c) Production of sperms in testis
7.  Sexual maturation	(d) Release of egg from the ovary
8.  Menstruation	(e) Attachment of zygote to uterine wall
10. Abortion	(f) Union-of sperm and egg
	(g) Monthly discharge from the vagina

Answer:
1 – d, 2 – f, 3 – e, 4 – b, 5 – g, 6 – a.

Column A	Column B
1. Testis	(a) Male sex cell
2. Scrotum	(b) Produces male germ cells
3. Ovary	(c) Female gamete
4. Sperm	(d) Product of fertilization
5. Ovum	(e) Houses testicle
	(f) Releases egg

Answer:
1 – b, 2 – e, 3- f, 4 – a, 5 – c.

Students can download Class 10 Science Chapter 12 Electricity Important Questions, KSEEB SSLC Class 10 Science Important Questions and Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka SSLC Class 10 Science Important Questions Chapter 12 Electricity

Question 1.
What does an electric circuit mean? Draw a simple electric circuit.

OR

Draw a schematic diagram of an electric circuit.
Answer:
A continuous and closed path of an electric current is called an electric circuit.

Question 2.
When do we say that a circuit is broken or complete? What is the device called that enables us to make or break a circuit?
Answer:
When the conducting path of electric current is complete, we say that the circuit is complete or closed. When the conducting path of electric current is broken somewhere, we say that the circuit is broken or open. The device that enables us to make or break a circuit is called switch.

Question 3.
When do we say that there is an electric current through a circuit?
When electric charges (usually electrons) flow through a conductor (circuit), we say that there is an electric current in it.

Question 4.
What is the S.I. unit of electric charge? Define it
Answer:
The S.I. unit of electric charge is called ‘coulomb’ (C). 1 coulomb of charge is the charge contained in nearly 6 × 10¹⁸ electrons.

Question 5.
What is the charge on an electron?
Answer:
An electron possesses a negative charge of 1.6 × 10^-19 C.

Question 6.
What is electric current? Mention its S.I. unit
Answer:
An orderly flow of electric charges (usually electrons) through a conductor constitutes electric current. The amount of electric charge flowing through a conductor in unit time is called electric current. In other words, the rate of flow of charges through a conductor is called electric current. Consider a conductor through which a charge Q flows in t seconds. Now, the electric current I is given by

Electric current = \(\frac{\text { Quantity of charge flown }}{\text { Time taken }}\)
Current, I = \(\frac{\mathrm{Q}}{\mathrm{t}}\)
The S.I. unit of electric current is called ‘ampere’ (A).

Question 7.
Define the unit of current.

OR

Define 1 A of current.
Answer:
The unit of current is ampere. When one coulomb charge flows through an electric circuit in one second, then the current flowing through the circuit is said to be one ampere (1 A).
1 ampere = \(\frac{1 \text { coulomb }}{1 \text { second }}\) = 1 Cs^-1
∴ 1 A = 1 Cs^-1.

Question 8.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
1.6 × 10^-19 C of charge is present in 1 electron.
∴ 1 C of charge is present in = ?
No. of electrons present in 1 C of charge = \(\frac{1}{1.6 \times 10^{-19} \mathrm{C}}\)
= 6 × 10¹⁸
Thus, there are 6 × 10¹⁸ electrons in 1 coulomb of charge.

Question 9.
Calculate the amount of charge that flows through a conductor when a current of 5A flows through it for 2 mins.
Answer:
Given, I = 5A, t = 2 min = 2 × 60 s = 120 s, Q = ?
Charge, Q = I × t
= 5 × 120 = 600 C.
Thus the amount of charge flowing through a conductor is 600 C.

Question 10.
A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
Answer:
Given: Current, I = 0.5 A; Time, t = 10 min = 600 s, Charge, Q = ?
Q = It
= 0.5 A × 600 s = 300 C
Charge of 300 C has flowed through the circuit in 10 minutes.

Question 11.
Name the device used to measure electric current flowing through a circuit.
Answer:
Electric current flowing through a circuit is measured using a device called ammeter.

Question 12.
Name the unit used to express small currents. Give their relationship with ampere.
Answer:
Small currents are expressed using a unit called ‘miiliampere’ or ‘microampere’.
1 miiliampere (1 mA) = 10^-3 A
OR
1 microampere (1 pA) = 10^-6 A.

Question 13.
How is an ammeter connected in a circuit?
Answer:
An ammeter is always connected in series in a circuit.

Question 14.
Draw a schematic diagram of a typical electric circuit comprising a cell, an electric bulb, an ammeter and a plug key.
Answer:

Question 15.
In any circuit, state the direction of electric current and the direction of electron current.
Answer:
In any circuit, electric current flows from the positive terminal of the cell towards the negative terminal. Electron current, however, flows from the negative terminal towards the positive terminal of the cell. This means that the direction opposite to the flow of electrons is taken as the direction of conventional current.

Question 16.
Define electrical potential at a point. Give an equation for the same.
Answer:
The electric potential at a point is defined as the work done in moving a unit positive charge from infinity to that point.

If W is the work done in moving a charge Q from infinity to a point, then, the electrical potential V at that point is given by

Electrical potential = \(\frac{\text { Work done }}{\text { Charge moved }}\)
V = \(\frac{\mathrm{W}}{\mathrm{Q}}\)

Question 17.
Define and explain potential difference. Give an equation for the same.
Answer:
Just as liquids flow from a higher level to a lower level, electricity also flows through a conductor from a point of higher electric potential to a point of lower electric potential. This means that a potential difference is necessary for the flow of current through a conductor.

The potential difference between any two points is defined as the work done in moving a unit positive charge from one of those points to the other. It is also called ‘voltage’.

Consider a charge Q moving from point B to point A. Let us say that the work done during the process is W. Let Vbe the potential difference between the two points. Now,
Potential difference = \(\frac{\text { Work done }}{\text { Charge moved }}\)
V = \(\frac{\mathrm{W}}{\mathrm{Q}}\)

Question 18.
What is the S.I. unit of ‘electric potential’ and ‘potential difference’?
Answer:
Both electrical potential and potential difference are expressed by the same unit namely ‘volt’. It is represented by the symbol V.

Question 19.
Define 1 V of electric potential
Answer:
The electric potential at a point is said to be 1 volt if 1 joule of work is done in moving a charge of positive 1 coulomb from infinity to that point.
1 volt = \(\frac{1 \text { joule }}{1 \text { coulomb }}\)
1 V = 1 JC^-1.

Question 20.
Define 1 V of potential difference.

OR

What is meant by saving that the potential difference between two points is 1 V?
Answer:
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving a charge of +1 C from one of those points to the other.
1 V = \(\frac{1 \text { joule }}{1 \text { coulomb }}\)
1 V = 1 JC^-1.

Question 21.
What is the device used to measure the potential difference across a conductor? How is a voltmeter connected in the circuit to measure the potential difference between two points?

OR

Name the device used to measure potential difference across a conductor. How should this device be connected in a circuit?
Answer:
The potential difference across a conductor is measured by using a device called voltmeter. A voltmeter is always connected in parallel in the circuit across the points between which the potential difference is to be measured.

Question 22.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
We can set up a potential difference by using an electric cell or a battery.

Question 23.
How does an electric cell help to maintain a potential difference across a conductor?
Answer:
The chemical reaction within a cell generates potential difference across the terminals of the cell. When the cell is connected to a circuit, the potential difference sets the charges in motion through the conductor and produces an electric current. The cell maintains the current in the circuit by expending the chemical energy stored in it.

Question 24.
In the circuit given below, the potential difference across the conductor AB is to be measured. Show by a diagram how you would connect a voltmeter to measure the potential difference across AB.

Answer:

A voltmeter should be connected in parallel between the points across which we want to measure the potential difference. Here, we desire to measure the potential difference across AB. The connection of the voltmeter is shown in the circuit diagram given below:

Question 25.
How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?
Answer:
Given: Charge moved, Q = 2 C; Potential difference, V= 12 V; Work done, W = ?
W= VQ
= 12 V × 2 C = 24 J.

Question 26.
The work done in moving 5 C of charge between two points in an electric field is 30 J. Calculate the potential difference between the two points.
Answer:
Given: Charge moved, Q = 5 C, Work done W= 30 J, Potential difference V= ?
V = \(\frac{w}{Q}\)
= \(\frac{30 \mathrm{J}}{5 \mathrm{C}}\) = 6 JC^-1
= 6V

Question 27.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Given: Charge moved, Q = 1 C; Potential difference, V = 6 V, Energy given, W = ?
The energy given is nothing but the work done in moving the charge.
W = VQ
= 6V × 1C = 6 J.
The energy given to each coulomb is 6 J.

Question 28.
Name the law that gives the relationship between current flowing through a conductor and the potential difference across it.
Answer:
The law that gives the relationship between current flowing through a conductor and the potential difference across it is Ohm’s law.

Question 29.
State Ohm’s law. Express it in the form of an equation.
Answer:
Ohm’s law states that ‘at constant temperature, the potential difference across a conductor is directly proportional to the current flowing through it’.

Consider a conductor AB. Let V be the potential difference between these points and I be the current flowing through it. Now, according to Ohm’s law,
Potential difference ∝ Current
V ∝ I
V = RI where R is a constant called ‘resistance’ of the conductor.
∴ \(\frac{\mathrm{V}}{\mathrm{I}}\) = Constant.

Question 30.
What is electrical resistance? Mention its S.I unit.

OR

What is resistance of a conductor?
Answer:
All conductors oppose the flow of electricity through them. The opposition offered by a conductor to the flow of current through it is called its electrical resistance. The S.I unit of electrical resistance is called ‘ohm’. It is represented by the symbol Ω.

Question 31.
Define 1 Ω of resistance.
Answer:
The resistance of a conductor is said to be 1 ? if 1 A of current flows through it when a potential difference of IV is applied across it.
1 Ω = \(\frac{1 \mathrm{V}}{1 \mathrm{A}}\)
1 Ω = 1 VA^-1

Question 32.
Draw a simple circuit diagram showing the arrangement of apparatus for the verification of Ohm’s law.

OR

Describe an experiment to show the relationship between potential difference and current.

OR

How do you verify Ohm’s law?
Answer:

Take an electrical conductor XY made of nichrome. Connect it in series with a battery, a key, a rheostat and an ammeter. Connect a voltmeter parallel to the given resistor. Close the key.

Adjust the rheostat for a small value of current. Note the current and the potential difference by noting down the readings in the ammeter and voltmeter respectively. Go on increasing the value of current in equal but small steps.

Note down the corresponding value of potential difference (V) for each value of current (I). Find the value of V/I in each case. Tabulate the readings as follows:

The value of V/I is found to be a constant. This shows that the potential difference across a conductor is directly proportional to the current flowing through it.
The graph of V against I:

Plot a graph of potential difference against current. The graph of V against I is found to be a straight line passing through the origin. This again shows that the potential difference is directly proportional to the current flowing through it. This verifies Ohm’s law.

Question 33.
Gowri, while studying the variation of current through a conductor with potential difference, prepared the following table. However, she has left a few blanks in the table. Complete the table using your knowledge of Ohm’s law.
Answer:

Question 34.
A student measured the potential difference across a nichrome wire for various values of current flowing through it. The values are given in the table below:
Answer:

Draw a graph of potential difference against current. Interpret the graph. Find the value of the nichrome wire used in this activity.
Answer:

The graph of potential difference (V) against current (I) is a straight line passing through the origin. This shows that the potential difference across the conductor is directly proportional to the current flowing through it.

In other words, the value of \(\frac{\mathrm{V}}{\mathrm{I}}\) is a constant. This constant value is nothing but the resistance of the given nichrome wire.
Resistance, R = \(\frac{\mathrm{V}}{\mathrm{I}}=\frac{0.4 \mathrm{V}}{0.1 \mathrm{A}}\) = 4 Ω.

Question 35.
Name the device used to provide variable resistance in a circuit.
Answer:
In an electric circuit, a device called rheostat is often used to provide continuously variable resistance. This device can be used to either increase or decrease the resistance of the circuit.

Question 36.
How do you show that the current flowing through an electric component depends upon its resistance?
Answer:
Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0-5 A range), a plug key and some connecting wires. Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in the figure below:

Connect the nichrome wire in the gap XY. Close the plug key. Note down the current in the circuit from the reading of the ammeter. Repeat the activity by replacing the nichrome wire with the torch bulb and the 10 W bulb one after the other.

Note down the ammeter reading in each case. Compare the current in the circuit in all the three cases. This is due to the fact that the resistance of each of three gadgets connected in the gap is different. From this we can conclude that the current through an electric component depends on its resistance.

Question 37.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

Plot a graph between V and I and calculate the resistance of the resistor.
Answer:
If we plot the values of current along the Y-axis and the potential difference (voltage) along X-axis for the values given, we get the following graph:

Resistance of resistor (R)

Question 38.
Why does the current flowing through a circuit depend on the resistance of the circuit? How does the current through a device vary with its resistance?
Answer:
Certain components offer an easy path for the flow of electric current while some others resist the flow. We know that motion of electrons in an electric circuit constitutes an electric current.

The electrons, however, are not completely free to move within a conductor. They are restrained by the attraction of the atoms among which they move. Thus, its resistance retards motion of electrons through a conductor. The current flowing through a device is inversely proportional to its resistance.

Question 39.
Let the resistance of an electrical component remain 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:
We know that the current flowing through a conductor is directly proportional to the potential difference across it. Therefore, when the potential difference across the component is reduced to half the initial value, the current will also get halved.

Question 40.
Distinguish between a good conductor, a resistor, a poor conductor and an insulator on the basis of resistance.
Answer:
A material, which offers low resistance to the flow of electrons or electric current in an electric circuit, is known as a good conductor.

A component in an electric circuit, which offers resistance to the flow of electrons constituting electric current, is known as resistor.

A material, which offers higher resistance to the flow of electrons or electric current in an electric circuit, is known as poor conductor.

A material, which offers very high resistance to the flow of electrons or electric current in an electric circuit, is known as insulator. Electric current does not flow through an insulator.

Question 41.
What potential difference must be applied across a 10 Ω wire in order that a current of 2.5 A flows through it?
Answer:
Given: Current I = 2.5 A; Resistance R= 10 Ω; Potential difference V = ?
V = RI (Ohm’s law)
= 10 Ω × 2.5 A = 25 V.

Question 42.
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:
Given: Potential difference, V = 12 V, Current, I = 2.5 mA = 0.0025 A, Resistance, R = ?
According to Ohm’s law, R = \(\frac{v}{1}\)
= \(=\frac{12 \mathrm{V}}{0.0025 \mathrm{A}}\) = 4800 Ω.

Question 43.

1. How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?
2. How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω?

Answer:
1. Given: Voltage, V= 220 V; Resistance, R = 1200 Ω; Current, I = ?

2. Given: Voltage, V= 220 V; Resistance, R = 100 Ω; Current, I = ?

Question 44.
The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?
Answer:
We are given, potential difference V= 60 V, current 7 = 4A.
V 60 V
According to Ohm’s law, R = \(\frac{V}{I}=\frac{60 \mathrm{V}}{4 \mathrm{A}}\) = 15 Ω.
When the potential difference is increased to 120 V the current is given by
Current = \(\frac{V}{R}=\frac{120 \mathrm{V}}{15 \Omega}\) = 8 A.
The current through the heater becomes 8A.

Question 45.
On what factors does the resistance of a conductor depend?
Answer:
The resistance of a conductor depends on the following factors:

1. Length of the conductor,
2. area of cross-section of the conductor, and
3. nature of the material of the conductor.

Question 46.
The resistance of a conductor varies with its length. How do you establish this fact?
Answer:
Take three nichrome wires L₁, L₂ and L₃ of same thickness but of lengths in the ratio 1:2:3 respectively. Complete an electric circuit consisting of a cell, an ammeter, nichrome wire L₁ and a plug key, as shown in the figure below.

Now, close the plug key. Note the current in the ammeter. Replace the nichrome wire L₁ with nichrome wire L₂. Note the ammeter reading. Now replace this wire with nichrome wire L₃. Note down the ammeter reading again.

Compare the values of the current through L₁, L₂ and L₃. The current is found to be in the ratio 3:2:1 respectively. This shows that the resistance of L₂ and L₃ is twice and thrice that of L₁ respectively. This establishes that the resistance of a conductor is directly proportional to its length.

Question 47.
How do you show experimentally that the resistance of a conductor varies with its area of cross section?
Answer:
Take three nichrome wires L₁, L₂ and L₃ of same length but of area of cross section in the ratio 1:2:3 respectively. Complete an electric circuit consisting of a cell, an ammeter, nichrome wire L₁ and a plug key, as shown in the figure.

Now, close the plug key. Note the current in the ammeter. Replace the nichrome wire L₁ with nichrome wire L₂. Note the ammeter reading. Now replace this wire with nichrome wire L₃.

Note down the ammeter reading again. Compare the values of the current through L₁, L₂ and L₃. The current is found to be in the ratio 1:2:3 respectively.

This shows that the resistance of L₂ and L₃ is half and one-third that of L₁ respectively. This establishes that the resistance of a conductor is inversely proportional to its area of cross section.

Question 48.
How does the resistance of a wire vary with its area of cross-section?
Answer:
Resistance of a wire is inversely proportional to its area of cross-section. This means that the resistance of a wire decreases in the same ratio as the increase in the area of its cross-section and vice- versa. Thus, if the wire is thick (large area of cross-section), then resistance is less. If the wire is thin (less area of cross-section), then resistance is large.

Question 49.
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:
The resistance of a conductor is inversely proportional to the area of cross-section of the conductor. The current will flow more easily through the thicker wire because the thicker wire has more area of cross-section hence has lesser resistance to the flow of current.

