Class 10

Balbharti Maharashtra State Board Class 10 History Solutions Chapter 2 Historiography Indian Tradition Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 History Solutions Chapter 2 Historiography Indian Tradition

Question 1.
(A) Choose the correct option from the given options and complete the statement.
(1) …………………..…….. was the first Director-General of the Archaeological Survey of India.
(a) Alexander Cunningham
(b) William Jones
(c) John Marshall
(d) Friedrich Max Muller
Answer:
(a) Sir Alexander Cunningham

(2) …………………..…….. translated the Sanskrit text of ‘Hitopadesh’ in the German language.
(a) James Mill
(b) Friedrich Max Muller
(c) Mountstuart Elphinstone
(d) Sir John Marshall
Answer:
(b) Friedrich Max Muller

(B) Identify and write the wrong pair in the following set.
(1) ‘Who were the Shudras’ – subaltern History
(2) ‘Stri-Purush Tulana’ – Feminist writing
(3) ‘The Indian War of Independence 1857’ – Marxist History
(4) Grant Duff – Colonial History.
Answer:
(1) Right Pair: ‘Who were the Shudras’ – subaltern History
(2) Right Pair: ‘Stri-Purush Tulana’ – Feminist writing
(3) Wrong Pair: ‘The Indian War of Independence 1857’ – Marxist History
(4) Right Pair: Grant Duff – Colonial History.

Question 2.
Explain the following with its reason.
(1) Writing of the regional history received a momentum.
Answer:

• Indian historians trained in the British educational system showed an inclination to restore the self-esteem of Indians They started writing Nationalistic historiography.
• Historians from the different regions of India criticised the prejudiced history of India written by the British officers.
• Their attention was drawn towards geographic conditions and history of different regions of India.
• The nationalistic historiography helped in triggering the independence movement.
• It provided a momentum to the writing of regional histories for e.g.
• Nationalistic writings in Maharashtra were inspired by Vishnu Shastri Chiplunkar.

Hence, writing of the regional history received momentum.

(2) Bakhar is an important type of historical documents.
Answer:

• Bakhars are important source of history like coins, inscriptions, texts, copper plates and travelogues.
• Bakhars contain eulogies of the heroes, historic events, stories of lives of great men and description of battles.
• It portrays a picture of prevalent society, religions and economic life.
• Bakhars are of various types which contain biographies of kings, dynastic history, description of events, sects, autobiographies, mythologies, administration of kings, etc.
• These descriptions are helpful for writing historiography.
• Among well-known Marathi. Bakhars ‘Sabhasad Bakhar’ written by Krishnaji Anant Sabhasad requires a special mention Bhausahebanchi Bakhar and Panipatachi Bakhar describe the ‘Battle of Panipat’.

Hence, Bakhars are important type of historical document.

Question 3.
Write detailed Answers the following questions in detail.
(1) What is Marxist History?
Answer:

1. In the post-independence era, new trends emerged in writing Indian historiography. Marxist historiography is prominent among them.
2. Historiography which is based on ideas of Karl Marx is known as Marxist Historiography. In Marxist Historiography, more importance was given to class struggles.
3. Marxist Historiography reflects concerns for means and modes of production as well as the industrial relations. These became centre of writings of the Marxist historians.
4. How every social event affects the life of people was analysed and it became the theme of Marxist historians Marxist historians studied the transitions that took place in the caste system.
5. Notable Indian historians who adopted Marxist ideologies were Damodar Kosambi, Comrade Shripad Dange, Sharad Patil and Ram Sharan Sharma.
6. Their writings reflected Marxist ideology. Comrade Dange was the founder member of the Indian Communist Party

(2) What is the contribution of Itihasacharya V.K. Rajwade to historiography?
Answer:

1. V.K. Rajwade is well-known for his writings in Marathi on subjects like history, linguistics, etymology and grammar. His contributions are as follows:
2. He compiled and edited 22 volumes of Marathyanchya Itihasachi Sadhane.
3. He stated that history is the all-inclusive image of the past societies.
4. V. K. Rajwade was of the view that history does not include only the stories of political images, conspiracies and Wars for seizing power.
5. He was of the firm opinion that we should write our own history and insisted that history . should be written only using the authentic sources.
6. He was of the opinion that description of any human event in historiography should have balanced combination of three factors namely – Time, Space and Personalities.

Question 4.
(a) Complete the following chart.

Historian	Name of the Book
James Mill	‘The History of British India’
James Grant Duff	……………………………………………
……………………………………………	‘The History of India’
S.A. Dange	……………………………………………
……………………………………………	‘Who were the Shudras’

Answer:

Historian	Name of the Book
James Mill	The History of British India
James Grant Duff	A History of Mahrattas
Mountstuart Elphinstone	The History of India
S. A. Dange	Primitive Communism to Slavery
Dr. Babasaheb Ambedkar	Who Were the Shudras

(b) Complete the following concept chart.

Answer:

Question 5.
Explain the following concepts.
(1) Orientalist historiography.
Answer:

1. In the later part of the eighteenth century many European scholars felt curious to know about civilisations and countries of the East.
2. These scholars who studied the Indian Civilization had respect and admiration for it. They were known as ‘Orientalists’.
3. They studied similarities between Sanskrit and some of the European languages, focussing more on Vedic traditions and Sanskrit literature.
4. It resulted in the notion that an ancient language could be the mother of all Indo-European languages.
5. Notable Orientalist scholars are Friedrich Max Muller and William Jones.

(2) Nationalistic historiography.
Answer:

1. English education started in India in the nineteenth century. The Indian historians who were trained in English education system realised the glory of their past.
2. They showed an inclination to restore the pride in the ancient glory of India and the self-esteem of the Indian readers.
3. They laid the foundation of Nationalistic writing. Nationalistic writings in Maharashtra were inspired by ‘Vishnushastri Chiplunkar’.
4. Nationalistic historians opposed the prejudiced history written by the British officers.
5. Mahadev Govind Ranade, Vinayak Damodar Savarkar, Ramesh Chandra Majumdar are renowned Nationalistic historians.
6. Nationalistic historiography inspired freedom struggle against the Britishers and gave momentum to write about regional history. While glorifying the past, at times they are blamed for ignoring to critically analyse the truth.

(3) Subaltern history.
Answer:

• History of deprived classes or communities who were deprived of all rights is known as subaltern history.
• The seeds of subaltern history can be traced in the Marxist history. Italian historian Antonio Gramsci has special importance in writing the subaltern history.
• He presented the idea that history should be written starting from the bottommost ranks of people in the society.
• Ranjit Guha, an Indian historian, has given prominence to subaltern history and played a major role in establishing it as an important academic school of historiography.
• Similar thoughts were expressed by Mahatma Phule and Dr. Babasaheb Ambedkar which can be found in their writings.

Project
Prepare an illustrated manuscript with the help of the internet giving more information about the historians mentioned in this lesson.
Answer:

Memory Map

Question 6.
Complete the sentences by choosing a correct option:
(a) The first written document found in India was inscribed during the rule of Emperor ……………….. .
(a) Akbar
(b) Harshavardhan
(c) Ashoka Maurya
(d) Aurangzeb
Answer:
(c) Ashoka Maurya

(b) Banabhatta wrote King Harsha’s biography titled ………………… in the 7th century CE.
(a) Meghdoot
(b) Rajtarangini
(c) Rasratnakar
(d) Harshacharit
Answer:
(d) Harshacharit

(c) In the 12th century C.E., ………………… wrote a book ‘Rajtarangini’ about history of Kashmir.
(a) Banabhatta
(b) Kalhana
(c) Ziquddin Barani
(d) Patanjali
Answer:
(b) Kalhana.

(d) The historians in the courts of Muslim rulers were influenced by ………………… historiography.
(a) Greek and Roman
(b) Arabic and Persian
(c) English and French
(d) Arabic and Urdu
Answer:
(b) Arabic and Persian

(e) We get information about the rule of Chhatrapati Shivaji Maharaj from ……………….. .
(a) Bhausahebanchi Bakhar
(b) Panipatachi Bakhar
(c) Sabhasad Bakhar
(d) Holkaranchi Bakhar
Answer:
(c) Sabhasad Bakhar

(f) ‘Bhausahebanchi Bakhar’ describes Battle of ……………….. .
(a) Raigad
(b) Panipat
(c) Plassey
(d) Vasai
Answer:
(b) Panipat

(g) The first book written by a British historian on Indian history is ……………….. .
(a) The History of India
(b) The History of Maharaj
(c) The History of British India
(d) The Sacred Books of the East
Answer:
(c) The History of British India

(h) Nilkanth Janardan Kirtane and V.K. Rajwade exposed the limitations in writings of ……………….. .
(a) Friedrich Max Muller
(b) Colonel Todd
(c) Grant Duff
(d) William Wilson Hunter
Answer:
(c) Grant Duff

(i) …………………, an Indian historian played a major role in establishing subaltern history as an important academic school of historiography.
(a) Comrade Sharad Patil
(b) Ranjit Guha
(c) Dharmanand Kosambi
(d) Ram Sharan Sharma
Answer:
(b) Ranjit Guha.

(j) The writings of British officials on India display strong influence of ……………….. .
(a) Marxism
(b) Colonalism
(c) Feminism
(d) Capitalism
Answer:
(b) Colonalism

(k) The Nationalistic Historiography in Maharashtra was inspired by works of ……………….. .
(a) Vinayak Damotfar Savarkar
(b) Ramkrishna Gopal Bhandarkar
(c) Vishnushastri Chiplunkar
(d) Justice Mahadev Govind Ranade
Answer:
(c) Vishnushastri Chiplunkar

(l) The two volumes written by ……………….. on Indian history show impartial treatment of the subject.
(a) Colonel Todd
(b) William Wilson Hunter
(c) Grant Duff
(d) James Mill
Answer:
(b) William Wilson Hunter

(m) ‘Primitive Communism to Slavery’, represents ……………….. the historiography.
(a) Colonial
(b) Orientalist
(c) Nationalistic
(d) Marxist
Answer:
(d) Marxist

(n) ‘Stripurush Tulana’ was written by ……………….. is acknowledged as first feminist book in India.
(a) Meera Kosambi
(b) Dr. Sharmila Rege
(c) Tarabai Shinde
(d) Pandita Ramabai
Answer:
(c) Tarabai Shinde.

Question 7.
Identify the wrong pair and write it:
(1)

Name of the Book	Types of Historiography
(1) Who were the	(a) Subaltern history
Shudras	(July ’19)
(2) Stripurush Tulana	(b) Feminist writing
(3) The Indian War of Independence	(c) Marxist History
(4) Grant Duff	(d) Colonial History

Answer:
Wrong pair: The Indian War of Independence – Marxist History.

(2)

Name of the Book	Historian
(1) The Rise of the Maratha Power	(a) Justice Mahadev Govind Ranade
(2) Marathyanchya Itihasachi Sadhane	(b) Vishnushastri Chiplunkar
(3) Maratha Riyasat	(c) Govind Sakharam Sardesai
(4) Gulamgiri	(d) Mahatma Phule

Answer:
Wrong pair: Marathyanchya Itihasachi Sadhane- Vishnushastri Chiplunkar.

(3)

Work	Scholars
(1)  Bharat Itihas Samshodhak Manda	(a)   V. K. Rajwade
(2)  Asiatic Society	(b)   Sir William Jones
(3)  Inspiration for Nationalistic historiography	(c)  Vishnushastri Chiplunkar
(4)  Discovery of Harappan Civilization	(d)  James Mill

Answer:
Wrong pair: Discovery of Harappan CIvilization – James Mill.

(4)

Historiography	Historian
(1) Nationalistic Historiography	(a) V. D. Savarkar
(2) Marxist Historiography	(b) Damodar Kosambi
(3) Subaltern Historiography	(c) Ramchandra Guha
(4) Feminist Historiography	(d) Sharmila Rege

Answer:
Wrong pair: Subaltern Historiography – Bamchandra Guha.

Question 8.
Do as directed:
(1)

Answer:

(2)

Answer:

(3) Prepare a flow chart on Literary work of Max Muller:
Answer:

Question 9.
Explain the concepts:
(a) Colonial Historiography:
Answer:

• The history written by British officers and Christian missionaries in support of British rule and colonialism is known as Colonial historiography.
• It reflected their prejudiced mindset about Indian culture. They ridiculed the Indian culture as being primitive.
• Their writings justified the colonial British rule.
• The distinct example of colonial historiography is ‘Cambridge History of India’ in five volumes published during 1922-1937.

Question 10.
Write short notes:
(a) Alexander Cunningham:
Answer:

• British rulers started archaeological exploration and research in India in the 20th century.
• They established the Archaeological Survey of India and appointed Sir Alexander Cunningham as its first Director General.
• He mainly chose the sites mentioned in the Buddhist texts for excavations.
• Many other ancient sites were excavated under his supervision.

(b) James Mill:
Answer:

• James Mill, a‘British historian, had written the monumental book ‘The History of British India’. He published it in three volumes in 1817.
• The book is important as it was the first book written by a British officer on Indian history.
• The book: lacks objective perspective and is prejudiced about various aspects of the Indian culture.

(c) Grant Duff:
Answer:

• Grant Duff, a British officer, wrote the book ‘A History of Mahrattas’, in three volumes.
• His name stands out among the British officers who wrote about Maratha history.
• The tendency of the British historians to condemn Indian culture and history is found in Grant Duff’s writing.
• Historians Nilkanth Janardan Kirtane and Vishwanath Kashinath Rajwade exposed the limitations of Grant Duff’s writings.

(d) Indian Feminist Historiography:
Answer:

1. Tarabai Shinde, Pandita Ramabai and Dr. Rukhamabai were feminist historians before independence.
2. They wrote about employment of women, their right to political equality, treatment meted out to women at their workplace and gender discrimination.
3. A lot of literature is available unfolding the viewpoint of Dalit women on problems like social class, caste, etc. In the post-independence era.
4. Dr. Sharmila Rege included her essays on the autobiographies of Dalit women in her book, ‘Writing Caste, Writing Gender: Reading Dalit Women’s Testimonies.’
5. Meera Kosambi’s book, ‘Crossing Thresholds: Feminist Essays in Social History’ contains essays on the life stories of women like Pandita Ramabai and Dr. Rukhamabai, the first practising lady Doctor of India.

Question 11.
Explain the following with reasons:
(a) The history of Indian civilization dates back to 3rd millennium BCE.
Answer:

• Indian civilisation is among the various ancient civilisations in the world.
• Excavations at Harappa took place in 1920 under the supervision of Sir John Marshall, then Director General of Archaeological Survey of India (ASI). This led to the discovery of Harappan Civilization.
• This discovery prqved helpful to trace the antiquity of Indian history which dates back 3rd millennium BCE or earlier.

(b) It became necessary to rethink of women’s position in history.
Answer:

• For a considerable period of time, mainly male scholars were involved in the writing of Indian history.
• It was obvious that the perspective was male-dominated.
• The work of the women historians was neglected for long. So it became important to highlight this fact.
• It was a huge challenge in front of the women: historians to compile the work and do research which would prove to be a medium to put forth their status in society.

Hence, it became necessary to rethink about the position of women in history.

(c) Govind Sakharam Sardesai was addressed as ‘Riyasatkar’.
Answer:

• Govind Sakharam Sardesai had a momentous achievement to his credit by publishing ‘Maratha Riyasat’.
• He published several volumes of Maratha history.
• His work became so popular that people began to address him as ‘Riyasatkar’.

Question 12.
Answer the following in 25-30 words:
(a) Which things are included in the description by Emperor Babur in bis autobiography?
Answer:
Babur, the founder of the Mughal empire, in his autobiography entitled Tuzuk-i-Babari describes:

• The battles fought by him.
• Recordings of minute observations of various regions and cities visited by him noting minute observations.
• Observation of the local customs, economy, fauna and flora of the region travelled by him.

(b) What is the contribution of Swatantryaveer Savarkar to nationalistic historiography?
Answer:

1. The writings of Swatantryaveer Savarkar were infused with nationalistic ideas. He looked at the revolt of 1857 as the First War of Independence and wrote the ‘book ‘The Indian War of Independence, 1857’.
2. His book gave inspiration to Nationalistic historiography which helped in triggering independence movement of the Indian people against the British. Even the writing of regional history got momentum.
3. The attention of historians was drawn towards the history and geographic conditions and variations in the South Indian region.

(c) Write about the contribution of Banabhatta in writing historiographies.
Answer:

• It was considered an important step in the Indian historiography to write biographies of kings and dynastic histories.
• Banabhatta wrote Harshacharit, King Harsha’s biography, in the 7th century C.E.
• It portrays a realistic picture of the social, economic, political, religious and cultural life during the period when King Harsha ruled.

(d) Write information on ‘Rajtarangini’.
Answer:

• Rajtarangini was written by Kalhana in the 12th century C.E. It traces the history of Kashmir.
• These eight books contain records of dynasties which ruled Kashmir, local traditions, coins, remains of monuments and inscriptions. Kalhana wrote the text after critically examining the various sources.
• The book is a good example to show how scientific technique is applied in writing history.

(e) Write about historiography in the post-independence period.
Answer:

• In the post-independence period, along with dynastic history, cultural, social and economic histories were also written.
• New schools of thought like Marxist, Subaltern and Feminist History were included in historiography.
• The scholars in this era began to feel the need of writing histories of various communities, sciences, economic systems, political ideologies, cultural aspects, etc.
• They presented the new schools of ideologies. These changes increased the scope of writing historiography.

(f) What are the objectives of writing history according to Ziauddin Barani?
Answer:

• Ziauddin Barani explained the ’ objectives of historiography. According to him, the duty of a historian does not end with recording the ruler’s valour and welfare policies.
• He should also write about the ruler’s failings and incorrect policies.
• He further adds that a historian should also take into account the impact of the teachings of the wise, the learned and the saints on the cultural life of people.

Thus, Ziauddin Barani holds important place as he expanded the scope of historiography.

(g) What are the salient features of writings of historians in the courts of Mughal emperors?
Answer:
The salient features of writings of historians in the courts of Mughal emperors are:

• Historians praised the Mughal emperors in their writings.
• Their loyalty towards the rulers is exhibited.
• The custom of adding suitable poetic quotes was introduced. The historians started to include beautiful pictures as well.

(h) Write about the importance of Tarabai Shinde in writing feminist historiography. *
Answer:

• Tarabai Shinde has a prominent place among the feminist historians who wrote about women in the nineteenth century, i.e. pre-independence era.
• She wrote a book Stripurush Tulana which was published in 1882. It is acknowledged as the first feminist book published in India.
• Her writings attacked the male-dominated social system and the caste system.

Question 13.
Read the given passage and answer the questions given below:
(a) Complete the graphical presentation:

Answer:

(b) What was considered more important by the historians in the Mughal Court?
Answer:
Praising the Mughal emperor and exhibition of loyalty was considered more 8 important by the historians in the Mughal Court.

(c) Why is Abbul Fazl’s writing considered realistic?
Answer:

• Abul Fazl’s Akbarnama reflects his critical viewpoint.
• He collected authentic historical documents for this book.
• He scrutinized these documents to affirm their reliability. Therefore, his method is looked upon as devoid of bias and hence realistic.

Question 14.
Write detailed answers to the following questions:
(a) Write information about historiography in the ancient period of India.
Answer:

• In the ancient period, stories of great deeds of ancestors, mythological lore, memories of social transitions were preserved through oral traditions.
• The art of writing was mastered by Indians as early as 3rd millennium B.C.E. But Harappan script is not deciphered yet.
• The earliest written documents found in India were in the form of inscriptions that are as old as 3rd century B.C.E. during the period of Emperor Ashok. His edicts are inscribed on natural rocks and stone pillars.
• Inscriptions started to occur on coins, metal images and sculptures and also on copper plates from the 1st century B.C.E.
• Epics like Ramayana, Mahabharat, Puranas, Jain and Buddhist texts are important sources of ancient Indian literature.
• Historical accounts by Indian authors and travelogues by foreign travellers are also important sources of history.

(b) Explain the works of Dr. Babasaheb Ambedkar and Mahatma Phule in Subaltern History.
Answer:

• History of the deprived classes expressed in folklore was known as ‘Subaltern History’.
• Similar thoughts were expressed by Mahatma Jotirao Phule and Dr. Babasaheb Ambedkar.
• Mahatma Phule unfolded the history of the Shudratishudra communities in his book ‘Gulamgiri’.
• He drew attention to show how women, Shudras and Atishudras were exploited under the name of religion.
• Dr. Ambedkar focused on the idea that in spite of having a significant role in shaping the cultural and political aspects of India, the role of Dalits was not duly acknowledged in colonial and nationalistic historiography.
• Injustice done to the Dalits was mostly written by him. ‘Who Were the Shudras’ and ‘The Untouchables’ are his books which can be cited as examples of subaltern history.

Question 15.
Observe the given picture and answer the question given below:

(a) Write a short note on Copper-plate inscriptions.

Answer:

• The copper-plate shown in the picture was found at Sohagaura in Gorakhpur district in Uttar Pradesh.
• It is supposed to be from the Mauryan period and the inscription on it is in Brahmi script.
• Many symbols are inscribed on it ‘Tree-in-railing’ and ‘Mountain’ at the beginning of inscription occur on punch marked coins.
• A structure created on four pillars is indicative of a granary.
• The inscriptions have a record of royal order that the grains stored in the granary should be distributed carefully.
• This order is suggestive of precautionary measures taken during famine.

Balbharti Maharashtra State Board Class 10 Political Science Solutions Chapter 3 Political Parties Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Political Science Solutions Chapter 3 Political Parties

Question 1.
Choose the correct option from the given options and complete the sentences.
(1) When people come together and participate in electoral process, to acquire political power, such organisations are called ………………………….. .
(a) Government
(b) Society
(c) Political parties
(d) Social organisations
Answer:
(c) Political parties

(2) National Conference is a party in ………………………….. this region.
(a) Odisha
(b) Assam
(c) Bihar
(d) Jammu and Kashmir
Answer:
(d) Jammu and Kashmir

(3) Justice Party- a non-Brahmin movement was transformed into ………………………….. Political Party.
(a) Assam Gan Parishad
(b) Shivsena
(c) Dravid Munnetra Kazhagam
(d) Jammu and Kashmir National Conference
Answer:
(c) Dravid Munnetra Kazhagam

Question 2.
State whether the following statements are true or false. Give reasons for your answer.
(1) Political parties act as a link between government and people.
Answer:
The above statement is True.

• The political parties convey the demands and complaints of the people to the government.
• Information about various government policies and schemes are conveyed to the people by the political parties.
• People’s reaction to the policies and their expectations are conveyed to the government too by the parties. In this way, political parties act as a link between government and the people.

(2) Political parties are social organisations.
Answer:
The above statement is True.

• People who have similar views come together voluntarily and form a party, so it can be said that political parties are an integral part of our society.
• The political parties try to solve the problems of the people.
• Some political parties represent the role of a particular community in society, its ideology and work to achieve it. Hence it can be said that political parties are social organizations.

(3) Coalition politics leads to instability.
Answer:
The above statement is False.

• As no single party got majority after the elections held in 1989, coalition government came to power at the centre and many states.
• Parties had to set aside their policies and programmes and concentrate on a common agenda to run the government.
• Both Congress Party and the Bharatiya Janata Party made an experiment of establishing a coalition government.
• The National Democratic Alliance formed by BJP and its partners and the Democratic Front government formed by Congress and its partners ruled at the centre and state level effectively.

Therefore, the belief that the coalition government brings instability was proved wrong.

(4) Shiromani Akali Dal is a national party.
Answer:
The above statement is False.
The Election Commission of India recognises a political party as National or Regional party on the basis of specific criteria namely:

• The number of seats secured in Parliament and Legislative Assembly or percentage of votes secured.
• Shiromani Akali Dal does not meet the criteria listed above.
• As its influence is limited to the state of Punjab and not in any other state.
• The Election Commission has given recognition to it as regional party.

Question 3.
Explain the following concepts.
(1) Regionalism
Answer:

• India is a vast country with great physical diversity. People belonging to different regions have their own language, culture and traditions.
• Along with the physical diversity, we find diversity in literature, education, history and movements.
• Each one feels an affinity for his language, traditions and history of social reforms. This gives rise to linguistic identity.
• This affinity towards all the above-mentioned aspects develops into identity consciousness and gives rise to excessive regional pride.
• People start giving preference only to the development of their language and region.
• This is known as Regionalism.

(2) National Parties
Answer:
A party which functions at the national level is known as National Party.

1. The Election Commission has laid certain criteria for a party to get recognised as a Political party. The political party should secure at least f 6% valid votes in four or more states in the earlier Lok Sabha. OR
2. State Assembly elections and minimum four members should be elected in the Lok Sabha from any state or states in the earlier elections. OR
3. The candidates of a party should be elected from a minimum three states or should be elected from a minimum 2% constituencies of the total Lok Sabha constituencies. OR
4. It is important to get recognition as a state party in at least four states.

Question 4.
Answer the following questions in brief.
(1) What are the major characteristics of political parties?
Answer:
The main features of political parties are as follows:

• Political parties aim at securing power by contesting elections.
• Every political party has some ideology.
• Every party has an agenda. It is based on the party’s ideology.
• The party that gets majority seats in election forms the government and the one that does not secure the majority act as opposition parties.
• Political parties are a link between the people and the government and work to gain the support of the people.

(2) What changes have taken place in the nature of political parties in India?
Answer:
The following changes were seen in the party system in India:

• Indian- National Congress was the only dominant party in states and at the centre after independence.
• All other parties came together and formed a coalition in 1977 and defeated the Congress.
• For the first time in history, the two-party system got importance over one-party system.
• The one-party system came to an end after the 1989 elections.
• Regional parties got importance and the coalition government was formed at the centre and in the states.

Project
(1) In a map of Maharashtra, point out the Loksabha constituency that includes the names of your parents.
(2) In a map of India, point out the places where national political parties have their influence.
Answer:

Memory Map

Question 5.
Choose the correct option from the given options and complete the sentences:
(a) The party which fails to secure majority is known as ………………………. party.
(a) Ruling
(b) Opposition
(c) Independent
(c) Coalition
Answer:
(b) Opposition

(b) ‘Bharatiya Janata Party’ which was formed in 1980 after a split with ………………………. a component of Janata Party.
(a) Jan Sangh
(b) Lok Dal
(c) Independent party
(b) Janata Dal
Answer:
(a) Jan Sangh.

(c) Social support received by a political party is called ………………………. .
(a) Democracy
(b) People’s rule
(c) Mass base
(d) Public relation
Answer:
(c) Mass base

(d) Promotion of the Marathi language is the objective of the ………………………. .
(a) Dravida Munnetra Kazhagam
(b) Shiromani Akali Dal
(c) Shiv Sena
(d) Assam Gana Parishad
Answer:
(c) Shiv Sena

(e) There are ………………………. recognised parties in India.
(a) five
(b) seven
(c) nine
(d) six
Answer:
(b) seven

(f) The main objective of the political party is to capture ………………………. and retain it.
(a) publicity
(b) victory
(c) power
(d) information
Answer:
(c) power

(g) The ideology of the Communist Party of India is based on ………………………. Philosophy.
(a) Capitalist
(b) Marxist
(c) Democracy
(d) Imperialist
Answer:
(b) Marxist

(h) Shiromani Akali Dal is a prominent regional party in ………………………. state.
(a) Odisha
(b) Assam
(c) Punjab
(d) Jammu and Kashmir
Answer:
(c) Punjab

Question 6.
State whether the following statements are true or false. Give reasons for your answer:

(a) Political parties do not need support of the people.
Answer:
The above statement is False.

• To secure power is the main aim of political parties.
• Political parties have to implement various policies and programmes for the welfare of people and to gain the support of the people.
• The party which solves the issues related to the welfare of the people get support.
• Without the support of the people they can never get a mandate.

Hence, all political parties require support of the people.

(b) The Communist Party of India was split.
Answer:
The above statement is True.

• The Communist Party of- India was established in 1925. The Communist Parties follow Marxist ideology.
• In 1962, due to differences in the leadership a rift was created in the party.
• The main reason for the split was about which political line to follow.

(c) Sometimes the sentiments of regionalism gets strengthened in the country.
Answer:
The above statement is True.

1. A country like India has diversity of government language, region and culture. This diversity creates affinity towards a region.
2. This affinity gives rise to excessive regional S pride and initiating movements to save it. The significance of such movements is claimed to be for preserving culture.
3. Asserting to be rightful claimant of material resources, job opportunities and development of the region these movements strengthen the regional pride.
4. Thus, this excessive regional pride turns morbid taking the form of regionalism.

(d) Regional parties have no influence on National Politics.
Answer:
The above statement is False.

• No single party got majority in the elections to Lok Sabha held in 1989.
• The national parties formed coalition government at the centre taking help of regional parties.
• As regional parties share power at the centre they have influence on national politics.

Question 7.
Explain the concept:
(a) Regional Parties:
Answer:

1. The groups having a proud feeling about the different identity of their region and who compete to capture power to bring about regional development are ‘Regional Parties’.
2. The influence of such parties are limited to their region. They give importance to regional problems. Development of one’s own region is the prime objective of these parties.
3. Sometimes they demand for independent state or autonomy. They influence the national politics because of their dominance in the region.
4. They have passed through various phases, from separatism to autonomy and have then merged with mainstream national politics.

(b) Political Party:
Answer:

1. A Political Party is a kind of social organisation of like-minded people. The main aim of any political party is to contest election and secure power.
2. The stand taken by various political parties on issues becomes their political ideology.
3. In democracy, political parties are at the centre of all political activities. They are a link that connects common people, democracy, representation and elections.
4. The party that secures a majority becomes the ruling party and the rest are known as the. opposition.

Question 8.
Write short note:
(a) Party System:
Answer:

1. If one party remains in power in a country for a long period and no other party has any kind of influence the party system it is called ‘One- Party System’.
2. When there are two equally strong parties which alternatively come to power. It is called ‘Two-Party System’.
3. In a country where many political parties exist and compete with each other to capture power and have more or less influence in different states is called ‘Multi-Party System’.
4. Democracy develops in Multi-Party System.

(b) Indian National Congress:
Answer:

• The Indian National Congress was established in 1885.
• The party represented the hopes and aspirations of Indians and led the freedom struggle.
• Groups having diverse ideologies became part of the Congress to fight for freedom.
• It believes in democratic socialism, social equality and international peace.
• After independence, the Congress Party was a dominant party for almost four decades and ruled at the centre and in the states.

(c) Communist Party of India:
Answer:

• The Communist Party of India was established in 1925. The Communist Party follows the Marxist ideology.
• The party aims at working to protect the interests of workers and labourers.
• The main objective of this party is to oppose capitalism and protect the interests of workers.
• The party split due to ideological differences in 1964. Those who walked out of the party formed Communist Party (Marxist).

(d) Nationalist Congress Party:
Answer:

• Nationalist Congress Party (NCP) is the splinter or breakaway party from the Congress Party.
• It considers the values like democracy, equality and secularism important.
• It became part of United Progressive Alliance (UPA) and came to power in Maharashtra.
• It was in power in Maharashtra from 1999 to 2014 and at the centre from 2004 to 2014.

(e) Trinamool Congress:
Answer:

• All India Trinamool Congress was established in 1998.
• The Election Commission of India gave it recognition in 2016.
• Democracy, Secularism and protection of the weaker sections of the society are some of the policies of the party.
• Presently (April 2019), this party is in power in the state of West Bengal.

(f) Shiv Sena:
Answer:

1. The Shiv Sena was founded in 1966.
2. The main objective of the party was to fight for the cause of Marathi-speaking people and preservation of Marathi language.
3. They resisted the people from other regions.
4. In 1995, it became the ruling party with BJP in Maharashtra and its strategic tie-up with BJP helped it to enter national politics.
5. It was in power at the centre with BJP from 1998 to 2004. Again it has shared power with the BJP at the centre and state since 2014.

(g) Shiromani Akali Dal:
Answer:

• Shiromani Akali Dal was formed in 1920.
• It is a dominating party in Punjab.
• It gives preference to nurture religious and for the regional pride.
• The party has retained power in Punjab for many years.

(h) Dravid Munnetra Kazhagam:
Answer:

• Justice party, a non-Brahmin movement, started in 1920. In 1944, it transformed into a political party called Dravid Kazhagam.
• In 1949, a group within the party got split and formed Dravid Munnetra Kazhagam (DMK).
• The party works for the protection of Tamil identity.
• This party implemented many policies for welfare of the people and therefore has support from all the sections of society.
• Another group separated from it in 1972 and formed All India Dravid Munnetra Kazhagham.
• It remained in power for a long time in Tamil Nadu and for sometime at the centre.

Question 9.
Do as directed
(a) Complete the concept map:

Answer:

Answer:

(3)

Answer:

(b) Complete the Time-line writing the name of the party or the year of establishment:

Answer:

Question 10.
Answer the following questions in brief:

(a) What is called “mass base”?
Answer:

• Every political party is formed with some ideology.
• Political parties take a certain stand towards public issues.
• Their ideology and their stand towards public issues are conveyed to the people through various programmes.
• If the people agree with their role and ideology, they get the support of the people.
• Such social support received by a political – party is called ‘mass base’.

(b) Write about the policies of Indian National Congress.
Answer:
The policies of the Indian National: Congress are as follows:

• Equal rights for the minorities and weaker sections of society.
• Secularism, all-round development and social welfare are its objectives.
• Introduction of socialist democracy.
• Belief in values like international peace and social equality.

(c) Explain the role of Bharatiya Janata Party.
Answer:
Bharatiya Janata Party was formed in 1980 with the following objectives:

• To preserve Indian culture and traditions.
• To create a strong and rich India.
• To give importance to economic reforms for development of India.

(d) Write about the policies of Communist Party of India (Marxist).
Answer:
Communist Party of India (Marxist) was formed in 1964 with the following objectives:

• The party advocates socialism, secularism and democracy.
• The main policy of the party is to work for the welfare of farmers, landless labourers and workers.
• The party opposes imperialism.

(e) Why was Bahujan Samaj Party formed?
Answer:
The Bahujan Samaj Party was formed in 1984 to achieve the following objectives:

• To bring socialist ideology into practice.
• To secure power for ‘majority’ comprising of scheduled castes, scheduled tribes and other backward classes.
• To give power in the hands of ‘majority’.

(f) What do the regional parties give preference for?
Answer:
The regional parties give preference to the following issues:

• Importance should be given to regional issues and bring about development.
• Regional issues should be handled* at regional level.
• The power rests in the hands of regional people. ‘
• People living in that region should get preference in administration and jobs. .

(g) What changes are seen in the nature and role of regional parties?
Answer:
The role of regional parties changed over period of time. The changes are as follows:

• After independence, regional identity gave rise to separatists movements and demands were put up by regional parties for separate state.
• They wanted to break away from the Indian union and form a separate state.
• The demands of the regional parties changed gradually and these parties started asking for autonomy.
• In the later period, for the development of their region and people the parties demanded representation of their people in the state and central government.
• Regional parties in the north eastern region also started demanding more autonomy after split up.
• The journey of the regional parties began with a split up and then a demand for autonomy going on to share power in federation and then entry into mainstream national politics.

(h) What are the objectives of Assam Gana Parishad?
Answer:
Assam Gana Parishad was formed in 1985 with the following objectives:

• To preserve the uniqueness of Assam’s culture, language and social identity.
• To make efforts for economical development of Assam.
• To resolve problems of displaced people.

(i) What are the criteria for a regional party to get recognition?
Answer:

• In the last general .elections to the State Assembly, the party should not secure less than 6% of the total number of valid votes polled.
• At least 2 members to be elected to the Legislative Assembly (Vidhan Sabha).
• 3% seats of the total seats of Legislative Assembly or at least 3 seats in Legislative Assembly.

(j) What work is done by political parties?
Answer:
The following work is done by political parties:

• To propagate policies and programmes of the party and contest elections.
• They implement various policies when they come to power.
• If they fail to secure power, they act as opposition party and try to gain support of the people.
• They communicate the demands and complaints of the people to the government and the policies and programmes of the government to the people.

Question 11.
Give your opinion:

(a) Imagine you are opposition party leader and you have observed that the ruling party has not done good work in the field of health. What will you do as a leader of opposition?
Answer:
As opposition leader, I will do the following to make the government realise its negligence in the health field:

• I will question the government in Parliament and point out their shortcomings and ask about the prolonged projects.
• I will create awareness among the people by writing articles in newspapers.
• I will give interviews,on radio and television and criticize the government.
• I will organise rallies and meetings to sensitize among the people.

(b) Think and write
What has to be done to bring partyless democracy in modern times?
Answer:

• The recognition of all the parties should be cancelled. ‘
• People should be given the right to make decisions.
• In order to improve the decision-making power, people should be educated on political issues.
• To improve the moral character of the people who want to take the leadership.
• To initiate direct democracy and prepare the people for the same.
• Stress should be given for decentralization.
• Mahatma Gandhi, Vinoba Bhave and Jayprakash Narayan put forth the idea of partyless democracy.
• What has to be done to bring such democracy in modern times?

Question 12.
What do you understand by the following news in newspapers? Explain in brief.

(a) Opposition parties held a meeting in Mumbai to resist the Ruling party. Will take up the issue of farmers?
Answer:

• The ruling party did not undertake any work for the welfare of farmers. Therefore all the opposition parties decided to trap the government on this issue.
• The opposition parties have kept aside their differences and come together for this cause.
• It is the duty of the opposition party to question the negligence of the government.
• This news makes us aware of the line of action taken by the opposition.
• Opposition parties held a meeting in Mumbai to resist the Ruling party. Will take up the issue of farmers?
• Ruling party organised ‘Samvad Yatra’ in rural areas.

(b) Ruling party has organised ‘Samvad Yatra’ in rural areas.
Answer:

• Political parties adopt different ways to popularise their party and influence the people.
• As the means of communication are limited in rural areas they organise the Yatra.
• It shows the efforts of the government to understand problems of people in rural areas.
• It is essential to have dialogue with people to know their problems and opinions.
• As the initiative is taken by the government, people will develop faith in government.
• The news conveys to us that the ruling party is carrying out its duties properly.
• The faith in government becomes profound when people feel that government is sensitive towards them.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 6 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 6 Statistics.

Problem Set 6 Algebra 10th Std Maths Part 1 Answers Chapter 6 Statistics

10th Geometry Problem Set 6 Question 1.
Find the correct answer from the alternatives given.

i. The persons of O – blood group are 40%. The classification of persons based on blood groups is to be shown by a pie diagram. What should be the measures of angle for the persons of O – blood group?
(A) 114°
(B) 140°
(C) 104°
(D) 144°
Answer:
Measure of the central angle = \(\frac { 40 }{ 100 } \) × 360° = 144°
(D)

ii. Different expenditures incurred on the construction of a building were shown by a pie diagram. The expenditure of ₹ 45,000 on cement was shown by a sector of central angle of 75°. What was the total expenditure of the construction?
(A) 2,16,000
(B) 3,60,000
(C) 4,50,000
(D) 7,50,000
Answer:
Measure of the central angle = \(\frac{\text { Expenditure of cement }}{\text { Total expenditure }} \times 360^{\circ}\)

(A)

iii. Cumulative frequencies in a grouped frequency table are useful to find.
(A) Mean
(B) Median
(C) Mode
(D) All of these
Answer:
(B)

iv. The formula to find mean from a grouped frequency table is \(\overline{\mathrm{X}}=\mathrm{A}+\frac{\sum f_{i} u_{i}}{\sum f_{i}} \times g\)
in the formula u_i = _________.

Answer:

(C)

v.

