Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process

Balbharti Maharashtra State Board Class 10 Political Science Solutions Chapter 2 The Electoral Process Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Political Science Solutions Chapter 2 The Electoral Process

Question 1.
Choose the correct option from the given options and complete the sentences.
(1) The Election Commissioner is appointed by the …………………………. .
(a) President
(b) Prime Minister
(c) Speaker of Loksabha
(d) Vice President
Answer:
(a) President

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(2) …………………………. was appointed as the first Chief Election Commissioner of independent India.
(a) Dr. Rajendra Prasad
(b) T.N. Sheshan
(c) Sukumar Sen
(d) Neela Satyanarayan
Answer:
(c) Sukumar Sen

(3) Constituencies are created by …………………………. committee of the Election Commission.
(a) Selection
(b) Delimitation
(c) Voting
(d) Timetable
Answer:
(b) Delimitation

Question 2.
State whether the following statements are true or false. Give reasons for your answer.
(1) The Elections Commission lays down the code of conduct during elections.
Answer:
The above statement is True. Reasons:

  • It ensures free and fair elections.
  • Maipractices during the election come under control.
  • Due to the strict observance of the code of conduct in the last few’ elections, the common voters have become confident.

(2) Under special circumstances the Election Commission holds re-elections in a particular constituency for a second time.
Answer:
The above statement is True. Reasons :

  • Sometimes, the representative of Lok Sabha, Vidhan Sabha or the local self governmènt resigns from his/her constituençy.
  • In some cases, death of the representative occurs.
  • In such special situations, the Election Commission has to conduct an election for a second time. It is called By-elections.

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(3) The state government decides as to when and in how many stages the elections would be held in a particular State.
Answer:
The above statement is False. Reasons :

  • The entire process of conducting elections is entrusted upon and managed by the Election Commission.
  • If this responsibility is given to the state government it may adopt a biased approach.
  • Hence, the Constitution has formed the Election Commission an independent body to carry out the responsibility.

Therefore, it is decided by the Election Commission as to when and in how many stages it will conduct elections.

Question 3.
Explain the concept.
(1) Reorganising the constituencies
Answer:
(1) The Election Commission of India formed constituencies for Lok Sabha and Legislative Assembly.
(2) The Election Commission had decided upon the constituencies before the first election. As the years passed, there was a lot of migration of the people for business and other activities from the villages to cities.
(3) This changed the demography to large extent. Number of voters in some constituencies reduced while in some it increased to a very great extent. This disturbed the ratio of- seats allotted as compared to population in those constituencies.
(4) Hence, the need to readjust the constituencies arose. The Delimitation Commission of the election commission does the work of reorganising or restructuring of constituencies.

(2) Midterm Elections

Question 4.
Complete the following picture.
Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 1
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 2
Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 3
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 4

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Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 5
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 6

Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 7
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 8

Question 5.
Answer in brief.
(1) Explain the functions of the Election Commission.
Answer:
The functions of the Election Commissipn are:
(1) Prepare the voters’ list.
(2) Decide election timetable and decide the entire process of holding elections.
(3) Scrutinize the applications of the candidates.
(4) Conduct free and fair elections and do all the work related to it.
(5) Give recognition and also de-recognize political parties.
(6) Resolve all the disputes and complaints regarding elections.

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(2) Write some additional information about post of the Election Commissioner.
Answer:
(1) The Election Commission in India has one Chief Election Commissioner and two other Chief Commissioners.
(2) All the commissioners are appointed by the President.
(3) The Chief Election Commissioner of India is usually a member of the Indian Civil Service or. Indian Administrative Service.
(4) The responsibility of conducting free and fair elections to the Parliament and State Legislatures lies with the Election Commissioner.
(5) In order do safeguard the independence of the Election Commissioner, he cannot be easily removed from the post for any political reasons.

(3) Explain the meaning of Code of Conduct.
Answer:
(1) After the announcement of elections till the declaration of results, the Election Commission enforces the Code of Conduct.
(2) It explains the rules to be followed by the government, political parties candidates and voters before and during elections.
(3) Code of conduct is adopted to control malpractices during elections. It ensure free and fair ecections.

Project
Organise a mock poll in the school to understand the process of voting.
Answer:

Memory Map
Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 10

Question 6.
Choose the correct option from the given options and complete the sentences:
(a) Article of Indian Constitution created the independent body of Election Commisšion.

(a) 351
(b) 370
(c) 324
(d) 301
Answer:
(c) 324

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(b) system exists in India.
(a) Single-party
(b) Two-party
(c) Multi-party
(d) No-party
Answer:
(c) Multi-party.

(c) The right to give recognition or de-recognize a political party lies with ……………….. .
(a) President
(b) Election Commission
(c) Parliament
(d) Vice-President
Answer:
(b) Election Commission

(d) There are constituencies of Lok Sabha at present.
(a) 288
(b) 350
(c) 500
(d) 543
Answer:
(d) 543

(e) from the present state of Himachal Pradesh was the first voter.
(a) Sukumar Sen
(b) Sham Sharan Negi
(c) Prem Kumar Ghumal
(d) P N. Chadda
Answer:
(b) Sham Sharan Negi

(f) Due to EVM, people can also vote easily.
(a) elder
(b) salaried
(c) Divyanga
(d) Transgender
Answer:
(c) Divyanga

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(g) The first elections in India were held in
(a) 1948-49
(b) 1949-50
(c) 1950-51
(d) 1951-52
Answer:
(d) 1951-52.

Question 7.
State whether following statements are True or False. Give reasons for your answer :
(a) There should be secrecy in Election process.
Answer:
The above statement is False. Reasons :

  • Election should be conducted in a free and fair environment.
  • If the elections are not held in free environment then there are chances of malpractices and corruption.
  • Then, it will be impossible to elect the honest and efficient candidates.

(b) The Election Commission has started awareness campaign for registration of voters.
Answer:
The above statement is True. Reasons :

  • The responsibility of preparing and updating electoral roll lies with the Election Commission.
  • The Election Commission starts an awareness campaign to create awareness among new eligible voters so that they register themselves in the voter’s list.
  • The Indian voter is not -much aware about the election process.
  • Special voter’s awareness campaign is run for voter’s registration.
  • For their awareness National Voter’s Day is celebrated every year.

(c) Every candidate who fills the nomination form can contest election.
Answer:
The above statement is False. Reasons :

  • Every candidate of a party or independent candidate has to be personally present to fill the nomination form.
  • It is necessary for him or her to give complete information in the nomination form as decided by the Election Commission.
  • The nomination forms are then scrutinized. If there are irregularities in a nomination paper and if the information is found to be false the nomination forms are rejected.

Therefore, it is not possible for every candidate who fills the nomination form to contest election.

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(d) Sometimes, the Election Commission has to conduct mid-term elections. OR Explain the concept : Mid-Term Elections.
Answer:
The above statement is True. Reasons :

  • If the elected government in power loses its majority before completing its term.
  • If no party gets complete majority, then two or more parties come together and form a coalition government.
  • Such coalition government collapses if any party withdraws the given support.
  • In such situations, the government is left with no option other than resigning.
  • If there is no alternative available to form government then the Parliament or Vidhan Sabha is dissolved before completing its term. In such a scenario, the Election Commission has to conduct mid-term elections.

Question 8.
Explain the following concepts :

(a) What is representation?
Answer:
Modern democracy is a representative democracy. In a democracy it is not possible to involve the entire population in the ecision-making process. This resulted in the starting of the practice of electing some people on behalf of entire population
as representatives who would run the government. The representatives who form the government are expected to be responsible to the people and give preference to the welfare of the people.

  • Direct and Indirect or representative democracy rire two types of democracy.
  • In modem nation-states; the population has increased to a great èxtent.
  • So it is impossible to involve all the people in decision-making process.
  • Thus, th practice of electing some people on behalf of entire population as representatives started.
  • The elected representatives form government and work for the welfare of the people.

(b) Election Commission :
Answer:
In India, the Election Commission is central to the process of elections. Art. 324 of the Indian Constitution has established this autonomous body which consists of one Chief Election Commissioner and two other commissioners.

  1. One of the most important features of a democratic nation is elections at regular intervals. Holding free and fair elections at regular intervals is essential for a democratic system.
  2. Under the Article 324 of the Constitution, Election Commission was formed in 1950. The President appoints one Chief Election Commissioner and two additional commissioners. It is an autonomous body.
  3. The rank and powers of all the three commissioners are the same. The declaration of dates of the elections to the announcement of the results the entire procedure is monitored by the Election Commission.
  4. The Election Commission does not have its own staff to carry out this procedure. So they carry out the work with help of government employees and teachers. Special provisions are made for all finances incurred by the Election Commission.

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Question 9.
Write short notes :
(a) Journey from Ballot box to EVM machine :
Answer:

  1. From the first election in 1951-52 till 1999, elections were held using ballot box. Twenty lakh ballot boxes were used in the first election. Voters used to cast his or her vote by stamping in front of the candidate’s name and put them in the metal boxes.
  2. Electronic Voting Machines (EVMs) were first used for 5 seats in Rajasthan, 5 seats in Madhya Pradesh and 6 seats in New Delhi 1998 in Legislative Assembly.
  3. EVM machines were used at all polling booths in the general elections held in 2004. It proved to be a very useful device.
  4. It has been improvised since its first use. Due to the use of EVMs the results are declared early and at a very fast rate.

(b) Recognition to Political Parties :
Answer:

  • India has a multi-party system with recognition accorded to national, state and regional level parties by the Election Commission.
  • Their recognition depends on the voting percentage received by them in the assembly elections and number of elected representatives of their party.
  • If any party does not fulfill these criteria, its recognition is cancelled.
  • The Election Commission allots appropriate symbols to parties and independent candidates. All political parties should have recognition of the Election Commission.

Question 10.
Complete the concept map :

(a) Prepare a flow chart on the process of election.
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 2 The Electoral Process 9

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(b) Which two conditions among following is the violation of code of conduct?
(1) The candidate distributes items of household use. –
(2) Promises made to resolve the water problem if elected.
(3) To go from door to door to meet voters and request them to vote.
(4) To appeal on the basis of caste and religion to get support.
Answer:
(1) The candidate distributes items of household use.
(2) To appeal on the basis of caste and religion to get support.

Question 11.
Answer in brief :
(a) Why is it important to conduct elections?
Answer:
It is important to conduct elections because of the following reasons :

  • The existence and working of democracy depends on elections.
  • All political parties get a chance to rule.
  • Elections help to bring a change in power through peaceful meAnswer:
  • It not only changes government policies but also society.

(b) What are the conditions for voting?
Answer:
The following are the conditions for voting:

  • The person should be a citizen of India.
  • He should have completed 18 years of age.
  • His name should appear in voters’ list.
  • The person should have photo identity card issued by the Election Commission of India.

(c) What action is taken by the Election Commission if disputes arise regarding elections?
Answer:

  • If any disputes arise regarding the elections, the Election Commission is empowered to take final decisions.
  • The Election Commission conducts a thorough inquiry about the said dispute.
  • If there is evidence of any malpractices during elections, in any constituency, it declares the elections invalid and announces re-polls.
  • If any candidate breaks the code of conduct and contests elections, he/she is barred by the Election Commission from contesting elections.

(d) What challenges are faced by the Election Commission to conduct free and fair elections?
Answer:
The following challenges are faced by the Election Commission tcx conduct free and fair elections :

  • Managing the large geographical landscape and huge electoral population.
  • To stop misuse of money and muscle power during elections.
  • Barring candidates with criminal background from contesting elections.
  • Conducting elections successfully in politically criminalised environment.
  • Conducting elections in spite of increasing instances of violence and making them a success.

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(e) What are the advantages of EVM machines?
Answer:
The battery operated Electronic Voting Machine (EVM) has more advantages than the ballot box. They are as follows :

  • It saves tonnes of paper used to make ballot paper.
  • So, it conserves the environment as it stops the reckless cutting of trees required to make paper.
  • If the voter does not wish to cast his vote in favour of any candidate contesting, he can make use of NOTA (None Of The Above).
  • It makes counting of the votes much faster which enables the election officer to declare result in a short time.
  • It is helpful for disabled (Divyanga) people to cast vote.

(f) Explain the features of procedures of voting during the first Lok Sabha Election.
Answer:

  • It was a challenge to prepare voters’ list at the time of the first election. Illiteracy rate was very high in our country. Therefore, the procedure to vote and making the voter list was a challenge.
  • 20 lac steel boxes were made and election symbols of political parties were stuck on it.
  • Blank ballot papers were given to the voters and they were supposed to drop in the box having the election symbols of the party they decide to vote for.
  • Even the illiterate people could vote because of this system.

Question 12.
Give your opinion :
(a) When candidates have only the condition of age, why should they give other information to Election Commission? Answer:

  • While filling the form candidates should reveal information about his property assets and if there are any criminal charges against him.
  • When candidates have only the condition of age as eligibility, why should they give other information to election commission?
  • Why are the candidates required to give the information of their property to Election Commission?
  • Such candidates if elected can misuse power and amass wealth with corrupt practices.
  • With criminal background they can even threaten voters to vote for him.
  • His nomination could get cancelled based on the information.

(b) Why is it so?
(A) Some constituencies are reserved for scheduled castes and scheduled tribes.
Answer:

  • It is difficult for the people of scheduled castes and scheduled tribes to get representation as they are scattered in different parts.
  • Without a representative it is difficult to discuss their problems in Parliament.
  • Lack of representative will hinder their progress. Hence some constituencies are reserved for scheduled castes and scheduled tribes.
  • Some constituencies are kept reserved for Scheduled caste and Scheduled tribes.
  • Every political party has an election symbol.
  • At the time of voting and counting of votes, the official representatives of political parties remain present.
  • Recognised parties have equal opportunity to present their side before media such as television and radio.

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(B) Why every political party has an election symbol?
Answer:

  • After independence, the literacy rate was quite low in India.
  • It was not possible for the voters to read the name of the candidate and vote.
  • Therefore, the Election Commission gave symbols to political parties and independent candidates which helped the voters to identify and decide whom to vote for.

(C) At the time of voting and counting of votes, the official representatives of political parties remain present.
Answer:

  • There are incidences of duplicate voters who register in multiple constituencies.
  • There are cases of rigging of EVM or booth capturing.
  • Such incidences are brought to light by representatives who are present at polling centres.
  • During the counting process, if the EVM machine looks tampered, the representative can raise an objection.

(D) All recognised parties should get an equal opportunity to express their opinion on media such as television and radio.
Answer:

  • All political parties should get a fair chance to express their agenda.
  • Their ideas and philosophy should reach the people.
  • Television and radio are owned by the government.
  • Political parties have equal right on both.
  • Hence, all the recognised parties can express their opinion on Doordarshan and Radio.

(c) Think!
(A) How political parties suffer due to family monopoly in the party? OR What are the disadvantages of dynasty rule?
Answer:

  1. If only one family has domination on the political party because of dynasty rule then others are not given leadership opportunity.
  2. It is impossible to have all the members of the family efficient. An inefficient heir can cause damage to the party.
  3. The growth and expansion of party comes to a halt because of such heir. His faults seep into the party making it weak in the long term.
  4. The nature of such a party become dictatorial. Opposing views are suppressed and the internal democracy in the party vanishes.
  5. If the heir does not have progressive thoughts then the party becomes regressive and of obsolete ideology.
  6. How political parties suffer due to family monopoly in the party?
  7. What do you understand by the system of ‘one vote one value’?

(B) What do you understand by the system ‘One Vote One Value’?
Answer:

  • There is great importance in political and social equality in democracy.
  • According to this ideology, ‘One Vote One Value’ is very important.
  • In a democracy, each vote has the same value. The value of the vote of a Prime Minister and a common man is same.
  • Under military rule or dictatorship or during monarchy the value of a vote for privileged classes was more. There was no importance given to the vote of the common man.
  • ‘One Man One Vote’ indicates all the people in the country have same status. This is the gift of democracy.

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(d) Voting is our duty as well as responsibility to vote.
Answer:

  • It is enshrined in the fundamental principles of our Constitution to vote.
  • It is not only our duty but also responsibility.
  • Democracy exists because of elections. People should elect honest and efficient representatives through election.
  • If voters show no interest in voting then the government will ignore people’s welfare.
  • Hence I feel it is not only the duty of every citizen to vote but also his responsibility.
  • Government has to observe the code of conduct declared by the Election commission.

(e) What measures should be taken to increase the credibility of elections?
Answer:
To increase the credibility of elections the following measures should be taken :

  • 50% seats should be reserved for women candidates by every party.
  • Candidates with a criminal background should be permanently barred from contesting any elections.
  • The misuse of money should be stopped during elections. The government should incur the expenditure.
  • Candidates who resort to malpractices should be immediately booked. A strict inquiry and action should be taken against them by the court.
  • Laws and regulations should be followed strictly by the political parties before giving election tickets.
  • If the political parties do not co-operate with the above terms, the Election Commission should cancel their recognition.