Question 50.
How does the resistance of a conductor vary with (a) its length, and (b) area of cross-section?
Answer:

1. The resistance of a conductor is directly proportional to its length.
2. The resistance of a conductor is inversely proportional to its area of cross-section.

Question 51.
Explain the factors that affect electrical resistance.
Answer:
1. The electrical resistance of a conductor is directly proportional to its length. This means that the resistance of a conductor increases in the same ratio as the increase in its length.

2. The resistance of a conductor is inversely proportional to its area of cross-section. This implies that the resistance decreases in the same ratio as the increase in its area of cross-section.

3. The resistance of a conductor depends bn its temperature. The resistance of most of the electrical conductors increases with the increase in temperature.

4. Finally, the resistance also depends on the material of the conductor. This means identical conductors made of different materials have different resistances.

Question 52.
Define ‘electrical resistivity’ of the material of a conductor. Give an expression for the resistivity of a material.
Answer:
The electrical resistivity of the material of a conductor is defined as the resistance of a conductor made of that material of length 1 m and area of cross-section 1 m².

Consider a conductor of length l and area of cross-section A. Let its resistance be R. We know that the resistance of a conductor is directly proportional to its length t and inversely proportional to its area of cross-section A.

where ρ is a constant called ‘resistivity’ of the material.
Resistivity, ρ = \(\frac{\mathrm{RA}}{l}\).

Question 53.
What is the S.I unit of resistivity? What is the other name for resistivity of a material?
Answer:
The S.I. unit of resistivity is called ‘ohm metre’. Its symbol is ‘Ω. m’. The resistivity of a material is also called its ‘specific electrical resistance’.

Question 54.
You are given three copper wires of length l, 2l and 3l having area of cross-section A, 2A and 3A. Which of these has highest resistivity?
Answer:
Resistivity of a conductor depends only on the material it is made of and not on its dimensions. Therefore, all the three wires have equal resistivity, as all of them are made of the same material namely copper.

Question 55.
A wire of resistivity ? is pulled to double its length. What will be its new resistivity?
Answer:
Since resistivity of a conductor depends only on the material it is made of and not on its dimensions, the resistivity ρ remains the same.

Question 56.
The electrical resistivity of copper is 1.62 × 10^-8? m. What is the meaning of this statement?
Answer:
The statement above means that a copper wire of length 1 m and area of cross-section 1 m² has resistance of 1.62 × 10^-18 Ω.

Question 57.
The resistivity of some materials is siven in the table below:

Material	Resistivity (Ω m)
1. Copper	1.62 x 10-8
2. Iron	10.0 x 10-8
3. Mercury	94.0 x 10-8
4. Silver	1.60 x 10-8

1. Which among iron and mercury is a better conductor?
2. Which material is the best conductor?

Answer:

1. Among iron and mercury, the better conductor is iron. This is because the resistivity of iron is lower than that of mercury.
2. Among the given materials, silver is the best conductor of electricity. This is because silver has the lowest value of resistivity.

Question 58.
Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature?
Answer:
Given: Resistance, R = 26 Ω; Diameter, d = 0.3 mm = 3 × 10^-4m, Length of the wire, l = 1 m.

Question 59.
A wire of given material having length l and area of cross-section A has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length l/2 and area of cross-section 2 A ?
Answer:
For the first wire: R₁ = ρ \(\frac{l}{A}\) = 4 Ω
Now for the second wire
R₂ = \(\frac{1 / 2}{2 A}=\frac{1}{4} \rho \frac{l}{A}\)
R₂ = \(\frac{1}{4}\) R
R₁ = 1 Ω.
The resistance of the new wire is 1 Ω.

Question 60.
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:
Area of cross-section of the wire, A = p \(\left(\frac{d}{2}\right)^{2}\)
Diameter = 0.5 mm = 0.0005 m
Resistance, R = 10 Ω

= 122.72 m
∴ Length of the wire = 122.72 m.
If the diameter of the wire is doubled, new diameter = 2 × 0.5 = 1 mm = 0.001 m.
Let new resistance be R’.
R’ = ρ \(\frac{l}{A}\)

= 250.2 × 10^-2
= 2.5 Ω
Therefore, the length of the wire is 122.7 m and the new resistance is 2.5 Ω.
Thus, the resistance of the conductor will become \(\frac{1}{4}\) th its initial value when the diameter is doubled.

Question 61.
Distinguish between resistance and resistivity of a conductor.
Answer:

Resistance	Resistivity
1. Resistance of a conductor is the obstruction offered by the conductor for the flow of current through it.	Resistivity of a material is the obstruction offered by a conductor of that material of length 1 m and area of cross-section 1 m2.
2. It depends on the material, length and area of cross-section of the conductor.	It does not depend on the length and area of cross-section of the conductor.
3. The S.I. unit of resistance is called ‘ohm’ (Ω).	The S.I. unit of resistivity is called ‘ohm-metre’ (Ωm).

Question 62.
When do we say that a group of resistances are connected in series? Show by a diagram the arrangement of three resistances in series.
Answer:
A group of resistances are said to be in series if they are connected from end to end so that each of them carries the same current.
Three resistances R₁, R₂ and R₃ are connected in series as shown below:

Question 63.
Draw the diagram of the electric circuit in which the resistors are connected in series.

OR

Three resistances R₁, R₂ and R₃ are connected in series. This combination is connected in series with a battery, a key and an ammeter. A voltmeter is connected to measure the potential difference across the combination. Draw a suitable circuit diagram for this.
Answer:

Question 67.
What is the equivalent resistance of a group of resistances connected in series? Give a suitable equation for it.
Answer:
The equivalent resistance of a group of resistances connected in series is equal to the sum of all the individual resistances.

Consider a number of resistances R₁, R₂, R₃………R_n connected in series. Let R be their equivalent resistance. Now,
R = R₁ + R₂ + R₃ + ……… + R_n.

Question 65.
Obtain an equation for the combined resistance of a group of resistances in series.
Answer:

A group of resistances are said to be in series if they are connected from end to end so that they carry the same current through each of them. Consider three resistances R₁, R₂ and R₃ connected in series.

Let V₁ be the potential difference across R₁, V₂ be the potential difference across R₂ and V₃ be the potential difference across R₃.

Let I be the current flowing through each of these resistors. Let V be the potential difference across the combination and R be the total resistance of the system.
Total potential difference V = V₁ + V₂ + V₃.
But, V = IR, V₁ = IR₁, V₂ = IR₂, and V₃ = IR₃.
Substituting these values in the equation above, we get,
IR = IR₁ + IR₂ + IR₃
∴ R_s = R₁ + R₂ + R₃
This shows that the combined resistance of a group of resistances connected in series is equal to the sum of all the individual resistances in the group.

Question 66.
How do you show that the same current flows through each resistor when a group of resistors are connected in series?
Answer:

Connect three resistors, say of 1 Ω, 2 Ω, 3 Ω, in series. Connect this combination with a battery of say 6 V, an ammeter and a plug key in series. Close the circuit. Note the ammeter reading. Change the position of the ammeter to anywhere in between the resistors and note the reading of the ammeter each time.

Each time the ammeter shows the same current. This shows that, whenever a number of resistors are connected in series, the current flowing through each part of the circuit is same. This is because the current has only one path to flow.

Question 67.
When do we say that a group of resistances are in parallel arrangement? Show by a diagram the arrangement of three resistances in parallel arrangement.
Answer:
A group of resistances are said to be in parallel arrangement when one end of each resistance is connected to one common terminal and the other end of each resistance is connected to another common terminal. In such an arrangement, the potential difference across each resistance would be the same as all of them have common end points. Three resistances R₁, R₂ and R₃ are connected, as shown below, in parallel arrangement:

Question 68.
Draw the diagram of the electric circuit in which the resistors R₁, R₂ and R₃ are connected in parallel including ammeter and voltmeter and mark the direction of current.
Answer:

Question 69.
What is the equivalent resistance of a group of resistances connected in parallel? Give a suitable equation for it.
Answer:
The reciprocal of the equivalent resistance of a group of resistances in parallel is equal to the sum of the reciprocals of all the individual resistances.

Consider a number of resistances R₁, R₂, R₃ ……….. R_n connected in parallel. Let R be their equivalent
resistance. Now,

Question 70.
Obtain an equation for the combined resistance of a group of resistances in parallel
Answer:

A group of resistances are said to be in parallel if one end of each resistance is connected to one common terminal and the other end is connected to another common terminal so that each resistor has the same potential difference across it.

Consider three resistances R₁, R₂ and R₃ connected in parallel. Let I₁ be the current through R₁, I₂ be the current through R₂ and I₃ be the current through R₃. Let I be the total current through the system, V be the potential difference across each resistor and R be the total resistance of the system.
Total current I = I₁ + I₂ + I₃.
Substituting these values in the equation above, we get,

Thus, the reciprocal of the combined resistance of a group of resistances in parallel is equal to the sum of the reciprocals of all the individual resistances.

Question 71.
Why is the series arrangement not used for domestic circuits?

OR

Mention the disadvantages of connecting electrical appliances in series in domestic wiring.
Answer:
In domestic circuits if gadgets are connected in series, voltage is divided. Each component of a series circuit receives a small voltage so that the amount of current decreases and the device becomes hot and does not work properly.

Secondly, when gadgets are in series arrangement, they cannot be operated independently. If one of them does not work, the circuit is broken and the other components also do not work. Hence, series arrangement is not used in domestic electric circuits.

Question 72.
Three resistors of 3 Ω, 4 Ω and 5 Ω are connected in

1. series and in
2. parallel Find the equivalent resistance in each case.

Answer:
R₁ = 3 Ω, R₂ = 4 Ω, R₃ = 5 Ω, combined resistance R = ?

1. When resistances are connected in series, the equivalent resistance R is given by
R = R₁ + R₂ + R₃
R = 3 Ω + 4 Ω + 5 Ω
= 12 Ω

2. When resistances are in parallel arrangement, the equivalent resistance R is given by

Question 73.
In the circuit diagram given below, suppose the resistors R₁, R₂ and R₃ have values 5 Ω, 10 Ω, 30 Ω? respectively, which are connected to a battery of 12 V. Calculate

1. The current through each resistor,
2. The total current in the circuit, and
3. The total circuit resistance.

Answer:

1. Current through 5 Ω resistor = \(\frac{V}{R}=\frac{12 \mathrm{V}}{5 \Omega}\) = 2.4 A
Current through 10 Ω resistor = \(\frac{V}{R}=\frac{12 \mathrm{V}}{10 \Omega}\) = 1.2 A
Current through 30 Ω resistor = \(\frac{V}{R}=\frac{12 \mathrm{V}}{30 \Omega}\) = 0.4 A

2. Total current through the circuit, I = 2.4 A + 1.2 A + 0.4 A = 4 A

3. Total resistance of the circuit, R = \(\frac{V}{I}=\frac{12 \mathrm{V}}{4 \mathrm{A}}\) = 3 Ω.

Question 74.
If, in the figure given below, R₁ = 10 Ω, R₂ = 40 Ω, R₃ = 30 Ω, R₄ = 20 Ω, R₅ = 60 Ω, and a 12 V battery is connected to the arrangement Calculate

1. The total resistance in the circuit, and
2. The total current flowing in the circuit

Answer:

1. Suppose we replace the parallel resistors R₁ and R₂ by an equivalent resistor of resistance R’ and resistors R₃, R₄ and R₅ by an equivalent single resistor R”. Then,

Now, R’ and R” are in series. Let their combined resistance be R. Now, R is the total resistance of the circuit.
R = R’ + R”
R = 8 Ω + 10 Ω = 18 Ω.

2. Current through the circuit, I = \(\frac{V}{R}=\frac{12 \mathrm{V}}{18 \Omega}\) = 0.67 A.

Question 75.
Four resistors 5 Ω, 6 Ω, 4 Ω and 8 Ω are connected in

1. series and in
2. parallel. Find the equivalent resistance in each case.

Answer:
R₁ = 5 Ω, R₂ = 6 Ω, R₃ = 4 Ω, R₄= 8Ω combined resistance R = ?

1. When the resistances are connected in series, the equivalent resistance R is given by
R = R₁ + R₂ + R₃ + R₄
R = 5 Ω + 6 Ω + 4 Ω + 8 Ω = 23 Ω.

2. When the resistances are in parallel arrangement, the equivalent resistance R is given by

Question 76.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?

OR

Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Answer:
Coils of electric toasters and electric irons are made of an alloy rather than a pure metal for the following reasons:

1. Alloys have higher resistivity than pure metals. This enables them to generate more heat.
2. Alloys, unlike pure metals, generally do not melt easily even at higher temperatures.

Question 77.
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 plus key, all connected in series.
Answer:

Question 78.
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 plus key, all connected in series. Put 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 readinss in the ammeter and the voltmeter?
Answer:

The resistors of 5 Ω, 8 Ω and 12 Ω are connected in series.
The resistance of the circuit = 5 Ω + 8 Ω + 12 Ω = 25 Ω.
Current through the circuit (Ammeter reading) = I = \(\frac{6 \mathrm{V}}{25 \Omega}\) = 0.24 A
Therefore, ammeter reading = 0.24 A.
Voltmeter reading (P.D across 12 Ω) = 12 Ω × 0.24 A = 2.88 V.
Therefore, voltmeter reading = 2.88 V.

Question 79.
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:

Total resistance of the circuit, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω
Current through the 12 Ω resistor, I = \(\frac{V}{R}=\frac{9 V}{13.4 \Omega}\) = 0.67 A
Each resistor in the circuit carries the same current. Therefore, the current flowing through the 12 Ω resistor is 0.67 A.

Question 80.
An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery as shown in the figure below. Calculate

1. The total resistance of the circuit,
2. The current through the circuit, and
3. The potential difference across the electric lamp and conductor.

Answer:

Given: Resistance of the electric lamp, R₁ = 20 Ω, Resistance of the conductor connected in series, R₂ = 4 Ω, Total resistance of the circuit, R = ?, Current, I = ?, Potential difference across the lamp V_L = ?, Potential difference across the 4 Ω resistor = ?

1. The total resistance in the circuit, R = R₁ + R₂
= 20 Ω + 4 Ω = 24 Ω.

2. Current through the circuit, I = \(\frac{V}{R}=\frac{6 V}{24 \Omega}\) = 0.25 A.

3. Potential difference across the lamp, V_L = 20 Ω × 0.25 A = 5 V
Potential difference across the conductor = 4 Ω × 0.25 A = 1 V.

Question 81.
Judee the equivalent resistance when the following are connected in parallel:

1. 1 Ω and 10⁶ Ω.
2. 1 Ω, 10³ Ω and 10⁶ Ω.

Answer:
1. When 1 Ω and 10⁶ Ω are connected in parallel, the equivalent resistance R is given by:

2. When 1 Ω, 10³ Ω and 10⁶ Ω are connected in parallel, the equivalent resistance R is given by:

Question 82.
How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a total resistance of

1. 4 Ω
2. 1 Ω?

Answer:
1. We must connect 3 Ω and 6 Ω resistors in parallel.
The resulting resistance would be \(=\frac{3 \Omega \times 6 \Omega}{3 \Omega+6 \Omega}\) = 2 Ω
Now, the combination above must be connected in series with the 2 Ω wire.
The resulting resistance would be = 2 Ω + 2 Ω = 4 Ω.

2. To get 1 Ω, we must connect 2 Ω, 3 Ω, and 6 Ω resistors in parallel, as shown below:

Let the combined resistance be R. Now,

Question 83.
What is

1. The highest,
2. The lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer:
1. The highest resistance can be secured by combining all the given four resistors in series as shown below:

The highest total resistance of the system = R₁ + R₂ + R₃ + R₄
= 4 Ω + 8 Ω + 12 Ω + 24 Ω
= 48 Ω.

2. The lowest total resistance is secured when the four resistors are in parallel arrangement as shown in the figure below.

Let the combined resistance be R. Now,

Question 84.
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of

1. 9 Ω,
2. 4 Ω.

Answer:
1. Two resistors in parallel and the combination is in series with the third:

Combined resistance of two 6 Ω resistors in parallel = \(\frac{6 \Omega \times 6 \Omega}{6 \Omega+6 \Omega}\) = 3 Ω
When this combination is in series with the other 6 ? resistor, the total resistance = 6 Ω + 3 Ω = 9 Ω.

2. Two resistors in series and the combination in series with the third:

Combined resistance of two 6 ? resistors in series = 6 Ω + 6 Ω = 12 Ω.
When this combination is in parallel with the other 6 Ω,
Total resistance =

Question 85.
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:
Let the number of 176 Ω resistors required to be connected in parallel to carry 5 A current on a 220 V line be n.

Thus, four resistors of 176 Ω (in parallel) are required to carry 5 A on a 220 V line.

Question 86.
An electric lamp of 100 ?, a toaster of resistance 50 ?, 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:
Given: Resistance of electric lamp, R₁ = 100 Ω, Resistance of toaster, R₂ = 50 Ω, Resistance of water filter, R₃ = 500 Ω, Potential difference of the source, V = 220 V.
Since all the resistances are in parallel, their equivalent resistance R is given by:

Current drawn by the three appliances, I

The resistance of electric iron which draws same current, R

Therefore, the resistance of the electric iron is 31.25 Ω and the current flowing through it is 7.04 A.

Question 87.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:
The following are the advantages of connecting electrical devices in parallel with the battery:

1. Voltage across each electrical device is same and the device can take current as per its resistance.
2. Separate on/off switches can be applied across each device. Hence, they can be operated independent of one another.
3. Total resistance in parallel circuit decreases, hence, greater current may be drawn from the cell as per the requirement of the devices.
4. Even if one electrical device fuses or fails, the other devices will continue to work.

