AP Inter 2nd Year Civics Study Material Chapter 8 State Judiciary

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 8th Lesson State Judiciary Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 8th Lesson State Judiciary

Long Answer Questions

Question 1.
Explain the powers and functions of the High Court.
Answer:
The constitution of India provides for a High Court for each state. But the 7th Amendment Act, of 1956 authorized the Parliament to establish a common High Court for two or more states and a Union Territory.

Articles 214 to 231 in Part-VI of the constitution deal with the organization, qualifications, appointment, independence, jurisdiction, powers, and procedures, etc., of the High Court.

Composition :
Every High Court consists of a Chief Justice and such other judges as the President may from time to time deem it necessary to appoint.

Appointment of Judges :
The Chief Justice of High Court is appointed by the President after consultation with the Chief Justice of India and the Governor of the concerned state. The other judges of High Court are appointed by the President with the consultation of the Chief Justice of High Court of the concerned State. In case of a common High Court for two or more States, the Governors of concerned States are cpnsulted by the President. ,

Qualifications of Judges :
A person to be appointed as a judge of High Court should possess the following qualifications.
a) He should be a citizen of India.
b) He should have held judicial office in the territory of India for atleast 10 years, or
c) He should have been an advocate of a High Court or of two or more such courts for 10 years period. However, the Constitution has not prescribed a minimum age for appointment as a judge of a High Court.

Salaries and Allowances :
The salaries, allowances, privileges, leave and pension of the judges of a High Court are determined by the Parliament from time to time. The Judge of a High Court gets a salary of ₹ 80,000/-per month and the Chief Justice gets ₹ 90,000/-. They are also entitled to get other allowances and are provided with free accommodation and other facilities like medical, car, telephone etc.

The salaries and allowances cannot be reduced except under financial emergency. The salaries and allowances are drawn from the Consolidated Fund of the State. Their retired Chief Justice and other judges are entitled to 50% of their last drawn salary as monthly pension.

Tenure :
Every Judge of a High Court including Chief Justice holds office until he attains the age of 62 years. The Judges including the Chief Justice will take oath in the presence of the Governor of the concerned State. He can resign for his office when he desire so by writing to the President to that effect.

Method of removal:
A Judge of a High Court can.be removed by the President on the grounds of proved misbehaviour or incapacity. The method of removal of a Judge of the High Court shall be the same as that of a Judge of the Supreme Court.

Powers and Functions of the High Court:
The following are the powers and functions of the High Court.

1) Original Jurisdiction:
Every High Court in India has original jurisdiction in regard to matters of admiralty, will, marriage, divorce, company laws, contempt of court and certain revenue cases. Every High Court is empowered to issue directions, orders or writs for the enforcement of any of the Fundamental Rights. Every High Court is empowered to settle disputes relating to election of members of Parliament and State Legislatures.

Under Article 226, the High Court is empowered to issue writs for enforcing Fundamental Rights. The High Court issue writs like Habeas-corpus, Certiorari, Mandamus, Quo-warranto and Injunction,for protecting the Fundamental Rights of the Indian Citizens.

2) Appellate Jurisdiction :
Every High Court hears appeals against the judgement rdinate courts. The appellate Jurisdiction of the High Court extends to both riminal Cases.

a) Civil Cases :
An appeal to the High Court on the civil side is either a first appeal or a second appeal. In civil cases, appeal to the High Court lies from the decision of a District Court. Appeals can also be made from the subordinate courts directly provided the dispute involves a value of more than ₹ 5,00,000/- (or) a question of fact of law. When a court subordinate to the High Court decides an appeal differing from the decision of an inferior court, a second appeal can be made to the High Court.

b) Criminal Cases :
In Criminal cases, it hears the appeals in which the accused has been sentenced to more than seven years imprisonment by the Sessions Judge. All cases involving capital punishment awarded by the Sessions Court come to High Court as appeals. Its approval is necessary for the imposition of death sentences by the District Sessions Judge. It also hears cases involving the interpretation of the Constitution.

3) Court of Record :
The State High Court acts as a Court of Record. It records all its decisions and judgements. Such records are of great significance. They carry evidentary value. They are taken as Judicial precedents to the Judges and Advocates in legal matters.

4) Power of Judicial Review :
The State High Court possesses the power of judicial review like the Supreme Court. It is the power of High Court to examine the constitutionality of legislature enactments and executive orders of both the Central and State Governments. On examination, if they are found to bp violated of the Constitution (Ultra Vires), they can be declared as illegal, unconstitutional and invalid (null and void) by the High Court. Consequently, they cannot be enforced by the government.

5) Power of Certification :
High Court certifies certain cases which can go to the Supreme Court. That appeals which go to Supreme Court depend up on the issue of a certificate by the High Court.

6) Advisory functions :
The High Court is consulted by the State Governor in the matters of appointment, posting and promotion of District Judges and in the appointment of personnel to the Judicial Services of the State (Other than District Judges). It deals with the matters of posting, promotion, grant of leave, transfers and discipline of the members of the Judicial Service of the State. It also, renders advice to the subordinate courts in the matters of public interest or of legal importance.

7) Administrative functions :
The High Court exercises certain administrative functions within its territorial jurisdictions.

  • Under Article 227, every High Court has the power of supervision over all Courts and Tribunals functioning in its territorial jurisdiction except Military Courts or Tribunals in the State.
  • It ensures the proper working of these courts. It exercises the power to make and issue general rules and regulations for securing the efficient working of the court.
  • The High Court can transfer any case from one court to another court under Article 228 and can even transfer the case itself and decide the same.
  • The High Court has the power to investigate or enquire into the records or other connected documents for any court subordinate to it.
  • It appoints its administrative staff and determines the salaries and allowances and other conditions of the personnel working in subordinate courts.
  • It is empowered to withdraw any case involving the interpretation of the constituion and dispose of the case itself.
  • The High Court is the Highest Court of Justice in the State. All other Courts and Tribunals (except Military Courts and Tribunals) in State function under the direct supervisions and control of the State High Court.

Other functions :
a) The High Court acts as the District Court where its head quarters are located.

b) The Chief Justice of the High Court acts as the Governor on the direction of the President tentatively whenever the vacancy arises in that office.

c) The High Court can admit Public Interest Litigation like the Supreme Court of India.

AP Inter 2nd Year Civics Study Material Chapter 8 State Judiciary

Question 2.
Write an essay on the District Level Courts.
Answer:
The state judiciary consists of a High Court and a hierarchy of Subordinate Courts also known as District Courts. The District Courts play an important role in judicial administration at the district level. The courts consist of District Judge and other Judges. They fullfill their obligations at the District, Town and Major Village levels. They hear civil and criminal cases. They are subject to the authority and control of the State government in administrative matters and to the State High Court in judicial matters.
There are two types of subordinate courts in a State namely :

  1. Civil Courts and
  2. Criminal Courts.

1) Civil Courts :
The Civil Courts deal with civil suits regarding the matter like marriages, divorce, inheritance, business etc. There will be District Civil Courts at the District Level. The District Judge acts as its head. He exercises control and supervision over other civil courts in the district.

There are some senior civil judge courts below the rank of the district civil courts. There are some other junior civil judge courts in addition senior civil judge court. Judicial officers of subordinate courts are given here under :

  1. Principal District Judge
  2. Family Court Judge
  3. SC & ST Act Court Judge
  4. Senior Civil Court Judge
  5. Junior Civil Court Judge

The Principal District Court admits the cases pertaining to an amount of Rupees 10 Lakhs and above worth of poverty and deliver the judgements. The Principal District Judge is appointed through direct as well as indirect recruitment (By promotion).

The Family Courts are presided by judicial officers of the cadre of District Judges. This court takes up cases under Hindu Marriage Act relating to divorce, ordering interim maintenance, ordering custody of children etc. In order to protect Scheduled Caste and Scheduled Tribes rights and to implement SC & ST Act strictly, there is a court for the entire district.

There are some courts namely Senior Civil Judge Courts which deal with the cases of property worth rupees above one lakh and below 10 lakhs and deliver the judgements.

Cases pertaining to property worth below one lakh will be taken up by Junior Civil Judge Court and the Judgements are delivered. There are some Nyaya Panchayats, Grama Kacheries, Adalati Panchayats and so on at the lowest level in the district to deal with local legal issues.

2) Criminal Courts :
The Sessions Court is the highest criminal court in the district. The Sessions court acts as the superior court at the district level in handling the criminal matters. The Sessions Judge delivers judgements according to the provisions mentioned in the Indian penal code and the criminal procedure code. The following judges deal with at the district level.
They are :

  1. District Sessions Judge
  2. Senior Assistant Sessions Judge
  3. Junior Civil Judge
  4. Special Judicial Magistrate

The Principal District Judge will act as District Sessions Judge, who deals with the cases relating to murder and motor vehicles act violation cases and delivers the judgement and imposes life imprisonment or death sentences which are to be confirmed by the State High Court. The Senior Assistant Sessions Judge will impose an imprisonment of five to seven years, depending on the nature of the Case.

If there is a Junior Civil Judge Court for the entire town the court acts as a civil as well as a criminal court and take up the cases and deliver judgement and impose imprisonment below three years. There are Second Class Magistrate Courts which deliver the judgement by imposing fine up to rupees five hundred or a sentence of one year or both.

Special Judicial Magistrate Courts will be established in every town which takes up petty cases and deliver the judgement by imposing fines below ₹ 500/- and impose imprisonment below six months.

Short Answer Questions

Question 1.
Explain briefly the composition of High Court.
Answer:
The Constitution of India provides for a High Court for each state. But the 7th Amendment Act, 1956 authorised the Parliament to establish a common High Court for two or more states and a Union Territory. *

Composition :
Every High Court shall consist of a Chief Justice and some other Judges. The President of India may appoint them from time to time. Besides, the President has the power to appoint Additional Judges for a temporary period not exceeding two years as an acting Judge, where a permanent Judge of a High Court is temporarily absent or unable to perform his duties.

Such judges hold office until the permanent Judge resumes his office. The number of Judges varies from 5 in Gauhati High Court to 48 in the Allahabad High Court. Our Constitution does not specify the exact strength of High Court judges and leaves it to the discretion of the President. Accordingly, the President determines the strength of a High Court from time to time depending upon its workload.

Question 2.
Write any two powers and functions of the State High Court. [Mar. 18, 16]
Answer:
The following are the two powers and functions of the High Court.
1) Original Jurisdiction:
Every High Court in India has original jurisdiction in regard to matters of admiralty, will, marriage, divorce, company laws, contempt of court and certain revenue cases. Every High Court is empowered to issue directions, orders or writs for the enforcement of any of the Fundamental Rights. Every High Court is empowered to settle disputes relating to election of members of Parliament and State Legislatures. The High Courts of Bombay, Calcutta and Madras possess original jurisdiction in Civil as well as Criminal cases arising within the presidency towns.

They are authorized to hear Civil Cases involving property of the value of ₹ 20,000/- or more. They enjoy exclusive privileges and authority in this regard. In fact this power of High Court was m vogue before independence. It has been retained in the new Constitution. The other High Courts also enjoy the same jurisdiction as was available to them before independence.

Under Article 226, the High Courts is empowered to issue writs for enforcing Fundamental Rights. The High Court issue writs like Habeas-corpus, Certiorari, Mandamus, Quo-warranto and Injunction for protecting the Fundamental Rights of the India Citizens.

2) Appellate Jurisdiction :
Every High Court hears appeals against the judgement of the subordinate courts. The appellate Jurisdiction of the High Court extends to both Civil and Criminal Cases.

a) Civil Cases :
An appeal to the High Court on the civil side is either a first appeal or a second appeal. In civil cases, appeal to the High Court lies from the decision of a District Court. Appeals can also be made from the subordinate courts directly provided the dispute involves a value of more than ₹ 5,00,000/- (or) a question of fact of law. When a court subordinate to the High Court decides an appeal differing from the decision of an inferior court, a second appeal can be made to the High Court.

b) Criminal Cases :
In Criminal cases, it hears the appeals in which the accused has been sentenced to more than seven years imprisonment by the Sessions Judge. All cases involving capital punishment awarded by the Sessions Court come to High Court as appeals. Its approval is necessary for the imposition of death sentences by the District Sessions Judge. It also hears cases involving the interpretation of the Constitution.

Question 3.
Explain the Administrative functions of the High Court
Answer:
The High Court exercises certain administrative functions within its territorial jurisdictions.

  1. Under Article 227, every High Court has the power of supervision over all Courts and Tribunals functioning in its territorial jurisdiction except Military Courts or Tribunals in the State.
  2. It ensures the proper working of these courts. It exercises the power to make and issue general rules and regulations for securing the efficient working of the court.
  3. The High Court can transfer any case from one court to another court under Article 228 and can even transfer the case itself and decide the same.
  4. The High Court has the power to investigate or enquire into the records or other connected documents of any court subordinate to it.
  5. It appoints its administrative staff and determines the salaries and allowances and other conditions of the personnel working in subordinate courts.
  6. It is empowered to withdraw any case involving the interpretation of the Constitution and dispose of the case itself.
  7. The High Court is the Highest Court of Justice in the State. All other Courts and Tribunals (except Military Courts and Tribunals) in State function under the direct supervision and control of the State High Court.

AP Inter 2nd Year Civics Study Material Chapter 8 State Judiciary

Question 4.
Explain the powers and functions of District Court.
Answer:
There will be District Civil Courts at the District Level. The District Judge acts as its head. He exercises control and supervision over other civil courts in the district.

There are some senior civil judge courts below the rank of the district civil courts. There are some other junior civil judge courts in addition senior civil judge courts. Judicial officers of subordinate courts are given here under :

  1. Principal District Judge
  2. Family Court Judge
  3. SC & ST Act Court Judge
  4. Senior Civil Court Judge
  5. Junior Civil Court Judge

The Principal District Court admits the cases pertaining to an amount of Rupees 10 Lakhs and above worth of property and deliver the judgements. The Principal District Judge is appointed through direct as well as indirect recruitment (By promotion).

The Family Courts are presided by judicial officers of the cadre of District Judges. This court takes up cases under Hindu Marriage Act relating to divorce, ordering interim maintenance, ordering custody of children etc. In order to protect Scheduled Caste and Scheduled Tribes rights and to implement SC & ST Act strictly, there is a court for the entire district. .

There are some courts namely Senior Civil Judge Courts which deal with the cases of property worth rupees above one lakh and below 10 lakhs and deliver the judgements.

Cases pertaining to property worth below one lakh will be taken up by Junior Civil Judge Court and the Judgements are delivered. There are some Nyaya Panchayats, Grama Kacheries, Adalati Panchayats and so on at the lowest level in the district to deal with local legal issues.

Question 5.
Discuss the powers and functions of State Advocate General. [Mar. 17]
Answer:
Every State in Indian Union shall have an Advocate General, an official corresponding to the Attorney-General of India. He performs similar functions for the State that of the Attorney-General of India. He is the highest law officer in the State.

Appointment:
The Advocate General is appointed by the Governor of the State under the Article 165 of the Constitution. A person to be appointed as Advocate General must possess the following qualifications.

  1. He should be a citizen of India.
  2. He must have held a judicial office for ten years or an advocate of a High Court for ten years.
  3. He must be a person who is qualified to be appointed a judge of a High Court.