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(f) Which rules would you include in Code of Conduct for voters?
Answer:
The following rules should be included in Code of Conduct for voters :

  • The voters who abstain from voting should be fined and government should suspend all the facilities given to them.
  • If it is proved that the voter has accepted money or any kind of gifts, he should be punished.
  • The action of voters should not instigate common people.
  • They should not involve in bogus voting.
  • They should not resort to illegal means for voting.
  • The candidate distributes items of household use.
  • Promise made to resolve the water problem if elected.
  • To go from door to door to meet voters and request them to vote.
  • To appeal on the basis of caste and religion to get support.

 

Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy

Balbharti Maharashtra State Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy

Question 1.
Choose the correct option from the given options and complete the sentences.
(1) In a democracy …………………….. participate in elections and get political power.
(a) political parties
(b) courts
(c) social organisations
(d) none of the above
Answer:
(a) political party

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(2) The major challenge faced by all democratic nations in the world is ……………………. .
(a) Religious conflicts
(b) Naxal activities
(c) Deepening the roots of democracy
(d) Importance to muscle power
Answer:
(c) Deepening the roots of democracy

Question 2.
State whether following statements are true or false. Give reasons for your answer.
(1) Alertness is required to sustain democracy.
Answer:
The above statement is True.

  1. In order to make democracy successful it should be practised by people in all aspects of their life. It should not remain limited at the government level.
  2. Conscious effort should be taken to keep the rights of the people intact.
  3. The different challenges faced by democracy like corruption,violence and criminalisation should be tackled timely and strictly. It is necessary that people and the government should remain alert for the sustenance of democracy.

(2) Importance of the problems of farmers and tribals has increased in the left extremist movement.
Answer:
The above statement is False.

  • Landlords confiscated lands of farmers and tribals.
  • Naxalite movement was started to remove injustice against the landless farmers and tribals.
  • But later the problems of farmers and tribals lost its focus and became violent.
  • They adopted violent ways like to attack army, police and oppose the government.
  • The importance of farmers and tribals decreased in the leftist movement.

(3) People may lose confidence in the democratic process due to corruption during elections.
Answer:
The above statement is True.

  • In order to strengthen democracy, it is important to conduct elections in a free and fair atmosphere.
  • There are instances of .the election process getting affected by corruption.
  • Bogus voting, bribing the voters, abducting voters and ballot boxes, distributing articles to lure voters and other such things take place.
  • All these things make people lose faith in the democratic process.

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Question 3.
Explain the concept.
(1) Left-extremism
(2) Corruption
Answer:

  • It is a form of dishonest or unethical conduct for personal gain by people at influential position.
  • It is found at all levels economic, political, social and at government levels. The misuse power is also corruption.
  • Bogus voting, giving bribe to voters, are examples of corruption in the election process.
  • Hoarding of goods and then selling them at price more than the market yalue is also a way of corruption.
  • Corruption in the public and private sector is the biggest problem in India.
  • People develop distrust and dissatisfaction about the entire system. They lose trust in democracy because of corruption.

Question 4.
Answer the following questions in brief.
(1) Which factors are required for the success of democracy in India?
Answer:
The following factors are required for the success of democracy in India:

  • Though democracy is the government of majority the opinions of minorities, religious, ethnic and lingustic should be included in the decision-making process.
  • Stringent laws must be implemented to stop criminalisation of politics.
  • Efforts should be made not only at the level of the government but also at social and personal level.
  • Decisions about important public policies should be taken after interaction with the people.. Participation of people should increase in all government undertakings.

(2) What are the effects of criminalisation of politics?
Answer:
The following effects are seen due to criminalisation of politics:

  • Money and muscle power gets undue importance.
  • Violence increases during elections and it becomes difficult to hold elections in free and fair environment.
  • Terror seizes the mind of people and their participation becomes less in administration.
  • People lose their tolerance and the development of democracy does not take place.

(3) What efforts are undertaken to bring transparency in political process?
Answer:
The following efforts are made to bring transparency in political process:

  • An independent Election Commission is appointed to hold free and fair elections.
  • Stringent laws are made to stop criminalisation of politics.
  • Laws are formed to stop corruption in politics and elections.
  • Courts have banned people who are corrupt and those having criminal background or criminal allegations from taking part in political process.

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Project
(1) What measures will you suggest to control corruption? Make a list of it.
(2) Organise a group discussion in your class on the problem of terrorism in India.
(3) Present a street play on ‘how to get rid of addiction?
Answer:

Memory Map
Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy 1

Question 5.
Choose the correct option from the given options and complete the sentences:
(a) ………………….. is a continuous and living process.
(a) Military rule
(b) Democracy
(c) Dictatorship
(d) Monarchy
Answer:
(b) Democracy

(b) The main objective of democracy is ………………….. .
(a) to hold elections
(b) public welfare
(c) public campaigns
(d) social work
Answer:
(b) public welfare

(c) Voting, elections, government structure, etc. is only ………………….. form of democracy.
(a) economic
(b) political
(c) social
(d) all-inclusive
Answer:
(b) political

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(d) In order to make democracy meaningful the Government of India adopted …………………. .
(a) internal party elections
(b) public welfare schemes
(c) decentralisation of power
(d) protection of democratic values.
Answer:
(c) decentralisation of power

(e) Naxalite movement began to remove injustice done to …………………. .
(a) workers and dalits
(b) middle and lower class
(c) poor and common class
(d) landless farmers and tribals
Answer:
(d) landless farmers and tribals

(f) The …………………. in India is making conscious efforts to make political process transparent,
(a) Parliament
(b) Government
(c) Judiciary
(d) Political parties
Answer:
(c) Judiciary

(g) Due to increase in ………………….., there is increase in religious conflict.
(a) communalism
(b) terrorism
(c) corruption
(d) casteism
Answer:
(a) communalism

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(h) In order to make democracy successful in India, participation of …………………. should increase.
(a) government
(b) people
(c) political parties
(d) social organization
Answer:
(b) people

Question 6.
Explain with reasons whether the following statements are true or false:

(a) Communalism does not cause much harm to the nation.
Answer:
The above statement is False.

  • Increase in communalism leads to religious conflict in the country.
  • Increasing conflicts in turn create frictions in society and destroy social unity.
  • It creates division in society resulting in terrorist activities.
  • People’s participation also reduces in the democratic process.
  • This ultimately creates instability in society and democracy comes in danger.

Hence communalism harms the nation to a great extent.

(b) The opinion of the minorities should not be taken into consideration.
Answer:
The above statement is False.

  1. Even though the government of the majority community comes to power, it should adhere to the democratic principle that government should work for the welfare of all communities.
  2. In democracy, opinion of all the communities should be valued instead of giving importance only to the majority community.
  3. All religious, linguistics, ethnic and caste groups should be part of the decision-making process of the government. Hence, to avoid injustice to minorities their opinion should also be taken into consideration.

(c) Democracy is the best form of governance.
Answer:
The above statement is True.

  • The freedom and rights of the people remain intact in democracy.
  • In democracy, values like liberty, equality, social justice, secularism and fraternity are nurtured in the real sense.
  • Welfare of the people is the main aim of’ democracy.
  • People do not enjoy so much freedom in any other form of government. Hence, democracy is the best form of governance.

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(d) Majority opinion has a lot of importance in Democracy.
Answer:
The above statement is True.

  • In a democracy, the political party getting majority of votes comes to power.
  • Parliament takes all the decisions by majority.
  • Democracy aims at the welfare of the majority of the people.

Therefore, majority opinion has a lot of importance in Democracy.

Question 7.
Explain the concept:
(a) Decentralisation:
Answer:

  • If complete power rests in the hands of government it gives rise to dictatorship.
  • Therefore division of power between legislature, executive and judiciary is essential.
  • This division of power is known as Decentralisation.
  • Decentralisation safeguards the freedom of the people.
  • People participate in the functioning of the government.
  • With increased participation, people become aware of their responsibilities.
  • Decentralisation has great importance in democracy.

(b) Criminalisation of politics:
Answer:

  • Participation of criminals in political process is criminalisation of politics. It is a serious problem and a threat to our democratic system.
  • Political parties or candidates spread terror among people using money and muscle power.
  • Political parties who give candidature to people with criminal background are responsible for violence during elections.
  • Such candidates after coming to power continue their criminal activities.

They create financial scams and trouble the opponents. Criminalisation of politics weakens democracy.

Question 8.
Write short notes:
(a) Communalism and Terrorism:
Answer:

  • Communalism and Terrorism causes great harm to nation. Communalism emerges out of narrow religious pride.
  • Increasing communalism triggers religious conflict in the country. Religious conflict hampers social stability.
  • Society gets divided on communal lines. Communalism gives rise to terrorism.
  • People’s participation in democratic process in reduced to a great extent due to terrorism.
  • It causes great damage to our nation.

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Question 9.
Do as directed:
(1)
Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy 2
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy 3

(2)
Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy 4
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy 5

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(3)
Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy 6
Asnwer:
Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy 7

(4)
Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy 8
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 5 Challenges faced by Indian Democracy 9

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Question 10.
Answer the following questions in brief:
(a) What challenges are faced at global level by democracy?
Answer:
The following challenges are faced by democracy at global level:

  • Many democratic countries in the world face the threat of military regime.
  • It becomes important to propagate democracy which safeguards people’s rights and freedom.
  • All-inclusive democracy is real democracy which should be adopted and practised instead of adopting just political form of democracy.

(b) What improvements are required for democracy to be deep rooted?
Answer:
The following improvements are required for democracy to be deep rooted:

  • Values like freedom, equality, fraternity, justice, peace and humanitarianism should be practised and nurtured by all sections in the society.
  • Autonomy should be given to various social organisation and assimilqje all the sections of the society.
  • Empowerment to the citizens, free and fair elections and independent judiciary is necessary.
  • Adoption of a form of democracy which gives preference to public welfare.

(c) What measures are adopted by India to strengthen democracy?
Answer:
In order to strengthen democracy following measures are adopted by India:

  • Decentralisation of power.
  • Reservation for minorities and women so that they get a share in power.
  • The values of liberty, equality, secularism and social justice adopted by our Constitution.
  • At administrative level various projects like Education for all, Clean Bharat campaign.
  • Gram Samruddhi Yojana, Mahatma Gandhi National Rural Employment Guarantee Scheme are undertaken.

Question 11.
Give your opinion:
(a) What is your opinion about participation of people in various undertakings of the government?
Answer:

  • There will be change in the public policies of the government.
  • There will be interaction between all sections of society and exchange of ideas.
  • There will be transparency in government administration and problems like corruption will be tackled.
  • More public welfare schemes will get implemented and no one will feel that injustice is done to them or they are left out.

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(b) Are internal elections held by political parties?
Answer:

  • It is mandatory to hold internal elections in a political party.
  • Elections are held for various posts like president, treasurer and secretary.
  • Elections are held after every 3 years as per the niles laid down by the Election Commission of Itdia.
  • The paity which does not abide by the rule loses recognition.
  • Because of such elections, there can never be dominance of any single person on a party.
  • Democracy sustains within the party. In Indian democracy the internal elections are held by all the parties regularly.
  • Another challenge before democratic nations is to ensure that democracy becomes deep rooted.
  • Freedom, equality, fraternity and justice, peace, development and humanitarianism are the values that should be practiced at all levels of the society.
  • The mass support for this purpose can be gathered only through democratic means.

(c) In spite of economic reforms China accepted dominance of only one party. Is China a democratic nation?
Answer:

  • After the Communist revolution in 1948 China became a republic.
  • After the formation of the republic, the Communist party of China became a dominant party.
  • All the offices in the party are elected through internal elections.
  • But in practice, there is dictatorship of the Communist party. Democracy is for namesake in China.
  • In such a situation, the people do not enjoy freedom as in true democracy.
  • However, China accepted economic reforms and became part of World Trade Organization.
  • Another challenge before democratic nations is to ensure that democracy becomes deep rooted.
  • Freedom, equality, fraternity and justice, peace, development and humanitarianism are the values that should be practiced at all levels of the society.
  • The mass support for this purpose can be gathered only through democratic means.

(d) Do you think there should be family monopoly in politics?
Answer:

  • There is no place for family monopoly in politics.
  • The elected representatives serve the people till their term lasts.
  • People vote out the inefficient representatives out of power in next election.
  • If there is monopoly of one family, the perspective of democracy becomes narrow, common people cannot share power.
  • If any inefficient heir comes to power, he or she could be a great loss for the party, people as well as the nation.
  • Family monopoly in politics is a major problem before democracy in India.
  • Monopoly of just one family in politics reduces democratic space. Common people cannot participate in the public sector.

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Hence there should be no family monopoly in politics.

Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History

Balbharti Maharashtra State Board Class 10 History Solutions Chapter 6 Entertainment and History Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 History Solutions Chapter 6 Entertainment and History

Question 1.
(A) Choose the correct option from the given options and complete the statement.
(1) ……… are supposed to be the first keertankar in Maharashtra.
(a) Saint Dnyanehshwar
(b) Saint Tukaram
(c) Saint Namdev
(d) Saint Eknath
Answer:
(c) Saint Namdev

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(2) Baburao painter made the movie, …………………….. .
(a) Pundalik
(b) Raja Harischandra
(c) Sairandhri
(d) Bajirao-Mastani
Answer:
(c) Sairandhri

(B) Identify and write the wrong pair in the following set.
(1) Raigadala Jevha Jag Yete – Vasant Kanetkar
(2) Tilak Ani Agarkar – Vishram Bedekar
(3) Sashtang Namaskar – Acharya Atre
(4) Ekach Pyala – Annasaheb Kirloskar
Answer:
(4) Wrong Pair: Ekach Pyala – Annasaheb Kirloskar

Question 2.
Complete the following chart.

Bhajan Keertan Lalit Bharud
Characteristics
Examples

Answer:

Sr.No. Point Bhajan Keertan Lalit Bharud
1 Characteristics Singing songs in praise of God and chanting his name accompanied by Taal (Cymbals), Pakhvaj and Mridangam (1) Naman and Nirupanacha Abhang and Nirupan comprise the Poorvarang
(2) Narration of a story to illustrate main theme is Uttarrang
(1) Deity is invoked during festival to fulfill desire
(2) It is performed in a theatrical style. Stories of Krishna, Rama and great devotees are presented during performances
It is a metaphorical song with spiritual and ethical teachings.
2 Examples Bhajans of Saint Tulsidas, Saint Tukadoji Maharaj and Saint Namdev Naraadiya Keertan and Mahatma Phule’s Keertan Popular in Konkan and Goa Bharuds of Saints Eknath, Namdev and Dnyaneshwar.

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Question 3.
Write short notes:
(1) Need of entertainment
Answer:

  • Entertainment of excellent quality is essential for healthy growth of a person as it is an integral part of one’s life?
  • To break the boredom of routine life and keep the mind lively and fresh we need gome entertainment.
  • It makes us feel more energetic and our efficiency at- work improves.
  • Hobbies and games are pursued for entertainment which eventually develops personality.
  • Entertainment refreshes our mind and helps to distress.
  • Lack of entertainment in one’s life will lead to monotonous life and boredom.

(2) Marathi Theatre
Answer:

  • Theatre is a place devoted to performances either solo or collective, of performing arts.
  • The 19th century saw a great development of the Marathi Theatre.
  • Vishnudas Bhave was known as the father of the Marathi Theatre.
  • In the initial years historical, mythological plays were performed along with light farcical plays.
  • The plays had no written script.
  • The tradition of having a complete written script began with the play ‘Thorale Madhavrao Peshwe’ in 1861.
  • At the end of 19th century,„ the tradition of musical plays started.
  • Historical themes and social problems were presented through these plays.
  • The popular plays by Acharya Atre like Udyacha Sansar, Gharabaher helped the Marathi theatre to sustain through a temporary decline. Vasant Kanetkar, Vishram Bedekar, Acharya Atre, enriched the Marathi theatre.

(3) Entertainment and professional opportunities
Answer:

  • There are many professions associated with theatre and cinema.
  • Professional hairstylists, costume designers, make-up artists, art directors who put up stage backdrops are required in theatre.
  • Directors, technicians, actors, lightmen, costume and jewellery designers and assistants are required too. Experts in music and script writers, singers are required.
  • Cinema requires all of them along with dance directors, singers, cameramen, dialogue writers and story writers. Scholars of history can work in this field as art directors.

Question 4.
Explain the following statements with reasons.
(1) Expertise in history is important in the film industry.
Answer:
It is essential to have knowledge of history while making films on historical’ events or a person. .