The median of the distances covered per litre shown in the above data is in the group
(A) 12 – 14
(B) 14 – 16
(C) 16 – 18
(D) 18 – 20
Answer:
(C)

vi.

The above data is to be shown by a frequency polygon. The coordinates of the points to show number of students in the class 4 – 6 are.
(A) (4, 8)
(B) (3,5)
(C) (5,8)
(D) (8,4)
Answer:
Class mark = 5
Frequency = 8
∴ Co-ordinates of the point = (5, 8)
(C)

Statistics Problem Set 6 Question 2.
The following table shows the income of farmers in a grape season. Find the mean of their income.

Solution:

∴ The mean of the income of the farmers is ₹ 52,500.

Statistics Problem Set Question 3.
The loans sanctioned by a bank for construction of farm ponds are shown in the following table. Find the mean of the loans.

Solution:

∴ The mean of the loans given by the bank is ₹ 65,400.

Question 4.
The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.

Solution:

∴ The mean of the weekly wages of the workers is ₹ 4250.

Problem Set 6 Algebra Class 9 Question 5.
The following frequency distribution table shows the amount of aid given to 50 flood affected families. Find the mean of the amount of aid.

Solution:

∴ The mean of the amount of aid given to families is ₹ 72,400.
[Note: The above problems are solved using direct method. Students can solve these problems by using other method.]

Problem Set 6 Algebra Class 10 Question 6.
The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distances.

Solution:

Cumulative frequency which is just greater than (or equal) to 125 is 180.
∴ The median class is 220 – 230.
Now, L = 220, f = 80, cf = 100, h = 10

∴ The median of the distances is 223.13 km (approx.).

Algebra 10th Class Problem Set 6 Question 7.
The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.

Solution:

Cumulative frequency which is just greater than (or equal) to 200 is 240.
∴ The median class is 20 – 40.
Now,L = 20, f = 100,cf = 140, h = 20

∴ The median of the prices of different articles is ₹ 32.

10th Algebra Problem Set 6 Question 8.
The following frequency table shows the demand for a sweet and the number of customers. Find the mode of demand of sweet.

Solution:


∴ The mode of the demand of sweet is 397.06 grams.

Question 9.
Draw a histogram for the following frequency distribution.

Solution:

Question 10.
In a handloom factory different workers take different periods of time to weave a saree. The number of workers and their required periods are given below. Present the information by a frequency polygon.

Solution:

Problem Set 6 Question 11.
The time required for students to do a science experiment and the number of students is shown in the following grouped frequency distribution table. Show the information by a histogram and also by a frequency polygon.

Solution:

Question 12.
Draw a frequency polygon for the following grouped frequency distribution table.

Solution:

Question 13.
The following table shows the average rainfall in 150 towns. Show the information by a frequency polygon.

Solution:

Question 14.
Observe the given pie diagram. It shows the percentages of number of vehicles passing a signal in a town between 8 am and 10 am.
i. Find the central angle for each type of vehicle.
ii. If the number of two-wheelers is 1200, find the number of all vehicles.

Solution:

∴ The total number of vehicles is 3000.

Problem Set 6 Geometry Class 10 Question 15.
The following table shows causes of noise pollution. Show it by a pie diagram.

Solution:

Question 16.
A survey of students was made to know which game they like. The data obtained in the survey is
presented in the given pie diagram. If the total number of students are 1000,
i. how many students like cricket?
ii. how many students like football?
iii. how many students prefer other games?

Solution:

∴ 225 students like cricket.

ii. Central angle for football (θ) = 63°

∴ 175 students like football.

iii. Central angle for other games (θ) = 72°

∴ 200 students like other games.

Question 17.
Medical check up of 180 women was conducted in a health centre in a village. 50 of them were short of hemoglobin, 10 suffered from cataract and 25 had respiratory disorders. The remaining women were healthy. Show the information by a pie diagram.
Solution:
Total number of women = 180

Question 18.
On an environment day, students in a school planted 120 trees under plantation project. The information regarding the project is shown in the following table. Show it by a pie diagram.

Solution:
Total number of trees planted = 120

Maharashtra Board Class 10 Maths Solutions

Balbharti Maharashtra State Board Class 10 English Solutions Unit 1.4 All the World’s a Stage Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 English Solutions Unit 1.4 All the World’s a Stage

Maharashtra State Board Class 10 English Solutions Unit 1.4 Warming Up Questions and Answers

Question 1.
Pair work

• Talk to your friend about all the things related to ‘Seven.’
  For example Seven wonders of the world.
• Pair up with your partner and name those given below all of the under :

(a) The seven wonders of the world
………………………………………………………
(b) The seven continents
………………………………………………………
(c) The seven colours of the rainbow
………………………………………………………
(d) The seven notes of the music
………………………………………………………
(e) The seven seas of the world
………………………………………………………
Answer:
(a) The Seven wonders of the world : The Great Wall of China, Christ the Redeemer Statue, Machu Picchu, Chichen Itza, The Roman Colosseum, The Taj Mahal and Petra.
(b) The Seven continents : Asia, Europe, Australia, Africa, North America, South America, Antarctica.
(c) The Seven colours of the rainbow : violet, indigo, blue, green, yellow, orange, red
(d) The Seven notes of the musical scale : sa-re-ga-ma-pa-da-ni./doh-re-me-fa-so-la-ti
(e) The Seven seas of the world : Arctic Ocean, Antarctic Ocean, North Atlantic Ocean, South Atlantic Ocean, Indian Ocean, North Pacific Ocean, South Pacific Ocean.

Question 2.
Life is often compared to many things. Write down 7 things that life can be compared to and justify the comparison. For example,
(a) Life is a keyboard, because if you press the right keys you have typed a good destiny.
(b) ………………………………………………………
(c) ………………………………………………………
(d) ………………………………………………………
(e) ………………………………………………………
(f) ………………………………………………………
(g) ………………………………………………………
Answer:
(a) Life is a keyboard; if you press the right keys, you have typed out a good destiny.
(b) Life is a river-always flowing.
(c) Life is like a seed; it will never grow unless planted, nourished and nurtured.
(d) Life is like an elevator, with lots of ups and downs. ‘
(e) Life is like an onion. You peel off layer after layer. Sometimes it makes you weep.
(f) Life is like a jigsaw puzzle. You have all the pieces, but you have to put them together correctly.
(g) Life is like a jack-in-the-box. You never know : when you are going to get a box on your chin.

Question 3.
Match the approximate ages with the stages.

No.	Age-group			Stages
1	Birth to 2 years		a	teenage/adolescence
2	3 years to 12 years		b	old age/second childhood
3	13 years to 17 years		c	middle-age
4	18 years to about 44 years		d	babyhood/infancy
5	About 45 years to 60 years		e	senior citizen/elderly person
6	65 years up to 75 to 80 years		f	adulthood
7	Above 80 years		g	childhood

Answer:

No.	Age-group	Stages
(1)	Birth to 2 years	(d) babyhood/infancy
(2)	3 years to 12 years	(g) childhood
(3)	13 years to 17 years	(a) teenage/adolescence
(4)	18 years to about 44 years	(f) adulthood
(5)	About 45 years to 60 years	(c) middle-age
(6)	65 years up to 75 to 80 years	(e) senior citizen/ elderly person
(7)	Above 80 years	(b) old age/second childhood

Maharashtra Board Class 10 English Kumarbharati Unit 1.4 Questions and Answers

Question 1.
Read the words in given clouds. Match them with what they signify.

Answer:
(1) Stage – Life
(2) Characters – Roles played by human beings
(3) Script – Story of life
(4) Dialogues – Conversation
(5) Entry – Birth
(6) Exit – Death

Question 2.
Read the poem carefully and complete the following table.

Ages of Man	Role	Qualities/Actions
1		1. 2.
2		1. 2.
3		1. 2.
4		1. 2.
5		1. 2.
6		1. 2.
7		1. 2.

Answer:

First	infant	(1)   frightened (2)   crying, puking
Second	schoolboy	(1)   unhappy (2)   whining, creeping unwillingly to school
Third	lover	(1)   woeful (2)   sighing, singing sad ballads
Fourth	soldier	(1)  jealous in honor, ambitious (2)   quarreling, facing danger

Question 3.
Write down in your own words the differences between the following stages of a man’s life.
2nd stage and 4th stage …………………………………
……………………………………………………………………
……………………………………………………………………

3rd stage and 5th stage …………………………………
……………………………………………………………………
……………………………………………………………………

1 st stage and 7th (last) stage …………………………………
……………………………………………………………………
……………………………………………………………………
Answer:
2nd stage and 4th stage:
The second stage is that of a school boy, complaining and crying. He goes to school early in the morning, with a well-scrubbed, shining face. He carries his school bag and creeps slowly, like a snail, unwillingly to school.

‘Full of strange oaths and bearded like the pard, Jealous in honour, sudden and quick in quarrel.’

3rd stage and 5th stage:
The third stage is that of a lover, who is’ breathless and sighing with passion. He sings sad Songs dedicated to the beauty of his lover.

The fifth stage is that of , a mature middle-aged man with a round belly. He is stern and formal and full of wisdom. He is full of wise sayings and gives modern examples.

1 st stage and 7th (last) stage:
The first stage is that of an infant, crying weakly and throwing up in his nurse’s arms. He is unaware of what is happening around him.The baby, at this stage, is without teeth, without vision, without taste and without anything.

The last stage of all, which ends one’s eventful life, is when man becomes senile and enters his second childhood. He is again unaware of what is happening around him. This final stage is when he is once more without teeth, without vision, without taste and without anything.

Question 4.
Pick out lines that contain Imagery (a picture created in the mind by using words) of the following people.
(a) School boy …………………………………
(2nd stage) …………………………………
(b) Soldier …………………………………
(4th stage) …………………………………
(c) Judge …………………………………
(5th stage) …………………………………
(d) Senior citizen …………………………………
(6th stage) …………………………………
Answer:
(a) a snail
‘The whining schoolboy, with his satchel and shining morning face, creeping like a snail unwillingly to school.’

(b) acts like the pard
‘Full of strange oaths and bearded like the pard, Jealous in honour, sudden and quick in quarrel.’

(c) In fair round belly with good capon lined
With eyes “severe and beard of formal cut

(d) His youthful hose, well-saved, a world too wide
For his shrunk shank

Question 5.
You will notice that there is no Rhyme-scheme in the poem. It appears similar to the poem 1.1 ‘Where the Mind is Without Fear by Tagore.

However Tagore’s poem has no steady rhythm/meter either it is called Free Verse. Shakespeare uses lines with a steady rhythm of 5 beats in each. It is termed as Blank Verse. (No rhyme-scheme but uniformity in rhythm) Copy the lines from “Ánd all the men and women merely players” to “sudden and quick in quarrel”. Put a stress mark on each of the syllables stressed in the lines as for example, And all the men and women merely players;
Answer:
The poem is written in free verse without any rhyme scheme or consistent metre. There are many figures of speech e.g. Repetition. Metaphor, Alliteration, etc. An Important figure of speech is Personification. ‘Where tireless striving stretches its arms towards perfection’. Here, we can actually visualize ‘tireless striv1ng stretching Its arms to reach its goal.

The poem is a prayer to God. Tagore addresses God as ‘my Father’ and asks Him to awaken his country Into a heaven of freedom, where there is total freedom of good thoughts, good words and good actions. He wishes for a country where people would be free from fear, where knowledge would be free to all individuals and people from all castes and religions would be united.

Question 6.
Think and write on your own.
(a) What is the theme/central idea of this poem?
(b) Which two stages of man, described by Shakespeare sound humorous? Say why.
(c) The last (7th) stage of life sounds very sad and miserable. How can you make old age also cheerful and happy?
Answer:
(a) The theme of this poem is the cycle of life.
It tells us how one starts out an infant, helpless and unaware of the surroundings, and ends the same way, i without being aware of what is happening around one.

(b) The second and third stages are humorous. One can just imagine the school boy, complaining and whining, creeping to school slowly with a well-scrubbed and shining face. The third stage, where the lover sighs loudly and sings sad poems and songs, is also humorous.

(c) We can make the last stage joyful by preparing for it beforehand. We know that most of us will reach this stage before our final exit. Hence, we must maintain good health, and keep good relations with our family and the people around us. We must save enough money to get us through this stage without any stress on ourselves or others.

Question 7.
(A) The poem is entirely metaphorical. Pick out the comparisons from the poem.
(a) world …………………………………
(b) actors …………………………………
(c) birth and death …………………………………
(d) school boy …………………………………
(e) the lover’s sigh …………………………………
(f) spotted leopard …………………………………
(g) last stage (old age) …………………………………
Answer:
(a) stage
(b) –
(c) entrances and exits
(d) a snail
(e) a furnace
(f) bearded soldier

(B) Pick out from the poem two examples of each.
(a) Simile
(1) …………………………………
(2) …………………………………
Answer:
Simile. Two dissimilar objects are compared using the word ‘like’.

(b) Onomatopoeia
(1) …………………………………
(2) …………………………………

(c) Alliteration
(1) …………………………………
(2) …………………………………

(d) Metaphor
(1) …………………………………
(2) …………………………………

(e) Inversion
(1) …………………………………
(2) …………………………………

(f) Transferred Epithet
(1) …………………………………
(2) …………………………………
Answer:

(a) Simile	(1)   Creeping like a snail (2)   Soldier bearded like the pard
(b) Onomatopoeia	(1)   And then the whining schoolboy (2)   Sighing like a furnace
(c) Alliteration	(1) They have -their exits and their entrances (2) His youthful hose, well- saved, a world too wide
(d) Metaphor	(1)   They have their exits and their entrances (2)   men and women merely players
(e) Inversion	(1)   His acts being seven ages. (2)   With eyes severe and beard of formal cut
(f) Transferred Epithet	(1)   with a woeful ballad (2)   Into the lean and slippered pantaloon

Question 8.
Read the summary of the play ‘As You Like It’ by William Shakespeare using the Internet. Find out which character has narrated the above poem and on what occasion. Also, make a list of all the characters of the play.
Answer:
The character who has narrated the above poem in the play ‘As You Like It’ by William Shakespeare is Jacques. He narrates it in Act II, Scene VII.

Some of the other characters in the play are Celia, Rosalind, Orlando, Oliver, Duke Senior, Duke Frederick, Touchstone, etc.

Question 9.
Read the poem again and write an appreciation of the poem in a paragraph format. (Refer to page no. 5)
Answer:
Point Format
(for understanding)
The title of the poem : All the World’s a Stage’
The poet : WIlliam Shakespeare
Rhyme scheme : blank verse I.e. no rhyme scheme, but there is a steady rhythm of five beats In each line.
Figures of speech : Metaphor, Simile, Alliteration, Repetition. etc.
Theme/Central idea : The theme of the poem Is the cycle of life.

Paragraph Format
The poem ‘All the World’s a Stage’ is by William Shakespeare. It is taken from Shakespeare’s play ‘As you like It’. It is a monologue by one of the characters in the play.

The poem is written in blank verse i.e. there is no rhyme scheme, but there Is a steady rhythm of five beats i.e. iambic pentameter in each line. There are many figures of speech, like Simile. Alliteration and Repetition. but the one that stands out Is Metaphor. In the lines ‘All the world’s a stage, And all men and women are merely players’, there is an implied comparison between two different things.

In this poem, Shakespeare compares life to a stage. He has divided life Into seven stages. each having its own varied qualities and features. The theme of the poem is the cycle of life. It tells us how one starts out as an Infant, helpless. without understanding. and ends the same way, without being aware of what Is happening around one.

Question 10.
Complete the following diagram:
Answer:

Question 11.
Match the following :

A	B
(1) Infant	(a) acts like the pard
(2) School boy	(b) mewling and puking
(3) Lover	(c) whining,
(4) Soldier ‘	(d) sighing like furnace

Answer:

(1) Infant	(b) mewling arjd puking
(2) School boy	(c) whining
(3) Lover	(d) sighing like furnace
(4) Soldier	(a) acts like the pard

Question 12.
Complete the following : (The answers are given directly and underlined.)
Answer:
(1) ‘Exits’ and ‘entrances’ refer to deaths and births.
(2) Reputation is like a bubble because one does useless things for one’s reputation, which can burst like a bubble in one instant.

Question 13.
Name and explain the poetic devices used in the following lines:
(a) Soldier bearded like the pard
Answer:
Simile. Two dissimilar objects are compared using the word ‘like’,

(b) Sighing like a furnace
Answer:
Simile. Two dissimilar objects are compared using the word ‘like’.

(c) men and women merely players
Answer:
Metaphor. Implicit comparison between two different things.

(d) They have their exits and their entrances
Answer:
Alliteration. Repetition of the sound of ‘t’ and ‘e’.
Metaphor. Implicit comparison between two different things.

(f) With a woeful ballad
Answer:
Alliteration. Repetition of the sound of ‘w’.
Transferred Epithet. It is not the ballad that is woeful but the lover.

Question 14.
Complete the following diagram :
Answer:

Question 15.
Write if the following statements are True or False :
Answer:

• Man in the fifth stage of life is full of wisdom. (True)
• Man in the sixth stage has a fair, round belly. (False)
• The last few lines are full of melancholy. (True)
• In the last stage, man is unaware of his surroundings. (True)

Question 16.
Write down in your own words the differences between the following stages of a man’s life :
(a) 5th and 6th stage :
Answer:
The fifth stage is that of a mature middle-aged man with a round belly. He is stern and formal and full of wisdom. He is full of wise sayings and gives modern examples.

In the sixth stage, man has become old, with thin legs in slippers and loose pants. He has spectacles on his nose and a pouch by his side. The close-fitting stockings, which he had saved from his younger days, are now too big for his thin legs, which have shrunk with age. His loud voice, which was once manly, has now become childish and shrill. There are whistling sounds when he talks.

Question 17.
Complete the following table based on the extract. (The answers are given directly.)
Answer:

Stages of Man	Role	qualities/Actions
Fifth	adult man	(1)   fat, serious and wise (2)   giving advice
Sixth	middle-aged man	(1)   thin and shrunken (2)   talking in a shrill-voice
Seventh	very old man	(1)   senile, child-like (2)   oblivious to his surroundings

Question 18.
The poem is entirely metaphorical. Pick out the comparison from the extract: last stage (old age)
Answer:
last scene

Question 19.
Name and explain the figures of speech used in the following lines :
(a) His youthful hose, well-saved, a world too wide
Answer:
Alliteration. Repetition of the sound of ‘w’.

(b) For his shrunk shank
Answer:
Alliteration. Repetition of the sound of ‘s’.

(c) Turning again towards childish treble
Answer:
Alliteration. Repetition of the sound of ‘t’.

(d) Sans teeth, sans eyes, sans taste, sans everything.
Answer:
Repetition. The word ‘sans’ is repeated for emphasis.

Maharashtra State Board Class 10 Social Science Book Solutions

Maharashtra State Board Class 10 History Solutions Answers

• Chapter 1 Historiography Development in the West
• Chapter 2 Historiography Indian Tradition
• Chapter 3 Applied History
• Chapter 4 History of Indian Arts
• Chapter 5 Mass Media and History
• Chapter 6 Entertainment and History
• Chapter 7 Sports and History
• Chapter 8 Tourism and History
• Chapter 9 Heritage Management

Maharashtra State Board Class 10 Political Science Solutions Answers

• Chapter 1 Working of the Constitution
• Chapter 2 The Electoral Process
• Chapter 3 Political Parties
• Chapter 4 Social and Political Movements
• Chapter 5 Challenges faced by Indian Democracy

Maharashtra State Board Class 10 Geography Solutions Answers

• Chapter 1 Field Visit
• Chapter 2 Location and Extent
• Chapter 3 Physiography and Drainage
• Chapter 4 Climate
• Chapter 5 Natural Vegetation and Wildlife
• Chapter 6 Population
• Chapter 7 Human Settlements
• Chapter 8 Economy and Occupations
• Chapter 9 Tourism, Transport and Communication

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 7 Lenses Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

Question 1.
Match the columns in the following table and explain them:

Column 1	Column 2	Column 3
Farsightedness	Nearby object can be seen clearly	Bifocal  lens
Presbyopia	Faraway object can be seen clearly	Concave  lens
Nearsightedness	Problem of old age	Convex  lens

Answer:

Column 1	Column 2	Column 3
Farsightedness	Faraway object can be seen clearly	Convex  lens
Presbyopia	Problem of old age	Bifocal  lens
Nearsightedness	Nearby object can be seen clearly	Concave  lens

1. Farsightedness:
Hypermetropia or farsightedness is the defect of vision in which a human eye can see distant objects clearly but is unable to see nearby objects clearly.
In this case the image of a nearby object would fall behind the retina instead of on the retina.

Possible reasons for hypermetropia:
(1) Curvature of the cornea and the eye lens decreases. Hence, the converging power of the eye lens becomes less.
(2) The distance between the eye lens and retina decreases (relative to the normal eye) and the focal length of the eye lens becomes very large due to the flattening of the eyeball.

2. Presbyopia:
Presbyopia is the defect of vision in which aged people find it difficult to see nearby objects
comfortably and clearly without spectacles.

Reason of presbyopia: The power of accommodation of eye usually decreases with ageing. The muscles near the eye lens lose their ability to change the focal length of the lens.

Therefore, the near point of the eye lens shifts rarther from the eye, This defect is corrected using a. convex lens of appropriate power. The lens converges light rays before they fall on the eye lens such that the action of the eye lens forms the image on the retina.

3. Nearsightedness:
Myopia or nearsightedness is the defect of vision in which a human eye can see nearby objects distinctly but is unable to see distant objects clearly as they appear indistinct.
In this case the image of a distant object is formed in front of the retina instead of on the retina.
[Figs. 7.29 (a), 7.29 (b)]

Possible reasons for myopia:
(1) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large. (2) The distance between the eye lens and the retina increases as the eyeball elongates.

Myopia is corrected using a suitable concave lens. Light rays are diverged by the concave lens before they strike the eye lens. A concave lens of proper focal length is chosen to produce the required divergence. Hence, after the converging action of the eye lens, the image is formed on the retina. [Fig. 7.29 (c)]

Question 2.
Draw a figure explaining various terms related to a lens.
Answer:
(1) Centre of curvature (C): The centres of the spheres whose parts form the surfaces of a lens are called the centres of curvature of the lens. A lens has two centres of curvature C₁, and C₂ for its two spherical surfaces.

(2) Radii of curvature (R₁, R₂): The radii of the spheres whose parts form surfaces of a lens are called the radii of curvature of the lens.

(3) Principal axis: The imaginary straight line passing through the two centres of curvature of a lens is called the principal axis of the lens.

(4) Optical centre (O): The point inside a lens on the principal axis, through which light rays pass without changing their path is called the optical centre (O) of the lens.

(5) Principal focus (F): When light rays parallel to the principal- axis are incident on a convex lens, they converge at a point on the principal axis. This point is called the principal focus (F) of the convex lens. Light rays travelling parallel to the principal axis of a concave lens diverge after refraction in such a way that they appear to be coming out of a point on the principal axis. This point is called the principal focus of the concave lens. A lens has two principal foci F₁ and F₂.
[Note: In this chapter, the terms focus and the principal focus are used in the same sense.]

(6) Focal length (f): The distance between the optical centre and the principal focus of a lens is called the focal length (f) of the lens.

C₁, C₂: Centres of curvature, R₁, R₂: Radii of curvature, O: Optical centre.
The cross sections of convex and concave lenses are shown in parts (a) and (b) of Fig. 7.4. The surface marked as 1 is part of sphere S₁ while the surface marked as 2 is part of sphere S₂.

P₁, P₂, P₃: Incident rays of light,
Q₁, Q₂, Q₃: Refracted rays of light, O: Optical centre

F₁, F₂: Principal foci of the lens, f: Focal length of the lens

Question 3.
At which position will you keep an object in front of a convex lens so as to get a real image of the same size as the object? Draw a figure.
Answer:
At 2F₁.

Question 4.
Give scientific reasons:
a. A simple microscope is used for watch repairs.
Answer:
(1) when an object is placed within the focal length of a magnifying glass or simple microscope (convex lens), its larger and erect image is obtained on the same side of the lens as that of the object.

(2) By adjusting the distance between the object and the lens, the image can be obtained at the minimum distance of distinct vision. Thus, a watch repairer can see the minute parts of a watch more clearly with the aid of a magnifying glass (a simple microscope) than with the naked eye, without any stress on the eye. Hence, watch repairers use a magnifying glass (a simple microscope) while repairing the watches.

b. One can sense colours only in bright light.
Answer:
(1) The retina in the eye is made of many light sensitive cells. The rod-shaped cells respond to the intensity of light while the cone-shaped cells j respond to various colours.
(2) The cone-shaped cells do not respond to faint light. They function only in bright light. Hence, one can sense colours only in bright, light.

c. We cannot clearly see an object kept at a distance less than 25 cm from the eye.
Answer:
(1) When we try to see a nearby object, the eye lens becomes more rounded and its focal length decreases. Then a clear image of the object is formed on the retina of the eye.
(2) The focal length of the eye lens cannot be decreased beyond some limit. Therefore we cannot clearly see an object kept at a distance less than 25 cm from the eye.

Question 5.
Explain the working of an astronomical telescope using refraction of light.
Answer:
Construction of a refracting telescope: It consists of two convex lenses called the objective lens (directed towards the object) and the eyepiece (directed towards the eye). The focal length and diameter of the objective lens are respectively greater than the focal length and diameter of the eyepiece. The objective lens is fitted at one end of a long metal tube.

A metal tube of smaller diameter is fitted in this metal tube and the eyepiece is fitted at the outer end of the smaller tube. With the help of a screw it is possible to change the distance between the eyepiece and the objective lens by sliding the tube fitted with the eyepiece. The principal axes of the objective lens and the eyepiece are along the same line. A telescope is usually mounted on a stand.

Working: When the objective lens is pointed towards the distant object to be observed, the rays of light from the distant object, which are almost parallel to each other, pass through the objective lens. The objective lens collects maximum amount of light as it is large in size. It forms a real, inverted and diminished image in the focal plane of the objective lens. Now, the position of the eyepiece is adjusted such that this image falls just within the focal length of the eyepiece and serves as the object for the eyepiece which works as a simple microscope.

The final image is highly magnified, virtual, on the same side as that of the object and inverted with respect to the original object. The final image can be observed by keeping the eye close to the eyepiece. If the image formed by the objective lens lies in the focal plane of the eyepiece, the final image is formed at infinity.

Question 6.
Distinguish between the following:
a. Farsightedness (Hypermetropia) and Nearsightedness (Myopia).
Answer:
Farsightedness:

1. In hypermetropia, a human eye can see distant distinctly but is unable to see nearby objects clearly.
2. In this case, the image of a nearby object would be formed behind the retina.
3. This defect can be corrected using a convex lens of appropriate power.

Nearsightedness:

1. In myopia, a human eye can see near objects distinctly, but is unable to see distant objects clearly.
2. In this case, the image of a distant object is formed in front of the retina.
3. This defect can be corrected using a concave lens of appropriate power.

b. Concave lens and Convex lens.
Answer:
Concave lens:

1. A concave lens has its surfaces curved inwards.
2. It is thicker at the edges than in the middle.
3. It can form only a virtual image.
4. It can form only a diminished image.

Convex lens:

1. A convex lens has its surfaces puffed up outwards.
2. It is thicker in the middle than at the edges.
3. It can form a real image as well as a % virtual image.
4. It can form a magnified, diminished or the same sized image (relative to the object) depending on the position of the object.

Question 7.
What is the function of the iris and the muscles connected to the lens in the human eye?
Answer:
When the incident light is very bright, the muscles of the iris stretch to reduce the size of the pupil. When the incident light is dim, the muscles of the iris relax to increase the size of the pupil. Thus, the iris controls the size of the pupil and thereby regulates the amount of light entering the eye. (Fig. 7.26)

When a distant object is to be observed, the ciliary muscles relax so that the eye lens becomes flat. This increases the focal length of the lens. Therefore, a sharp image of the distant object is formed on the retina.
Thus, we can see a distant object clearly. (Fig. 7.27)

when an object closer to the eye is to be observed. the ciliary muscles contract increasing the curvature. of the eye lens. The eye lens, therefore, becomes rounded. This decreases the focal length of the lens. Therefore, a sharp image of the nearby object is formed on the retina. Thus, we can see a nearby object clearly. (Fig. 7.27)

Question 8.
Solve the following examples:
i. Doctor has prescribed a lens having I power + 1.5 D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?
Solution:
Data: P = + 1.5 D, f = ?
Focal length of the lens, f = \(\frac{1}{P}=\frac{1}{1.5 \mathrm{D}}\)
= \(\frac{10}{15}\) m = 0.6667 m = 0.67 m
P is positive. This shows that the lens is convex. The defect of vision is farsightedness (hypermetropia).

ii. 5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.
Solution:
Data: Converging lens, f = 10 cm
u = – 25 cm, h₁ = 5 cm, v = ?, h₂ = ?


The height of the image = -3.3 cm (inverted image ∴ minus sign).
(iii) The image is real, inverted and smaller than the object.

iii. Three lenses having powers 2, 2.5 and 1.7 D are kept touching in a row. What is the total power of the lens combination?
Solution:
Data: P₁ = 2 D, P₂ = 2.5 D, P₃ = 1.7 D, P = ?
Total power of the lens combination,
P = P₁ +P₂ + P₃
= 2 D + 2.5 D + 1.7 D
= 6.2 D.

iv. An object kept 60 cm from a lens gives a virtual image 20 cm in front of the lens. What is the focal length of the lens? Is it a converging lens or diverging lens?
Solution:
Data: u = -60 cm, v = -20 cm, f = ?

∴ The focal length of the lens, f = – 30 cm. As f is negative, it is a diverging lens.

Project:

Question 1.
Make a Powerpoint presentation about the construction and use of binoculars. (Do it your self)

Can you recall? (Text Book Page No. 80)

Question 1.
Indicate the following terms related to spherical mirrors in figure 7.1: pole, centre of curvature, radius of curvature, principal focus.

Answer:

Question 2.
How are concave and convex mirrors constructed?
Answer:
The given part of a hollow spherical glass can be converted into a concave mirror by (i) polishing (silvering) its inner side (inner surface or concave surface) to make it reflecting or (ii) coating its outer side with a thin layer of silver and painting it with red colour to protect the silver coating.
[Note: Case (i) corresponds to the front surface silvered concave mirror.]

The given part of a hollow spherical glass can be converted into a convex mirror by (i) polishing (silvering) its outer side (outer surface or convex surface) to make it reflecting or (ii) coating its inner side with a thin layer of silver and painting it with red colour to protect the silver coating.
[Note: Case (i) corresponds to the front surface silvered convex mirror. ]

Use your brain power! (Text Book Page No. 85)

Question 1.
From equations (1) and (2) what is the relation between h₁, h₂, u and v?
Answer:
M = \(\frac{h_{2}}{h_{1}}\) …….(1)
Also, M = \(\frac{v(\text { image distance })}{u \text { (object distance) }}\)……(2)
\(\frac{h_{2}}{h_{1}}\) = \(\frac{v}{u}\)

Try this (Text Book Page No. 88)

(1) Try to read a book keeping it very far from your eyes.
(2) Try to read a book keeping it very close to your eyes.
(3) Try to read a book keeping it at a distance of 25 cm from your eyes.
At which time do you see the alphabets clearly? Why?
Answer:
Minimum distance of distinct vision: Though the focal length of the eye lens is adjustable, it cannot be decreased below a certain limit. Hence, if an object is very close to the eye, it cannot be seen clearly. For a normal human eye, the minimum distance from the eye at which an object is clearly visible without stress on the eye, is called the minimum distance of distinct vision. For the normal human eye, it is 25 cm.

Use your brain power! (Text Book Page No. 89)

Question.
(1) Why do we have to bring a small object near the eyes in order to see it clearly?
(2) If we bring an object closer than 25 cm from the eyes, when can we not see it clearly even though it subtends a bigger angle at the eye?
Answer:
(1) when a small object is brought near the eyes, its apparent size increases. Therefore, it is
seen clearly.

(2) Minimum distance of distinct vision: Though the focal length of the eye lens is adjustable, it cannot be decreased below a certain limit. Hence, if an object is very close to the eye, it cannot be seen clearly. For a normal human eye, the minimum distance from the eye at which an object is clearly visible without stress on the eye, is called the minimum distance of distinct vision. For the normal human eye, it is 25 cm.

Try this (Text Book Page No. 91)

Question 1.
Take a burning incense stick in your hand and rotate it fast along a circle.
Answer:
A circle of red light is seen.

Question 2.
Draw a cage on one side of a cardboard and a bird on the other side. Hang the cardboard with the help of a thread. Twist the thread and leave It. What do you see and why?
Answer:
The bird appears to be inside the cage. This happens due to persistence of vision.
Persistence of vision: We see an object when its image is formed on the retina. The image disappears when the object is removed from our sight. But this is not instantaneous and the image remains imprinted on the retina for about \(\frac{1}{16}\) th of a second after the removal of the object. The sensation on the retina persists for a while. This effect is known as the persistence of vision.

It is due to persistence of vision that we continue to see the object in its position for about \(\frac{1}{16}\) th of a second after it is removed.
Example: When a burning stick of incense is moved fast in a circle, a circle of red light is seen.

Can you tell? (Text Book Page No. 91)

Question 1.
How do we perceive different colours?
Answer:
(1) In nature we field objects of various colours. Perception of colour means to be able to respond to colour.
(2) We can distinguish between various colours due to perception of colour.
(3) The cone-shaped cells on the retina of the eye respond to the various colours when light is bright and communicate to the brain about the colours of the image formed on the retina. This gives us the proper idea about the colours of the object.
(4) If, in the retina of a person, the cone-shaped cells responding to certain specific colours are absent, the person is unable to distinguish between the colours. As a result, he lacks perception of colour.

Fill in the blanks and rewrite the statements:

Question 1.
The focal length of a………..lens is positive.
Answer:
The focal length of a convex lens is positive.

Question 2.
The focal length of a………..lens is negative.
Answer:
The focal length of a concave lens is negative.

Question 3.
The magnification produced by a………..lens is always positive.
Answer:
The magnification produced by a concave lens is always positive.

Question 4.
The power of a………..lens is positive.
Answer:
The power of a convex lens is positive.

Question 5.
The power of a………..lens is negative.
Answer:
The power of a concave lens is negative.

Question 6.
The focal length of a lens with power 2.5 D is………..
Answer:
The focal length of a lens with power 2.5 D is 40 cm (0.4 m).

Question 7.
The power of a lens with focal length 20 cm is………..
Answer:
The power of a lens with focal length 20 cm is 5D.

Question 8.
The minimum distance of distinct vision for a normal human eye is………..
Answer:
The minimum distance of distinct vision for a normal human eye is 25 cm.

Question 9.
If two lenses with focal lengths 10 cm and 20 cm respectively are kept in contact with each other, the effective power of the combination is………..
Answer:
If two lenses with focal lengths 10 cm and 20 cm respectively are kept in contact with each other, the effective power of the combination is 15 D.

Question 10.
A………..lens is used as a simple microscope.
Answer:
A convex lens is used as a simple microscope.

Rewrite the following statements by selecting the correct options:

Question 1.
Inside water, an air bubble behaves………..
(a) like a flat plate
(b) like a concave lens
(c) like a convex lens
(d) like a concave mirror
Answer:
Inside water, an air bubble behaves like a concave lens.

Question 2.
…………represents the lens formula.

Answer:
(b) \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) represents the lens formula.

Question 3.
The power of a convex lens of focal length 25 cm is………..
(a) +4.0 D
(b) 0.25 D
(c) -4.0 D
(d) -0.4D
Answer:
The power of a convex lens of focal length 25 cm is +4.0 D

Question 4.
A lens does not produce any deviation of a ray of light passing through………..
(a) it’s centre of curvature
(b) it’s optical centre
(c) it’s principal focus
(d) an axial point at a distance 2F from its centre
Answer:
A lens does not produce any deviation of a ray of light passing through its optical centre.

Question 5.
The image formed by a concave lens is always………..
(a) virtual and erect
(b) real and erect
(c) virtual and inverted
(d) real and inverted
Answer:
The image formed by a concave lens is always virtual and erect.

Question 6.
A convex lens forms a virtual image of an object placed………..
(a) at infinity
(b) at a distance 2F from the lens
(c) at a distance F from the lens
(d) between the principal focus and the optical centre of the lens.
Answer:
A convex lens forms a virtual image of an object placed between the principal focus and the optical centre of the lens.

Question 7.
When an object is placed at 2F₁ of a convex lens, its image is formed………..
(a) at F₁
(b) at 2F₂
(c) beyond 2F₂
(d) on the same side as the object
Answer:
When an object is placed at 2F₁ of a convex lens, its image is formed at 2F₂

Question 8.
To obtain an image of the same size as that of an object with the help of a convex lens, the object should be placed………..
(a) at infinity
(b) beyond F₁
(c) between F₁ and 2F₁
(d) at 2F₁
Answer:
To obtain an image of the same size as that of an object with the help of a convex lens, the object should be placed at 2F₁

Question 9.
When an object is placed between O and F₁ in front of a convex lens, the image formed is ………..
(a) enlarged and erect
(b) diminished and erect
(c) real and enlarged
(d) diminished and inverted
Answer:
When an object is placed between O and F₁ in front of a convex lens, the image formed is enlarged and erect.

Question 10.
When an object is placed at any finite distance from a concave lens, the image is formed ………..
(a) between F₁ and 2F₁
(b) beyond 2F₁
(c) at F₁
(d) between F₁ and O on the same side as the object.
Answer:
When an object is placed at any finite distance from a concave lens, the image is formed between F₁ and O on the same side as the object.

Question 11.
A student obtained a clear image of window grills on the screen. But the teacher told him to get the image of a tree far away, instead of the window. To get a clear image, the lens must be ……….. (Practice Activity Sheet – 2)
(a) moved towards the screen
(b) moved away from the screen
(c) moved behind the screen
(d) moved far away from the screen
Answer:
A student obtained a clear image of window grills on the screen. But the teacher told him to get the image of a tree far away, instead of the window. To get a clear image, the lens must be moved towards the screen.

Question 12.
The image obtained while finding the focal length of a convex lens is ……….. (Practice Activity Sheet – 3)
(a) real and erect
(b) virtual and erect
(c) real and inverted
(d) virtual and inverted
Answer:
The image obtained while finding the focal length of a convex lens is real and inverted.

Question 13.
Yash found out f₁ and f₂ of a symmetric convex lens experimentally. Then which of the following conclusions is true? (Practice Activity Sheet – 4)
(a) f₁ = f₂
(b) f₁ > f₂
(c) f₁ < f₂
(d) f₁ ≠ f₂
Answer:
(a) f₁ = f₂

State whether the following statements are true or false: (If a statement is false, correct it and rewrite it.)

Group (A)

Question 1.
Power of a lens, P = \(\frac{1}{f}\).
Answer:
True.

Question 2.
If the power of a lens is 2 D, its focal length = 0.5 m.
Answer:
True.

Question 3.
A concave lens is a converging lens. (March 2019)
Answer:
False. (A concave lens is a diverging lens.)

Question 4.
A convex lens is a diverging lens.
Answer:
False. (A convex lens is a converging lens.)

Question 5.
A concave lens always forms a virtual image.
Answer:
True.

Question 6.
A convex lens always forms a virtual image.
Answer:
False. (A convex lens forms a real image or a virtual image depending on the object distance.)

Question 7.
Due to the light sensitive cells in the eye, we get information about the brightness or dimness of the object and the colour of the object.
Answer:
True.

Question 8.
The focal length of a concave lens is negative.
Answer:
True.