Question 88.
Why is it impracticable to connect an electric bulb and an electric heater in series?
Answer:
We have seen that, in a series circuit, the current is constant throughout the electric circuit.
However, gadgets such as an electric bulb and an electric heater need currents of widely different values to operate properly.

This cannot be provided when they are connected in series. Secondly, an electric bulb and an electric heater cannot be operated independently when they are put in series. Therefore, it is impracticable to connect an electric bulb and an electric heater in series. These problems can be overcome by putting these gadgets in parallel connection.

Question 89.
Why is parallel arrangement used in domestic wiring?

OR

It is advantageous to connect electric devices in parallel instead of connecting them in series. Why?
Answer:
In parallel arrangement,

1. All gadgets get the same voltage which is equal to the supply voltage itself,
2. Gadgets can be operated independent of each other, and
3. Even if one of the gadgets goes out of order, the other gadgets will continue to work as the circuit is not broken.

Question 90.
What is meant by heating effect of electric current?
Answer:
When an electric current flows through a conductor, a part of the electrical energy will dissipate in the form of heat. This phenomenon is known as heating effect of electric current.

Question 91.
Name some devices that work on the principle of heating effect of electric current.
Answer:
Electric heater, electric bulb, electric iron, electric room heater, electric hair drier, electric fuse etc., are some of the devices that work on the principle of heating effect of electric current.

Question 92.
What happens to the power consumed by filament lamps?
Answer:
Most of the power consumed by filament lamps is wasted in the form of heat. A small part of the power appears as light.

Question 93.
What is electric fuse? State the principle on which it works.
Answer:
A safety device provided in electric circuits to protect the circuit and the gadgets from overloading (current exceeding a safe limit) is called an electric fuse. An electric fuse works on the principle of heating effect of electric current.

Question 94.
What is an electric fuse made of? How does it protect electric circuits from overloading?
Answer:
An electric fuse is a short wire made of an alloy of low melting point and low resistance. It is inserted in series in an electric circuit usually inside a porcelain box. When current exceeds the safe limit of the circuit, the heat generated will melt the fuse wire and break the circuit. This is how fuse protects circuits and gadgets in the event of overloading.

Question 95.
An electric iron consumes 1 kW electric power when operated at 220 V. What is the rating of the fuse wire to be used in the circuit to protect the electric iron from overloading?
Answer:
Given: Power, P = 1 kW = 1000 W; Voltage, V = 220 V; Current, I = ?
Current, I

Therefore, a fuse wire of rating 5 A must be used.

Question 96.
Fuse wire should have higher resistance than the other wires in the circuit but it should not have high melting point. Why?
Answer:
An electric fuse is a device that protects electric circuits from the dangers of overloading. It works on the principle of heating effect of electric current. The fuse wire should have high resistance and low melting point.

This is because its high resistance helps to produce more heat and hence high temperature. Because of its low melting point, the fuse wire melts in the event of overloading and protects the circuit.

Question 97.
Obtain an expression for the heat produced in a conductor carrying current I and operating under potential difference V.
Answer:

Consider a current 1 flowing through a resistor of resistance R. Let the potential difference across it be V as shown in the figure. Let t be the time during which a charge Q flows across. The work done in moving the charge Q through a potential difference V is VQ. Therefore, the source must supply energy equal to VQ in time t. Hence the power input to the circuit by the source is
P = V \(\frac{Q}{t}\) = VI.
The energy supplied to the circuit by the source in time t, E = P × t.
E = VIt
This energy gets dissipated in the resistor as heat. Thus, for a steady current I, the amount of heat H produced in time t is H = VIt.
But, according to Ohm’s law, V = RI.
∴ H = I² Rt.

Question 98.
State Joule’s law of heating effect of electric current. Express it in the form of an equation.

OR

According to Joule’s law of heating, mention the factors on which heat produced in a resistor depends. According to this law, write the formula used to calculate the heat produced.
Answer:
According to Joule’s law of heating effect of electric current, the heat generated in a conductor carrying a steady current is directly proportional to:

1. The square of the current for a given resistance,
2. The resistance of the conductor for a given current, and
3. The time for which current flows through the resistor.

Consider a conductor of resistance R carrying current I when the potential difference across it is V.
The heat generated in the conductor in time t second is given by:
H ∝ I²Rt
H = K I²Rt
Here, K is the constant of proportionality. When I is in ampere, R is in ohm, heat produced H is in joule and time t is in second, the value of K becomes equal to 1.
∴ H = I²Rt.

Question 99.
Compute the heat generated while transferrins 96000 coulomb of charge in one hounthroueh a potential difference of 50 V.
Answer:
Given: Charge transferred, Q = 96000 C; Time, t= 1 hour = 60 × 60 = 3600 s; Potential difference, V= 50 V; Heat generated, H = ?

Heat generated, H = VIt
= 50 V × 26.66 A × 3600 s
= 4800000 J = 4800 kJ
4800 kJ is generated while transferring the charge.

Question 100.
100 J of heat is produced each second in a 4 Ω resistance. Find the potential difference across the resistor.
Answer:
Given: Heat generated, H = 100 J; Resistance, R = 4 Ω, Time, t = 1 s; Potential difference, V = ?
From the equation H = I²Rt we have the current through the resistor as

Thus the potential difference across the resistor V is
V=IR
= 5 A × 4 Ω = 20 V.

Question 101.
Why does the cord of an electric heater not slow while the heatine element does?
Answer:
The heating element of the heater is usually made up of alloys, which have very high resistance. Therefore, such heating elements produce high amount of heat and become very hot and glow. The resistance of a conducting cord is usually made of copper or aluminium. These materials have very low resistance. Hence heat produced is also low. Hence they do not glow.

Question 102.
An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
Given: Resistance, R = 20 Ω; Current, I = 5 A, Time taken, t = 30 s; Heat generated, H = ?
Heat generated, H = I²Rt
H = (5 A)² × 20 Ω × 30 s
= 15000J = 15 kJ.
15 kJ heat is developed in 30 s.

Question 103.
An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Answer:
Given, R = 8 Ω, I = 15 A, t = 2h = 7200 s.
Heat generated, H = I²Rt
Rate at which heat is generated, \(\frac{\mathrm{H}}{\mathrm{t}}\) = I²R
= 15 A × 15 A × 8 Ω
= 1800 J.

Question 104.
What determines the rate at which energy is delivered by a current?
Answer:
Electric power determines the rate at which energy is delivered by a current.

Question 105.
Define electric power. Mention its S.I. unit.
Answer:
The rate at which energy is dissipated (consumed) or the rate at which work is done by electric energy is called electric power. It is represented by the letter P.

The S.I. unit of electric power is called ‘watt’ (W).

Question 106.
Give an expression for electric power in terms of voltage and current.
Answer:
Electric power is given by the product of voltage and current.
Power = Voltage (Potential difference) × Current P = VI.

Question 107.
Write any two formulae used to calculate electric power.
Answer:
1. Power = Voltage × Resistance
P = VI

2.

Question 108.
Define 1 watt of power.
Answer:
A body is said to have power of 1 watt if it does work at the rate of 1 joule in 1 second. 1 watt of power is the power consumed by a device that carries 1 A of current when operated at a
potential difference of 1 V.
1 watt = \(\frac{1 \mathrm{J}}{1 \mathrm{s}}\) = 1 Js^-1.

Question 109.
What is the commercial unit of electric power? Define it.
Answer:
The commercial unit of electric power is called ‘kilo-watt hour’ (kWh). One kilo-watt hour is the amount of energy consumed by an agent in one hour working at a constant rate of one kilo-watt, i.e., at the rate of 1000 J per second.

Question 110.
An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
Answer:
Given: Voltage, V= 220 V; Current drawn, I = 0.5 A; Power, P = ?
Power, P = VI
= 220 V × 0.50 A
= 110 J/s = 110 W.
Power of the lamp is 110 W.

Question 111.
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:
Given: Current, I = 5 A; Voltage, V = 220 V, Time, t = 2 h = 7200 s; Power, P = ?; Energy consumed, E = ?
Electric power, P = VI
P = 220 V × 5 A
= 1100 w
Energy consumed, E = Power × Time
= 1100 × 7200J
= 7920000 J
= 7920 kJ or 7.92 × 10⁶ J.

Question 112.
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:

Current, I = \(\frac{P}{V}\)
Current drawn by 100 W lamp = \(\frac{100 \mathrm{W}}{220 \mathrm{V}}\) = 0.45 A.
Current drawn by 60 W lamp = \(\frac{60 \mathrm{W}}{220 \mathrm{V}}\) = 0.27 A.
Total current drawn by the two lamps from the mains = 0.45 A + 0.27 A = 0.72 A.

Question 113.
Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?
Answer:
Energy consumed = Power × Time
Energy consumed by a 250 W television set in 1 hour = 250 W × 3600 s
= 900000 J = 900 kJ
Energy consumed by a 1200 W toaster in 10 minutes = 1200 W × 600 s
= 720000 J = 720 kJ.
Thus, a TV set consumes more energy than the toaster.

Question 114.
A person connects a bulb having resistance 1200 Ω and an electric heater having resistance 100 Ω in series. But they do not work properly. When he connects the same appliances in parallel in the circuit, they function normally. Explain the reason for this.
Answer:
Electrical gadgets of different powers need different currents for their proper functioning. This means the current drawn by an appliance depends on its power. A bulb of 1200 Ω and a heater of 100 Ω cannot work properly when same current flows through each of them.

The bulb needs smaller current while the heater requires larger current. These gadgets can draw the required current from the source only when they are in parallel arrangement. When they are in series, the same current flows through each of them and hence they do not work properly.

Question 115.
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:
Resistance Ri of the bulb is given by the expression R = \(\frac{V^{2}}{P}\)
Supply voltage, V = 220 V
Maximum allowable current, I = 5 A
Rating of electric bulb P = 10 watts
Given, power = 10W
Potential difference, V= 220 V
Electric current, I = 5 A
To calculate total number of bulbs, total resistance has to be calculated.

Let the number of bulbs connected in parallel = n.
Let the total effective resistance, when n bulbs are connected in parallel = R₁.

Number of electric bulbs that can be connected in parallel is 110.

Question 116.
An electric iron consumes energy at the rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?
Answer:
Power input, P = VI.
Thus the current I = \(\frac{P}{V}\).

1. When heating is at the maximum rate:
Current, I = \(\frac{840 \mathrm{W}}{220 \mathrm{V}}\) = 3.28 A.
Resistance of the electric iron, R = \(\frac{V}{I}=\frac{220 \mathrm{V}}{3.82 \mathrm{A}}\) = 57.60 ?

2. When heating is at the minimum rate:
Current, I = \(\frac{360 \mathrm{W}}{220 \mathrm{V}}\) = 1.64 A.
Resistance of the electric iron, R = \(\frac{V}{I}=\frac{220 \mathrm{V}}{1.64 \mathrm{A}}\) = 134.15 Ω

Question 117.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:
Given: Potential difference, V=220 V; Resistance of each coil, R = 24 Ω; Current drawn, I = ?

1. When current is drawn separately by the two coils:
Current drawn by each of the coils, I = \(\frac{V}{R}=\frac{220 \mathrm{V}}{24 \Omega}\) = 9.16 A.

2. When the two coils are in series:
Their combined resistance = 24 Ω + 24 Ω = 48
Current drawn by the combination, I = \(\frac{V}{R}=\frac{220 \mathrm{V}}{48 \Omega}\) = 4.58 A.

3. When the two coils are in parallel:
Their combined resistance

Current drawn by the combination,

Question 118.
Compare the power used in the 2 Ω resistor in each of the following circuits:

1. A 6 V battery in series with 1 Ω and 2 Ω resistors, and
2. A 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer:
1. The circuit shown below has resistance connected in series combination.

Current in the circuit,

Power used = I²R = (2)² × 2 = 2 × 2 × 2 = 8W.

2. The circuit is as shown below:

In parallel combination, potential across each resistor is same and equal to the potential applied to the circuit.
Potential across 2 Ω resistor, V = 4 V.
Power used, \(\frac{V^{2}}{R}=\frac{4 \times 4}{2}\) = 8 W.
Power used in both the cases is same.

Question 119.
What is the commercial unit of energy? Define it. Give its relationship to S.I. unit of power.
Answer:
The commercial unit of energy is called ‘kilo-watt hour’ (kW h).
1 kW h is the energy consumed in one hour at the rate of 1000 joule per second.
1 kilowatt hour = 1 kilowatt × 1 hour
= 1 kW × 1 h
= 1000 W × 1 h
= 1000 J s^-1 × 3600 s
1 kW h = 3600000 J = 3600 kJ.

Question 120.
An electric refrigerator rated 400 W is used for 8 hours/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kWh?
Answer:
Total energy consumed = Power in watts × No. of hours
= 400 W × 8.0 × 30 days = 96000 W h
= 96 kW h.
The cost of energy = 96 kWh × Rs 3.00 = Rs. 288.00.

Question 121.
An electric refrigerator rated 400 W is used for 8 hours a day. An electric iron box rated 750 W is used for 2 hours a day. Calculate the cost of using these appliances for 30 days, if the cost of 1 kWh is Rs. 3/-.
Answer:
Total energy consumed by the refrigerator in 30 days
= 400 × 8 × 30 = 96000 Wh = 96 kWh
Total energy consumed by the iron box in 30 days
= 750 × 2 × 30 = 45000 Wh = 45 kWh
Total energy consumed by the refrigerator and iron box is
= 96 kWh + 45 kWh = 141 kWh
The bill amount for 141 kWh at the rate of Rs. 3 per 1 kWh is
= 141 × 3 = Rs.423.

Question 122.
A potential difference of220 V is applied across a resistance of 440 Ω in an electrical appliance. Calculate the current drawn and the heat energy produced in 20 seconds.
Answer:

Given: V = 220V, R = 440 Ω
Heat energy produced, H = VIt
= 220 × 0.5 × 20 = 2200 J.

Question 123.
Explain the following:

1. Why is tungsten used almost exclusively in filament of electric lamps?
2. Why are copper and aluminium wires usually employed for electricity transmission?

Answer:
1. Tungsten is a metallic chemical element that has very high melting point and very high resistivity. Therefore, it produces high temperatures but does not melt at such temperatures. It retains as much of heat generated, so that it becomes very hot and emits light. Therefore, it is an ideal material for making filament of lamps.

2. Copper and aluminium are good conductors of electricity and have low resistivity. So, they conduct electric current without heavy heat losses. They are highly ductile. This is why they are usually used for making wires for electricity transmission.

Fill In The Blanks

1. The rate of flow of charges through a conductor is called current
2. The device used to measure potential difference between two points of a conductor is called voltmeter
3. A device that provides continuously variable resistance is called rheostat
4. The S.I. unit of resistivity of a material is ohm-metre
5. The resistance of a conductor is inversely proportional to its area of cross-section
6. Power of an appliance is given by the product of current and voltage
7. A safety device, which protects circuits from overloading, is electric fuse
8. The commercial unit of electrical energy is called kilo-watt hour

Multiple Choice Questions

Question 1.
A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross-section of the filament in 16 seconds would be roughly
(A) 10²⁰
(B) 10¹⁶
(C) 10¹⁸
(D) 10²³
Answer:
(A) 10²⁰

Question 2.
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

Question 3.
Which of the following terms does not represent electrical power in a circuit?
(A) I²R
(B) IR²
(C) VI
(D) V²/R
Answer:
(B) IR²

Question 4.
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
Answer:
(D) 25 W

Question 5.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in 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

Question 6.
Resistivity of a wire depends on
(A) its length
(B) its material
(C) its cross-section area
(D) All of the above
Answer:
(B) its material

Question 7.
Resistance of a wire is ‘r’ ohm. If the wire is stretched to double its length, then its resistance in ohm is
(A) \(\frac{r}{2}\)
(B) 4 r
(C) r
(D) \(\frac{r}{4}\)
Answer:
(B) 4 r

Question 8.
The unit ‘ampere second’ could be the unit of
(A) power
(B) conductance
(C) energy
(D) electric charge
Answer:
(D) electric charge

Question 9.
The commercial unit of electrical energy is
(A) kilo-watt
(B) joule
(C) kilo-watt hour
(D) kilo-joule
Answer:
(C) kilo-watt hour

Question 10.
Two resistances R₁ and R₂ give combined resistance of 4.5 Ω when in series and 1 Ω when in parallel The resistances are
(A) 1.5 Ω and 0.5 Ω
(B) 1.5 Ω and 3 Ω
(C) 2 Ω and 6 Ω
(D) 3 Ω and 9 Ω
Answer:
(B) 1.5 Ω and 3 Ω

Question 11.
The amount of work done in joule, when +1 C of electric charge moves from one point to another point in an electric circuit is called
(A) electric current
(B) electric resistance
(C) electrical resistivity
(D) potential difference
Answer:
(D) potential difference

Question 12.
One kilowatt hour of electrical energy is the same as
(A) 36 × 10⁵ joule
(B) 36 × 10⁵ watt
(C) 36 × 10⁸ watt
(D) 1000 kilo-joule
Answer:
(A) 36 × 10⁵ joule

Question 13.
Two bulbs marked 200 watt -220 volt and 100 watt -220 volt are joined in series to a 220 volt supply. Power consumed in the circuit is about
(A) 33 watt
(B) 67 watt
(C) 100 watt
(D) 300 watt.
Answer:
(B) 67 watt

Question 14.
The equation form of Ohm’s law is
(A)V = \(\frac{I}{R}\)
(B) R = VI
(C) V = RI
(D) I = VR²
Answer:
(C) V = RI

Question 15.
When a 40 V battery is connected across an unknown resistor, there is a current of 100 mA in the circuit. The value of the resistance is
(A) 5000 Ω
(B) 800 Ω
(C) 0.8 Ω
(D) 400 Ω
Answer:
(D) 400 Ω

Question 16.
In a circuit, there are two unequal resistances in parallel arrangement. Now,
(A) both resistances carry the same current
(B) higher current flows through higher resistance
(C) potential difference across each resistance is same
(D) the resultant resistance of the circuit increases.
Answer:
(C) potential difference across each resistance is same

Question 17.
The rate of flow of electric charges is called
(A) electric current
(B) electric potential
(C) electrical resistance
(D) resistivity
Answer:
(A) electric current

Question 18.
What is the maximum resistance that can be made using five resistors each of \(\frac{1}{5}\) Ω?
(A) \(\frac{1}{5}\) Ω
(B) 10 Ω
(C) 5 Ω
(D) 1 Ω
Answer:
(D) 1 Ω

Question 19.
Which of the following statements is not true, regarding the circuit set-up for the verification of Ohm’s law?
(A) The voltmeter is connected in parallel with the resistance.
(B) The ammeter is connected in series in the circuit.
(C) Use of a power supply in the circuit is optional.
(D) A single key is used in the electric circuit.
Answer:
(C) Use of a power supply in the circuit is optional.