Tenure and Removal :
The Constitution of India did not mention the tenture of Advocate General. Further, the Constitution does not contain the procedure and grounds for his removal. He holds office during the pleasure of the Governor. He may be removed by the Governor at any time. He may also quit his office by submitting his resignation to the Governor. Conventionally, he resigns when the government resigns or is replaced, as he is appointed on the advice of the government.

Salary :
His remuneration is not fixed by the Constitution. He receives such remuneration as the Governor may decide from time to time.

Powers and functions :
As the highest law officer of the State Government, he exercises the following powers and functions.

  • He advises the State Government upon such legal matters which are referred to him by the Governor.
  • He performs such other duties of a legal character that are assigned to him by the Governor.
  • He discharges the functions and conferred on him by the Constitution.
  • He appeared before any court of law within the State.
  • He has a right to speak and to take part as member in the proceedings of the house (s) but no right to vote.
  • He can also attend any of the Standing Committee meetings of State Legislature.

Very Short Answer Questions

Question 1.
Appointment of High Court Judges. [Mar. 18, 17]
Answer:
The Chief Justice of High Court is appointed by the President after consultation with the Chief Justice of India and the Governor of the concerned State. The other judges of High Court are appointed by the President with the consultation of the Chief Justice of High Court of the concerned State. In case of a common High Court for two or more States, the Governors of concerned States are consulted by the President.

Question 2.
Qualifications of High Court Judges. [Mar. 16]
Answer:
A person to be appointed as judge of High Court should possess the following qualifications.

a) He should be a citizen of India.
b) He should have held a judicial office in the territory of India at least 10 years, or
c) He should have been an advocate of a High Court or of two or more such courts for 10. years period. However, the constitution has not prescribed a minimum age for appointment as a judge of a High Court.

Question 3.
High Court as a Court of Record.
Answer:
The State High Court acts as a court of Record. It records all its decisions and judgements. Such records are of great significance. They carry evidentary value. They are taken as Judicial precedents to the judges and Advocates in legal matters.

AP Inter 2nd Year Civics Study Material Chapter 8 State Judiciary

Question 4.
Advisory functions of High Court.
Answer:
The High Court is consulted by the State Governor in the matters of appointment, posting and promotion of District Judges and in the appointment of personnel to the Judicial Services of the State (Other than District Judges). It deals with the matters of posting, promotion, grant of leave, transfers and discipline of the members of the Judicial Service of the State. It also renders advice to the subordinate courts in the matters of public interest or of legal importance.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 7th Lesson State Legislature Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 7th Lesson State Legislature

Long Answer Questions

Question 1.
Explain the composition, powers, and functions of the State Legislative Assembly.
Answer:
The Constitution provides for a Legislature for every State on the model of the Parliament. As per Article 168, the State Legislature consists of the Governor and one or two Houses. In India, while some States have Bicameral Legislatures, others have Unicameral Legislatures. Andhra Pradesh at present possesses Unicameral Legislature.

The Lower House of the State Legislature is known as the Legislative Assembly or Vidhana Sabha and the Upper House is the Legislative Council or Vidhana Parishad.

Composition of Legislative Council (Vidhana Parishad) :
The Upper House of the State Legislature is known as Legislative Council. The Constitution lays down that a Legislative Council shall have not less than 40 members and not more than [latex]\frac{1}{3}[/latex]rd of the total membership of the State Assembly. The Legislative Council consists of both nominated and elected members.

The election is conducted through the indirect method by means of proportional representation with a single transferable vote.

Distribution of Seats :

  1. [latex]\frac{1}{3}[/latex]rd are elected by the members of the State Assembly.
  2. [latex]\frac{1}{3}[/latex]rd are elected by members of local bodies.
  3. [latex]\frac{1}{12}[/latex]th are elected by teachers.
  4. [latex]\frac{1}{12}[/latex]th are elected by graduates.
  5. The remaining [latex]\frac{1}{6}[/latex]th members are nominated by the Governor from among persons who have distinguished themselves in the fields of Literature, Science, Arts, Social Services etc.

Qualifications :
The members of the Council 1) must be citizens of India, 2) must have completed 30 years of age and 3) must possess such other qualifications as may be prescribed by the Legislature.

Term :
the members are elected for a period of 6 years. But [latex]\frac{1}{3}[/latex]rd of them retire for every 2 years. The Council is a permanent body. It cannot be dissolved by the Governor.

Chairman and Deputy Chairman:
The Council has Chairman and a Deputy Chairman who are elected by the members of the Council from among themselves. The Chairman presides over the meetings of the Council.

Legislative Assembly (Vidhana Sabha) :
Composition of the Legislative Assembly :
The Legislative Assembly is the popular and powerful chamber of the State Legislature. It is the lower house and resembles more or less the Lok Sabha at the Centre. It consists of representatives directly elected by the people of the State on the basis of universal adult franchise. It’s maximum strength is fixed at 500 and minimum strength at 60. Only the Legislative Assembly of Sikkim has less than 60 because of it’s small population. Those who become members of State Legislative Assembly must be citizens of India and must be above 25 years of age.

Term of Office :
The normal term of Assembly is 5 years. It may be dissolved earlier by the Governor on the advice of the Chief Minister. The Parliament may extend it’s term by one year, when National Emergency is in force.

Presiding Officers :
The Presiding officer of Assembly is known as Speaker. He is elected by the members of the Assembly. The Assembly also elects a Deputy Speaker to conduct the business of the House in the absence of the Speaker.

Powers and Functions of the State Legislature :
The State Legislature has the following powers.

1) Legislative Powers and Functions :
The State Legislature has the power to make laws on all the subjects included in the State List. It has also the power to make laws in respect of subjects included in the Concurrent List. However, such a law should not disagree with a law already made by the Parliament on the same subject. In the making of laws, Legislative Assembly has been given more powers than the Legislative Council. The Legislative Council at the most may delay the legislation for a period of 4 months. Later the Assembly sends the bill to the Governor for his assent.

2) Constitutional Powers and Functions :
Even though, the Legislature has no powers to move the Constitution amendment bills, it’s consent is required for amending certain provisions of the Constitution. Such bills have to be referred to it after they are approved by the Parliament.

3) Executive Powers and Functions :
The State Legislature exercises control over the Council of Ministers. It’s members make the Ministers individually and collectively responsible to the Legislature. The Council of Ministers is collectively responsible to the State Legislative Assembly. The Legislature can expose the actions of Executive, through questions, debates and adjournment motions. In controlling the Executive, the Legislative Assembly has more powers than the Council. The Ministry has to resign when the Legislative Assembly passes no confidence motion against the Government.

4) Financial Powers and Functions:
The State Legislature exercises complete control over the finances of the State Government. It sanctions money to the State Government to enable it to run the administration. It may pass, reduce or reject the demands for grants presented to it by the Government. It may accept or reject proposals for taxation and borrowings presented to it by the Government. In financial matters the Assembly is more powerful than the Council. Because all money bills, including the Budget, shall be introduced first only in the Assembly. It can accept or reject any recommendations made by the Council.

5) Electoral Powers :
The elected members of the Assembly participate in the election of the President. They also elect the representatives of the State to the Rajya Sabha and l/3rd members of the Legislative Council if the State Legislature is bicameral. They also elect the presiding officers and deputy presiding officers of Assembly and Council.

Miscellaneous Powers : The state legislature :

  1. Safeguards the dignity and privileges of its members.
  2. Suspends, expels or terminates their membership.
  3. Examines the report of the State Public Service Commission and the Comptroller and Auditor General etc.

Conclusion:
The State Legislature plays an important role in the State Administration. It makes necessary laws for the welfare of the people of the State. It controls the Executive by making it responsible for their actions.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 2.
Write briefly the composition, powers and functions of the. State Legislative Council.
Answer:
The Upper House of the State Legislature is known as Legislative Council. The Constitution lays down that a Legislative Council shall have not less than 40 members and not more than [latex]\frac{1}{3}[/latex]rd of the total membership of the State Assembly. The Legislative Council is a body of partly nominated and partly elected members.

The election is conducted through indirect method by means of proportional representation with a single transferable vote.
Composition of the Council:

  1. [latex]\frac{1}{3}[/latex]rd members are elected by the Legislative Assembly.
  2. [latex]\frac{1}{3}[/latex]rd members are elected by members of local bodies.
  3. [latex]\frac{1}{12}[/latex]th members are elected by teachers.
  4. [latex]\frac{1}{12}[/latex]th members are elected by graduates.
  5. The remaining [latex]\frac{1}{6}[/latex]rd members are nominated by the Governor from among persons who have distinguished themselves in the fields of Literature, Science, Arts, Social Services etc.

Tenure :
The Legislative council is a quasipermanent House. l/3rd of the members of this House retire every two years. But the term of each member is six years. New members are elected in the place of retired members. All the members of the House do not retire at a time as it is a permanent House.

Powers and Functions of State Legislative council:
State Legislative council is the primary law making body along with the Legislative Assembly. The State Legislative Council has the following powers and Functions.

a) Legislative powers and Functions :
The Legislative Council does not possess equal powers and functions when compared to counterpart, State Legislative Assembly. It is said that the Legislative Council enjoys equal status and not power. However it exercises the following powers and functions. All the bills, other than money bills may be introduced in either of the House. They will be sent to the assent of the Governor only with the approval of both the Houses. The Council may reject any bill and spnt it back for the reconsideration of the Assembly. However, incase of a disagreement between two Houses, the decision of the Assembly will be supreme.

The Council must approve all the bills sent by the Assembly with in a period of three months or at the maximum of four months. It implies that the Council can withhold its assent over the bills sent by the Assembly for a maximum period of four months. Thus, the Legislative Council can only delay the bills but the Legislative Assembly can override it.

b) Executive Powers and Functions :
The State Legislative Council has very limited executive powers when compared to that of the Assembly. The council of Ministers headed by the Chief Minister is responsible for its acts only to the Assembly and not to the Council. The Council cannot decide the future of the Council of Ministers. However, the Council can influence the policies and programmes of the ministers by asking questions and supplementary questions by drawing the call attention motion etc., but they cannot force the Council of Ministers to resign.

c) Financial Powers and Functions :
The Legislative Council has only limited powers in the financial matters. Money bills cannot at first be introduced in the Legislative Council first. The Council must accept all money bills with or without recommendation within fourteen days of the receipt of the bill. The Assembly possess the discretion powers either to accept or reject these recommendations. If the Council does not return the Money Bill to the Assembly within 14 days, then the bill is deemed to have been passed by both the Houses. It is clear that in the financial field the Legislative Council has a subordinate status and that Legislative Assembly has dominant position.

d) Electoral Functions:
The Legislative Council elects a Chairman and Deputy Chairman to preside over its meetings in a dignified manner. Some of its members are elected to various legislative committees like Public Accounts Committee, Estimates Committee and Public Undertakings committee etc.

e) Other Functions :
The Legislative council acts as the best means for formulating and consolidating public opinion. It discusses technical and other contemporary matters, as there are experts in various fields.

Question 3.
Explain the role and responsibilities of the Speaker of Legislative Assembly.
Answer:
The members of State Legislative Assembly elect one among them as a speaker to conduct the business of the House. His term of office is five years.

The speaker is the guardian of the Rights and Liberties of the members of the House.

Role and Responsibilities (or) Powers and Functions of the Speaker :
The powers and functions of the speaker of State Legislative Assembly are almost the same as those of the Speaker of Lok Sabha. His powers and functions are as follows.

  1. The speaker preserves order and decorum in the House for conducting legislative business.
  2. He allocates time for different kinds of business in the House.
  3. He interprets the rules and procedure.
  4. He puts matters to vote and announces the results.
  5. He has the right of casting vote in case of a tie.
  6. He admits motions, resolutions and points of order.
  7. He is empowered to adjourn the meeting of the House in the absence of a quorum.
  8. He can order for removal of indecent and incriminatory references from the records.
  9. He allows the members to speak in the House.
  10. He may name a member and ask him to leave the House in case of disorderly behavior.
  11. He can adjourn the House in case of grave disorder or serious matter.
  12. He accepts and rejects the resignation of a member of the House after ascertaining whether it was submitted under due process or not.
  13. He appoints the Chairmen of all the committee of the assembly and supervises their functioning. He himself is a Chairman of Business Advisory Committee, Rules Committee and the General Purpose Committee.
  14. He decides where a bill is a Money Bill or not. His decision on this question is final.

Short Answer Questions

Question 1.
Write a note on the Legislative Assembly.
Answer:
Legislative Assembly is the Lower House of the State Legislature. The Members of Legislative Assembly are called M.L.As. According to Article 170of the Indian Constitution it consists of not more than 500 members and not less than 60 members. It means that it’s strength depends on the population and size of the state. But small states have been allowed to have less number of members. Thus Goa and Mizoram have only 40 members, while Sikkim has 32 Members.

Composition of the Legislative Assembly:
The Legislative Assembly is the popular and powerful chamber of the State Legislature. It is the lower house and resembles more or less the Lok Sabha at the Centre. It consists of representatives directly elected by the people of the State on the basis of universal adult franchise. Those who become members of State Legislative Assembly must be citizens of India and must be above 25 years of age.

Term of office :
The normal term of Assembly is 5 years. It may be dissolved earlier by the Governor on the advice of the Chief Minister. The Parliament may extend it’s term by one year, when National Emergency is in force.

Presiding officers :
The Presiding Officer of Assembly is known as Speaker. He is elected by the members of the Assembly. The Assembly also elects a Deputy Speaker to conduct the business of the House in the absence of the speaker.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 2.
Write a note on Estimates Committee.
Answer:
According to the Rules of Procedure and Conduct of Business in the State Legislature, the Estimates committee consists of 20 members. Among them 15 members belong to Assembly. The remaining 5 members belong to Legislative Council. The members hold office for a period of one year. They are elected through in indirect election.

Functions :
The functions of the Estimates Committee in the State Legislature are the same as that of Estimates Committee of Lok Sabha. These are given here under.

  1. Estimates Committee exercises control over public expenditure.
  2. It suggests fiscal reforms in organization, the efficiency or administration reforms consistent with the policy underlying estimates.
  3. It advises alternative policies for securing efficiency and economy in administration.
  4. It examines whether the money is well laid out within the limits of the policy implied in the estimates.
  5. It also suggests the form in which the estimates shall be presented to the Assembly.

Question 3.
What do you know about Public Accounts Committee? [Mar.-18, 17]
Answer:
Public Accounts Committee consists of 20 members out of which 15 members belong to Assembly and 5 members belong to Legislative Council. They are elected through indirect election by following the principle of proportional representation Jor a period of one year. The Chairman is normally the member of Opposition Party. The Ministers of Cabinet cannot be member of Public Accounts Committee.

Functions :
Public Account Committee performs the following functions.

  1. The committee examines the accounts showing the appropriation of sums granted by the house for expenditure of the state government.
  2. It scrutinizes the appropriation accounts of the state and the reports of the Comptroller and Auditor General.
  3. It shall be the duty of the Public Accounts Committee to examine such a trading, manufacturing, and profit and loss accounts and balance sheets and the accounts of the state government and also to consider the report of the Comptroller and Auditor General.
  4. The committee carefully considers the accounting and audit procedures.
  5. The committee is not concerned with the question of policy approved by the legislature.
  6. The committee investigates expenditure after it has already incurred. An overall, this committee is generally described as a ‘post-mortem committee’.

Question 4.
Write the powers and functions of Vidhcma Sabha Speaker. [Mar. 16]
Answer:
The powers and functions of the Speaker of State Legislative Assembly are almost the same as those of the Speaker of Lok Sabha. His powers and functions are as follows.