  • If the movie has a historical theme then art directors are required to create backdrop designs showcasing the atmosphere of that period.
  • To write movie dialogues, the knowledge of the culture and language as spoken in that period is necessary.
  • It is important to have knowledge of appropriate hairstyles, costumes, jewellery make¬up of that era.
  • Scholars of history are required who can work as art directors or as consultants to the art director.
  • Experts in field of history can find many professional opportunities.

(2) Bharuds composed by Saint Eknath are popular in Maharashtra.
Answer:

  • Saint Eknath composed Bharuds with the purpose of educating people on various aspects of life.
  • Bharuds composed by Saint Eknath had a wide range of subjects, dramatic quality, easy rhythm and humour.
  • People liked the way it was performed.
  • A message was given in a humorous way.

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Question 5.
Answer the following questions in detail.
(1) Why is Maharashtra known as the land that nurtured the Indian film industry?
Answer:

  1. The contribution of Madanrao Madhavrao Pitale, the Patwardhan family of Kalyan and Harishchandra Sakharam Bhatvadekar is very important in the development of Indian movies.
  2. Dadasaheb Torane, A. E Karandikar, S. N. Patankar, V. E Divekar sought help from foreign technicians and made a movie entitled Pundalik. It was released in Mumbai in 1912.
  3. ‘Raja Harischandra’ was the first movie to be processed completely in India. It was released in Mumbai in 1913.
  4. The credit of making a full-length movie goes to Maharashtra.
    Therefore Maharashtra is known as the land that nurtured the Indian film industry.

(2) What is Powada?
Answer:

  1. Powada is a dramatic narration by altematingly reciting poetry and prosaic extracts. Powada narrates great deeds of heroic men and women in a very forceful and inspiring style.
  2. The Powada composed by – Adnyandas, a contemporary poet of Chhatrapati Shivaji Maharaj which narrated the incident of Afzal Khan’s death and battle of Simhgarh composed by Tulsidas are very famous.
  3. In the British period, Powadas narrating the stories of Umaji Naik, Chaphekar brothers and Mahatma Gandhi were composed.
  4. During the Samyukta Maharashtra Movement the Powadas were used as medium of creating public awareness.

Project
Get the lyrics of any one of Saint Eknath’s Bharud, and enact it in the cultural programme of your school.

Question 6.
Complete the sentences by choosing the correct option:
(a) In the 18th century, ………………………. started a Phad of Dashavatara artists which used to perform all over Maharashtra.
(a) Saint Gadge Maharaj
(b) Adnyandas
(c) Tulsidas
(d) Shyamiji Naik Kale
Answer:
(d) Shyamji Naik Kale

(b) Traditionally, ………………………. is supposed to be the founder of keertan tradition.
(a) Saint Namdev
(b) Saint Eknath
(c) Naradmuni
(d) Saint Gadge Maharaj
Answer:
(c) Naradmuni

(c) The Powada composed by the poet ………………………. on the incident of the killing of Afzal Khan is well-known.
(a) Adnyandas
(b) Tulsidas
(c) Ramdas
(d) Surdas
Answer:
(a) Adnyandas.

(d) Compositions of ………………………. are not part of Bhajans sung in Karnataka.
(a) Purandardas
(b) Surdah
(c) Bodhendraguruswami
(d) Thyagraj
Answer:
(b) Surdas

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(e) Varkari sect has developed a glorious tradition of ………………………. by chanting God’s name.
(a) Powada
(b) Dashavatari Natak
(c) Bhajan-Keertan
(d) Bharud
Answer:
(c) Bhajan-Keertan

(f) Powada composed by Tulsidas on the battle of ………………………. is very well known.
(a) Panhala
(b) Raigarh
(c) Pratapgad
(d) Simhgarh
Answer:
(d) Simhgarh

(g) Powadas composed by ………………………. were not the part of Samyukta Maharashtra Movement.
(a) Amarsheikh
(b) Patthe Bapurao
(c) Annabhau Sathe
(d) Gawankar
Answer:
(b) Patthe Bapurao

(h) ………………………. is known as the Father of Marathi theatre.
(a) V. J. Kirtane
(b) Dattopant Patwardhan
(c) Vishnudas Bhave
(d) Annasaheb Kirloskar
Answer:
(c) Vishnudas Bhave.

(i) started the tradition of having a complete written script.
(a) V. J. Kirtane
(b) Vishnudas Bhave
(c) Shripad Krishna Kolhatkar
(d) Govind Ballal Deval
Answer:
(a) V. J. Kirtane

(j) In the latter half of the 19th century, made special efforts to introduce classical khyal music in Maharashtra.
(a) Bhimsen Joshi
(b) Kumar Gandharva
(c) Kishori Amonkar
(d) Balkrishnabuva Ichalkaranjikar
Answer:
(d) Balkrishnabuva Ichalkaranjikar

(k) In India, is the first one to make a cine camera.
(a) Anandrao Painter
(b) Baburao Painter
(c) Dadasaheb Torne
(d) Dadasaheb Phalke
Answer:
(a) Anandrao Painter

(l) In 1925, made a movie Bajirao Mastani, which was later on banned by the British Government under the suspicion of spreading nationalistic sentiments.
(a) Sanjay Leela Bhansali
(b) Dadasaheb Phalke
(c) Baburao Painter
(d) Bhalaji Pendharkar
Answer:
(d) Bhalaji Pendharkar

(m) was the first woman producer of: Marathi movies.
(a) Kamalabai Mangarulkar
(b) Devika Rani
(c) Amirbai Karnataki
(d) Kanandevi
Answer:
(a) Kamalabai Mangalurkar.

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

Name of the Play Playwright
(1) Raygadala Jevha Jag Yete Vasant Kanetkar
(2) Tilak Ani Agarkar Vishram Bedekar
(3) Sashtang Namaskar Acharya Atre
(4) Ekach Pyala Annasaheb Kirloskar

Answer:
Wrong pair: Ekach Pyala – Annasaheb Kirloskar

(2)

First release Movie
(1) First full length movie released in India Raja Harishchandra
(2) First historical film in India Simhgarh
(3) Movie dealing with real social issues Savkari Pash
(4) Indian movie which got international acclaim Saint Dnyaneshwar

Answer:
Wrong pair: Indian movie which got international acclaim — Saint Dnyaneshwar

(3)

Film producer Produced Biographical Movies on
(1) Acharya Atre Ram Shastri
(2) Vishram Bedekar Vasudev Balwant Phadke
(3) Dinakar D. Patil Dhanya te Santaji Dhanaji
(4) Prabhakar Pendharkar Bal Shivaji

Answer:
Wrong Pair: Acharya Atre – Ram Shashtri

(4)

(1) Keechakvadh Krishnaji Prabhakar Khadilkar
(2) Ekach Pyala Ram Ganesh Gadkari
(3) Ithe Oshalala Mrutyu Vasant Kanetkar
(4) Natasamrat Vijay Tendulkar

Answer:
Wrong Pair: Natasamrcrt -Vijay Tendulkar

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(5)

Work Saint
(1) Gave momentum to the Bhakti movement in Gujarat Saint Kabir
(2) First Keertankar of Maharashtra Saint Namdev
(3) Popularised Khanjiri Bhajan Saint Tukdoji Maharaj
(4) Tradition of Rashtriya Keertan was started Dattopant Patwardhan

Answer:
Wrong Pair: Gave momentum to Bhakti movement in Gujarat — Saint Kabir

(6)

(1) The first play in Marathi Seetaswayamvar
(2) First play having  complete written script Thorale  Madhavrao
(3) Metaphorical Drama Udyacha Sanskar
(4) Play based on

Shakespeare’s King Lear

Natasamrat

Answer:
Metaphorical Drama-Udyacha Sanskar.

(7)

Name of the Play Playwright
(1) Thorale Madhavrao Peshwe V. J. Kirtane
(2) Ekach Pyala Annasaheb Kirloskar
(3) Sangeet Sharada Shripad Krishna Kolhatkar
(4) Sangeet Manapaman Krishnaji Prabhakar Khadilkar

Answer:
Wrong Pair: Sangeet Sharada – Shripad Krishna Kolhatkar

Question 8.
Complete the graphical presentation:
(a) Prepare concept map on:
(1) Types of Puppets:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 1
Answer:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 2

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(2) Saints who popularised Bhajans:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 3
Answer:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 4

(3) Plays by famous Playwrights:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 5
Answer:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 6

(4) Saints who popularised Bhajans in North India:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 7
Answer:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 8

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(b) Prepare a flow chart on the development of Marathi Theatre:
Answer:
Development of Marathi Theatre:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 9

Question 9.
Explain the concept:
(1) Dashavatara Theatre:
Answer:

  • The stories presented in Dashavatara shows are based on the ten incarnations of Vishnu.
  • The method of acting, make-up, costumes in Dashavatara show is set by the tradition.
  • It is mostly a musical show but there may be a few spontaneous dialogues.
  • At the beginning of the show, Sutradhar, the narrator invokes Lord Ganesha, for its successful run.
  • Dashavatara is part of the folk theatre in Maharashtra which has its origin in mythological plays.

(2) Bhajan:
Answer:

  • Singing songs in praise of God and chanting God’s name accompanied by instruments like taal (cymbals), mridangam, pakhavaj is known as Bhajan.
  • Bhajan is an important element of devotional music for those who are on the path of devotion.
  • Varkari sect made Bhajans accessible to all.
  • There are two types of Bhajans, Chakri and Songi BhajAnswer:
  • Devotees keep moving in circular fashion and without break in Chakri Bhajan.
  • In Songi Bhajan, singer-actors act as devotees and deliver dialogues in the form of devotional songs.

(3) Bharud:
Answer:

  • Bharud can be described as a metaphorical song that has spiritual and ethical lessons. Bharud is similar to road show.
  • Bharud is popular because of its wide range of subjects, humorous presentation, dramatic quality and easy rhythm.
  • Bharuds are composed with the purpose of educating people on various aspects of life.
  • Even though Bharuds of Saint Eknath are famous, bharuds were composed by many saints including Saint Dnyaneshwar.

(4) Keertan.
Answer:

  • Keertan involves oratory, singing, acting, dancing and story telling.
  • Naradmuni is assumed to be the founder of Keertan tradition.
  • It is pure glorification of god. It is also a medium to educate the masses about good values of life and very purpose of human life.
  • There are two parts in Naraadiya keertan Poorvarang and Uttarrang. Poorvarang comprises of Naman.
  • Nirupanacha Abhang and Nirupan; Uttarrang comprises of narration of a story to illustrate the main theme.
  • Keertan has two traditions in Maharashtra – Naraadiya and Varakari.

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(5) Documentaries: (You would like to know this: Textbook page 44)
Answer:

  • A film which gives information, inspires and educates people and is a short film is a documentary.
  • Documentaries were made on freedom struggle, national leaders, social issues and superstitions, forts, animal species, sports, etc.
  • They were aimed at creating public awareness about various issues.
  • They are shown in the cinema theatres before the start of the main movie.

Question 10.
Write short notes:
(a) Means of Entertainment:
Answer:

  • Entertainment is an integral part of man’s life. Man has developed many means of entertainment since ancient times.
  • Ancient times saw the rise of festivals, fairs, sports, dance-music, etc.
  • The means of entertainment changed with times.
  • Television, mobiles, video games and movies and such other modem means of entertainment were introduced.
  • Folk music, classical music, plays, books, newspapers, magazines are some mediums of entertainment which are available. Different types of sports, hobbies and travel too are means of entertainment.

(b) Lalit:
Answer:

  1. Lalit is an old form of entertainment popular in Konkan, Maharashtra and Goa. It belongs to the tradition of Naaradiya Keertan.
  2. It is presumed that the presiding deity is present on the throne. It is invoked by the people as it is widely believed that the deity fulfils all the wishes.
  3. Stories of Krishna, Rama and of great devotees are presented during the performance.
  4. Lalit forms a part of the backdrop of modem Marathi theatre.

(c) Keechakvadh: (Do You Know? Textbook page 43)
Answer:

  • Krishnaji Prabhakar Khadilkar wrote Keechakvadh in the pre-independence era. It was a metaphorical drama.
  • It was based on the incident of Keechakvadh described in the epic, Mahabharata.
  • Draupadi represented helpless Mother India, while Yudhishthira represented the moderates and Bheem the extremists.
  • Keechak represented the insolent Viceroy Lord Curzon.
  • The audience used to perceive characters in this fashion and feel* enraged about the imperialistic British rule.

(d) Natashmrat: (Do You .Know? Textbook page 43)
Answer:

  • The renowned author-poet Vishnu Waman Shirwadkar, also knpwn as Kusumagraj wrote Natasctmrat.
  • It is styled after Shakespeare’s well known play ‘King Lear’.
  • Ganpatrao Belvalkar, the tragic protagonist of Natasamrat represents a blend of two well-known personalities of early Marathi stage Ganpatrao Joshi and Nanasaheb.
  • The traits of both great actors are found in the main character of Natasamrat.
  • Natasamrat is a tragic story of an aging actor who gives his entire wealth to his sons and is humiliated by them.
  • This play was very popular and created history on stage performance and in playwriting.

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(e) Tamasha (Folk theatre):
Answer:

  1. Tamasha is a Persian word which means a pleasing sight. Tamasha emerged as an independent form in the 18th century absorbing the traits of folk theatre and classical arts.
  2. Tamasha is classified into ‘Sangeet Bari’ and ‘Dholakicha Phad’. Dance and music are more important than drama in Sangeet Bari. Tamasha with drama as main part was developed later. It included Vag, the dramatic part a little later.
  3. The show begins by singing the praise of Lord Ganesha, known as Gana. It is followed with the presentation of Gavalan.
  4. The second part of Tamasha presents the Vag. The plays like ‘Vichchha Mazi Puri Kara’ or ‘Gadhavache Lagna’ were very popular.

Question 11.
Explain the following statements with reasons:
(a) Puppetry is an oldest form of entertainment.
Answer:

  • Puppet show was an important form of entertainment.
  • The remains of clay dolls have been found in the excavations at the archaeological sites of Harappa, Egypt and Greece civilisations.
  • Information on puppets is found in the texts like Mahabharata and in Panchatrantra stories.
  • The ancient text Mahabharat has a mention that puppetry was one of the 64 arts.

(b) Vishnudas Bhave is known as the Father of Marathi Theatre.
Answer:

  • The origins of the Marathi theatre can be traced to Dashavatara tradition. Lalit forms a backdrop of Marathi theatre.
  • Seetaswayamvar, the first play, written and presented by Vishnudas Bhave was very successful.
  • The movement of stage plays started by Vishnudas Bhave was followed in Maharashtra by historical, mythological and also light farcical stage plays.
  • The farcical plays dealt with social issues in a humorous way. Therefore, he is known as the Father of Marathi theatre.

Question 12.
Answer the following questions in 25 – 30 words:
(a) Make a list of various types of entertainment and classify them into different categories. (Try to do it: Textbook Page 39)
Answer:

  • Entertainment can be classified into two categories, active and passive.
  • Active entertainment means an individual’s mental-physical participation. In passive entertain-ment, a person may not be an actual participant.
  • To play cricket is active form of entertainment but to watch a cricket match is passive entertainment.
  • To participate in festivals, fairs, celebrations is active entertainment but to watch as audience is passive entertainment.

(b) Write about Dashavatara form of folk theatre.
Answer:

  • The stories presented in Dashavatara are based on the 10 incarnations of Lord Vishnu.
  • The method of acting, make-up, costumes in Dashavatara shows is set by the tradition.
  • The show is mostly musical but sometimes there may be a few spontaneous dialogues.
  • The characters representing gods use wooden masks. At the start of the show, the sutradhara invokes Lord Ganesha.
  • The show ends by breaking dahihandi, followed by aarati, praising the God.
  • This is part of folk theatre in Maharashtra.
  • Dashavatara shows are presented in the regions of Konkan and Goa after the harvesting season is over.

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(c) What is required to be a Keertankar?
Answer:
The following qualities are required to be a Keertankar:

  • A Keertankar also known as Haridas or Kathekaribuva needs to be very well informed.
  • He should have wide experience and knowledge of the world.
  • He should be well-versed in mythological and social subjects.
  • He needs to train himself in qratory, singing, musical instruments, dance and humour. .
  • He should dress in a traditional way.

(d) Write about the contributions of Bhosale family to drama.
Answer:

  • The Bhosale family of Tanjore were successors of Chhatrapati- Shivaji Maharaj. They were patrons of arts.
  • The rulers of the Bhosale family encouraged dramas in Marathi and in southern languages.
  • Some of them have written a few plays and also translated Sanskrit plays.