Question 9.
The magnification produced by a concave lens is positive or negative depending on the object distance.
Answer:
False. (The magnification produced by a concave lens is always positive.)

Question 10.
The magnification produced by a convex lens is positive or negative depending on the object distance.
Answer:
True.

Question 11.
A concave lens is used as a magnifying glass.
Answer:
False. (A convex lens is used as a magnifying glass.)

Question 12.
A convex lens is used as a simple microscope.
Answer:
True.

Question 13.
A concave lens is used to correct myopia.
Answer:
True.

Question 14.
A convex lens is used to correct hypermetropia.
Answer:
True.

Group (B)

Question 1.
When red light falls on the eyes, the cells responding to red light get excited more than those responding to other colours and we get the sensation of red colour.
Answer:
True.

Question 2.
When an object is placed in front of a concave lens, its image is obtained on the opposite side of the object.
Answer:
False. (When an object is kept in front of a concave lens, its image is obtained on the same side of the lens as the object.)

Question 3.
The image formed by a concave lens is always virtual.
Answer:
True.

Question 4.
The principal focus of a convex lens is virtual.
Answer:
False. (The principal focus of a convex lens is real.)

Question 5.
An object of height 2 cm forms an image of height 3 cm when placed in front of a concave lens.
Answer:
False. (An object of height 2 cm forms an image of height less than 2 cm when placed in front of a concave lens.)

Question 6.
Absence of rod like cells results in colour blindness.
Answer:
False. (Absence of conical cells results in colour-blindness.)

Question 7.
Nearsightedness can be corrected using spectacles having convex lenses.
Answer:
False. (Nearsightedness can be corrected using spectacles having concave lenses.)

Question 8.
Farsightedness can be corrected using spectacles having convex lenses of suitable focal length.
Answer:
True.

Question 9.
As one grows old, ciliary muscles become weak.
Answer:
True.

Question 10.
In a simple microscope, the object is placed within the focal length of the convex lens.
Answer:
True.

Question 11.
A compound microscope forms an erect and real image of a small object.
Answer:
False. (A compound microscope forms an inverted and virtual image of a small object.)

Question 12.
In a compound microscope, a real image acts as an object for the eyepiece.
Answer:
True.

Question 13.
In television, we see a continuous picture due to persistence of vision.
Answer:
True.

Question 14.
The conical cells can respond differently to red, green and blue colours.
Answer:
True.

Question 15.
The rod like cells respond to colours and communicate the presence of colours in the retinal image of the brain.
Answer:
False. (The rod like cells respond to the intensity of light and communicate the degree of brightness and darkness, to the brain.)

Question 16.
The conical cells respond to the intensity of light and communicate the degree of brightness and darkness to the brain.
Answer:
False. (The conical cells respond to colours and communicate the presence of colours in the retinal image to the brain.)

Question 17.
Generally, using the same objective lens, but different eyepieces, different magnification can be obtained.
Answer:
True.

Find the odd one out and give the reason:

Question 1.
Simple microscope, Compound microscope, Telescope, Myopia.
Answer:
Myopia. It is a defect of vision; others are instruments.

Question 2.
Myopia, Presbyopia, Hypermetropia, Spectrometer.
Answer:
Spectrometer. It is an instrument; others are defects of vision.

Question 3.
Presbyopia, Retina, Nearsightedness, Farsightedness.
Answer:
Retina. It is a part of the eye; others are defects of vision.

Question 4.
Compound microscope, Kaleidoscope, Simple microscope, Astronomical telescope.
Answer:
Kaleidoscope. Others are optical instruments.

Question 5.
TV, Motion picture, Complete circle formed by a revolving burning incense stick, Colour blindness.
Answer:
Colour-blindness. Others are examples of persistence of vision.

Question 6.
Planets, Stars, Satellites, Rainbow.
Answer:
Rainbow. Others are celestial bodies.

Considering the correlation between the words of the first pair, pair the third word accordingly with proper answer:

Question 1.
Nearsightedness: Elongated eyeball :: Farsightedness:………
Answer:
Flattened eyeball

Question 2.
Convex lens : Converging :: Concave lens :………..
Answer:
Diverging

Question 3.
Object at 2F₁ of a convex lens : Image at 2F₂ :: Object at F₁ :………..
Answer:
Image on the opposite side at infinity

Question 4.
Magnification positive : Erect image :: Magnification negative :………..
Answer:
Inverted image

Question 5.
Convex lens : Positive power of the lens :: Concave lens:
Answer:
Negative power of the lens.

Question 6.
\(\frac{1}{f(\text { in metre })}\) : Power of the lens (in dioptre) :: \(\frac{\text { Image distance }}{\text { Object distance }}\)
Answer:
Magnification.

Question 7.
Focal length : Metre :: Power of a lens :………..
Answer:
Dioptre.

Question 8.
Iris : Pupil :: Ciliary muscles :……….
Answer:
Eye lens

Question 9.
Nearsightedness : Concave lens :: Farsightedness :………..
Answer:
Convex lens

Question 10.
Nearsightedness : Image in front of the retina :: Farsightedness :………..
Answer:
Image behind the retina

Question 11.
Observation of stars and planets : Telescope :: Repairing a watch :……….
Answer:
Simple microscope

Question 12.
Cinema : Persistence of vision :: Rainbow :………
Answer:
Refraction, dispersion and internal reflection of light.

Match the following:

Question 1.

Column A	Column B
(1) Conical cells	(a) Intensity of light
(2) Rod like cells	(b) Colour of an image
(3) Pupil	(c) Iris
(4) Cornea	(d) Aperture
	(e) Transparent

Answer:
(1) Conical cells – Colour of an image
(2) Rod like cells – Intensity of light
(3) Pupil – Aperture
(4) Cornea – Transparent.

Question 2.

Column A	Column B
(1) Magnification	(a) \(\frac{1}{f}\)
(2) Power of a lens	(b) \(\frac{h_{2}}{h_{1}}\)
(3) Focal length	(c) f
(4) Distance of an object from a lens	(d) u
	(e) \(\frac{h_{1}}{h_{2}}\)

Answer:
(1) Magnification: \(\frac{h_{2}}{h_{1}}\)
(2) Power of a lens: \(\frac{1}{f}\)
(3) Focal length: f
(4) Distance of an object from a lens: u.

Question 3.

Answer:
(1) Lens: \(\frac{1}{f}: \frac{1}{v}-\frac{1}{u}\)
(2) Magnification: \(\frac{h_{2}}{h_{1}}\)
(3) Refractive index: \(\frac{\sin i}{\sin r}\)

Question 4.

Column A	Column B
(Convex lens)	(a) Image virtual, erect and enlarged
(1) Object at 2F1	(b) Image real, inverted and of the same size
(2) Object between F1 and 2F1	(c) Image real, inverted and highly diminished
(3) Object between O and F1	(d) Image real, inverted and highly enlarged
(4) Object at infinity	(e) Image real, inverted and enlarged

Answer:
(1) Object at 2F₁ – Image real, inverted and of the same size
(2) Object between F₁ and 2F₁ – Image real, inverted and enlarged
(3) Object between O and F₁ – Image virtual, erect and enlarged
(4) Object at infinity – Image real, inverted and highly diminished

Question 5.

Column A	Column B
(1) Nearsightedness	(a) Ciliary muscles become weak
(2) Farsightedness	(b) Image in front of the retina
(3) Presbyopia	(c) Colour-blindness
	(d) Image behind the retina

Answer:
(1) Nearsightedness – Image in front of the retina
(2) Farsightedness – Image behind the retina
(3) Presbyopia – Ciliary muscles become weak

Question 6.

Column A	Column B
(1) Convex lens	(a) To see small objects clearly
(2) Astronomical telescope	(b) To observe minute objects
(3) Compound microscope	(c) To observe astronomical objects such as stars, planets, etc.
(4) Simple microscope	(d) Presbyopia
	(e) Power of a lens

Answer:
(1) Convex lens – Presbyopia
(2) Astronomical telescope – To observe astronomical objects such as stars, planets, etc.
(3) Compound microscope – To observe minute objects
(4) Simple microscope – To see small objects clearly.

Question 7.

Column A	Column B
(1) Persistence of vision	(a) Lenses and mirrors are used
(2) Reflecting telescope	(b) To see objects far away from us
(3) Telescope	(c) Motion picture
(4) Compound microscope	(d) To observe blood
	(e) Convex lens

Answer:
(1) Persistence of vision – Motion picture
(2) Reflecting telescope – Lenses and mirrors are used
(3) Telescope – To see objects far away from us
(4) Compound microscope – To observe blood corpuscles.

Name the following:

Question 1.
Name the lens which forms a real image or a virtual image depending on the position of the object.
Answer:
A convex lens.

Question 2.
Name the lens which produces magnification always less than 1.
Answer:
A concave lens.

Question 3.
Name the lens which always forms an image virtual and smaller than the object.
Answer:
A concave lens.

Question 4.
Name the lens used to obtain the image on a screen.
Answer:
A convex lens.

Question 5.
Name the lens for which the image always lies between the object and the lens.
Answer:
A concave lens.

Question 6.
Name the instrument used to observe bacteria.
Answer:
A compound microscope.

Question 7.
Name the instrument used to observe planets.
Answer:
An astronomical telescope.

Answer the following questions in one sentence each:

Question 1.
An object is placed at 60 cm from a convex lens of focal length 20 cm. State the nature and size of the image relative to that of the object.
Answer:
The image is real, inverted and smaller than the object.

Question 2.
If an object is placed at 50 cm from a convex lens of focal length 25 cm, what will be the image distance?
Answer:
The image distance will be.50 cm.

Question 3.
An object is placed at 40 cm from a convex lens of focal length 20 cm. State the nature and the size of the image relative to that of the object.
Answer:
The image is real, inverted and of the same size as that or the object.

Question 4.
An object is placed at 30 cm from a convex lens of focal length 20 cm. State the nature and the size of the image relative to that of the object.
Answer:
The image is real, inverted and larger than the object.

Question 5.
An object is placed at 15 cm from a convex lens of focal length 25 cm. State the nature and size of the image relative to that of the object.
Answer:
The image is virtual, erect and larger than the object.

Question 6.
State the type of lens that can be used to burn paper in sunlight at noon.
Answer:
A convex lens can be used to burn paper in sunlight at noon.

Question 7.
State the type of lens used to correct myopia.
Answer:
A concave lens is used to correct myopia.

Question 8.
State the type of lens used to correct hypermetropia.
Answer:
A convex lens is used to correct hypermetropia.

Question 9.
If two lenses with focal lengths 10 cm and – 20 cm respectively are kept in contact with each other, what will be the effective power of the combination of the lenses?
Answer:
The effective power of the combination of the lenses will be + 5 D.

Question 10.
If two lenses with focal lengths – 10 cm and 40 cm respectively are kept in contact with each other, what can you say about the behaviour of the combination of the lenses?
Answer:
The combination of the lenses will behave as a concave lens.

Answer the following questions:

Question 1.
What is a lens?
Answer:
A lens is a transparent material bound by two surfaces, out of which at least one surface is spherical.
[Note: A lens is normally made of glass or plastic.]

Question 2.
In which instruments have you seen a lens?
Answer:
We have seen a lens in a microscope and a telescope.

Question 3.
How is a lens different from a mirror?
Answer:
A mirror has one reflecting surface. By reflection of light, it forms an image of the object placed in front of it. A mirror is not transparent. A lens has two surfaces that form an image by refraction of light. A lens is transparent.

Question 4.
Make a list of optical devices you know.
Answer:
Microscope, telescope, binoculars, camera, projector.

Question 5.
Do you know which is the natural optical device?
Answer:
Yes. The eye is the natural optical device.

Question 6.
What is a convex lens?
Answer:
A lens having both spherical surfaces puffed up outwards is called a convex lens or double convex lens or biconvex lens. It is thicker in the middle than at the edges.
[Note: A convex lens is also called a converging lens.]

Question 7.
What is a concave lens?
Answer:
A lens having both spherical surfaces curved inwards is called a concave lens or double concave lens or biconcave lens. It is thicker at the edges than in the middle.
[Note: A concave lens is also called a diverging lens.]

Question 8.
Draw neat labelled diagrams: Types of lenses.
Answer:

[Note: Positive meniscus behaves as a convex lens as it is thicker in the middle than at the edges. Negative meniscus behaves as a concave lens as it is thicker at the edges than in the middle.]

Question 9.
In general, when a ray of light passes through a lens, there occurs a change in its direction of propagation. Why?
Answer:
The working of a lens is similar to that of a triangular prism. When a ray of light passes through a lens, it is refracted twice: When entering the lens and when emerging from the lens. There is a change 5 in its direction of propagation every time and as both the changes occur in the same sense, the direction of propagation of the emergent ray is different from that of the incident ray.

Question 10.
State the rules used for drawing ray diagrams for the formation of an image by a convex lens.
Answer:
Rules used for drawing ray diagrams for the formation of an image by a convex lens:
(1) When the incident ray is parallel to the principal axis, the refracted ray passes through the principal focus.

(2) When the incident ray passes through the principal focus, the refracted ray is parallel to the principal axis.

(3) When the incident ray passes through the optical centre of the lens, it passes without changing its direction.

Question 11.
In the case of a convex lens, show the path of the refracted ray when the incident ray of light (1) is parallel to the principal axis of the lens (2) passes through the focus of the lens (3) passes through the optical centre of the lens.
Answer:
1.

2.

3.

Question 12.
Draw neat and well labelled ray diagrams for image formation by convex lens when an object is (1) at infinity (2) beyond 2F₁ (3) at 2F₁ (4) between F₁ and 2F₁ (5) at focus F₁ (6) between focus F₁ and optical centre O. Also, in each case, state the position, nature and size of the image relative to that of the object.
Answer:
(1) Object at infinity:

In this case, the image is formed at focus F₂ of the convex lens. It is real, inverted and highly diminished (point-sized).

(2) Object beyond 2F₁:

In this case, the image is formed between F₂ and 2F₂. It is real, inverted and diminished.

(3) Object at 2F₁:

In this case, the image is formed at 2F₂. It is real, inverted and of the same size as that of the object.

(4) Object between F₁ and 2F₁:

In this case, the image is formed beyond 2F₂. It is real, inverted and magnified (enlarged).

(5) Object at focus F₁:

In this case, the image is formed at infinity. It is real, inverted and infinitely large (highly magnified).

(6) Object between focus F₁ and optical centre O:

In this case, the image is formed on the same side of the lens as the object. It is virtual, erect and larger than the object.
[Note: Here, the image is virtual. Hence, it is shown by a dotted line.]

Question 13.
Observe the following figure and complete the table: (March 2019)

Answer:

Point	Answer
(i) Position of the object	Between F1 and O
(ii) Position of the image	On the same side of the lens as the object
(iii) Size of the image	Very large
(iv) Nature of the image	Virtual and erect

Question 14.
At which position will you keep an object in front of a convex lens to get a real image smaller than the object? Draw a figure.
Answer:
The object should be placed beyond 2F₁.

Question 15.
State the rules used for drawing ray diagrams for the formation of an image by a concave lens.
Answer:

1. When the incident ray is parallel to the principal axis, the refracted ray, when extended backwards, passes through the principal focus.
2. When the incident ray is directed towards the principal focus F₂, the refracted ray is parallel to the principal axis.
3. When the incident ray passes through the optical centre of the lens, it passes without changing its direction.

Question 16.
State the characteristics of an image formed by a concave lens.
Answer:
The image formed by a concave lens is always virtual, erect and smaller than the object. It is on the same side of the lens as the object. Generally, it is formed between the optical centre of the lens and the principal focus F₁. If the object is at infinity, the image is a point image formed at F₁.

Question 17.
In the case of image formation by a concave lens, what can you say about the position, nature and size of the image relative to the size of the object?
Answer:
Image formation by a concave lens :
(1) If the object is at infinity, the image is formed at the focus of the lens, on the same side of the lens as the object. It is virtual, erect and much smaller than the object (point image).

(2) If the object is at any finite distance from the lens, the image is formed on the same side of the lens as the object and between the focus and the optical centre of the lens. It is virtual, erect and smaller than the object. The image distance is less than the object distance.

Question 18.
Draw a ray diagram to show image formation by a concave lens.
Answer:

PQ : Object
P’Q’ : Image (virtual, therefore shown by a dotted line),
O : Optical centre,
F₁ : Principal focus,
f : Focal length of the lens
[Note: If in a Board examination, incomplete diagram (as shown below) is given, students should complete it and label its parts as shown in Figure.]

Question 19.
State the Cartesian sign convention for refraction of light (image formation) by a lens.
Answer:
Cartesian sign convention for refraction of light (image formation) by a lens:
In this case, the optical centre (O) of the lens is taken as the origin and the principal axis of the lens is taken as X-axis of the coordinate system.

(1) The object is always placed at the left of the lens.
All distances parallel to the principal axis are measured from the optical centre of the lens.
(2) All distances measured to the right of the origin are taken as positive while distances measured to the left of the origin are taken as negative.
(3) Distances measured perpendicular to and above the principal axis are taken as positive.
(4) Distances measured perpendicular to and below the principal axis are taken as negative.
(5) The focal length of a convex lens is positive and that of a concave lens is negative. (Fig. 7.21)

Question 20.
(i) What is a lens formula? (ii) State it.
Answer:
(i) The relationship between the object distance (u), image distance (v) and focal length (J) of a lens is called the lens formula.
(ii) It is \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
[Note: The lens formula holds good for all values of u and v and is applicable to a convex lens as well as a concave lens. The sign convention for u, v and f must tie used in solving numerical examples.]

Question 21.
What is meant by the magnification produced by a lens? State the formulae for it.
Answer:
The magnification (M) produced by a lens is the ratio of the height of the image (h₂) to the height of the object (h₁).
M = \(\frac{h_{2}}{h_{1}}\)……….(1)
Also M = \(\frac{v(\text { image distance })}{u \text { (object distance) }}\)……………(2)

Question 22.
When is the magnification produced by a lens (1) positive (2) negative?
Answer:
The magnification produced by a lens is
(1) Positive when the image is virtual (as it is erect)
(2) Negative when the image is real (as it is inverted).

Question 23.
Express the magnification produced by a lens in terms of the focal length of the lens and (1) the object distance (2) the image distance.
Answer:
Magnification (M) produced by a lens

where f is the focal length of the lens:

(2) From eq. (2), we have

Question 24.
An object is kept in front of a lens of focal length + 10 cm. Describe the nature of the image in the following cases: (1) The object” distance is 25 cm. (2) The object distance is 5 cm.
Answer:
Since, the focal length of the lens ( +10 cm) is positive, it is a convex lens.
(1) If an object is kept at 25 cm from the lens, the image will be real, inverted and smaller than the object.
(2) If an object is kept at 5 cm from the lens, the image will be virtual, erect and larger than the object.

Question 25.
Anu and Anand have concave and convex lenses respectively. They took lenses in sunlight and tried to burn two pieces of paper of equal areas and temperature. State which lens will burn the paper. Give the reason. Explain with the help of a diagram, why the other paper did not burn.
Answer:
(1) The convex lens will burn the paper. See Fig. 7.22 for reference. The ray of sunlight will converge at the principal focus of the lens. Hence, if the paper is held at the focus, it will burn due to concentration of heat energy.

(2) The paper held in front of the concave lens, will not burn. For reference, see Fig. 7.23. The concave lens will diverge the rays of sunlight falling on it. Hence, the paper will not burn.

Question 26.
To obtain a magnified real image of a small film strip, which type of lens is used? Where is the film strip placed to obtain the image on the screen?
Answer:
To obtain a magnified real image of a small film strip, a convex lens is used. The film strip is placed between F₁ and 2F₁ and the screen is placed on the other side of the lens.

Question 27.
When an object of height 2 cm is placed in front of a convex lens, the height of the image is found to be 3 cm. State the nature and position of the image giving reason.
Answer:
When an object is placed between the optical centre and the principal focus of a convex lens, the image formed by the lens is virtual and larger than the object. When an object is placed between F₁ and 2F₁ the image formed by the lens is real and larger than the object.

In the above case, if the image is virtual, it will be erect and on the same side of the lens as that of the object. If the image is real, it will be inverted and beyond 2F₂ on the other side of the lens with respect to the object.

Question 28.
You are given a lens which gives a virtual, erect and enlarged image. What type of lens is it?
Answer:
Since the lens gives a virtual, erect and enlarged image, it must be a convex lens.

Question 29.
When an object of height 3 cm is placed in front of a concave lens, the height of the image is found to be 6 cm. State, giving the reason, whether the given statement is true or false.
Answer:
When an object is placed in front of a concave lens, the image formed by the lens is always smaller than the object. In the statement given in the question, the height of the image is reported as greater than that of the object. Hence, the statement given in the question is false.

Question 30.
State two uses of a concave lens.
Answer:

1. A concave lens is used to correct myopia (nearsightedness).
2. In some optical instruments, a combination of a concave lens and a convex lens is used.

Question 31.
State two uses of a convex lens.
Answer:
A convex lens is used (1) to read words in small print (2) to correct hypermetropia (Far-sightedness).

Question 32.
An object is kept in front of a lens of j focal length – 20 cm. Describe the nature of the image when the object distance is 25 cm.
Answer:
Since the focal length of the lens (- 20 cm) is negative, it is a concave lens.
If an object is kept at 25 cm from the lens, the image will be virtual, erect and smaller than the object.
[Note: The nature of the image is independent of the object distance as it is a concave lens.]

Question 33.
An object is placed in front of a convex lens of focal length 20 cm. If the object distance is changed from 60 cm to 40 cm, what can you say about the size of the image relative to that of the object?
Answer:
In this case, the focal length (f) of the lens is 20 cm.
∴ 2f = 40 cm.
When the object distance is 60 cm (which is greater than 2f), the image will be smaller than the object. When the object distance becomes 40 cm (which is equal to 2f), the image will be of the same size as that of the object.

Question 34.
What is the power of a lens?
Answer:
The capacity of a lens to converge or diverge incident rays is called its power. The power (P) of a lens is the inverse of the focal length (f) of the lens.
P = \(\frac{1}{f}\)

Question 35.
What is the unit of power of a lens? Define it.
Answer:
The unit of power of a lens is the dioptre (D).
One dioptre is the power of a lens whose focal length is one metre.
1 dioptre (D) = \(\frac{1}{1 \text { metre }(\mathrm{m})}\)
[Note: The dioptre, the SI unit of power of a lens, is denoted by D.]

Question 36.
What is the sign of the power of (i) a convex lens (ii) a concave lens?
Answer:
The power of a convex lens is positive while that of a concave lens is negative.

Question 37.
If there is an increase or decrease in the focal length of a lens, what will be the effect on the power of the lens?
Answer:
The power of a lens is the inverse of its focal length. Hence, if there is an increase in the focal length of a lens, the power of the lens will decrease accordingly. Similarly, if there is a decrease in the focal length of a lens, the power of the lens will increase accordingly.

Question 38.
If two lenses of focal lengths f₁ and f₂ are kept in contact with each other, state the formula for the focal length of the combination. If P₁ and P₂ are the powers of these lenses, state the formula for the power of the combination.
Answer:
If two lenses of focal lengths f₁ and f₂ are kept in contact with each other, the focal length (f) of the combination is given by \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

[Two lenses kept in contact with each other]
If P₁ and P₂ are the powers of these lenses, the power (P) of the combination is given by P = P₁ + P₂.
[Note: The figures are given only for reference.]

Question 39.
Draw a neat labelled diagram to show the structure of the human eye.
Answer:

Question 40.
What is cornea?
Answer:
The cornea is a thin and transparent cover (membrane) on the human eye through which light enters the eye. Maximum refraction of light rays entering the eye occurs at the cornea.

Question 41.
What is iris?
Answer:
The iris is a dark fleshy screen (muscular diaphragm) behind the cornea in the human eye. Its colours are different for different people.

Question 42.
What is pupil?
Answer:
The pupil is a small circular opening of changing diameter at the centre of the iris in the human eye.

Question 43.
What is the use of the pupil in the human eye?
Answer:
The pupil in the human eye is useful for controlling and regulating the amount of light entering the eye. The pupil contracts in the presence of too much light and dilates when light is insufficient, thus changing the amount of light entering the eye.

Question 44.
With reference to the functioning of the pupil in the human eye, what is adaptation?
Answer:
The tendency of the pupil in the human eye to adjust the opening for light, depending on the intensity of incident light, to control and regulate the amount of light entering the eye is called adaptation.

Question 45.
What is the shape and the size of the human eyeball?
Answer:
The human eyeball is approximately spherical in shape with a diameter of about 2.4 cm.

Question 46.
Name the part of the human eye that forms a transparent bulge on the surface of the eyeball.
Answer:
The cornea forms a transparent bulge on the surface of the eyeball.

Question 47.
Which part of the human eye is located just behind the pupil?
Answer:
A transparent biconvex crystalline lens is located just behind the pupil in the human eye.

Question 48.
What is retina?
Answer:
The retina is a light sensitive screen consisting of a delicate membrane with a large number of light sensitive cells.

Question 49.
What is the nature of the eye lens and what does the eye lens do?
Answer:
The eye lens is a double convex transparent crystalline lens, just behind the pupil. The eye lens provides small adjustment of focal length to form a real and inverted sharp image on the retina.

Question 50.
What happens when light falls on the retina?
Answer:
When light falls on the retina, light sensitive cells of the retina are activated. They generate electrical signals which are passed by optic nerves to the brain. The brain interprets the signals and processes the information such that we perceive the object as it is.

Question 51.
What are ciliary muscles?
Answer:
The muscles which hold the eye lens in its position, and bring about changes in the shape (curvature) of the eye lens, and hence of focal length are known as ciliary muscles.

Question 52.
What is the focal length of the eye lens of a normal eye in relaxed position of eye muscles?
Answer:
The focal length of the eye lens of a normal eye in relaxed position of eye muscles is about 2 cm.

Question 53.
Where does the second focal point of the eye lens of a normal eye in relaxed position of eye muscles lie?
Answer:
The second focal point of the eye lens of a normal eye in relaxed position of eye muscles lies on the retina.

Question 54.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye lens to adjust its focal length is called the power of accommodation of the eye.

Question 55.
Explain the term power of accommoda¬tion of the eye.
(OR)
Write a short note on the power of accommodation of the eye.
Answer:
Power of accommodation of the eye: The eye lens is held in its position by the ciliary muscles. When we look at a nearby object, the ciliary muscles compress the eye lens so that it becomes rounded. Hence, the focal length of the eye lens decreases. Therefore, the image is formed on the retina of the eye and hence the nearby object is seen clearly.

When we look at a distant object, the ciliary muscles relax so that the eye lens becomes flat. Hence, the focal length of the eye lens increases. Therefore, the image is formed on the retina of the eye and hence the distant object is seen clearly. This ability of the eye lens to adjust its focal length is called the power of accommodation of the eye.

Question 56.
What is meant by accommodation? How is it brought about?
Answer:
The process of focusing the eye on objects at different distances is called accommodation. It is brought about by changing the curvature of the f elastic eye lens making it thinner or thicker.

Question 57.
The human eye is very similar to a photographic camera. The figure given shows the main parts of a photographic camera. Now answer the following questions:

(1) Name the parts of the human eye similar to the following parts of the photographic camera :
(a) Photographic film (b) Aperture.
(2) State one difference between the human eye lens and camera lens.
(3) Name the muscles which adjust the curvature of the eye lens.
(4) Which phenomenon of light is responsible for the working of the eye?
Answer:
(1) (a) The retina in the human eye is similar to the photographic film in a camera.
(b) The pupil in the human eye is similar to the aperture in a camera.
(2) In a photographic camera, the focal length of the lens changes when the position of the lens is changed. In the human eye, the focal length of the eye lens is changed by the ciliary muscles and the distance of the image from the eye lens is fixed.
(3) The ciliary muscles adjust the curvature of the eye lens.
(4) The refraction of light is responsible for the working of the eye.

Question 58.
Have you seen a photographic camera in which a film is used? Compare the human eye with it. State similarities between them. State the points of difference between them.
Answer:
Yes, we have seen a photographic camera in which a film is used.
Cameras, in general, have various shapes and sizes. Some cameras are much bigger than the human eye while some are smaller than the human eye. Here we shall consider a simple camera.

Similarities: In the case of a camera as well as the human eye, it is possible to control the amount of incoming light with the help of a diaphragm and an aperture. Both use a convex lens for focusing. The photographic film in a camera is coated with a photosensitive material. The retina in the eye consists of a large number of light sensitive cells. The photographic film in a camera is processed using chemicals and then prints (photographs) can be obtained using the appropriate paper.

In the human eye, the electrical signals generated by light sensitive cells are passed by optic nerves to the brain which interprets them.

Differences: Cameras come in a variety of sizes and shapes unlike the human eye. Unlike the human eye, a wide variation in exposure time is possible in the case of cameras. The human eye is sensitive in the visible region (red to violet) of the electromagnetic spectrum, while a much wider range of the electromagnetic spectrum can be covered with cameras designed for specific J purposes. In comparison with the human eye, a wider view and range can be covered by a camera.

In comparison with the human eye, a wider intensity (of light) range can be covered with a camera. The retina is indispensable in the human eye, while cameras without a photographic film have been designed with the help of photosensitive materials and are in current use.

[Note: With advances in technology, improved cameras are designed all the time, and the list of differences between the human eye and a camera in general would be practically endless.]

Question 59.
What is meant by the minimum distance of distinct vision?
Answer:
The minimum distance from the normal eye, at which an object is clearly visible without stress on the eye is called the minimum distance of distinct vision.

Question 60.
Explain the term minimum distance of distinct vision.
(OR) Write a short note on distance of distinct vision.
Answer:
Minimum distance of distinct vision; Though the focal length of the eye lens is adjustable, it cannot be decreased below a certain limit. Hence, if an object is very close to the eye, it cannot be seen clearly. For a normal human eye, the minimum distance from the eye at which an object is clearly visible without stress on the eye, is called the minimum distance of distinct vision. For the normal human eye, it is 25 cm.

Question 61.
State four reasons related to problems of vision.
Answer:
Problems of vision are related to (i) weakening of ciliary muscles (ii) change in the size of the eyeball (iii) irregularities on the surface of cornea (iv) formation of a membrane over the eye lens.

Question 62.
What is myopia or nearsightedness? What are the possible reasons of myopia? How is myopia corrected? Explain with diagrams.
Answer:
Myopia or nearsightedness is the defect of vision in which a human eye can see nearby objects distinctly but is unable to see distant objects clearly as they appear indistinct.
In this case the image of a distant object is formed in front of the retina instead of on the retina. [Figs. 7.29 (a), 7.29 (b)]

Possible reasons of myopia: (1) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large. (2) The distance between the eye lens and the retina increases as the eyeball elongates.

Myopia is corrected using a suitable concave lens. Light rays are diverged by the concave lens before they strike the eye lens. A concave lens of proper focal length is chosen to produce the required divergence. Hence, after the converging action of the eye lens, the image is formed on the retina. [Fig. 7.29 (c)]

Question 63.
Observe the following diagram and answer the questions.
(a) Which eye defect is shown in this diagram?
(b) What are the possible reasons for this eye defect?
(c) How is this defect corrected? Write it in brief.

Answer:
(a) Myopia or Nearsightedness

(b) Possible reasons for the defect:
(i) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large.
(ii) The eyeball elongates so that the distance between the lens and the retina increases.

(c) Correction of the defect: This defect can be corrected using spectacles with concave lenses.
A concave lens diverges the incident rays and these diverged rays can be converged by the lens in the eye to form an image on the retina.

Question 64.
What is the sign of the power of the lens used to correct myopia?
Answer:
The power of the lens used to correct myopia is negative.
[Note: It is a concave lens. Negative focal length Negative power.]

Question 65.
In a Std. X class, out of 40 students, 10 students use spectacles, 2 students have ( positive power and 8 students have negative power of lenses in their spectacles.
Answer the following questions:
(1) What does the negative power indicate?
(2) What does the positive power indicate?
(3) Generally which type of spectacles do most of the students use?
(4) What defect of eyesight do most of the students suffer from?
(5) Give two possible reasons for the above defect.
Answer:
(1) The negative power indicates a concave lens or myopia.
(2) The positive power indicates a convex lens or hypermetropia.
(3) Generally, most of the students use spectacles with concave lenses.
(4) Most of the students suffer from myopia.
(5) Two possible reasons for myopia:

1. The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large.
2. The distance between the eye lens and the retina increases as the eyeball elongates.

Question 66.
What is hypermetropia or farsightedness? What are the possible reasons of hypermetropia? How is hypermetropia corrected? Explain with figures.
Answer:
Hypermetropia or farsightedness is the defect of vision in which a human eye can see distant objects clearly but is unable to see nearby objects clearly.
In this case the image of a nearby object would fall behind the retina instead of on the retina.
[Figs. 7.31 (a), 7.31 (b)]

Possible reasons of hypermetropia:
(1) Curvature of the cornea and the eye lens decreases. Hence, the converging power of the eye lens becomes less. (2) The distance between the eye lens and retina decreases (relative to the normal eye) and the focal length of the eye lens becomes very large due to the flattening of the eyeball.

Hypermetropia is corrected using a suitable convex lens. Light rays are converged by the convex lens before they strike the eye lens. A convex lens of proper focal length is chosen to produce the required convergence. Hence, after the converging action of the eye lens, the image is formed on the retina. [Fig. 7.31 (c)]

Question 67.
What is the sign of the power of the lens used to correct hypermetropia?
Answer:
The power of the lens used to correct hypermetropia is positive.
[Note: It is a convex lens. Positive focal length ∴ Positive power.]

Question 68.
Given below is a diagram showing a defect of human eye.

Study it and answer the following questions:
(1) Name the defect shown in the figure. (Practice Activity Sheet – 3)
(2) Give two possible reasons for this defect of eye in human beings.
(3) Name the type of lens used to correct the eye defect.
(4) Draw a labelled diagram to show how the defect is rectified by using the lens.
Answer:
A lens having both spherical surfaces puffed up outwards is called a convex lens or double convex lens or biconvex lens. It is thicker in the middle than at the edges.
[Note: A convex lens is also called a converging lens.]

Question 69.
Observe the following figures and complete the table. (Practice Activity Sheet – 1)
Answer:

Question 70.
What is presbyopia? State the reason for this defect. How is presbyopia corrected?
Answer:
Presbyopia is the defect of vision in which aged people find it difficult to see nearby objects comfortably and clearly without spectacles.

Reason of presbyopia: The power of accommodation of eye usually decreases with ageing. The muscles near the eye lens lose their ability to change the focal length of the lens.
Therefore, the near point of the eye lens shifts farther from the eye.

This defect is corrected using a convex lens of appropriate power. The lens converges light rays before they fall on the eye lens such that the action of the eye lens forms the image on the retina.

Question 71.
What is a bifocal lens?
Answer:
A bifocal lens is a lens of which the upper part is a concave lens to correct myopia and the lower part is a convex lens to correct hypermetropia.

[Note: A person suffering from myopia as well as hypermetropia, uses a bifocal lens. Nowadays, the defects of vision such as myopia and hypermetropia can be corrected using contact lenses or by laser surgery.]

Question 72.
(A) Anil cannot see the blackboard writing clearly, but he can see nearby objects clearly.
(i) What is the eye defect he is suffering from?
(ii) How is it corrected?
(B) Anil’s uncle cannot see nearby objects clearly, but he can see distant objects clearly.
(i) What is the eye defect he is suffering from?
(ii) How is it corrected?
Answer:
(A) Anil cannot see the blackboard writing clearly, but he can see nearby objects clearly.
(i) This defect is called myopia (nearsightedness).
(ii) It is corrected using spectacles having concave lenses of appropriate power.

(B) Anil’s uncle cannot see nearby objects clearly, but he can see distant objects clearly.
(i) This defect is called hypermetropia (farsightedness).
(ii) It is corrected using, spectacles having convex lenses of appropriate power.

Question 73.
When are bifocal lenses used in spectacles?
Answer:
When a person cannot see nearby objects as well as distant objects clearly, bifocal lenses are used in spectacles.

Question 74.
Aniket from Std. X uses spectacles. The power of the lenses in his spectacles is -0.5 D. Answer the following questions:
(1) State the type of’ lenses used in his spectacles.
(2) Name the defect of vision Aniket is suffering from.
(3) Find the focal length of the lenses used in his spectacles.
Answer:
(1) Concave lenses are used in the spectacles used by Aniket.
(2) Aniket is suffering from myopia (near-sightedness).
(3) Focal length of the lenses used in his spectacles

= -2 m (Concave lens ∴ Minus sign)

Question 75.
Sunita from Std. X uses spectacles. Her spectacle number is -1.5 D. Answer the following questions:
(1) Name the defect of eye from which she is suffering.
(2) What type of lens is she using?
(3) Find the focal length of the lens.
Answer:
(1) Myopia.
(2) Concave.

(Concave lens ∴ minus sign)
This is the focal length of the lens.

Question 76.
Surabhi from Std. X uses spectacles. The power of the lenses in her spectacles is 0.5 D. Answer the following questions from the given information: (March 2019)
(i) Identify the type of lenses used in her spectacles.
(ii) Identify the defect of vision Surabhi is suffering from.
(iii) Find the focal length of the lenses used in her spectacles.
Answer:
(i) Convex lenses are used in the spectacles used by Surabhi.
(ii) Surabhi is suffering from hypermetropia (farsightedness).
(iii) Focal length of the lenses used in her spectacles

Question 77.
My grandfather uses a bifocal lens in his spectacles. Explain why.
Answer:
In old age, people usually suffer from both myopia and hypermetropia. Therefore, they need spectacles having bifocal lenses.

The upper part of a bifocal lens is a concave lens to correct myopia. The lower part of a bifocal lens is a convex lens to correct hypermetropia.

Question 78.
State uses of concave lens.
Answer:

1. Concave lenses are used for proper working of medical equipment, scanner, CD player – the instruments that employ laser rays.
2. One or more concave lenses are used in a small safety device, fitted in the peep hole in a door, due to which we can see a large area outside the door.
3. Concave lenses are used in spectacles to correct nearsightedness (myopia).
4. A concave lens is used to spread light emitted by the small bulb in a torch over a wide area.
5. A concave lens is used in front of the eyepiece or inside the eyepiece fitted in a camera, telescope and microscope – the instruments employing convex lenses.

Question 79.
State uses of a convex lens.
Answer:
Convex lenses are used in a simple microscope, compound microscope, refracting telescope, camera, projector, spectroscope, spectacles for correcting farsightedness (hypermetropia) and binoculars.

Question 80.
What is meant by the apparent size of an object? With a neat and labelled diagram, explain the relation between the apparent size of an object and the angle subtended by the object at the eye.
Answer:
An object appears small or big depending upon the size of its image formed on the retina of the eye. The size of an object as perceived by the eye is called the apparent size of the object. Consider two objects of the same size, one held near the eye and the other away from the eye as shown in the following figure (Fig. 7.34). The nearby object (PQ) appears larger than the distant object (P₁Q₁).

Also, the angle a subtended by the nearby object at the eye is larger than the angle β subtended by the distant object at the eye. This shows that the apparent size of an object depends upon the angle subtended by the object at the eye. The greater the angle subtended by the object at the eye, the greater is the apparent size of the object. Similarly, the smaller the angle subtended by the object at the eye, the smaller is the apparent size of the object.

Question 81.
With a neat labelled diagram, explain the working of a simple microscope. State uses of a simple microscope.
(OR)
What does a simple microscope consist of? What is the order of magnification obtained by a simple microscope? What is a simple microscope used for?
Answer:
A simple microscope consists of a convex lens of short focal length, usually fixed in a suitable frame with a handle or mounted on a stand.