  1. The speaker preserves order and decorum in the House for conducting legislative business.
  2. He allocates time for different kinds of business in the House.
  3. He interprets the rules and procedure.
  4. He puts matters to vote and announces the results.
  5. He has the right of easting vote in case of a tie.
  6. He admits motions, resolutions and points of order.
  7. He is empowered to adjourn the meeting of the House in the absence of a quorum.
  8. He can order for removal of indecent arid incriminatory references from the records.
  9. He allows the members to speak in the House. ‘
  10. He may name a member and ask him to leave the House in case of disorderly behavior.
  11. He can adjourn the House in case of grave disorder or serious matter.
  12. He accepts and rejects the resignation of a member of the House after ascertaining whether it was submitted under due process or not.
  13. He appoints the Chairmen of all the committees of the assembly and supervises their functioning. He himself is a Chairman of Business Advisory Committee, Rules Committee and the General Purpose Committee.
  14. He decides where a bill is a Money Bill or not. His decision on this question is final.

Very Short Answer Questions

Question 1.
Qualifications of M.L.A. [Mar. 18, 16]
Answer:
A person who wishes to contest for the membership of the State Legislative Assembly must be possess the following qualifications.

  1. He should be a citizen of India.
  2. He should have completed the age of 25 years.
  3. He should possess such other qualifications as prescribed by any act of Parliament.
  4. However, no person can simultaneously be a member of any House of the Parliament and of a State Legislature.

Question 2.
Qualifications of M.L.C.
Answer:
A person who wishes to contest for the membership of the State Legislative Council must possess the following qualifications.
a) He should be a citizen of India.
b) He should have completed 30 years of age.
c) He should possess such other qualifications as laid down by an Act of Parliament.

Question 3.
Quorum.
Answer:
Quorum is the minimum number of members required to be present in the house before it can transact any business. According to Article 188 of the constitution, the Quorum for conducting,the State Legislative Assembly meeting was fixed at 1/10th of the total membership. However, in some states, where the strength of the State Legislative Assembly is very less, the quorum will be a minimum number of 10. The speaker decides whether there is a quorum or not on a particular day.

Question 4.
Salaries and Allowances of M.L.A.
Answer:
The salary of MLA is decided by the respective State Legislature as per the Article 164 of the Indian constitution. The members of Andhra Pradesh State Legislative Assembly receive a monthly salary of ₹ 90,000/- which includes a basic pay of ₹ 15,000/- and constituency allowance of ₹ 75,000/-. Those legislators who are not provided government accommodation will get an additional ₹ 10,000/- as H.R.A members also get daily allowance of ₹ 800/- when the state legislature is in session.

Question 5.
Privileges of State Legislature.
Answer:
Privileges of a State Legislature are a sum of special rights, immunities and exemptions enjoyed by the State Legislatures. They are necessary in order to secure independence and effectiveness of their actions. The Houses cannot maintain the authority, dignity and honour without these privileges. They can protect their members from any obstructions in the discharge of their legislative responsibilities.

  1. Collective privileges
  2. Individual privileges

i) Collective Privileges :
The legislature has the right to publish its reports, debates and proceedings and also to prohibit others publishing the same.

ii) Individual Privileges:
The privileges belonging to the members of state legislature individually. They can not be arrested during the session of the state legislature or 40 days before and after the end of the session.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 6.
Brief History of AP legislature.
Answer:
The Andhra state was formed on October 1, 1953 Andhra State legislature initially had 140 MLAs. Elections were held to the Andhra State Legislative Assembly for the first time in 1955.

As per the recommendations of states re-organization committee, Hyderabad State was merged with Andhra State on linguistic basic and formed into Andhra Predesh State which had 245 MLAs [Including 150 MLAs of Hyderabad State.] Elections were held to the Andhra Pradesh legislative Assembly in 1957.

The State Legislative Council was established on July 1,1958. Since then it continued to exist till June 1, 1985, before being abolished. Again on March 30, 2007 the Andhra Pradesh Legislature became again bicameral after the revival of the legislative council.

Question 7.
Chairman of Legislative council. [Mar. 16]
Answer:
There will be a chairman in the Legislative council for conducting the meetings. He is elected by the members of the Legislative council among themselves. Dr. A. Chakrapani Yadav is the Present Chairman of Legislative council of Andhra Pradesh.

Question 8.
Deputy Speaker.
Answer:
The members of State Legislative Assembly elect one among them as a Deputy Speaker. His term of office is 5 years. The deputy speaker performs the duties in the absence of the speaker.

Question 9.
Deputy Chairman of Legislative Council.
Answer:
The Members of State Legislative Council elect one among them as a Deputy Chairman. The Deputy Chairman performs the duties in the absence of the Chairman.

AP Inter 2nd Year Civics Study Material Chapter 7 State Legislature

Question 10.
Types of committees. [Mar. 18]
Answer:
The committees are of two types i.e., Standing committees and Ad-hoc committees.
i) Standing Committees:
Standing committees deal with specific business (financial matters) Ex : Estimates committee, Public accounts committee and Committee on public undertakings.

ii) Ad-hoc Committees :
Ad-hoc committees are concerned with the matters of temporary nature. They cease to exist after completion of the work. They perform some specific functions assigned to them from time to time.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 6th Lesson State Executive Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 6th Lesson State Executive

Long Answer Questions

Question 1.
Discuss the powers and functions of the Governor. [Mar. 18]
Answer:
The Governor is the constitutional head of the state government. Article 153 of the Indian constitution provides for the office of the governor in the states. The administration of a state is carried on in the name of the governor

Appointment:
The president appoints the governor on the advice of the Prime Minister. In this Contest, the president generally follows two conventions which are mentioned below for the appointment of the governor.

  1. Consulting the Chief Minister of the State Concerned.
  2. Choosing an eminent person not belonging to the state concerned.

Qualifications :
Article 157 of our constitution lays down the following qualifications for the appointment of a person as a governor.

  1. He shall be a citizen of India.
  2. He should have completed the age of 35 years.
  3. He should not hold any office of profit.
  4. He should not be a member of either union or state legislature.
  5. He should not be an insolvent declared by any court of Law.

Pay and Allowances :
The governor is entitled to receive a monthly salary of ₹ 1,10,000/-. He resides in official rent-free building “Raj Bhavan”. Besides, he is entitled to many other allowances and privileges.

Oath of Office :
The Chief Justice of the state High Court administers the oath of office to the governor.

Tenure :
The Governor holds office as a convention for a term of five years. However, He holds office during the pleasure of the president.

Powers and Functions of the Governor:
The Governor exercises six important powers and functions. They are explained as follows.

1) Legislative Powers and Functions :
Article 163 describe that the Governor is an intregral part of the Stage Legislature. In that capacity he exercises certain powers and performs functions related to the State Legislature.

  1. The Governor inaugurates the first sessions of the State Legislative Assembly after the general elections are over.
  2. He also addressing the first session of State Legislative Assembly every year i.e. budget session.
  3. He appoints Pro-tern Speaker of the State Legislative Assembly.
  4. He summons and prorogues the sessions of the two Houses of the State Legislature.
  5. He addresses the Members of the state Legislature and sends messages in relation to the state legislature.
  6. The Governor gives his assent to the bills passed by the state Legislature.
  7. He may return a bill sent by the state Legislature for its reconsideration.
  8. He dissolves the State Legislative Assembly when he feels no party is in a position to form a stable and viable Government and the advice of the Chief Minister.
  9. He may promulgate Ordinances to meet an emergency which require immediate action during the recess of the State Legislature.
  10. He nominates members of Anglo-indian community to the Legislative Assembly of the state if he feels that community is not represented in the house!
  11. The Governor nominates 1/6 of the total members of the State Legislative Council.

2) Executive powers and functions:
Article 154 of our constitution vests the governor with the executive powers of the state. The governor exercises these powers either directly of through officers subordinate to him. The governor has the following executive powers.

  1. The Governor appoints, the Chief Minister and the members of the Council of Ministers on the advice of the Chief Minister.
  2. He allocates Portfolios among the ministers and reshuffles their portfolios.
  3. He removes the Ministers on the advice of the Chief Minister.
  4. He appoints the Vice-Chancellors of the Universities in the State. He acts as the Chancellor of the universities.
  5. He appoints the Chief Secretary and Advocate General of the State Government.
  6. He appoints the Chairmen and other members of the State Commissions such as a) State Public Service Commission, b) State Election Commission, c) Official Language Commission, d) Commission for Women, e) Minorities Commission, f) Backward Classes Commission and g) SC & ST Commission.
  7. He regulates the postings and transfers of the All India Services personnel working in the state.

3) Judicial Powers and Functions :
The Governor also exercises the following judicial powers and functions.

  1. The Governor renders advice to the President of India in the appointment of Chief Justice and other judges of the High Court of the State.
  2. The Governor appoints the Advocate General of the State.
  3. He makes appointments, postings and promotions of the District Judges in consultation with the Chief Justice of High Court of the State.
  4. He also appoints persons to the judicial services of the state (Other than the district courts) in consultation with the Chief Justice of High Court and State Public Service Commission.
  5. He can grant pardon, retrieve, remit and commute the sentence of any person convicted of any offence against any law of the concerned state.

4) Financial Powers and Functions :
The Governor will have the following powers and functions of financial nature.

  1. The Governor sees that the Annual Financial Statement,(i.e. Budget) is laid before the State Legislature.
  2. No Money bill shall be introduced in the prior permission of the Governor.
  3. No Demand for Grant can be made except on his recommendation.
  4. He maintains the Contingency Fund of the State. He can make advances out of the Contingency Fund to meet any unforeseen expenditure.
  5. He constitutes a Finance Commission for every five years to review the financial position of village Panchayats and Municipalities.
  6. He sees that reports of various financial committees are laid before the State Legislature.

5) Miscellaneous Powers and Functions :
The Governor receives the Annual Report of the State Public Service Commission and passes it on to the Council of Ministers for comments. Thereafter, he passes on the report on these comments to the Speaker of the Assembly for placing it before the legislature. He receives the report of the Auditor-General regarding income and expenditure made by different departments working under the State Government. On behalf of the President, he runs the administration as the real head of the state through the enforcement of law and policies during the period of President’s Rule.

6) Discretionary Powers:
Under Article 163 (1) of the constitutions the Governor has some discretionary powers which are discharged by him. His decisions in this regard are final, These are mentioned as below.

  1. Playing a decisive role in appointing the new Chief Minister in a situation when no single party has a clear majority in the state Legislative Assembly.
  2. Dismissing a Minister When it refuses to resign even after losing majority support in the House.
  3. Dissolution of the Assembly on the advice of the Chief Minister who lost the majority members support.
  4. Rendering advice to the President for the imposition of the President’s Rule in the State.
  5. Reserving a Bill for the consideration and approval of the President.
  6. Seeking instructions from the president before promulgating ordinance on some important matters.
  7. Sending back a bill passed by the state legislature for its reconsideration, .except money bills.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Question 2.
Explain the powers and Functions of the Chief Minister.
Answer:
Articles 163 and 164 of our constitution deal with the office of the chief minister. The chief minister is the real executive head of the State Government. He plays a decisive role and occupies a key position in the State Government.

Appointment :
The Chief Minister is appointed by the governor under article 164. After general elections, the governor normally invites the leader of the majority party in the Legislative Assembly to form the government and appoints him as the Chief Minister.

Powers and Functions of Chief Minister:
The Chief Minister has high authority and heavy responsibility in discharging his powers and related functions. His powers and functions are related to the following heads.

i) Formation of the Ministry :
The first and foremost responsibility of the Chief Minister is the formation of Ministry of his choice. The CM has a free hand in the selection and appointment of Ministers. He chooses some members of his party (or coalition partners in the case of a coalition) and recommends their names to the Governor to be appointed as Ministers. He advises the Governor to allocate portfolios among the Ministers.

ii) Leader of the State Council of Ministers :
The Chief Minister is the head of the Council of Ministers. As such he occupies a position of exceptional authority. He is the Chairman of the State Council of Ministers. The Chief Minister decides the time, venue and the agenda of Gabinet meetings. The CM presides over such meetings, discussions are carried under his direction. He guides, directs, controls and co-ordinates the activities of the Ministers.

iii) Link between the Governor and the State Council of Ministers:
The Chief Minister is the principal channel of communication between the Governor and the State Council of Ministers. As part of his Constitutional duty he communicates all the administrative decisions and legislative proposals of the State Council of Ministers to the Governor. It is his responsibility to furnish any information related to the actives of the Ministers as the Governor may call for. No minister shall meet the Governor without the consent of the Chief Minister.

iv) Leader of the Legislative Assembly:
As the Chief Minister enjoys the confidence and support of the majority Legislators he acts as the leader of the Assembly. In that capacity he extends complete co-operation to the Presiding Officers for the smooth conduct of the Business of the House. He ensures discipline .of his party members in the Assembly. The CM helps other Ministers in case they are unable to satisfy the House with their replies or when a situation goes out of control in the Assembly. He announces the Government policies on the floor of the Legislative Assembly.

v) Chief Spokes Person :
The Chief Minister is the chief spokesperson of the Government. He announces the major policies and programs of the State Government. His statements in and outside of the State Legislature will carry much legitimacy and influence in the State. The Members in the State Legislature demand for clarification and statements on particular issues of the State from the Cheif Minister. So, he maintains much restrain without making controversial statements.

vi) Leader of the party in power:
The Chief Minister is the leader of party in power at the State level, he participates in the meetings organized by his party. He informs the party members about the policies and programs initiated by the State Government to fulfill the poll promises of his party. He seeks the co-operation and support of the party members for the effective implementation of the government policies and successful function pf the Government. He brings co-ordination between the party in power and the Government.

If he happens to be the President or the General Secretary of the party he gains control over his party. He utilizes the services of the senior, experienced and prominent party leaders in improving the image and efficiency of the State Government. He sees that his party members do not make controversial and embarrassing comments that may land the executive in the troubled waters. –

vii) Leader of the people :
He tries to know and understand the needs and interests, aspirations and expectations of the people in the State. For this purpose he frequently makes visits to different places and addresses the public gatherings. He invites petitions from the people and patiently listens to them. He informs the people about the welfare measures and developmental programs taken up by the Government. He motivates the people to take active participation in the implementation of various welfare schemes. He undertakes relief „ measures and consoles the people affected during the natural calamities. He maintains good rapport with the people and wins their confidence and trust, as their prominent leader of the people.

viii) Chief Advisor to the Governor :
It is the Constitutional obligation of the CM to render advice to the Governor on all matter of the State Government. His advice is binding over the Governor in the matter of appointment of ministers, allocation of portfolios, reshuffling of the Ministry and accepting the resignation of Ministers. It is a rate privilege and opportunity of the Chief Minister to advise the Governor to dissolve the State Legislative Assembly when he still has majority support of the members in the Assembly.

ix) Cordial relations with the Union Government:
The Chief Minister, being the real head of the. State administration, has the main responsibility of maintaining harmonious relations with the Union Government. He should develop cordial and amicable relations with the Prime Minister and the Union Ministers. He will have to interaction with several Union Ministers particularly of Home, Finance, Industry, Agriculture, Education, and Rural Development etc.