(e) What is the contribution of Vishnudas Bhave and V. J. Kirtane to Marathi theatre?
Answer:

  1. Vishnudas Bhave presented the first play, Seetaswyamwar, on stage. Initially no written scripts were used for plays. Only the lyrics were written and dialogues were spontaneous.
  2. The movement started by him was followed in Maharashtra by historical, mythological and also light farcical stage plays. He is known as the Father of the Marathi theatre.
  3. V. J. Kirtane was the first author who wrote the script of Thorale Madhavrao Peshwe in 1861 and its printed copy was made available.
  4. It was the beginning of the tradition of having a complete written script ready before staging the play.

(f) Explain the nature of Rashtriya Keertan.
Answer:

  • During the independence movement, a new type of Keertan was developed known as Rashtriya Keertan.
  • It is performed in the same way as Naradiya Keertan.
  • It placed more importance on creating awareness by narrating the life stories of great leaders of the Indian independence movement, scientists, social reforms, etc.
  • Dattopant Patwardhan of Wai started Rashtriya Keertan.

Question 13.
Read the following passage and answer the questions:
(a) Who presented the play ‘Seetaswayamvar’?
Answer:
‘Seetaswayamvar’ was the first play presented by Vishnudas Bhave.

(b) Who wrote the musical play ‘Sharada’?
Answer:
Govind Ballal Deval wrote the musical play Sharada.

(c) How can plays bring about social awakening?
Answer:

  • As theatre is an audio-visual medium, it creates a strong impact on the audience.
  • They commented on evil customs, traditions, superstitions in our society. This started the reformation process.
  • Sharada, a musical play, written by Govind Ballal Deval shed light on the evil custom of marrying young girls to aged men in a humorous style.
  • ‘Ekach Pyala’ by Ram Ganesh Gadkari made the society aware about the evil effects of drinking.

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Question 14.
Answer the following questions in detail:
(a) Write about the art of Puppetry.
Answer:

  • The Kathputali is a traditional art of puppetry which has two styles.
  • One that developed in Rajasthan and the other in South India.
  • In ancient India, materials like wood, wool, leather, horns and ivory were used to make puppets.
  • The role of the narrator known as Sutradhar is very crucial in stage show.
  • The stage for this puppetry show is very small but the puppeteers use light and sound effects in an ingenious way.
  • Shadow puppets, hand puppets, string puppets and wooden puppets are used in Kathputali shows.
  • The artists who perform Kathputali shows are found in Uttar Pradesh, Maharashtra, Rajasthan, Telangana, Karnataka and Kerala.

(b) Write about the development of Indian film industry.
Answer:

  1. Cinema is a medium that brings together art and technology. With the advent of the technology of motion pictures the film industry came into being. It gave rise to the era of silent movies.
  2. The technology of sound recording paved the way for talkies. Dadasaheb Torane, A. P Karandikar, S. N. Patankar and V. E Divekar made the movie Pundalik with help from foreign techniciAnswer: This was a great step in the development of the art.
  3. Dadasaheb Phalke made a full length movie, completely processed in India. He made silent movies and documentaries also.
  4. Baburao Painter’s cousin, Anandrao Painter made the first cine-camera. Baburao Painter made many historical movies and a movie on realistic social issues. Bhalaji Pendharkar made movies invoking nationalist sentiments.
  5. Kamalabai Mangarulkar was the first woman producer, who made movies in Marathi as well as Hindi.
  6. Prabhat Film Company made many religious, historical, mythological and social movies. Production studios like Bombay Talkies, Rajkamal Productions, R. K. Studios, Navketan played significant role in development of the Indian film Industry. Period from 1961 to 1981 is the golden period of Indian film industry.

Question 15.
Identify the given picture and write about his contribution:
Maharashtra Board Class 10 History Solutions Chapter 6 Entertainment and History 10
Answer:

  1. The given picture is of Dadasaheb Phalke who is known as the Father of Indian Film Industry.
  2. He released the first movie ‘Raja Harishchandra’ in Mumbai in 1913. He directed the movie which was entirely processed in India for the first time.
  3. He made silent movies named as Mohini- Bhasmasur, Savitri-Satyavana.
  4. He also made documentaries on the rock cut caves of Verul and pilgrim centres of Nashik and Tryambakeshwar. Later, he made historical and mythological movies. Maharashtra Board Solutions
  5. The Government of India has honoured him by instituting Dadasaheb Phalke Award given for lifetime contribution to cinema, which is considered one of the most prestigious awards.

Maharashtra Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution

Balbharti Maharashtra State Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution

Question 1.
Choose the correct option from the given options and complete the sentences.
(1) In Maharashtra …………………… seats are reserved for women in local self-governing institutions.
(a) 25%
(b) 30%
(c) 40%
(d) 50%
Answer:
(d) 50%

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(2) Which of the following laws created a favorable environment for women to secure freedom and self-development?
(a) Right to Information Act
(b) Dowry Prohibition Act
(c) Food Security Act
(d) None of the above
Answer:
(b) Dowry Prohibition Act

(3) The essence of democracy is ……………………
(a) universal adult franchise.
(b) decentralisation of power.
(c) policy of reservation of seats.
(d) judicial decisions.
(d) Judicial decisions
Answer:
(b) decentralization of power

Question 2.
State whether the following statements are true or false. Give reasons for your answer.

(1) Indian democracy is considered the largest democracy in the world.
Answer:
The above statement is True.

  • Indian Constitution cancelled all the conditions which were put for voting before independence thereby increasing the number of voters.
  • The Constitution adopted adult suffrage which has facilitated all adult men and women to cast their vote.
  • The age limit to vote was reduced to 18 years from 21 years which gave opportunity of political participation to the young generation. No other democratic country in the world has voters- in such large numbers.

Hence, Indian democracy is the largest democracy in the world.

(2) Secrecy in the working of Government has increased due to the Right to Information.
Answer:
The above statement is False.

  • To strengthen democracy and increase mutual trust between the government and the people, it is very important that the people should know about ‘the functioning of the government.
  • Transparency and accountability are the hallmarks of good governance.
  • With Right to Information given to the citizens, Government became more transparent. Thus, the Right to Information has reduced element of secrecy in administration.

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(3) The nature of the Constitution is seen as a living document.
Answer:
The above statement is True.

  • Parliament has the powers to make changes in the Constitution according to the changing circumstances and conditions.
  • However, it should be done without tampering or changing the basic structure of the Constitution.
  • As it has kept itself abreast with the changes, S the Constitution became a live and dynamic document instead of a rigid one

Question 3.
Explain the following concepts.

(1) Right based approach
Answer:

  • India adopted democracy after independence. Each government which came to power took efforts to make the democratisation process more profound.
  • In the initial stages, these governments considered citizens as ‘beneficiaries’.
  • After the year 2000, the approach towards citizens changed. Democratic reforms were considered as ‘rights’ of citizens.
  • Hence, the Right to Information, Education and Food Security was granted not as beneficiaries but as rights of the citizens. This approach is known as Rights Based Approach.

(2) Right to information
Answer:

  • In order to bring transparency in the administration and make it accountable, Indian citizens are given Right to Information.
  • Right to Information helped in promoting harmony between government and people and empowered the citizens.
  • It brought transparency in administration, made the government realise that they are answerable to people.
  • It has helped to reduce the element of secrecy which surrounded the functioning of the government. It made the government open and transparent.

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(3) Women representation in the Loksabha.
Answer:

  • The Constitution of India empowers women by granting them equal status with men and equal political rights.
  • 22 women were elected in the Lok Sabha elections held in 1951-52. Their number increased to 66 in 2014 elections.
  • Reservation has been increased to 50% in local self-governing institutions in Maharashtra and in many other states.
  • If the number of women representative increased in Lok Sabha, it will help in reducing incidents of violence against women and increase their involvement in the decision-making process.

Question 4.
Answer in brief.
(1) What are the effects of reducing the voting age from 21 years to 18 years?
Answer:

  • Earlier the Indian Constitution had fixed the voting age of 21 years.
  • The voting age was lowered to 18 years.
  • The young voters secured the right to vote and the increased number of voters was unparalleled. ‘
  • They got the right to decide how their representative should be. ,
  • Moreover, it provided to the unrepresented youth an opportunity to become part of political process being literate and politically conscious.
  • It increased the scope of democracy by including the youth of the country thus making it the largest democracy in the world.

(2) What is meant by establishment of social justice?
Answer:
Establishing social justice means:

  • To eliminate the practices and beliefs which are responsible for injustice towards a person or a community and which hampers collective progress of society should be eliminated.
  • Government policies should be all inclusive which means it should aim at accommodating different sections of society.
  • There should not be any discrimination based on caste, creed, religion, gender, language, property, region or place of birth.
  • All should get equal opportunities for development.

(3) Which decision of the Court has resulted in protection of honour and dignity of women?
Answer:

  • The apex court has given several judgments which have helped in protection of honour and prestige of women.
  • Court gave judgement on Right to alimony as well as Right to equal remuneration.
  • Women have an equal share in the property of husband and father. This gave them financial security. Dowry prohibition Act was a measure for women empowerment.
  • The Act against sexual harrassment. Domestic Violence Prohibition Act are also very important in the direction of women empowerment.
  • All these acts emphasised the need to protect women and protect their self-esteem and dignity rejecting the traditional forms of domination and authoritarianism.

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Project
(1) Which information can be secured with the help of the right to information? Find out with the help of your teachers.
(2) Make a list of concessions given by the Government for the students of minority communities?
(3) Visit the official website of the National election commission and collect more information about it.
(4) Take an interview of women representations from local self-governing institutions from your area.

Memory Map
Maharashtra Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution. 7

Q. Complete the sentences by choosing the correct option:

1. Right to Information came into force from
(a) 2002
(b) 2004
(c) 2005
(d) 2006
Answer:
(c) 2005

2. The Indian government has been working in accordance with principle of the
(a) Parliament
(b) President
(c) Constitution
(d) Party
Answer:
(c) Constitution

3. Men and women above years of age can vote in India.
(a) 15
(b) 18
(c) 21
(d) 25
Answer:
(b) 18

4. can make amendments to the Constitution.
(a) President
(b) Prime Minister
(c) Council of Ministers
(d) Parliament
Answer:
(d) Parliament

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5. got recognition because of 73rd and 74th Amendments to the Constitution.
(a) Parliament
(b) Local self-governing institutions
(c) Legislative council
(d) Cooperative societies
Answer:
(b) Local self- governing institutions

6. ………………….. has made the government more transparent.
(a) Equality
(b) Freedom
(c) Right to Information
(d) Social Justice
Answer:
(c) Right to Information

7. and are two features of good governance.
(a) Popular, populist
(b) Efficient, democratic
(c) Transparent, responsible
(d) Equality, decentralization
Answer:
(c) Transparency, responsible.

Q. State whether the following statements are true or false:

1. Indian democracy is evidently unsuccessful.
Answer:
The above statement is False.

  • India has the largest number of voters as compared to any other democratic nation in the world.
  • Free and fair elections which are held regularly is the key for successful democracy.
  • Recurring elections have helped in understanding the political process. As the voting age was reduced from 21 to 18 years, the political participation has increased.
  • Increasing participation of the people in the political process and political contest shows that Indian democracy is successful.

2. There is less friction in all inclusive democracy.
Answer:
The above statement is True.

  • To establish social justice and equality is the aim of democracy. .
  • If all the sections of society are given equal opportunities without any discrimination then all 8 components come into the main stream.
  • In fact, democracy is the process of accommodating different sections of society which ultimately reduces the social conflict.

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Q. Explain the following concepts:

1. Decentralisation:
Answer:

  • Division of power within, a country is known as Decentralisation.
  • Under dictatorship and military rule, the power is centred around one person or a group of individuals.
  • But in democracy the power is divided among Centre, State and Local Self-governing institutions. Decentralisation is the core of democracy.
  • Decentralisation can stop the misuse of power and facilitate common people to participate in democracy.

2. Provisions regarding Minorities:
Answer:

  • Constitution has adopted several measures for the protection of the rights of minorities.
  • Several policies have been adopted by the government to provide them the opportunities in education and employment.
  • The Constitution has prohibited any form of discrimination on the basis of caste, creed, religion, language and region.
  • The Constitution provided rights to the minorities to protect and conserve their language, culture and establish educational institutions

3. Policy of Reservation of Seats:
Answer:

  • Some sections of Indian society were denied social justice.
  • They were deprived of educational and employment opportunities. It was essential to bring them in the main stream of society.
  • The policy was adopted to give reservation to Scheduled Castes and Scheduled Tribes in educational institutions and public employment.
  • Seats were also reserved for Other Backward Classes. Reservation policy gave the deprived classes justice and opportunities for development.

Q. Complete the Concept Map:

1. Complete the Concept Map:
Maharashtra Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution. 1
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution. 2

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2. important to understand:
Maharashtra Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution. 3
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution. 4

3. please understand:
Maharashtra Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution. 5
Answer:
Maharashtra Board Class 10 Political Science Solutions Chapter 1 Working of the Constitution. 6

Q. Answer the following questions in brief:

1. What are the provisions made in the Directive Principles regarding decentralization?
Answer:
The Directive Principles of the Indian Constitution has made provisions for decentralization of power. It means:

  • Indian Constitution has divided the power between the centre, state and local self-governing institutions.
  • The guidelines about empowering the local self-governing bodies are given in the Constitution.
  • After independence, 73rd and 74th Amendment to the Constitution in 1992 has given constitutional status to the local self-governing bodies.
  • Due to this, there is increase in the participation by the people at grass root level. Democracy was put into practice which eventually got strengthened. Decentralisation curbed the misuse of power.

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2. What were the measures taken to increase the number of women in politics?
Answer:

  • The 73rd and 74th Constitutional Amendment provided reservation of 33% seats for women in local self governing institutions.
  • This reservation has been increased to 50% in Maharashtra and many other states.
  • One-third seats and the offices (posts) of Sarpanch, Mayor, etc. are reserved for women. Several policies have been adopted to remove illiteracy among women and to make available opportunities for their development.
  • A bill is tabled in the house to reserve 50% seats in Legislative Assembly and Parliament.

3. The Judiciary in India has played an important role in strengthening democracy. Explain.
Answer:

  • The Judiciary has always taken into consideration the fundamental objectives of the Constitution and also the intention of its framers.
  • While interpreting the Constitution, it has taken care that its basic structure is not altered as Constitution is foundation of our democracy.
  • The important role played by judiciary is in strengthening and helping democracy to achieve the objectives of social justice and equality.

4. Which particular Laws/Acts created a favourable environment for the protection of freedom of women and secure their development?
Answer:

  1. Several policies have been adopted in the post-independence period for empowering women.
  2. Provisions were made in the Constitution to provide opportunities for progress.
  3. Many laws were passed to empower them. They are:
    • Right to have equal share in the property of father and husband.
    • Dowry Prohibition Act.
    • Act against Sexual Harassment.
    • Domestic Violence Prohibition Act.

Q. Give your opinion:

1. What measures should be adopted to increase the number of women in representative democracy?
Answer:
Several restrictions were imposed in the name of traditions and practices, making’ the workspace of women limited to home.

  • To curb injustice the representation of women should increase in all institutions. The 73rd and 74th Amendment reserved 33% of seats for women in local self governing institutions.
  • Access to more and more social and political fields should be made available for women. They should be involved in the decision-making process and work for their betterment.
  • This would ultimately lead to eliminating injustice done to them and will enhance their self respect and status.

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2. Do you agree that Indian democracy has become profound?
Answer:
I entirely agree that Indian democracy has become profound.

  • The Indian Constitution has laid down representative structure of democracy.
  • The actual practice of the principles of democracy is the essence of representative system.
  • People have direct representation in Lok Sabha, Rajya Sabha and Local Self-Government Institutions.
  • Free and fair elections are held at regular intervals to elect representatives.
  • Citizens cast their votes weighing the public issues and policies related to it. All the above factors show that Indian democracy has become profound.

3. Do you think that citizens in India should 8 have the right to employment? (Discuss – Textbook page 70)
Answer:
I agree that all Indian citizens should get I employment.

  • If they are deprived of employment opportunities their families would face hunger and starvation.
  • Incident of crime will increase in society.
  • Democracy will collapse leading to chaos.
  • More and more employment opportunities S should be generated for their progress.

4. According to you, if everyone gets the right to shelter, how will it affect democracy in 8 India?
Answer:

  • Food, clothing and shelter are the basic necessities of a man.
  • Shelter is not only his necessity but also his right.
  • A permanent need is to get settled in life. 8 If a man gets a home, a large part of his struggle 8 in life will come to an end.
  • The financial burden will be low and he will x work with honesty and will contribute in nation’s X progress.
  • Home for all creates a healthy society. It has an all-round social, economical and psychological effect to strengthen democracy and make it 8 profound.