The object is placed in front of the convex lens of short focal length such that the object distance is less than the focal length. The image is virtual and larger than the object. It is formed on the same side of the lens as the object.

A maximum magnification of about 20 can be obtained by a simple microscope. A simple microscope is used by watch repairers to observe small parts of a watch and by jewellers to examine ornaments. A simple microscope (also called a magnifying glass) is also used to read words in small print.

Question 82.
With a neat labelled diagram, explain the construction and working of a compound microscope.
Answer:
Construction of a compound microscope:

(1) A compound microscope consists of a metal tube fitted with two convex lenses at the two ends. These lenses are called the objective lens (the lens directed towards the object) and the eyepiece (the lens directed towards the eye). Both the lenses are small in size, but the cross section of the objective lens is less than that of the eyepiece. The objective lens has a short focal length. The focal length of the eyepiece is more than that of the objective lens.

(2) The metal tube is mounted on a stand. The principal axes of the objective lens and the eyepiece are along the same line. The distance between the object and the objective lens can be changed with a screw. It is possible to change the distance between the objective lens and the eyepiece.

Working :
(1) The object to be observed is illuminated and placed in front of the objective lens, slightly beyond the focal length of the objective lens. Its real, inverted and enlarged image is formed by the objective lens on the other side.

(2) This intermediate image lies within the focal length of the eyepiece. It serves as an object for the eyepiece. The eyepiece works as a simple microscope. The final image is virtual, highly enlarged and inverted with respect to the original object. It can be formed at the minimum distance of distinct vision from the eyepiece. The final image is observed by keeping the eye close to the eyepiece.

Question 83.
State two uses of a compound microscope.
Answer:
Uses of a compound microscope:

1. It is used to observe blood corpuscles, plant and animals cells, microorganisms like bacteria, etc.
2. It is used in a pathological laboratory to observe blood, urine, etc.
3. It is a part of a travelling microscope used for measurement of very small distance.

Question 84.
What will happen if in a compound microscope, the objective lens is large in size and has a focal length?
Answer:
If the objective lens of a compound microscope is large in size, in addition to the light coming from an object, other unwanted light will be incident on the objective lens. Hence, the image will not be seen clearly. If the objective lens has a large focal length, the magnification produced by it will be less.

Question 85.
(a) In which type of microscope do you find the lens arrangement as shown in the following diagram? (Practice Activity Sheet – 1)

(b) Write in brief, the working of this microscope.
(c) Where is this microscope used?
Answer:
(a) Compound microscope.

(b)
An object appears small or big depending upon the size of its image formed on the retina of the eye. The size of an object as perceived by the eye is called the apparent size of the object. Consider two objects of the same size, one held near the eye and the other away from the eye as shown in the following figure (Fig. 7.34). The nearby object (PQ) appears larger than the distant object (P₁Q₁).

Also, the angle a subtended by the nearby object at the eye is larger than the angle β subtended by the distant object at the eye. This shows that the apparent size of an object depends upon the angle subtended by the object at the eye. The greater the angle subtended by the object at the eye, the greater is the apparent size of the object. Similarly, the smaller the angle subtended by the object at the eye, the smaller is the apparent size of the object.

(c) A simple microscope is used by watch repairers to observe small parts of a watch and by jewellers to
examine ornaments. A simple microscope (also called a magnifying glass) is also used to read words in small print.

Question 86.
(i) Which type of microscope has the arrangement of lenses shown in the following figure?

(ii) Label the figure correctly.
(iii) Write the working of this microscope.
(iv) Where is this microscope used?
(v) Suggest a way to increase the efficiency of this microscope. (Practice Activity Sheet – 2)
Answer:
(i) Compound microscope.

(ii)

(iii) Working:
(1) The object to be observed is illuminated and placed in front of the objective lens, slightly beyond the focal length of the objective lens. Its real, inverted and enlarged image is formed by the objective lens on the other side.

(2) This intermediate image lies within the focal length of the eyepiece. It serves as an object for the eyepiece. The eyepiece works as a simple microscope. The final image is virtual, highly enlarged and inverted with respect to the original object. It can be formed at the minimum distance of distinct vision from the eyepiece. The final image is observed by keeping the eye close to the eyepiece.

(iv)

1. It is used to observe blood corpuscles, plant and animals cells, microorganisms like bacteria. etc.
2. It is used in a pathological laboratory to observe blood, urine, etc.
3. It is a part of a travelling microscope used for measurement of very small distance.

(v) Lenses with appropriate focal lengths should be selected.

Question 87.
State the use of a telescope.
Answer:
A telescope is used to observe a distant object such as mountain, moon, planet, star in the magnified form.

Question 88.
Observe the following figure and answer the questions. (Practice Activity Sheet – 3)
(a) Which optical instrument shows arrangement of lenses as shown in the figure?
(b) Write in brief the working of this optical instrument.

(c) How can we get different magnifications in this optical instrument?
(d) Draw the figure again and labelled it properly.
Answer:
(a) Refracting telescope.

(b) working: When the objective lens is pointed towards the distant object to be observed, the rays
of light from the distant object, which are almost parallel to each other. pass through the objective lens. The objective lens collects maximum amount of light as it is large in size. It forms a real, inverted and diminished image in the focal plane of the objective lens. Now, the position of the eyepiece is adjusted such that this image falls just within the focal length of the eyepiece and serves as the object for the eyepiece which works as a simple microscope.

The final image is highly magnified, virtual, on the same side as that of the object and inverted with respect to the original object. The final image can be observed by keeping the eye close to the eyepiece. If the image formed by the objective lens lies in the focal plane of the eyepiece, the final image is formed at infinity.

(c) We can get different magnifications by using the eyepiece with different focal lengths.

(d)

Question 89.
What is persistence of vision? Give one example of persistence of vision.
Answer:
Persistence of vision: We see an object when its image is formed on the retina. The image disappears when the object is removed from our sight. But this is not instantaneous and the image
remains imprinted on the retina for about \(\frac{1}{16}\) th of a second after the removal of the object.

The sensation on the retina persists for a while. This effect is known as the persistence of vision. It is due to persistence of vision that we continue to see the object in its position for about \(\frac{1}{16}\) th of a second after it is removed.
Example: When a burning stick of incense is moved fast in a circle, a circle of red light is seen.

Question 90.
Name two devices whose working is based on the phenomenon of persistence of vision.
(OR)
Name any two applications based on persistence of vision.
Answer:
The working of a television set and motion picture is based on the phenomenon of persistence of vision.
[Note: These are the examples of persistence of vision in daily life.

Question 91.
How is the phenomenon of persistence of vision used in motion pictures?
Answer:
In motion pictures, photographs of a moving object are taken at the rate of more than sixteen pictures per second. These photographs are projected on the screen at the same rate.
Each picture is slightly different from the other. As a result of persistence of vision, we get the impression of observing the object in continuous motion.

Question 92.
Name the two types of light sensitive cells present in the retina of the human eye. What are their functions?
Answer:
(1) The retina of the human eye contains a large number of light sensitive cells. These cells are of two shapes : (i) rods and (ii) cones.
(2) The rod-like cells respond to the intensity of light.
(3) The conical cells respond to various colours of light. They respond differently to red, green and blue colours. They do not respond to faint light.

Question 93.
When do you say that a person is colour blind?
Answer:
When a person is unable to distinguish between certain colours, he is said to be colour blind.
[Note: (i) Except for being colour-blind, their eyesight is normal, (ii) Rod-shaped cell ≡ rod-like cell, cone-shaped cell = Conical cell.]

Question 94.
Explain the perception of colour in the human eye.
(OR)
Explain in short perception of colour.
(OR)
Write a note on perception of colour.
Answer:
(1) In nature we firld objects of various colours. Perception of colour means to be able to respond to colour.
(2) We can distinguish between various colours due to perception of colour.
(3) The cone-shaped cells on the retina of the eye respond to the various colours when light is bright and communicate to the brain about the colours of the image formed on the retina. This gives us the proper idea about the colours of the object.
(4) If, in the retina of a person, the cone-shaped cells responding to certain specific colours are absent, the person is unable to distinguish between the colours. As a result, he lacks perception of colour.

Question 95.
What is colour-blindness?
Answer:
(1) The retina of the human eye contains a large number of light sensitive cells. These cells are of two shapes : (i) rods and (ii) cones.
(2) The cone-shaped cells respond to various colours of light when light is bright.
(3) Thus, the perception of colour is due to the presence of the cone-shaped cells in the retina.
(4) In the retina of some persons, cone-shaped cells responding to certain specific colours are absent. Hence, these persons are unable to distinguish between certain colours, i.e., they are colour-blind. This defect is known as colour¬blindness.

Question 96.
Why are some persons colour-blind? What is the cause of this defect?
Answer:
In the retina of some persons, cone-shaped cells responding to certain specific colours are absent. Hence, these persons are unable to distinguish between certain colours, i.e., they are colour-blind.

Question 97.
What are the difficulties faced by a colour-blind person?
Answer:
(1) A colour-blind person cannot distinguish between different colours. For example, he cannot distinguish between red and green colours. Also he cannot distinguish between blue and green colours. Red and green, both appear grey. Since a colour¬blind person cannot distinguish between red and green colours, it is difficult for him to cross a road. There is a possibility of an accident while crossing a road.

(2) A colour-blind person cannot distinguish between two objects of different colours, which are otherwise identical, e.g., clothes.

(3) A colour-blind person may have an inferiority complex and hence may find it difficult to mix with other persons.

Give scientific reasons:

Question 1.
A convex lens is known as a converging lens.
Answer:
When rays of light parallel to the principal axis of a convex lens pass through the lens, they converge to a point on the principal axis. Hence, a convex lens is known as a converging lens.

Question 2.
A concave lens is called a diverging lens.
Answer:
When rays of light parallel to the principal axis of a concave lens pass through the lens, they appear to diverge from a point on the principal axis. Hence, a concave lens is called a diverging lens.

Question 3.
In old age, a bifocal lens is necessary for some persons.
Answer:
(1) Some people, in old age, suffer from myopia (nearsightedness) as well as hypermetropia (farsightedness).
(2) Myopia is corrected using a concave lens of appropriate power. Hypermetropia is corrected using a convex lens of appropriate power. Therefore, they need a bifocal lens.

Question 4.
A person suffering from myopia (nearsightedness) uses spectacles of concave lenses.
Answer:
(1) A person suffering from myopia can see nearby objects clearly as the image of a nearby object is formed on the retina, but cannot see distant objects clearly as the image of a distant object is formed in front of the retina instead of on the retina.

(2) A concave lens diverges the rays of light passing through it. When spectacles of concave lenses of appropriate power are used, the parallel rays coming from a distant object are diverged to proper extent before they are incident on the eye lens. Therefore, after the converging action of the eye lens, the image of a distant object is formed on the retina of the eye and hence the distant object can be seen clearly.

Question 5.
A person suffering from hypermetropia (farsightedness) uses spectacles of convex lenses.
Answer:
(1) A person suffering from hypermetropia can see distant objects clearly as the image of a distant object is formed on the retina, but cannot see nearby objects clearly as the image of a nearby object would be formed behind the retina instead of on the retina.

(2) A convex lens converges the rays of light passing through it. When spectacles of convex lenses of appropriate power are used, the rays of light coming from a nearby object are converged to proper extent before they are incident on the eye lens. Therefore after the converging action of the j eye lens, the image of a nearby object is formed on j the retina of the eye and hence the nearby object | can be seen clearly.

Question 6.
You cannot enjoy watching a movie from a very short distance from the screen in a cinema hall.
Answer:
(1) The less the distance between the screen in a cinema hall and the person watching the movie, the more is the intensity of light falling on the eye.
(2) This results in great contraction of the pupil of the eye causing a strain. Hence, you cannot enjoy watching a movie from a very short distance from the screen in a cinema hall.

Question 7.
The rays of light travelling through the optical centre of a lens pass without changing their path.
Answer:
The portion of a lens near the optical centre is like a very thin slab of glass. Hence, the rays of light travelling through the optical centre of a lens pass without changing their path.

Question 8.
A convex lens converges the rays of light falling on it.
Answer:

• A convex lens can be regarded as made of a very large number of portions of triangular prisms. The bases of these prisms are towards the central thicker portion of the lens.
• A ray of light passing through a prism bends towards its base. Hence, a convex lens converges the rays falling on it.

Question 9.
A concave lens diverges the rays of light falling on it.
Answer:

• A concave lens can be regarded as made of a very large number of portions of triangular prisms. The bases of these prisms are towards the edges of the less, i.e, away from the central thinner portion of the lens.
• A ray of light passing through a prism bends towards its base. Hence, a concave lens diverges the rays of light falling on it.

Question 10.
When a burning stick of incense is moved fast in a circle, a circle of red light is seen.
Answer:
The impression of the image on the retina lasts for about \(\frac{1}{16}\) th of a second after the removal of the object. If a burning stick of incense is moved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Question 11.
Colour-blind persons are unable to distinguish between different colours.
Answer:
(1) The cone-shaped cells in the retina of a person respond to colours. This makes the perception of colours possible.
(2) In the retina of colour-blind persons, cone-shaped cells responding to certain specific colours are absent. Hence, they are unable to distinguish between different colours.

Question 12.
It is risky to issue a driving license to a person suffering from colour-blindness.
Answer:
A colour-blind person cannot distinguish between different colours. If a driver is colour-blind, he will not be able to distinguish between the colours of the signal and the colours on different sign boards. This will lead to an accident. Hence, it is risky to issue a driving license to a person suffering from colour-blindness.

Distinguish the following:

Question 1.
Real image and Virtual image.
Answer:
Real image:

1. A real image is formed when the light rays starting from an object meet after reflection or refraction.
2. It can be projected on a screen.
3. It is inverted with respect to the object.

Virtual image:

1. A virtual image is formed when the light rays starting from an object (when extended backward) appear to meet after reflection or refraction.
2. It cannot be projected on a screen.
3. It is erect with respect to the object.

Question 2.
Simple microscope and Compound microscope.
Answer:
Simple microscope:

1. In a simple microscope, only one convex lens is used.
2. In this case, the object is placed within the focal length of the convex lens.
3. Its magnifying power is much less than that of a compound microscope.
4. It is used to observe minute parts of a watch, to read words in small print, etc.

Compound microscope:

1. In a compound microscope, two convex lenses, objective and eyepiece, are used.
2. In this case, the object is placed beyond the focal length of the objective lens.
3. Its magnifying power is much greater than that of a simple microscope.
4. It is used to observe blood corpuscles, plant and animal cells, etc.

Question 3.
Compound microscope and Astronomical refracting telescope.
Answer:
Compound microscope:

1. In a compound microscope, the focal length and cross section of the objective lens are respectively smaller than the focal length and cross section of the eyepiece.
2. In this case, to observe the object, the distance between the object and the objective lens is adjusted.
3. It forms a magnified image of a small object.
4. It is used to observe blood corpuscles, plant and animal cells, etc.

Astronomical refracting telescope:

1. In an astronomical refracting telescope, the focal length and cross section of the objective lens are respectively greater than the focal length and cross section of the eyepiece.
2. In this case, to observe the object, the distance between the objective lens and eyepiece is adjusted.
3. It forms a near image of a distant object.
4. It is used to observe sateulites, planets, stars, etc.

Question 4.
Simple microscope and Astronomical refracting telescope.
Answer:
Simple microscope:

1. In a simple microscope, only one convex lens is used.
2. In this case, the object is placed within the focal length of the convex lens.
3. In this case, the image is erect.
4. It is used to observe minute parts of a watch, to read words in small print, etc.

Astronomical refracting telescope:

1. In an astronomical refracting telescope, two convex lenses, objective lens and eyepiece are used.
2. In this case, the object is far away from the objective lens.
3. In this case, the image is inverted.
4. It is used to observe satellites, planets, stars, etc.

Read the following paragraph and answer the questions given below it:

Construction of a compound microscope:
(1) A compound microscope consists of a metal tube fitted with two convex lenses at the two ends. These lenses are called the objective lens (the lens directed towards the object) and the eyepiece (the lens directed towards the eye). Both the lenses are small in size, but the cross section of the objective lens is less than that of the eyepiece. The objective lens has a short focal length. The focal length of the eyepiece is more than that of the objective lens.(2) The metal tube is mounted on a stand. The principal axes of the objective lens and the eyepiece are along the same line. The distance between the object and the objective lens can be changed with a screw. It is possible to change the distance between the objective lens and the eyepiece.
Working:
(1) The object to be observed is illuminated and placed in front of the objective lens, slightly beyond the focal length of the objective lens. Its real, inverted and enlarged image is formed by the, objective lens on the other side.
(2) This intermediate image lies within the focal length of the eyepiece. It serves as an object for the eyepiece. The eyepiece works as a simple microscope. The final image is virtual, highly enlarged and inverted with respect to the original object. It can be formed at the minimum distance of distinct vision from the eyepiece. The final image is observed by keeping the eye close to the eyepiece.
Use: This microscope is used to observe blood cells, microorganisms, etc.

Question 1.
In a compound microscope, which lens has greater focal length?
Answer:
In a compound microscope, the eyepiece has greater focal length.

Question 2.
Where do you place the object to be observed with a compound microscope?
Answer:
In a compound microscope, the object to be observed is placed in front of the objective lens, slightly beyond the focus of the objective lens.

Question 3.
State which distance is adjusted to observe the object with a compound microscope.
Answer:
To observe the object with a compound microscope, the distance between the object and objective lens is adjusted.

Question 4.
State the nature of the final image in compound microscope relative to the object.
Answer:
In a compound microscope, the final image is highly enlarged, inverted and virtual relative to the object.

Question 5.
State the use of a compound microscope.
Answer:
A compound microscope is used to observe blood cells, microorganisms, etc.

Fill in the blanks for a convex lens:

Question 1.

f (m)	0.2	—————–	0.1
P (D)	—————	2	——————

Answer:
[P (D) = \(\frac{1}{f(\mathrm{m})}\)]

f (m)	0.2	0.5	0.1
P (D)	5	2	10

Question 2.

h1 (cm)	—————-	5	10
h2 (cm)	-30	-20	—————-
M	-2	—————–	-0.5

Answer:
[M = \(\frac{h_{2}}{h_{1}}\)]

h1 (cm)	15	5	10
h2 (cm)	-30	-20	-5
M	-2	-4	-0.5

Solve the following numerical problems:

Problem 1.
An object Is kept at 60 cm in front of a convex lens. Its real image is formed at 20 cm from the lens. Find the focal length or the lens.
Solution:
Data: Convex lens, u = -60 cm,

The focal length of the lens = 15 cm.

Problem 2.
The focal length of a convex lens is 20 cm. If an object of height 2 cm is placed at 30 cm from the lens, find (i) the position and nature of the Image (ii) the height of the image (iii) the magnification produced by the lens.
Solution:
Data: Convex lens, f = 20 cm,
u = -30 cm, h₁ = 2 cm, v = ?, h₂ = ?, M = ?

The image will be formed at 60 cm from the lens and on the other side of the lens with respect to the object. It is a real image.

h₂ is negative. This shows that the image is inverted.
The height of the image -4 cm.
(iii) M = \(\frac{h_{2}}{h_{1}}=\frac{-4 \mathrm{cm}}{2 \mathrm{cm}}\) = -2
M is negative, indicating that the image is inverted.
The magnification produced by the lens = -2.

Problem 3.
When a pm of height 3 cm is fixed at 10 cm from a convex lens, the height or the virtual image formed is 12 cm. Find the focal length of the lens.
Solution:
Data: Convex lens, h₁ =3 cm,
h₂ = 12 cm (virtual image), u = -10 cm, f = ?

The focal length of the lens = 13.33 cm [approximately]

Problem 4.
At what distance from a convex lens of focal length 2.5 m should a boy stand so that his image is half his height?
Solution:
Data: Convex lens, f= 2.5 m,
M= –\(\frac{1}{2}\), u=?

∴ u = -3f = -3 × 2.5 m = -7.5 m
This is the object distance.
The boy should stand at 7.5 m from the convex lens so that his image is half his height.

Problem 5.
A convex lens forms a real image or a pencil at a distance of 40 cm from the lens. The image formed is of the same size as the object. Find the focal length and power of the lens. At what distance is the pencil placed from the lens?
Solution:
Data: Convex lens, v = 40 cm,
M = -1, f = ?, h₁ = ?, u = ?
M = = -1 = \(\frac{v}{u}\)
∴ u = -v = -40 cm (object distance)
The pencil is placed at 40 cm from the convex lens.

The focal length of the lens 20 cm.

The power of the lens = 5 D.

Problem 6.
A spherical lens is used to obtain an image on a screen. The size of the image is four times the size of the object. What is the type of lens and at what distance is the screen placed from the lens?
Solution:
Data: M = -4, type of lens? v = ?
As the image formed by the lens is obtained on a screen, it is a real image. The lens is, therefore, a convex 1ens.

The distance of the screen from the lens = 5f.

Problem 7.
An object of height 5 cm Is held 20 cm away from a converging lens of focal length 10 cm. Find the position, nature and size of the image formed.
Solution:
Data: Converging lens (convex lens),
f = 10 cm, h₁ = 5 cm, u = -20 cm, v = ?, h₂ = ?

The image is real and inverted. it is formed at 20 cm from the lens and on the other side of the lens relative to the object.

The height of the image, h₂ = -5 cm
Thus, it is numerically the same as the height of the object.

Problem 8.
An object is placed at 10 cm from a convex lens of focal length 12 m. Find the position and nature of the image.
Solution:
Data: Convex lens, u = -10 cm,
f = 12 cm, v = ?


∴ v = -60 cm
It is negative.
The image is formed at 60 cm from the lens and on the same side of the lens relative to the object. It is virtual, erect, and enlarged.

Problem 9.
An object of height 4 cm is placed in front of a concave lens of focal length 40 cm. If the object distance is 60 cm, find the position and height of the image.
Solution:
Data: f = -40 cm (concave lens),
u = -60 cm, h₁ = 4 cm, v = ?, h₂ = ?

The image Is formed at 24 cm from the lens.
It is on the same side as the object.

The height of the image is 1.6 cm.

Problem 10.
What is the power of a convex lens having focal length 0.5 m?
Solution:
Data: Convex lens, f= 0.5 m, P = ?
P = \(\frac{1}{f}=\frac{1}{0.5 \mathrm{m}}\) = 2D
The power of the lens = 2D.

Problem 11.
The power of a convex lens is 2.5 dioptres. Find its focal length.
(OR)
Calculate the focal length of a corrective lens having power +2.5 D.
Solution:
Data: Convex lens, P = +2.5 D, f = ?
P = \(\frac{1}{f}\)
∴ 2.5 D = \(\frac{1}{f}\)
∴ f = \(\frac{1}{2.5 \mathrm{D}}\) = 0.4 cm = 40 cm
The focal length of the lens = 40 cm.

Problem 12.
Two convex lenses of focal length 20 cm each are kept in contact with each other. Find the power of their combination.
Solution:
Data: f₁ = 20 cm = 0.2 m,
f₂ =20 cm = 0.2 m, P (combination) = ?

∴ Focal length of the combination or the lenses, f = 0.1 m.
P = \(\frac{1}{f}=\frac{1}{0.1 \mathrm{m}}\) = 10 D
The power of the combination of the lenses, P = 10 D.

Problem 13.
Two convex lenses of equal focal lengths are kept in contact with each other. If the power of their combination is 20 D, find the focal length of each convex lens.
Solution:
Data: Convex lens, P = 20 D, f₁ = f₂ = ?
The focal length (f) of the combination of the lenses is given by

This gives the focal length or each convex lens.

Problem 14.
If a convex lens of focal length 10 cm and a concave lens of focal length 50 cm are kept in contact with each other, (i) what will be the focal length of the combination? (ii) what wiil be the power of the combination? (iii) what will be the behaviour of the combination (behaviour as a convex lens/concave lens)?
Solution:
Data: f₁ = +10 cm = +0.1 m (convex lens),
f₂ = -50 cm = -0.5 m (concave lens),
f (combination) = ?, P (combination) = ?

The focal length or the combination of the lenses = 0.125 m = 12.5 cm.

The power of the combination of the lenses 8.D.
(iii) The focal length of the combination of the lenses is positive. This shows that the combination will behave as a convex lens.

Numerical problems for practice:

Problem 1.
Find the focal length of a convex lens which produces a real image at 60 cm from the lens when an object is placed at 40 cm in front of the lens.
Answer:
24 cm

Problem 2.
Find the focal length of a convex lens which produces a virtual image at 10 cm from the lens when an object is placed at 5 cm from the lens.
Answer:
10 cm

Problem 3.
A real image is obtained at 30 cm from a convex lens of focal length 7.5 cm. Find the distance of the object from the lens.
Answer:
u = -10 cm

Problem 4.
An object is kept at 20 cm in front of a convex lens and its real image is formed at 60 cm from the lens. Find (1) the focal length of the lens (2) the height or the image if the height of the object is 6 cm.
Answer:
(1) 15 cm
(2) h₂ = -18 cm]

Problem 5.
An object is kept at 10 cm in front of a convex lens. Its image is formed on the screen at 15 cm from the lens. Calculate (1) the focal length of the lens (2) the magnification produced by the lens.
Answer:
(1) 6 cm
(2)M = -1.5

Problem 6.
An object is kept at 60 cm in front of a convex lens of focal length 15 cm. Find the image distance and the nature of the image. Also find the magnification produced by the lens.
Answer:
v = 20 cm. The image is real, inverted and smaller than the object. M = –\(\frac{1}{3}\)]

Problem 7.
An object of height 2 cm is kept at 30 cm from a convex lens. Its real image is formed at 60 cm from the lens. Find the focal length and power of the lens.
Answer:
f = 20 cm, P = 5 D

Problem 8.
If the power of a lens is 4 dioptres, find its focal length.
Answer:
25 cm

Problem 9.
Find the power of a convex lens of focal length 40 cm.
Answer:
2.5 D

Problem 10.
Find the power of a convex lens of focal length 12.5 cm.
Answer:
8 D

Problem 11.
If for a lens, f = – 20 cm, what is the power of the lens?
Answer:
-5 D

Problem 12.
An object of height 4 cm is kept in front of a concave lens of focal length 20 cm. If the object distance is 30 cm, find the position and the height of the image.
Answer:
v = -12 cm, h₂ = 1.6 cm

Problem 13.
If two convex lenses of focal lengths 10 cm and 5 cm are kept in contact with each other, what is their combined focal length?
Answer:
\(\frac{10}{3}\) cm [approximately 3.33 cm]

Problem 14.
If a convex lens of focal length 20 cm and a concave lens of focal length 30 cm are kept in contact with each other, (i) What will be the focal length of the combination? (ii) What will be the power of the combination? (iii) What will be the behaviour of the combination?
Answer:
(i) f = 60 cm
(ii) P = \(\frac{5}{3}\) D = 1.6667 D (approximately)
(iii) The combination will behave as a convex lens.

Problem 15.
A concave lens of focal length 12 cm and a convex lens of focal length 20 cm are kept in contact with each other, (i) Find the focal length of the combination, (ii) What will be the behaviour of the combination?
Answer:
(i) f = -30 cm
(ii) The combination will behave as a concave lens.

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light

Question 1.
Fill in the blanks and explain the completed statements:
a. Refractive index depends on the………….of light.
Answer:
Refractive index depends on the velocity of light.
It is an experimental fact. (There is no question of explanation.)

b. The change in…………of light rays while going from one medium to another is called refraction.
Answer:
The change in the direction of propagation of light rays while going from one medium to another is called refraction. This is definition of refraction. It is assumed that the ray of light passes obliquely from one medium to another. (There is no question of explanation.)

Question 2.
Prove the following statements:
a. If the angle of incidence and angle of emergence of a light ray falling on a glass slab are i and e respectively, prove that i = e. (Practice Activity Sheet – 4)
Answer:
In the following figure, SR || PQ and NM is the refracted ray. Hence, r = i₁.
Now _gn_a = sin i/sin r and _an_g = sin i₁/ sin e.

Also gna = \(\frac{1}{{ }_{\mathrm{a}} n_{\mathrm{g}}}\)
∴ \(\frac{\sin i}{\sin r}=\frac{\sin e}{\sin i_{1}}\)
As r = i₁, it follows that sin i = sin e
∴ i = e.

b. A rainbow is the combined effect (an exhibition) of the refraction, dispersion, and total internal reflection of light (taken together). (Practice Activity Sheet – 1)
(OR)
With a neat labelled diagram, explain how the formation of rainbow occurs.
Answer:
(1) The formation of a rainbow in the sky is a combined result of refraction, dispersion, internal reflection and again refraction of sunlight by water droplets present in the atmosphere after it has rained.

Here, for simplicity only violet and red colours are shown. The remaining five colours lie between these two.

(2) The sunlight is a mixture of seven colours: violet, indigo, blue, green, yellow, orange and red. After it has stopped raining, the atmosphere contains a large number of water droplets. When sunlight is incident on a water droplet, there is (i) refraction and dispersion of light as it passes from air to water (ii) internal reflection of light inside the droplet and (iii) refraction of light as it passes from water to air.

(3) The refractive index of water is different for different colours, being maximum for violet and minimum for red. Hence, there is dispersion of light (separation into different colours) as it passes from air to water. [ See above Figure for reference.]

(4) The combined action of different water droplets, acting like tiny prisms, is to produce a rainbow with red colour at the outer side and violet colour at the inner side. The remaining five colours lie between these two.
The rainbow is seen when the sun is behind the observer and water droplets in the front.

Question 3.
Mark the correct answer in the following questions :
A. What is the reason for the twinkling of stars?
(i) Explosions occurring in stars from time to time
(ii) Absorption of light in the earth’s atmosphere
(iii) Motion of stars
(iv) Changing refractive index of the atmospheric gases
Answer:
Changing refractive index of the atmospheric gases.

B. We can see the Sun even when it is little below the horizon because of
(i) reflection of light
(ii) refraction of light
(iii) dispersion of light
(iv) absorption of light
Answer:
refraction of light

C. If the refractive index of glass with respect to air is 3/2, what is the refractive index of air with respect to glass?
(i) \(\frac{1}{2}\)
(ii) 3
(iii) \(\frac{1}{3}\)
(iv) \(\frac{2}{3}\)
Answer:
(iv) \(\frac{2}{3}\)

Question 4.
Solve the following examples:
a. If the speed of light in a medium is 1.5 × 10⁸ m/s, what is the absolute refractive index of the medium? (Practice Activity Sheet – 1 and 4)
Solution:
Data: v = 1.5 × 10⁸ m/s,
c = 3 × 10⁸ m/s, n = ?
n = \(\frac{c}{v}=\frac{3 \times 10^{8} \mathrm{m} / \mathrm{s}}{1.5 \times 10^{8} \mathrm{m} / \mathrm{s}}\) = 2
This is the absolute refractive index of the medium.

b. If the absolute refractive indices of glass and water are \(\frac{3}{2}\) and \(\frac{4}{3}\) respectively, what is the refractive index of glass with respect to water?
Solution:

This is the refractive index of glass with respect to water.

Project:

Question 1.
Using a laser and soap water. study the refraction of light under the guidance of your teacher. (Do it your self)

Can you recall? (Text Book Page No. 73)

Question 1.
What is meant by reflection of light?
Answer:
Reflection of light: When light is incident on the surface of an object, in general, it is deflected in different directions. This process is called reflection of light.

Question 2.
What are the laws of reflection?
Answer:
Laws of reflection of light:

1. The incident ray and the reflected ray of light are on the opposite sides of the normal to the reflecting surface at the point of incidence and all the three are in the same plane.
2. The angle of incidence j and the angle of reflection are equal in measure.

Can you recall? (Text Book page No. 75)

Question 1.
If the refractive index of the second medium with respect to the first medium is ₂n₁ and that of the third medium with respect to the second medium is ₃n₂, what and how much is ₃n₁.
Answer:
₃n₁ is the refractive index of the third medium with respect to the first medium.

∴ ₃n₁ = ₂n₁ × ₃n₂.

[Suppose medium 1 = air, medium 2 ≡ ice and medium 3 ≡ diamond. Then, ₂n₁ ÷ 1.31, ₃n₂ = 1.847
∴ ₃n₁ = ₂n₁ × ₃n₂ = 1.31 × 1.847 = 2.42 which is the refractive index of diamond with respect to air.]

Can you tell? (Textbook page No. 76)

Question 1.
Have you seen a mirage which is an illusion of the appearance of water on a hot road or in a desert?
Answer:
Due to the changes in refraction of light, the light rays coming from a distant object appear to be coming from the image of the object inside the ground. This is called a mirage. When the earth’s surface is heated by the sun, the temperature of air increases. This produces a layer of hot air of lower density (mass per unit volume) and lower refractive index at the surface.

Hot air works as an optically rarer medium relative to cool air. When the temperature changes rapidly in the vertical direction, as refraction of light takes place, the angle of refraction changes continuously. The rays of light from the top of an object such as a car or tree cross the rays from the bottom of the object on their way to the observer’s eye.

Hence, an inverted image is formed below the object’s true position and downward towards the surface in the direction of air at higher temperature. In this case, some rays of light bend back up into the denser air above figure. Mirage produces an impression of water near the hot ground.

Question 2.
Have you seen that objects beyond and above a holy fire appear to be shaking? Why does this happen?
Answer:
The temperature of air beyond and above a holy fire changes all the time. Hence, the density of air also changes constantly. Hence, the direction of propagation of the rays of light approaching us from the objects beyond and above the holy fire changes constantly. Therefore, those objects appear to be shaking.

Use your brain power! (Text Book Page No. 77)

Question 1.
From incident white light how will you obtain white emergent light by making use of two prisms?
Answer:

• Take a prism. Allow white light to fall on it.
• Obtain a spectrum.
• Take a second identical prism. Place it parallel to the first prism in an upside down position with the first prism [as shown in Figure]
• Allow the colours of the spectrum to pass through the second prism.
• Obtain the beam of light emerging from the other side of the second prism.

The beam of light emerging from the other side of the second prism is a beam of white light.

Explanation: White light is made up of seven colours. The first prism produces dispersion of white light while the second prism combines light of different colours to produce white light again. The net deviation of a ray of light is zero.
[Note: This experiment is due to sir Isaac Newton. It proved that it was not the prism which added colours to the white light but a property of the white light itself.]

Question 2.
You must have seen chandeliers having glass prisms. The light from a tungsten bulb gets dispersed while passing through these prisms and we see coloured spectrum. If we use an LED light instead of a tungsten bulb, will we be able to see the same effect?
Answer:
Light emitted by LED (light-emitting-diode) does not have all wavelengths in the region 400 nm to 700 nm. Hence, its spectrum is not the same as that of light from a tungsten bulb or as that of sunlight.

Fill in the blanks and rewrite the statements:

Question 1.
The phenomenon of change in the………..of light when it passes obliquely from one transparent medium to another is called refraction.
Answer:
The phenomenon of change in the direction of propagation of light when it passes obliquely from one transparent medium to another is called refraction.

Question 2.
The refractive index depends upon the…………of propagation of light in different media.
Answer:
The refractive index depends upon the velocity of propagation of light in different media.

Question 3.
The process of separation of light into its component colours while passing through a medium is called………..
Answer:
The process of separation of light into its component colours while passing through a medium is called dispersion of light.

Question 4.
When a light ray travels obliquely from air to water, it bends………the normal at the point of incidence.
Answer:
When a light ray travels obliquely from air to water, it bends towards the normal at the point of incidence.

Question 5.
When a light ray travels obliquely from benzene to air, it bends…………the normal at the point of incidence.
Answer:
When a light ray travels obliquely from benzene to air, it bends away from the normal at the point of incidence.

Question 6.
In glass, the speed of red ray is……violet ray.
Answer:
In glass, the speed of red ray is greater than that of violet ray.

Question 7.
The speed of light in glass is………in water.
Answer:
The speed of light in glass is less than that in water.

Question 8.
The speed of light in water is…………in benzene.
Answer:
The speed of light in water is greater than that in benzene.

Question 9.
Rainbow occurs due to refraction, dispersion,……….and again refraction of sunlight by water droplets.
Answer:
Rainbow occurs due to refraction, dispersion, internal reflection and again refraction of sunlight by water droplets.

Question 10.
In dispersion of sunlight by a glass prism,………..ray is deviated the least.
Answer:
In dispersion of sunlight by a glass prism, red ray is deviated the least.

Rewrite the following statements by selecting the correct options:

Question 1.
The change in the direction of propagation of light when it passes obliquely from one transparent medium to another is called………
(a) dispersion
(b) scattering
(c) refraction
(d) reflection
Answer:
(c) refraction

Question 2.
When a ray of light travels from air to glass slab and strikes the surface of separation at 90°, then it…………
(a) bends towards the normal
(b) bends away from the normal
(c) passes unbent
(d) passes in zigzag way
Answer:
(c) passes unbent

Question 3.
If a ray of light passes from a denser medium to a rarer medium in a straight line, the angle of incidence must be…………
(a) 0°
(b) 30°
(c) 60°
(d) 90°
Answer:
(a) 0°

Question 4.
A ray of light strikes a glass slab at an angle of 50° with the normal to the surface of the slab. What is the angle of incidence?
(a) 50°
(b) 25°
(c) 40°
(d) 100°
Answer:
(a) 50°

Question 5.
If a ray of light propagating in air strikes a glass slab at an angle of 60° with the surface of the slab, the angle of refraction is…………
(a) more than 30 °
(b) less than 30 °
(c) 60°
(d) 30°
Answer:
(b) less than 30 °

Question 6.
A ray of light gets deviated When it passes obliquely from one medium to another medium because………..
(a) the colour of light changes
(b) the frequency of light changes
(c) the speed of light changes
(d) the intensity of light changes
Answer:
(c) the speed of light changes

Question 7.
The speed of light in turpentine oil is 2 × 10⁸ m/s. The absolute refractive index of turpentine oil is about……..[Speed of light in vacuum ≈ 3 × 10⁸ m/s]
(a) 1.5
(b) 2
(c) 1.3
(d) 0.67
Answer:
(a) 1.5

Question 8.
LASER stands for………..
(a) light amplification by stimulated emission of radiation
(b) light and sound energy radiation
(c) light and simulated energy radiation
(d) light amplification by sound energy radiation
Answer:
(a) light amplification by stimulated emission of radiation

Question 9.
Out of the following……….has the highest absolute refractive index.
(a) fused quartz
(b) diamond
(c) crown glass
(d) ruby
Answer:
(b) diamond

Question 10.
The absolute refractive index…………
(a) is expressed in dioptre
(b) is expressed in m/s
(c) of air is about \(\frac{4}{3}\)
(d) has no unit
Answer:
(d) has no unit

Question 11.
The speed of light in a medium of refractive index n is………., where c is the speed of light
in vacuum.
(a) \(\frac{c}{n}\)
(b) nc
(c) \(\frac{n}{c}\)
(d) \(\sqrt{\frac{c}{n}}\)
Answer:
(a) \(\frac{c}{n}\)

Question 12.
The speed of light in a transparent medium having absolute refractive index 1.25 is……….[Speed of light in vacuum ≈ 3 × 10⁸ m/s]
(a) 1.25 × 10⁸ m/s
(b) 2.4 × 10⁸ m/s
(c) 3.0 × 10⁸ m/s
(d) 1.5 × 10⁸ m/s
Answer:
(b) 2.4 × 10⁸ m/s

Question 13.
…………light is deviated the maximum in the spectrpm of white light obtained with a glass prism.
(a) Red
(b) Yellow
(c) Violet
(d) Blue
Answer:
(c) Violet

Question 14.
………..light is deviated the least in the spectrum of white light obtained with a glass prism.
(a) Red
(b) Yellow
(c) Violet
(d) Blue
Answer:
(a) Red

Question 15.
A ray of light makes an angle of 50° with the surface S1 of the glass slab. Its angle of incidence will be………….(March 2019)
(a) 50°
(b) 40°
(c) 140°
(d) 0°
Answer:
(a) 50°

Question 16.
A glass slab is placed in the path of convergent light. The point of convergence of light:
(a) moves away from the slab
(b) moves towards the slab
(c) remains at the same point
(d) undergoes a lateral shift
Answer:
(a) moves away from the slab

Question 17.
In refraction of light through a glass slab, the directions of the incident ray and the refracted ray are………… (Practice Activity Sheet – 1)
(a) perpendicular to each other
(b) non-parallel to each other
(c) parallel to each other
(d) intersecting each other
Answer:
(c) parallel to each other

Question 18.
If we gradually increase the angle of incidence of a ray of light passing through a prism, then………….. (Practice Activity Sheet – 4)
(a) the angle of deviation goes on decreasing
(b) the angle of deviation decreases but after certain value of incident angle, deviation angle increases
(c) the angle of deviation goes on increasing
(d) the angle of deviation increases but after certain value of incident angle, deviation angle decreases
Answer:
(b) the angle of deviation decreases but after certain value of incident angle, deviation angle increases

State whether the following statements are True or False. (If a statement is false, correct it and rewrite it.):

Question 1.
The incident ray and the refracted ray of light are on the opposite sides of the normal at the point of incidence.
Answer:
True.