How much Union support a State gets in the form of financial grants to the centrally sponsored schemes depend on the Chief Minister’s influence on and the rapport with the Union Ministers. The State’s representation in the Union Cabinet also influences the quantum and quality of the support to the State.

x) Relations with Party in Opposition :
The Chief Minister maintains good relations with the Presidents, Floor Leaders, and MLAs of the Opposition Parties. Good contacts, healthy relations and cordial approach the Chief Minister in securing constructive co-operation from the Opposition. He takes the Opposition parties into confidence on crucial issues of the State. He organizes all party meetings and takes delegation of all parties to the Union Government for communicating issues of the state.

xi) Related to the Constitution:
The Indian Constitution confers all the powers of real executive on the Chief Minister. He owes his position to the Constitution. He has to exercise his authority and discharge his responsibilities in accordance with the provisions of Constitution. He must uphold the democratic norms and Constitutional principles in running the State administration.

Question 3.
Describe the powers and functions of the State Council of Ministers.
Answer:
Article 163(1) of the Indian Constitution provides for the State Council Ministers with the Chief Minister at its head, to aid and advise the Governor in the exercise of his powers and in running the State administration.

Composition :
The State Council of Ministers is generally a three-tier body. It consisting of.
1. Cabinet Ministers 2. Ministers of State and 3. Deputy Ministers. There will be some only Parliamentary Secretaries in some states on rare occasions. It constitutes the fourth wing of the hierarchy of the Council of Ministers.

Qualifications :

  1. They should be members of either House of the Legislature (if it is bi-cameral)
  2. If the Ministers are not the members of the State Legislature, they should be elected to the State Legislature within six months from the date of assuming their office. Otherwise they cease to hold their office.
  3. They must possess such other qualifications as*is determined by the Parliament from time to time.

Appointment:
All the Ministers are appointed by the governor (Article 164) on the advice and recommendation of the chief minister.

Powers and Functions of the Council of Ministers :
i) Policy Formulation :
The State Council of Ministers formulates policies suitable for the progress of the people and development of the State. It is an intellectual and laborious process. The Cabinet Ministers meet frequently under the leadership of the Chief Minister, discuss thoroughly various matters of the State adihinistration and finalize the policies along with the necessary decisions.

ii) Enactment of Laws :
The State Council of Ministers takes Legislative initiation on different matters of State Government. It is the Council Ministers that drafts and finalizes the public Bill and pilots them in the State Legislature at different stages in order to get them approved by the Legislature. Once the bills are approved by the Legislature, the Council of Ministers advises the Governor to assent them so that they become laws. The Council of Ministers may propose amendment to the existing laws or enactment Of new laws for the administrative convenience.

iii) Provision of Good Administration :
The State Council of Ministers i.e., the real executive is voted to power to provide good administration and promote the well-being of the .people of the State. The chief responsibility of the Council of Ministers is running the administration in accordance with the Constitutional cardinals and democratic doctrines. The.total administrative work is divided into different ministries. Each minister has one or more departments under his control and is responsible for the effective and the transparent administration of such departments. It formulates and implements different developmental programs and welfare schemes.

iv) Co-ordination of Governmental Activities :
The State Council of Ministers is responsible and the authority for coordinating the functions of different government departments. Without proper co-ordination among the departments the success of the State administration cannot be ensured. The Chief Minister guides and takes lead in coordinating the cabinet discussions and government activities.

v) Appointment Power:
The State Council of Ministers plays a key role in all important appointments to various offices in the State. It makes all appointments in the name of the Governor to various higher offices like the Chief secretary. Advocate General, D.G.R Principal Secretaries and other Heads of the Departments etc.

vi) Financial Fucntions :
The State Council of Ministers wields control over the Finances of the State. It determines fiscal policy and deals with the matters concerning the State Revenue, Expenditure, Investment and Audit of Accounts. It prepares the budget proposals of the State Government and places it before the State Legislature for its consideration and approval. It manages the Finances of the State according to the policy and budget as approved by the Legislature. Its role is that of a trustee.

vii) Miscellaneous Functions :
The State Council of Ministers finalizes strategies for the overall development of the State in the sphere of Agriculture, Irrigation, Industry, Transport, Education, Planning, IT etc. It proclaims ordinances in the name of the Governor during the recess of the State Legislature.

Short Answer Questions

Question 1.
Explain any three Powers and Functions of the Governor.
Answer:
1) Legislative Powers and Functions :
Article 168 describes that the Governor is an intregral part of the State Legislature. In that capacity he exercises certain powers and performs functions related to the State Legislature.

  1. The Governor inaugurates the first sessions of the State Legislative Assembly after the general elections are over.
  2. He also addressing the first session of State Legislative Assembly every year i.e. budget session.
  3. He appoints Pro-tem Speaker of the State Legislative Assembly.
  4. He summons and prorogues the sessions of the two houses of the State Legislature.
  5. He addresses the Members of the state legislature and sends messages in relation to the state legislature.
  6. The Governor gives his assent to the bills passed by the State Legislature.
  7. He may return a bill sent by the State Legislature for its reconsideration.
  8. He dissolves the State Legislative Assembly when he feels no party is in a position to form a stable and viable Government and the advice of the Chief Minister.
  9. He may promulgate Ordinances to meet an emergency which require immediate action during the recess of the State Legislature.
  10. He nominates members of Anglo-Indian community to the Legislative Assembly of the state if he feels that community is not represented in the house.
  11. The Governor nominates 1/6 of the total members of the State Legislative Council.

2) Executive Powers and Functions:
The Governor has the following executive powers.

  1. The Governor appoints, the Chief Minister and the members of the Council of Ministers on the advice of the Chief Minister.
  2. He allocates Portfolios among the ministers and reshuffles their portfolios.
  3. He removes the Ministers on the advice of the Chief Minister.
  4. He appoints the Vice-Chancellors of the Universities in the State. He acts as the Chancellor of the universities.
  5. He appoints the Chief Secretary and Advocate General of the State Government.
  6. He appoints the Chairmen and other members of the State Commissions such as a) State Public Service Commission, b) State Election Commission, c) Official Language Commission, d) Commission for Women, e) Minorities Commission, f) Backward Classes Commission and g) SC & ST Commission.
  7. He regulates the postings and transfers of the All India Services personal working in the state.

3) Judicial Powers and Functions :
The Governor also exercises the following judicial powers and functions.

  1. The Governor renders advice to the President of India in the appointment of Chief Justice and other judges of the High Court of the State.
  2. The Governor appoints the Advocate General of the State.
  3. He makes appointments, postings and promotions of the District Judges in consultation with the Chief Justice of High Court of the State.
  4. He also appoints persons to the judicial services of the state (other than the district courts) in consultation with the Chief Justice of High court and State Public Service Commission.
  5. He can grant pardon; retrieve, remit and commute the sentence of any person convicted of any offence against any law of the concerned state.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Question 2.
What are the Discretionary Functions of the Governor?
Answer:
Under Article 163(1) of the constitution the governor has been armed with certain discretionary powers. In the exercise of his discretionary powers, the governor is not bound by the advice of his ministers or even to seek such advice. His actions shall not be called into question on the ground that he ought or ought not to have acted in his discretion.

The discretionary powers of governor are mentioned below.

  1. Playing a decisive role in appointing the new Chief Minister in a situation when single party has a clear majority in the State Legislative Assembly.
  2. Dismissing a Ministry when it refuses to resign even after losing majority support in the House.
  3. Dissolution of the Assembly on the advice of the Chief Minister who lost the majority members support.
  4. Rendering advice to the President for the imposition of the President’s Rule in the State.
  5. Reserving a Bill for the consideration and approval of the President.
  6. Seeking instructions from the President before promulgating ordinance on some important matters.
  7. Sending back a bill passed by the state legislature for its reconsideration, except money bills.
  8. Seeking information from the Chief Minister with regard to the administrative and legislative matters of the State.

Question 3.
What are the Differences between the Governor of a State and the President of India?
Answer:

Governor of a State President of India
1) The Governor is a nominated person. 1) The President is an elected person.
2) He has no security of Tenure. His Tenure depends upon the Pleasure of the President. 2) The President has a fixed tenure of Office of five years in general.
3) The Governor can be removed easily by the President on the advice of the Union Council of Ministers headed the Prime Minister. 3) He can be removed only by the difficult process of Impeachment by the Parliament.
4) The Governor has discretionary power. 4) The President has no discretionary, powers.
5) The Governor does not have Military and diplomatic powers. 5) He has Military and Diplomatic powers.
6) Pardoning power of the Governor is limited. He cannot pardon death sentence and any sentence inflicted by the Martial Court. 6) Pardoning power of the president is absolute. He can pardon even the death sentence and sentence of Martial Court.
7) The Governor does not have emergency powers. He can only suggest for the imposition of President’s Rule. 7) The President can Promulgate orders for the declaration of all the three types of Emergencies.
8) The Governor has no power to remove the Chairman and Members of the State Public Service Commission though he appoints them. 8) The President can remove the Chairman and the Members of the Union Public Service Commission on the grounds stipulated by the constitution.
9) The Governor sometimes may reserve a bill for the consideration of the President. 9) The President need not reserve any bill for the consideration of any other authority before giving his assent.
10) The Governor cannot issue ordinance without instructions from the President on the matters (a) which might affect the powers of the Union (b) affecting powers of the High Court (c) Imposing reasonable restrictions upon Inter¬State Trade or Commerce. 10) The President can promulgate any ordinance on the advice of Council of Ministers of the Union.

Question 4.
What is the position and significance of the Governor in the State?
Answer:
The Constitution of India provides for the Parliamentary System of Government both at Centre and in the States. While the Governor is only a nominal executive, the real executive constitutes the Council of Ministers headed by the Chief Minister. The Constitution has assigned a dual role to the office of a Governor in the Indian federal system. He is the Constitutional head of the State Government as well as the representative of the Union Government.

As the Constitutional head of the state government, he must positively contribute to the progress and development of the State. He has to see that the political and administrative heads of the State Government strive for the promotion of the interests of the people. The Governor has to ensure that the ministers and bureaucrats must observe the constitutional and democratic norms. It is the responsibility of the Governor to see that the affairs of the government are carried on in accordance with Constitutional provisions. The Governor has to maintain close and harmonious relations with the real executive heads of the Union and State Governments.

The Governor is not supposed to run a parallel government in the State. His role is that of a good counselor, mediator and arbitrator than an active politician. He shall abide by the advice of the State Council of Ministers. This does not mean that he should accept all proposals immediately. He can reserve Bills for reconsideration and prevent hasty decisions. Great caution and restrainment must be exercised while reporting to the President under Article 356. Otherwise, his image as guardian of the State Government would tarnished. He should keep himself away from active politics. If he identifies himself with a political party, he cannot inspire the total trust of the people.

Being the representative of the Centre, the Governor has the responsibility of informing through reports whether the State is complying with the directives issued by the Union from time to time. It is his constitutional obligation to inform the Union whether the constitutional machinery is functioning smoothly in the state or not.

The Centre-State relations largely depend upon the action and performance of the Governor. He can make or mar the healthy relations between the Union and the State. The Constitution has given certain discretionary powers to the Governor. If the Governor makes use of these powers sparingly, judiciously and impartially, tensions between the Centre and the States would certainly be reduced. If he acts with bias and at the behest of the Central Government, the tensions between the Centre and State would undoubtedly be enhanced. The role of the Governor in the formation or dissolution of the Ministry or imposing of President’s Rule will have far reaching implications and consequences in the healthy and harmonious Centre-State relations.

Question 5.
Explain any three Powers and Functions of the Chief Minister. [Mar. 16]
Answer:
The following are the three important powers and functions of the Chief Minister.
i) Formation of the Ministry :
The first arid foremost responsibility of the Chief Minister is the formation of Ministry of his choice. The CM has a free hand in the selection and appointment of Ministers. He chooses some members of his party (or coalition partners in the case of a coalition) and recommends their names to the Governor to be appointed as Ministers. He advises the Governor to allocate portfolios among the Ministers.

ii) Leader of the State Council of Ministers :
The Chief Minister is the head of the Council of Ministers. As such he occupies a position of exceptional authority. He is the Chairman of the State Council of Ministers. The Chief Minister decides the time, venue and the agenda of Cabinet meetings. The CM presides over such meetings, discussions are carried under his direction. He guides, directs, controls and co-ordinates the activities of the Ministers.

iii) Link between the Governor and the State Council of Ministers :
The Chief Minister is the principal channel of communication between the Governor and the State Council of Ministers. As part of his Constitutional duty he communicates all the administrative decisions and legislative proposals of the State Council of Ministers to the Governor. It is his responsibility to furnish any information related to the actives of the Ministers as the Governor may call for. No minister shall meet the Governor without the consent of the Chief Minister.

Question 6.
Explain the Composition of the State Council of Ministers.
Answer:
Article 163(1) of the Indian Constitution provides for the State Council Ministers with the Chief Minister at its head, to aid and advise the Governor in the exercise of his powers and in rurining the State administration.

Composition :
The State Council of Ministers is generally a three-tier body. It consisting of.
1. Cabinet Ministers 2. Ministers of State and 3. Deputy Ministers. There will be some only Parliamentary Secretaries in some States on rare occasions. It constitutes the fourth wing of the hierarchy of the Council of Ministers.

i) Cabinent Ministers :
The cabinent is a small body consisting of ministers holding the most important portfolios such as Home, Finance, Planning and Industries etc. They enjoy independence in taking and implementing decisions concerning their ministry. They attend the Cabinet meetings, concerned by the Chief Minister. Some times the Ministers of state and deputy ministers may attend the cabinet meetings, in case their presence is needed during deliberations. They meet frequently and determine the policies of the State Government under the stewardship of the Chief Minister.

ii) Ministers of State :
The Ministers of State hold portfolios of less importance compared to the Cabinent Ministers. They may be attached to the individual Cabinent Ministers or might be given independent charge of crucial departments’in the major Ministries. In such a case they enjoy independence. They are answerable directly to the Chief Minister. They are not subject to the control of Cabinet Ministers.

iii) Deputy Ministers :
The Deputy Ministers are attached to the Cabinet Ministers. They performs such functions which are assigned by the Cabinet Ministers. His role is mainly to relieve the burden of the Cabinet Minister. He assists the Cabinent Minister in the administrative and legislative affairs of the Ministry the Constitution (91st Amendment) Act 2003 fixes a ceiling on the size of the Council of Ministers. The total number of the Ministers cannot be more than 15% of the total strength of the State Legislative Assembly.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Question 7.
Point out any three Powers of the State Council of Ministers. [Mar. 17]
Answer:
The three of the following are the important powers of the state council of Ministers.
i) Policy Formulation :
The State Council of Ministers formulates policies suitable for the progress of the people and development of the State. It is an intellectual and laborious process. The Cabinent Ministers meet frequently under the leadership of the Chief Minister, discuss throughly various matters of the State administration, and finalize the policies along with the necessary decisions.

ii) Enactment of Laws :
The State Council of Ministers takes Legislative initiation on different matters of State Government. It is the Council of Ministers that drafts and finalizes the public Bills and pilots them in the State Legislature at different stages in order to get them approved by the Legislature. Once the bills are approved by the Legislature, the Council of ministers advises the Governor to assent them so that they become laws. The Council of Ministers may propose amendment to the existing laws or enactment of new laws for the administrative convenience.

iii) Provision of Good Administration :
The State Council of Ministers i.e., the real executive is voted to power to provide good administration and promote the wellbeing of the people of the State. The chief responsibility of the Council of Ministers is running the administration in accordance with the Constitutional cardinals and democratic doctrines.

The total administrative work is divided into different ministries. Each minister has one or more departments under his control and is responsible for the effective and the transparent administration of such departments. It formulates and implements different developmental programs and welfare schemes.