5. What steps should be taken to stop injustice done to the Backward Classes? OR What efforts should be made to prevent x atrocities?
Answer:
The backward classes have suffered for thousands of years. I feel the following measures x will remove injustice done to the backward classes:

  • Atrocities laws should be made stringent.
  • Fast courts should be set up to handle such cases.
  • Stringent punishment should be given if 8 found guilty.
  • Efforts should be made to improve economic 8 status of the backward classes.
  • The Government should make efforts to establish social justice and equality. .

6. What efforts should be made to bring in all the features of good governance in democracy?
Answer:
For the successful functioning of democracy, good governance is very essential.

  • People should elect good and professional candidates.
  • People should keep watch on the work done by them.
  • Corrupt candidates should not be elected or re-elected.
  • People should respond to various policies which are beneficial for society.
  • People should pressurise the government to start various developmental policies for country’s progress.

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Q. B Can you tell the reasons for the following changes?

1. Some seats were reserved for women to increase their participation in political process.
Answer:

  • Women empowerment movement started after independence.
  • All the countries in the world started increasing number of women representatives.
  • The 73rd and 74th constitutional amendment reserved 33% seats for women in local self- governing institutions.
  • Some seats are kept reserved for women to increase their participation in political process.
  • Some seats are kept reserved for weaker sections of the society so that they can get a share in political power.
  • The State Election Commission has been established. The 11th and 12th schedule was added to the constitution.

2. Some seats are kept reserved for weaker sections of the society so that they can get a share in political power.
Answer:

  • The weaker sections suffered injustice for thousand of years.
  • The opportunities of education and employment had been denied to them.
  • In order to establish social justice and equality, reservations are now given to the weaker sections.

3. The State Election Commissions were set up.
Answer:

  • Elections for the Parliament and State Assemblies are conducted by the National Election Commission.
  • It was impossible to put upon the local self government bodies the responsibility of conducting elections. Hence the State Election Commissions were formed.

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4. 11th and 12th Schedule was added to the Constitution.
Answer:

  • The 11th schedule of Indian Constitution was added in 1992 by the 73rd Constitution Amendment Act. This schedule contains 29 subjects related to panchayat.
  • The 74th Amendment to the Constitution added 12th schedule and covered 18 subjects related to the Municipalities.
  • ”It gave constitutional status to Municipalities and Panchayats and aimed to strengthen rural and urban governments so that they can function efficiently.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Balbharti Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 1.
Remake the table taking into account relation between entries in three columns.

I II III
Coal Potential energy Wind electricity plant
Uranium Kinetic energy Hydroelectric plant
Water reservoir Nuclear energy Thermal plant
Wind Thermal energy Nuclear power plant

Answer:

I II III
Coal Thermal energy Thermal plant
Uranium Nuclear energy Nuclear power plant
Water reservoir Potential energy Hydroelectric plant
Wind Kinetic energy Wind electricity plant

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 2.
Which fuel is used in thermal power plant? what are the problems associated with this type of power generation?
Answer:
(1) The fuel used in the thermal power plant is coal. Coal contains chemical energy. Upon burning it releases heat energy. This heat is used for generation of electricity in the thermal power plants.

(2) Problems associated with power 8enerations by thermal power plant:
(a) Air pollution: Due to burning of coal, there is emission of carbon dioxide, carbon monoxide, sulphur dioxide and nitrogen dioxide gases. These are harmful and toxic to health.
(b) Soot particles emitted during combustion can cause severe respiratory problems such as asthma.

Question 3.
Other than thermnl power plant. which power plants use thermal energy for power generation? In what different ways is the thermal energy obtained?
Answer:
(1) The power plant based on natural gas and the nuclear power plants also used thermal energy for the power generation. Apart from these, solar energy is also used to produce heat and thereby create the power.
(2) In nuclear power plant, the energy is released by carrying out fission of nuclei of atoms like Uranium or Plutonium. This energy is used to generate the steam of high temperature and high pressure. The steam rotates the turbine. The kinetic energy in steam drives the turbine and turbine in turn drives the generator.
(3) The combustion of natural gas produces gas, which is used to run the turbine. This gas is under high pressure and high temperature. This is used to produce thermal energy.
(4) In solar thermal power plant, thermal energy is generated with the help of solar radiation. For this reflectors and absorbers are used for concentrating solar radiation and converting it into thermal energy.

Question 4.
Which type/types of power generation involve maximum number of steps of energy conversion? In which power generation is the number minimum?
Answer:
The steps of energy conversion are maximum in the thermal power generation. They are minimum in wind energy generation.

Question 5.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 1
a. Maximum energy generation in India is done using …………… energy.
Answer:
Maximum energy generation in India is done using thermal energy.

b. ………. energy is a renewable source of energy.
Answer:
wind energy is a renewable source of energy.

c.Solar energy can be called ……. energy.
Answer:
Solar energy can be called clean energy.

d. ……. energy of wind is used in wind mills.
Answer:
kinetic energy of wind is used in wind mills.

e. ………. energy of water in darns is used for generation of electricity.
Answer:
Potential energy of water in darns is used for generation of electricity.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 6.
Explain the difference.
a. Conventional and Non-conventional Sources of energy.
Answer:
Conventional Sources of energy:

  1. Conventional sources of energy are largely polluting, they release lot of carbon through its emissions.
  2. Conventional sources of energy are not eco¬friendly.
  3. The fuels produced from the conventional sources of energy are comparatively costlier.
  4. Conventional energy power plants require less area and its management cost is also less.
  5. Conventional source of energy are non-renewable.
  6. Conventional sources of energy are in the form of limited reserves. After few years they will be completely over. e.g. Fossil fuels, coal, crude oil, diesel, petrol, natural gas, etc.

Non-conventional Sources of energy:

  1. Non-conventional sources of energy are not polluting, They do not release carbon or other toxic gases.
  2. Non-conventional sources of energy are eco-friendly.
  3. The energy obtained from the non-conventional sources of energy are comparatively cheaper.
  4. Non-conventional energy power plants require more area and its management cost is also more.
  5. Non-conventional source of energy are renewable.
  6. Non-conventional energy sources qre in abundance on the earth. They are persistent and sustainable. Thus they will not get over. e.g. Solar energy, wind energy, etc.

b. Thermal electricity generation and Solar thermal electricity generation.
Answer:
Thermal electricity generation:

  1. After burning the coal, the heat that is produced is used in the generation of thermal electricity.
  2. For producing heat, the coal is burnt in the boilers.
  3. The combustion of coal produces heat. This heat converts water into steam, which is under very high temperature and pressure. By its force the turbines move. The turbines in turn are connected to generator which rotates and produces energy.
  4. Thermal energy is polluting and not eco-friendly.
  5. The fuel here is coal, its reserves are limited.

Solar thermal electricity generation:

  1. Solar radiations are used in solar thermal electricity production.
  2. For production of heat, many reflectors are used which reflect the radiations of the sun into the absorbent.
  3. Sun’s heat convert the water into steam that rotates the turbine. The turbines then rotate the generators. This generates the electricity.
  4. Solar energy is not polluting, it is eco-friendly.
  5. The solar radiations are in abundance and are sustainable and persistent.

Question 7.
What is meant by green energy? Which energy sources can be called green energy sources and why? Give examples.
Answer:
(1) Green energy means eco-friendly form of energy which does not cause environmental problems and are non-exhaustible, perpetual and sustainable.
(2) These sources of energy do not produce toxic gases or other pollutants, therefore they are safe.
(3) Examples of green energy: (i) Hydroelectric energy (ii) Wind energy (iii) Solar energy (iv) Energy obtained biofuels.

Question 8.
Explain the following sentences.
a. Energy obtained from fossil fuels is not green energy.
Answer:
Fossil fuels like petrol, diesel or natural gas when burnt, emit toxic gases and soot particles. Thus, fossil fuels cause air pollution. Burning of fossil fuels cause increased levels of carbon dioxide, carbon monoxide and nitrogen dioxide. The increased carbon dioxide emission results in global warming. Nitrogen oxide results later in acid-rain. Soot particles generated through burning of fuels cause respiratory problems iike asthma.

Moreover, the fossil fuels are non-renewable and exhaustible fuels. They have to be explored from the
deeper layers of the earth causing lots of environmental problems. Green energy is sustainable, renewable and abundant. It never creates any environmental problems and is non-polluting. Thus, energy obtained from fossil fuels is not at all a green energy.

b. Saving energy is the need of the hour.
Answer:
In modern civilization, continuous energy supply is needed for the technology and development. The energy has become a basic need for man. Most of the energy used in India is obtained from thermal power plant. For this energy generation, various fuels are used. The coal and fossil fuels are limited. Due to over-exploitation, these reserves are getting fast depleted. Use of fossil fuels is also resulting in pollution and climate change.

Nuclear energy can be very hazardous. Lot of research is being done in the field of green energy, but the tremendous human population always is in need of more energy. Therefore, each and every person should save the energy, as saving energy is the need of the hour.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 9.
Answer the following questions.
a. How can we get the required amount of energy by connecting solar panels?
Answer:

  • The photovoltaic solar cells can be connected in a series or in parallel to make a solar panel.
  • When solar cells are connected in a series, the potential difference of individual cells are added in the combination, however the currents from individual cells are not added.
  • When solar cells are connected in parallel, the currents of the individual cells are added in the combination, but the potential differences from individual cells are not added.
  • Through such connections the required potential difference and current can be obtained.
  • Many such solar panels are connected in series and in parallel to generate required current and potential difference.
  • When many solar panels connected in series they form a solar string. Many solar strings connected in parallel make a solar array. In such manner we can get the required amount of energy by connecting solar panels.

b. What are the advantages and limitations of solar energy?
Answer:
I. Advantages:

  • While generating the power through solar radiations, no fuel is burnt.
  • Solar energy generation thus does not create any type of pollution. The technology can be completely utilized in regions with abundant sunlight.
  • Solar energy is eco-friendly, green energy.

II. Limitations:

  • Sunlight is available only during day time. Thus solar cells can generate power only during day.
  • In rainy season and in cloudy conditions, solar power generation suffers.
  • The power present in the solar cells is DC while most of the domestic equipments work on AC.

Question 10.
Explain with diagram step-by-step energy conversion in
a. Thermal power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 2a
In thermal power plant the turbines are rotated using steam. Here the coal is burnt. The heat energy liberated from this burning is used to heat the water in the boiler. This water produces steam of very high temperature and pressure. The kinetic energy in the steam rotates the turbines. The rotation of turbines produces its own mechanical kinetic energy.

The generators connected to turbines produce electrical energy. The steam is condensed in a condenser and converted back into water. In this way in thermal power plant, thermal energy to kinetic energy, kinetic energy into mechanical energy and mechanical energy to electrical energy, are the conversions that take place.

b. Nuclear power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 3a
In nuclear power plant, the energy is releasdd by fission of nuclei of atbms like Uranium or Plutonium. This energy is used to generate the steam or high temperature ind high pressure. The kinetic energy in the steam rotates the turbine. The turbine in turn drives the generator to produce electricity.

c. Solar thermal power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 4a
Solar radiation is used to produce thermal energy. For this purpose, many reflectors are used which concentrate the solar radiation on absorbers. The heat energy created due to solar radiations is used to make steam. The steam possesses kinetic energy. This kinetic energy drives turbine and generator. The electrical energy is thus created from this kinetic energy.

d. Hydroelectric power plant:
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 5a
In hydroelectric plant the water stored in the reservoir is used as a source or potential energy. This water is made to fall at a great speed and hence there is production of kinetic energy in flowing water. This fast flowing water ralling down from the reservoir is brought to the turbine at the lower levels. The kinetic energy of the flowing water in turn drives the turbine, The turbine then drives the generator and electrical energy is produced.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 11.
Give scientific reasons:
a. The construction of turbine is different for different types of power plants.
Answer:

  • Generators work on the principles of electromagnetic induction.
  • For this the generator must be rotated.
  • For this purpose, there is a turbine for each generator.
  • This rotation needs energy. The turbines are different according to the type of energy source that is used for its rotation.
  • Therefore, the construction of turbine is different for each power plant.

b. It is absolutely necessary to control the fission reaction in nuclear power plants.
Answer:

  • Nuclear fission reaction is a type of chain reaction.
  • In nuclear power plants these reactions are closely controlled.
  • If these reactions are not managed properly, there can be more production of neutrons in an uncontrolled way.
  • Each released neutron further causes fission of 3 Uranium (U-235) atoms, such uncontrolled reactions can cause hazardous accidents, hence ft is absolutely necessary to control the fission reaction in nuclear power plants.

c. Hydroelectric energy, solar energy and wind energy are called renewable energies. (July ’19, Board’s Model Activity Sheet)
Answer:

  • Hydroelectric energy, solar energy and wind energy is obtained respectively from flowing water, solar radiations and flowing wind.
  • These sources, i.e. water reservoirs, sun and the wind are inexhaustible and sustainable. They will not be finished.
  • On the contrary, the conventional energy sources such as coal and fossil fuels have limited reserves.
  • They cannot be renewed and may get exhausted in future. Hydroelectric energy, solar energy and wind energy can be replenished and hence they are called renewable.

d. It is possible to produce energy from mW to MW using solar photovoltaic cells.
Answer:

  1. Solar panels can be constructed by connecting solar photovoltaic cells in either series or in parallels.
  2. The combinations are done in such a way that it can give the desired potential difference and the current.
  3. Solar strings are then made by joining solar panels in a series.
  4. When solar strings are joined in parallel; they form solar array.
  5. Therefore, by proper combinations, it becomes possible to produce energy from mW to MW using solar photovoltaic cells.

Question 12.
Draw a Schematic diagram of Solar thermal electric energy generation.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 6

Question 13.
Give your opinion about whether hydroelectric plants are environment-friendly or not?
Answer:

  • Hydroelectric plants are advantageous in some respect while in some aspects it does create problems.
  • Hydroelectric power generation does not need burning of fuels. Therefore, there is no problem regarding combustion of fuels and release of toxic pollutants.
  • Electricity can be obtained as and when required if there is enough water in the reservoir.
  • Water is replenished every time when there is sufficient rainfall.
  • All the above facts give an impression that hydroelectric power generation is eco-friendly but it is not.
  • Many villages and settlements are submerged when a dam and reservoir is constructed. The displaced people are given re-settlement, but it causes lot of emotional trauma to people.
  • Biodiversity is affected as forest lands is submerged. The river flow is obstructed by the dam which affects the aquatic organisms residing in such water.
  • Due to excessive pressure of water on land, it is said that the region gets prone to earthquakes.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 14.
Draw a neat labelled diagrams.
a. Energy transformation in solar thermal electric energy generation.
Answer:

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 7a

b. One solar panel produces a potential difference of 18 V and current of 3A. Describe how you can obtain a potential difference of 72 Volts and current of 9 A with a solar array using solar panels. You can use sign of a battery for a solar panel.
Answer:
Given Potential difference is 18 V and current is 3A. The requirement is potential difference of 72 V and current is 9A Voltage remains the same if connected in parallel and gets added it they are connected in series. Current remains the same if connected in series but adds if connected in parallel.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 8

Question 15.
Write a short note on Electrical energy generation and environment.
Answer:
The energy obtained through the fossil fuels as well as nuclear energy can cause degradation of the environment. If such energy sources are used, they can cause harm to the environment.

(1) The burning of fossil fuels cause air pollution. The incomplete combustion of fossil fuels cause release of carbon monoxide. Some more toxic gases and soot particles cause various respiratory diseases. The carbon dioxide produced is creating global warming and climate change. The nitrogen dioxide released through burning is responsible for acid rains.

(2) Fossil fuels are limited. They are getting fast depleted. It has taken millions of years for the fossil fuels to form. The exploration of such fuels also cause environmental degradation and marine pollution too.

(3) In production of nuclear energy, there is a great risk of accidents. The safe disposal of nuclear waste is also a problem.

(4) Hydroelectric power from water reservoirs, wind power from wind, solar energy from sun and electricity from biofuels are eco-friendly alternatives.

Projects: (Do it your self)

Project 1.
Gather information about solar light, solar water heating system and solar cooker.

Project2.
Gather information about a power plant near your locality by visiting the plant.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Can you recall? (Text Book Page No. 47)

Question 1.
What is Energy?
Answer:
The capacity to do work is called energy.

Question 2.
What are different types of Energy?
Answer:
Potential energy and kinetic energy are the two types of energy.

Question 3.
What are different forms of Energy?
Answer:
Heat, light, electric energy, solar energy, chemical energy, nuclear energy, mechanical energy, etc. are different forms of energy.

Use your brain power! (Text Book Page No. 54)

Question 1.
The schematic of hydroelectric plant is shown in Figure 5.17 on text book page no. 54. Water from about middle of the total height of the dam is taken to the turbine, as shown by point B in the diagram.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 9
(i) With reference to point B, potential energy of how much water reservoir in the dam will be converted into kinetic energy?
Answer:
When the sluice gate at point B is opened, the water from reservoir will start flowing. The potential energy of the stored water will become kinetic energy of the quantity of water that is let out through the sluice gates.