Question 2.
The refractive index of a medium (such as glass) does not depend on the wavelength of light.
Answer:
False. (The refractive index of a medium depends on the wavelength of light.)

Question 3.
When a light ray travels obliquely from an optically rarer medium to an optically denser medium, it bends away from the normal.
Answer:
False. (When a light ray travels obliquely from an optically rarer medium to an optically denser medium, it bends towards the normal.)

Question 4.
When a light ray travels obliquely from glass to air, it bends towards the normal.
Answer:
False. (When a light ray travels obliquely from glass to air, it bends away from the normal.)

Question 5.
If the angle of incidence is 0°, the angle of refraction is 90°.
Answer:
False. (If the angle of incidence is 0°, the angle of refraction is also 0°.)

Question 6.
In dispersion of white light by a glass prism, yellow colour is deviated the least.
Answer:
False. (In dispersion of white light by a glass prism, red colour is deviated the least.)

Question 7.
In vacuum, the speed of light does not depend upon the frequency of light.
Answer:
True.

Question 8.
In glass, the speed of violet ray is less than that of red ray.
Answer:
True.

Question 9.
In a material medium, the speed of light depends on the frequency of light.
Answer:
True.

Question 10.
The velocity of light is different in different media.
Answer:
True.

Question 11.
Wavelength of red light is close to 700 nm.
Answer:
True.

Question 12.
Wavelength of orange light is greater than that of blue light.
Answer:
True.

Find the odd one out and give the reason:

Question 1.
Reflection, Neutralization, Refraction, Dispersion.
Answer:
Neutralization. It is associated with a chemical reaction between an acid and an alkali; others are phenomena associated with light.

Answer the following questions in one sentence each:

Question 1.
Mention any two phenomena in nature where refraction of light takes place.
Answer:
Mirage and twinkling of a star.

Question 2.
What is the angle of refraction when the angle of incidence is 0°?
Answer:
When the angle of incidence is 0°, the angle of refraction is also 0°.

Question 3.
In refraction of light, \(\frac{\sin i}{\sin r}\) = constant in sin a particular case. What is this constant called?
Answer:
The constant \(\frac{\sin i}{\sin r}\) (in a particular case) is called the refractive index of the second medium with respect to the first medium.

Question 4.
If the refractive index of medium 2 with respect to medium 1 is 5/3, what is the refractive index of medium 1 with respect to medium 2?
Answer:
The refractive index of medium 1 with respect to medium 2 is 0.6.

Question 5.
In dispersion of sunlight by a glass prism, which colour is deviated the least?
Answer:
In dispersion of sunlight by a glass prism, red colour is deviated the least.

Question 6.
In dispersion of sunlight by a glass prism, which colour is deviated the most?
Answer:
In dispersion of sunlight by a glass prism, violet colour is deviated the most.

Question 7.
What is the wavelength of violet light?
Answer:
The wavelength of violet light is (about) 400 nm.

Question 8.
State the relation between ₂n₁ and critical angle.
Answer:
₂n₁ = sin i, where i is the critical angle.

Answer the following questions:

Question 1.
What is meant by refraction of light?
Answer:
The change in the direction of propagation of light when it passes obliquely from one transparent medium to another is called refraction of light.

Question 2.
Why is there a change in the direction of propagation of light when it passes obliquely from one transparent medium to another?
Answer:
The velocity of light is different in different media. Hence, there is a change in the direction of propagation of light when it passes obliquely from one transparent medium to another.

Question 3.
In the case of refraction of light through a glass slab, the emergent ray is parallel to the incident ray, but it is displaced sideways. Why does this happen?
Answer:

The first refraction takes place as light passes obliquely from air to glass. In this case, the ray of light bends towards the normal at point N. The second refraction takes place as light passes obliquely from glass to air. In this case, the ray of light bends away from the normal at point M. The faces PQ and SR of the glass slab are parallel. Hence, the extent of bending of light at SR is equal in magnitude but opposite in sense relative to the bending of light at PQ. Hence, the emergent ray of light (MD) is parallel to the incident ray of light (AN), but it is displaced sideways as shown in Figure.

Question 4.
Define angle of incidence and angle of refraction.
Answer:
(1) The angle made by the incident ray of light with the normal to the surface at the point of incidence is called the angle of incidence.

(2) The angle made by the refracted ray of light with the normal to the surface at the point of incidence is called the angle of refraction.

[Note: The angle e in Fig. 6.3 is also called the angle of emergence as it is the angle made by the emergent ray with the normal to the surface at the point of emergence. ]

Question 5.
Repeat the activity “Refraction of light passing through a glass sl^b” by replacing the glass slab by a transparent plastic slab.
(i) What similarity do you observe?
(ii) What difference do you notice?
Answer:
(i) Similarity: The emergent ray is parallel to the incident ray, but it is displaced sideways.
(ii) Difference: For a given angle of incidence, the extent of refraction (bending) is different (in general, less) for a transparent plastic slab relative to the glass slab.

Question 6.
State the laws of refraction of light.
Answer:
Laws of refraction of light:
(1) The incident ray and the refracted ray are on the opposite sides of the normal to the surface at the point of incidence and all the three, i.e., the incident ray, the refracted ray and the normal are in the same plane.

(2) For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant (Snell’s law). This constant is called the refractive index of the second medium with respect to the first medium.
[Note: Here, a ray means a ray of light.]

Question 7.
How is refraction of light related to refractive index?
Answer:
When a ray of light travels obliquely from an optically rarer medium (lower refractive index) to an optically denser medium (higher refractive index), the ray bends towards the normal. When a ray of light travels obliquely from an optically denser medium to an optically rarer medium, the ray bends away from the normal. For a given angle of incidence (i ≠ 0), the extent of refraction (bending) of light is different in different media.

If the refractive index of the second medium with respect to the first medium is greater than 1, the greater the refractive index, the greater is the bending of the ray of light towards the normal. If the refractive index of the second medium with respect to the first medium is less than 1, the greater the refractive index, the lesser is the bending of the ray of light away from the normal.

Question 8.
Define the refractive index of the second medium with respect to the first medium.
(OR)
What is meant by refractive index?
Answer:
The refractive index of the second medium with respect to the first medium is defined as the ratio of the sine of the angle of incidence to the sine of the angle of refraction when the ray of light is obliquely incident at the boundary separating the

two media and travels from the first medium to the second medium. (See Fig. 6.4.)
(OR)
The refractive index of the second medium with respect to the first medium is defined as the ratio of (the magnitude of) the velocity of light in the first medium to (the magnitude of) the velocity of light in the second medium.

[Note: Velocity is a vector, i.e., it has magnitude and direction. In definition of refractive index, we consider only the magnitude of velocity of light (speed of light). Velocity of light in a medium depends on the physical condition of the medium as well as the frequency of light. Velocity of light is different in different media. For a given medium, the refractive index depends on the colour of light (frequency of light.)]

Question 9.
State the formulae for the refractive index of the second medium with respect to the first medium.
Answer:
The refractive index of the second medium with respect to the first medium,
₂n₁ = \(\frac{\sin i}{\sin r}=\frac{v_{1}}{v_{2}}\)
where i is the angle of incidence, r is the angle of refraction (as the ray of light passes obliquely from the first medium to the second medium), v₁ is the magnitude of the velocity (speed) of light in the first medium and v₂ is the magnitude of the velocity of light in the second medium.

Question 10.
Define absolute refractive index.
Answer:
The absolute refractive index of a medium is defined as the ratio of the magnitude of the velocity of light in vacuum to the magnitude of the velocity of light in the medium.

[Note: The speed of light is maximum in vacuum, about 3 × 10⁸ m/s. When light travels from one medium to another, there occurs a change in its speed and wavelength (A). But its frequency (v) remain the same.]

Question 11.
Obtain the relation between the refractive index of the second medium with respect to the first medium and the refractive index of the first medium with respect to the second medium.
Answer:
Let v₁ = speed of light in the first medium, v₂ = speed of light in the second medium, ₂n₁ = refractive index of the second medium With respect to the first medium and ₁n₂ = refractive index of the first medium with respect to the second medium.
By definition, ₂n₁ = \(\frac{v_{1}}{v_{2}}\) and ₁n₂ = \(\frac{v_{2}}{v_{1}}\)
Hence,

(OR)
₁n₂ × ₂n₁ = 1.

Question 12.
If the refractive index of a certain material with respect to air is 1.5, what is the refractive index of air with respect to that material?
Answer:
As the refractive index of the given material with respect to air is 1.5, the refractive index of air with respect to the material is
\(\frac{1}{1.5}=\frac{1}{3 / 2}=\frac{2}{3}\) = 0.6667 (approximately)

Question 13.
Explain the terms optically rarer medium and optically denser medium with examples.
Answer:
When we consider two media (such as air and glass), the medium with lower refractive index is called the optically rarer medium (in the present case, air) and the medium with higher refractive index is called the optically denser medium (glass, in the present case).

The higher density does not necessarily mean higher refractive index. For example, the density of water is greater than that of kerosene, but the absolute refractive index of water is less than that of kerosine. Thus, when we consider water and kerosine, water is an optically rarer medium while kerosine is an optically denser medium.

If we consider kerosene and benzene, kerosine is an optically rarer medium while benzene is an optically denser medium.

Question 14.
A ray of light is incident obliquely at a boundary separating two media. What is its behaviour if (1) the refractive index of the second medium is greater than that of the first medium (2) the refractive index of the first medium is greater than that of the second medium? Draw the corresponding neat and labelled diagrams.
Answer:
Consider a ray of light incident obliquely at a boundary separating two media.
(1) If the refractive index of the second medium is greater than that of the first medium, the ray bends towards the normal at the point of incidence as it travels from the first medium (optically rarer medium) to the second medium (optically denser medium). The angle of refraction (r) is less than the angle of incidence (i). (Fig. 6.6)

Fig. 6.6: A ray of light travelling from a rarer medium to a denser medium (Schematic diagram)

Fig. 6.7: A ray of light travelling from a denser medium to a rarer medium (Schematic diagram)

(2) If the refractive index of the first medium is greater than that of the second medium, the ray bends away from the normal at the point of incidence as it travels from the first medium (optically denser medium), to the second medium (optically rarer medium). The angle of refraction (r) is greater than the angle of incidence (i). (Fig. 6.7)

[Note In this chapter, a rarer medium means an optically rarer medium and a denser medium means optically denser medium unless stated otherwise.]

Question 15.
Observe the following figure and write accurate conclusion regarding refraction of light. (Practice Activity Sheet – 2)

Answer:
When a light ray passes obliquely from a rarer medium to a denser medium, it bends towards the normal.

Question 16.
What happens when a ray of light is incident normal to the interface between two media? Draw the corresponding neat and labelled diagram.
Answer:
When a ray of light is incident normal to the interface between two media, the ray propagates undeviated as it travels from the first medium to the second medium irrespective of the refractive indices of the two media. In this case, the angle of incidence (i) is zero and so also the angle of refraction (r).

Fig. 6.9: A ray of light incident normal to the interface between two media propagates without any change in its direction of propagation

Question 17.
Draw a neat and labelled diagram to show the path of a ray of light in air and glass when the ray is incident obliquely on a glass slab. Show the (i) incident ray (ii) refracted ray (iii) emergent ray (iv) angle of incidence (v) angle of refraction (vi) angle of emergence in the diagram.
(OR)
Draw a neat and labelled diagram to show refraction of light through a glass slab.
Answer:

Fig. 6.10: The path of the ray of light in air and glass when the ray is incident obliquely on a glass slab
In Fig. 6.10, i = angle of incidence, r = angle of refraction and e = angle of emergence.

Question 18.
Observe the given figure and name the following rays:

(i) ray AB
(ii) Ray BC
(iii) ray CD
Answer:
(i) The ray AB is the incident ray.
(ii) The ray BC is the refracted ray.
(iii) The ray CD is the emergent ray.

Question 19.
A plane mirror is kept at the bottom of a trough with water in it as shown in the following figure (Fig. 6.12). The ray of light emerging from a source at the point S outside the trough reaches the point A on the surface of water. Draw a neat ray diagram to show the subsequent path of light and complete the ray diagram.

Answer:

Question 20.
Give two examples of the effect of atmospheric refraction on a small scale in local environment.
Answer:

1. The occurrence of a mirage
2. Flickering of an object seen through a turbulent stream of hot air rising above the Holi fire are examples of the effect of atmospheric refraction on a small scale in local environment.

Question 21.
What is a mirage? With a neat labelled diagram, explain the conditions under which it is seen.
Answer:
Due to the changes in refraction of light, the light rays coming from a distant object appear to be coming from the image of the object inside the ground. This is called a mirage.

When the earth’s surface is heated by the sun, the temperature of air increases. This produces a layer of hot air of lower density (mass per unit volume) and lower refractive index at the surface. Hot air works as an optically rarer medium relative to cool air. When the temperature changes rapidly in the vertical direction, as refraction of light takes place, the angle of refraction changes continuously.

The rays of light from the top of an object such as a car or tree cross the rays from the bottom of the object on their way to the observer’s eye. Hence, an inverted image is formed below the object’s true position and downward towards the surface in the direction of air at higher temperature. In this case, some rays of light bend back up into the denser air above figure. Mirage produces an impression of water near the hot ground.

Question 22.
Explain in brief the flickering of an object seen through a turbulent stream of hot air rising above the Holi fire.
Answer:
During the Holi fire, the temperature of the air just above the fire becomes much greater than that of the air further up. The hot air has lower density (mass per unit volume) and lower refractive index. It becomes an optically rarer medium. The cool air has higher density and higher refractive index. It is an optically denser medium relative to hot air. Hence, in refraction of light, the angle of refraction changes continuously due to a continuous variation in refractive index.

As the physical conditions of air change rapidly, the apparent position of an object fluctuates rapidly. This gives rise to the flickering of an object seen through a turbulent stream of hot air rising above the Holi fire.

Question 23.
With a neat labelled diagram, explain twinkling of a star. Also explain why a planet does not twinkle.
Answer:
(1) As a star is far away from the earth, it appears as a point source of light. The density of air decreases with height above the earth’s surface. Hence, the refractive index of air also decreases with height. When starlight enters the earth’s atmosphere, it undergoes refraction continuously in the medium with gradually changing refractive index. The bending of starlight occurs towards the normal as it passes from the optically rarer part of the medium to the optically denser part.

(2) Hence, when a star is observed near the horizon, its apparent position is slightly higher than the actual position (See below figure).

(3) Further, the apparent position varies with time as the medium is not stationary due to mobility of air and change in temperature. When more light is refracted towards the observer the star appears bright. When less light is refracted towards the observer, the star appears dim.

(4) Thus there is fluctuation in the brightness of a star when observed from the earth. This is called twinkling of a star.

(5) Compared to stars, planets are relatively closer to the earth. Hence, a planet appears as a collection of a large number of point sources. Due to the changes in the refractive index of air, there is a change in the position and brightness of these point sources.

There is an increase in intensity of light coming from some point sources while there is a decrease in intensity of light coming from equal number of other point sources, on an average. The average brightness of a planet remain the same. Also, there is no change in the average position of a star. Hence, a planet does not twinkle.

Question 24.
What is the correct reason for blinking/flickering of stars? Explain it.
(a) The blasts in the stars.
(b) Absorption of star light by the atmosphere.
(c) Motion of the stars.
(d) Changing refractive index of gases in the atmosphere. (Practice Activity Sheet – 2)
Answer:
(d) Changing refractive index of the gases in the atmosphere results in blinking/flickering of stars.

Explanation:
(1) As a star is far away from the earth, it appears as a point source of light. The density of air decreases with height above the earth’s surface. Hence, the refractive index of air also decreases with height. When starlight enters the earth’s atmosphere, it undergoes refraction continuously in the medium with gradually changing refractive index. The bending of starlight occurs towards the normal as it passes from the optically rarer part of the medium to the optically denser part.

(2) Hence, when a star is observed near the horizon, its apparent position is slightly higher than the actual position (See below figure).

(3) Further, the apparent position varies with time as the medium is not stationary due to mobility of air and change in temperature. When more light is refracted towards the observer the star appears bright. When less light is refracted towards the observer, the star appears dim.

(4) Thus there is fluctuation in the brightness of a star when observed from the earth. This is called twinkling of a star.

(5) Compared to stars, planets are relatively closer to the earth. Hence, a planet appears as a collection of a large number of point sources. Due to the changes in the refractive index of air, there is a change in the position and brightness of these point sources.

There is an increase in intensity of light coming from some point sources while there is a decrease in intensity of light coming from equal number of other point sources, on an average. The average brightness of a planet remain the same. Also, there is no change in the average position of a star. Hence, a planet does not twinkle.

Question 25.
With a neat labelled diagram, explain advanced sunrise and delayed sunset.
Answer:
(1) The sunrise (the appearance of the sun above the horizon) is advanced due to atmospheric refraction of sunlight. An observer on the earth sees the sun two minutes before the sun reaches the horizon. A ray of sunlight entering the earth’s atmosphere follows a curved path due to atmospheric refraction before reaching the earth. This happens due to a gradual variation in the refractive index of the atmosphere.

For the observer on the earth, the apparent position of the sun is slightly higher than the actual position. Hence, the sun is seen before the sun reaches the horizon.

(2) Increased atmospheric refraction of sunlight occurs also at the sunset (the sun disappearing below the horizon). In this case, the observer on the earth continues to see the setting sun for two minutes after the sun has dipped below the horizon, thus delaying the sunset.

The advanced sunrise and delayed sunset increases the duration of day by four minutes.

Question 26.
Water in a swimming pool or water tank appears shallower than its depth. Why?
Answer:
When light rays travel obliquely from an optically denser medium (water, in this case) to an optically rarer medium (air, in this case), they bend away from the normal at the point of incidence. As a result, the bottom of a swimming pool or water tank appears raised to an observer standing near the edge of the pool or the tank. Therefore, the swimming pool or water tank appears shallower than its depth.

Question 27.
Place a coin at the bottom of a glass jar containing water. Now tilt the jar suitably. When viewed at a suitable angle, the coin appears to be floating. Why?
Answer:
When light rays travel obliquely from an optically denser medium (water, in this case) to an optically rarer medium (air, in this case), they bend away from the normal at the point of incidence. As a result, the coin appears raised. Therefore, when the jar is tilted suitably and observed at a suitable angle, the coin appears to be floating.

Question 28.
State the wavelength range of electromagnetic radiation to which our eyes are sensitive.
Answer:
Our eyes are sensitive to light (electromagnetic radiation). Its wavelength range is 400 nm to 700 nm.
[Note: Wavelength (λ) goes on decreasing and frequency (ν) goes on increasing from red (λ ≃ 700 nm) → orange → yellow → green → blue → indigo → violet (A ≃ 400 nm). c = vλ, where c is the speed of light in vacuum.]

Question 29.
What do you mean by dispersion of light? What is a spectrum of light? Name the different colours of light in the proper sequence in the spectrum of white light.
(OR)
What do you mean by dispersion? Name the different colours of light in the proper sequence in the spectrum of white light.
Answer:
The process of separation of light into its component colours while passing through a medium is called dispersion of light. The band of coloured components of a light beam is called spectrum.
The different colours of light in the spectrum of white light are violet, indigo, blue, green, yellow, orange and red.

Question 30.
What is a prism?
Answer:
A prism is a transparent medium bound by two plane surfaces inclined at an angle. Normally it is made of glass and has triangular cross section.

Question 31.
With a neat labelled diagram, describe the experiment to demonstrate dispersion of sunlight (white light) by a prism.
Answer:
Experiment:
(1) Procedure: Keep a glass prism on a table in a dark room. Hold a plane mirror outside the room so that it reflects a beam of sunlight into the room. Allow this beam to pass through a narrow slit made in cardboard and then fall on the prism. Place a white screen on the other side of the prism as shown in the following figure. [Fig. 6.17]

(2) Observations:

1. A pattern of various colours is observed on the screen. This pattern is called the spectrum.
2. It is found that in dispersion, the ray corresponding to violet colour deviates the most.
3. The ray corresponding to red colour deviates the least.
4. The deviation of rays corresponding to other colours is intermediate.

(3) Conclusion: When sunlight (white light) is incident on a prism, dispersion of light takes place, forming a spectrum.

[Notes: (1) This experiment is due to Sir Isaac Newton (1642 – 1727), English physicist and mathematician. (2) If in a Board examination, incomplete diagram (as shown in Fig. 6.18) is given, students should complete it and label its parts as shown in Fig. 6.17.]

Question 32.
How does the dispersion of white light take place when it passes through a glass prism?
Answer:
When rays of light are incident on a prism, they are refracted twice, while travelling from air to glass and then from glass to air. Even when the incident rays are directed away from the base of the prism, the emergent rays bend towards the base of the prism, as the prism is triangular. Thus, the rays are deviated as they pass through the prism.

The refractive index of glass is different for different colours. Therefore, the rays corresponding to different colours are deviated to different extents. White light is a mixture of seven colours : violet, indigo, blue, green, yellow, orange and red. Hence, when white light is incident on a prism, a spectrum of seven colours is obtained.

The refractive index of glass is maximum for violet light and minimum for red light. Hence, violet light is deviated the most and red light is deviated the least. The deviation of rays corresponding to other colours is intermediate. In this manner, the dispersion of light takes place when it passes through a glass prism. [For reference, see Fig. 6.17.]

Question 33.
What is a spectrum? Why do we get a spectrum of seven colours when while light is dispersed by a prism?
(OR)
Explain how a spectrum is formed.
Answer:
A band of coloured components of a light beam is called a spectrum. When white light is incident on a prism, the rays corresponding to different colours bend through different angles on refraction.

Of the various colours in the visible region, red light bends the least and violet light bends the most. Each colour emerges through the prism along a different path and becomes, distinct. Hence, we get a spectrum of seven colours.

Question 34.
What is partial reflection of light?
Answer:
When light travels from a denser medium to a rarer medium, it is partially reflected, i.e., part of light comes back into the denser medium as per the laws of reflection. This is called partial reflection of light.

[Note: Partial reflection of light occurs even when light travels from a rarer medium to a denser medium. The rest of light is refracted.]

Question 35.
Explain the terms total internal reflection and critical angle.
Answer:
Figure 6.20 shows passage of light from water (denser medium) to air (rarer medium).

The ray of light incident at the boundary separating the two media bends away from the normal on refraction. Here, the angle of refraction r, is greater than the angle of incidence i.
Now _an_w = \(\frac{\sin i}{\sin r}\) < 1. Here, _an_w is the refractive index of sin r air with respect to water.

As _an_w is constant, r increases as i increases. For r = 90°, the ray travels along the boundary. If i is increased further, as r cannot be greater than 90°, light does not enter air. There is no refraction of light and all the light enters water on reflection. This is called total internal reflection.
For r = 90°, _an_w = \(\frac{\sin i}{\sin 90^{\circ}}\) = sin i. This angle i is sin 90° called the critical angle.

Question 36.
Swarali has got the following observations while doing an experiment. Answer her questions with the help of observations. (Practice Activity Sheet – 2)
Swarali observed that the light bent away from the normal, while travelling from a denser medium to a rarer medium. When Swarali increased the values of the angle of incidence (i). the values of the angle of refraction (r) went on increasing. But at a certain angle of incidence, the light rays returned into the denser medium.
So, Swarali has some questions. Answer them.
(a) Name this certain value of i. What is the value of r at that time?
(b) Name this process of returning light in the denser medium. Explain the process.
Answer:
(a) Critical angle r = 90°
(b) Total internal reflection.

As light goes from a denser to rarer medium, if the value of the angle of incidence increases, then the value of the angle of refraction also increases. But after a specific angle of incidence called the critical angle, the light gets reflected back into the denser medium.

The ray of light incident at the boundary separating the two media bends away from the normal on refraction. Here, the angle of refraction r, is greater than the angle of incidence i.
Now _an_w = \(\frac{\sin i}{\sin r}\) < 1. Here, _an_w is the refractive index of sin r air with respect to water. As _an_w is constant, r increases as i increases. For r = 90°, the ray travels along the boundary. If i is increased further, as r cannot be greater than 90°, light does not enter air. There is no refraction of light and all the light enters water on reflection. This is called total internal reflection.

For r = 90°, anw = \(\frac{\sin i}{\sin 90^{\circ}}\) = sin i. This angle i is sin 90° called the critical angle.

Question 37.
The observations made by Swarali while doing the experiment are given below. Based on these write answers to the questions:
Swarali found that the light ray travelling from the denser medium to a rarer medium goes away from the normal. If the angle of incidence (i) is raised by Swarali, the angle of refraction (r) went on increasing. However, after certain value of the angle of incidence, the light ray is seen to return back into the denser medium. (March 2019)
(i) What is the specific value of ∠i called?
(ii) What is the process of reflection of incident ray into a denser medium called?
(iii) Draw the diagrams of three observations made by Swarali.
Answer:
(i) Critical angle
(ii) Total internal reflection
(iii)

Question 38.
Define total internal reflection of light.
Answer:
When light travels from a denser medium to a rarer medium, if the angle of incidence is greater than the critical angle, there is no refraction of light and all the light is reflected in the denser medium. This is called total internal reflection of light.

Question 39.
Define critical angle.
Answer:
When light travels from a denser medium to a rarer medium, the angle of incidence for which the angle of refraction becomes 90°, is called the critical angle.

Question 40.
If the refractive index of a rarer medium with respect to a denser medium is 0.5, what is the critical angle?
Answer:
₂n₁ = 0.5 = sin i
∴ Critical angle i = 30°.

Question 41.
Name the devices in which total internal reflection of light is used.
Answer:

1. Total internal reflecting prisms are used in a camera, binoculars, periscope.
2. Total internal reflection of light is used in optical fibres.

[Note: Total internal reflection of light plays an important role in sparkling brilliance of a diamond.]

Question 42.
Explain why an empty test tube held obliquely in water appears shiny to an observer looking down.
Answer:
When an empty test tube is held obliquely in Water in a beaker, some light rays passing from water to air are incident at an angle greater than the critical angle. They are, thus, totally internally reflected as shown, and the surface of the test tube has a silvery shine.

Question 43.
Observe the given figure and answer the following questions. (Practice Activity Sheet – 3)

(a) Identify and write the natural process shown in the figure.
(b) List the phenomena which are observed in this process.
(c) Redraw the diagram and show the above phenomena in it.
Answer:
(a) The natural process shown in the figure is formation of rainbow.
(b) The phenomena observed in this process are refraction, internal reflection and dispersion of light.
(c)

Write a short note on the following:

Question 1.
Refraction observed in the atmosphere.
Answer:
When a ray of light passes obliquely from an optically rarer medium to an optically denser medium, it bends towards the normal at the point of incidence. If opposite is the case, the ray bends away from the normal.

Atmosphere is never static. Air is mobile and its density and temperature are not uniform. As a result, in general, the path of a ray of light through atmosphere of varying refractive index is a curve. The refractive index of cool air is greater than that of hot air.

Atmospheric refraction of light results in many interesting optical phenomena such as twinkling of a star, advanced sunrise and delayed sunset, mirage and flickering of an object seen through a turbulent stream of hot air rising from a fire.

Question 2.
Dispersion of light.
Answer:
The process of separation of light into its component colours while passing through a medium is called dispersion of light. When white light passes through a glass prism, it spreads out into a band of different colours (components) called the spectrum of light. The colours in the spectrum of white light are violet, indigo, blue, green, yellow, orange and red.

Formation of a rainbow is an example of dispersion of light in nature. In this case, raindrops are responsible for dispersion of sunlight.

Dispersion takes place because the refractive index of a material such as glass or water, is different for different colours. It is maximum for violet colour and minimum for red colour. Hence, in the spectrum of white light (sunlight) obtained with a prism, violet light is deviated the most while red light is deviated the least. The deviation of light corresponding to other colours lies in between.

Give scientific reasons:

Question 1.
A coin kept in a bowl is not visible when seen from one side. But, when water is poured in the bowl, the coin becomes visible.
Answer:
(1) When the bowl is empty, the rays of light coming from the coin are obstructed by the side of the bowl, and hence the coin is not visible when seen from one side of the bowl.

(2) When water is poured in the bowl, the rays of light coming from the coin travel from water (denser medium) to air (rarer medium). Hence, they bend away from the normal on refraction. Therefore, the coin appears to be raised and becomes visible when observed from one side of the bowl.

Question 2.
A pencil dipped in water obliquely appears bent at the surface of water.
(OR)
When a pencil is partly immersed in water and held in a slanting position, it appears to be bent at the boundary separating water and air.
Answer:
(1) When a pencil is partly immersed in water and held in a slanting position, the rays of light coming from the immersed part of the pencil emerge from water (a denser medium) and enter air (a rarer medium). During this propagation, they bend away from the normal on refraction.

The pencil appearing bent at the boundary of water and air (schematic diagram)

(2) As a result, the immersed part of the pencil does not appear straight with respect to the part outside the water, but appears to be raised. Hence, a pencil dipped obliquely in water appears bent at the surface of the water.

Question 3.
The shadow of the edge of an empty vessel is formed due to the slanting rays of the sun. When water is poured in the vessel, the shadow is shifted.
Answer:
(1) When the slanting rays of the sun are obstructed by the edge of the empty vessel, the shadow of the edge is formed.

(2) When water is poured in the vessel, the slanting rays of the sun travel from air (rarer medium) to water (denser medium). During this propagation, they bend towards the normal on refraction. Hence, some part in the region of the shadow is now illuminated and the shadow appears to have shifted.

Question 4.
The bottom of a pond appears raised.
Answer:
(1) The rays of light coming from the bottom of a pond bend away from the normal as they travel from water (denser medium) to air (rarer medium).

(2) Hence, they appear to come from a point above the actual point from which they come.
Therefore, the bottom of the pond appears raised.

Question 5.
While shooting a fish in a lake, the gun is aimed below the apparent position of the fish.
Answer:
(1) The rays of light coming from the fish bend away from the normal as they travel from water (denser medium) to air (rarer medium).
(2) Hence, the position of the fish in water appears to be above Its real position. Therefore, while shooting a fish in a lake, the gun is aimed below the apparent position of the fish.

Question 6.
The sun is seen on the horizon a little before sunrise.
(OR)
The sun is seen on the horizon for sometime even after sunset.
Answer:
(1) The earth is surrounded by an atmosphere which is denser near the surface of the earth. When the rays of light from the sun enter the earth’s atmosphere from outer space, they travel from a rarer medium to a denser medium. Hence, they bend towards the normal on refraction.

(2) Hence, even when the sun is below the horizon while rising or setting, its rays reach us due to refraction and it appears to be on the horizon. Therefore, the sun is seen on the horizon a little before sunrise as well as for some time even after sunset.

Distinguish between:

Question 1.
Reflection of light and Refraction of light:
Answer:

Reflection of light	Refraction of light
1. The rays of light, before and after reflection, travel in the same medium.	1. In refraction of light, the rays travel from one medium to another medium.
2. In reflection, the angle of incidence and the angle of reflection are equal.	2. In refraction, when the rays travel obliquely from one medium to another medium, the angle of incidence and the angle of refraction are not equal.
3. In reflection, there is no change in the speed and wavelength of light.	3. In refraction, there occurs a change in the speed and wavelength of light.
4. In reflection, there is no dispersion of light.	4. Generally, in refraction, there occurs dispersion of light.

[Note: The frequency of light remains the same in reflection and refraction.]

Complete the following or Solve and fill in the blanks :

Question 1.

Speed of light in the first medium (v1)	Speed of light in the second medium (v2)	Refractive index 2n1	Refractive index 2n1
3 × 108 m/s	1.2 × 108 m/s	————————–	————————–
————————–	2.25 × 108 m/s	4/3	————————–
2 × 108 m/s	————————–	————————–	1.5

Answer:

Speed of light in the first medium (v1)	Speed of light in the second medium (v2)	Refractive index 2n1	Refractive index 2n1
3 × 108 m/s	1.2 × 108 m/s	2.5	0.4
3 × 108 m/s	2.25 × 108 m/s	4/3	0.75
2 × 108 m/s	3 × 108 m/s	2/3	1.5

Formulae:
₂n₁ = v₁/v₂, ₁n₂ = v₂/v₁

Solve the following examples/numerical problems:
c = 3 × 10⁸ m/s

Problem 1.
The speed of light in a transparent medium is 2.4 × 10⁸ m/s. Calculate the absolute refractive index of the medium.
Solution:
Data: c = 3 × 10⁸ m/s,
v = 2.4 × 10⁸ m/s, n = ?

The absolute refractive index of the medium = 1.25.

Problem 2.
The velocity of light in a medium is 2 × 10⁸ m/s. What is the refractive index of the medium with respect to air, if the velocity of light in air is 3 × 10⁸ m/s?
Solution:
Data: v₁ = 3 × 10⁸ m/s,
v₂ = 2 × 10⁸ m/s, ₂n₁ = ?
₂n₁ = \(\frac{v_{1}}{v_{2}}\)
\(=\frac{3 \times 10^{8}}{2 \times 10^{8}}\)
= 1.5
The refractive index of the medium with respect to air is 1.5.

Problem 3.
Light travels with a velocity 1.5 × 10⁸ m/s in a medium. On entering second medium its velocity becomes 0.75 × 10⁸ m/s. What is the refractive index of the second medium with respect to the first medium? (Practice Activity Sheet – 3)
Solution:
Given: Velocity of light in the first medium = v₁ = 1.5 × 10⁸ m/s,
velocity of light in the second medium = v₂ = 0.75 × 10⁸ m/s,
refractive index of the second medium with respect to the first medium = ₂n₁ = ?
₂n₁ = \(\frac{v_{1}}{v_{2}}\)
₂n₁ = \(\frac{1.5 \times 10^{8}}{0.75 \times 10^{8}}\) = 2
Hence, the refractive index of the second medium with respect to the first medium is 2.
[Note : The absolute refractive index of the second medium = \(\frac{3 \times 10^{8} \mathrm{m} / \mathrm{s}}{0.75 \times 10^{8} \mathrm{m} / \mathrm{s}}\) = 4 (greater than that of diamond, not likely).]

Problem 4.
The refractive index of water is 4/3 and the speed of light in air is 3 × 10⁸ m/s. Find the speed of light in water.
Solution:
Data: ₂n₁ = 4/3, v₁ = 3 × 10⁸ m/s, v₂ = ?

The speed of light in water = 2.25 × 10⁸ m/s.

Problem 5.
The speed of light in water and glass is 2.2 × 10⁸ m/s and 2 × 10⁸ m/s respectively. What is the refractive index of (i) water with respect to glass (ii) glass with respect to water?
Solution:
Data: u_w = 2.2 × 10⁸ m/s,
v_g= 2 × 10⁸ m/s, _wn_g = ?, _gn_w = ?

The refractive index of water with respect to glass = 0.909 (approximately).

The refractive index of glass with respect to glass = 1.1 (approximately).

Numerical Problems For Practice:
(Given: C = 3 × 10⁸m/s)

Problem 1.
The speed of light in a transparent medium is 2 × 10⁸ m/s. Find the absolute refractive index of the medium.
Solution:
1.5

Problem 2
The absolute refractive index of a transparent medium is 5/3. Find the speed of light in the medium.
Solution:
1.8 × 10⁸ m/s

Problem 3.
The absolute refractive index of a transparent medium is 2.4 and the speed of light in that medium is 1.25 × 10⁸ m/s. Find the speed of light in air.
Solution:
3 × 10⁸ m/s

Problem 4.
The speed of light in water is 2.25 × 10⁸ m/s and that in glass is 2 × 10⁸ m/s. Find the refractive index of (i) the glass with respect to water (ii) water with respect to the glass.
Solution:
(i) 1.125
(ii) 0.889 (approximately)

Problem 5.
If the refractive index of a certain glass with respect to water is 1.25, find the refractive index of water with respect to the glass.
Solution:
0.8

Problem 6.
If the absolute refractive index of glass is 1.5 and that of water is \(\frac{4}{3}\), find the refractive index of water with respect to glass.
Solution:
\(\frac{8}{9}\)

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 7 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Problem Set 7 Geometry 10th Std Maths Part 2 Answers Chapter 7 Mensuration

Problem Set 7 Question 1. Choose the correct alternative answer for each of the following questions.

i. The ratio of circumference and area of a circle is 2 : 7. Find its circumference.
(A) 14 π
(B) \(\frac{7}{\pi}\)
(C) 7π
(D) \(\frac{14}{\pi}\)
Answer:

(A)

ii. If measure of an arc of a circle is 160° and its length is 44 cm, find the circumference of the circle.
(A) 66 cm
(B) 44 cm
(C) 160 cm
(D) 99 cm
Answer:

(D)

iii. Find the perimeter of a sector of a circle if its measure is 90° and radius is 7 cm.
(A) 44 cm
(B) 25 cm
(C) 36 cm
(D) 56 cm
Answer:

(B)

iv. Find the curved surface area of a cone of radius 7 cm and height 24 cm.
(A) 440 cm²
(B) 550 cm²
(C) 330 cm²
(D) 110 cm²
Answer:

(B)

v. The curved surface area of a cylinder is 440 cm² and its radius is 5 cm. Find its height.
(A) \(\frac{44}{\pi}\) cm
(B) 22π cm
(C) 44π cm
(D) \(\frac{22}{\pi}\)
Answer:

(A)

vi. A cone was melted and cast into a cylinder of the same radius as that of the base of the cone. If the height of the cylinder is 5 cm, find the height of the cone.
(A) 15 cm
(B) 10 cm
(C) 18 cm
(D) 5 cm
Answer:

(A)

vii. Find the volume of a cube of side 0.01 cm.
(A) 1 cm
(B) 0.001 cm³
(C) 0.0001 cm³
(D) 0.000001 cm³
Answer:
Volume of cube = (side)³
= (0.01)³ = 0.000001 cm³
(D)

viii. Find the side of a cube of volume 1 m³
(A) 1 cm
(B) 10 cm
(C) 100 cm
(D) 1000 cm
Answer:
Volume of cube = (side)³
∴ 1 = (side)³
∴ Side = 1 m
= 100 cm
(C)

Problem Set 7 Geometry Class 10 Question 2. A washing tub in the shape of a frustum of a cone has height 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub? = (π = \(\frac { 22 }{ 7 } \))
Given: For the frustum shaped tub,
height (h) = 21 cm,
radii (r₁) = 20 cm, and (r₂) = 15 cm
To find: Capacity (volume) of the tub.
Solution:
Volume of frustum = \(\frac { 1 }{ 3 } \) πh (r₁² + r₂² + r₁ × r₂)

∴ The capacity of the tub is 20.35 litres.