Question 8.
Estimate the relationship between the Chief Minister and the Governor.
Answer:
In a Parliamentary Democracy like India the real executive of the state plays a pivotal role in the state administration. The Cheif Minister as the real executive head in the State is responsible ultimately to the state electorate. The Chief Minister has also the obligation to facilitate the exercise of powers of the Governor by providing necessary information about the affairs of the administration of the State. The Governor has a right to seek any information on administrative and legislative activities of the state Council of Ministers through the Chief Minister.

However, this right does not allow permit the Governor to become a parallel centre in this state. It may be noted that the nature of the power available to the Governor is persuasive and not authoritarian. So he cannot under the grab of this right start over riding or vetoing the decisions or proposals of the state Council of Ministers.

The founding fathers of our Constitution have laid great emphasis on the need for harmonious relations between the Governor and his Council of Ministers headed by the Chief Minister. This was the main idea behind abandoning the proposal for an elected Governor and adopting for his nomination by the President.

The Sarkaria Commission in its report emphasized that for the proper working of the Parliamentary system there needs to be a good personnel rapport between the Governor and the Chief Minister of a State. For fostering good personnel relationship, the Sarkaria Commission suggested that the Union Government has to consult the concerned Chief Minister before appointing the Governor of the State.

Pandit Jawaharlal Nehru during his speeches in the Constituent Assembly stated that the Governor should be acceptable to the Chief Minister. Both the Chief Minister and Governor must work together in mutual co-operation to promote the development of the State and safeguard the interests of the people of the State.

Very Short Answer Questions

Question 1.
Qualifications of Governor. [Mar. 18]
Answer:
Article 157 of our Constitution lays down the following qualifications for the appointment of a person as a Governor.

  1. He shall be a citizen of India.
  2. He should have completed the age of 35 years.
  3. He should not be a member of either house of Parliament or state legislature.
  4. He should not hold any other office of profit.
  5. He should not be an insolvent declared by any court of law.

Question 2.
Special responsibilities of the Governor. [Mar. 17]
Answer:
The Governor has certain special responsibilities to discharge according to the directives issued by the President under Articles 371 (Z) 371 (A) (1) b, 371 (C) in case special responsibility through the Governor is to constilt the Council of Ministers the final decision shall be in his individual judgement which no court can question.

The Governor of Assom. Maharashtra, Gujarat, Wougaland, Manipur and Sikkim have special responsibility on specific matters related to their respective states. For example : 1. The Governor of Assom shall in his discretion determine the amount payable by the state of Assom to district council as the royalty accruing from licences of minerals decides the amount of money received from mineral resources and which has to be allocated to the District Council.

Question 3.
State Executive. [Mar. 18]
Answer:
Articles 153 to 167 deal with the matters of the state executive. The state executive consists of (a) the Governor (b) the Chief Minister and (c) Members of the State Council of Ministers. In our Parliamentary system Governor is the titular or constitutional head of the state. The Chief Minister is the real executive head of the Government. The Chief Minister and the Ministers being represents the people.

Question 4.
Immunities of the Governor.
Answer:
Our Constitution provides certain legal Immunities to the office of the Governor to enable him to discharge his constitutional functions in a free and fair manner, to ensure the state government works constitutionally. He shall not be held responsible for any act done or purporting to have been done in his official capacity. No criminal proceedings can be initiated against the Governor during his term of office. No proceedings for his arrest or imprisonment can be taken by any court of law.

Question 5.
Any two executive powers of the Governor.
Answer:

  1. The Governor appoints, the Chief Minister and the members of the Council of Ministers on the advice of the Chief Minister.
  2. He allocates portfolios among the Ministers and reshuffles their portfolios.

Question 6.
Two discretionary powers of the Governor.
Answer:

  1. Playing a decisive role in appointing the new Chief Minister in a situation when no single party has a clear majority in the State Legislative Assembly.
  2. Dismissing a Ministry when it refuses to resign even after losing majority support in the house.

Question 7.
The Chief Minister. [Mar. 17]
Answer:
The Chief Minister is the centre of the real executive authority at the state level. He plays a decisive role and occupies a key position in the State Government. The progress of the people and development of the state largely depends upon the Cabinet, Personality, Preservance and political stature of the Chief Minister.

Question 8.
Cabinet Ministers. [Mar. 16]
Answer:
The Cabinet is a small boy consisting of Ministers holding the most important portfolios such as Home, Finance, Planning and Industries etc. They enjoy independence in taking and implementing decisions concerning their Ministry. They attend the Cabinet meetings, concerned by the Chief Minister. Some times the Ministers of state and Deputy Ministers may attend the Cabinet meetings, incase their presence is needed during deliberations. They met frequently and determine the policies of the State Government under the Stewardship of the Chief Minister.

Question 9.
Composition of the State Council of Ministers. [Mar. 18]
Answer:
The State Council of Ministers is generally a three-tier body. It consisting of.

  1. Cabinet Ministers
  2. Ministers of state and
  3. Deputy Ministers. There will be some only Parliamentary Secretaries in some states on rare occassions. It constitutes the fourth wing of the hierarchy of the Council of Ministers. .

Question 10.
Deputy Ministers.
Answer:
The Deputy Ministers are attached to the Cabinet Ministers. They performs such functions which are assigned by the Cabinet Ministers. His role is mainly to relieve the burden of the Cabinet Minister. He assists the Cabinet Minister in the .administrative and legislative affairs of the Ministry.

AP Inter 2nd Year Civics Study Material Chapter 6 State Executive

Question 11.
State Ministers.
Answer:
The Ministers of State hold portfolios of less importance compared to the Cabinet Ministers. They may be attached to the individual Cabinet Ministers or might be given independent charges of crucial departments in the major minorities. In such a case they enjoy independence. They are answerably directly to the Chief Minister. They are not subject to the control of Cabinet Ministers.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 5th Lesson Union Judiciary Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 5th Lesson Union Judiciary

Long Answer Questions

Question 1.
Write an essay on the Supreme Court of India.
Answer:
The Supreme Court of India is the Highest Court of Justice in India.

Part V of the Indian constitution from Articles 124 to 147 deals with the composition, Appointment, Qualifications of Judges, powers and Functions of the Supreme Court.

Article 124 provides for the establishment of the Superme Court.

The Supreme Court of India was established in New Delhi on January 26, 1950 with the inauguration of our constitution.

Composition :
The Supreme Court Consists of the Chief Justice and 30 other Judges. There may be some Ad-hoc Judges and retired Judges on temporary basis.

1) Seat of the Supreme Court:
The Head quarters of the Superme Court is situated at New Delhi. The Supreme Court ordinarily shall sits at New Delhi.

  • All general cases are adjudicated by a Division Bench Comprising two or more Judges.
  • Cases involving the constitutional matters are heard by a constitutional Bench consisting Five Judges.
  • For considering special cases larger benches consisting of Five or more than Five Judges are constituted.

2) Appointment:
The judges are appointed by the President. While appointing the Chief Justice, the President consults the retiring Chief Justice and the Prime Minister. He appoints the remaining judges on the advice of the Chief Justice.

3) Qualification :
A judge of the Supreme Court 1) must be a citizen of India. 2) must have worked as a judge of any High Court for atleast 5 years or must have 10 years of experience as an advocate either in the Superme court or any High court or must be a legal expert in the opinion of the President.

4) Salaries :
The Chief Justice gets a monthly salary of Rs. 1,00,000/- and each other judge gets 90,000 per month. They also get allowances. Their salaries should not be reduced to their disadvantage during their term of office.

5) Term of Office :
The judges hold office till they reach 65 years of age. However, they can be removed from office by Parliament through a resolution known as impeachment before the end of their term or they may submit their resignation.

Removal:
A Judge of the Supreme Court can be removed from his position only on the grounds of proved misbehaviour or in capacity by an order of the president after on an address by the Parliament by a Majority of not less than 2/3 votes.

Immunities of Judges :
According to Article 121 of the constitution, No discussion shall take place in the Parliament with respect to the conduct of any Judge of the Supreme Court in the discharge of his duties except upon a motion of impeachment.

Powers and Functions of Supreme Court:
1) Original Jurisdiction :
According to the original Jurisdiction, the Supreme Court hears directly any dispute (a) between the Government of India and one or more States, (b) between the Government of India and any State or States on one side and one or more States on the other (c) between two or more States. The Supreme Court also decides all disputes and doubts regarding the election of the President and Vice President. It protects the Fundamental Rights guaranteed to the citizens by issuing several writs.

2) Appellate jurisdiction:
The Supreme Court is the highest Court of appeal in India. All appeals from all other courts can be heard by Supreme Court. The appellate jurisdiction extends to four types of cases namely civil, criminal, constitutional and special. But, the High Court should give certificate for appealing in the Supreme Court in the first three kinds of cases. But in special cases the certificate of High Court is not required.

3) Advisory Jurisdiction :
The Supreme Court offers it’s advice to the President on those matters of legal or public importance which are referred to him (Art. 143). It’s opinion is purely advisory and not binding on the President. The Supreme Court may refuse to give its opinion to the President. In October 1994 when Dr.S.D.Sharma, the then President of India asked the Supreme Court to give it’s advisory opinion on the Ayodhya issue, the court refused to give it’s opinion.

4) Judicial Review:
The Supreme Court has the power of Judicial Review. It examines the validity of laws passed by the legislatures or the orders issued by the Executive and declares them as ultra vires or unconstitutional if they are against the provisions of the constitution.

5) Court of Record:
The Supreme Court acts as a court of record. All the judgements and interpretations of the Supreme Court are recorded for future references. .

6) Other Powers :
The Supreme Court.

  1. reviews it’s own decisions.
  2. supervises the working of the State High Courts and other Subordinate Courts.
  3. recruits it’s own personnel for its maintenance.
  4. interprets the constitution and acts as it’s guardian.
  5. initiates contempt proceedings against those who criticise or defy it’s judgements etc.

Conclusion :
From the above account it is evident that our Supreme Court unique position in the judicial system of the country.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 2.
Explain the Powers and Functions of the Supreme Court of India.
Answer:
The supreme Court of India is the Highest Court of Justice in India. Article 124 of the Indian constitution provides for the establishment of the Supreme Court.

The Supreme Court of India was established in New Delhi on January 26, 1950. The Supreme Court consists of the Chief Justice and 30 other Judges.

Powers and Functions of Supreme Court:
1) Original Jurisdiction :
According to the original jurisdiction, the Supreme Court hears directly any dispute

  • between the Government of India and one or more States.
  • between the Government of India and any State or States on one side and one or more States on the other.
  • between two or more State. The Supreme Court also decides all disputes and doubts regarding the election of the President and Vice President. It protects the Fundamental Rights guaranteed to the citizens by issuing several writs.

2) Appellate Jurisdiction:
The Supreme Court is the highest Court of appeal in India. All appeals from all other courts can be heard by Supreme Court. The appellate jurisdiction extends to four types of cases namely civil, criminal, constitutional and special. But, the High Court should give certificate for appealing in the Supreme Court in the first three kinds of cases. But in special cases the certificate of High Court is not required.

3) Advisory Jurisdiction :
The Supreme Court offers it’s advice to the President on those matters of legal or public importance which are referred to him (Art. 143). It’s opinion is purely advisory and not binding on the President. The Supreme Court may refuse to give its opinion to the President. In October, 1994 when Dr. S.D. Sharma, the then President of India asked the Supreme Court to give it’s advisory opinion on the Ayodhya issue, the court refused to give it’s opinion.

4) Judicial Review:
The Supreme Court has the power of Judicial Review. It examines the validity of laws passed by the legislatures or the orders issued by the Executive and declares them as ultravires or unconstitutional if they are against the provisions of the constitution.

5) Court of Record:
The Supreme Court acts as a court of record. All the judgements and interpretations of the Supreme Court are recorded for future references.

6) Other Powers :
The Supreme Court.

  1. reviews it’s own decisions.
  2. supervises the working of the State High Courts and other Subordinate Courts.
  3. recruits its own personnel for its maintenance.
  4. Interprets the constitution and acts as it’s guardian.
  5. initiates contempt proceedings against those who criticise defy it’s judgements etc.-

7) Review of Judgement :
The Supreme court is empowered to review its own Judgements. It can uphold, modify or nullify its previous judgements. For instance it, while pronouncing its judgement in Golak nathcase vs. Punjab state case in 1967, declared the Parliament has no powers sb amend any of the provisions of fundamental rights of Indian Citizens.

Conclusion :
From the above account it is evident that our Supreme Court enjoys unique position.

Question 3.
Describe Judicial Review.
Answer:
Judicial Review is perhaps the most important power of the Supreme Court. The purpose of Judicial Review is to maintain the supremacy of the Constitution. There is no explicit mention of Judicial Review in the Indian Constitution. The higher courts derive this power from the provisions of Article 13 of the Constitution.

This Article empowers the Supreme Court to validate those laws and executive orders which infringe upon the Fundamental Rights. The makers of our constitution adopted this concept from the American Constitution keeping in view the written nature of the Indian Constitution and federal character of Indian polity.

Judicial Review means the power of the Supreme. Court or High Courts to examine the Constitutional validity of the legislative enactments and executive actions of both Central and State Governments and to declare them ‘null and void’ if found repugnant, of the provisions of the Constitution. As M.V Pylee stated, “Judicial Review is the competence of a court of law to declare the Constitutionality or otherwise of a legislative enactment”.

Article 13 declares all laws that are inconsistent with or direspectful of the fundamental rights, or void to the extent of their inconsistency. Hence, the Supreme Court being responsible for protecting fundamental rights, can declare any legislative act or executive decision that is inconsistent with provisions on fundamental rights as ultra vires or null and void, meaning unconstitutional and inapplicable. Besides, in case of federal relations, the Supreme Court can avail this power if a law is inconsistent with the provisions concerning the distribution of powers between the governments as laid down by the Constitution.

In this context the Supreme Court and High courts reviews legislations on the grounds that (a) they violate fundamental rights or (b) they violate the federal distribution of powers. The Supreme Court’s power of Judicial Review extends to the

  • Laws passed by the union and state legislatures,
  • Executive actions of the union and states,
  • Decisions of the public sector undertakings and
  • Constitutional Amendments. The Supreme Court for the first time utilized this power in 1950 itself by declaring Section 14 of the Preventive Detention Act as unconstitutional.

It may be noted that the Supreme Court of India is prominent in the world by exercising the power to determine the validity of the Constitutional Amendment Acts. However, Judicial Review is inevitable due to the following reasons.

  1. The Supreme Court has to uphold the supremacy of the Constitution.
  2. It has to maintain the federal equilibrium.
  3. It has to protect the fundamentals rights of the citizens.

Apart from the above, the power of Judicial Review is a resultant of the position of the Supreme Court as the; guardian of the Constitution. As such it has the final say in interpretation of the Constitution and by such an interpretation, the Supreme Court has extended its power of Judicial Review to almost all the provisions of the Constitution.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 4.
What is Judicial Activism? What are its merits and demerits?
Answer:
Generally, Judicial Activism is construed to be the over willingness of the judiciary to jump into the arena of executive or legislative functions. Judicial Activism, in fact, is not a distinctly separate concept from usual judicial activities. In general parlance, the expression “activism” means “being active”, “doing things with decision” and the expression “activist” should mean “one who favours intensified activities”. In this sense every judge is an activist.