Question 20.
A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of length 21 and resistance R of the same material has area of cross-section equal to
(A) 2 A
(B) \(\frac{\mathrm{A}}{2}\)
(C) \(\frac{3 \mathrm{A}}{2}\)
(D) 3 A
Answer:
(A) 2 A

Question 21.
Substances that have a large number of free electrons and offer a low resistance are
(A) electrical insulators
(B) good conductors of electricity
(C) materials of high resistivity
(D) poor conductors of electricity
Answer:
(B) good conductors of electricity

Question 22.
Electrical resistivity of a given metallic wire depends upon
(A) its length
(B) its thickness
(C) its shape
(D) nature of the material.
Answer:
(D) nature of the material.

Question 23.
The purpose of a rheostat in a circuit is to
(A) increase the potential difference supplied by the battery
(B) decrease the magnitude of current only
(C) increase or decrease the resistivity of the resistors
(D) change the current in the circuit by changing the resistance
Answer:
(D) change the current in the circuit by changing the resistance

Question 24.
The resistance of a 100 W, 200 V lamp is
(A) 100 Ω
(B) 200 Ω
(C) 400 Ω
(D) 1600 Ω
Answer:
(C) 400 Ω

Question 25.
In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W, respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?
(A) Brightness of all the bulbs will be the same.
(B) Brightness of bulb B will be more than that of A.
(C) Brightness of bulb A will be the maximum.
(D) Brightness of bulb C will be less than that of B.
Answer:
(B) Brightness of bulb B will be more than that of A.

Question 26.
The SI unit of electric current is
(A) Ohm
(B) Volt
(C) Ampere
(D) Watt
Answer:
(C) Ampere

Match The Following

Question 1.

Column A	Column B
1. Electric current	a. ohm-metre
2. Electric charge	b. coulomb-metre
3. Potential difference	c. ohm
4. Power	d. volt
5. Resistance	e. kilo-watt hour
6. Energy	f. ampere
7. Resistivity	g. coulomb
	h. watt

Answer:
1 – f, 2 – g, 3 – d, 4 – h, 5 – c, 6 – e, 7 – a.

Students can download Class 10 Science Chapter 13 Magnetic Effects of Electric Current Important Questions, KSEEB SSLC Class 10 Science Important Questions and Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka SSLC Class 10 Science Important Questions Chapter 13 Magnetic Effects of Electric Current

Question 1.
What is magnetic effect of electric current? Who discovered this phenomenon?
Answer:
A current flowing through a conductor produces a magnetic field in the region surrounding the conductor. This phenomenon is known as magnetic effect of electric current. H.C. Oersted discovered the phenomenon of magnetic effect of electric current.

Question 2.
Describe an experiment to show the magnetic effect of electric current.
Answer:
Take a straight thick copper wire AB. Connect the ends of the conductor in series with a battery and a tap key as shown in the figure. Place a small compass needle horizontally just below the copper wire. Note the position of its needle. Now, pass current through the circuit by closing the key. Suddenly we see a deflection in the compass needle.

On stopping the current, the needle comes back to its original direction. On reversing the direction of current, the direction of deflection of the compass needle will also get reversed. This means that the electric current flowing through the copper wire has produced a magnetic effect.

Question 3.
How many magnetic poles does a magnet usually have? Name them.
Answer:
A magnet usually has two magnetic poles: a north pole and a south pole.

Question 4.
How do you identify the north pole and the south pole of a bar magnet or a compass needle?
Answer:
When a bar magnet is suspended freely, it aligns itself approximately in north-south direction. Similarly, the needle of a compass aligns itself in north-south direction. The end of the magnet that points in the southern direction is called the south pole and the end that points in the northern direction is called the north pole of the magnet.

Question 5.
State the rule regarding attraction and repulsion between magnetic poles.
Answer:
The rule regarding magnetic poles states that ‘like magnetic poles repel each other and unlike magnetic poles attract each other.’

Question 6.
Why does a compass needle set deflected when brought near a bar magnet?
Answer:
A compass needle is a small bar magnet. When it is brought near a bar magnet, its poles interact with those of the bar magnet. Like magnetic poles repel and unlike magnetic poles attract. Hence, a compass needle shows a deflection when brought near the bar magnet.

Question 7.
What is magnetic field?
Answer:
The region surrounding a magnet where the influence of the magnet can be experienced is called a magnetic field. It is the region where a magnetic force exists.

Question 8.
Why is magnetic field called a vector quantity?
Answer:
Magnetic field has both magnitude and direction and we need to mention both for its complete specification. This is why magnetic field is called a vector quantity.

Question 9.
What is a magnetic line of force?
Answer:
The path taken by an imaginary unit north pole in a magnetic field is called a magnetic line of force. .

Question 10.
What is the direction of a magnetic line of force?
Answer:
A magnetic line of force always moves away from North Pole and moves towards South Pole.

Question 11.
Draw magnetic field lines around a bar magnet.
Answer:

Question 12.
How do you obtain the pattern of magnetic lines of force around a bar magnet using iron filings?
Answer:
Fix a sheet of white paper on a drawing board using some adhesive. Place a bar magnet in the centre. Sprinkle some iron filings uniformly around the bar magnet. Now tap the board gently. The iron filings arrange themselves in a pattern shown in the figure along the magnetic field lines. This will give us an idea of the pattern of the magnetic field around a bar magnet.

Question 13.
List two methods of producing magnetic field.
Answer:
Three methods of producing magnetic field are as follows:

1. Magnetic field can be produced by placing a permanent magnet or a horse-shoe magnet at the place where magnetic field is required.
2. Magnetic field is produced around a current carrying straight conductor or a current carrying coil.
3. Magnetic field is also produced due to flow of current in a solenoid.

Question 14.
How can we obtain the pattern of magnetic lines of force around a bar magnet using a compass needle?
Answer:
Take a small compass and a bar magnet. Fix a drawing sheet on a table with the help of drawing pins. Place the bar magnet in the middle and mark the boundary of the magnet. Place the compass near the north pole of the magnet. The south pole of the needle points towards the north pole of the magnet and the north pole of the compass is directed away from the north pole of the magnet. Mark the position of the two ends of the needle.

Now move the needle to a new position such that its south pole occupies the position previously occupied by its north pole. In this way, proceed step by step till you reach the south pole of the magnet. Join the points marked on the paper by a smooth curve.

This curve represents a field line. Repeat the procedure above and draw as many lines as possible on either side of the magnet. The lines represent the magnetic field around the magnet. These are known as magnetic field lines.

Question 15.
List the properties of magnetic lines of force.

OR

Mention the properties of magnetic field lines.
Answer:
The following are the properties of magnetic lines of force:

1. The magnetic lines of force will always start from north pole and end at south pole.
2. The direction of the lines of force inside the magnet move from south pole towards north pole.
3. Magnetic lines of force never cross each other.
4. Magnetic lines of force are denser in the region where magnetic field is stronger.

Question 16.
Why don’t two magnetic lines of force intersect each other?
Answer:
No two magnetic field-lines are found to cross each other. If they did, it would mean that at the point of intersection, the compass needle would point towards two directions at the same time, which is impossible.

Question 17.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
Uniform magnetic field:

The field lines in a uniform magnetic field arc parallel straight lines. In such a field, an isolated unit north pole will experience the same force in the same direction at all points.

Question 18.
Draw a neat diagram showing the pattern of concentric circles indicating the field lines of a magnetic field around a straight conducting wire.
Answer:

Question 19.
Describe an experiment to obtain the pattern of the magnetic field around a straight conductor carrying current. Describe the pattern.
Answer:
Take a rectangular cardboard having a small hole in the middle. Insert a straight conductor AB through the hole. Connect the conductor in series with a battery, a key and a rheostat. Sprinkle iron filings uniformly on the cardboard.

Ensure that the conductor is at right angles to the plane of the cardboard. Now close the key and pass current through the conductor. Gently tap the cardboard a few times. Iron filings arrange themselves along the field lines. The circles are found to become larger as we move away from the conductor.

The magnetic lines of force around a straight conductor carrying current are in the form of concentric circles with the conductor as centre. The direction of the magnetic lines of force can be known by the Right Hand Thumb Rule.

Question 20.
State Right Hand Thumb Rule.

OR

Draw the diagram showing the magnetic field lines around a current-carrying conductor. Label the following:

• The direction of magnetic field lines.
• The direction of electric current.

Answer:
Right Hand Thumb Rule:

Imagine you are holding a current-carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then, the fingers will wrap around the conductor in the direction of the field lines of the magnetic field. This is known as Right Hand Thumb Rule.

Question 21.
A current through a horizontal power line flows in east to west direction. What is the direction of magnetic field at a point directly below it and at a point directly above it?
Answer:
The current is in the east-west direction. Applying the right-hand thumb rule, we get that the magnetic field (at any point below or above the wire) turns clockwise in a plane perpendicular to the wire, when viewed from the east end, and anti-clockwise, when viewed from the west end.

Question 22.
Describe an experiment to show the pattern of the magnetic field around a circular conductor carrying current.
Answer:
Take a rectangular cardboard having two small holes in it. Insert a circular coil of wire through the two holes. Connect the ends of the coil in series with a battery, a key and a rheostat. Sprinkle iron filings uniformly on the cardboard.

Ensure that the plane of the cardboard is at right angles to the plane of the coil. Pass current through the coil by closing the key. Gently tap the cardboard a few times. The iron filings arrange themselves along the field lines in a pattern shown in the figure.

The magnetic lines of force around a circular conductor carrying current are in the form of concentric circles near the conductor and are in the form of parallel straight lines at the centre of the circular coil. The direction of the lines of force can be known by the Right Hand Thumb Rule.

Question 23.
Consider a circular loop of wire Ivins in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:
In the diagram the current is flowing clockwise. If we are applying right hand thumb rule to the left side of the loop then the direction of magnetic field lines inside the loop are going into the table while outside the loop they are coming out of the table. If we are applying right hand thumb rule to the right side of the loop then the direction of magnetic field lines inside the loop are again going into the table while outside the loop they are coming out of the table.

Question 24.
Why do we say that the magnetic field at the centre of a circular coil carrying current is uniform?
Answer:
The magnetic field lines at the centre of a circular coil carrying current are in the form of parallel straight lines. This shows that the magnetic field formed there is uniform.

Question 25.
What are the factors on which the field around a circular conductor carrying current depends?
Answer:
The magnetic field around a circular conductor carrying current depends on:

1. The amount of current flowing through the wire.
2. Radius of the circular wire.
3. Number of turns in the circular wire.

Question 26.
How can the strength of the magnetic field produced by a current-carrying circular coil be increased?
Answer:
The strength of the magnetic field produced by a current-carrying circular coil can be increased by

1. Increasing the number of turns of wire in the coil.
2. Increasing the current flowing through the coil.
3. Decreasing the radius of the coil.

Question 27.
What is a solenoid? Draw a neat diagram showing the pattern of the magnetic field around a current carrying solenoid. Describe the pattern of the magnetic field produced around a current carrying solenoid.
Answer:
A coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder is called a solenoid. The pattern of the magnetic field lines around a current-carrying solenoid is similar to the magnetic field around a bar magnet.

In fact, one end of the solenoid behaves as the magnetic north pole, while the other behaves as the south pole. The field lines outside the solenoid move away from the end that behaves like north pole towards the end that behaves like south pole.

The field lines inside the solenoid are in the form of parallel straight lines. This indicates that the magnetic field is the same at all points inside the solenoid. That is, the field is uniform inside the solenoid. The field is stronger at the ends of the solenoid.

Question 28.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar masnet? Explain.
Answer:
A solenoid begins to behave like a magnet when electric current flows through it. We know that any current-carrying conductor creates a magnetic field around it and that is what happens in the case of solenoid.

In order to determine the polarities of its ends, place it on a brass hook and suspend it with a long thread so that it moves freely. Bring the north pole of a bar magnet near one of its ends. In case the solenoid moves towards the bar magnet, the end of the solenoid is its south pole and in case the solenoid moves away from the magnet, the end of the solenoid is its north pole. The polarity of the other end of the solenoid can similarly be determined.

Question 29.
Identify the region where the magnetic field around a current-carrying solenoid is uniform.
Answer:
The magnetic field is uniform inside the solenoid.

Question 30.
The magnetic field lines inside a solenoid carrying current are in the form of parallel straight lines. What does this indicate?
Answer:
The magnetic field lines being in the form of parallel straight lines inside a solenoid carrying current indicates that a uniform magnetic field is produced there.

Question 31.
What is an electromagnet?
Answer:
A strong magnetic field produced inside a solenoid can be used to magnetise a piece of magnetic material, like soft iron, when placed inside the coil. A soft metal core made into a magnet by the passage of electric current through a coil surrounding it is called an electromagnet.

Question 32.
How do you magnetize a steel rod and convert it into an electromagnet?
Answer:
Take a small rod made of soft iron. Put several turns of an insulated copper wire around the soft iron rod. Connect the ends of the wire to a battery and a key as shown in the figure. On closing the key, the soft iron core becomes a powerful magnet only as long as current flows through the coil. On stopping the flow of current, the magnetic properties of the core disappear. This is an electromagnet.

Question 33.
What happens when a current-carrying conductor is placed in a magnetic field?
Answer:
A current-carrying conductor experiences a mechanical force when placed in a magnetic field.

Question 34.
Mention the factors on which the mechanical force acting on a current-carrying conductor in a magnetic field depends.
Answer:
The mechanical force on a current-carrying conductor in a magnetic field depends on:

1. The strength of the magnetic field,
2. The strength of the current flowing through the conductor, and
3. The angle between the direction of current and the direction of the magnetic field.

Question 35.
How do you show that a mechanical force acts on a conductor carrying current in a magnetic field?
Answer:
Take a small aluminium rod AB about 5 cm long. Using two connecting wires suspend it horizontally from a stand, as shown in the figure. Place a strong horse-shoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. For this put the North Pole of the magnet vertically below and South Pole vertically above the aluminium rod.

Connect the aluminium rod in series with a battery, a key and a rheostat. Now pass current through the aluminium rod from end B to end A. It is observed that the rod is displaced towards the left. When the direction of current is reversed, the aluminium rod is now towards the right.

This suggests that a mechanical force is exerted on the current-carrying aluminium rod when it is placed in a magnetic field. It also suggests that the direction of the mechanical force is also reversed when the direction of current through the conductor is reversed.

Question 36.
In the activity on the mechanical force on a conductor carrying current in a magnetic field, how do you think the displacement of rod AB will be affected if

1. Current in rod AB is increased;
2. A stronger horse-shoe magnet is used: and
3. Length of the rod AB is increased?

Answer:

1. When the current flowing through rod AB is increased, the rod AB gets displaced more due to an increase in the magnitude of the mechanical force acting on it.
2. When a stronger horse-shoe magnet is used in the activity, the mechanical force on the rod AB will increase and hence it gets displaced to a higher extent.
3. When the length of the rod is increased, the displacement of the rod will also get increased.

Question 37.
What are the factors on which the direction of the mechanical force acting on a current carrying conductor placed in a magnetic field depends?
Answer:
The direction of the mechanical force acting on a current-carrying conductor placed in a magnetic field depends on the

• Direction of flow of current, and the
• Direction of the magnetic field.

Question 38.
When will the magnitude of the mechanical force acting on a current-carrying conductor placed in a magnetic field become

1. Highest and
2. Lowest?

Answer:
The magnitude of the mechanical force acting on a current-carrying conductor placed in a magnetic field:

1. Becomes highest when the direction of current is at right angles to the direction of the magnetic field.
2. Becomes lowest (zero) when the direction of current is parallel to the direction of the magnetic field.

Question 39.
When is the force experienced by a current-carrying conductor placed in a maenetic field largest?
Answer:
The force experienced by a current-carrying conductor in a magnetic field is largest when the direction of current is perpendicular to the direction of the magnetic field.

Question 40.
Name the law that helps to find out the direction of the mechanical force on a current-carrying conductor in a magnetic field.
Answer:
When the direction of current is at right angles to the direction of the magnetic field, the mechanical force acting on a current-carrying conductor in a magnetic field can be known by Fleming’s left hand rule.