10th Geometry Problem Set 7 Question 3. Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube?
Given: For the cylindrical tube,
height (h) = 90 cm,
outer radius (R) = 30 cm,
thickness = 2 cm
For the plastic spherical ball,
radius (r₁) = 1 cm
To find: Number of balls melted.
Solution:

Inner radius of tube (r)
= outer radius – thickness of tube
= 30 – 2
= 28 cm
Volume of plastic required for the tube = Outer volume of tube – Inner volume of hollow tube
= πR²h – πr²h
= πh(R² – r²)
= π × 90 (30² – 28²)
= π × 90 (30 + 28) (30 – 28) …[∵ a² – b² = (a + b)(a – b)]
= 90 × 58 × 2π cm³

∴ 7830 plastic balls were melted to make the tube.

Problem Set 7 Geometry Question 4.
A metal parallelopiped of measures 16 cm × 11cm × 10cm was melted to make coins. How many coins were made if the thickness and diameter of each coin was 2 mm and 2 cm respectively?
Given: For the parallelopiped.,
length (l) = 16 cm, breadth (b) = 11 cm,
height (h) = 10 cm
For the cylindrical coin,
thickness (H) = 2 mm,
diameter (D) 2 cm
To find: Number of coins made.
Solution:
Volume of parallelopiped = l × b × h
= 16 × 11 × 10
= 1760 cm³
Thickness of coin (H) = 2 mm
= 0.2 cm …[∵ 1 cm = 10 mm]
Diameter of coin (D) = 2 cm

∴ 2800 coins were made by melting the parallelopiped.

Mensuration Problem Question 5.  The diameter and length of a roller is 120 cm and 84 cm respectively. To level the ground, 200 rotations of the roller are required. Find the expenditure to level the ground at the rate of ₹ 10 per sq.m.
Given: For the cylindrical roller,
diameter (d) =120 cm,
length = height (h) = 84 cm
To find: Expenditure of levelling the ground.
Solution:
Diameter of roller (d) = 120 cm

Now, area of ground levelled in one rotation = curved surface area of roller
= 3.168 m²
∴ Area of ground levelled in 200 rotations
= 3.168 × 200 =
633.6 m²
Rate of levelling = ₹ 10 per m²
∴ Expenditure of levelling the ground
= 633.6 × 10 = ₹ 6336
∴ The expenditure of levelling the ground is ₹ 6336.

Question 6.
The diameter and thickness of a hollow metal sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm3. Find the outer surface area and mass of the sphere, [π = 3.14]
Given: For the hollow sphere,
diameter (D) =12 cm, thickness = 0.01 m
density of the metal = 8.88 gm per cm³
To find: i. Outer surface area of the sphere
ii. Mass of the sphere.

Solution:
Diameter of the sphere (D)
= 12 cm
∴ Radius of sphere (R)
= \(\frac { d }{ 2 } \) = \(\frac { 12 }{ 2 } \) = 6 cm
∴ Surface area of sphere = 4πR²
= 4 × 3.14 × 6²
= 452.16 cm²
Thickness of sphere = 0.01 m
= 0.01 × 100 cm …[∵ 1 m = 100 cm]
= 1 cm
∴ Inner radius of the sphere (r)
= Outer radius – thickness of sphere
= 6 – 1 = 5 cm
∴ Volume of hollow sphere
= Volume of outer sphere – Volume of inner sphere

∴ The outer surface area and the mass of the sphere are 452.16 cm² and 3383.19 gm respectively.

Question 7.
A cylindrical bucket of diameter 28 cm and height 20 cm was full of sand. When the sand in the bucket was poured on the ground, the sand got converted into a shape of a cone. If the height of the cone was 14 cm, what was the base area of the cone?
Given: For the cylindrical bucket,
diameter (d) = 28 cm, height (h) = 20 cm
For the conical heap of sand,
height (H) = 14 cm
To find: Base area of the cone (πR²).
Solution:
Diameter of the bucket (d) = 28 cm


The base area of the cone is 2640 cm².

Question 8.
The radius of a metallic sphere is 9 cm. It was melted to make a wire of diameter 4 mm. Find the length of the wire.
Given: For metallic sphere,
radius (R) = 9 cm
For the cylindrical wire,
diameter (d) = 4 mm
To find: Length of wire (h).
Solution:

∴ The length of the wire is 243 m.

Question 9.
The area of a sector of a circle of 6 cm radius is 157t sq.cm. Find the measure of the arc and length of the arc corresponding to the sector.
Given: Radius (r) = 6 cm,
area of sector = 15 π cm²
To find: i. Measure of the arc (θ),
ii. Length of the arc (l)
Solution:

∴ The measure of the arc and the length of the arc are 150° and 5π cm respectively.

Question 10.
In the adjoining figure, seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion.

(π = 3.14, \(\sqrt { 3 }\) = 1.73)
Given: Radius (r) = PA = 8 cm,
PC = 4 cm
To find: Area of shaded region.
Solution:

Similarly, we can show that, ∠BPC = 60°
∠APB = ∠APC + ∠BPC …[Angle sum property]
∴ θ = 60° + 60° = 120°


Area of shaded region = A(P-ADB) – A(∆APB)
= 66.98 – 27.68
= 39.30 cm²
∴ The area of the shaded region is 39.30 cm².

Question 11.
In the adjoining figure, square ABCD is inscribed in the sector A-PCQ. The radius of sector C-BXD is 20 cm. Complete the following activity to find the area of shaded region.

Solution:
Side of square ABCD
= radius of sector C-BXD = [20] cm
Area of square = (side)² = 20² = 400 cm² ….(i)
Area of shaded region inside the square = Area of square ABCD – Area of sector C-BXD

Radius of bigger sector
= Length of diagonal of square ABCD
= \(\sqrt { 2 }\) × side
= 20 \(\sqrt { 2 }\) cm
Area of the shaded regions outside the square
= Area of sector A-PCQ – Area of square ABCD
= A(A – PCQ) – A(꠸ABCD)

Alternate method:

□ABCD is a square. … [Given]
Side of □ABCD = radius of sector (C-BXD)
= 20 cm
Radius of sector (A-PCQ) = Diagonal
= \(\sqrt { 2 }\) × side
= \(\sqrt { 2 }\) × 20
= 20 \(\sqrt { 2 }\) cm

Now, Area of shaded region
= A(A-PCQ) – A(C-BXD)
= 628 – 314
= 314 cm²
∴ The area of the shaded region is 314 cm².

Question 12.
In the adjoining figure, two circles with centres O and P are touching internally at point A. If BQ = 9, DE = 5, complete the following activity to find the radii of the circles.

Solution:
Let the radius of the bigger circle be R and that of smaller circle be r.
OA, OB, OC and OD are the radii of the bigger circle.
∴ OA = OB = OC = OD = R
PQ = PA = r
OQ + BQ = OB … [B – Q – O]
OQ = OB – BQ = R – 9
OE + DE = OD ….[D – E – O]
OE = OD – DE = [R – 5]
As the chords QA and EF of the circle with centre P intersect in the interior of the circle, so by the property of internal division of two chords of a circle,
OQ × OA = OE × OF
∴ (R – 9) × R = (R – 5) × (R – 5) …[∵ OE = OF]
∴ R² – 9R = R² – 10R + 25
∴ -9R + 10R = 25
∴ R = [25units]
AQ = AB – BQ = 2r ….[B-Q-A]
∴ 2r = 50 – 9 = 41
∴ r = \(\frac { 41 }{ 2 } \) = 20.5 units

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 5 Heat Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 5 Heat

Question 1.
Fill in the blanks and rewrite the sentences:
a. The amount of water vapour in air is determined in terms of its………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

b. If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their………….
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

c. When a liquid is getting converted into solid, the latent heat is……….  (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Question 2.
Observe the following graph. Considering the change in volume of water as its temperature is raised from 0 °C, discuss the difference in the behaviour of water and other substances. What is this behaviour of water called?

Answer:
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C. It is minimum at 4 °C. The volume of water goes on increasing in the range 4 °C to 10 °C.

In general, when a substance is heated, its volume goes on increasing with temperature. Thus, in the range 0 °C to 4 °C, behaviour of water is different from other substances. It is called anomalous behaviour of water.

Question 3.
What is meant by specific heat capacity?
How will you prove experimentally that different substances have different specific heat capacities?
Answer:
The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C is called the specific heat capacity of that object.

Question 4.
While deciding the unit for heat, which temperatures interval is chosen? why?
Answer:
While deciding the unit for heat, the temperature interval chosen is 14.5 °C to 15.5 °C. For the reason, see the information given in the following box.

Question 5.
Explain the following temperature vs time graph:

(Practice Activity Sheet – 1 and 4; March 2019)
Answer:
The graph shows what happens when a mixture of ice and water is heated continuously. The temperature of the mixture remains constant (0 °C) till all the ice melts as shown by the line AB. This temperature is the melting point of ice. On further heating, the temperature rises steadily from 0 °C to 100 °C as shown by the line BC, At 100 °C water starts converting into steam. This temperature is the boiling point of water. Further heating does not change the temperature and the conversion waters steam continues as shown by the line CD.

Question 6.
Explain the following:
a. the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contracting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature of the water at the surface continues to fall to 0 °C. Finally, the water at the surface is converted into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

b. How can you relate the formation of water droplets on the outer surface of a bottle taken out of a refrigerator with formation of dew?
Answer:
At a given temperature, there is a limit on how much water vapour the given volume of air can hold. The lower the temperature, the lower is the capacity of air to hold water vapour.

The temperature of a bottle kept in a refrigerator is lower than room temperature. Hence, when the bottle is taken out of the refrigerator, the temperature of the air surrounding the bottle is lowered. Therefore, the capacity of the air to hold water vapour becomes less. Hence, the excess water vapour condenses to form water droplets (like dew) on the outer surface of the bottle.

c. In cold regions in winter, the rocks crack due to anomalous expansion of water.
Answer:
Sometimes water enters into crevices of the rocks. When the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts a tremendous pressure on the rocks which crack and break up into small pieces.

Question 7.
a. What is meant by latent heat? How will the state of matter transform if latent heat is given off?
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.
In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).
(Note: change of state = change of phase)

b. Which principle is used to measure the specific heat capacity of a substance?
Answer:
The principle of heat exchange is used to measure the specific heat capacity of a substance. This principle is as follows: If a system of two objects is isolated from the environment by keeping it inside a heat resistant box, then no energy can leave the box or enter the box. In this situation, heat energy lost by the hot object = heat energy gained by the cold object.

c. Explain the role of latent heat in the change of state of a substance.
Answer:
When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.
In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).
(Note: change of state = change of phase)

d. what basis and how will you determine whether air is saturated with vapour or not?
Answer:
Whether the air is saturated with water vapour or not is determined on the basis of the extent of water vapour present in the air. If the relative humidity is 100%, the air is saturated with water vapour. In that case, we can see the formation of water droplets on the leaves of plants/grass.
If the relative humidity is less than 100%, the air is not saturated with water vapour.

Question 8.
Read the following paragraph and answer the questions:
If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy.

The change in temperature continues till the temperatures of both the objects attain the same value. In this process, the cold object gains heat energy and the hot object loses’ heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat resistant box (meaning that the energy exchange takes place between the two objects only), then no energy can flow from inside the box or come into the box.
(1) Heat is transferred from where to where?
(2) Which principle do we learn about from this process?
(3) How will you state the principle briefly?
(4) Which property of the substance is measured using this principle?
Answer:
(1) Heat is transferred from a hot object to a cold object.
(2) This process shows the principle of heat exchange.
(3) In this process, the cold object gains heat energy and the hot object loses energy. If a system of two objects is isolated from the surroundings, heat energy lost by the hot object = heat energy gained by the cold object.
(4) This principle is used to measure the specific heat capacity of a substance.

Question 9.
Solve the following problems:
a. Equal heat is given to two objects A and B of mass 1 g. The temperature of A increases by 3 °C and B by 5°C. Which object has more specific heat? And by what factor?
Solution:
Data: m = 1 g, Δ T₁ = 3 °C, Δ T₂ = 5 °C,
Q same
Here, Q = mc₁ ΔT₁ = mc₂ ΔT₂

Thus, c₁ > c₂
The specific heat of A is more than that of B and

b. Liquid ammonia is used in ice factory for making ice from water. If water at 20 °C is to be converted into 2 kg, ice at 0 °C, how many grams of ammonia is to be evaporated?
(Given: The latent heat of vaporization of 1 ammonia = 341 cal/g)
Solution:
Data : m₁ = 2kg, ΔT₁=20 °C – 0 °C
= 20 °C, c₁ = 1 kcal/kg·°C, L₁ (ice) = 80 kcal/kg,
L₂ (vaporization of ammonia) = 341 cal/g = 341 kcal/kg, m₂ =?
Q₁ (heat lost by water) = m₁c₁ ΔT₁ + m₁L₁
= 2kg × 1 kcal/kg·°C × 20 °C + 2 kg × 80 kcal/kg
=40 kcal + 160 kcal = 200 kcal
Q₂ (heat absorbed by ammonia) = m₂L₂
= m₂ × 34l kcal/kg
According to the principle of heat exchange, Q₁ = Q₂
∴ 200 kcal = m₂ × 341 kcal/kg
∴ m₂ = \(\frac{200}{341}\) kg = 0.5864 kg = 586.4 g
586.4 g of ammonia are to be evaporated.

c. A thermally insulated pot has 150 g ice at temperature 0 °C. How much steam of 100 °C has to he mixed to it, so that water of temperature 50 °C will be obtained?
(Given: Latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)
Solution:
Data: m₁ = 150 g, ΔT₁ = 50 °C – 0 °C
= 50 °C, c_w = 1 cal/g.°C, L₁ = 80 cal/g, L₂ = 540 cal/g,
Δ T₂ = 100°C – 50 °C = 50 °C, m₂ = ?
Q₁ (heat absorbed by ice) = m₁L₁
= 150 g × 80 cal/g = 12000 cal
Q₂ (heat absorbed by water formed on melting of ice) =m₁ c_w ΔT₁
= 150 g × 1 cal/g·°C × 50 °C = 7500 cal
Q₃ (heat given out by steam) = m₂L₂
= m₂ × 540 cal/g
Q₄ (heat given out by water formed on condensation of steam)
= m₂ c_w ΔT₂ = m₂ × 1 cal/g·°C × 50 °C
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ 12000 cal + 7500 cal = m₂ × 540 cal/g + m₂ × 50 cal/g
∴ 19500 cal = m₂ (540 + 50) cal/g
∴ m₂ = \(\frac{19500}{590}\) g
33.5 g of steam is to be mixed.

d. A calorimeter has mass 100 g and specific heat 0.1 kcal/kg ·°C. It contains 250 g of liquid at 30 °C having specific heat of 0.4 kcal/kg·°C. If we drop a piece of ice of mass 10 g at 0 °C into the liquid, what will be the temperature of the mixture?
Solution:
Data: m₁ = 100 g, c₁ = 0.1 kcal/kg·°C,
= 0.1 cal/g·°C, T₁ = 30 °C, m₂ = 250 g,
c₂ = 0.4 kcal/kg·°C = 0.4 cal/g·°C, T₂ = 30 °C,
m₃ = 10 g, T₃ = 0 °C, L = 80 cal/g,
c (water) = 1 cal/g·°C, T = ?
Q₁ (heat lost by calorimeter) = m₁c₁ (T- T₁),
Q₂ (heat lost by liquid) = m₂c₂ (T – T₂),
Q₃ (heat absorbed by ice) = m₃ L,
Q₄ (heat absorbed by water formed on melting of ice) = m₃c (T – 0 °C)
According to the principle of heat exchange,
Q₁ + Q₂ = Q₃ + Q₄
∴ m₁c₁ (T₁ – T) + m₂c₂ (T₂ – T) = m₃L + m₃c (T – 0 °C)
∴ m₁c₁T₁ – m₁c₁T + m₂c₂T₂ – m₂c₂T = m₃L + m₃c (T – 0°C)
∴ m₁c₁T₁ + m₂c₂T₂ = m₃L + (m₁c₁ + m₂c₂ + m₃c)T
∴ 100g × 0.1 cal/g°C × 30 °C + 250g × 0.4 cal/g.°C × 30 °C J
= 10 g x× 80 cal/g + (100 g × 0.1 cal/g.°C + 250 g × 0.4 cal/g.°C + 10 g × 1 cal/g.°C) T
∴ (10 + 100 + 10) T = (300 + 3000 – 800)°C
∴ 120 T = 2500 °C
∴ T = \(\frac{2500}{120}\) °C = \(\frac{125}{6}\) °C = 20.83 °C
This is the temperature of the mixture.

Project:
Take help of your teachers to make a working model of Hope’s apparatus and perform the experiment. Verify the results you obtain. [Do it your self]

Can you recall? (Text Book Page No. 62)

Question 1.
What is the difference between heat and temperature?
Answer:
Heat is a form of energy. Particles of matter (atoms, molecules, etc.) possess potential energy and kinetic energy. Total energy (potential energy + kinetic energy) of all particles of matter in a given sample is called it’s thermal energy. When two bodies at different temperatures are in thermal contact with each other, there is transfer of thermal energy from a body at higher temperature to a body at lower temperature. This energy in transfer is called heat. It is expressed in joule, calorie and erg.

Temperature is a quantitative measure of degree of hotness or coldness of a body. It is expressed in °C, °F or K (kelvin). Temperature determines the direction of energy transfer.

Question 2.
What are the different ways of heat transfer?
Answer:
Ways of heat transfer: conduction, convection and radiation.
[Note: heat ≡ heat energy. In the textbook, both the terms are used.]

Use your brain power! (Text Book Page No. 63)

Question 1.
Is the concept of latent heat applicable during transformation of gaseous phase to liquid phase and from liquid phase to solid phase?
Answer:
Yes.

Question 2.
Where does the latent heat go during these transformations?
Answer:
During these transformations, the latent heat is given out by the substance to the surroundings.

Use your brain power! (Text Book Page No. 64)

Question 1.
In the above experiment, the wire moves through the ice slab. However, the ice slab does not break. Why?
Answer:
When the thin wire with two equal weights attached to its ends is hung over the block of ice, it exerts pressure on the ice below it. Due to this, the melting point of the ice below the wire is lowered and some ice melts. The wire passes through the water so formed.

The water above the wire is no longer under pressure and, therefore, refreezes. Once again the ice below the wire melts, and the wire passes through it, and the process continues. In this way, due to alternate melting of ice and refreezing of water, the wire cuts right through the block of ice leaving the block intact.

Question 2.
Is there any relationship of latent heat with regelation?
Answer:
Yes. when the ice melts, heat is absorbed, but the temperature does not change. Also, when water refreezes, heat is given out, but the temperature does not change. This heat absorbed or given out is the latent heat.

Question 3.
You know that as we go higher than the sea level, the boiling point of water decreases. What would be the effect on the melting point of a solid?
Answer:
As we go higher than the sea level, the melting point of solids (i) that expand on melting is lowered due to a decrease in pressure (ii) that contract on melting is raised due to a decrease in pressure.

[The wire used in the experiment is made of a metal (usually copper). Metals are good conductors of heat. Hence, exchange of heat between the portion of the ice above the wire and that below the wire takes place readily.]

Can you tell? (Text Book Page No. 64)

Question 1.
We feel that some objects are cold, and some are hot. Is this feeling related in some way to our body temperature?
Answer:
Yes. If the temperature of the object is lower than our body temperature, e.g., ice, we feel the object is cold. If the temperature of the object is higher than our body temperature, e.g., hot water, we feel the object is hot.

Use your brain power! (Text Book Page No. 66)

Question 1.
How will you explain the following statements with the help of the anomalous behaviour of water?
(1) In regions with cold climate, the aquatic plants and animals can survive even when the atmospheric temperature goes below 0 °C.
(2) In cold regions in winter the pipes for water supply break and even rocks crack.
Answer:

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contraeting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface.

The temperature or the water at the surface continues to fall to 0 °c. Finally, the water at the surface is converted Into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

(2) Sometimes water enters into crevices of the rocks. when the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts tremendous pressure on the rocks which crack and break up Into small pieces.

In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. when the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is rormed, there is an increase in the volume.

As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Fill in the blanks and rewrite the sentences:

Question 1.
The amount of water vapour in air is determined in terms of its…………..
Answer:
The amount of water vapour in air is determined in terms of its absolute humidity.

Question 2.
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their……………
Answer:
If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

Question 3.
When a liquid is getting converted into solid, the latent heat is…………. (Practical Activity Sheet – 1 and 2)
Answer:
When a liquid is getting converted into solid, the latent heat is released.

Rewrite the following statements by selecting the correct options:

Question 1.
……….is used to study the anomalous behaviour of water.
(a) Calorimeter
(b) Joule’s apparatus
(c) Hope’s apparatus
(d) Thermos flask
Answer:
(c) Hope’s apparatus

Question 2.
When water boils and is converted into steam, then………..
(a) heat is taken in and temperature remains constant
(b) heat is taken in and temperatures rises
(c) heat is given out and temperature lowers
(d) heat is given out and temperature remains constant
Answer:
(a) heat is taken in and temperature remains constant

Question 3.
When steam condenses to form water,………..
(a) heat is absorbed and temperature increases
(b) heat is absorbed and temperature remains the same
(c) heat is given out and temperature decreases
(d) heat is given out and temperature remains the same
Answer:
(d) heat is given out and temperature remains the same

Question 4.
The temperature of ice can be decreased below 0 °C by mixing………..in it. (Practice Activity Sheet – 3)
(a) saw dust
(b) sand
(c) salt
(d) coal
Answer:
(c) salt

Question 5.
Ice/water is a substance that………..
(a) expands on melting and contracts on freezing
(b) contracts on melting and does not undergo change in volume on freezing
(c) contracts on melting and expands on freezing
(d) does not undergo any change in volume on melting or freezing
Answer:
(c) contracts on melting and expands on freezing

Question 6.
Heat absorbed when 1 g of ice melts at 0 °C to form 1 g of water at the same temperature is………..cal.
(a) 80
(b) 800
(c) 540
(d) 54
Answer:
(a) 80

Question 7.
The latent heat of vaporization of water is………..
(a) 540 cal/g
(b) 800 cal/g
(c) 80 cal/g
(d) 54 cal/g
Answer:
(a) 540 cal/g

Question 8.
The latent heat of fusion of ice is………..
(a) 540 cal/g
(b) 80 cal/g
(c) 800 cal/g
(d) 4cal/g
Answer:
(b) 80 cal/g

Question 9.
If the temperature of water is decreased from 4 °C to 10 °C, then its………..
(a) volume decreases and density increases
(b) volume increases and density decreases
(c) volume decreases and density decreases
(d) volume increases and density increases
Answer:
(b) volume increases and density decreases

Question 10.
At 4 °C, the density of water is………..
(a) 10 g/cm³
(b) 4g/cm³
(c) 4 × 10³ kg/m³
(d) 1 × 10³ kg/m³
Answer:
(d) 1 × 10³ kg/m³

Question 11.
The density of water is maximum at………..
(a) 0 °C
(b) – 4 °C
(c) 100 °C
(d) 4 °C
Answer:
(d) 4 °C

Question 12.
………..heat is needed to raise the temperature of 1 kg of water from 14.5 °C to 15.5 °C.
(a) 4180 J
(b) 103 J
(c) 1 cal
(d) 4180 cal
Answer:
(a) 4180 J

Question 13.
………..heat is needed to convert 1 g of water at 0 °C and at a pressure of one atmosphere into 1 g of steam under the same conditions.
(a) 80 cal
(b) 540 cal
(c) 89 J
(d) 540 J
Answer:
(b) 540 cal

Question 14.
Water expands on reducing its temperature below………..°C. (March 2019)
(a) 0
(b) 4
(c) 8
(d) 12
Answer:
(b) 4

State whether the following statements are true or false. (If a statement is false, correct it and rewrite it.):

Question 1.
Specific latent heat of fusion is expressed in g/cal.
Answer:
False. (Specific latent heat of fusion is expressed in cal/g.)

Question 2.
If the temperature of water is raised from 0 °C to 10 °C, its volume goes on increasing.
Answer:
False. (If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C and then goes on increasing in the range 4 °C to 10 °C.)

Question 3.
At dew point relative humidity is 100%.
Answer:
True.

Question 4.
1 kcal = 4.18 joules.
Answer:
False. (1 kcal = 4180 joules.)

Question 5.
Specific heat capacity is expressed in cal/g·°C
Answer:
True.

Question 6.
Latent heat of fusion, Q = mL.
Answer:
True.

Question 7.
If the relative humidity is more than 60%, we feel that the air is humid.
Answer:
True.

Question 8.
If the relative humidity is less than 60%, we feel that the air is dry.
Answer:
True.

Question 9.
Relative humidity has no unit.
Answer:
True.

Question 10.
Absolute humidity is expressed in kg/m3.
Answer:
True.

Identify the odd one and give the reason:

Question 1.
Temperature, conduction, convection, radiation.
Answer:
Temperature. It is a physical quantity. Others are modes of transfer of heat.

Question 2.
The joule, The erg, The calorie, The newton.
Answer:
The newton. It is a unit of force. Others are units of energy (as well as work.)

Question 3.
cal/g, cal/g·°C, k cal/kg·°C, erg/g·°C.
Answer:
cal/g. It is a unit of specific latent heat. Others are units of specific heat capacity.

Match the columns:

Column A	Column B
1. Latent heat	a. Q = mc ΔT
2. Specific heat capacity	b. Q = mL
3. Heat absorbed or given out by a body when its temperature changes.	c. kcal
	d. cal/g·°C

Answer:
(1) Latent heat – Q = mL
(2) Specific heat capacity – cal/g·°C
(3) Heat absorbed or given out by a body when its temperature changes – Q = mc ΔT.

Answer the following questions in one sentence each:

Question 1.
State units of temperature.
Answer:
Units of temperature: °C, °F and K (kelvin).

Question 2.
State units of energy.
Answer:
Units of energy: the erg, the joule, the calorie.

Question 3.
State the relation between the joule and the calorie.
Answer:
1 calorie = 4.18 joules.

Question 4.
State the relation between the erg and the joule.
Answer:
1 joule = 10⁷ ergs.

Question 5.
State the relation between the erg and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10¹⁰ ergs.

Question 6.
State the relation between the joule and the kilocalorie.
Answer:
1 kilocalorie = 4.18 × 10³ joules.

Question 7.
When heat energy is absorbed by an object, ΔT represents the rise in temperature. What would ΔT represent if the object loses heat energy? (Practice Activity Sheet – 4)
Answer:
If the object loses heat energy, ΔT would represent the decrease in temperature.

Answer the following questions:

Question 1.
Define latent heat of fusion.
(OR)
What is latent heat of fusion? State its units.
Answer:
When a solid is converted into liquid at constant temperature (melting point of the substance) the amount of heat absorbed by it is called the latent heat of fusion.
Heat is a form of energy. Hence, latent heat is expressed in units joule, erg, calorie or kilocalorie.

Question 2.
Define specific latent heat of fusion.
(OR)
What is specific latent heat of fusion? State its units.
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a solid to convert into liquid phase is called the specific latent heat of fusion.
It is expressed in units J/kg, erg/g, cal/g, kJ/ kg and kcal/kg.

[Note: Specific latent heat (L) = \(\frac{\text { latent heat }(Q)}{\text { mass of the substance }(m)}\)
:. SI unit of specific latent heat = SI unit of energy / SI unit of mass = J/kg]

Question 3.
Explain the term latent heat of vaporization.
Answer:
When a liquid is heated continuously, initially, its temperature increases. Later, at a certain stage, its temperature does not increase even when heat is supplied to it. At this temperature, heat absorbed by the liquid is used for breaking the bonds between its atoms or molecules, i.e., for doing work against the forces of attraction between the atoms or molecules and conversion into gaseous phase.

This heat is called the latent heat of vaporization and the constant temperature at which this change of state occurs is called the boiling point of the liquid.

Question 4.
Define boiling point of a liquid.
(OR)
What is boiling point of a liquid?
Answer:
The constant temperature at which a liquid transforms into gaseous state is called the boiling point of the liquid.
[Note: On application of pressure, the boiling point of a liquid is raised. On reducting the pressure, the boiling point is lowered.]

Question 5.
Define specific latent heat of vaporization.
OR
What is specific latent heat of vaporization?
Answer:
The amount of heat energy absorbed at constant temperature by unit mass of a liquid to convert into gaseous phase is called the specific latent heat of vaporization.

Question 6.
The specific latent heat of fusion of ice is 80 cal/g. Explain this statement.
Answer:
When 1 g of ice at a pressure of one atmosphere and at a temperature 0 °C is converted into 1 g of water, heat absorbed by the ice is 80 cal.

Question 7.
The specific latent heat of fusion of silver is 88.2 kJ/kg. Explain this statement.
Answer:
When 1 kg of silver at a pressure of one atmosphere and at a temperature of 962 °C (melting point of silver) is converted into 1 kg of silver in liquid phase, heat absorbed by the silver is 88.2 kJ.

Question 8.
The specific latent heat of vaporization of water is 540 cal/g. Explain this statement.
Answer:
When 1 g of water at a pressure of one atmosphere and at a temperature of 100 °C is converted into 1 g of steam, heat absorbed by the water is 540 cal.

Question 9.
Define regelation.
(OR)
What is regelation?
Answer:
The phenomenon in which the ice converts to liquid due to applied pressure and then re-converts to ice once the pressure is removed is called regelation.

Question 10.
The terms hot and cold are used in relative context. Explain.
Answer:

(1) Take three large bowls, P, Q and R. Fill bowl P with cold water, bowl Q with lukewarm water, and bowl R with hot water.
(2) Immerse your right hand in bowl P, and left hand in bowl R for about five seconds.
(3) Now, immerse both the hands in bowl Q at the same time.
(4) You will find that the water in bowl Q appears warm to your right hand, and cold to your left hand. Thus, the hand immersed in cold water for some time finds the lukewarm water hot while the one immersed in hot water finds the same lukewarm water cold. This experiment shows that the terms hot and cold are relative.

Question 11.
Draw a neat labelled diagram of Hope’s apparatus. Explain how this apparatus can be used to demonstrate anomalous behaviour of water. Draw a graph of temperature of water against time.
Answer:

The figure shows Hope’s apparatus. Initially, the cylindrical container in Hope’s apparatus is filled with water at about 12 °C and the flat bowl is filled with a freezing mixture of ice and salt.

The temperature of water in the upper part of the container (T₂) is recorded by thermometer T₂ and that of water in the lower part of the container (T₁) is recorded by thermometer T₁. Figure shows variation of temperature of water with time.

Initially, both the thermometers show the same temperature (say, 12 °C). In a short time, the temperature shown by the lower thermometer starts decreasing, while the temperature shown by the upper thermometer does not change very much.

This process continues till the temperature shown by the lower thermometer falls to 4 °C and remains constant thereafter. This shows that in the temperature range 12 °C to 4 °C, the density of the water in the central part of the container goes on increasing and hence the water sinks to the bottom. It means that water contracts, i.e., its volume decreases as its temperature falls from 12 °C to 4 °C.

As the temperature of the water in the central part of the container becomes less than 4 °C, the temperature shown by the upper thermometer begins to fall rapidly to 0 °C. But the temperature shown by the lower thermometer remains constant (4 °C). Later, the heading shown by the lower thermometer decreases to 0 °C.

In the temperature range 4 °C to 0 °C, the water moves upward. This shows that the density of water goes on decreasing in this range. It means that water expands, i.e. its volume increases as its temperature falls from 4 °C to 0 °C.

Thus, the volume of a given mass of water is minimum at 4 °C, i.e., the density of water is maximum at 4 °C.

In the above figure, the point of intersection of the two curves shows the temperature at which the density of water is maximum. This temperature is 4 °C.

Question 12.
A mountaineer climbing on the Everest, experienced the following facts. Explain each fact with the scientific reason : (1) He found j fishes alive below the ice (2) Time required for cooking was more as he went higher (3) He saw many times cliffs falling suddenly (4) He saw tubes carrying water broken.
Answer:
Explanation:
(1) Water expands as its temperature decreases from 4 °C to 0 °C. Water is converted into ice at 0 °C. The density of water is more than that of ice. Fishes can remain alive in the water (at 4 °C) below the ice.
(2) At high altitudes, atmospheric pressure is low and hence water boils at a temperature lower than its normal boiling point. Therefore, the time required for cooking food is more at higher altitudes.
(3) Water expands while freezing. Hence, the water present in the crevices of the rocks exerts a tremendous pressure on the rocks, while freezing. Therefore, the cliffs fall.
(4) Water expands while freezing. Hence, the water in the tube exerts a large pressure on the tube, while freezing. Therefore, the tube carrying water breaks.

Question 13.
What is humidity?
Answer:
The moisture, i.e., the presence of water vapour, in the atmosphere is called humidity.

Question 14.
When is air said to be saturated with water vapour?
Answer:
When air contains maximum possible water vapour, it is said to be saturated with water vapour at that temperature.

Question 15.
What does the amount of water vapour needed to saturate air depend on?
Answer:
The extent of water vapour needed to saturate air depends on the temperature. The greater the temperature, the greater is the amount of water vapour needed to saturate air.

Question 16.
When is air said to be unsaturated with water vapour?
Answer:
When air contains water vapour less than its capacity to hold water vapour at that temperature, it is said to be unsaturated with water’vapour.

Question 17.
What is dew point temperature?
(OR)
Define dew point temperature.
Answer:
If the temperature of unsaturated air is decreased, a temperature is reached at which the air becomes saturated with water vapour. This temperature is called the dew point temperature.

Question 18.
Name the physical quantity used to express the amount of water vapour present in air.
Answer:
Absolute humidity.

Question 19.
Define absolute humidity.
(OR)
What is absolute humidity? State its unit.
Answer:
The mass of water vapour present in a unit volume of air is called absolute humidity. Generally it is expressed in kg/m³.

Question 20.
Define relative humidity.
(OR)
What is relative humidity? Write the formula for % relative humidity.
Answer:
The ratio of the actual mass of water vapour content in the air for a given volume and temperature to that required to make the same volume of air saturated with water vapour at the same temperature is called the relative humidity.

% Relative humidity = [the actual mass of water vapour content in the air for a given volume and temperature ÷ the mass of water vapour required to make the same volume of air saturated with water vapour at the same temperature] × 100%.

Question 21.
What is the value of relative humidity at the dew point temperature?
Answer:
At the dew point temperature, relative humidity is 100%.

Question 22.
The mass of water vapour in air enclosed in a certain space is 60 g and the mass of water vapour needed to saturate the same air with water vapour under the same conditions is 100 g. What is the corresponding % relative humidity?
Answer:
Here, % relative humidity = (\(\frac{60 \mathrm{g}}{100 \mathrm{g}}\)) × 100% = 60%

Question 23.
During winter, sometimes we see a white trail at the back of a flying aeroplane in a clear sky. Explain why.
Answer:
In winter, air temperature is low. Hence, when an aeroplane flies, the vapour released by its engine condenses and forms white clouds. If the relative humidity of the air surrounding the plane is high, we see this white trail at the back of the plane for a long time before it disappears. If.the relative humidity is low, the white trail is short and disappears quickly. If the relative humidity is very low, there is no formation of the white trail.

Question 24.
State two effects of humidity present in atmosphere.
Answer:
Effects of humidity present in atmosphere: When the temperature of air falls below the dew point, dew and fog are formed.

Question 25.
Explain how dew and fog are formed.
(OR)
Write a short note on formation of dew and fog.
Answer:
At a particular temperature, a given volume of air can contain a certain maximum amount of water vapour. Normally, the temperature of air during the day is such that air is not saturated with water vapour present in it.

As the temperature falls, the capacity of air to hold water vapour becomes less. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed. If the water vapour condenses on the fine dust particles present in the atmosphere, mist or fog is formed.

Question 26.
State the units of heat.
Answer:
Units of heat: joule, erg, calorie, kilocalorie.

Question 27.
Define the kilocalorie.
Answer:
The amount of heat necessary to raise the temperature of 1 kg of water by 1 °C from 14.5 °C to 15.5 °C is called one kilocalorie.

Question 28.
Define the calorie.
Answer:
The amount of heat necessary to raise the temperature of 1 g of water by 1 °C from 14.5 °C to 15.5 °C is called one calorie.

Question 29.
State the relation between the kilocalorie and the calorie.
Answer:
1 kilocalorie = 10³ calories.

Question 30.
Study the following procedure and answer the questions below:  (Practice Activity Sheet – 2)
1. Take 3 spheres of iron, copper and lead of equal mass.
2. Put all the 3 spheres in boiling water in a beaker for some time.
3. Take 3 spheres out of the water. Put them immediately on a thick slab of wax.
4. Note the depth that each sphere goes into the wax.
(i) Which property of a substance can be studied with this procedure?
(ii) Describe that property in minimum words.
(iii) Explain the rule of heat exchange with this property.
Answer:
(i) Specific heat.
(ii) Specific heat: The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C.
(iii) According to the rule/principle of heat exchange, heat energy lost by the hot object = heat energy gained by the cold object.

In this activity, heat absorbed by the iron sphere is transmitted more in the wax, hence the sphere goes deepest into the wax, while the lead sphere absorbs less heat, resulting in less transmission of heat in the wax, hence, the sphere goes the least depth into the wax.

Question 31.
Write the symbol for specific heat capacity. State the units of specific heat capacity.
Answer:
Symbol for specific heat capacity: c.
Units of specific heat capacity: J/kg·°C,
erg/g·°C, cal/g·°C, kcal/kg·°C.
[ Notes: (1) Specific heat capacity

In SI, heat is expressed in joule (J), mass in kg and temperature in kelvin(K).
∴ SI unit of specific heat capacity = \(\frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\). (2) The specific heat capacity of a substance depends upon its constituent particles (atoms, molecules, etc.), interaction between them, structure of the substance (atomic/molecular arrangement), temperature of the substance, etc.]

Question 32.
Explain the principle of heat exchange. Ans. Suppose two objects A and B at different temperature T₁ and T₂ respectively are enclosed in a box of heat resistant material as shown in figure.

Let m₁ = mass of A, m₂ = mass of B, c₁ = specific heat capacity of A, c₂ = specific heat capacity of B and T = common temperature attained by A and B after the heat exchange between A and B. Here, no heat leaves the box or enters the box from outside. Hence, if T₁ > T₂, heat energy lost by A (Q₁) = heat energy gained by B (Q₂).
∴ m₁c₁ (T₁ – T) = m₂c₂ (T – T₂)
[Note: If m₁, c₁, T₁, T, m₂ and T₂ are known, c₂ can be determined.]

Question 33.
The specific heat capacity of silver is 0.056 kcal/kg·°C. Explain this statement.
Answer:
The amount of heat needed to raise the temperature of 1 kg of silver by 1 °C is 0.056 kcal.

Question 34.
Explain how the specific heat capacity of a solid can be determined (measured) by the method of mixture.
Answer:
A hot solid is put in water in a calorimeter. The mixture is stirred continuously and the maximum temperature of the mixture is measured with a thermometer. Heat exchange between the hot solid, water and calorimeter results in sill bodies attaining the same temperature after some time. Hence, according to the principle of heat exchange, heat lost by the solid = heat gained by the water in the calorimeter + heat gained by the calorimeter.