Judicial Activism is a policy making in competition with policy making by legislature and executive. The essence of true Judicial Activism is rendering of decisions which are in tune with the temper and tempo of the times. The nature of Judicial Activisim is that it furthers the cause of social change or articulates concepts like liberty, equality or justice.

Judiciary which is an institution that traditionally confined to responding to cases brought before it, began considering many cases merely on the basis of Newspaper reports and postal complaints received by the court. But most of the cases of Judicial Activism have occurred through Public Interest Litigatiqn in the sphere of Public Health, Child Labour, Environment, Corruption etc. Therefore, the Judicial Activism became the most popular description of the role of Judiciary.
Causes for Judicial Activism :
The following are the causes for the emergence of Judicial Activism in India.

  1. Expansion of the rights of hearing in the administrative process.
  2. Excessive delegation without limitation.
  3. Judicial Review over administration.
  4. Promotion of open government.
  5. Indiscriminate exercise of contempt of power.
  6. Exercise of jurisdiction when non – exist.
  7. Over extending the standard rules of interpretation in its search for socio-economic and educational objectives.
    8) Breakdown of other machinery of the government.

Merits of Judicial Activism :

  1. Judicial Activism has democratized the judicial system by giving, access to the courts not just to individuals but also to groups.
  2. It has enforced executive accountability.
  3. It made an attempt to make the electoral system more free and fair.
  4. It is due to the impact of Judicial Activism during elections the candidates who tender affidavits disclosing their assets, income, educational qualifications, criminal record etc. This enable the people to elect better candidates.

De – Merits:

  1. Judicial Activism has blurred the line of distinction between the executive and legislature on the one hand and the judiciary on the other hand.
  2. Some felt that Judicial Activism led to the worsening of relations and balance among the three organs of government.
  3. Democracy is based on the principle that each organ of government will respect the powers and jurisdiction of others. But Judicial Activism may negate this democratic principle.

Short Answer Questions

Question 1.
Write about the composition of the Supreme Court.
Answer:
The Supreme Court of India is the highest court of Justice in India part V of the India Constitution from Articles 124 to 147 deals with the composition, Appointment, Qualifications of Judges, powers and Functions of the Supreme Court.

The Supreme Court of India was established in New Delhi on January 26, 1950.

Composition :
The Supreme Court consists of the chief Justice of India and such number of other Judges as is provided by the law. The parliament is authorised to determine the number of Judges in the Supreme Court. At present, there are a chief Justice and 30 other Judges in the Supreme Court. There may be some Adhoc Judges and retired Judges on temporary basis in the Supreme Court. All general cases are adjudicated by division Bench comprising two or more judges.

Cases involving the constitutional matters are heard by a constitution bench consisting Five Judges.

For considering special cases larger benches consisting of Five or more than Five Judges are constituted.

Question 3.
Mention any two Jurisdictions of the Supreme Court.
Answer:
1. Original Jurisdiction :
The original jurisdiction of the Supreme Court is purely of federal in nature. This power is confined to disputes between (a) the Government of India and any of the States in India, (b) The Government of India and any State of States on one side and other States on the other side or (c) two or more States. This power exclusively belongs to the Supreme Court and no other court in India is empowered to entertain any such suit. However disputes arising out of any treaty agreement, convenant, engagement etc., do not come under this Jurisdiction unless referred to by the President for advisory opinion. The Supreme Court can directly hear the disputes concerning the election of the President and the Vice-President.

2. Appellate Jurisdiction:
The Supreme Court is the highest court of appeal in India. Its appellate Jurisdiction may be divided into three heads.

  1. Cases involving interpretation of constitution,
  2. Civil cases and.
  3. Criminal cases.

i) The Supreme Court hears cases involving a substantial question of Law as the interpretation of. the constitution. It requires the certificate of the High Court for hearing such a case. Sometimes the Supreme Court can take up the appeal if it is satisfied that the case has do with certain intepretation of the constitution.

ii) In case where no constitutional question is involved, the Supreme Court hears appeals on the basis of a certificate of the High Court. Such cases, in the opinion of High Court involve (a) a substantial question of law and (b) the decision of the Supreme Court,

iii) In case of the criminal matters the Supreme Court hears appeals from any Judgement, whether they are in the form of final order or sentence of the High Court. It hears two specified cases namely.
a) Where the High Court has on an appeal reversed on order of acquittal of an accused and sentenced him to death and b) Where the High Court has tried the appeals from any of its subordinate courts convicted the accused and sentenced him to death.

Question 3.
What are the powers of Appellate Jurisdiction of the Supreme Court?
Answer:
The Supreme Court is the highest court of appeal in India. Its appellate Jurisdiction may be divided into three heads.

  1. Cases involving interpretation of constitution,
  2. Civil cases and
  3. Criminal cases.

i) The Supreme Court hears cases involving a substantial question of Law as to the interpretation of the constitution. It requires the certificate of the High Court for hearing such a case. Sometimes the Supreme Court can take up the appeal if it is satisfied that the ease has do with certain intepretation of the constitution.

ii) In case where no constitutional question is involved, the Supreme Court hears appeals on the basis of the High Court. Such cases, in the opinion of High Court involve (a) a substantial question of law and (b) the decision of the Supreme Court.

iii) In case of the criminal matters the Supreme Court hears appeals from any Judgement, whether they are in the form of final order or sentence of the High Court. It hears two specified cases namely.

a) Where the High Court has on an appeal reversed on order of acquittal of an accused and sentenced him to death and b) Where the High Court has tried the appeals from any of its subordinate courts convicted the accused and sentenced him to death.

The Supreme Court also hears appeals by special leave on any judgement of the High Court when the latter certifies that the case is fit for hearing by the Supreme Court. Besides, the Supreme Court as per Article 136 hears appeals over the cases that remain outside the purview of the ordinary law.

Question 4.
Explain the Advisory Jurisdiction of the Supreme Court.
Answer:
Under Article 143, the Supreme Court has advisory jurisdiction. Accordingly, the supreme Court offers its advice to the President on those matter of. legal or public or constitutional importance, when the President seeks such advice. It also reports its opinion over the disputes referred to it by the President, arising out of any treaty, agreement which was made or executed before the commencement of the Constitution. So far the Supreme Court rendered its advice to the President on eight occasions. The president, in the recent past, sought the advice of the Supreme Court on the ‘Ayodhya Issue’.

These are excluded by Article 131. However, the Supreme Court is not bound to render advice on such matters and the president is not bound to accept such an advice.

One may immediately question about the utility of the advisory powers of the Supreme Court. The utility is twofold. In the first place, it allows the government to seek legal opinion on a matter of importance before taking, action on it. This may prevent unnecessary litigations later. Secondly, in the light of the advice of the Supreme Court, the government can make suitable changes in its action or legislations.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 5.
Write about the Writ Jurisdiction.
Answer:
The word ‘writ’ literally means ‘order’ in written form. Article 32 of our Constitution confers authority upon the Supreme Court, to issue a constitutional writ for the enforcement of Fundamental Rights of the citizens. Any person, whose fundamental rights have been violated, can directly move the Supreme Court for remedy. The Supreme Court issues Habeas Corpus, Mandamus, Prohibitions, Quo-Warranto and Certiorari, for enforcing the Fundamental rights.

i) Habeas Corpus :
Literally means ‘to have the body of. It is issued by the court to affect the release of a person who has not been detained legally. Under this writ the court issues orders to the concerned authority to produce the person before-the court. The failure to abide by the writ order is met with punishment for contempt of court.

ii) Mandamus :
It means ‘we command’. The writ is a command issued by the court to a public official to do a duty which he has failed to do. This writ cannot be issued against private persons.

iii) Prohibition :
It means ‘to forbid’. It is issued by a higher court to a lower court to prevent the latter from exceeding its jurisdiction that it does not posses. This writ can be issued only against judicial and quasi-judical authorities.

iv) Certiorari:
It means ‘to be certified’ or ‘to be informed’. This writ is issued against the lower courts by the Supreme Court or High Courts, if the lower former courts violate their jurisdiction.

v) Quo Warranto :
It means ‘by what authority’ or ‘warrant’. If the court finds that a person is holding a public office which is not entitled to hold that office, it issues this writ for restricting that person from acting in that office. This writ is also not issued with respect to the private offices.

Besides the above, many other mechanism have been established for protecting of human rights. The National Commission for Women, the National Commission for Scheduled Castes & Scheduled Tribes, the National Human Rights Commission etc are some examples in this regard.

Question 6.
Describe Judicial Activism in India.
Answer:
It is the collective responsibility of the legislative, executive and the judiciary to accomplish the goals of the Constitution. Social Justice is the prime goal of the Constitution. Judiciary plays a vital role in achieving this goal. In order to meet the basic needs of the poor the oppressed and suppressed classes of the society, the Supreme Court entertains and also encourages the Public Interest Litigation (PIL) in its expanded, role of Judicial Activism.

As a result the apex court has evolved, developed new techinique, discovered and applied new remedies for violation of Fundamental Rights, and attempted to fill the vacuum arising out of executive and legislative inaction.

Due to the negligible attitude of the legislature and lack of edicts from the executive, the vulnerable classes of the society are sometimes denied social justice. In such circumstances, Social Action Groups, Civil Liberties Organizations, Voluntary Organizations etc., have come forward to their rescue through Public Interest Litigation.

As Chief Justice A.S. Ananad remarked that “the expanded concept of Public Interest Litigation by judicial interpretation from time to time has expanded the judicial limits of the courts exercising Judicial Review. This expanded role has been given the title of Judicial Activism by those who are critical of this expanded role of the Judiciary”.

Question 7.
What is meant Public Interest Litigation (PIL). [Mar. 17]
Answer:
The institution of Public Interest Litigation originated in USA during the mid 1960s. PIL or Social Action Litigation is an offshoot of liberalized rules of locus – standi. The traditional rule of locus-standi was based on the fact that judicial remedy can be sought only by those who have suffered an injury on account of violation of legal right by some public authority. The PIL choose liberalize this rule by making it clear that any person who suffer an injury but is unable to reach the court can take help of public minded citizens to reach the court to seek justice.

Public Interest Litigation Movement in India emerged during post-emergency years intending to make the judicial system accessible to the socially and economically lower sections of the society. In most of cases, Judicial Activism has occurred through public interest litigation. In public interest litigation any person or group can approach the Supreme Court and High Court for the redressal of grievances on behalf of the victim or victims who were incapable of approaching the court.

Under this new arrangement, a destitute citizen can file a writ petition even through a simple letter written on the post card. This derives authenticity from the “right be heard” as implied by Article 32 of the Constitution. But the court has to ensure that the petitioner who approaches the court with PIL, is acting bona- fide and not for personal gains private profit, political or other oblique considerations. The court should not allow this process to be abused by politicians and others to delay legitimate administrative action or to gain a political objective.

Question 8.
What is the Meaning of Independence of Judiciary? How is it ensured by the constitution?
Answer:
The Judiciary performs its Functions independently. The Legislature or the executive shall not interfere in the working of the Judiciary. The Judiciary carries on its obligations according to the constitutional norms and democratic principles.

The following measures have been taken by the constitution to ensure the independence of Judiciary in India.

Measures ensuring for Independence of Judiciary
1. The Legislature is not involved in the process of appointment of judges. Thus, it is believed that party politics would not play a role in the process of appoinments. In order to be appointed as judge, a person must have experience as a advocate and / or must be well versed in law. Political opinion of the person or his/her political loyalty should not be the criteria for appointments to judiciary.

2. The judge have a fixed tenure. They hold office till reaching the age of retirement. Only in exceptional cases judges may be removed as per the procedure prescribed in the Constitution. This measure ensures that judge could function without fear % or favor.

3. The Judiciary is not financially dependent on either the executive or legislature. The Constitution provides that the salaries and allowances of the judges are not subjected to the approval of the legislature.

4. The actions and decisions of the judges are immune from personal criticism. The judiciary has the power to penalize those who are found guilty of contempt of court. This authority of the court is seen as an effective protection to the judges from unfair criticism.

5. Judiciary in India is neither a branch Of the executive nor a hand-made of the legislature. It has an independent identity under the Constitution.

6. Our Constitution prescribes specific and high qualifications for the judges. Thus, only those persons who have specific qualifications and experience as prescribed by the Constitution can be appointed as Judges of Supreme Court.

7. Security of the service for the Judges is an essential quality for securing the independence of Judiciary. No Judge can be removed from the office except by impeachment and only on the grounds of proven misbehavior or incapacity.

8. The Judiciary in India enjoys the vast jurisdiction. It is no way subordinate to the other organs of the government. Its decisions bind all. Such a powerful position helps the Judiciary to maintain its independence.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 9.
What are the Power and Functions of the Attorney-General of India? [Mar. 18, 16]
Answer:
Article 76 of our Constitution provided for the office of the Attorney General of India. The Attorney General is the highest law officer of the Union Government. He is appointed by the President. He holds the office during the pleasure of the President. He is entitled to all privileges and immunities allowed to a Member of Parliament. When he attends sessions of the House, he occupies a seat on the treasury government benches.

Qualifications :
The Attorney General of India Possess the same qualifications that are necessary for a judge of the Supreme Court. They are as follows.

  1. He must be a citizen of India.
  2. He must have served as a judge in some High Court for a period of at least five years.
  3. He must have served as an advocate in some High Court for a period of at least ten years.
  4. He must be a distinguished jurist in the opinion of the President.

Pay and Allowances :
The Attorney General of India is paid not a salary but a remuneration that is determined by the President. The remuneration of Attorney General is equal to salary of a judge of the Supreme Court.

Removal:
He may quit his office by submitting his resignation to the President. He can be removed by the President in case a special address is passed charging him with ‘proved misbehavior’ or ‘incapacity’ by each House of Parliament with its absolute majority and with two-thirds majority of the members present and voting.

Powers and functions :
The Constitution assigned some specific powers and functions to the Attorney General of India. They are mentioned as follows :

  1. The Attorney General of India render advice to the Union Government upon such legal matters which are referred to him by the President.
  2. He performs such other functions of legal character that are assigned to him by the President from time to time.
  3. He discharges the Functions conferred on him by the constitution or any other Laws.
  4. He appears in any court of Law on behalf of the union government in all. cases.
  5. He represents the government in any reference made by the president to the Supreme Court.
  6. He appears in any High, Court on behalf of the union government.

Very Short Answer Questions

Question 1.
Qualifications of Judges of Supreme Court.
Answer:
A person to be appointed as a judge of the supreme court shall possess the following qualifications :

  1. He should be a citizen of India.
  2. He should have continuously worked as a judge in one or more High Courts at least for a period of 5 years.
  3. He should have continuously worked as an advocate of one or more High Courts for not less than 10 years or
  4. He should be a distinguished jurist in the opinion of the president of India.

Question 2.
Removal of the Judges of Supreme Court.
Answer:
A judge of the Supreme Court can be removed from his position only on the grounds of proved misbehavior or incapacity. He can be removed from his office by an order of the president, after an address from each house of Parliament, supported by a majority of the total membership of that house and by a majority of not less than 2/3 notes of the members present and voting passed.

Question 3.
Judicial Review. [Mar. 16]
Answer:
Judicial Review means the power of the Supreme Court or High Court to Examine the constitutional validity of the legislative enactments and executive actions of both central and state governments and to declare them ‘null and void’ if found repugnant of the provisions of the constitution.

Question 4.
Court of Record.
Answer:
According to Article 141, Supreme Court acts as court of Record. Being the highest court of the land, its proceedings acts and judgements are kept in record for perpetual memory and further verification and reference.