Question 41.
State Fleming’s left-hand rule.
Answer:

According to Fleming’s left-hand rule, ‘If we stretch the thumb, forefinger and middle finger of the left hand such that they are perpendicular to each other, if the forefinger points in the direction of magnetic field and the middle finger shows the direction of current, then, the thumb will point in the direction of motion of conductor.

Question 42.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from the back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of the magnetic field?
Answer:
The direction of the magnetic field is vertically downward. The direction of current is from the front wall to the back wall because negatively charged electrons are moving from the back wall to the front wall. The direction of the magnetic force is rightward. Hence, using Fleming’s left hand rule, it can be concluded that the direction of the magnetic field inside the chamber is downward.

Question 43.
What is an electric motor? What is the principle of an electric motor?
Answer:
A device that converts electrical energy into mechanical energy is called an electric motor. An electric motor works on the principle that a conductor carrying current experiences a mechanical force when placed in a magnetic field.

Question 44.
Draw a neat, labelled diagram of an electric motor.

OR

Draw the diagram of an electric motor and label the following parts:

1. Split rings,
2. Armature.

Answer:


B₁ and B₂ – Carbon brushes
Ba – Battery

Question 45.
What is a DC motor?
Answer:
An electric motor that works on direct current (unidirectional current) is called a DC motor.

Question 46.
With the help of a neat diagram, explain the construction and working of a DC motor.

OR

Draw a labelled diagram of an electric motor. Explain its principle and working.
Answer:
1. Construction:
A simple DC motor consists of a rectangular coil of wire ABCD mounted on an axle between the opposite poles N and S of a strong horseshoe magnet. One of the ends of the coil is soldered to a half of a split ring, say R₁ while the other end is connected to the other half of the split ring R₂ as shown in the figure. Two conducting carbon brushes B₁ and B₂ are kept pressed on R₁ and R₂ respectively. B] and B2 are connected to a source of direct current. The coil together with the split ring is called armature.

2. Working:
When a current is passed through the armature coil, a mechanical force acts on the limb AB and CD of the coil. These forces are equal in magnitude and opposite in direction. These two forces constitute a couple which rotates the coil between the magnetic poles.

As per the diagram, a current flows from the positive of the battery to the brush B₁ and from B₁ to R₁. Therefore, current flows through the coil in the direction ABCD. The direction of the mechanical force acting on the limbs AB and CD can be known by Fleming’s left hand rule.

Obviously, the coil, during the first half of its cycle, rotates in anticlockwise direction. In the next half of its rotation, the brushes will be in contact with the other half of the split ring. This ensures the rotation of the coil in the same direction.

Question 47.
What is the function of a split ring in an electric motor?
Answer:
In an electric motor, after every half rotation, the direction of current in the coil gets reversed due to change in orientation of the magnetic field. To ensure a continuous rotation, a split ring is attached to the coil so that the polarity of the coil changes after every half rotation. The split ring thus changes the direction of current and ensures continuous rotation of the armature in one direction.

Question 48.
Identify the diagram given. What is the direction in which the armature rotates? Name the rule that helped you find out the answer to the question above.
Answer:
The diagram is that of a DC motor. In this, the armature rotates in clockwise direction. The rule that helps to answer this question is Fleming’s left hand rule.

Question 49.
How is the power of commercial electric motors enhanced?
Answer:
The power of commercial electric motors is enhanced by:

1. Using an electromagnet in place of permanent magnet,
2. Taking large number of turns of the conducting wire in the current-carrying coil, and
3. Winding the coil on a soft iron core.

Question 50.
Name some devices in which electric motors are used.
Answer:
Electric motor is used in all such devices where electric energy is required to be converted into ‘ mechanical energy. Such devices include electric fans, mixer grinders, coolers, air conditioners, washing machines, etc.

Question 51.
Who discovered that a changing magnetic field linked with a closed coil induces an electric current in the coil?
Answer:
Michael Faraday discovered that a changing magnetic field linked with a closed coil induces an electric current in the coil.

Question 52.
Name the phenomenon discovered by Michael Faraday to produce electric current from a changing magnetic field.
Answer:
Michael Faraday discovered a phenomenon called electromagnetic induction in which an electric current was produced in a coil using a changing magnetic field.

Question 53.
What is a galvanometer? How does it work?
Answer:
A galvanometer is an instrument that can detect the presence of current in a circuit. In a galvanometer, gradations are made on either side of zero (the centre of the scale). The pointer remains at zero for zero current flowing through it.

It can deflect either to the left or to the right of the zero mark depending on the direction of current. Thus, a galvanometer not only detects the presence of current but also indicates the direction of flow of current.

Question 54.
Describe Faraday’s magnet and coil experiment which led to the discovery of electromagnetic induction.

OR

Explain Faraday’s experiment to illustrate the phenomenon of electromagnetic induction.
Answer:
Michael Faraday’s magnet and coil experiment consisted of an insulated copper coil having a large number of turns. He connected the ends of the coil to the terminals of a galvanometer. He thrust a pole of a strong bar magnet into the coil quickly.

He observed a momentary deflection in the galvanometer. He pulled the magnet out of the coil, The galvanometer now showed deflection in the opposite direction. He repeated this experiment with the other pole. The same effect was observed in the opposite direction.

However, there was no deflection in the galvanometer when the magnet was held at rest inside the coil. He kept the magnet at rest and moved the coil itself around the magnet. Even now, the galvanometer showed a deflection. Through this experiment, Faraday demonstrated that a changing magnetic field linked with a coil can produce electricity in the coil.

Question 55.
Which was the generalization drawn from Faraday’s magnet and coil experiment on electromagnetic induction?
Answer:
Faraday’s magnet and coil experiment showed that ‘whenever there is a relative motion between a closed coil and a magnetic field linked with it, an emf is induced inside the coil’.

Question 56.
You are given a coil of insulated copper wire and a bar magnet. How do you induce an electric current in the coil?
Answer:
Connect the ends of the coil to a galvanometer. Now push a pole of the given magnet into the coil. The galvanometer shows a momentary deflection. On withdrawing the magnet from inside the coil, a deflection is seen in the galvanometer in the opposite direction. The deflection in the galvanometer shows that a current is induced in the coil.

Question 57.
A bar magnet is placed inside a closed coil Both the magnet and the coil are made to move in the same direction with the same speed. Does this motion induce an emf in the coil? Why?
Answer:
When a magnet and a coil move in the same direction with the same speed, the magnetic field linked with the coil does not change. Therefore, no emf is induced in the coil.

Question 58.
Can a stationary coil and stationary magnet produce induced current? Give reason.
Answer:
A stationary coil and a stationary magnet cannot produce induced current. For current to get induced in a coil, the magnetic field linked with the coil should change. In the given situation, the magnetic field linked with the coil does not change. This is why no current is induced in the coil.

Question 59.
A coil of insulated copper wire is connected to a galvanometer. What would happen if a bar magnet is

1. Pushed into the coil,
2. Withdrawn from inside the coil,
3. Held stationary inside the coil?

Answer:

1. Due to change in magnetic field in the coil, the galvanometer shows deflection.
2. Due to change in magnetic field in the coil, the galvanometer shows deflection in the opposite direction. It indicates the current in the coil is in opposite direction.
3. As the magnet is stationary, there is no change in magnetic field in the coil. Therefore, the galvanometer shows no deflection.

Question 60.
Explain different ways to induce current in a coil
Answer:
Current can be induced in a coil by

1. Moving the coil in a magnetic field or moving the coil rapidly between the two poles of a horse-shoe magnet.
2. Bringing a bar magnet close to the coil or taking away from it.
3. Rotating the coil in a uniform magnetic field.

Question 61.
What is electromagnetic induction?
Answer:
The process by which a changing magnetic field in a conductor induces a current in another conductor is called electromagnetic induction.

Question 62.
Name the device that works on the principle of electromagnetic induction.
Answer:
An electric generator works on the principle of electromagnetic induction.

Question 63.
What is meant by induced emf and induced current?
Answer:
The emf produced in a coil during electromagnetic induction is called induced emf. The current that is induced in the coil is called induced current.

Question 64.
Mention the factors on which the emf induced in a coil by electromagnetic induction depends.
Answer:
The current induced in a coil due to the changing magnetic field increases with the increase in:

1. The number of turns in the coil, and
2. The rate of change of magnetic field linked with the coil.

Question 65.
Electricity is induced in a coil by thrusting a bar magnet into the coil. What are the ways to increase the induced electric current in the coil?
Answer:
The current induced in the coil can be increased by

1. Increasing the number of turns in the coil,
2. Taking a magnet of higher strength, and
3. Increasing the speed of motion of magnet inside the coil.

Question 66.
How do you demonstrate that any change in current in a coil will induce a current in another coil placed close to it?

OR

Describe Faraday’s coil and coil experiment.
Answer:
Take two different coils (coil-1 and coil-2) of copper wire having turns 100 and 50 respectively. Insert them over a non-conducting cylinder as shown in the figure. Connect coil-1 in series with a battery, a rheostat and a plug key.

Also connect the other coil-2 to the terminals of a galvanometer. Close the key. Observe the galvanometer. The needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in coil-2.

Disconnect coil-1 from the battery. The needle momentarily moves to the opposite side. It means that now the current flows in the opposite direction in coil-2. When the current in coil-1 is steady, no deflection is observed in the galvanometer. Whenever current through coil-1 is changed, a potential difference is induced in coil-2.

Question 67.
In the figure, as the current changes in coil-1, the galvanometer connected to coil-2 shows deflection. Explain the phenomenon that causes this effect. Name and state the law used to know the direction of current in the device that works due to this phenomenon.
Answer:
The process by which a changing magnetic field in a conductor conductor is called electromagnetic induction. Whenever the electric current through coil-1 changes, the magnetic field associated with it also changes. Thus the magnetic field lines associated with coil-2 also change. Hence the change in magnetic field lines associated with coil-2 is the cause of induced electric current in it.

The law used to know the direction of current in the device that works on this phenomenon is called Fleming’s right hand rule.

Rule:
Fleming’s right-hand rule states that, if the fore finger, middle finger and the thumb of the right hand are stretched at right angles to each other, with the fore finger in the direction of magnetic field and the thumb in the direction of motion of the conductor, then the middle finger gives the direction of flow of induced current.

Question 68.
Why does current get induced in a coil when current flowing through another coil placed close to it is changed?
Answer:
As the current in a coil changes, the magnetic field associated with it also changes. Thus the magnetic field lines around the second coil placed close to the first also change. Hence the change in magnetic field lines associated with the secondary coil is the cause of induced electric current in it.

Question 69.
Two circular coils A and B are placed close to each other. If the current in coil A is changed, will some current be induced in coil B? Give reason.
Answer:
When the current in coil A is changed, some current will definitely be induced in coil B. When the current in coil A is changed, the magnetic field associated with it also changes. As a result, the magnetic field around coil B also changes. This change in magnetic field lines around coil B induces an electric current in it. This process is known as mutual (electromagnetic) induction.

Question 70.
What rule helps us to find the direction of induced current?

OR

For what purpose is Fleming’s right hand rule used?
Answer:
Direction of induced current in electromagnetic induction can be known by Fleming’s Right Hand Rule. This rule is also known as generator rule.

Question 71.
When can we apply Fleming’s Right Hand Rule?
Answer:
In electromagnetic induction, the induced current would be the highest when the direction of the magnetic field is perpendicular to the direction of motion of the coil. In this situation, Fleming’s Right Hand Rule can be applied to find out the direction of induced current.

Question 72.
State Fleming’s Right Hand Rule.
Answer:
Extend the thumb, the fore finger and the middle finger of the right hand perpendicular to one another. If the fore finger shows the direction of magnetic field and the thumb shows the direction of motion of the coil, then, the middle finger will show the direction of induced current.

Question 73.
How can you increase the induced current in a coil in the magnet and coil experiment on electromagnetic induction?
Answer:
Induced current in a closed coil can be increased when

1. The magnet moves at a faster speed in and out of the coil
2. A stronger magnet is used
3. The number of turns in the coil is increased.

Question 74.
State the purpose for which the following rules are used:

1. Right hand thumb rule.
2. Fleming’s left hand rule.
3. Fleming’s right hand rule.

Answer:

1. Right hand thumb rule is used to find the direction of the lines of magnetic force in a current-carrying wire.
2. Fleming’s left hand rule gives the direction of magnetic force acting on a conductor.
3. Fleming’s right hand rule shows the direction of induced current when a conductor moves in a magnetic field.

Question 75.
State the rule to determine the direction of

1. A magnetic field produced around a straight conductor carrying current,
2. A force experienced by a current-carrying straight conductor placed in a masnetic field which is perpendicular to it, and
3. A current induced in a coil due to its rotation in a magnetic field.

Answer:
1. Right hand thumb rule:
The rule states that, if you are holding a current-carrying straight conductor in your right hand such that the thumb points towards the direction of current, then, the fingers wrapped around the conductor will show the direction of the field lines of the magnetic field.

2. Fleming’s left hand rule:
‘If we stretch the thumb, fore finger and middle finger of the left hand at right angles to each other such that the fore finger points in the direction of magnetic field and the middle finger shows the direction of current, then, the thumb will point in the direction of the mechanical force acting on the conductor’.

3. Fleming’s right hand rule:
Extend the thumb, the fore finger and the middle finger of the right hand perpendicular to one another such that the fore finger shows the direction of magnetic field and the thumb shows the direction of motion of the coil. Then, the middle finger will show the direction of induced current.

Question 76.
What is alternating current? Represent alternating current graphically.
Answer:
An electric current whose direction keeps on changing regularly is called alternating current. It is also called oscillating current. An alternating current is represented graphically as shown below:

Question 77.
Which sources produce alternatim current?
Answer:
Dynamo or generators with slip rings will produce alternating current.

Question 78.
What is direct current? Represent it graphically.
Answer:
Electric current that flows through a conductor in only one direction is called direct current. The following is the graphical representation of direct current:

Question 79.
Name some sources of direct current.
Answer:
Battery, cells and generators that use split rings are some of the sources of direct current.

Question 80.
What is an electric generator? State the principle of an electric generator.
Answer:
A device that converts mechanical energy into electrical energy is called an electric generator. A generator works on the principle of electromagnetic induction. It states that ‘whenever a magnetic field linked with a closed coil changes, an emf is induced inside the coil’.

Question 81.
Which are the two types of generators?
Answer:
The two well-known types of generators are

1. AC generator and
2. DC generator.

Question 82.
Distinguish between alternating current (AC) and direct current (DC).
Answer:
Current which changes direction after equal intervals of time is called alternating current. Current which does not change its direction with time is called direct current.

Question 83.
What is AC generator?
Answer:
A generator that produces alternating current is called an AC generator.

Question 84.
Draw a neat, labelled diagram of an AC generator.
Answer:

• ABCD – Rectangular coil of wire
• N and S – Poles of a horse shoe magnet
• R₁ and R₂ – Thick metal rings
• B₁ and B₂ – Conducting brushes
• G – Galvanometer

Question 85.
List the major components of a simple AC generator.
Answer:
A simple AC generator consists of a powerful horseshoe magnet, a rectangular coil of wire, two thick metal rings, two conducting brushes and a load.

Question 86.
What is the rotating part of a generator called? What does it consist of?
Answer:
The rotating part of a generator is called armature. It consists of a rectangular coil of wire and two thick metal rings. Both of these are mounted on an axle between the opposite poles of the horseshoe magnet.

Question 87.
Explain with the help of a neat, labelled diagram, the construction and working of a simple AC generator.

OR

Explain the underlying principle and working of an electric generator by drawing a labelled diagram.
Answer:

1. Construction:
A simple AC generator consists of a rectangular coil of wire ABCD mounted on an axle between the opposite poles N and S of a strong horseshoe magnet. One of the ends of the coil is soldered to a thick metal ring, say R₁ while the other end is connected to another similar ring R₂ as shown in the figure.

These rings are called slip rings. Two conducting carbon brushes B₁ and B₂ are kept pressed on R₁ and R₂ respectively. B₁ and B₂ are connected to an external circuit. The coil together with the slip rings is called armature.

2. Working:
A generator works on the principle of electromagnetic induction. The armature is rotated between the poles of the magnet. When the armature rotates, the magnetic field linked with the coil changes. This induces an electric current in the coil.

However, the current induced is different at different locations of the coil. During the first half rotation, a current is induced along DCBA if the rotation of the armature is anticlockwise. During this half cycle, current in the external circuit flows from B₂ to B₁.

During the second half rotation, current is induced along ABCD. Therefore, current in the external circuit flows from B₁ to B₂. Thus, the current in the external circuit keeps on changing its direction for every half rotation of the armature. Such a current is known as alternating current (AC) and the generator is known as AC generator.

Question 88.
What is the function of brushes in generators and motors?
Answer:
The brushes in generators and motors conduct electric current between the stationary wires and the rotating parts of a motor or generator. In the case of generators, the brushes maintain contact between the slip rings (or split rings) and pass on the current to the external circuit. In the case of motors, the brushes pass the current from the source to the slip rings (or split rings).

Question 89.
What is meant by frequency of AC?
Answer:
The number of complete rotations per second made by the armature of an AC generator is called the frequency of rotation of the armature. This is also the frequency of AC produced by it.

Question 90.
Mention the frequency of the electric current produced in India.
Answer:
The frequency of electric current produced in India is 50 hertz (Hz).

Question 91.
The armature of an AC generator rotates 100 times in 2 second. What is the frequency of rotation of the armature? What is the frequency of AC produced by this generator? How many times does the direction of AC change per second? Why?
Answer:
The frequency of rotation is the number of complete rotations made by the armature in one second. Therefore, the frequency of rotation of the armature is 50 Hz. The frequency of rotation of the armature and the frequency of AC produced are same. Thus, the frequency of AC produced by the generator is also 50 Hz. The current changes its direction 100 times per second. This is because the direction of current changes twice for every rotation of the armature.