Now, heat lost by the solid (Q) = mass of the solid × its specific heat capacity × decrease in its temperature, heat gained by the water (Q₁) = mass of the water × its specific heat capacity × increase in its temperature and heat gained by the calorimeter (Q₂) = mass of the calorimeter × its specific heat capacity × increase in its temperature.

Heat lost by the hot object = heat gained by the calorimeter + heat gained by the water. Q = Q₂ + Q₁
Using this equation, the specific heat capacity of the solid can be determined (measured) when the other quantities are known.

Give scientific reasons:

Question 1.
Even though heat is supplied to boiling water, there is no increase in its temperature.
Answer:
Once water starts boiling, all the heat supplied to it is used in conversion of water into steam at the boiling point of water. Hence, there is no rise in its temperature.

Question 2.
Burns from steam are worse those from boiling water at the same temperature.
Answer:
1. A given quantity of steam contains more heat than the same quantity of boiling water at the same temperature.
2. When steam comes in contact with one’s body, it releases extra heat of 540 calories per gram and causes a more serious burn than that caused by boiling water.

Question 3.
In winter, the pipelines carrying water burst in cold countries.
Answer:
1. In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. When the temperature of water falls below 4 °C, it expands. Hence, the water in pipes expands. Even if ice is formed, there is an increase in the volume.

2. As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.

Question 4.
If crushed ice is pressed and then the pressure is released, a lump of ice is formed.
Answer:
1. When crushed ice is pressed, its melting point is lowered and some ice melts to form water.
2. When pressure is released, the melting point becomes normal and the water freezes to form ice forming a lump.

Question 5.
In cold countries, in winter, even when the water of lakes freezes, aquatic animals and plants can survive.
Answer:
1. In cold countries, in winter, a layer of ice is formed on the surface of lakes when the atmospheric temperature falls below 0 °C. However, below this layer, there is water at 4 °C.
2. Ice, being a bad conductor of heat, does not allow transfer of heat from this water to the atmosphere. Hence, aquatic animals and plants can survive in this water.

Question 6.
Water droplets are seen on’ the outer surface of a cold drink bottle.
Answer:
1. The temperature of the outer surface of a cold drink bottle is less than that of the atmosphere.
2. Therefore, the excess of water vapour from the air condenses to form droplets on the outer surface of the cold drink bottle.

Question 7.
During cold nights, sometimes dew is formed.
Answer:
1. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. 2. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed.

Question 8.
When you enter a warm room after being outside on a frosty early morning, your spectacles ‘steam up’.
Answer:
1. On a frosty early morning, the temperature of air outside a warm room is lower than the dew point.
2. Hence, when you enter the room from outside, some water vapour in the room condenses on the glass of your spectacles, i.e., the spectacles ‘steam up’.

Question 9.
A plastic bottle, completely filled with water, when kept in a freezer, is likely to break.
Answer:
The temperature of air in the freezer (deep freeze) compartment of a refrigerator is less than 0 °C. 2. When a plastic bottle, completely filled with water, is kept in this compartment, the temperature of water falls below 4 °C and the water expands. Even when water freezes and ice is formed, there is an increase in the volume. It exerts a large pressure on the sides of the bottle and hence the bottle is likely to break.

Question 10.
The outer surface of a beaker containing ice cubes becomes wet in a short while.
Answer:
1. When ice cubes are placed in a beaker, ice starts melting. The heat required for melting is absorbed from the surrounding air and also from the beaker to some extent.
Hence, the temperature of the air and beaker falls.

2. The capacity of air to hold water vapour depends upon the temperature of the air, and this capacity decreases as the temperature decreases. At a certain low temperature, the surrounding air becomes saturated with water vapour present in it. As the temperature falls further, the air is unable to hold all the water vapour.

Hence, the extra water vapour starts condensing on the cold outer surface of the beaker in the form of minute drops. Therefore, the outer surface of the beaker containing ice cubes becomes wet in a short while.

Distinguish between the following:

Question 1.
Absolute humidity and Relative humidity.
Answer:
Absolute humidity:

1. Absolute humidity is the mass of water vapour present in a unit volume of air.
2. It is commonly expressed in kg/m³.

Relative humidity:

1. Relative humidity is the ratio of the mass of water vapour in a given volume of air at a given temperature to the mass of water vapour required to saturate the same volume of air at the same temperature.
2. It does not have unit.

Solve the following examples/Numerical problems:
[Use the data given in the Tables on pages 130 and 131.]

Problem 1.
Calculate the amount of heat required to convert 5 g of ice of 0 °C into water at 0 °C. (Specific latent heat of fusion of ice = 80 cal/g)
Solution: Here, m = 5 g, L = 80 cal/g; Q = ?
Amount of heat required, Q = mL
= 5 g × 80 cal/g
= 400 calories.

Problem 2.
Find the amount of heat required to convert 10 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution: Here, m = 10 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 10 g × 540 cal/g
= 5400 calories.

Problem 3.
Calculate the amount of heat required to convert 15 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
m = 15 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 15 g × 540 cal/g
= 8100 calories.

Problem 4.
How many calories of heat will be absorbed when 3 kg of ice at 0 °C melts?
Solution:
m = 3 kg = 3000 g; L = 80 cal/g; Q = ?
Quantity of heat absorbed, Q = mL
= 3000 g × 80 cal/g
∴ Q = 240000 calories.

Problem 5.
Calculate the amount of heat required to convert 10 g of water at 30 °C into steam at 100 °C. (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
Here, m = 10 g; c = 1 cal/g·°C
T₂ – T₁ = 100 °C – 30 °C = 70 °C; L = 540 cal/g; Q = ?
Amount of heat required, Q = mc (T₂ – T₁) + mL
= 10 g × 1 cal/g·°C × 70 °C + 10 g × 540 cal/g
= 700 cal + 5400 cal
∴ Q = 6100 calories.

Problem 6.
If water of mass 80 g and temperature 45 °C is mixed with water of mass 20 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
Data : m₁ = 80 g, T₁ = 45 °C, m₂ = 20 g,
T₂ = 30 °C, T = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T1 – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₁T₁ + m₂T₂ = (m1 + m₂)T
∴ Maximum temperature of the mixture,

= (36 + 6) °C
= 42°C.

Problem 7.
When water of mass 70 g and temperature 50 °C is added to water of mass 30 g, the maximum temperature of the mixture is found to be 41 °C. Find the temperature of water of mass 30 g before hot water was added to it.
Solution:
Data : m₁ = 70 g, T₁ = 50 °C, m₂ = 30 g, T = 41 °C, T₂ = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
∴ m₁c (T₁ – T) = m₂c (T – T₂)
∴ m₁T₁ – m₁T = m₂T – m₂T₂
∴ m₂T₂ = (m₁ + m₂) T – m₁T₁

This is the required temperature.

Problem 8.
Find the heat needed to raise the temperature of a silver container of mass 100 g by 10 °C. (c = 0.056 cal/g.°C)
Solution:
Data: m = 100 g, ΔT = 10 °C, c = 0.056 cal/g·°C
Heat needed to raise the temperature of the container = mc ΔT
= 100 g × 0.056 cal/g·°C × 10 °C .
= 56 calories.

Problem 9.
If steam of mass 100 g and temperature 100 °C is released on an ice slab of temperature 0 °C, how much ice will melt?
Solution:
Data: m₁ = 100 g, L₁ = 540 cal/g,
T₁ = 100 °C, mass of ice, m = ?, L₂ = 80 cal/g, c (water) = 1 cal/g·°C
According to the principle of heat exchange, heat lost by hot body = heat gained by cold body. Conversion of steam into water:
Q₁ = m₁L₁ = 100 g × 540 cal/g = 54000 cal
Decrease in the temperature of this water to 0 °C:
Q₂ = m₁c × (T₁ – 0 °C) = 100 g × 1 cal/g·°C × (100 °C – 0 °C) = 10000 Cal
Melting of ice: Q₃ = mL₂
= m × 80 cal/g
Now, Q₁ + Q₂ = Q₃
∴ (54000 + 10000) cal = m × 80 cal/g
∴ m = \(\frac{64000}{80}\) = 800 g
800 g of ice will melt.

Numerical problems for practice:

Problem 1.
Calculate the amount of heat required to convert 80 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
6400 cal

Problem 2.
Find the heat required to convert 20 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
1600 cal

Problem 3.
Calculate the quantity of heat released during the conversion of 10 g of ice cold water (temperature 0 °C) into ice at the same temperature. (Specific latent heat of freezing of water = 80 cal/g)
Solution:
800 cal

Problem 4.
How many calories of heat will be absorbed when 2 kg of ice at 0 °C melts? (Specific latent heat of fusion of ice = 80 cal/g)
Solution:
160000 cal

Problem 5.
How much heat will be required to convert 20 g of water at 100 °C into steam at 100 °C? (Specific latent heat of vaporization of water = 540 cal/g)
Solution:
10800 cal

Problem 6.
Find the heat absorbed by 25 g of water at 100 °C to convert into steam at the same temperature. (Specific latent heat of vaporization of water = 540 cal/g.)
Solution:
13500 cal

Problem 7.
If water of mass 60 g and temperature 50 °C is mixed with water of mass 40 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution:
42 °C

Problem 8.
If water of mass 60 g and temperature 60 °C is mixed with water of mass 60 g and temperature 40 °C, what will be the maximum temperature of the mixture?
Solution:
50 °C

Problem 9.
Find the heat needed to raise the temperature of a piece of iron of mass 500 g by 20 °C. (c = 0.110 cal/g·°C)
Solution:
1100 cal

Problem 10.
Water of mass 200 g and temperature 30 °C is taken in a copper calorimeter of mass 50 g and temperature 30 °C. A copper sphere of mass 100 g and temperature 100 °C is released into it. What will be the maximum temperature of the mixture? [c (water) = 1 cal/g·°C, c (copper) =0.1 cal/g·°C]
Solution:
33.26 °C

Problem 11.
A copper calorimeter of mass 100 g and temperature 30 °C contains water of mass 200 g and temperature 30 °C. If a piece of ice of mass 40 g and temperature 0 °C is added to it, what will be the maximum temperature of the mixture? [c (copper) = 0.1 cal/g·°C, c (water) = 1 cal/g·°C, L = 80 cal/g]
Solution:
12.4 °C

Balbharti Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

Question 1.
Match the pairs.

Group A	Group B
a. C2H6	1. Unsaturated hydrocarbon
b. C2H2	2. Molecular formula of an alcohol
c. CH4O	3. Saturated hydrocarbon
d. C3H6	4. Triple bond

Question 2.
Draw an electron dot structure of the following molecules. (Without showing the circles)
a. Methane.
Answer:
Molecular formula: CH₄

b. Ethene.
Answer:
Molecular formula: H₂C = CH₂

c. Methanol.
Answer:
Molecular formula: H₃C – OH

d. Water.
Answer:
Molecular formula: H₂O

Question 3.
Draw all possible structural formulae of compounds from their molecular formula given below.
a. C₃H₈
b. C₄H₁₀
c. C₃H₄
Answer:
a. C₃H₈ Propane:

b. C₄H₁₀ Butane:

c. C₃H₄ Propyne:

Question 4.
Explain the following terms with example.
a. Structural isomerism.
Answer:
The phenomenon in which compounds having different structural formulae have the same molecular formula is called structural isomerism. Butane is represented by two different compounds as their structural formulae are different. The first compound is a straight chain compound and the second compound is a branched chain compound. These two different structural formulae have the same molecular formula i.e. C₄H₁₀.

b. Covalent bond.
Answer:
The chemical bond formed by sharing of two valence electrons between the two atoms is called covalent bond.
Example:
1. Hydrogen molecule formation: The atomic number of hydrogen is 1, its atom contains 1 electron in K shell. It requires one more electron to complete the K shell and attain the configuration of helium (He). To meet this requirement two hydrogen atoms share their electrons with each other to form H₂ molecule. One covalent bond, i.e. a single bond is formed between two hydrogen atoms by sharing of two electrons.

2. Formation of oxygen molecule:
(1) The atomic number of oxygen is 8. The electronic configuration of oxygen is (2, 6). Oxygen has 6 electrons in the outermost shell.
(2) It requires 2 electrons to complete the L shell and attain the configuration of neon (Ne).
(3) Each oxygen atom shares its valence electron with the valence electron of another oxygen atom to give two shared pairs of electrons which results in the formation of oxygen molecule.
(4) Thus, two electron pairs are shared between two oxygen atoms, forming double covalent bond (=).

c. Hetero atom in a carbon compound.
Answer:
Carbon compounds are formed by formation of bonds of carbon with other elements such as halogens, oxygen, nitrogen, sulfur. The atoms of these elements substitute one or more hydrogen atoms in the hydrocarbon chain and thereby the tetravalency of carbon is satisfied. The atom of the element which is substitute for hydrogen is referred to as a heteroatom. Sometimes hetero atoms are not alone but exist in the form of certain groups of atoms.

d. Functional group.
Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the functional groups.

e. Alkane.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: In methane, four hydrogen atoms are bonded to carbon atom by four single covalent bonds.

f. Unsaturated hydrocarbons.
Answer:
The carbon compounds having a double bond or triple bond between two carbon atoms are called unsaturated hydrocarbons. The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
e.g. Ethene (CH₂ = CH₂), Propene (CH₃ – CH = CH₂).
The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes e.g. Ethyne (CH ≡ CH).

g. Homopolymer.
Answer:
The polymers formed by repetition of single monomer are called homopolymer. e.g. polyethylene (CH₂ – CH₂)n.

h. Monomer.
Answer:
The small unit that repeats regularly to form a polymer is called monomer.
Example: Ethylene.

i. Reduction.
Answer:
In a chemical reaction, removal of oxygen from a compound or addition of hydrogen to a compound is called a reduction.

j. Oxidant.
Answer:
An oxidant is a reactant that oxidizes or removes electrons from other reactants during a redox reaction. An oxidant may also be called an oxidizer or oxidizing agent. When the oxidant includes oxygen, it may be called an oxygenation reagent or oxygen-atom transfer (OT) agent.
Examples of oxidants include:

1. Hydrogen peroxide
2. Ozone
3. Nitric acid
4. Sulfuric acid

Question 5.
Write the IUPAC names of the following structural formulae.
a. CH₃ – CH₂ – CH₂ – CH₃
Answer:
The number of carbon atopic in the longest chain: 4
Parent alkane: Butane IUPAC name: n-Butane

b. CH₃ – CHOH – CH₃ (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3

Parent alkane: Propane
Functional group: -OH (ol)
Assign the number: 2
The carbon atom to which the -OH group is attached is numbered as C₂. If the carbon chain of the compound contains a -OH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘ol’. (ol stands for alcohol)
Parent suffix: Propan-2-ol
IUPAC name: Propan-2-ol

c. CH₃ – CH₂ – COOH (Practice Activity Sheet – 3)
Answer:
The number of carbon atoms in the longest chain: 3
Parent alkane: Propane
Functional group: -COOH (-oic acid)
If the carbon chain of the compound contains a -COOH group then change the ending of the parent name, i.e., ‘e’ of propane is replaced by ‘oic acid’.
Parent suffix: Propanoic acid
IUPAC name: Propanoic acid

d. CH₃ – CH₂ – NH₂
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -NH₂ (amine)
If the carbon chain of the compound contains a -NH₂ group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘amine’.
Parent suffix: Ethanamine
IUPAC name: Ethanamine.

e. CH₃ – CHO
Answer:
Number of carbon atoms: 2
Parent alkane: Ethane
Functional group: -CHO (al)
If the carbon chain of the compound contains a -CHO group, then change the ending of the parent name, i.e., ‘e’ of ethane is replaced by ‘al’.
Parent suffix: Ethanal
IUPAC name: Ethanal

f. CH₃ – CO – CH₂ – CH₃
Answer:
Number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -CO- (one)
Assign the number:

In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the function group.
If the carbon chain of the compound contains a (-CO-) group, then change the ending of the parent name, i.e., ‘e’ of butane is replaced by ‘one’.
Parent suffix: Butan-2-one
IUPAC name: Butan-2-one

Question 6.
Identify the type of the following reaction of carbon compounds.
1. CH₃ – CH₂ – CH₂ – OH + (O) → CH₃ – CH₂ – COOH
2. CH₃ – CH₂ – CH₃ + O₂ → 3CO₂ + 4H₂O
3. CH₃ – CH = CH – CH₃ + Br₂ → CH₃ – CHBr – CHBr – CH₃
4. CH₃ – CH₃ + Cl₂ → CH₃ – CH₂ – Cl + HCl
5. CH₃ – CH₂ – CH₂ – CH₂ – OH → CH₃ – CH₂ – CH = CH₂ + H₂O
6. CH₃ – CH₂ – COOH + NaOH → CH₃ – CH₂ – COONa⁺ + H₂O
7. CH₃ – COOH + CH₃ – OH → CH₃ – COO – CH₃ + H₂O
Answer:

Question 7.
Write the structural formulae for the following IUPAC names:
a. Pent-2-one
Answer:
Pent-2-one.
(1) Pent stands for 5 carbon atoms in a chain.
Number the carbon atoms in a chain as 1, 2, 3,…..

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

b. 2-Chlorobutane
Answer:
(1) In 2-chlorobutane, butane is parent alkane stands for 4 carbon atoms and number the carbon atoms in a chain as 1, 2, 3,….

(2) Chloro (Halo) is the prefix and the number assigned for prefix (chloro) is 2. Show the chloro atom at C₂

(3) Now satisfy the valencies of each carbon atom

c. Propan-2-ol
Answer:
(1) Propan stands for 3 carbon atoms in a chain. Number the carbon atom in a chain as 1, 2, 3.

(2) ‘-ol’ stands for (-OH) hydroxyl group. The number assigned for the hydroxyl group is 2. Show the -OH group at C₂.

(3) Now satisfy the valencies of each carbon atom

d. Methanal
Answer:
(1) Meth – stands for one carbon atom and assigned the number ‘1’ to carbon in the functional group -CHO.
(2) ‘-al’ stands for functional group (-CHO) aldehyde.
(3) Now satisfy the valencies of carbon in -CHO.

e. Butanoic acid
Answer:
(1) But stands for 4 carbon atoms in a chain. Number the carbon atoms in a chain as 1, 2, 3,….

‘-oic acid’ stands for functional group -COOH. Assign the number 1 to carbon in the functional group -COOH.

Now satisfy the valencies of each carbon atom

f. 1-Bromopropane.
Answer:
(1) In 1-bromopropane, propane is parent alkane stands for 3 carbon atoms and number the carbon atoms in a chain as 1, 2, 3…….

(2) Bromo (Halo) is the prefix and the number assigned for prefix (bromo) is 1, show the bromine atom at C₁.

(3) Now satisfy the valencies of each carbon atom

g. Ethanamine
Answer:
(1) Eth stands for 2 carbon atoms in a chain and the parent alkane is ethane.
– C – C –
(2) ‘amine’ stands for (- NH₂) amino group. Show the amino (-NH₂) at any carbon atom.
– C – C – NH₂
(3) Now satisfy the valencies of each carbon atom

h. Butanone.
Answer:
(1) But stands for 4 carbon atoms in a chain and the parent alkane is butane. Number the carbon atoms in a chain 1, 2, 3,….

(2) ‘one’ stands for functional group

ketone. The number assigned for the ketone group is 2. Show the ketone group at C₂.

(3) Now satisfy the valencies of each carbon atom

Question 8.
a. What causes the existance of very large number of carbon compound?
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms; this results in formation of big molecules. This property of carbon is called catenation power. The carbon compounds contain open chains or closed chains of carbon atoms. An open chain can be a straight chain or a branched chain. A closed chain is a ring structure. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

(2) One, two or three covalent bonds can bond together two carbon atoms. These bonds are called single covalent bond, double covalent bond and triple covalent bond respectively. Due to the ability of carbon atoms to form multiple bonds as well as single bonds, the number of carbon compounds increases. For example, there are three compounds, namely, ethane (CH₃ – CH₃), ethene (CH₂ = CH₂) and ethyne (CH = CH) which contain two carbon atoms.

(3) Carbon being tetravalent, one carbon atom can form bonds with four other atoms (carbon or any other). This results in formation of many compounds. These compounds possess different properties as per the atoms to which carbon is bonded. For example, five different compounds are formed using one carbon atom and two monovalent elements hydrogen and chlorine: CH₄, CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄. Similarly carbon atoms form covalent bonds with atoms of elements like O, N, S, halogen and P to form different types of carbon compounds in large number.

(4) Isomerism is one more characteristic of carbon compound which is responsible for large number of carbon compounds.

b. Saturated hydrocarbons are classified into three types. Write these names giving one example each.
Answer:
In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called saturated hydrocarbons. Methane molecule contains only one carbon atom. In methane, four hydrogen atoms are bonded to carbon atom by four covalent bonds.

c. Give any four functional groups containing oxygen as the heteroatom in it. write name and structural formula of one example each.
Answer:

d. Give names of three functional groups containing three different heteroatoms. Write name and structural formula of one example each.
Answer:

e. Give names of three natural polymers. write the place of their occurance and names of monomers from which they are formed.
Answer:

1. Poly saccharide is a natural polymer. It occurs in starch/carbohydrates. It is formed from monomer glucose.
2. Protein is a natural polymer. It occurs in muscles, hair, enzymes, skin, egg. It is formed from alpha amino acids.
3. Rubber is a natural polymer. It occurs in latex of rubber tree. It is formed from monomer isoprene.

f. What is meant by vinegar and gasohol? What are their uses?
Answer:
(1) Vinegar is a 5 – 8% aqueous solution of acetic acid. It is used as preservative in pickles. It is used to cook meat. 1t is used as a salad dressing.
(2) To increase the efficiency of petrol, it is mixed with 10% anhydrous ethanol, such a fuel is called gasohol. It is used as a fuel in cars and other vehicles.

g. what is a catalyst ? write any one reaction which is brought about by use of catalyst?
Answer:
Catalyst is a substance, which changes the rate of reaction, without causing any disturbance to it. Vegetable oil (unsaturated compound) undergoes addition reaction with hydrogen in the presence of nickel catalyst to form vanaspati ghee (saturated compound).

Project:
Prepare a chart giving detailed information of carbon compounds in everyday use. Display it in the cluss and discuss.

Can you recall? (Text Book Page No. 110)

Question 1.
What are the types of compounds?
Answer:
Organic and inorganic compounds are the two important types of compounds.

Question 2.
Objects in everyday use such as foodstuff, fibres, paper, medicines, wood, fuels are made of various compounds. Which constituent elements are common in these compounds?
Answer:
The constituent elements common in these compounds are carbon (C), hydrogen (H) and oxygen (O).

Question 3.
To which group in the periodic table does the element carbon belongs? Write down the electronic configuration of carbon and deduce the valency of carbon.
Answer:
The element carbon belongs to group 14 and its electronic configuration is 2, 4. The valency of carbon is 4.

Use your brain Power! (Text Book Page No. 115)

Question 1.
The molecular formula of ethyne is C₂H₂. From this draw its structural formula and electron-dot structure.
Answer:
Ethyne: Molecular formula: C₂H₂

Question 2.
How many bonds have to be there in between the carbon atoms in ethyne so as to satisfy their tetra valency?
Answer:
To satisfy their tetravalency, three double bonds have to be there in between two carbon atoms in ethyne.

Use your brain power! (Text Book Page No. 116)

Question 1.
Draw the electron-dot structure of cyclohexane.
Answer:
Cyclohexane: Molecular formula: C₆H₁₂

Use your brain power! (Text Book Page No. 112)

Question 1.
Atomic number of chlorine is 17. What is the number of electrons in the valence shell of the chlorine?
Answer:
There are seven electrons in the valence shell of the chlorine.

Question 2.
Molecular formula of chlorine is Cl₂. Draw electron-dot and line structure of a chlorine molecule.
Answer:

Question 3.
The molecular formula of water is H₂O. Draw electron-dot and line structures for triatomic molecule. (Use dots for electron of oxygen atom and crosses for electrons of hydrogen atoms.)
Answer:

Question 4.
The molecular formula of ammonia is NH₃. Draw electron-dot and line structures for ammonia molecule.
Answer:

Question 5.
The molecular formula of carbon dioxide is CO₂. Draw the electron-dot structure (without showing circle) and line structure for CO₂.
Answer:

Question 6.
With which bond C atom in CO₂ is bonded to each of the O atoms ?
Answer:
In CO₂, carbon atom is bonded to each of the O atoms by double bond.

Question 7.
The molecular formula of sulfur is S₈ in which eight sulphur atoms are bonded to each other to form one ring. Draw electron-dot structure for S₈ without showing the circles.
Answer:

The above S₈ molecule of sulphur has crown shaped structure. One molecule of sulpur is made up of eight atoms of sulphur.

Use your brain power! (Text Book Page No. 113)

Question 1.
Hydrogen peroxide decomposes of its own by the following reaction:
2H – O – O – H → 2H – O – H + O₂
From this, what will be your inference about the strength O – O covalent bond ?
Tell from the above example whether oxygen has catenation power or not.
Answer:
In hydrogen peroxide (H₂O₂), the O – O covalent bond is not strong as oxygen has no catenation power.

Name	Molecular formula	Condensed Structural formula	Number of carbon atoms	Number of -CH2– units	Boiling point ° C
Ethene	C2H4	CH2 = CH2	2	0	-102
Propone	C3H6	CH3 – CH = CH2	3	1	-48
1-Butene	C4H8	CH3 – CH2 – CH = CH2	…	…	-6.5
1-Pentene	C5H10	CH3 – CH2 – CH2 – CH = CH2	…	…	30

Use your brain power! (Text Book Page No. 120)

Question 1.
The above table shows the homologous series of alkenes. Inspect the molecular formulae of the members of this series. Do you find any relationship, in the number of carbon atoms and the number of hydrogen atoms in the molecular formulae?
Answer:
In the above homologous series, if we observe the molecular formulae of alkenes then the number of carbon atoms are half the number of hydrogen atoms.

Question 2.
If the number of carbon atoms in the molecular formulae of alkenes is denoted by ‘n’ what will be the number of hydrogen atoms?
Answer:
If the number of carbon atoms in the molecular formulae of alkenes is denoted by ‘n’ then the number of hydrogen atoms would be 2n.

Question 3.
What would be the general formula for the molecular formulae of the members of the homologous series of alkanes? What would be the value of ‘n’ for the first member of this series?
Answer:
The general formula for the homologous series of alkane is C_nH_{2n + 2}. The value of ‘n’ for the first member of homologous series is 1.
C_nH_2n+2 = C₁H_{2 × 1} + 2 = CH₄

Question 4.
The general molecular formula for the homologous series of alkynes is C_nH_{2n – 2}. Write down the individual molecular formulae of the first, second and third members by substituting the values 2, 3 and 4 respectively for ‘n’ in this formula.
Answer:
The general molecular formula for the homologous series of alkynes is C_nH_{2n – 2}
n = 2 C₂H_{2 × 2 – 2} = C₂H₂ Ethyne
n = 3 C₃H_{2 × 3 – 2} = C₃H₄ Propyne
n = 4 C₄H_{2 × 4 – 2} = C₄H₆ Butyne

Question 5.
Write down structural formulae of the first four members of the various homologous series formed by making use of the functional groups.
Answer:

Functional group Halo – X (Cl, Br, -I)	Functional group Aldehyde – CHO	Functional group Carboxylic acid – COOH	Functional group Amine -NH2
CH3Cl Chloromethane	HCHO Methanal	HCOOH Methanoic acid	CH3NH2 Methenamine
CH3 – CH2 – Cl Chloroethane	CH3CHO Ethanal	CH3COOH Ethanoic acid	CH3CH2NH2 Ethanamine
CH3 – CH2 – CH2 – Cl 1-Chloropropane	CH3CH2CHO Propanal	CH3CH2COOH Propanoic acid	CH3CH2CH2NH2 Propanamine
CH3 – CH2 – CH2 CH2 – Cl 1-Chlorobutane	CH3CH2CH2 CHO Butanal	CH3CH2CH2COOH Butanoic acid	CH3CH2CH2CH2NH2 Butanamine

Question 6.
General formula of the homologous series of alkanes is C_nH_{2n + 2}. Write down the molecular formula of the 8th and 12th member using this.
Answer:
General formula of alkanes is C_nH_{2n + 2}
n = 8 C₈H_{2 × 8 + 2} = C₆H₁₈ Octane
n = 12 C₁₂H_{2 × 12 + 2} = C₁₂H₂₆ Dodecane

Use your brain power! (Text Book Page No. 121)

Question 1.
Draw three structural formulae having molecular formula C₅H₁₂. Give the names n-pentane, i-pentane and neo-pentane to the above structural formulae.
Answer:

Question 2.
Draw all possible structural formulae having molecular formula C₆H₁₄. Give names to all the isomers.
Answer:

Try This! (Text Book Page No. 124)

Question 1.
Light a bunsen burner. Open and close the air hole at the bottom of the burner by means of the movable ring around it. When do you get yellow sooty flame? When do you get blue flame?
Answer:
When the air hole at the bottom of the burner is open, sufficient oxygen is mixed gaseous fuel for complete combustion and a clean blue flame is obtained. When the air hole is partially blocked by means of the movable ring around it, the air supply is limited which results in incomplete combustion. Hence, yellow sooty flame is produced.

(Text Book Page No. 126)

Question 1.
The names of four fatty acids separated from vegetable oils are given in the table. Identify the number of carbon – carbon double bonds from their structure and molecular formula from the below fatty acids which one when reacts with iodine will make the colour of iodine disappear.
Answer:

Name	Molecular Formula	Number of C = C double bonds	Will it decolorise I2?
Stearic acid	C17H35COOH	———————–	yes/no
Oleic acid	C17H33COOH	One double bond	yes/no
Plamitic acid	C15H31COOH	———————–	yes/no
Linoleic acid	C17H31COOH	Two double bonds	yes/no

Use your brain power! (Text Book Page No. 128)

Question 1.
Explain by writing a reaction, what will happen when pieces of sodium metal are put in n-propyl alcohol.
Answer:
n-Propyl alcohol reacts with pieces of sodium metal, sodium propoxide and hydrogen gas are obtained.

Question 2.
Explain by writing a reaction, which product will be formed on heating n-butyl alcohol with concentrated sulphuric acid.
Answer:
When n-butyl alcohol is heated with concentrated sulphuric acid, one molecule of water is removed from its molecule to form 1-butene.

Use your brain power! (Text Book Page No. 129)

Question 1.
Which one of ethanoic acid and hydrochloric acid is stronger?
Answer:
Hydrochloric acid is stronger acid.

Question 2.
Which indicator paper out of blue litmus paper and pH paper is useful to distinguish between ethanoic acid and hydrochloric acid ?
Answer:
pH paper is useful to distinguish between ethanoic acid and hydrochloric acid.

Use your brain power! (Text Book Page No. 130)

Question 1.
Explain why does the lime water turns milky in the reaction of acetic acid with sodium carbonate.
Answer:
In the reaction of acetic acid with sodium carbonate, carbon dioxide gas is evolved which turns lime water milky resulting in the formation of insoluble calcium carbonate.

Question 2.
Explain the reaction that would take place when a piece of sodium metal is dropped in ethanoic acid.
Answer:
When a piece of sodium metal is dropped in ethanoic acid, sodium acetate and hydrogen gas is formed.

Question 3.
Two test tubes contain two colourless liquids ethanol and ethanoic acid. Explain by writing reaction which chemical test you would perform to tell which substance is present in which test tube.
Answer:
Ethanol does not react with sodium bicarbonate, while ethanoic acid reacts with sodium bicarbonate to form carbon dioxide gas.

Use your brain power! (Text Book Page No. 131)

Question 1.
When fat is heated with sodium hydroxide solution, soap and glycerin are formed. Which functional groups might be present in fat and glycerin? What do you think?
Answer:
The functional group carboxylic acid (-COOH) is present in fat whereas the functioned group hydroxyl group (-OH) is present in glycerin.

Can you tell? (Text Book Page No. 131)

Question 1.
What are the chemical names of the nutrients that we get from the foodstuff, namely, cereals, pulses and meat?
Answer:
The nutrients that we get from the foodstuff, namely cereals, pulses and meat are alpha amino acids.

Question 2.
What are the chemical substances that make cloth, furniture and elastic objects?
Answer:
The chemical substances that make cloth, furniture and elastic objects are cellulose and rubber.

Use your brain power! (Text Book Page No. 133)

Question 1.
Structural formulae of some monomers are given below. Write the structural formula of the homopolymer formed from them.

Answer:

Question 2.
From the given structural formula of polyvinyl acetate, that is used in paints and glues, deduce the name and structural formula of the corresponding monomer.

Answer:

(Text Book Page No. 133)

Question 1.
Complete the following table by writing their Structural formulae and Molecular formulae.
Answer:
(Answer is given directly in bold letters.)

Fill in the gaps in the table: (Text Book Page No. 119)
(Answer is given directly in bold letters.)
a. Homologous series of alkanes.
Answer:

b. Homologous series of alcohol.
Answer:

c. Homologous series of alkenes.
Answer:

(Text Book Page No. 123)

Question 1.
Complete the table by writing the IUPAC names in the third column.
(Answer is directly given with underline.)
Answer:

Common name	Structural formula	IUPAC name
Ethylene	CH2 = CH2	Ethene
Acetylene	HC = CH	Ethyne
Acetic acid	CH3 -COOH	Ethanoic acid
Methyl alcohol	CH3 – OH	Methanol
Ethyl alcohol	CH3 – CH2 – OH	Ethanol

Use your brain power! (Text Book Page No. 119)

Question 1.
By how many -CH₂– (methylene) units do the formulae of the first two members of homologous series of alkanes, methane (CH₄) and ethane (C₂H₆) differ? Similarly, by how many -CH₂– units do the neighbouring members ethane (C₂H₆) and propane (C₃H₈) differ from each other?
Answer:
The first two members of homologous series of alkanes, methane (CH₄) and ethane (C₂H₆) differed by one -CH₂– unit. Similarly, ethane (C₂H₆) and propane (C₃H₈) differed by -CH₂– unit.

Question 2.
How many methylene units are extra in the formula of the fourth member than the third member of the homologous series of alcohols?
Answer:
There is only one, methylene unit extra in the formula of the fourth member and the third member of the homologous series of alcohols.

Question 3.
How many methylene units are less in the formula of the second member than the third member of the homologous series of alkenes?
Answer:
There is only one methylene unit less in the formula of the second member of and the third member of the homologous series of alkenes.

Fill in the blanks and rewrite the completed statements:

Question 1.
The organic compounds having double or triple bonds in them are termed as …………
Answer:
The organic compounds having double or triple bonds in them are termed as unsaturated hydrocarbons.

Question 2.
The general formula of alkane is ……………
Answer:
The general formula of alkane is C_nH_{2n + 2}.

Question 3.
The compounds of homologous series have the same ………….. group.
Answer:
The compounds of homologous series have the same functional group.

Question 4.
A double bond is formed between carbon atoms by ………… pairs of electrons.
Answer:
A double bond is formed between carbon atoms by two pairs of electrons.

Question 5.
The compounds having different structural formulae having the same molecular formula is called ……….
Answer:
The compounds having different structural formulae having the same molecular formula is called structural isomerism.

Question 6.
The functional group of ether is …………..
Answer:
The functional group of ether is -O-.

Question 7.
The general formula of alkene is …………
Answer:
The general formula of alkene is C_nH_2n.

Question 8.
The bond between two atoms of nitrogen is a ………… bond.
Answer:
The bond between two atoms of nitrogen is a triple bond.

Question 9.
Benzene ring is made up of ………….. carbon atoms.
Answer:
Benzene ring is made up of six carbon atoms.

Question 10.
Due to …………., vegetable oil is converted into vanaspati ghee.
Answer:
Due to hydrogenation, vegetable oil is converted into vanatspati ghee.

Question 11.
………….. control the heredity at molecular level.
Answer:
Nucleic acids control the heredity at molecular level.

Question 12.
The regular repetition of a small unit is called …………..
Answer:
The regular repetition of a small unit is called polymer.

Question 13.
The structural formula of polypropylene is ……………….
Answer:
The structural formula of polypropylene is

Question 14.
The monomers of proteins are ……………..
Answer:
The monomers of proteins are alpha amino acids.

Question 15.
The monomer of cellulose is …………
Answer:
The monomer of cellulose is glucose.

Question 16.
…………. have sweet odour.
Answer:
Esters have sweet odour.

Choose the correct alternative and rewrite the statement:

Question 1.
The property of direct bonding between atoms of the same element to form a chain is called ………..
(a) catenation
(b) isomerism
(c) dehydration
(d) polymerization
Answer:
The property of direct bonding between atoms of the same element to form a chain is called catenation.

Question 2.
The molecular weight of two adjacent members in homologous series of an alkane differ by ………. units.
(a) 16
(b) 20
(c) 14
(d) 12
Answer:
The molecular weight of two adjacent members in homologous series of an alkane differ by 14 units.

Question 3.
Consecutive members of a homologous series differ by ………. group.
(a) -CH
(b) -CH2
(C) -CH3
(d) -CH4
Answer:
Consecutive members of a homologous series differ by CH₂ group.

Question 4.
……….. is used to prepare carbon black.
(a) Methane
(b) Ethene
(c) Propane
(d) Butane
Answer:
Methane is used to prepare carbon black.

Question 5.
……….. is the general formula of alkene.
(a) C_nH_2n
(b) C_nH_{2n + 2}
(c) C_nH_{2n – 2}
(d) C_nH_{n – 2}
Answer:
C_nH_2n is the general formula of alkene.

Question 6.
The reaction of methane with chlorine in the presence of sunlight is called ………..
(a) pyrolysis
(b) an elimination reaction
(c) a substitution reaction
(d) an addition reaction
Answer:
The reaction of methane with chlorine in the presence of sunlight is called a substitution reaction.

Question 7.
The general formula for alkynes is ………….
(a) C_nH_2n
(b) C_nH_{2n + 2}
(c) C_nH_{2n – 2}
(d) C_nH_{2n – 1}
Answer:
The general formula for alkynes is C_nH_{2n – 2}

Question 8.
The reaction of ………… with ethanol is a fast reaction.
(a) calcium
(b) magnesium
(c) sodium
(d) aluminum
Answer:
The reaction of sodium with ethanol is a fast reaction.

Question 9.
Ethylene has …………. bond between two carbon atoms.
(a) a single
(b) a double
(c) a triple
(d) an ionic
Answer:
Ethylene has a double bond between two carbon atoms.

Question 10.
The saturated hydrocarbons are those in which carbon atom are linked by ………….
(a) a single bond
(b) a double bond
(c) a triple bond
(d) an ionic bond
Answer:
The saturated hydrocarbons are those in which carbon atom are linked by a single bond.

Question 11.
C₇H₁₆ is ………….
(a) hexane
(b) octane
(c) methane
(d) heptane
Answer:
C₇H₁₆ is heptane.

Question 12.

(a) carboxylic acid group
(b) aldehyde group
(c) ketonic group
(d) alcohol group
Answer:
(a) carboxylic acid group

Question 13.
The possible isomers for C₅H₁₂ are ……………
(a) 2
(b) 4
(c) 1
(d) 3
Answer:
The possible isomers for C₅H₁₂ are 3.

Question 14.
………. contains alcoholic functional group.

Answer:
(d) all of these

Question 15.
Oxygen molecule has ………… bond between two oxygen atoms.
(a) a double
(b) a single
(c) a triple
(d) an ionic
Answer:
Oxygen molecule has a double bond between two oxygen atoms.

Question 16.
Some acetic acid is treated with solid NaHCO₃. The resulting solution will be ………..
(a) colourless
(b) blue
(c) green
(d) yellow
Answer:
Some acetic acid is treated with solid NaHCO₃. The resulting solution will be colourless.