Question 5.
Judicial Activism.
Answer:
Generally, Judicial Activism is construed to be the over willingness of the judiciary to jump into the arena of executive or legislative functions. In general parlance, the expression “activism” means “being active”, “doing things with decision” and the expression “activist” should mean “one who favours intensified activities”. Judicial Activism is a policy making in competition with policy making by legislature and executive.

The essence of true Judicial Activism is rendering of decisions which are in tune with the temper and tempo of the times. The nature of Judicial Activism is that it furthers the cause of social change or articulates concepts like liberty equality or justice.

Question 6.
PIL.
Answer:
In Public Interest Litigation any person or group can approach the Supreme Court and High Court for the redressal of grievances on behalf of the victim or victims who were incapable of approaching the court. Under this new arrangements a destitute citizen can file a writ petition even through a simple letter written on the post card.

Question 7.
Independence of Judiciary.
Answer:
The Judiciary perform its functions independently. The legislature or the executive shall not interfere in the working of the judiciary. The Judiciary carries on its obligations accordingly to the constitutional norms and democratic principles.

Question 8.
Habeas corpus.
Answer:
Literally means ’to have the body of. It is issued by the court to affect the release of a person who has not been detained legally. Under this writ the court issues orders to the concerned authority to produce the person before the court. The failure to abide by the writ order is. met with punishment for contempt of court.

Question 9.
Seat of the Supreme Court. [Mar. 18, 17]
Answer:
The Supreme Court ordinarily shall sit at New Delhi. The Supreme Court of India was inaugurated on January 28, 1950. All general cases are adjudicated by a division Bench comprising two or more judges. Cases involving the constitutional matters are heard by a constitution bench consisting five judges. For considering special causes larger benches consisting of five or more than five judges are constituted.

AP Inter 2nd Year Civics Study Material Chapter 5 Union Judiciary

Question 10.
Writs.
Answer:
The word ‘writ’ literally means ‘order’ in written form. Article 32 of our constitution, confers authority upon the supreme court, to issue a constitutional writ for the enforcement of fundamental rights of the citizens. Any person whose fundamental rights have been violated, can directly move the supreme court for remedy. The Supreme Court issues harbeas corpus mapdomus prohibition quo-warranto and certiorari for enforcing the fundamental rights.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 4th Lesson Union Legislature Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 4th Lesson Union Legislature

Long Answer Questions

Question 1.
Describe the powers and functions of the speaker. [Mar. 18]
Answer:
Articles 93 to 97 of the Indian Constitution deal with the office of the Speaker of the Lok Sabha. The Speaker acts as head of the Lok Sabha, guardian of members, and principal spokesman of the house. He enjoys Supreme Authority and power on the floor of the house.

Election :
The members of the Lok Sabha elect the Speaker from among themselves. According to the Parliamentary convention the speaker is unanimously elected or chosen by the members on the request of the Prime Minister.

A person elected as the speaker must be a member of the Lok Sabha.

Tenure:
The speaker continues in office for five years. Though the Lok Sabha is dissolved the speaker continues in office until the new Lok Sabha elects its speaker. (Article 94).

Removal:
The speaker can be removed from office by a majority members, resolution, proceeded by a 14-day prior notice to that effect.

Salary and. allowances :
At present, the speaker receives a monthly salary of ₹ 1,40,000. Besides he is provided with rent free accommodation, Medical, travelling, and telephone facilities.

Powers and functions of the speaker :

  1. The speaker presides over the meetings of the Lok Sabha. He conducts the meetings with dignity, order and efficiency.
  2. He allots time to the members to express their views on the bills, conducts voting if necessary and announces the results.
  3. He sends bills to the Rajya Sabha after they are approved by the Lok Sabha. On the receipt of the Bills from the Rajya Sabha, he certifies and sends them to the President of India for his consent.
  4. He acts as the representative of the Lok Sabha. He sends messages and directives to the members on behalf of the Lok Sabha.
  5. He takes steps for safeguarding the rights and privileges of the members and for upholding the respect of the house.
  6. He has the privilege of determining whether a bill is money bill or not.
  7. He accords permission to the members for introducing various bills in the house. He gives his signature on the bill approved by the house.
  8. He is empowered to permit the members to move a No-confidence motion against the government, postpone the meetings of the house and decide the Quorum in the house.
  9. He constitutes various house committees and appoints their chairpersons.
  10. He presides over the joint session of the Parliament.
  11. He exercises his casting vote in case of a tie over a bill. ,
  12. He conducts the election of the Deputy Speaker in case of a vacancy.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 2.
Explain the powers and functions of the Union Legislature. [Mar. 16]
(Or)
Write about the composition powers and functions of the Indian Parliament.
Answer:
Introduction :
The Union Legislature (Parliament) is the highest legislative organ of the Union government. Articles 79 to 129 in part V of Indian Constitution deals with the composition, organization, powers, and functions of the Indian Union Legislature.

Composition :
Indian Parliament consists of the

  1. President
  2. Rajya Sabha (Council of states)
  3. Lok Sabha (House of people)

The upper house Rajya Sabha represents the states and union territories.

The lower house Lok Sabha represents the people.

The President of India has the power to summon or prorogue the two houses of Parliament though he is not a member of either house.

He can dissolve the Lok Sabha on the advice of the Union Council of Ministers headed by the Prime Minister.

Powers and functions of the Union Legislature (or) Indian Parliament :
The Parliament enjoys extensive powers and performs variety of functions. These powers and functions are under the following points.

1) Legislative Powers and Functions:
The main function of the Indian Parliament is law making. It makes laws on all the subjects mentioned in the Union List and Concurrent List. Under certain circumstances it also makes laws on the subjects mentioned in the State List. Further, it also makes laws on the matters that are not included in any of the three Lists i.e., on residuary matters.

2) Executive Powers and Functions :
Another important function of the Indian Parliament is controlling the Executive (Union Council of Ministers). Parliament controls the Executive through various ways, such as by asking questions, supplementary questions, and by introducing adjournment motions and no confidence resolutions against the Ministry. Hence the survival of the Government depends upon the will of the members in the Lower House. The executive remains in office so long as it enjoys the confidence of the Lok Sabha.

3) Financial Powers and Functions :
The Parliament controls the financial resources of the nation. It accepts the budget and other money bills required by the government. Its permission is needed for the government for imposing and collecting tax and for revising the existing tax rates. In this regard the Lok Sabha has more financial powers than the Rajya Sabha. All money, bills shall at first be introduced in the Lok Sabha. The Rajya Sabha has to accept all money bills sent by the Lok Sabha within 14 days. It approves the railway budget, appropriation bill, and other money bills.

4) Judicial Powers and Functions :
The Parliament has certain judicial powers and functions. It has the power to remove the President and Vice President. The procedure is called impeachment. It has also the power to recommend to the President the removal of the higher officials of the country such as the Chief justice and Judges of Supreme Court, High Court and the Chairman, and other members of U.P.S.C., Chief Election Commissioner etc., for violation of certain principles.

5) Constitutional Powers and Functions:
The Parliament takes initiative for changing the provisions of the Constitution according to the changing times. Bills relating to the Constitution amendments may be introduced in either House. The State legislatures also join with the Parliament in accepting certain important Constitutional amendment bills. There are three methods of amending the Constitution.

6) Electoral Powers:
The Parliament also serves as an electoral college. It participates in the election of the President and Vice President. The Speaker and Deputy Speaker who act as the presiding officers are elected by the members of Lok Sabha. The Deputy Chairman is elected by the members of the Rajya Sabha.

7) Deliberative Powers and Functions :
The Parliament acts as the highest forum and direct agency of public opinion. Its members discuss various issues of national and international significance. They demand the government to solve the people’s problems.

8) Miscellaneous Powers :
In addition to the above, the Indian Parliament has also the power to a) create or abolish Legislative Councils b) change the names and boundaries of the States, etc.

Conclusion:
A look at the powers and functions of the Indian Parliament it is the centre of legislative activity and political activity of our country.

Question 3.
Examine the role of financial committees in Parliament.
Answer:
The Financial committees of Parliament endeavour of undertake the task of detailed scrutiny of governmental spending and performance, there by securing the accountability of the administration to the Parliament in financial matters.
There are three financial committees in Indian Parliament. They are :

  1. Public Accounts committee
  2. Estimates committee and
  3. Committee on public undertakings

1) Public Accounts Committee :
Public Accounts Committee was set up on 1921. It consists of 22 members. Out of them, 15 members belong to the Lok Sabha and 7 members to the Rajya Sabha. Their term of office is one year. They are elected by means of proportional representation and single transferable vote. The speaker nominates one of the members as the chairman of the committee. It became a convention to appoint one of the members of the opposition in the Lok Sabha as its chairman since 1967-68. The Committee performs the following functions :

  1. The Committee examines the annual audit reports of the Comptroller and Auditor General (CAG) of India.
  2. It examines Public Expenditure not only from legal and formal point of view to discover technical irregularities but also from the point of view of economy prudence, wisdom, and property.
  3. It brings out the appropriation accounts and the finance accounts of the Union Government and any other accounts laid before the Lok Sabha.
  4. It examines whether the public funds are disbursed properly.
  5. It examines the accounts of Autonomous and Semi-Autonomous bodies, the audit of which is conducted by the CAG.
  6. It investigates the money spent on any service during the financial year in the excess of the amount granted by the Lok Sabha for that purpose.

The Comptroller and Auditor General renders assistance to this committee. The members of this committee carry out a country-wide tour and meet the concerned officers non-officials, people and receive petitions from them and it submits a final report to the Speaker of the Lok Sabha.

2) Estimates Committee :
The origin of this committee can be traced to the Standing Financial Committee setup in 1921. The Estimates Committee was at first constituted in April 1950 in free India. It consists of 30 members of the Lok Sabha. The Rajya Sabha has no representation in this Committee. These members are elected by the Lok Sabha every year from among its own members according to the principle of proportional representation by means of single transferrable vote. The members hold their office for a year.

The Speaker ha appoints the Chairman of the Committee. If the Duty Speaker is a member ttee, the Speaker appoints him as the Chairman of the Committee. One third of the total members belong to newly elected members. The Chairman of the Committee will be appointed invariably from the ruling party. Mr. M. Ananthasayanam Ayyangar acted the first chairman of this committee in the first Lok Sabha.

The members of this-committee may visit different projects and hold discussions with the officers, non-officials, business groups and receive suggestions from them. The committee functions on permanent basis.

The committee performs the following four important functions :

  1. It offers suggestions in regard to the economy in expenditure, improvement in organization and efficiency of the Union Government.
  2. It examines as to whether the public funds are disbursed as per the estimates.
  3. It also examines the matters assigned to it by the Speaker of the Lok Sabha.
  4. It examines whether the money is well laid out within the, limits of the policy implied in the estimates. Hence, it has been described as a ‘continuous economy committee’.

3) Committee on Public Undertakings :
The Committee on Public Undertakings was created in 1964 on the recommendations of Krishan Menon Committee. It consists of 22 members out of which 15 are from Lok Sabha and 7 from Rajya Sabha. The members of the committee are elected by the Parlianient every year from amongst its own members according to the principle of proportional representation by means of single transferable vote.

The purpose of the committee is to lighten the burden of Public Accounts Committee. The Chairman of the committee is appointed by the speaker from amongst its members who are drawn from the Lok Sabha only. The members of Rajya Sabha are not be appointed as a Chairman of the committee. The functions of the committee are :

  1. To examine the reports and accounts of Public Undertakings.
  2. To examine the reports of the Comptroller and Audit of General on undertakings.
  3. To examine whether the affairs of the public undertakings are being managed in accordance with sound business principles and practices.
  4. To exercise such other functions vested in the Public Accounts Committee and the Estimates Committee.

Short Answer Questions

Question 1.
Write about the composition of the Lok Sabha.
Answer:
The Lok Sabha or the House of the people is the lower house in Indian Parliament. Maximum strength of the Lok Sabha envisaged by the constitution is now 552 (530 members to represent states, 20 to represent union territories and 2 members of Anglo-Indian community, to be nominated by the President).

At present there are 545 members in the Lok Sabha out of them,

  • 530 members are elected from the states.
  • 13 members are elected from the union territories and the remaining.
  • 02 members are nominated by the President from the Anglo Indian community.

Out of 543 elected seats, 79 seats are reserved for the scheduled castes and 41 for the scheduled tribes.

The election is through direct franchise.
The tenure of the Lok Sabha is normally 5 years.

A person who wishes to contest as a candidate for the membership of the Lok Sabha must
A) Be an Indian citizen.
B) Have completed 25 years of age.
C) Not hold any office of profit in union, state or local governments.
D) Possess such other qualifications as prescribed by the Parliament.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 2.
Explain the election of the speaker of the Lok Sabha.
Answer:
The members of the Lok Sabha elect the Speaker from among themselves. According to the Parliamentary convention, the speaker is unanimously elected or chosen by the members on the request of the Prime Minister.

When no single party secures majority or when a coalition Ministry is formed, the coalition Ministry is formed, the coalition partners will make efforts for deciding the candidature for the office of the speaker. Sometimes coalition partners may hand over that office to a candidate selected by the parties that declare support from outside. A person elected as the speaker must be a member of the Lok Sabha.

Question 3.
What do you know about the composition and qualifications of members . of the Rajya Sabha?
Answer:
The Rajya Sabfya is the upper chamber in Indian Parliament. Article 80 stipulates that the Rajya Sabha shall consist of
A) 12 members nominated by the President and
B) Not more than 238 representatives of the states and of the union territories.

Thus its maximum membership shall be 250 only.
At present there are 245 members in the Rajya Sabha of them

  • 229 members belong to the elected from the 29 states.
  • 3 members belong to the National capital territory of Delhi.
  • 1 member represents the union territory of Pondicherry and the remaining.
  • 12 members nominated by the President having practical experience in respect of matters such as literature, science, arts and social service.

The members are elected in accordance with the proportional representation by means of single transferable vote system.

Qualifications :

  1. He shall be a citizen of India.
  2. He shall have completed 30 years of age.
  3. He should not hold any office of profit under union, state or local government.
  4. He shall possess such other qualifications as prescribed by the Parliament.

The Raja Sabha is a permanent house of which l/3rd of its total members shall retire for every two years.

Question 4.
Write a note on the chairman and Deputy chairman of the Rajya Sabha.
Answer:
Chairman of Rajya Sabha:
The Presiding Officer of Rajya Sabha is popularly known as the Chairman. The Vice President of India acts as the Ex-Chairman of the Rajya Sabha. He is not a member of the House. The members of Parliament elected him for every five years as the Vice-President of India. It implies that both the members of the Lok Sabha and the Rajya Sabha cast their vote in the Vice-Presidential election.

At present he is paid ₹ 1,40,000/- towards monthly salary and allowances. His salaries and allowances are charged on the Consolidated Fund of India. The Chairman of Rajya Sabha vacates his office only if he is removed from the office of the Vice President.

Deputy Chairman:
The Deputy Chairman of Rajya Sabha is elected by the members of the Rajya Sabha amongst the members. The Deputy Chairman receives a monthly salary of ₹ 90,000/-. In the absence of the Chairman, the Deputy Chairman acts as Chairman and presides over the meetings of the Rajya Sabha. Whenver the office of the Deputy Chairman falls vacant, the members of Rajya Sabha will elect another member to fill the vacancy.