Question 92.
In a generator, when is the value of induced current lowest? When is the value of induced current highest? Give reason.
Answer:
In a generator, the current induced is lowest when the coil is parallel to the direction of the magnetic field. The value of induced current is highest when the coil is perpendicular to the direction of the magnetic field.

When the plane of the coil is parallel to the direction of the magnetic field, the field lines do not intercept the coil. Therefore, the current induced is zero. When the plane of the coil is perpendicular to the direction of the magnetic field, the highest number of field lines intercept the coil. Therefore, the current induced is highest.

Question 93.
Draw a neat schematic diagram of a simple domestic electric circuit and name the parts.
Answer:

Question 94.
Describe briefly domestic electric circuit.
Answer:
In our homes electricity is brought from the electric power station by two insulated cables made of aluminium or copper. Out of these two wires, one wire is in red insulation and is called ‘live wire’. The other wire is in black insulation and is called ‘neutral wire’.

The electricity coming into our house is actually alternating current of voltage 220 V. Both the live wire and the neutral wire enter in a box where a main fuse is put on the live wire. These two wires then pass through the electricity meter.

After the electric meter, these wires enter the distribution board through the main switch. The main switch is used to switch off the electricity supply whenever required. After the main switch, the live wire is connected to live wires of two separate circuits present in the house to supply electricity.

Out of these two circuits, one circuit of 5A rating, is used to run electric appliances with low power such as tube lights, bulbs and fans. The other circuit is of 15 A rating which is used to run electric appliances with higher power such as air conditioners, refrigerators, electric iron and heaters.

It should be noted that all the circuits in the house are connected in parallel so that switching off of one circuit does not affect the other circuit. One more advantage of parallel connection of circuits is that each electric appliance gets equal voltage.

A third wire called ‘earth wire’, having green insulation, is usually connected to the body of metallic electric appliances. The earth wire sends the current from the body of the appliance to the earth whenever a live wire accidentally touches the body of metallic electric appliances.

Thus, the earth wire gives protection from severe electric shocks. Each electric appliance has a separate switch to switch on or switch off the flow of current to it.

Question 95.
Name two safety measures commonly used in electric circuits and appliances.
Answer:
Two of the safety measures commonly used in electric circuits and appliances are:

• Earthing the metallic body of electrical appliances to protect from electric shocks.
• Use of electric fuse to protect circuit and appliances from the consequences of overloading.

Question 96.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
The earth wire is connected as a safety measure for all electrical appliances that have metallic body. It protects us from electric shocks due to accidental leakage of charges. The earth wire provides a low resistance-conducting path for electric current and passes the current to the earth.

Question 97.
What is overloading of a circuit? Name the safety device used to protect circuits and gadgets from the dangers of overloading.
Answer:
All circuits are designed to carry a certain maximum value of current safely. The amount of current that can be safely drawn in a circuit depends on the power rating of the wire used there. When the amount of current passing through the circuit exceeds the maximum permissible limit, the wires get heated to such an extent that fire may be caused.

This is known as overloading. Electric circuits and gadgets are protected from the danger of overloading by using a device called electric fuse.

Question 98.
When does an electric short-circuit occur? Explain.
Answer:
An electric short-circuit occurs when the live wire and the neutral wire come in direct contact with each other. When this happens, the current flowing in the circuit rises alarmingly and short-circuiting occurs.

Short-circuiting happens when the insulation of wires is damaged and the live wire comes into contact with the neutral wire. When this happens, the resistance becomes too small leading to flow of large amount of current through the wires. Such a large current heats up the wires to such an extent that fire may be caused in the building or the gadgets used in the circuit may get damaged.

Question 99.
What precautions should be taken to avoid overloading of domestic electric circuits?
Answer:
The following precautions may be taken to avoid overloading:

• Two separate circuits should be used in domestic wiring, one for lighting (5 A) and the other for heating (15 A).
• Fuse should be installed in both lighting and heating circuits.
• Parallel circuits should be used to operate gadgets.
• Current should not be drawn by plugging too many gadgets to a single plug point.

Question 100.
An electric oven of 2 kW power ratine is operated in a domestic electric circuit (220 V) that has a current ratine of 5 A, What result do you expect? Explain.
Answer:
Given: Power P = 2 kW = 2000 W; Voltage = 220 V; Current rating of wires = 5 A.
Power, P = VI.
Current drawn by oven, I = \(\frac{\text { Power }}{\text { Voltage }}=\frac{2000 \mathrm{W}}{220 \mathrm{V}}\) = 9.09 A.

Since the current drawn by the oven (9.09 A) is higher than the safe limit (5 A), there would be overloading. When the oven is operated, the fuse would blow off and the circuit would be broken. If no fuse is used in the circuit, there may be fire in the circuit.

Question 101.
How does overload and short-circuit occur in it an electric current? Explain. What is the function of fuse during this situation?
Answer:
All electric circuits are designed to carry a certain maximum value of current safely. Over-loading occurs when the current value exceeds the safe limit of the circuit. Over-loading can occur due to short- circuiting or faulty electric appliance.

Short-circuit occurs when the live wire and the neutral wire come into direct contact instead of passing through a load leading to a huge upsurge of current. This is also an instance of overloading of the circuit. A fuse protects the circuit and its appliances in the event of overloading as well as short circuit.

Question 102.
State whether the following statements are true or false:

1. An electric motor converts mechanical energy into electrical energy..
2. An electric generator works on the principle of electromagnetic induction.
3. The field at the centre of a lone circular coil carrying current will be parallel straight lines.
4. A wire with a green insulation is usually the live wire of an electric supply.

Answer:

1. False
2. True
3. True
4. False.

Fill In The Blanks

1. The phenomenon of electromagnetic induction was discovered by Faraday
2. A coil of many turns wound in the form of a cylinder is called solenoid
3. The direction of magnetic field around a current-carrying conductor can be known by Right hand thumb rule
4. The maximum load of current in heating circuits in our homes is 15 A
5. A device that converts electrical energy into mechanical energy is called electric motor
6. A generator works on the principle of electromagnetic induction
7. The insulation colour of the earth wire in domestic circuits is green
8. A device that detects electric current in a circuit is galvanometer

Multiple Choice Questions

Question 1.
The magnetic field inside a long straight solenoid carrying current
(A) is zero.
(B) decreases as we move towards its end.
(C) increases as we move towards its end.
(D) is same at all points.
Answer:
(D) is same at all points.

Question 2.
An electron enters a magnetic field at right angles to it as shown in the figure. The direction of force acting on the electron will be

(A) to the right.
(B) to the left.
(C) out of the page.
(D) into the page.
Answer:
(D) into the page.

Question 3.
Which of the following properties of a proton can change while it moves freely in a magnetic field? (There may be more than one correct answer.)
(A) mass
(B) speed
(C) velocity
(D) momentum
Answer:
(D) momentum

Question 4.
A positively-charged particle (alpha particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is
(A) towards south
(B) towards east
(C) downward
(D) upward
Answer:
(C) downward

Question 5.
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each
(A) one-fourth revolution
(B) half revolution
(C) one revolution
(D) two revolutions
Answer:
(B) half revolution

Question 6.
Which of the following correctly describes the magnetic field near a long straight wire?
(A) The field consists of straight lines perpendicular to the wire.
(B) The field consists of straight lines parallel to the wire.
(C) The field consists of radial lines originating from the wire.
(D) The field consists of concentric circles centred on the wire.
Answer:
(D) The field consists of concentric circles centred on the wire.

Question 7.
The phenomenon of electromagnetic induction is
(A) the process of charging a body.
(B) the process of generating magnetic field due to a current passing through a coil.
(C) the process of producing induced current in a coil due to relative motion between magnet and the coil.
(D) the process of rotating the coil of an electric motor.
Answer:
(C) the process of producing induced current in a coil due to relative motion between magnet and the coil.

Question 8.
The most important safety method used for protecting home appliances from short-circuiting or overloading is
(A) earthing
(B) use of fuse
(C) use of stabilizers
(D) use of electric meter
Answer:
(B) use of fuse

Question 9.
The device used for producing electric current is called a
(A) generator.
(B) galvanometer.
(C) ammeter.
(D) motor.
Answer:
(A) generator.

Question 10.
The essential difference between an AC generator and a DC generator is that
(A) AC generator has an electromagnet while a DC generator has permanent magnet.
(B) DC generator will generate a higher voltage.
(C) AC generator will generate a higher voltage.
(D) AC generator has slip rings while the DC generator has a commutator.
Answer:
(D) AC generator has slip rings while the DC generator has a commutator.

Question 11.
At the time of short-circuit, the current in the circuit
(A) reduces substantially.
(B) does not change.
(C) increases heavily.
(D) varies continuously.
Answer:
(C) increases heavily.

Question 12.
Commercial electric motors do not use
(A) a soft iron core on which the coil is wound.
(B) a permanent magnet to rotate the armature.
(C) an electromagnet to rotate the armature.
(D) effectively large number of turns of conducting wire in the current-carrying coil.
Answer:
(B) a permanent magnet to rotate the armature.

Question 13.
Choose the incorrect statement from among the following:
(A) Fleming’s right-hand rule is a simple rule to know the direction of induced current.
(B) The right-hand thumb rule is used to find the direction of magnetic field due to current carrying conductor.
(C) The difference between direct and alternating currents is that direct current always flows in one direction, whereas alternating current reverses its direction periodically.
(D) In India, the AC changes direction after every 1/50 second.
Answer:
(D) In India, the AC changes direction after every 1/50 second.

Question 14.
Choose the incorrect statement from the following regarding magnetic field lines:
(A) If magnetic field lines are parallel and equidistant, they represent zero field strength.
(B) Magnetic field lines are closed curves.
(C) Relative strength of magnetic field is shown by the degree of closeness of the field lines.
(D) The direction of magnetic field at a point is taken to be the direction in which the north pole of a magnetic compass needle points.
Answer:
(A) If magnetic field lines are parallel and equidistant, they represent zero field strength.

Question 15.
Which of the following has magnetic field like that of the one around a bar magnet?
(A) Current-carrying wire
(B) Current-carrying ring
(C) Current-carrying solenoid
(D) Current-carrying rectangular loop
Answer:
(C) Current-carrying solenoid

Question 16.
Who gave the principle of electromagnetic induction?
(A) Faraday
(B) Oersted
(C) Volta
(D) Ampere
Answer:
(A) Faraday

Question 17.
Which is the direction of mechanical force acting on a current-carrying wire placed in a magnetic field?
(A) Along magnetic field and electric field.
(B) Along the electric current but opposite the magnetic field.
(C) Perpendicular to magnetic field and the direction of current.
(D) Opposite to magnetic field.
Answer:
(C) Perpendicular to magnetic field and the direction of current.

Question 18.
To convert an AC generator into a DC generator
(A) split-ring type commutator must be used
(B) slip rings and brushes must be used
(C) a stronger magnetic field has to be used
(D) a rectangular wire loop has to be used
Answer:
(A) split-ring type commutator must be used

Question 19.
The upper limit of current that can safely flow through the lighting and heating circuit in our houses is respectively
(A) 5 A, 10 A
(B) 5 A, 15 A
(C) 10 A, 15 A
(D) 15 A, 5 A
Answer:
(B) 5 A, 15 A

Question 20.
A commutator changes the direction of current in the coil of
(A) a DC motor only
(B) a DC motor and an AC generator
(C) a DC motor and a DC generator
(D) an AC generator only.
Answer:
(C) a DC motor and a DC generator

Match The Following

Question 1.

Column A	Column B
1. Right hand thumb rule	a. Detects electric current in a circuit
2. Electric fuse	b. Direction of induced current
3. Fleming’s right hand rule	c. Direction of mechanical force
4. Fleming’s left hand rule	d. A cylindrical coil having many turns
5. Solenoid	e. Protects circuits from overloading
	f. Direction of magnetic field around a current-carrying conductor

Answer:
1 – f, 2 – e, 3 – b, 4 – c, 5 – d.

Students can download Class 10 Science Chapter 1 Chemical Reactions and Equations Important Questions, KSEEB SSLC Class 10 Science Important Questions and Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka SSLC Class 10 Science Important Questions Chapter 1 Chemical Reactions and Equations

Question 1.
What happens in each of the following situations when:

1. Milk is left at room temperature during summers.
2. An iron tawa/pan/nail is left exposed to humid atmosphere.
3. Grapes get fermented.
4. Food is cooked.
5. Food gets digested in our body.
6. We respire?

Answer:
1. When milk is left at room temperature during summer season, the milk undergoes chemical change. The sugar present in milk is converted into an acid. This chemical change spoils the milk.

2. When an iron material such as a nail is left exposed to humid atmosphere, the iron undergoes oxidation (a chemical change) and begins to rust. A new substance called iron oxide is formed.

3. When grapes get fermented, the sugar present in grapes undergoes chemical change and a new substance such as alcohol is formed.

4. When food is cooked, the components present in food undergo chemical change resulting in the formation of new substances.

5. When food gets digested in our body, the various constituents of food will undergo chemical change and are broken down into simpler substances. For example, carbohydrates present in the food we consume changes into glucose during the process of digestion.

6. When we respire, the food present in the body cells will undergo chemical change. This process produces energy and carbon dioxide.

All the reactions above involve chemical changes and in each of the changes, one or more new substances are formed.

Question 2.
How do you show the formation of magnesium oxide from magnesium?
Answer:
Clean a magnesium ribbon about 2 cm long by rubbing it with sandpaper. Hold it with a pair of tongs. Heat it using a spirit lamp. Magnesium ribbon burns with a dazzling white flame and changes into a white powder.

Collect this powder in a watch-glass. The ash-like powder so formed is magnesium oxide. Burning of magnesium is a chemical reaction. During this process, magnesium combined with oxygen to form magnesium oxide.

Question 3.
Why should a magnesium ribbon be cleaned before burning in air while preparing magnesium oxide?
Answer:
When magnesium ribbon is exposed to moist air, a layer of magnesium oxide is formed on its surface. This hinders the process of burning. This is why magnesium ribbon must be cleaned with sand paper before heating it while preparing magnesium oxide.

Question 4.
Explain the reaction between lead nitrate solution and potassium iodide.
Answer:
Take lead nitrate solution in a test tube. Add potassium iodide into it and shake well. We observe the formation of a new yellow solid substance called lead iodide. During this process, another new substance called potassium nitrate is also formed.

This compound remains in solution. The reaction between lead nitrate and potassium iodide is a chemical change that results in the formation of new substances.

Question 5.
Explain the reaction between zinc and dilute sulphuric acid.
Answer:
Take a few pieces of zinc granules in a small conical flask. Pour dilute sulphuric acid into the flask. The reaction begins immediately forming zinc sulphate and hydrogen gas. This is an example of a chemical change because new substances are formed during this reaction.

Question 6.
What is a chemical change? Give two examples of chemical changes in daily life.
Answer:
Any change in a substance or substances that results in the formation of one or more new substances is called a chemical change. Burning of fuels, respiration, rusting of iron etc., are examples of chemical change.

Question 7.
Describe the reaction between calcium oxide and water.
Answer:
Take a small amount of calcium oxide (quick lime) in a beaker. Slowly add water to it. A vigorous reaction takes place and the beaker becomes wanner. This is a chemical reaction during which calcium hydroxide is formed along with the liberation of heat.

Thus we may say that water reacts with calcium oxide forming calcium hydroxide along with the liberation of heat.

Question 8.
What changes can be generally observed during a chemical reaction?
Answer:
One or more of the following changes are generally observed during a chemical reaction:

1. change in state
2. change in colour
3. evolution of a gas
4. change in temperature.

Question 9.
What are reactants and products of a chemical reaction? Explain with an example.
Answer:
The substance or the substances that take part in a chemical reaction are called reactants. The substance or the substances formed during a chemical reaction are called products.

For example, lead nitrate reacts with potassium iodide forming lead iodide and potassium nitrate. In this reaction, the starting materials namely lead nitrate and potassium iodide are the reactants. The substances that are formed during the reaction namely lead iodide and potassium nitrate are the products.

Question 10.
What is a chemical equation?
Answer:
The representation of a chemical reaction in the form of symbols and formulae of the substances involved in it, wherein the reactants are written on the left-hand side and the products on the right-hand side is called a chemical equation.

Question 11.
What is the simplest way of representing a chemical reaction?
Answer:
A chemical reaction is represented in the simplest manner using a word equation.

Question 12.
What is a word equation in Chemistry? Explain with an example.
Answer:
A chemical reaction that is represented by writing the complete names of the reactants and products rather than their symbols or formulae is known as a word equation. It is simply a chemical equation written in words.

In a word equation, the names of the reactants are written on the left hand side and the names of the products are written on the right hand side. The reactants and products are separated by an arrow directed to the right (towards the products).

In case the reaction has more than one reactant, + sign is put between the reactants. This sign on the reactants’ side indicates ‘reacts with’ or ‘combines with’. Similarly, + sign is put between the products when there is more than one product. Here the + sign indicates ‘and’. The arrow sign indicates ‘produces’ or ‘yields’.

Consider the reaction between zinc and dilute hydrochloric acid. During this reaction, zinc chloride and hydrogen gas are formed. This can be represented by a word equation as follows:

Question 13.
What does the arrow in a chemical equation indicate? What does a chemical equation represent?
Answer:
The arrow in a chemical reaction indicates the direction of the reaction. A chemical equation represents a chemical reaction.