Question 17.
Ethanoic acid has a ……… odour.
(a) rotten eggs
(b) pungent
(c) mild
(d) vinegar-like
Answer:
Ethanoic acid has a vinegar-like odour.

Question 18.
Acetic acid ………..
(a) turns red litmus blue
(b) has pungent odour
(c) is red in colour
(d) is odourless
Answer:
Acetic acid has pungent odour.

Question 19.
When acetic acid reacts with sodium metal ……….. gas is formed.
(a) oxygen
(b) hydrogen
(c) chlorine
(d) nitrogen
Answer:
When acetic acid reacts with sodium metal hydrogen gas is formed.

Question 20.
The molecular formula of acetic acid (ethanoic acid) is …………
(a) HCOOH
(b) CH₃COOH
(c) C₂H₅COOH
(d) C₃H₇COOH
Answer:
The molecular formula of acetic acid (ethanoic acid) is CH₃COOH.

Question 21.
When sodium bicarbonate solution is added to dilute acetic acid …………
(a) a gas is evolved
(b) a solid settles at the bottom
(c) the mixture becomes warm
(d) the colour of the mixture becomes yellow
Answer:
When sodium bicarbonate solution is added to dilute acetic acid a gas-is evolved.

Question 22.
2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in ………..
(a) test tube A
(b) test tube B
(c) test tube C
(d) all the test tubes
Answer:
2 ml of ethanoic acid was taken in each of test tubes A, B, C and 2 ml, 4 ml, 6 ml of water was added respectively to them. A clear solution is obtained in all the test tubes.

Question 23.
In the presence of acid catalyst, ethanoic acid reacts with ethanol and ……….. ester is produced.
(a) ethanol
(b) ethanoic
(c) ethyl ethanoate
(d) ethyl ethanol (Practice Activity Sheet – 1)
Answer:
In the presence of acid catalyst, ethanoic acid reacts with ethanol and ethyl ethanoate ester is produced.

Question 24.
The following structural formula belongs to which carbon compound?

(a) Camphor
(b) Benzene
(c) Starch
(d) Glucose (Practical activity sheet- 2)
Answer:
(b) Benzene

Question 25.
What type of reaction is shown below?
\(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \stackrel{\text { Sunlight }}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{Cl}+\mathrm{HCl}\)
(a) Addition
(b) Substitution
(c) Decomposition
(d) Reduction (Practice Activity Sheet – 3)
Answer:
(b) substitution

Question 26.
The carbon compound is used in daily life is ………..
(a) edible oil
(b) salt
(c) carbon dioxide
(d) baking soda (March 2019)
Answer:
The carbon compound is used in daily life is edible oil

State whether the following statements are true or false. (If a statement is false, correct it and rewrite it.):

Question 1.
Generally the melting and boiling points of carbon compounds are high.
Answer:
False. (Generally the melting and boiling points of carbon compounds are low.)

Question 2.
Till now the number of known carbon compounds is about 10 million.
Answer:
True.

Question 3.
Unsaturated hydrocarbons are less reactive than saturated hydrocarbons.
Answer:
False. (Unsatured hydrocarbons are more reactive than saturated hydrocarbons.)

Question 4.
Benzene is an aromatic compound.
Answer:
True.

Question 5.
The carbon-carbon double and triple bonds are also recognised as functional groups.
Answer:
True.

Question 6.
The general formula of alkyne is C_nH_2n.
Answer:
False. (The general formula of alkyne is C_nH_{2n – 2})

Question 7.
Naphthalene burns with a yellow flame.
Answer:
True.

Question 8.
When vegetable oil and tincture iodine react, the color of iodine does not change.
Answer:
False. (When vegetable oil and tincture iodine react, the colour of iodine changes.)

Question 9.
Saturated fats are healthy.
Answer:
False. (Saturated fats are harmful to health.)

Question 10.
Aqueous solution of ethanol is found to be neutral.
Answer:
True.

Question 11.
Denatured ethanol is used as industrial solvent.
Answer:
True.

Question 12.
Vinegar is a 12-15 % aqueous solution of acetic acid.
Answer:
False. (Vinegar is a 5-8 % aqueous solution of acetic acid.)

Question 13.
The functional group of ethanoic acid is a carboxylic group.
Answer:
True.

Question 14.
Sodium hydroxide is used in the preparation of soap from fats and oils.
Answer:
True.

Question 15.
Rubber is a manmade macromolecule.
Answer:
False. (Rubber is a natural macromolecule.)

Question 16.
Polyvinyl chloride is used in the manufacture of P.V.C. pipes and bags.
Answer:
True.

Question 17.
Polyethylene is a homopolymer.
Answer:
True.

Question 18.
The chemical bonds in carbon compounds do not produce ions.
Answer:
True.

Find the odd one out:

Question 1.
Propane, methane, ethene, pentane
Answer:
Ethene. (Others are saturated hydrocarbons.)

Question 2.
Methane, butane, benzene, sodium chloride
Answer:
Sodium chloride. (Others are organic compounds.)

Question 3.
CH₄, C₂H₆, C₃H₈, CaCO₃
Answer:
CaCO₃. (Others are organic compounds.)

Question 4.
C₂H₂, C₃H₈, C₂H₆, CH₄
Answer:
C₂H₂. (Others are saturated hydrocarbons.)

Question 5.
C₂H₄, C₄H₁₀, C₃H₈, CH₄
Answer:
C₂H₄. (Others are saturated hydrocarbons.)

Question 6.
Polyethylene, Polysaccharide, Polystyrene, Polypropylene
Answer:
Polysaccharide (Others are manmade polymers.)

Question 7.
-NH₂, -COOH,-SO₄, -Br
Answer:
-SO₄ (Others are functional groups.)

Question 8.
Methane, Ethane, Propene, Propane, Butane
Answer:
Propene (Others are members of homologous series of alkanes.)

Match the columns:

Question 1.

Column I	Column II
(1) CH4	(a) CH2 = CH2
(2) Ethane	(b) CnH2n – 2
(3) Alkene	(c) Methane
(4) Alkyne	(d) C2H6
	(e) C3H8

Answer:
(1) CH₄ – Methane
(2) Ethane – C₂H₆
(3) Alkene – CH₂ = CH₂
(4) Alkyne – C_nH_{2n – 2}.

Question 2.

Column I	Column II
(1) Aromatic hydrocarbon	(a) Propyne
(2) Alkane	(b) Benzene
(3) Alkyne	(c) Saturated hydrocarbon
(4) Alkene	(d) CnH2n
	(e) C n H2n – 1 OH

Answer:
(1) Aromatic hydrocarbon – Benzene
(2) Alkane – Saturated hydrocarbon
(3) Alkyne – Propyne
(4) Alkene – C_nH_2n.

Question 3.

Column I	Column II
(1) Cyclohexane	(a) CH3COOH
(2) Methanol	(b) CH3Cl
(3) Acetaldehyde	(c) CH2Cl2
(4) Ethanoic acid	(d) CH3OH
	(e) C6H12
	(f) CH3CHO

Answer:
(1) Cyclohexane – C₆H₁₂
(2) Methanol – CH₃OH
(3) Acetaldehyde – CH₃CHO
(4) Ethanoic acid – CH₃COOH.

Question 4.

Answer:
(1) (-OH) – Alcohol
(2) (-COOH) – Carboxylic acid
(3) (-CHO) – Aldehyde

Question 5.

Column I	Column II
(1) Ethyne	(a) C2H6
(2) Ethene	(b) C2H2
(3) Ethane	(c) C3H6
(4) Propyne	(d) C2H4
	(e) C3H4

Answer:
(1) Ethyne – C₂H₂
(2) Ethene – C₂H₄
(3) Ethane – C₂H₆
(4) Propyne – C₃H₄.

Question 6.

Column I	Column II
(1) Cellulose	(a) P.V.C. pipes, bags
(2) R.N.A	(b) Blankets
(3) Polyacrylonitrile	(c) Wood
(4) Polyvinyl chloride	(d) Chromosomes of plants

Answer:
(1) Cellulose – Wood
(2) R.N.A. – Chromosomes of plants
(3) Polyacrylonitrile – Blankets
(4) Polyvinyl chloride – P.V.C. pipes, bags.

Consider the relation between Column I and II. Fill in Column IV to match Column III.

Column I	Column II	Column III	Column IV
(1) Ethylene	Polyethylene	Tetrafluoroethylene	—————–
(2) Poly­propylene	Propylene	Polystyrene	—————–
(3) Poly­saccharide	Glucose	Proteins	—————–
(4) Rubber	Isoprene	D.N.A.	—————–
(5) Wood	Cellulose	Chromosomes of plants	—————–

Answer:
(1) Teflon
(2) Styrene
(3) Alpha aminoacid
(4) Nucleotide
(5) R.N.A.

Define the following:

Question 1.
Define Alkane
Answer:
Alkane: In hydrocarbon, the four valencies of carbon atom are satisfied only by the single bonds, such compounds are called alkane.
Example: Methane (CH₄), Ethane (C₂H₆)

Question 2.
Define Alkene.
Answer:
Alkene: The unsaturated hydrocarbons containing a carbon-carbon double bond are called alkenes.
Example : Ethene (CH₂ = CH₂)

Question 3.
Define Alkyne.
Answer:
Alkyne: The unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes.
Example: Ethyne C₂H₂ (CH ≡ CH).

Question 4.
Define Addition reaction.
Answer:
Addition reaction: When a carbon compound combines with another compound to form a product that contains all the atoms in both the reactants; it is called an addition reaction.

Question 5.
Define Substitution reaction.
Answer:
Substitution reaction: The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms, is called substitution reaction.

Question 6.
Define Esterification.
Answer:
Esterification: A carboxylic acid reacts with an alcohol in presence of an acid catalyst, an ester is formed. The reaction is known as esterification.

Question 7.
Define Saponification.
Answer:
Saponification: When an ester reacts with the alkali, i.e. sodium hydroxide, the corresponding alcohol and sodium salt of carboxylic acid are obtained. This reaction is called saponification reaction. It is used in the preparation of soap.
Ester + Sodium hydroxide → Sodium salt of carboxylic acid + Alcohol.

Question 8.
Define Polymerization.
Answer:
Polymerization: The reaction by which monomer molecules are converted into a polymer is called polymerization.

Name the following:

Question 1.
The higher homologue of hexane.
Answer:
Keptane.

Question 2.
The number of double bonds in benzene.
Answer:
Three.

Question 3.
The functional group in ether and halogen.
Answer:
Functional groups:
Ether: – O –
Halogen: – X (-Cl, -Br, -I).

Question 4.
Polymer of tetrafluoroethylene.
Answer:
Teflon.

Question 5.
The monomer of polysaccharide.
Answer:
Glucose.

Question 6.
The Polymer of nucleotide.
Answer:
D.N.A./R.N.A.

Question 7.
The Monomer of rubber.
Answer:
Isoprene.

Question 8.
Two oxidising compounds.
Answer:
Potassium permanganate, Potassium dichromate.

Question 9.
IUPAC name of sodium acetate.
Answer:
Sodium ethanoate.

Question 10.
The main component of natural gas.
Answer:
Methane.

Question 11.
Two isomers of butane.
Answer:
n-butane and i-butane.

Question 12.
A nomenclature system based on the structure of the compounds and it was accepted all over the world.
Answer:
International Union of Pure and Applied Chemistry (IUPAC).

Answer the following questions in one sentence each:

Question 1.
State the atomic number and electronic configuration of carbon.
Answer:
The atomic number of carbon is 6 and the electronic configuration of carbon is (2, 4).

Question 2.
State number of electrons in the outermost orbit of carbon and valency of carbon.
Answer:
Four electrons are present in the outermost orbit of carbon and the valency of carbon is 4.

Question 3.
What are hydrocarbons? Give one example.
Answer:
The compounds containing only carbon and hydrogen are called hydrocarbons. These compounds are known as organic compounds. E.g. Methane, Ethane.

Question 4.
What is the molecular formula and structural of methane?
Answer:
The molecular formula of methane is CH₄.

Question 5.
How many atoms of carbon and hydrogen are present in methane?
Answer:
The molecule of methane has one carbon atom and four hydrogen atoms.

Question 6.
State the general formula of alkane.
Answer:
The general formula of an alkane is C_nH_{2n + 2}.

Question 7.
Give two examples of alkanes.
Answer:
Methane (CH₄) and ethane (C₂H₆) are alkanes.

Question 8.
Give two examples of alkenes.
Answer:
Ethene (CH₂ = CH₂) and propene (CH₃ – CH = CH₂) are alkenes.

Question 9.
Give two examples of alkynes.
Answer:
Ethyne (HC ≡ CH) and propyne (CH₃ – C ≡ CH) are alkynes.

Question 10.
Write the name and molecular formula of a higher homologue of propane.
Answer:
Butane (C₄H₁₀) is a higher homologue of propane.

Question 11.
Write the structure and molecular formula of ethane.
Answer:
Structure of ethane:

Molecular formula of ethane: C₂H₆

Question 12.
What is meant by catenation power?
Answer:
Carbon has a unique ability to form strong covalent bonds with other carbon atoms, this result in formation of big molecules. This property of carbon is called catenation power.

Question 13.
State the structural and molecular formula of benzene.
Answer:

Question 14.
Which functional groups are present in aldehyde and ketone?
Answer:
The functional group -CHO is present in aldehyde and the functional group

is present in ketone.

Question 15.
Which functional group is present in

Answer:

Question 16.
Compare: The proportions of carbon atoms in ethanol (C₂H₅OH) and naphthalene (C₁₀H₈).
Answer:
Ethanol contains two carbon atoms while naphthalene contains 10 carbon atoms. Ethanol is a saturated hydrocarbon and naphthalene is an unsaturated hydrocarbon.

Question 17.
What are the products of combustion of methane?
Answer:
Carbon dioxide (CO₂) and water (H₂O) are the products of combustion of methane.

Question 18.
Which gas is evolved when ethanol reacts with sodium?
Answer:
Hydrogen gas (H₂) is evolved when ethanol reacts with sodium.

Question 19.
Compare: How is the transformation of ethanol into ethanoic acid on oxidation reaction?
Answer:
The transformation of ethanol into ethanoic acid is an oxidation process, in which ethanol accepts oxygen.

Question 20.
Complete the following:

Answer:

Question 21.
Which of the following hydrocarbons undergo addition reactions:
C₂H₆, C₃H₈, C₂H₄, C₃H₈?
Answer:
C₂H₄ (ethene) and C₃H₆ (propene) undergo addition reactions.

Question 22.
How many covalent bonds are there in a molecule of cyclohexane?
Answer:
A molecule of cyclohexane contains 18 covalent bonds.

Question 23.
Give the IUPAC name for CH₃COOH.
Answer:
The IUPAC name for CH₃COOH is ethanoic acid.

Question 24.
Write the IUPAC name of CH₃COONa.
Answer:
IUPAC name of CH₃COONa is sodium ethanoate.

Question 25.
What is meant by denatured alcohol?
Answer:
Ethanol is the important commercial solvent. To prevent the misuse of this solvent, it is mixed with the poisonous methanol. Such ethanol is called denatured spirit.

Question 26.
What is meant by glacial acetic acid ?
Answer:
The melting point of pure acetic acid is 17 °C. Therefore, during winter in old countries acetic acid freezes at room temperature itself and looks like ice. Therefore it is named glacial acetic acid.

Question 26.
Which useful components of hydro¬carbon are obtained by fractional distillation of crude oil?
Answer:
Various useful components of hydrocarbon such as CNG, LPG, petrol (gasoline), rockel, diesel, engine oil, lubricant, etc. are obtained by separation of crude oil using fractional distillation.

Question 28.
Which functional groups are present in ester and amine?
Answer:
Ester: -COO-
Amine: -NH₂–

Question 29.
Give two examples of natural macromolecules.
Answer:
Examples: Polysaccharide, protein and nucleic acid.

Question 30.
Write the structure of polystyrene and give its uses.
Answer:
Structure:

Polystyrene is used to make thermocoal articles.

Question 31.
Write the name and the structure of monomer of polyacrylonitrile.
Answer:
The name and structure of monomer: Acrylonitrile CH₂ = CH – CN

Question 32.
Write the name and the structure of monomer of teflon and its uses.
Answer:
The name and structure of monomer: Tetrafluro ethylene CF₂ = CF₂
Teflon is used to make nonstick cookware.

Question 33.
What is meant by copolymers?
Answer:
The polymers formed from two or more monomers are called copolymers.
Examples: Poly ethylene terephthalate.

Answer the following questions:

Question 1.
How is hydrogen molecule formed?
Answer:

The atomic number of hydrogen is 1, its atom contains 1 electron in K shell. It requires one more electron to complete the K shell and attain the configuration of helium (He). To meet this requirement two hydrogen atoms share their electrons with each other to form H₂ molecule. One covalent bond, i.e. a single bond is formed between two hydrogen atoms by sharing of two electrons.

Question 2.
Describe the formation of oxygen molecule (O₂).
Answer:

(1) The atomic number of oxygen is 8. The electronic configuration of oxygen is (2, 6). Oxygen has 6 electrons in the outermost shell.
(2) It requires 2 electrons to complete the L shell and attain the configuration of neon (Ne).
(3) Each oxygen atom shares its valence electron with the valence electron of another oxygen atom to give two shared pairs of electrons which results in the formation of oxygen molecule.
(4) Thus, two electron pairs are shared between two oxygen atoms, forming double covalent bond (=).

Question 3.
Describe the formation of nitrogen molecule.
Answer:

(1) The atomic number of nitrogen is 7. The electronic configuration of nitrogen is (2, 5). Nitrogen has 5 electrons in the outermost shell.
(2) It requires three more electrons to complete the L shell and attain the configuration of neon (Ne).

(3) Two nitrogen atoms come close together and share three pairs of electrons with each other, resulting in the formation of a triple bond.
(4) Thus, two nitrogen atoms are bound with a triple bond (=) to form a nitrogen molecule.

Question 4.
How is the methane molecule formed?
Answer:
(1) The electronic configuration of carbon is (2, 4). Carbon has four electrons in the outermost shell, hence it is tetravalent.
(2) The electronic configuration of hydrogen is 1, hence it is monovalent.
(3) Carbon needs four electrons to complete the L shell and attain the configuration of neon (Ne).
(4) Four atoms of hydrogen share 1 electron each with 4 electrons of carbon.
(5) A single covalent bond is formed by sharing of two electrons.

Thus, the methane molecule contains four single bonds between the carbon and hydrogen atoms.

Question 5.
State the various compounds and its formulae formed by a single atom of carbon with monovalent hydrogen and chlorine.
Answer:

Compounds	Names
CH4	Methane
CH3Cl	Methyl chloride
CH2Cl2	Methylene dichloride
CHCl3	Methylene trichloride
CCl4	Carbon tetrachloride

Question 6.
Observe the straight chain hydrocarbons given below and answer the following questions:

(i) Which of the straight chain compounds from A and B is saturated and unsaturated straight chains?
(ii) Name these straight chains.
(iii) Write their chemical formulae and number of -CH₂ units. (Practice Activity Sheet – 2)
Answer:
(i) A is a saturated hydrocarbon, B is an unsaturated hydrocarbon.
(ii) A = Propane, B = Propene
(iii) The chemical formula of A = C₃H₈ and number of -CH₂ units are 3.
The chemical formula of B = C₃H₆ and number of -CH₂ unit is 1.

Question 7.
Draw electron-dot and line structure of an ethane molecule.
Answer:
The molecular formula of ethane is C₂H₆.

Question 8.
The molecular formula of propane is C₃H₈. From this draw its structural formula. (Practice Activity Sheet – 3)
Answer:
The structural formula of propane:

Question 9.
Draw the structure and carbon skeleton for cyclohexane.
Answer:

Question 10.
Classify into saturated and unsaturated hydrocarbons: (1) Methane (2) Ethene (3) Ethane (4) Ethyne (5) Propene (6) Propyne (7) Butane (8) Cyclohexene (9) Cyclopentane (10) Heptane.
Answer:
(i) Saturated hydrocarbons: (1) Methane (2) Ethane (3) Butane (4) Cyclopentane (5) Heptane.
(ii) Unsaturated hydrocarbons: (1) Ethene (2) Ethyne (3) Propene (4) Propyne (5) Cyclohexene.

Question 11.
Classify into alkanes, alkenes and alkynes: (1) Ethane (2) Ethene (3) Methane (4) Propene (5) Ethyne (6) Propyne (7) Butane (8) Pentane.
Answer:
Alkanes: (1) Ethane (2) Methane (3) Butane. (4) Pentane
Alkenes: (1) Ethene (2) Propene
Alkynes: (1) Ethyne (2) Propyne

Question 12.
Classify into straight chain carbon compounds, branched chain carbon compounds and ring carbon compounds:
(1) Propene (2) Butane (3) Iso-butane (4) Cyclopentane (5) Benzene (6) Isobutylene.
Answer:
Straight chain carbon compounds:

1. Propene
2. Butane.

Branched chain carbon compounds:

1. Iso-butane
2. Isobutylene.

Ring carbon compounds:

1. Cyclopentane
2. Benzene.

Question 13.
Draw chain and ring structures of organic compound having six carbon atoms in it.
Answer:
Chain structures of an organic compound having six carbon atoms:

Ring structures of an organic compound having six carbon atoms:

Question 14.
Explain the structure of benzene.
Answer:
The molecular formula of benzene is C₆H₆. It is a cyclic unsaturated hydrocarbon. Benzene ring is made of six carbon atoms. In benzene, each carbon atom is linked to two other carbon atoms, on one side by a single bond and on the other side by a double bond, i.e. three alternate single bonds and double bonds in the six membered ring structure of benzene. The compound having this characteristic unit in their structure are called aromatic compounds.

Question 15.
Draw the structures of isomers of pentane (C₅H₁₂).
Answer:

Question 16.
Recognize the carbon chain type for each of the following:

(Practice Activity Sheet – 1)
Answer:
In the reaction of acetic acid with sodium carbonate, carbon dioxide gas is evolved which turns lime water milky resulting in the formation of insoluble calcium carbonate.

Question 17.
What is meant by functional group? Give examples.
(OR)
Explain the term functional group with example.
Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the functional groups.
Example: Methyl alcohol, acetic acid.

In methane (CH₄), when one hydrogen atom is replaced by an -OH group, methyl alcohol (CH₃OH), is formed. The -OH is known as the alcoholic functional group.
Similarly, from methane (CH₄) when one hydrogen atom is replaced by -COOH group, acetic acid (CH₃COOH) is formed. The -COOH group is known as the carboxylic acid functional group.

Question 18.
Define functional group and complete the following table:

Functional group	Compound	Formula
——————–	Ethyl alcohol	——————–
——————–	Acetaldehyde	——————–

Answer:
The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of length and nature of the carbon chain in that compound. Therefore these hetero atoms or groups of atoms containing hetero atoms are called the
functional groups.

Functional group	Compound	Formula
-OH	Ethyl alcohol	C2H5OH
-CHO	Acetaldehyde	CH3CHO

Question 19.
What is meant by homologous series?
Answer:
The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH₂– (methylene) unit and their mass differ by 14 units.

The homologous series of straight chain alkanes can be represented by the general formula CnH_{2n + 2}. The members of this series are as follows:

Methane – CH4 Ethane – C2H6	– These differ by – CH2 units
Ethane – C2H6 Propane – C3H8	– These differ by – CH2 units
Butane – C4H10 Pentane – C5H12	– These differ by – CH2 units

Question 20.
State the four characteristics of homologous series.
Answer:
Characteristics of Homologous series:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit ( -CH₂– ) gets added
(b) molecular mass increases by 14 u (c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) While going in an increasing order of the length there is gradation in the physical properties i.e. the boiling and melting points.

Question 21.
Write names of first four homologous series of alcohols: (March 2019)

Answer:
First four homologous series of alcohols are

1. Methanol CH₃ – OH
2. Ethanol C₂H₅ – OH
3. Propanol C₃H₇ – OH
4. Butanol C₄H₉ – OH

Question 22.
Describe the IUPAC rules of naming organic compounds.
Answer:
IUPAC nomenclature system: International Union for Pure and Applied Chemistry (IUPAC) put forth a nomenclature system based on the structure of the compounds and it was accepted all over the world. There are three units in the IUPAC name of any carbon compound: parent, suffix and prefix. These are arranged in the name as follows:

Prefix-parent-suffix:
An IUPAC name is given to a compound on the basis of the name of its parent alkane. The name of the compound is constructed by attaching appropriate suffix and prefix to the name of the parent-alkane. The steps in the IUPAC nomenclature of straight chain compounds are as follows:

Step 1: Draw the structural formula of the straight chain compound and count the number of carbon atoms in it. The alkane with the same number of carbon atoms is the parent alkane of the concerned compound. Write the name of this alkane.

In case the carbon chain of concerned compound contains a double bond, change the ending of the parent name from ‘ane’ to ‘ene’. If the carbon chain in the concerned compound contains a triple bond, change the ending of the parent name from ‘ane’ to ‘yne’.

Sr. No.	Structural formula	Straight chain	Parent name
1.	CH3 – CH2 – CH3	C – C – C	propane
2.	CH3 – CH2 – OH	C – C	ethane
3.	CH3 – CH2 – COOH	C – C – C	propane
4.	CH3 – CH2 – CH2 – CHO	C – C – C – C	butane
5.	CH3 – CH = CH2	C – C = C	propene
6.	CH3 – C ≡ CH	C – C ≡ C	propyne

Step 2: If the structural formula contains a functional group, replace the last letter ‘e’ from the parent name by the condensed name of the functional group as the suffix. (Exception: The condensed name of the functional group ‘halogen’ is always attached as the prefix.)

Step 3: Number the carbon atoms in the carbon chain from one end to the other. Assign the number T to carbon in the functional group -CHO or -COOH, if present. Otherwise, the chain can be numbered in two directions. Accept that numbering which gives smaller number to the carbon carrying the functional group. In the final name, a digit (number) and a character (letter) should be separated by a small horizontal line.

Question 23.
Write the IUPAC names of the following structural formulae.
a. CH₃ – CH₂ – CH = CH₂
Answer:
Number of carbon atoms in the longest chain: 4
Parent alkane: Butene
Functional group: double bond
Assign the number:

In the longest chain, the numbering of carbon atom starts from the carbon atom nearest to the double bond and the other c-atoms are numbered accordingly.
Parent suffix: But-1-ene
IUPAC name: But-1-ene

b. CH₃ – C ≡ C – H
Answer:
Number of carbon atoms in the longest chain: 3
Parent alkane: Propyne
Functional group: triple bond
Parent suffix: Propyne
IUPAC name: Propyne

c.

Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Prefix functional group: Chloro
Assign the number: 2

The carbon atom to which the -Cl atom is attached is numbered as C₂ and the other C atoms are numbered accordingly.
Prefix parent: 2-Chloropentane
IUPAC name: 2-Chloropentane

d. CH₃ – CH₂ – CH₂ – Br
Answer:
The number of carbon atoms in the longest chain: 3
Parent alkane : Propane Prefix functional group: Bromo
Assign the number:

The carbon atom to which the -Br atom is attached is numbered as C₁ and the other C atoms are numbered accordingly.
Prefix parent: 1-Bromopropane
IUPAC name: 1-Bromopropane

e.

Answer:
The number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -OH (ol)
Assign the number:

The carbon atom to which the -OH group is attached is numbered as C₂.
If the carbon chain of the compound contains a -OH group, then change the ending ‘e’ of the parent name, i.e. ,‘e’ of butane is replaced by ‘ol’ (ol for alcohol).
Parent suffix: Butan-2-ol
IUPAC name: Butan-2-ol

f.

The number of carbon atoms: 3
Parent alkane: Propane
Functional group: -NH₂ (amine)
Assign the number:

If the carbon chain of the compound contains a -NH2 group then change the ending of the parent name, i.e. ‘e’ of propane is replaced by ‘amine’.
Parent suffix: 2-Propanamine
IUPAC name: 2-Propanamine

g. HCOOH
Answer:
The number of carbon atoms: 1
Parent alkane: Methane
Functional group: -COOH (-oic cid)
If the carbon chain of the compound contains a -COOH group, then change the ending of the parent name, i.e. ‘e’ of methane is replaced by ‘-oic acid’.
Parent suffix: Methanoic acid
IUPAC name: Methanoic acid

h. CH₃ – CH₂ – CH₂ – CHO
Answer:
The number of carbon atoms in the longest chain: 4
Parent alkane : Butane Functional group: -CHO (al)
Assign the number: 1
Assign the number ‘1’ to carbon in the functional group

If the carbon chain of the compound contains a -CHO group then change the ending of the parent name, i.e. ‘e’ of the butane is replaced by ‘al’.
Parent suffix: Butanal
IUPAC name: Butanal

i.

Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Functional group:

Assign the numbering:

In the longest chain, the numbering of carbon atom starts from the carbon nearest to the functional group (both the numbering equivalent).
If the carbon chain of compound contains a

group, then change the ending of the parent name i.e. ‘e’ of pentane is replaced by ‘one’.
Parent suffix: Pentan-3-one
IUPAC name: Pentan-3-one.

Question 24.
What happens when methane is burnt in air? Write the balanced chemical equation for the same.
Answer:
When methane burns in air, carbon dioxide and water are formed. The reaction is exothermic with release of large amount of heat and light.

Question 25.
What happens when ethanol is burnt in air?
Answer:
When ethanol is burnt in air, it burns with a clean blue flame, carbon dioxide and water are formed. In this reaction, release of large amount of heat and light takes place.

Question 26.
What happens when ethanol is treated with alkaline potassium permanganate? Write the balanced chemical equation for the same.
Answer:
When ethanol is treated with alkaline potassium permanganate, ethanol gets oxidised by alkaline potassium permanganate to’ form ethanoic acid.

Question 27.
What happens when vegetable oil is hydrogenated? Write the balanced chemical equation.
Answer:
When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, vanaspati ghee (saturated) compound is formed.

Question 28.
What happens when chlorine is treated with methane?
(OR)
Describe the action of chlorine on methane.
(OR)
Write a note on chlorination of methane.
Answer:
Methane reacts rapidly with chlorine in the presence of sunlight to form four products. In this reaction, chlorine atoms replace, one by one, all the hydrogen atoms in the methane.

The reaction in which the place of one type of atom/group in a reactant is taken by another atom/group of atoms is called substitution reaction. Chlorination of methane is a substitution reaction.

Question 29.
What happens when ethanol is reacted with sodium?
Answer:
When ethanol is reacted with sodium at room temperature, sodium ethoxide is formed and hydrogen gas is liberated.

Question 30.
What happens when ethanol is heated at 170 °C with excess of conc. sulphuric acid?
Answer:
When ethanol is heated at 170 °C with excess of conc. sulphuric acid, one molecule of water is removed from its molecule to form ethene (unsaturated compound).

Question 31.
What happens when ethanoic acid is treated with sodium hydroxide? Write the balanced equation for the same.
Answer:
When ethanoic acid is treated with sodium hydroxide, neutralization takes place to form sodium acetate (sodium ethanoate) and water.

Question 32.
What happens when ethanoic acid is treated with sodium carbonate?
Answer:
When ethanoic acid is treated with sodium carbonate, sodium ethanoate, carbon dioxide and water is formed.

Question 33.
What happens when ethanoic acid is treated with sodium bicarbonate?
Answer:
When ethanoic acid is treated with sodium bicarbonate, sodium ethanoate, water and carbon dioxide is formed.

Question 34.
What happens when ethanoic acid is treated with ethanol? Write the balanced equation for the same.
Answer:
When ethanoic acid is treated with ethanol in the presence of an acid catalyst, an ester, i.e., ethyl ethanoate is formed.

Question 35.
What happens when ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst?
Ans. When ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst, it polymerizes to form polyethylene or polythene (plastic).

Question 36.
State the physical properties of ethyl alcohol ethanol.
Answer:

1. Ethanol is a colourless liquid and it is soluble in water in all proportions and has pleasant odour.
2. The boiling point of ethanol is 78 °C and the freezing point is -114 °C.
3. It is combustible and burns with a blue flame.
4. An aqueous solution of ethanol is neutral to litmus paper.

Question 37.
State the properties of ethanoic acid.
Answer:

1. Ethanoic acid is a colourless liquid with boiling point 118 °C and melting point 17 °C. It has a pungent odour.
2. Its aqueous solution is acidic and turns blue litmus red.
3. A 5-8% aqueous solution of acetic acid is used as vinegar.
4. It is a weak acid.

Write short notes:

Question 1.
Catenation power.
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms; this results in formation of big molecules. This property of carbon is called catenation power.

(2) Carbon shows catenation. Two or more carbon atoms can share their valence electrons and bond with each other. Thus, carbon chains can be straight or branched or closed chain ring structure forming large molecules. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

(3) Hence, carbon atoms can form an unlimited number of compounds.

Question 2.
Characteristics of Carbon.
Answer:
(1) Carbon has a unique ability to form strong covalent bonds with other carbon atoms: this results in formation of big molecules. This property of carbon is called catenation power. The carbon compounds contain open chains or closed chains of carbon atoms. An open chain can be a straight chain or a branched chain. A closed chain is a ring structure. The covalent bond between two carbon atoms is strong and therefore stable. Carbon is bestowed with catenation power due to the strong and stable covalent bonds.

Question 3.
Functional group.
Answer:
(1) The compound acquire specific chemical properties due to these hetero atoms or the groups of atoms that contain hetero atoms, irrespective of the length and nature of the carbon chain in that compound. Therefore these hetero atoms or the groups of atoms containing hetero atoms are called functional groups.

All organic compounds are derivatives of hydrocarbons. The derivatives are formed by replacing one or more H-atom/atoms of hydrocarbon by some other hetero atom or groups of atoms containing hetero atoms. After replacement, a new compound is formed which has properties different from the parent hydrocarbon.

Examples: For methane, if one hydrogen atom is replaced by an – OH group, then a compound is methyl alcohol (CH₃OH). The -OH group is known as the alcoholic functional group.
Functional group is organic compound:
1. Alcohol: – OH (hydroxy group)

4. Carboxylic acid : -COOH

Question 4.
Homologous series.
Answer:
The length of the carbon chains in carbon compounds is different their chemical properties are very much similar due to the presence of the same functional group in them. The series of compounds formed by joining the same functional group in place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. Two adjacent members of the series differ by only one -CH₂– (methylene) unit and their mass differ by 14 units.

The homologous series of straight chain alkanes can be represented by the general formula C_nH_{2n + 2}
The members of this series are as follows:

Methane – CH4 Ethane – C2H6	– These differ by – CH2 units
Ethane – C2H6 Propane – C3H8	– These differ by – CH2 units
Butane – C4H10 Pentane – C5H12	– These differ by – CH2 units

Characteristics:
(1) In homologous series while going in an increasing order of the length of carbon chain
(a) one methylene unit (-CH₂-) gets added
(b) molecular mass increases by 14 u
(c) number of carbon atoms increases by one.
(2) Chemical properties of members of a homologous series show similarity due to the presence of the same functional group in them.
(3) Each member of the homologous series can be represented by the same general molecular formula.
(4) while going in an increasing order of the length there is gradation in the physical properties i.e., the boiling and melting points.

Question 5.
Polymerization.
Answer:
(1) The reaction by which monomer molecules are converted into a polymer is called polymerization. A macromolecule formed by regular repetition of a small unit is called polymer. The small unit that repeats regularly to form a polymer is called monomer. The important method of polymerization is to make a polymer by joining alkene type of monomers.

(2) When ethylene gas is heated at high pressure and high temperature in the presence of suitable catalyst, it polymerizes to form polyethylene or polythene (plastic).

(3) The polymer polystyrene is used to make thermocoal articles. The polymer polyvinyl chloride is used to make P.V.C. pipes, doormats, etc. The polymer teflon is used to make nonstick cookware. The polymer polypropylene is used to make injection syringe, furniture, etc.

Give scientific reasons:

Question 1.
Carbon atoms are capable of forming an unlimited number of compounds.
Answer:

1. Carbon has the property of catenation. Two or more carbon atoms can share some of their valence electrons to form (single, double and triple) bonds.
2. The straight chains or branched chains or rings may have different shapes and sizes. This results in formation of many compounds. Hence, carbon atoms are capable of forming an unlimited number of compounds.

Question 2.
Ethylene is an unsaturated hydrocarbon.
Answer:
(1) Ethylene (CH₂ = CH₂) contains a double bond between carbon atoms.
(2) Thus, the valencies of the two carbon atoms are not fully satisfied by single covalent bonds. Hence, ethylene is an unsaturated hydrocarbon.

Question 3.
Naphthalene burns with a yellow flame.
Answer:
(1) Naphthalene is an unsaturated compound. In unsaturated hydrocarbon the proportion of carbon is larger than that of saturated hydrocarbon. As a result, some unburnt carbon particles are also formed during combustion of unsaturated compounds.

(2) In the flame. these unburnt hot carbon particles emit yellow light and therefore the flame appears yellow. Hence, naphthalene burns with a yellow flame.

Question 4.
The colour of iodine disappears in the reaction between vegetable oil and iodine.
Answer:
(1) Vegetable oils (unsaturated compound) contains a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.
(2) The addition reaction of vegetable oil with iodine takes place instantaneously at room temperature. The colour of iodine disappears in this reaction. This iodine test indicates the presence of a multiple bond in vegetable oil.

Question 5.
The hydrogenation of vegetable oil in the presence of nickel catalyst forms vanaspati ghee.
Answer:
(1) The molecules of vegetable oil contain long and unsaturated carbon chains. These unsaturated hydrocarbons contain a multiple bond as their functional group. They undergo addition reaction to form a saturated compound as the product.

(2) When vegetable oil (unsaturated compound) is hydrogenated in the presence of nickel catalyst, the addition reaction takes place, vanaspati ghee (saturated compound) is formed.

Distinguish between the following:

Question 1.
Saturated hydrocarbons and Unsaturated hydrocarbons.
Answer:
Saturated hydrocarbons:

1. In saturated hydrocarbons, the carbon atoms are linked to each other only by single covalent bonds.
2. They contain only a single bond.
3. They are chemically less reactive.
4. Substitution reaction is a characteristic property of these hydrocarbons.
5. Their general formula is C_nH_{2n + 2}.

Unsaturated hydrocarbons:

1. In unsaturated hydrocarbons, the valencies of carbon atoms are not fully satisfied by single covalent bonds.
2. They contain carbon to carbon double or triple bonds.
3. They are chemically more reactive.
4. Addition reaction is a characteristic property of these hydrocarbons.
5. Their general formula is C_nH_2n or C_nH_{2n – 2}

Question 2.
Open chain hydrocarbons and closed chain hydrocarbons.
Answer:
Open chain hydrocarbons:

1. A hydrocarbon in which the chain of carbon atoms is not cyclic is called an open chain hydrocarbon.
2. All aliphatic hydrocarbons contain open chains.

Closed chain hydrocarbons:

1. A hydrocarbon in which the chain of carbon atoms is present in a cyclic form or ring form is called a closed chain hydrocarbon.
2. All aromatic hydrocarbons contain closed chains.

Question 3.
Alkane and Alkene.
Answer:
Alkane

1. Alkanes in which the carbon atoms are linked to each other only by single bonds.
2. The general formula of an alkane is C_nH_{2n + 2}
   
3. They are chemically less reactive.

Alkene:

1. Alkenes in which carbon atoms are linked to each other by double bonds.
2. The general formula of an alkene is C_nH_2n.
3. They are chemically more reactive.

Project:

Question 1.
Prepare a list of carbon compounds which occur in nature and discuss their uses in daily life.