(ii) What will be the effect on electricity generation, if the channel taking water to turbine starts at point A?
Answer:
If the channel taking water to turbine starts at point A, then the water will flow with a greater speed. Since point A is at hSight, water will acquire speed. This will result into more efficient rotation of the blades of turbine. The electricity generation will thus become more efficient.

(iii) What will be the effect on electricity generation, if the channel taking water to turbine starts at point C?
Answer:
If the channel taking water to turbine starts at point C, it will affect the electricity generation adversely. Point C is on the lower height as compared to the channel that carries water to the turbine. The flow of the water thus will be affected resulting into improper rotation of blades of turbine. This will certainly affect the electricity generation.

Choose the correct alternative and write its alphabet against the sub-question number:

Question 1.
Large ……….. are used in commercial power generation plants.
(a) machines
(b) generators
(c) turbines
(d) pannels
Answer:
(b) generators

Question 2.
The principle of electromagnetic ……….. was invented by Michael Faraday.
(a) induction
(b) attraction
(c) repulsion
(d) expulsion
Answer:
(a) induction

Question 3.
………… is used to rotate the magnet in the generator.
(a) fan
(b) Generator
(c) Turbine
(d) Panels
Answer:
(c) Turbine

Question 4.
In thermal power plants, the ………… energy in the coal is converted into electrical energy through several steps.
(a) physical
(b) biological
(c) kinetic
(d) chemical
Answer:
(d) chemical

Question 5.
At ………. in Andhra Pradesh power plant based on natural gas has been installed.
(a) Hyderabad
(b) Vishakhapatnam
(c) Samaralkota
(d) Kakinada
Answer:
(c) Samaralkota

Question 6.
Burning of coal may cause serious health problems related to ……….. system.
(a) digestive
(b) respiratory
(c) nervous
(d) excretory
Answer:
(b) respiratory

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 7.
Incomplete combustion of fuels leads to formation of ……….
(a) carbon dioxide
(b) carbon monoxide
(c) carbon tetrachloride
(d) All the above
Answer:
(b) carbon monoxide

Question 8.
Solar cells are made of a special type of material called semiconductor such as ………..
(a) silicon
(b) uranium
(c) borosilicate
(d) hydrogen
Answer:
(a) silicon

Question 9.
……….. of the following is eco-friendly energy resource. (Board’s Model Activity Sheet)
(a) Coal
(b) Hydroelectric power
(c) Fossil fuel
(d) Atomic energy
Answer:
(b) Hydroelectric power

Question 10.
Which is the most abundant and renewable energy?
(a) Thermal power
(b) Solar energy
(c) Fossil fuels
(d) Atomic power
Answer:
(b) Solar energy

Question 11.
What are the two technologies for harnessing solar energy?
(a) Solar photovoltaics and solar thermal
(b) Solar cooker and solar lamp
(c) Heat capturing and Heat conversation
(d) Active and passive technologies
Answer:
(a) Solar photovoltaics and solar thermal

Question 12.
Which of the following is used in solar cooker to harvest the solar energy?
(a) Solar panels
(b) Silicon cell
(c) Mirrors
(d) Glass lid
Answer:
(c) Mirrors

Question 13.
Which of the following is not the source of green energy?
(a) Wind
(b) Natural gas
(c) Sunlight
(d) Fossil fuel
Answer:
(d) Fossil fuel

Question 14.
The solar lamp uses the energy.
(a) Heat
(b) Wind
(c) Light
(d) Sound
Answer:
(c) Light

State whether the following statements are true or false with proper explanation:

Question 1.
In thermal power plants, the turbines work on solar energy.
Answer:
False. (In thermal power plant, the turbines work on steam. The turbines working on solar energy are not used.)

Question 2.
How to dispose the nuclear waste safely is a big challenge before the scientists.
Answer:
True. (Nuclear waste disposal is the greatest problem. It produces highly toxic effects in any ecosystem. Therefore, disposing such radioactive substances becomes a major challenge.)

Question 3.
The efficiency of power generation using coal plant is higher than that of power generation plant based on natural gas.
Answer:
False. (The efficiency of power generation using natural gas plant is higher than that of power generation plant based on coal.)

Question 4.
Energy obtained from nuclear fission is eco-friendly.
Answer:
False. (Energy obtained from nuclear fission is not eco-friendly, because if accidents happen it leads to hazardous accidents.)

Question 5.
In hydroelectric power plant, the kinetic energy in water stored in dam is converted into potential energy of water.
Answer:
False. (In hydroelectric power plant, the potential energy in water stored in dam is converted into kinetic energy of water. The forceful downpour of flowing water causes this kinetic energy.)

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 6.
The turbine is connected to electric generator, therefore the magnet rotates and electric energy is thus produced.
Answer:
True. (The rotating wheels of turbine cause mechanical energy. This energy helps to produce electrical energy.)

Question 7.
Use of energy is unavoidable in our daily life, but we must use it carefully and only in the required amount.
Answer:
True. (The energy Supply for everyday use results into lot of pollution. This causes harmful effects in the surrounding environment. Therefore, energy should be used in minimal amount and with great care.)

Question 8.
The machine which converts the potential energy of wind to electrical energy is called wind-turbine.
Answer:
False. (When wind blows, the kinetic energy is present in it. This kinetic energy is converted into electricity. The flowing wind never has a potential energy.)

Question 9.
The potential difference available from a solar cell is independent of its area.
Answer:
True. (The potential difference available from a solar cell is independent of its area. However, it is dependent on the way in which solar cells are connected.)

Question 10.
The power available from the solar cells is AC.
Answer:
False. (The power available from solar cells is always DC while the domestic appliances that we use work on AC.)

Match the columns:

Question 1.

Column I Column II
(1) Polluting energy (a) Soot particles
(2) Eco-friendly energy (b) Thermal energy
(c) Nuclear energy
(d) Wind energy

Answer:
(1) Polluting energy – Thermal energy
(2) Eco-friendly energy – Wind energy

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 2.

Column I Column II
(1) Pollutants (a) Soot particles
(2) Hazard to ecosystem (b) Thermal energy
(c) Nuclear energy
(d) Wind energy

Answer:
(1) Pollutants – Soot particles
(2) Hazard to ecosystem – Nuclear energy

Question 3.

Type of energy Problem
(1) Nuclear energy (a) Rehabilitation of displaced people
(2) Natural gas (b) Rainy season and darkness
(c) Limited reserves
(d) Disposal of wastes

Answer:
(1) Nuclear energy – Disposal of wastes
(2) Natural gas – Limited reserves

Question 4.

Type of energy Problem
(1) Solar energy (a) Rehabilitation of displaced people
(2) Hydroelectric  energy (b) Rainy season and darkness
(c) Limited reserves
(d) Disposal of wastes

Answer:
(1) Solar energy – Rainy season and darkness
(2) Hydroelectric energy – Rehabilitation of displaced people

Find the odd one out:

Question 1.
Kudankulam, Tarapur, Ravatabhata, Anjanvel
Answer:
Anjanvel. (All others are places having nuclear power plants.)

Question 2.
Samaralkota, Kudankulam, Bavanaa, Kondapalli
Answer:
Kudankulam. (All others are places having power plants based on natural gas.)

Question 3.
Tehari, Koyana, Srishailam, Tarapur
Answer:
Tarapur. (All others are places having hydroelectric projects.)

Question 4.
Edible oil, crude oil, LPG, CNG
Answer:
Edible oil. (All others are fossil fuels.)

Question 5.
Hydroelectric energy, Solar energy, Nuclear energy, Wind energy
Answer:
Nuclear energy. (All others are eco-friendly green energy types.)

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Explain with diagram step-by-step energy conversion in:

Question 1.
Power plant based on natural gas.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 10a
In a power plant based on natural gas, there are three main sections of the plant. There is combustion chamber with compressor in which the steam under pressure is introduced. The natural gas burns in the presence of air in this combustion chamber. This results in a production of a gas which is at very high temperature and pressure. This generated gas from the chamber runs the turbine. The kinetic energy of the turbine drives the generator. The generator produces electrical energy.

Question 2.
Power plant based on wind energy.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 11a
Wind energy is used for moving turbines. The wind with specific speed is used to rotate the large fins of wind turbine. The kinetic energy in these fins is transferred to generator which then produces electrical energy.

Explain the following questions in detail:

Question 1.
What are the advantages of hydroelectric power generation? (March 2019)
Answer:

  1. Hydroelectric energy does not cause pollution.
  2. Generation of hydroelectric energy does not involve burning of fossil fuel.
  3. If sufficient water storage is available then electricity generation can be done as per requirement.
  4. Rainwater can replenish the water storage and power generation can thus be done uninterrupted.

Question 2.
How is nuclear fission reaction carried out in nuclear power plants?
Answer:

  • In nuclear power plants neutrons are bombarded on atom of Uranium – 235.
  • This causes conversion of Uranium – 235 into its isotope U – 236.
  • U-236 is very unstable and thus forms atoms of Barium and Krypton by nuclear fission. This forms 3 neutrons and 200 MeV energy.
  • In a similar way three more Uranium – 235 atoms are subjected to nuclear fission which then releases energy.
  • The neutrons released are again used for further nuclear fission reactions. In this way nuclear fission reactions are carried out in controlled manner in nuclear power plants.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 3.
Draw schematic of power plant based on natural gas and answer the following questions: (July 2019)
(a) At which place natural gas power plant is situated in Maharashtra?
(b) How is pollution reduced in natural gas based power plant?
(c) Give two examples of eco-friendly electricity process.
Answer:
(a) Natural gas power plant is situated at Anjanvel in Maharashtra.
(b) Natural gas does not contain sulfur. Burning of such natural gas does not produce pollution.
(c) Solar energy and wind energy are two examples of eco-friendly energy.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 30

Complete the paragraph by choosing the appropriate words given in the brackets:

Question 1.
(marginal, array, cell, panel, string, current, power station, potential difference).
Many solar panels are connected in series and in parallel to generate required ………… and ……… Solar …………. is the basic unit in solar electric plant. Many solar cells come together to form a solar …………… Many solar panels connected in series form a solar ………., and many solar strings connected in parallel form a solar …………. As we can obtain as much electrical power as needed, they are used in applications which need ……….. power (e.g. calculators that run on solar energy) to ……….. of MW capacity.
Answer:
Many solar panels are connected in series and in parallel to generate required current and potential difference. Solar cell is the basic unit in solar electric plant. Many solar cells come together to form a solar panel. Many solar panels connected in series form a solar string, and many solar strings connected in parallel form a solar array. As we can obtain as much electrical power as needed, they are used in applications which need marginal power (e.g. calculators that run on solar energy) to power station of MW capacity.

Read the paragraph and answer the questions given below:

1. Renewable energy is, energy produced from sources that do not deplete or can be replenished within a human’s life time. The most common examples include wind, solar, geothermal, biorhass, and hydroelectric power. This is in contrast to non-renewable sources such as fossil fuels. Most renewable energy is derived directly or indirectly from the sun. Sunlight can be captured directly using solar technologies. The sun’s heat drives winds, whose energy is captured with turbines. Plants also rely on the sun to grow and their stored energy can be utilized for bioenergy. Not all renewable energy sources rely on the sun. For example, geothermal energy utilizes the Earth’s internal heat, tidal energy relies on the gravitational pull of the moon, and hydroelectric power relies on the flow of water.

Questions and Answers :
Question1.
What is renewable energy?
Answer:
Renewable energy is energy that is produced from sources which will not get exhausted within a human’s life time.

Question 2.
Give the examples of renewable energy.
Answer:
Wind, solar, geothermal, biomass and hydroelectric power are some examples of renewable energy.

Question 3.
Why will energy from fossil fuel be over soon?
Answer:
Fossil fuels are exhaustible in their amount. We have been using these extensively in the past 100 years and hence it may get over soon. It is a non-renewable resource.

Question 4.
Name the renewable sources of energy which are not dependent on sun. What are they dependent upon?
Answer:
Geothermal energy, tidal energy and hydroelectric power are renewable energy resources which are not dependent on sun. Geothermal energy utilizes the Earth’s internal heat, tidal energy relies on the gravitational pull of the moon, and hydropower relies on the flow of water.

Question 5.
Which type of energy do we mostly use in India?
Answer:
The most used energy resource is coal, i.e. fossil fuel based energy followed by hydroelectric energy.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

2. Read the information given below and solve the questions based on it.
Electric energy is produced in various ways like hydroelectric, wind power, solar energy, bio-fuel, etc. These energy sources are inexhaustible, sustainable. Besides, it does not cause any environmental problem.

Questions and Answers:
Question 1.
Above information is about which type of energy?
Answer:
From the above information, we understand about green energy.

Question 2.
Whether the fossil fuel is an example of this energy?
Answer:
Fossil fuels are not green energy.

Question3.
Draw the flow chart of production of electric energy.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 12

Diagram based questions:

Question 1.
Observe the connections of cells shown in the following images.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 13
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 14
(i) Which connection will give maximum potential difference?
Answer:
The solar cells shown in the diagram 5.19 (a) are connected in series. This gives maximum potential difference.

(ii) Give one advantage and one disadvantage of this energy.
Answer:
Advantage of Solar energy: Solar energy is eco-friendly which does not create pollution. It is boundless source.
Disadvantage of solar energy: Solar energy is available only when sun is in the sky. Therefore, it has to be stored in batteries.

Question 2.
Answer the following questions:
(a) Write the name of the device shown in the above diagram.
Answer:
Steam turbine is the device shown in the above diagram.

(b) Write briefly the work of this device.
Answer:
Turbine is a device with the blades. When the flow of liquid or gases is directed on the blades of the turbine, they rotate. The rotation produces kinetic energy. This turbine is then used to rotate the magnet in the electric generator. For this purpose, turbines are connected with the generators. The magnets rotate and produce electric energy by electromagnetic induction. The turbines working on steam are used in large commercial power generation plants.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 3.
Label the given diagram of Electromagnetic induction.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 15

Question 4.
Answer the questions with the help of picture.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 16
(a) Which type of energy is produced?
(b) This power plant is based on which energy source.
(c) Is this power plant eco-friendly? How?
Answer:
(a) In the picture, it is shown that using wind energy electricity is produced.
(b) The power plant shown here is based on kinetic energy of wind which is converted to electric energy by utilizing kinetic energy from rotating turbines.
(c) This power plant is eco-friendly because it does not cause pollution. Wind energy is green energy which is non-exhaustible and perpetual.

Question 5.
Observe the figure and answer the questions given below.
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 17
Answer:
(a) Name the reaction.
Answer:
The reaction shown in nuclear fission or chain reaction.

(b) Where is this reaction used?
Answer:
This reaction is used in nuclear power plants where electricity is generated.

(c) Which element is used in it?
Answer:
Uranium-235 is used in the nuclear fission reactions.

(OR)

Identify the process shown in figure and name it. (March 2019)
Answer:
The above figure shows nuclear fission chain reaction of Uranium – 236.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 6.
Observe the diagram and answer the questions : (March 2019)
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 18
(a) Which energy is generated from the power plant?
Answer:
The diagram shows electricity generated from natural gas.

(b) State its source.
Answer:
The energy is generated from natural gas.

(c) Which is more eco-friendly – Power generation from coal or Power generation from natural gas? Why?
Answer:
Power generation from natural gas is more eco-friendly. Natural gas does not contain sulfur and hence its burning does not cause major pollution by forming sulphur dioxide. The efficiency of power generation by natural gas is also high.

Question 7.
Write the names of apparatus that is used in thermal power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 19

Question 8.
Label correctly the diagram of Nuclear power plant.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 20

Question 9.
Label correctly the diagram of power plant baded on natural gas.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 21

Question 10.
Label correctly the structures seen in Windmill.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 22

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Question 11.
Sketch two ways in which solar cells can be connected. Also draw the diagrams to show the arrangement of solar cells to form solar? panel and solar array.
a. Solar cells in series.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 23

(b) Solar cells in parallel.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 24

(c) Solar panel.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 25

(d) Solar array.
Answer:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 26

Question 12.
Observe the figure given below and answer the given questions:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 27
(a) Identify the type of energy generation process shown in this picture.
(b) Name any four equipments which use this type of energy. (Board’s Model Activity Sheet)
Answer:
(a) In this figure solar energy is converted into electrical energy. Solar energy is also called clean energy.
(b) Solar energy is used in following equipment:

  • Solar cooker
  • Solar heater
  • Calculator
  • Solar Photovoltaic cell.

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Activity based questions.