Question 14.
Distinguish between a skeletal equation and a balanced equation.
Answer:
1. Skeletal equation:
A chemical equation in which the number of atoms of different elements is not equal on the two sides of the equation is called an unbalanced chemical equation. It is also called skeletal equation.

2. Balanced equation:
A chemical equation in which the number of atoms of different elements is equal on the two sides of the equation is called a balanced chemical equation.

Question 15.
What is a balanced chemical equation? Why should chemical equations be balanced?
Answer:
A balanced chemical equation is one in which the total number of atoms of each element are equal on both sides of the equation. A chemical equation is balanced so that the numbers of atoms of each type involved in a chemical reaction are the same on the reactant and product sides of the equation.

A balanced chemical equation satisfies the law of conservation of mass, i.e., the total mass of the reactants is equal to the total mass of the products.

Question 16.
Write the steps involved in writing a balanced chemical equation.
Answer:
Step 1:
Know the reactants and products of the given chemical reaction and write down the word equation for the reaction. Note that the reactants must be written on the left hand side while the products must be written on the right hand side.

Step 2:
The second step is to write the skeletal equation. Write down the symbol and formula of each of the reactants and products.

Step 3:
Enclose the formula of each reactant and product in separate boxes and do not change the formula or the subscripts of anything inside the boxes while balancing the equation.

Step 4:
List the number of atoms of different elements on LHS (reactants) and RHS (products) separately.

Step 5:
Start balancing with the compound (either on the reactants side or on the products side) that contains the maximum number of atoms. After choosing the compound, select the element that has the maximum number of atoms. To balance the atoms, put a small whole number coefficient before the formula of the compound.

Step 6:
Balance all the elements one by one similarly. Check for the correctness of the balanced equation by counting and comparing the number of atoms of each element on both sides of the equation.

Step 7:
Write the physical state of each of the reactants and products.

Question 17.
Iron reacts with water (steam) forming iron oxide and hydrogen gas. Write a balanced chemical equation for this reaction.
Answer:
Step 1:
Let us write the word equation:
Iron + Water → Iron oxide + Hydrogen

Step 2:
Let us write the skeletal equation for the reaction using symbols and formulae:
Fe + H₂O → Fe₃O₄ + H₂

Step 3:
Let us put all the formulae/symbols in separate boxes.

Step 4:
Let us list the number of atoms of each element on both sides of the equation:

Step 5:
We shall start with the compound having the highest number of atoms namely iron oxide. In this compound, let us start with oxygen as the number of oxygen atoms is highest in the compound.

There are four oxygen atoms in iron oxide on the RHS. However, there is only one oxygen atom on the LHS. To balance this, we shall put 4 as the coefficient for water (H₂O).

Now let us balance the iron atoms. This can be done by putting 3 as the coefficient for Fe on the LHS. Now let us balance the hydrogen atoms. This is done by putting 4 as the coefficient for H₂ on the RHS. Now the equation becomes 3 Fe + 4 H₂O → Fe₃O₄ + 4 H₂

Step 6:
Let us now check for the correctness of the balance by comparing the number of atoms on either side of the equation.

Step 7:
Since each type of atom is balanced on both sides, the equation is balanced. Let us now mention the physical state of each of the reactants and products.

The gaseous, liquid, aqueous and solid states of reactants and products are represented by the notations (g), (l), (aq) and (s), respectively. The word aqueous (aq) is written if the reactant or product is present as a solution in water.
3 Fe (s) + 4 H₂O (g) → Fe₃O₄ (s) + 4 H₂ (g)
Note that the symbol (g) is used with H₂O to indicate that in this reaction water is used in the form of steam.

Question 18.
Write the balanced equation for the following chemical reactions:

1. Hydrogen + Chlorine → Hydrogen chloride
2. Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride
3. Sodium + Water → Sodium hydroxide + Hydrogen.

Answer:

1. H₂ (g) + Cl₂ (g) → 2 HCl (g)
2. 3 BaCl₂ (aq) + Al₂(SO₄)₃ (aq) → 3 BaSO₄ (s) + 2 AlCl₃ (aq)
3. 2 Na (s) + 2 H₂O (l) → 2 NaOH (aq) + H₂ (g).

Question 19.
Write balanced chemical equations with symbols for the following reactions:

1. Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.
2. Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.

Answer:
1. Barium chloride + Sodium sulphate → Barium sulphate + Sodium chloride
BaCl₂ (aq) + Na₂SO₄ (aq) → BaSO₄ (s) + 2 NaCl (aq)

2. Sodium hydroxide + Hydrochloric acid → Sodium chloride + Water
NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)

Question 20.
Translate the following statements into chemical equations and then balance them:

1. Hydrogen gas combines with nitrogen to form ammonia.
2. Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
3. Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
4. Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.

Answer:
1. Hydrogen + Nitrogen → Ammonia
3 H₂ (g) + N₂ (g) → 2 NH₃ (g)

2. Hydrogen sulphide + Oxygen → Water + Sulphur dioxide
2 H₂S (g) + 3 O₂ (g) → 2 H₂O (l) + 2 SO₂ (g)

3. Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride
3 BaCl₂(s) + Al₂(SO₄)₃ (s) → 3 BaSO₄ (s) + 2 AlCl₃ (s)

4. Potassium + Water → Potassium hydroxide + Hydrogen
2 K (s) + 2 H₂O (l) → 2 KOH (aq) + H₂ (g)

Question 21.
Balance the following chemical equations:

1. HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + H₂O
2. NaOH + H₂SO₄ → Na₂SO₄ + H₂O
3. NaCl + AgNO₃ → AgCl + NaNO₃
4. BaCl₂ + H₂SO₄ → BaSO₄ + HCl.

Answer:

1. 2 HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2 H₂O
2. 2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O
3. NaCl + AgNO₃ → AgCl + NaNO₃
4. BaCl₂ + H₂SO₄ → BaSO₄ + 2 HCl.

Question 22.
Write the balanced chemical equations for the following reactions:

1. Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
2. Zinc + Silver nitrate → Zinc nitrate + Silver
3. Aluminium + Copper chloride → Aluminium chloride + Copper
4. Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride.

Answer:

1. Ca(OH)₂ (aq) + CO₂ (g) → CaCO₃ (s) + H₂O (l)
2. Zn (s) + 2 AgNO₃ (aq) → Zn(NO₃)₂ (aq) + 2 Ag (s)
3. 2Al (s) + 3 CuCl₂ (aq) → 2 AlCl₃ (aq) + 3 Cu (s)
4. BaCl₂ (aq) + K₂SO₄ (aq) → BaSO₄ (aq) + 2 KCl (aq)

Question 23.
A solution of a substance ‘X’ is used for white washing.

1. Name the substance ‘X’ and write its formula.
2. Write the reaction of the substance ‘X’ named in (i) above with water.

Answer:

1. The substance ‘X’ is known as calcium oxide (quick lime). Its formula is CaO.
2. Calcium oxide reacts with water forming calcium hydroxide.

Calcium oxide + Water → Calcium hydroxide
CaO (s) + H₂O(l) → Ca(OH)₂ (aq)

Question 24.
List the various types of chemical reactions.
Answer:
The different types of chemical reactions are:

1. Chemical combination
2. Chemical decomposition
3. Chemical displacement
4. Double displacement.

Question 25.
What are exothermic and endothermic chemical reactions? Give an example each.
Answer:
1. Exothermic reactions:
Those chemical reactions during which heat is released are known as exothermic chemical reactions.
E.g.: Burning of fuels and respiration are examples of exothermic reactions.

2. Endothermic reactions:
Those chemical reactions during which heat is absorbed are known as endothermic reactions.
E.g.: Reaction between nitrogen and oxygen that results in nitric oxide is an example of an endothermic reaction.

Question 26.
What is meant by chemical combination or combination reaction? Give an example.
Answer:
A chemical reaction in which two or more substances (reactants) chemically combine to form a product is called chemical combination. It is also called combination reaction.
E.g.: Magnesium on heating bums in air forming magnesium oxide. This is a combination reaction.
2Mg + O₂ → 2MgO.

Question 27.
Explain the reaction that occurs when calcium oxide is made to react with water. Write the balanced chemical equation for the reaction. What type of chemical reaction is this? Why is it called an exothermic reaction?
Answer:
Take a small amount of calcium oxide (quick lime) in a beaker. Slowly add water to this. Touch the beaker. The beaker will have become warmer. During this reaction, calcium oxide reacts with water forming calcium hydroxide. This reaction also releases heat. This reaction can be represented by the following equation:
CaO (s) + H₂O(l) (Quick lime) → Ca(OH)₂ (aq) (Slaked lime) + Heat.
In this reaction two substances are combining chemically forming a single compound. This is an example of chemical combination. The reaction above produces heat. Therefore, it is an exothermic reaction.

Question 28.
Distinguish between quick lime and slaked lime.
Answer:
Quick lime is the common name of a compound called calcium oxide. Its molecular formula is CaO. Slaked lime is the common name of a chemical compound called calcium hydroxide. Its molecular formula is Ca(OH)₂.

Question 29.
Why do walls appear shiny two to three days after they are white washed with calcium hydroxide? Write the chemical equation for the reaction.
Answer:
Walls are white washed with calcium hydroxide. Calcium hydroxide slowly reacts with carbon dioxide present in air forming calcium carbonate. Calcium carbonate gives shiny finish to the walls. The reaction that occurs here can be represented by the following equation:
Ca(OH)₂ (aq) + CO₂ (g) → CaCO₃ (s) + H₂O (l).

Question 30.
Represent the reaction involving burning of coal by a balanced chemical equation. What type of chemical reaction is this? Why? Why is this reaction called an exothermic reaction?
Answer:
Coal, on heating, combines with oxygen present in air, forming carbon dioxide. This reaction also releases plenty of heat.
C (s) + O₂ (g) → CO₂ (g) + Heat
In this reaction two chemical substances combine to form a single compound. Therefore, this is an instance of chemical combination reaction. This reaction releases heat. Therefore, it is an exothermic reaction.

Question 31.
Why is respiration considered an exothermic reaction? Explain.
Answer:
During respiration, glucose reacts with oxygen forming carbon dioxide and water.
C₆H₁₂O₆ (aq) + 6O₂ (aq) → 6CO₂ (aq) + 6H₂O (l) + energy
This reaction also releases heat. This is why respiration is considered an exothermic reaction.

Question 32.
Hydrogen combines with oxygen to form water. What type of reaction is this? Why? Write a balanced chemical equation to represent this reaction.
Answer:
In the reaction above, two substances namely hydrogen and oxygen chemically combine to form a single compound namely water. Therefore, this is a case of chemical combination. The reaction can be represented by the following equation:
2H₂ (g) + O₂ (g) → 2H₂O (l)

Question 33.
What is meant by decomposition reaction? Explain with an example.
Answer:
A chemical reaction in which a single compound breaks up into two or more products is called a decomposition reaction. It is also called chemical decomposition.

For example, ferrous sulphate decomposes on heating forming anhydrous ferrous sulphate and water. Anhydrous ferrous sulphate on further heating decomposes to form ferric oxide, sulphur dioxide and sulphur trioxide. In this reaction, there are two decomposition reactions. The first one is:
FeSO₄.7H₂O (Ferrous sulphate crystals) (s) → FeSO₄  (Anhydrous ferrous sulphate) (s) + 7H₂O
Anhydrous ferrous sulphate on further heating decomposes to give the following reaction:

Question 34.
Describe an experiment to show the decomposition of ferrous sulphate crystals. What are the products forrtied during the reaction?
Answer:
Take about 2 g of ferrous sulphate crystals in a dry boiling tube. Note the green colour of the ferrous sulphate crystals. Heat the boiling tube over a flame using a spirit lamp. Observe the changes.

Now the green ferrous sulphate changes its colour on losing water present in its crystal and forms anhydrous ferrous sulphate. On further heating, ferrous sulphate decomposes to form a reddish brown residue. This is ferric oxide. Gases with a characteristic smell are also produced during the reaction. The products formed in this reaction are ferric oxide, sulphur dioxide and sulphur trioxide.

Question 35.
Suma took a pale green substance A in a test tube and heated it over the flame of a burner. A brown coloured residue B was formed along with the evolution of two gases with burning smell of sulphur. Identify A and B. Write the chemical reaction involved.
Answer:
Substance A is ferrous sulphate (FeSO₄). Two gases, sulphur dioxide and sulphur trioxide evolved.
Substance B is ferric oxide (Fe₂O₃).
The reaction involve is

Question 36.
Explain the thermal decomposition of lead nitrate. Write a balanced chemical equation for the reaction. Why is this reaction called decomposition reaction?

OR

Which coloured fumes are obtained when lead nitrate is heated? Write the balanced chemical equation for this reaction. Name the type of this chemical reaction.

OR

Name the brown fumes liberated when lead nitrate is heated. Write the balanced chemical equation for this reaction.
Answer:
Take about 2 g of lead nitrate powder in a boiling tube. Hold the boiling tube with a pair of tongs and heat it over a flame using a spirit lamp. Now, brown fumes of nitrogen dioxide are produced. A yellow residue remains in the boiling tube. This is lead oxide. This reaction can be represented by the following equation:

In this reaction, a single compound breaks down chemically forming three new substances (products). Hence, it is called decomposition reaction.

Question 37.
Limestone on heating decomposes to form quick lime and carbon dioxide. Represent this reaction by a balanced chemical equation.
Answer:
The chemical name of limestone is calcium carbonate. This compound on heating decomposes to give calcium oxide (quick lime) and carbon dioxide.

Question 38.
Draw a neat, labelled diagram to show the electrolytic decomposition of water.

OR

Draw the diagram of the apparatus used to show that water is a compound of hydrogen and oxygen. Label the following parts:

1. The part where oxygen is collected.
2. The part where hydrogen is collected.

Answer:

Question 39.
Describe an experiment to show the electrolytic decomposition of water. Write a suitable balanced equation for the reaction.
Answer:
Take a plastic mug and drill two holes at its base. Fit a rubber stopper in each of these holes. Insert carbon (graphite) electrodes in these rubber stoppers. Connect the electrodes to a 6-volt battery. Fill the mug with water such that the electrodes get fully immersed. Add a few drops of dilute sulphuric acid to the water to make it conducting.

Invert a test tube filled with water over each of the two carbon electrodes. Switch on the current and leave the apparatus undisturbed for some time. Now, water decomposes to form hydrogen and oxygen.

Hydrogen gets collected in the test tube inverted over the cathode while the oxygen gets collected in the test tube inverted over the anode. It is also found that the volume of hydrogen is twice the volume of oxygen. This reaction is represented by the following equation:
2 H₂O (l) → 2H₂(g) + O₂(g).

Question 40.
During the electrolytic decomposition of water, why is the amount of sas collected in one of the test tubes double of the amount collected in the other? Name this eas.
Answer:
In water, hydrogen and oxygen are present in the ratio 2:1 by volume. During decomposition, hydrogen and oxygen are produced in the same ratio. Therefore, the amount of gas in one of the test tubes is double the amount collected in the other. The gas present in higher volume is hydrogen and the other gas is oxygen.

Question 41.
What is photo-decomposition? Give an example.
Answer:
The decomposition of a chemical compound in presence of light is known as photo-decomposition. For example, silver chloride decomposes in presence of sunlight into silver and chlorine.

Question 42.
Silver chloride decomposes in the presence of sunlight. How do you show this? Write a balanced chemical equation for the reaction.
Answer:
Take about 2 g of silver chloride in a china dish. Place this china dish in sunlight for some time. Observe the changes. The white crystals of silver chloride turn grey in sunlight. In the presence of sunlight, silver chloride decomposes into silver and chlorine.
2AgCl(s) → 2Ag (s) + Cl₂ (g)

Question 43.
Silver chloride is usually stored in dark coloured bottles. Give reason.
Answer:
Silver chloride decomposes into silver and chlorine when exposed to light. In order to prevent photo-decomposition, silver chloride is stored in dark coloured bottles.

Question 44.
Write one equation each for decomposition reactions where energy is supplied in the form of

1. heat
2. light and
3. electricity.

Answer:
1. CaCO₃ (s) → CaO (s) + CO₂ (g)
In the reaction above, energy for decomposition is supplied in the form of heat.

2. 2AgCl (s) → 2Ag (s) + Cl₂ (g)
In the reaction above, energy for decomposition is supplied in the form of sunlight.

3. 2H₂O (l) → 2H₂ (g)+ O₂ (g)
In the reaction above, energy for decomposition is supplied in the form of electricity.

Question 45.
Identify the reactant which is oxidized and that which is reduced in the following chemical
reaction: CuO + H₂ → Cu + H₂O.
Answer:
In this chemical reaction, the reactant that is reduced is CuO and the reactant that is oxidised is H₂.

Question 46.
Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.
Answer:
In decomposition reaction a single substance decomposes to form two or more substances. This is the exact opposite of combination reaction in which two or more reactants combine to form a single product. This is why composition reactions are called the opposite of combination reactions.
General equation for a combination reaction: A + B → AB.
General equation for a decomposition reaction: AB → A + B.

Question 47.
What is meant by displacement reaction? Give an example.
Answer:
A chemical reaction in which a more reactive element displaces a less reactive element from its compound is known as a displacement reaction.

For example, iron is more reactive than copper. Therefore, iron displaces copper from copper sulphate solution.

Question 48.
Why does the colour of copper sulphate solution change when an iron nail is dipped in it?
Answer:
Copper sulphate solution is blue in colour. When an iron nail is dipped in a solution of copper
sulphate, iron displaces copper from copper sulphate and forms iron sulphate. Iron sulphate is greenish. Therefore, the colour of copper solution changes to green when an iron nail is dipped in it.