Question 1.
Make a table: (Text Book Page No. 47)
Make a table based on forms of energy and corresponding devices.
Answer:

Forms of energy Devices based on this type of energy
(1) Electric Electric iron, Geyser, Heater, Oven, Refrigerator, Fans, Lights, Elevator.
(2) Mechanical Sewing machine, Car, Bicycle, Different machines.
(3) Thermal Chulha, Furnace, Steam engine
(4) Solar Solar cooker, Solar heater.

Question 2.
Let’s Think: (Text Book Page No. 52)
Which electricity generation process is eco-friendly and which not?
Answer:
Electricity generated through solar energy and wind energy are truly eco-friendly. Though it is said that hydroelectricity is non-polluting and eco-friendly, it is not true. Hydroelectric project cause destruction of biodiversity and displacement of the local people. Thermal energy, nuclear energy and energy obtained through natural gas are not at all eco-friendly.

Question 3.
Find out: (Text Book Page No. 55)
What is lake tapping? Why it takes place?
Answer:
A lake tap involves excavating a tunnel at the bottom of the lake. Dynamites are planted therein and blasted carefully. The water flows with greater force through the tunnel after such blasting is done. This increased flow of water is then driven to the hydroelectric power generation plant for increased electricity production. This technique is done to establish waterways for hydropower, for making drinking water available, for irrigation water purposes and also for the landing of oil and gas pipes from offshore fields.

Question 4.
Get information: (Text Book Page No. 56)
Get information about major wind-power stations in India and their capacity. Make a table of their location, state and their power generation capacity in MW.
Answer:

Location State Power generation capacity in MW
Muppandal, Kanyakumari Tamil Nadu 7,684.31
Dhule, Satara, Sangli, Dhalgaon Maharashtra 4,664.08
Bhuj Gujarat 4,227.31
Dangiri Wind Farm Jaisalmer Wind Park Rajasthan 4,123.35
Jogmatti BSES Karnataka 3,082.45
Bhopal at Nagda Hills near Dewas Madhya Pradesh 2,288.60
Tirumala hills Andhra Pradesh 1,866.35
Telangana 98.70
Kanjikode in Palakkad Kerala 43.50
Others 4.30
Total 28, 082.95

Question 5.
Find out: (Text Book Page No. 58)
Gather information about major solar photovoltaic power generating plants and their capacity in India.
Answer:
List of solar power stations:
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 28a
Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy 29

Maharashtra Board Class 10 Science Solutions Part 2 Chapter 5 Towards Green Energy

Project:

Project 1.
Let’s Discuss: (Text Book Page No. 47)
Make a list of the work that we do in our day-to-day life using energy. Which forms of energy do we use to do this work? Discuss with your friends.

Project 2.
Compare: (Text Book Page No. 51)
Observe the schematic of thermal power plant and the nuclear power plant. Discuss what are the similarities and differences between the two.

Project 3.
Use of ICT: (Text Book Page No. 49)
Prepare a presentation about thermal power plant using computerized presentation, animation, video, pictures, etc. Send it to others and upload on YouTube.

Project 4.
Internet is my friend: (Text Book Page No. 51)
Complete the following table for some important nuclear power plants in India.

Maharashtra Board Class 10 History Solutions Chapter 3 Applied History

Balbharti Maharashtra State Board Class 10 History Solutions Chapter 3 Applied History Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 History Solutions Chapter 3 Applied History

Question 1.
(a) Choose the correct option from the given options and complete the statement.
(1) The earliest museum in the world was discovered in the excavations at the city of ……………..………. .
(a) Delhi
(b) Harappa
(c) Ur
(d) Kolkata
Answer:
(c) Ur

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(2) The National Archives of India is in ……………..………. .
(a) New Delhi
(b) Kolkata
(c) Mumbai
(d) Chennai
Answer:
(a) New Delhi

(b) Identify and write the wrong pair in the following set.
(1) Kootiyattam- Sanskrit theatre, Kerala
(2) Ramman- Dance form in West Bengal
(3) Ramlila- Traditional Performance of the Ramayana in Uttar Pradesh
(4) Kalbelia- Folk songs and dances of Rajasthan.
Answer:
(2) Wrong pair: Ramman – Dance form in West Bengal

Question 2.
Explain the following concepts :
(1) Applied History
Answer:

  •  When one subject is applied to different fields and conclusions are derived, it is known as application of that subject.
  • When objectives of History are applied to other subjects; new conclusions are obtained.
  • This is known as ‘Applied History’. It is also known as Public History.
  • We get insights of the events that took place in the past through history.
  • Applied History is concerned with application of knowledge of history to provide guidance in finding solutions.
  • Contemporary social issues and include them in social planning.
  • Thus Applied History’ is a field of study concerned with the application of history for the benefit of people in contemporary and future times.

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(2) Archives
Answer:

  •  A place where historical documents are preserved is called an ‘Archive’.
  • Many old documents, official records, old films, records of treaties are kept at this place.
  • We get references of original documents because of Archives.
  • A study of historical events of a particular period is possible.
  • We come to know about language, script used in a particular period/era.
  • The National Archives of India is in Delhi. It is the largest Archive in Asia.
  • Every state in India maintains its archives independently.

Question 3.
Complete the following concept chart.
Maharashtra Board Class 10 History Solutions Chapter 3 Applied History 1
Answer:
Maharashtra Board Class 10 History Solutions Chapter 3 Applied History 2

Question 4.
Explain the following statements with reasons.
(1) It is essential to study the history of technology.
Answer:

  • Man acquired different skills at various stages in his evolution process.
  • From mxking stone tools to developing new techniques of agriculture production, he learnt many skills and entered in the age of science.
  • Agricultural production, commodity production, architecture, engineering, etc.
  • underwent several changes. Production increased manifold bcause of technology.
  • Hence, it is necessary to know the history of technology in order to understand the development in mechanisation and mutual dependence between science and technology.

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(2) The list of world Heritage Sites is announced by UNESCO.
Answer:

  • Cultural heritage is a form of human creation. It is a cultural treasure that has been inherited from our ancestors.
  • We feel connected to it and hence preserve and conserve it.
  • This heritage is our asset that binds us to our past and hence we should preserve it for the benefit of future generations.
  • In order to save this heritage, UNESCO has announced some directives.
  • On the basis of those directives, list of sites, and traditions are declared as ‘World Heritage’.
  • It has helped us in promoting these sites as tourist places.

Question 5.
Write elaborate answers
(1) How is the method of history useful in the research of
(a) Science
(b) Arts
(c) Management Studies
Answer:
Each field has its own history of building’ knowledge. The direction of development in every field is dependent on the stcrte of available knowledge in that field. The method of history has proved valuable in the research of various fields.

(a) Science: Scientific discoveries /invenfions take place because of humdn efforts to satisfy needs and curiosity. These efforts are based on scientific knowledge that already existed. Knowledge of the history of science helps in understanding the reasons, chronology and factors that facilitated a scientific invention/discovery. Knowledge of history of science helps in every field.

(b) Art: Development of any style of art lies in their expression through intellectual, emotional and cultural traditions. While studying any art form we have to understand the history of its traditions, the key to the expressions in it, and emotional temperament of the artist. Prevalent art style in the given period can be understood with the help of cultural history.

(c) Management Studies: All components like means of production, human resources, processes of production, the chain of market and sales, etc. are interrelated and require management.

Different social and economic institutions are involved in these industrial and commercial processes. Knowledge of similar functional systems of the past is essential to bring about changes, improvements and make the management smooth and efficient. It becomes necessary to know its history.

(2) How can we correlate applied history with our present?
Answer:
History gives us knowledge about the events that happened in past. Applied history helps us in understanding how the knowledge can be put to use for our benefit as well as for the benefit of future generations. Applied history is correlated with our present in the following ways:

(1) Knowledge of our past helps us in deciding our course of action in present. The.heritage of our ancestors exists in tangible and intangible- form. We have the curiosity to know more about our part because they represent the creative thoughts and traditions of that period.

(2) With the help of applied history we can not only come to know about the heritage but also conserve and preserve it.

(3) We can be better equipped to face social challenges in the present because the knowledge of history can provide guidance in finding solutions to contemporary social issues and incorporate them in the ’social planning.

(4) The direction of future development is S decided when we rightly analyse our present with our knowledge of the past acquired through applied history.

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(3) Suggest at least 10 solutions for the preservation of the sources of history.
Answer:
There are three types of sources which are used to study history. They are written, oral and material sources. These sources are preserved in different ways. According to me, the following measures should be taken to preserve the sources:

  1. A regular repair and maintenance work of forts, memorials and palaces should be carried out. Timely action should be taken to avoid vandalisation of historic and public places.
  2. Historical coins, weapons should be handled with precaution and utmost care. They should be kept in a safe place to avoid the possibility of theft.
  3. Social awareness regarding the conservation and preservation of our cultural and natural heritage should be created. People will develop affinity towards the heritage and feel connected.
  4. Variations in environment and climatic conditions pose a threat to monumental structures. Humidity, dampness, moulds and pollution cause irrevocable damage to manuscripts, rare artefacts, statues, coins, etc.
  5. To protect them, special arrangements have to be made like rooms with controlled temperature or sanitising and detoxing the rooms.
  6. Oral literature like owis, folk songs should be compiled and written down.
  7. Public training programmes should be conducted for understanding the importance of history.
  8. Experts from different fields should be involved in preservation and conservation projects. S Stringent laws should be enacted to protect the monuments.
  9. Above all, people’s participation in conservation and preservation process is paramount.

(4) What objectives can be fulfilled through the heritage projects.
Answer:
The following objectives can be fulfilled through the heritage projects:

  • Preservation, conservation and development of heritage sites can be done without any change in its original state/ structure.
  • The local social structure and psychology of the local people, challenges they face in the present situations and their expectations can be surveyed.
  • In the process of completing the project of conservation, preservation and development of a heritage site due precaution should be taken to avoid hurting the sentiments of the local people.
  • Participation of the local people in the proposed project can be facilitated.
  • In order to engage and employ the local skills in a creative way and create better opportunities of livelihood, it becomes easier to design systematic plans.

Project
Show the Indian heritage sites on the map of India.
Answer:

Memory Map
Maharashtra Board Class 10 History Solutions Chapter 3 Applied History 3

Question 6.
Complete the sentences by choosing the correct option:
(a) ……………………… Institute of Art, Design and Technology at Bengaluru has an independent department named ‘Centre for Public History’.
(a) Pitch
(b) INTACH
(c) Shrushti
(d) AIMS
Answer:
(c) Shrushti

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(b) ……………………… is the mother of science and all other branches of knowledge.
(a) Philosophy
(b) History
(c) Technology
(d) Mythology
Answer:
(a) Philosophy

(c) In order to preserve and conserve cultural and natural heritage for the benefit of future generations ……………………… has given directives.
(a) UNICEF
(b) UN
(c) UNESCO
(d) INTACH
Answer:
(c) UNESCO

(d) The ……………………… in the Satara District of the Western Ghats is included in the list of World Natural Heritage.
(a) Balaghat Mountain
(b) Melghat
(c) Masai Plateau
(d) Kas Plateau
Answer:
(d) Kas Plateau

(e) The main office of the National Film Archives of India is situated at ……………………… .
(a) Delhi
(b) Mumbai
(c) Pune
(d) Nagpur
Answer:
(c) Pune

(f) Excavations at the city of ‘Ur’ in Mesopotamia were conducted by ……………………… .
(a) Nathaniel Wallick
(b) Ennigaldi
(c) Leonard Wolley
(d) Sir John Marshall
Answer:
(c) Leonard Wolley.

Question 7.
Identify the wrong pair in the following, and write it:

(1) Kootiyattam (a) Sanskrit theatre, Kerala
(2) Ramman (b) Dance form in West Bengal
(3) Ramlila (c) Traditional performance of Ramayana in Uttar Pradesh
(4) Kalbelia (d) Folk songs and dances of Rajasthan

Answer:
Wrong pair: Ramman – Dance form in West Bengal

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(2)

Cultural Heritage Place
(1) Red Fort (a) Udaipur
(2) Jantar Mantar (b) Jaipur
(3) Brihadeeshvara Temple (c) Thanjavur
(4) Capital Complex (d) Chandigarh

Answer:
Wrong pair: Red Fort – Udaipur

Question 8.
Complete the graphical presentation:
(1)

Answer:

(2)

Answer:

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(3)

Answer:

Question 9.
Explain the following concept:
(a) Natural Heritage:
Answer:

  • In Cultural heritage what we inherit is man-made but Natural heritage is bestowed upon us by Nature.
  • The concept of natural heritage gives importance to the thought of biodiversity.
  • The following things are included in natural heritage:
    • Fauna
    • Flora
    • Ecology required for sustaining animal and plant life, flora and fauna of a particular region.
    • Geomorphic characteristics, sanctuaries, mountain ranges, river valleys, lakes and dams are all part of natural heritage.

Question 10.
Write short notes:
(a) Importance of Applied History:
Answer:

  • Applied History is a field of study which helps us understand our ancient heritage.
  • We realise the importance of preserving our heritage.
  • The process of preservation and conservation of tangible and intangible heritage sites generate employment opportunities.
  • The study of Applied History help us to get a better understanding of the present and can provide guidance for the future.

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(b) Preservation of cultural heritage of India:
Answer:
Conservation and preservation of India’s cultural heritage is done at different levels:

  1. Archaeological Survey of India (ASI) is an important body of Indian government which mainly does the work of conservation and preservation along with State Departments of Archaeology.
  2. INTACH (Indiqn National Trust for Art and Cultural Heritage) is actively working in this field.
  3. Local self-government bodies, social institutions all over India, people who have love for history are seen involved in the work of preservation of our heritage.
  4. The work of conservation and preservation of cultural and natural heritage requires participation of experts from various fields as well as local people.

(c) Indian Museum:
Answer:

  • The Indian Museum at Kolkata is the oldest museum and one of the important heritage sites of India.
  • Nathaniel Wallich, a Danish botanist, founded the museum in 1814 C.E. He was also its first curator.
  • The museum has three main departments Arts, Archaeology and Anthropology.
  • Other affiliated departments are conservation, publication, photography, exhibition- presentation, model-making, training, library and security.

(d) National Film Archives:
Answer:

  • The main office of the National Rim Archives of India is located at Pune and was established in 1964.
  • It functions as the Media Unit of the Ministry of Information and Broadcasting of the Indian Government.
  • It was established with the view of three objectives. They are:
    • To search and obtain the rare Indian films and preserve our heritage for the benefit of future generations.
    • To categorise important aspects of films, create documentation and do research.
    • To establish a centre for dissemination of ‘films culture’.

Question 11.
Explain the following statements giving reasons:
(a) It is essential to study the history of philosophy.
Answer:

  • With the passage of time, many schools of thought came into existence with different ideologies.
  • These ideologies had their effect on people and society as a whole.
  • What led to the origin of various ideologies, the intellectual tradition giving rise to those ideologies and their historical development needs to be researched.
  • To know about the progress of those ideologies, their development and expansion or downfall we need to have knowledge of the history of philosophy.

(b) It is important to know the history of Industry and Commerce.
Answer:

  • With the growth of industry and trade, the field of mutual social transactions also expand.
  • It promoted development of the network of cultural interactions.
  • As the nature of market and commerce changed sp did the nature of human relationships and the social organisation.
  • This series of changes is influenced by prevalent culture, social structure and economic system.
  • All this is an integral part of the industrial and commercial management.

Therefore, it becomes important to know the history of Industry and Commerce.

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(c) It is necessary to preserve and conserve our natural and cultural heritage.
Answer:

  • The visible and invisible relics of the past exist in the present.
  • They represent the thought and traditions of our ancestors.
  • We nurture some kind of curiosity and attraction towards them.
  • The history of our heritage links us with our origin.

Therefore, it becomes necessary to preserve and conserve for our future benefit as well as for the benefit of future generations.

Question 12.
Answer the following in 25-30 words:
(a) Explain the concept of ‘Public History’.
Answer:

  • Public History also known as ‘Applied History’ is a field of study concerned with the application of history for the benefit of people in the contemporary and future life.
  • It provides solutions to the contemporary social issues and incorporate them in social planning.
  • It helps to overcome misgivings about history and makes history meaningful in everyday life connecting people to history.

(b) What is the role of UNESCO in the context of heritage sites?
Answer:
The role of UNESCO, the global agency of United Nations, is as follows:

  • For preservation and conservation of natural and cultural heritage which is in a ruined state because of years of neglect, UNESCO has announced some directive principles.
  • The organisation announces list which includes the names of sites and traditions which meet the criteria.
  • On the basis of those directives it declares a list of site and traditions of world heritage.