Question 49.
Describe a simple experiment to show that a more active element displaces a less active element from the solution of its compound. Represent this reaction by a suitable equation.
Answer:
Take two iron nails and clean them by rubbing with sand paper. Take about 10 mL of copper sulphate solution in a test tube. Tie one of the nails with a thread and carefully immerse it inside the copper sulphate solution. Keep the other nail for comparison. After about 15-20 minutes, take out the nail from the copper sulphate solution.

A brownish substance is formed on the iron nail and the solution will have turned greenish. In this reaction, iron displaces copper from copper sulphate. The brownish colour on the iron nail is due to the deposition of displaced copper. The greenish colour of the solution is due to the formation of iron sulphate.
Iron + Copper sulphate → Iron sulphate + Copper
Fe (s) + CuSO₄ (aq) → FeSO₄ (aq) + Cu (s)
This experiment shows that iron is more reactive than copper.

Question 50.
Give an example of a double displacement reaction.
Answer:
Zinc is a more reactive element than copper. Zinc displaces copper from copper sulphate solution. The reaction is represented by the following equation:
Zinc + Copper sulphate → Zinc sulphate + Copper
Zn(s) + CuSO₄ (aq) → ZnSO₄ (aq) + Cu (s)

Question 51.
Describe the reaction between lead and copper chloride solution. What type of reaction is this? Between lead and copper, which is more reactive? Write a balanced chemical equation for this reaction.
Answer:
When a lead rod is dipped in a solution of copper chloride, lead displaces copper from copper chloride. During this reaction lead chloride and copper are formed as products. This is a displacement reaction.

Between lead and copper, lead is more reactive. This is why lead displaces copper from copper chloride solution.
The reaction above is represented by the following equation:
Lead + Copper chloride → Lead chloride + Copper
Pb (s) + CuCl₂ (aq) → PbCl₂ (aq) + Cu(s)

Question 52.
Zinc liberates hydrogen gas when it reacts with dilute hydrochloric acid whereas copper does not. Explain why.
Answer:
Zinc is chemically more reactive than hydrogen. Therefore, zinc can displace hydrogen from dilute hydrochloric acid. However, copper is less reactive than hydrogen. Therefore, copper cannot displace hydrogen from dilute hydrochloric acid.

Question 53.
What is meant by double displacement reaction? Give an example.
Answer:
A chemical reaction in which the two reactant compounds exchange their positive ions to form two new compounds is known as a double displacement reaction.

For example, silver nitrate reacts with sodium chloride forming silver chloride and sodium nitrate. Silver nitrate + Sodium chloride → Silver chloride + Sodium nitrate.
AgNO₃ (aq) + NaCl (aq) → AgCl (s) + NaNO₃ (aq).

Question 54.
Describe an experiment to show double displacement reaction.
Answer:
Take about 3 mL of sodium sulphate solution in a test tube. Add about 3 mL of barium chloride solution to it. Shake well to ensure proper mixing of the two solutions. Soon we observe the formation of a white substance, which is insoluble in water.

This insoluble substance formed is known as a precipitate. This is barium sulphate. Another product formed during this reaction is sodium chloride. In this reaction, the chloride ion present in barium chloride and sulphate ion in sodium sulphate are exchanged to form two new compounds. This is an instance of double displacement reaction.
Sodium sulphate + Barium chloride → Barium sulphate + Sodium chloride
Na₂SO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) + 2NaCl (aq)

Question 55.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal Write down the reaction involved.
Answer:
Silver nitrate + Copper → Copper nitrate + Silver
2AgNO₃ (aq) + Cu (s) → Cu(NO₃)₂ (aq) + 2Ag (s)

Question 56.
What is a precipitation reaction? Give an example.
Answer:
An insoluble substance formed during a chemical reaction is called a precipitate. Any reaction that produces a precipitate can be called a precipitation reaction. It is represented by a downward arrow (↓).

For example, sodium sulphate reacts with barium chloride to form barium sulphate and sodium chloride. Here barium chloride is a precipitate. Therefore, this reaction is a precipitation reaction.

Question 57.
When the solutions of lead (II) nitrate and potassium iodide are mixed,

1. What is the colour of the precipitate formed? Name the compound precipitated,
2. Write the balanced chemical equation for this reaction,
3. Is this also a double displacement reaction?

Answer:
1. When lead nitrate and potassium iodide are mixed, a yellow precipitate is formed.’ The compound formed in the form of a precipitate is lead iodide (PbI₂).

2. Pb(NO₃)₂ (aq) + 2KI (aq) → PbI₂ (s) + 2KNO₃ (aq).

3. Yes, this reaction is a double displacement reaction.

Question 58.
What is the difference between displacement and double displacement reactions? Write equations for these reactions.
Answer:
When a more reactive element displaces a less reactive element from its salt solution, then it is called displacement reaction. When there is an exchange of ions between the reactants to produce new substances, it is called double displacement reaction.

In a displacement reaction one displacement takes place, while in a double displacement reaction two displacements take place.

Example for a displacement reaction:
Mg (s) + 2HCl (aq) → MgCl₂ + H₂ (g).

Example for a double displacement reaction:
2KBr (aq) + BaI₂ (aq) → 2KI (aq) + BaBr₂ (aq).

Question 59.
Write the balanced chemical equations for the following and identify the type of reaction in each case:

1. Potassium bromide (aa) + Barium iodide (aa) → Potassium iodide (aq) + Barium bromide(s).
2. Zinc carbonate (s) → Zinc oxide (s) + Carbon dioxide (g)
3. Hydroeen fg) + Chlorine (fg) → Hydrogen chloride (s)
4. Magnesium (s) + Hydrochloric acid (aq) → Magnesium chloride (aq) + Hydrogen (s).

OR

Write the balanced chemical equations for the following chemical reactions:

• Potassium bromide reacts with Barium iodide.
• Zinc carbonate is heated.

Answer:
1. 2KBr (aq) + BaI₂ (aq) → 2KI (aq) + BaBr₂ (aq).
This is a double displacement reaction.

2. ZnCO₃ (s) → ZnO (s) + CO₂ (g).
This is a decomposition reaction.

3. H₂ (g) + Cl₂ (g) → 2HCl (g).
This is a combination reaction.

4. Mg (s) + 2HCl (aq) → MgCl₂ (aq) + H₂ (g).
This is a displacement reaction.

Question 60.
Balance the following chemical equations and identify the type of chemical reaction:

1. Mg (s) + Cl₂ (g) → MgCl₂ (s)
2. HgO (s) → Hg (l) + O₂ (g)
3. Na (s) + S (s) → Na₂S (s)
4. TiCl₄ (l) + Mg (s) → Ti (s) +MgCl₂(s)
5. CaO (s) + SiO₂ (s) → CaSiO₃ (s)
6. H₂O₂ (l) → H₂O (l) + O₂ (g).

Answer:
1. Mg (s) + Cl₂ (g) → MgCl₂ (s).
This reaction is a chemical combination.

2. 2 HgO (s) → 2 Hg (l) + O₂ (g).
This reaction is a chemical decomposition.

3. 2 Na (s) + S (s) → Na₂S (s).
This reaction is a chemical combination.

4. TiCl₄ (l) + Mg (s) → Ti (s) + MgCl₂ (s).
This reaction is a chemical displacement.

5. CaO (s) + SiO₂ (s) → CaSiO₃ (s).
This reaction is a chemical combination.

6. H₂O₂ (l) → H₂O (l) + O₂ (g).
This reaction is a chemical decomposition.

Question 61.
What is oxidation and reduction? Give an example for each.

OR

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

1. Oxidation,
2. Reduction.

Answer:
1. Oxidation:
A chemical reaction that involves gain of oxygen or loss of hydrogen is known as oxidation.
E.g.: Consider the reaction in which magnesium ribbon burns in air forming magnesium oxide.
2Mg + O₂ → 2MgO
In this reaction, oxygen is added to magnesium and therefore, magnesium is oxidized. This is an example of oxidation.

2. Reduction:
A chemical reaction that involves gain of hydrogen or loss of oxygen is known as reduction.
E.g.: Zinc oxide reacts with carbon forming zinc and carbon monoxide.
ZnO + C → Zn + CO
In this reaction, oxygen is removed from zinc oxide. Therefore, zinc oxide is reduced to zinc. This is a reduction reaction.

Question 62.
How do you show the oxidation of a metal into its metal oxide?
Answer:
Take about 1 g of copper powder in a china dish. Heat the china dish. The surface of copper powder becomes dark due to the formation of black copper (II) oxide. In this reaction, oxygen is added to copper and copper oxide is formed. We say that copper is oxidised to copper oxide.
2CU + O₂ → 2CuO.

Question 63.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Write the name of the element ‘X’ and the black coloured compound formed.
Answer:
The shiny brown coloured element is the copper metal (Cu). When heated in air, copper reacts with the atmospheric oxygen to form a black coloured compound called copper oxide.
2Cu (s) + O₂ (g) → 2CuO (s).

Question 64.
What happens when hydrogen is passed over hot copper (II) oxide? What type of reaction is this? Write a balanced chemical equation for this reaction.
Answer:
If hydrogen gas is passed over heated copper (II) oxide (CuO), the black coating on the surface turns brown. During this reaction, oxygen is removed from copper oxide. Therefore, this is a reduction reaction. This is represented by the following equation:
CuO + H₂ → Cu + H₂O.

Question 65.
Identify the substances that are oxidised and the substances that are reduced in the following reactions:

1. 4 Na(s) + O₂ → 2 Na₂O(s)
2. CuO(s) + H₂ (g) → Cu(s) + H₂O (l).

Answer:

1. In this reaction oxygen is added to sodium. Therefore, sodium is oxidised to sodium oxide. This is an example of oxidation reaction.
2. In this reaction, oxygen is removed from copper oxide. Therefore, copper oxide is reduced to copper. This is an example of reduction reaction.

Question 66.
What are redox reactions? Explain with an example.
Answer:
A chemical reaction between two substances in which one substance is oxidized and the other reduced is known as reduction-oxidation reaction. Such reactions are also called redox reactions.
E.g.: Consider the reaction between zinc oxide and carbon. During this process, zinc and carbon monoxide is formed.
ZnO + C → Zn + CO.
In the reaction above, oxygen is removed from zinc oxide and oxygen is added to carbon. Therefore, zinc oxide is getting reduced to zinc while carbon is getting oxidised to carbon monoxide. This is an example of a redox reaction where one of the reactants is oxidised while the other is reduced.

Question 67.
When hydrogen is passed over hot copper oxide, we get copper and water. Write a balanced chemical equation for this reaction. State the substance that is being oxidised and the one that is getting reduced.
Answer:
The balanced chemical equation for the reaction is: CuO + H₂ → Cu + H₂O.
In the reaction above, copper oxide is losing oxygen. Therefore, CuO is reduced to copper. Oxygen is being added to hydrogen. Therefore, hydrogen is oxidised to water.

Question 68.
Why is the following reaction called a redox reaction:
MnO₂ + 4 HCl → MnCl₂ + 2 H₂O + Cl₂? Explain.
Answer:
In the reaction above, HCl loses hydrogen and forms chlorine. Thus, HCl is oxidised to Cl₂ whereas MnO₂ loses oxygen and is reduced to MnCl₂. In this reaction, one of the reactants is getting oxidised while the other is being reduced. Therefore, this reaction is called a redox reaction.

Question 69.
What is corrosion? Give two examples. What is corrosion due to?
Answer:
The process of slow eating up of metals due to reaction with atmospheric gases such as oxygen, water vapour, carbon dioxide etc., is known as corrosion.

When silver is exposed to atmospheric air, a black coating appears on silver. This is an instance of corrosion. Similarly, when copper is exposed to atmospheric air, a green coating appears on its surface. This is another instance of corrosion. Most instances of corrosion are due to oxidation.

Question 70.
Which is the most common form of corrosion? Give a chemical equation to represent the reaction involved in this case.
Answer:
The most common form of corrosion is rusting of iron. When iron articles are exposed to moist air, the surface of iron turns brown due to the formation of a flaky substance called rust. This is nothing but hydrated ferric oxide. Iron reacts with atmospheric oxygen in presence of water vapour to form ferric oxide (rust).
4 Fe + 3 O₂ + 2 H₂O → 2 Fe₂O₃.H₂O

Question 71.
What is rancidity? Why does it occur?
Answer:
The oxidation of oils and fats present in food substances resulting in bad smell or bad taste that makes the food unsuitable for consumption is known as rancidity.

Rancidity occurs due to the oxidation of fats and oils present in food due to exposure to atmospheric air.

Question 72.
How can rancidity of food materials be prevented?
Answer:

• Keeping the food substances in air-free containers (vacuum packing),
• Adding anti-oxidants, or
• Replacing air inside the containers with nitrogen can prevent rancidity. These measures help to reduce the oxidation of fats and oils present in food.

Question 73.
Oil and fat-containing food items are flushed with nitrogen. Why?
Answer:
Food items containing oil and fat become rancid when exposed to air, which gives the food an unpleasant smell and taste. In order to prevent rancidity, food items are flushed with nitrogen. Nitrogen does not react with oils or fats. Therefore, rancidity is prevented.

Question 74.
Which substances get oxidized in rancidity of food?
Answer:
The fats and oils present in food get oxidized when food turns rancid.

Question 75.
Which substance is flushed in bags of potato chips to prevent rancidity?
Answer:
In order to prevent rancidity, chips bags are flushed with nitrogen gas.

Question 76.
Explain the disadvantages of corrosion.
Answer:
Corrosion causes damage to car bodies, bridges, iron railings, ships and to all objects made of metals, especially those made of iron. Corrosion of iron is a serious problem. Every year an enormous amount of money is spent to replace damaged iron or for the maintenance of iron structures.

Question 77.
Why do we apply paint on iron articles?
Answer:
Iron articles get rusted when exposed to moist air forming a brownish compound called iron (III) oxide. When iron articles are painted, their surface does not come in contact with moisture and air. Therefore rusting does not occur.

Fill In The Blanks

1. A compound of calcium used for white washing walls is Calcium hydroxide
2. The chemical name of rust is Iron oxide
3. The number of products in a combination reaction is one
4. A chemical reaction in which an atom, molecule or ion loses oxygen is called reduction
5. The chemical name of quick lime is Calcium oxide

Multiple Choice Questions

Question 1.
Which of the following is not a physical change?
(A) Boiling of water to give water vapour
(B) Melting of ice to give water
(C) Dissolution of salt in water
(D) Combustion of liquefied petroleum gas (LPG)
Answer:
(D) Combustion of liquefied petroleum gas (LPG)

Question 2.
Which of the statements about the reaction below arc correct:
2PbO (s) + C (s) → 2Pb (s) + CO₂ (g) ?
(i) Lead is getting reduced.
(ii) Carbon dioxide is getting oxidised.
(iii) Carbon is getting oxidised.
(iv) Lead oxide is getting reduced.
(A) (i) and (ii)
(B) (i) and (iii)
(C) (ii) and (iii)
(D) (iii) and (iv)
Answer:
(D) (iii) and (iv)

Question 3.
Electrolysis of water is a decomposition reaction. The mole ratio of hydrogen and oxygen gases liberated during electrolysis of water is
(A) 1:1
(B) 2:1
(C) 4:1
(D) 1:2
Answer:
(B) 2:1

Question 4.
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
The reaction above is an example of a
(A) Combination reaction.
(B) Double displacement reaction.
(C) Decomposition reaction.
(D) Displacement reaction.
Answer:
(D) Displacement reaction.

Question 5.
What happens when carbon dioxide is passed in limewater?
(A) Limewater turns red because of formation of permanganate.
(B) Limewater turns milky because of formation of calcium carbonate.
(C) Limewater turns, milky because of formation of water.
(D) Limewater turns red, because of formation of copper sulphate.
Answer:
(B) Limewater turns milky because of formation of calcium carbonate.

Question 6.
What happens when dilute hydrochloric acid is added to iron filings?
(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.

Question 7.
Which of the following is an endothermic reaction?
(i) Dilution of sulphuric acid
(ii) Sublimation of dry ice
(iii) Condensation of water vapour
(iv) Evaporation of water.
(A) (i) and (iii)
(B) (ii) only
(C) (iii) only
(D) (ii) and (iv)
Answer:
(D) (ii) and (iv)

Question 8.
The product formed when ferrous sulphate is heated is
(A) Ferric oxide
(B) Sulphur dioxide
(C) Sulphur trioxide
(D) All of the above
Answer:
(D) All of the above

Question 9.
A compound formed when lead metal reacts with a solution of copper chloride is
(A) No reaction takes place
(B) Copper chloride
(C) Lead chloride
(D) Copper-lead chloride
Answer:
(C) Lead chloride

Question 10.
What happens when copper metal is dipped in a solution of zinc sulphate?
(A) Copper sulphate is formed
(B) Oxygen gas is formed
(C) Zinc metal is separated
(D) No reaction takes place
Answer:
(D) No reaction takes place

Question 11.
Which of the following products is formed when calcium oxide reacts with water?
(A) Slaked lime
(B) Carbon dioxide
(C) Calcium oxide
(D) Oxygen gas.
Answer:
(A) Slaked lime

Question 12.
Rancidity of food substances can be prevented by
(A) Keeping food items in the open
(B) Adding antioxidants in food
(C) Adding more oxygen to food
(D) All of the above
Answer:
(B) Adding antioxidants in food

Question 13.
The substance that is oxidised in the following chemical reaction is
MnO₂ + 4 HCl → MnCl₂ + 2 H₂O + Cl₂
(A) HCl
(B) MnO₂
(C) MnCl₂
(D) H₂O
Answer:
(A) HCl