(c) Find out the cultural heritage sites located in Maharashtra and write their names.
Answer:
The following places in Maharashtra are included in the list of cultural heritage sites declared by UNESCO:

  • Ajanta Caves
  • Verul (Ellora) Caves and Kailas Temple
  • Elephanta Caves
  • Chhatrapati Shivaji Maharaj Terminus, Mumbai.

Many forts like the Raigad, Devgiri, Janjira and Sindhudurg are part of the cultural heritage Of Maharashtra though they are not included in the list.

(d) What are the misgivings about history among the people?
Answer:
There are many misgivings about history among the people:

  • History is a subject only for historians and students who pursue higher studies in history.
  • It has no relevance or applicability in our day-to-day life.
  • History is nothing but piece of information about the battles fought by kings and politics.
  • History cannot be connected to an economically productive field.

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(e) Where can we trace the roots of philosophy?
Answer:

  • Ancient people all over the world tried to speculate the relationship between the universe and human existence.
  • This gave rise to various myths about origin of this world, universal order, human life.
  • Imagination about gods and goddesses, rituals to please them and philosophical explanation of rituals.
  • Thus, the roots of philosophy can be traced in these speculations and ponderings of mythological stories.

(f) What are the opportunities created by affiliated professional fields of Applied History?
Answer:
Affiliated fields of Applied History which include museums, archives, tourism ‘and hospitality, entertainment, mass media, etc. needs the following professionals:

  • Experts like Historians, Archaeologists, Sociologists, legal experts, etc. are assigned new projects.
  • Officers like Managers, Secretaries and Directors are taken into service.
  • Technicians like engineers, architects, skilled photographers, laboratory assistants, archive management professionals are hired.
  • Tourist guides, boarding and lodging, food services, entertainment and related business get a wide scope for flourishing.

(g) What are the main objectives of ‘National Film Archives’?
Answer:
The main objectives of ‘National Rim Archives’ are as follows:

  • To research and obtain the rare Indian films.
  • To preserve such films for the benefit of future generations.
  • To categorise the important aspects of films to create documentation, catalogue them -and carry out research in the field.
  • To establish a centre for dissemination of films culture.

(h) How can we increase people’s participation in the field of applied history?
Answer:

  • The field of applied history includes museums, archives, collection and maintenance of artefacts.
  • Tourism creates interest in history.
  • Awareness is created among people to preserve ancient monuments and sites in their vicinity.
  • This realisation increases people’s participation in the field of applied history.

(i) Write information on World Heritage sites in Maharashtra.
Answer:
Maharashtra is endowed with many natural and historical sites. Among them the prominent ones are Ajanta caves, Verul caves, Gharapuri/Elephanta caves and Chhatrapati Shivaji Maharaj Terminus Railway Station at Mumbai.

Chhatrapati Shivaji Maharaj Terminus is finest example of Indo-Gothic architecture emerged in India during British rule. It stands witness to the British empire, freedom struggle and transformation of our country into a modern nation. It needs conservation as it a architectural marvel which stood the test of times recently when there was a terror attack on Mumbai city.

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(j) What is the role of Applied History?
Answer:
The role of Applied History is as follows:

  • In public training programmes for understanding the importance of history.
  • Using knowledge of history, to create awareness regarding the conservation and preservation of our cultural and natural heritage.
  • To facilitate participation of people in various projects and programmes related to applied history even in their capacity as tourists.
  • To enhance professional skills of people and to develop industrial commercial field.
  • To carry out preparatory work and plan for effective implementation of heritage projects.

Question 13.
Read the given passage carefully and answer the questions given below:
(a) Who built this museum?
Answer:
This museum was built by Ennigaldi, the princess of Mesopotamia.

(b) Where was the earliest museum in the world discovered? Who discovered it?
Answer:
The earliest museum in the world was discovered in the city of ‘Ur’ in Mesopotamia. It was discovered by Leonard Woolley.

(c) What is a noteworthy feature of this museum?
Answer:
A noteworthy feature of this museum is the clay tablets inscribed with the descriptions of exhibited artefacts.

Question 14.
Write elaborate answers: OR Answer in detail:
(a) Who does the work of conservation and preservation of the cultural heritage of Indian?
Answer:
The preservation and conservation of Indian cultural heritage is done at various stages:

  • The work of preservation and conservation primarily falls under jurisdiction of the Archaeological Survey of India.
  • In addition, INTACH (Indian National Trust for Art and Cultural Heritage) is also actively working .in this field.
  • Experts from various fields and local people too participate in the work of conservation and preservation of natural and cultural heritage.

Question 15.
Observe the picture and write information about it:
Maharashtra Board Class 10 History Solutions Chapter 3 Applied History 5
Answer:

  • This picture is of the largest Indian rock-cut ancient Hindu temple located in the Ellora cave.
  • This famous rock-cut Shiva temple was constructed during the reign of Rashtrakuta king Krishna I.
  • It is regarded as a marvel of architecture.
  • It bears an eloquent testimony to the high level skill attained by India in the arts of sculpture and architecture under the Rashtrakuta patronage.
  • A sculpture of an elephant and a pillar can be seen in the picture.
  • The famous rock cut temple in Maharashtra was declared as a world heritage site by UNESCO in 1983.

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Brain Teaser

Hints:
Across:
(1) Queen’s stepwell at Patan
(2) Dancing, singing, drumming ritual of Manipur
(3) Religious festival and ritual theatre of Garhwal
(4) Archaeological site of Nalanda University
(5) World famous monument at Agra
(6) Famous city built by emperor Akbar marking his victory
(7) Sanskrit Theatre: Kerala

Down:
(1) The traditional performance of Ramayan in Uttar Pradesh
(2) Ellora caves also known as caves
(3) Observatory at Jaipur, Rajasthan
(4) Group of monuments at (place) in Karnataka
(5) Archaeological park at Pavagadh
(6) Folk songs and dances of Rajasthan
(7) A ritual theatre of Kerala
Maharashtra Board Class 10 History Solutions Chapter 3 Applied History 6

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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

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)]
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 2

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 3
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 4
P1, P2, P3: Incident rays of light,
Q1, Q2, Q3: Refracted rays of light, O: Optical centre
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 5
F1, F2: Principal foci of the lens, f: Focal length of the lens

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 6

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

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 8
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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 9

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 = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 10
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 11
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 = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 12
∴ 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)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 13
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 14

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)
\(\frac{h_{2}}{h_{1}}\) = \(\frac{v}{u}\)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 15
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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 16
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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

Question 8.
Draw neat labelled diagrams: Types of lenses.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 17
[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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 18
(2) When the incident ray passes through the principal focus, the refracted ray is parallel to the principal axis.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 19
(3) When the incident ray passes through the optical centre of the lens, it passes without changing its direction.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 20

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 21

2.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 22

3.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 23

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 24
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 25
In this case, the image is formed between F2 and 2F2. It is real, inverted and diminished.

(3) Object at 2F1:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 26
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 27
In this case, the image is formed beyond 2F2. It is real, inverted and magnified (enlarged).

(5) Object at focus F1:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 28
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 29
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.]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

Question 13.
Observe the following figure and complete the table: (March 2019)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 30
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 31

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

Question 18.
Draw a ray diagram to show image formation by a concave lens.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 32
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.]
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 33

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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 34
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 35

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.]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 36
where f is the focal length of the lens:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 37
(2) From eq. (2), we have
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 38

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 39
(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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 40

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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).

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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 \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 41
[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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 42

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 43
(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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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)]
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 44

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 45
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.]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 46
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 47
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.]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

Question 69.
Observe the following figures and complete the table. (Practice Activity Sheet – 1)
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 48

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 49
= -2 m (Concave lens ∴ Minus sign)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 50
(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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 51

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 52
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 53
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 54
(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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 55
(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).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 56
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?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 57
(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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 58

(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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 59
(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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 60

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 61
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 = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 62
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 63
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 = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 64
The focal length of the lens = 13.33 cm [approximately]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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=?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 65
∴ 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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 66
The focal length of the lens 20 cm.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 67
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 68
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 = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 69
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 70
The height of the image, h2 = -5 cm
Thus, it is numerically the same as the height of the object.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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 = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 71
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 72
∴ 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 = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 73
The image Is formed at 24 cm from the lens.
It is on the same side as the object.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 74
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) = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 75
∴ 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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 76
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) = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 77
The focal length or the combination of the lenses = 0.125 m = 12.5 cm.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses 78
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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses

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.

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

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.)

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

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 gna = sin i/sin r and ang = sin i1/ sin e.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 1
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 2
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) \(\frac{2}{3}\)

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

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 3
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 4
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 5
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.

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

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.

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

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

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

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

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

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

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.)

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

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.

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

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 7
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. ]

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

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 8
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.

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

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,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 9
(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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 10
Fig. 6.6: A ray of light travelling from a rarer medium to a denser medium (Schematic diagram)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 11
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.]

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

Question 15.
Observe the following figure and write accurate conclusion regarding refraction of light. (Practice Activity Sheet – 2)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 12
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).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 13
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 14
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 15
(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.

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

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 16
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 17

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 18
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.

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

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 19
(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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 20
(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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 21
The advanced sunrise and delayed sunset increases the duration of day by four minutes.

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

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]
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 22
(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.]
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 23

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

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).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 24
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 anw = \(\frac{\sin i}{\sin r}\) < 1. Here, anw is the refractive index of sin r air with respect to water.

As anw 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 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 anw = \(\frac{\sin i}{\sin r}\) < 1. Here, anw is the refractive index of sin r air with respect to water. As anw 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.

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

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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 25

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 26

Question 43.
Observe the given figure and answer the following questions. (Practice Activity Sheet – 3)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 27
(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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 28

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 29
(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.

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

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 30
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 31
(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).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 32
(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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 33

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

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 = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 34
The absolute refractive index of the medium = 1.25.

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

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 = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 35
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: uw = 2.2 × 108 m/s,
vg= 2 × 108 m/s, wng = ?, gnw = ?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 36
The refractive index of water with respect to glass = 0.909 (approximately).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 6 Refraction of Light 37
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)

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

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:
\(\frac{8}{9}\)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 1
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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

Question 5.
Explain the following temperature vs time graph:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 2
(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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 3
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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 4
Thus, c1 > c2
The specific heat of A is more than that of B and
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 5

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, cw = 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 cw 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 cw 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]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 6
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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

(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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 7
(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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 8
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 9
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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 10
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 11
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.]

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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,
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 12
= (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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat 13
This is the required temperature.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat

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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 1

b. Ethene.
Answer:
Molecular formula: H2C = CH2
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 2

c. Methanol.
Answer:
Molecular formula: H3C – OH
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 3

d. Water.
Answer:
Molecular formula: H2O
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 4

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 5

b. C4H10 Butane:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 6

c. C3H4 Propyne:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 7

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 8

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 9

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 (=).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 10

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 11

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 12

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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 13
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 14
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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 15
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 16

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,…..
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 17
(2) ‘one’ stands for functional group
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 18
ketone. The number assigned for the ketone group is 2. Show the ketone group at C2.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 19
(3) Now satisfy the valencies of each carbon atom
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 20

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,….
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 21
(2) Chloro (Halo) is the prefix and the number assigned for prefix (chloro) is 2. Show the chloro atom at C2
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 22
(3) Now satisfy the valencies of each carbon atom
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 23

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 24
(2) ‘-ol’ stands for (-OH) hydroxyl group. The number assigned for the hydroxyl group is 2. Show the -OH group at C2.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 25
(3) Now satisfy the valencies of each carbon atom
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 26

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 27

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,….
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 28
‘-oic acid’ stands for functional group -COOH. Assign the number 1 to carbon in the functional group -COOH.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 29
Now satisfy the valencies of each carbon atom
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 30

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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…….
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 31
(2) Bromo (Halo) is the prefix and the number assigned for prefix (bromo) is 1, show the bromine atom at C1.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 32
(3) Now satisfy the valencies of each carbon atom
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 33

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 34

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,….
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 35
(2) ‘one’ stands for functional group
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 36
ketone. The number assigned for the ketone group is 2. Show the ketone group at C2.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 37
(3) Now satisfy the valencies of each carbon atom
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 38

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 39

c. Give any four functional groups containing oxygen as the heteroatom in it. write name and structural formula of one example each.
Answer:

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 40

d. Give names of three functional groups containing three different heteroatoms. Write name and structural formula of one example each.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 41

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 42

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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 43

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

Question 2.
Molecular formula of chlorine is Cl2. Draw electron-dot and line structure of a chlorine molecule.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 44

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 45

Question 4.
The molecular formula of ammonia is NH3. Draw electron-dot and line structures for ammonia molecule.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 46

Question 5.
The molecular formula of carbon dioxide is CO2. Draw the electron-dot structure (without showing circle) and line structure for CO2.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 47

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 48
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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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 CnH2n + 2. The value of ‘n’ for the first member of homologous series is 1.
CnH2n+2 = C1H2 × 1 + 2 = CH4

Question 4.
The general molecular formula for the homologous series of alkynes is CnH2n – 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 CnH2n – 2
n = 2 C2H2 × 2 – 2 = C2H2 Ethyne
n = 3 C3H2 × 3 – 2 = C3H4 Propyne
n = 4 C4H2 × 4 – 2 = C4H6 Butyne

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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 CnH2n + 2. Write down the molecular formula of the 8th and 12th member using this.
Answer:
General formula of alkanes is CnH2n + 2
n = 8 C8H2 × 8 + 2 = C6H18 Octane
n = 12 C12H2 × 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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 49

Question 2.
Draw all possible structural formulae having molecular formula C6H14. Give names to all the isomers.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 50
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 51

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 52

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 53

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 54

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 55

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 56

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 57
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 58

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 59
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 60

(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.)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 61
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 62

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 63

b. Homologous series of alcohol.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 64

c. Homologous series of alkenes.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 65

(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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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 CnH2n + 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 CnH2n.

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

Question 13.
The structural formula of polypropylene is ……………….
Answer:
The structural formula of polypropylene is
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 66

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) CnH2n
(b) CnH2n + 2
(c) CnH2n – 2
(d) CnHn – 2
Answer:
CnH2n 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) CnH2n
(b) CnH2n + 2
(c) CnH2n – 2
(d) CnH2n – 1
Answer:
The general formula for alkynes is CnH2n – 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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 67
(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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 68
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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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?
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 69
(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.)

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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 CnH2n.
Answer:
False. (The general formula of alkyne is CnH2n – 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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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 – CnH2n – 2.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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 – CnH2n.

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 70
Answer:
(1) (-OH) – Alcohol
(2) (-COOH) – Carboxylic acid
(3) (-CHO) – Aldehyde
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 71

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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).

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 72

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 73

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 74

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 75

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).

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 76

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 CnH2n + 2.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 77
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 78

Question 14.
Which functional groups are present in aldehyde and ketone?
Answer:
The functional group -CHO is present in aldehyde and the functional group
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 79
is present in ketone.

Question 15.
Which functional group is present in
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 80
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 81

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 82
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 83

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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 84
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 85
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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 86
(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 (=).

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

Question 3.
Describe the formation of nitrogen molecule.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 87
(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).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 88
(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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 89
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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

Question 6.
Observe the straight chain hydrocarbons given below and answer the following questions:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 90
(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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 91

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 92

Question 9.
Draw the structure and carbon skeleton for cyclohexane.
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 93

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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 94
Ring structures of an organic compound having six carbon atoms:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 95

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 96

Question 15.
Draw the structures of isomers of pentane (C5H12).
Answer:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 97

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

Question 16.
Recognize the carbon chain type for each of the following:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 98
(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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 99

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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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 CnH2n + 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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 100
Answer:
First four homologous series of alcohols are

  1. Methanol CH3 – OH
  2. Ethanol C2H5 – OH
  3. Propanol C3H7 – OH
  4. Butanol C4H9 – OH

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 101

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 102
Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Prefix functional group: Chloro
Assign the number: 2
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 103
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

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 104
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 105
Answer:
The number of carbon atoms in the longest chain: 4
Parent alkane: Butane Functional group: -OH (ol)
Assign the number:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 106
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 107
The number of carbon atoms: 3
Parent alkane: Propane
Functional group: -NH2 (amine)
Assign the number:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 108
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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 109
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 110
Answer:
The number of carbon atoms in the longest chain: 5
Parent alkane: Pentane
Functional group:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 111
Assign the numbering:
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 112
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
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 113
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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 114

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 115

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 116

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 117

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 118
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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 119

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).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 120

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 121

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 112

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 123

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 124

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).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 125

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.
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 126

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon 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)
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 127
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 CnH2n + 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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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).
Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds 128

(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.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

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 CnH2n + 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 CnH2n or CnH2n – 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 CnH2n + 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 CnH2n.
  3. They are chemically more reactive.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 9 Carbon Compounds

Project:

Question 1.
Prepare a list of carbon compounds which occur in nature and discuss their uses in daily life.