Question 5.
Mention any three powers and functions of Indian Parliament.
Answer:
The Indian Parliament, the law body in our country, has extensive powers and performs a variety of functions. There are as follows : ‘

1) Legislative Powers :
The main function of the Indian Parliament is law making. It makes laws on all the subjects mentioned in the Union List and Concurrent List. Under certain circumstances it also makes laws on the subjects mentioned in the State List. Further, it also makes laws on the matters that are not included in any of the three Lists i.e., on residuary matters.

2) Executive Powers :
Another important function of the Indian Parliament is controlling the Executive (Union Council of Ministers) . The members exercise control over the Executive by asking questions, supplementary questions, and by introducing adjournment motions and no-confidence resolutions against the ministry. The ministers are collectively responsible for their actions, to the Lower House of the Parliament i.e., Lok Sabha. They will be in office as long as they enjoy the confidence of the majority of members in the Lok Sabha.

3) Financial Powers :
The Parliament controls the financial resources of the nation. It accepts the budget and other money bills required by the government. It’s permission is needed for imposing and collecting taxes and for revising the existing tax rates. In this regard the Lok Sabha has more financial powers than Rajya Sabha.

Question 6.
Write a note on the types of bills.
Answer:
A Bill is a proposed law under consideration by a legislature. A bill does not become law until it is passed by the legislature. Once a bill has been enacted into law, it is called on Act or statute.

Bills introduced in the Parliament are of two typek 1) Public bill (Government bill) and 2) Private bills (Private member’s bills) the public bills are introduced by the Ministers in the Parliament whereas private bills are introduced by any member of Parliament other than a Minister. The bills introduced in the Parliament may also be classified into four categories.

1) Ordinary bill:
Ordinary bills are concerned with any matter other than financial subjects.

2) Money bill:
Money bills are concerned with the financial matters like taxation, public expenditure etc.

3) Finance bill:
The financial bills are also concerned with financial matters but are different from money bills. These bills deal with fiscal matters i.e., revenue of, expenditure.

4) Constitutional Amendment Bill: Constitutional Amendment bills are concerned with the provisions of the constitution.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 7.
Explain briefly about the stages of law-making procedure in Indian Parliament.
Answer:
Law making is an important function of Indian Parliament. The Parliament has the power to pass all acts. A bill becomes an act after it receives the assent of the President. Every bill has to pass through different stages as described below.

1) First Reading:
A bill may be introduced by any member of Parliament. One has to ask for the leave of the House to introduce a bill. The title of the bill is to be read out. If the bill is voted for it is deemed to have been read first time and is published in the Gazette of India.

2) Second Reading:
The printed copies of the bill are distributed to all the members at this stage. It may be moved that the bill be referred to a select committee or that the bill be circulated for public opinion. There will be a general discussion on the main principles of the bill at this stage.

3) Committee Stage :
If the House approves the principles, the bill is referred to and examined by the select committee. Then it is thoroughly discussed clause by clause.

4) Report Stage :
The report of the committee with suggestions is presented to the House. If the House agrees to consider the bill as reported by the select committee the bill is taken up for clause by clause discussions and members may move amendments.

5) Third Reading:
The bill enters for the third reading. If it is approved by the House, only oral amendments are allowed at this stage. If the House accepts the bill it is deemed to have been passed by the House.

6) Consideration by the other House :
When the bill is passed by the House, it is sent to the other House for consideration. The procedure in one House is repeated in the other House. If the second House disagrees, a joint sitting is arranged to resolve the differences.

7) Assent by President:
If the bill is passed by both the Houses of Parliament, it is sent to the President for his assent. After the bill is given assent, it becomes an act which will be implemented by the Executive.

The President sometimes send a bill passed by the Parliament for reconsideration. The suggestions sent by the President along with the bill have to be taken up for discussion by the Parliament immediately. If the bill is passed second time by the Parliament, then the President has to give his assent to the bill.

Question 8.
What do you know about Public Accounts Committee.
Answer:
Public Account’s Committee was set up in 1921. It consists of 22 members. Out of them 15 members belong to the Lok Sabha and 7 members to the Rajya Sabha. Their term of office is one year. They are elected by means of proportional representation and single transferable vote. It became a convention to appoint one of the members of the Opposition in the Lok Sabha as its chairman since 1967-68. The Committee performs the following functions :

  1. The Committee examines the annual audit reports of the Comptroller and Auditor General (CAG) of India.
  2. It examines Public Expenditure not only from legal and formal point of view to discover technical irregularities but also from the point of view of economy prudence, wisdom and property.
  3. It brings out the appropriation accounts and the finance accounts of the Union Government and any other accounts laid before the Lok Sabha.
  4. It examines whether the public funds are disbursed properly.
  5. It examines the accounts of Autonomous and Semi-Autonomous bodies, the audit of which is conducted by the CAG.
  6. It investigates the money spent on any service during the financial year in the excess of the amount granted by the Lok Sabha for that purpose.

Question 9.
Describe the composition and functions of Estimates Committee.
Answer:
The Estimates Committee was setup in 1950. It consists of 30 members of the Lok Sabha. The members hold their office for a year. The Speaker of the Lok Sabha appoints the Chairman of the Committee. If the Duty Speaker is a member of this Committee, the

Speaker appoints him as the Chairman of the Committee. The members of this committee may visit different projects and hold discussions with the officers, non-officials, business groups and receive suggestions from them. The committee functions on permanent basis.

The committee performs the following four important functions :

  1. It offers suggestions in regard to the economy in expenditure, improvement in organization and efficiency of the Union Government.
  2. It examines as to whether the public funds are disbursed as per the estimates.
  3. It also examines the matters assigned to it by the Speaker of the Lok Sabha.
  4. It examines whether the money is well laid out within the limits of the policy implied in the estimates. Hence, it has been described as a “continuous economy committee”.

Very Short Answer Questions

Question 1.
Composition of Indian Parliament.
Answer:
Indian Parliament consists of the i) President ii) Rajya Sabha (Council of states ) iii) Lok Sabha (House of people)
The upper house, Rajya Sabha represents the states and union territories.

The lower house, Lok Sabha represents the people.

The President of India has the power to summon or prorogue the two houses of Parliament though he is not a member of either house.

He can dissolve the Lok Sabha on the advice of the Union Council of Ministers headed by the Prime Minister :

Question 2.
Qualifications of Rajya Sabha member.
Answer:

  1. He shall be a citizen of India.
  2. He shall have completed 30 years of age.
  3. He should not hold any office of profit under union, state or local government.
  4. He shall possess such other qualifications as prescribed by the Parliament.

Question 3.
Quorum of Lok Sabha. [Mar. 18, 16]
Answer:
Quorum implies minimum attendance of members required for conducting the meetings of the Lok Sabha. Quorum is fixed at l/10th of the total membership. The speaker determines whether there is Quorum on a particular day for conducting the meetings.

Whenever there is no Quorum, he postpones the meetings for an hour or two or for the next day. There are several instances where in the meetings of the Lok Sabha were deffered due to lack of Quorum.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 4.
The Speaker of Lok Sabha.
Answer:
The office of the Speaker has great dignity, honour and authority. As the speaker acts as the chairman of the Lok Sabha and as the member of the Lok Sabha represents the people directly. He represents the whole Nation. He acts as the friend, philosopher and guide to the members.

Question 5.
Deputy Speaker of Lok Sabha.
Answer:
There will be a Deputy Speaker for conducting the proceedings of the Lok Sabha in the absence of the Speaker. The Deputy Speaker is elected by the members of the Lok Sabha from among themselves. The Deputy Speaker while acting as the presiding officer, enjoys all the powers and priveleges of the Speaker.

Question 6.
Committee on Public Undertakings. [Mar. 17]
Answer:
The Committee on Public Undertakings was created in 1964. It consists of 22 members out of which 15 are from Lok Sabha and 7 from Rajya Sabha. It examines whether the autonomy and efficiency of public sector undertakings are being managed in accordance with Sound business principles and prudent commercial practices.

Question 7.
Panel of Speakers.
Answer:
The Speaker nominates some of the members of the Lok Sabha as panel speakers. Maximum strength of panel chairpersons will be 10. If both the Speaker and Deputy Speaker are absent at particular time, one of the members from the panel of chairpersons will act as the Speaker. ,

Question 8.
Pro-tem Speaker.
Answer:
The President appoints the pro-tem Speaker for presiding over the meeting of the] first session of the Parliament after general elections. The pro-tem Speaker administers the oath of office on the elected members. Election to the office of the Speaker is held later. Pro-tem Speaker post is dissolved soon after the election of the new Speaker.

Question 9.
Question Hour. [Mar. 18, 16]
Answer:
In both houses of Parliament first hour is allotted to question hour. The members, by giving notice to the presiding officer, can ask questions pertaining to public issues, administrative inefficiency or about the role of the government.

Question 10.
Adjournment Motion.
Answer:
Adjournment Motion is tabled in the Parliament to draw attention of the house to a definite matter of urgent public importance and needs the support of 50 members to be admitted. If any member wants to introduce adjournment motion he should give in writing to the speaker, the Minister concerned and the Secretary General of Parliament before 10 A.M. on that day.

Question 11.
Whip [Mar. 17]
Answer:
Every political party whether ruling or opposition has its own whip in the Parliament. He is appointed by the concerned party to serve as an assistant floor leader. He is charged with the responsibility of ensuring the attendance of his party members in large numbers. He regulates and monitors their behaviour in the Parliament. The members are supposed to follow the directives given by whip, otherwise, disciplinary action can be taken against those members.

AP Inter 2nd Year Civics Study Material Chapter 4 Union Legislature

Question 12.
No Confidence Motion.
Answer:
According to Article 75 of the constitution, No Confidence Motion can be tabled in the Parliament when the Cabinet behaves in an irresponsible manner and if the ruling party does not enjoy majority. No confidence motion is introduced by the opposition parties through written notice supported by atleast 50 members and there will be a discussion on the motion. After the discussion there will be a voting. If the No-Confidence Motion is passed or approved In the house the cabinet has to resign.

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 3rd Lesson Union Executive Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 3rd Lesson Union Executive

Long Answer Questions

Question 1.
Explain the powers and functions of the President of India.
Answer:
Introduction:
The President of India is the constitutional head of the Indian Republic. He is the First citizen of India. He administers the affairs of the union Government either himself or through the officers subordinate to him. (Articles 52 and 53)

Qualifications :
A person to be eligible to contest the office of the President shall possess the following qualifications :

  1. He should be a citizen of India.
  2. He should have completed the age of 35 years.
  3. He should be qualified for election as a member of the Lok Sabha.
  4. He should not hold any office of profit under the union, state or local Governments (Article 59 (i))
  5. Possess such other qualifications as prescribed by the Parliament.

Election Procedure :
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both Houses of Parliament. State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Oath of office :
The person who is elected as President assumes office only after he takes oath of office and secrecy by the Chief Justice of India.

Term of office :
The President continues in office for five years from the date of his assumption of office.

Salary and Allowances :
The President now gets a monthly salary of ₹ 1,50,000/-. His official residence is Rashtrapathi Bhavan at New Delhi. On retirement, he will get a monthly pension of ₹ 75,000/-.

Removal (or) Impeachment :
The President can be removed from the office by a process of impeachment for violation of the constitution. Impeachment is a quasi-judicial procedure adopted by the Parliament.

Powers and Functions :
The President shall exercise his powers with the help of the Council of Ministers headed by the Prime Minister. His powers may be analysed under the following heads :

1. Executive Powers :
An executive action of the Union Government shall be expressed in the name of the President. The President appoints the Prime Minister and other Ministers, Attorney General, Comptroller and Auditor General of India, State Governors, Judges of the Supreme Court and State High Courts, Finance Commission, Chairman and members of U.P.S.C., Election Commission, and Chief Commissioners of Unit Territories. He allocates portfolios to the Ministers.

2. Legislative Powers :
The President is an integral part of Parliament (Art. 79) and as such enjoys extensive legislative powers.
They are :

  1. He summons from time to time each House of Parliament, adjours, and prorogues either or both the Houses.
  2. He addresses either House separately or both the Houses Jointly.
  3. He can dissolve the Lok Sabha on the advice of the Prime Minister.
  4. He opens the first session of Parliament after the General Elections and at the commencement of the first session of each year.
  5. He can send messages to the Parliament.
  6. He arranges a joint session of both the Houses when there is a dead-lock over an ordinary bill.
  7. All bills passed by Parliament require his assent for becoming in acts.
  8. He nominates 12 members to Rajya Sabha and two Anglo Indian members to Lok Sabha.
  9. He promulgates ordinances when the Parliament is not in session.
  10. He sends the annual reports of Finance Commission, U.P.S.C etc., for the consideration and approval of Parliament.

3. Financial Powers :
The President also enjoys some financial powers. They are :

  1. He recommends the financial bills to be introduced by the members in parliament. The Budget is caused to be laid down before the Parliament by the President.
  2. He operates the Consolidated Fund of India.
  3. He determines the shares of States in the proceeds of Income Tax.
  4. No Money Bill can be introduced in the Parliament except on his recommendations.
  5. He constitutes a Finance Commission for every five years etc.

4. Judicial Powers :

  1. The President can grant pardons, reprieves, respites or remission of punishments.
  2. He appoints the judges of the Supreme Court and State High Courts.
  3. He can also remove them on an address by the Parliament.

5. Military Powers:
The President is the Supreme Commander of the Defence Forces of the Union. He appoints the Chiefs on the Staff for Army, Navy, and Air Force. He can declare war and conclude peace. But he has to take the approval of Parliament.

6. Diplomatic Powers :
The president appoints Ambassadors to foreign countries to represent India. He receives the credentials of the Ambassadors appointed in India. He represents our Nation in International affairs. He makes treaties and agreements with other countries subject to the ratification by the Parliament.

7. Emergency Powers :
In extraordinary conditions, the President can proelaim emergency to safeguard the security, integrity, and independence of our country. They are of three types :

  • Emergency caused by war or external aggression or armed rebellion (Article 352).
  • Emergency due to failure of Constitutional machinery in the States (Article 356)
  • Emergency due to threat to the financial stability of India (Article 360).

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Question 2.
Write briefly the Emergency powers of the President of India. [Mar. 16]
Answer:
Articles 352 to 360 of part XVIII of Indian constitution deals with three types of emergency powers of the Indian President. They are :

  1. National Emergency,
  2. Constitutional Emergency,
  3. Financial Emergency

They may be explained as follows.
1) National Emergency : (Article 352)
The President exercises this power during the period of war, external aggression or armed rebellion. He declares emergency if he is satisfied that the sovereignty and security of India or any part thereof is threatened by external aggression.

When National Emergency is in force, the federal provisions of our constitution ceases to operate. So far, National Emergency was proclaimed on four occasions in our country. They are : 1. Chinese Aggression (1962), 2. Indo – Pak war (1965), 3. Indo – Pak war in the context of Bangladesh Liberation Movement (1971), 4. Opposition’s call for blocking Parliament (1975).

2) Constitutional Emergency : (Article 356)
Article 356 of Indian constitution empowers the President to proclaim the constitutional emergency. If the President, on receipt of a report from the Governor or other wise; is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency. It is also called as the President’s Rule. So far this type of emergency was proclaimed for over 100 times.

3) Financial Emergency : (Article 360)
If the President is satisfied that a situation has arisen where by the financial stability or credit of India is threatended, then he may proclaim financial emergency in the country. During the period of financial emergency, the President enjoys the following powers.

  • The President may reserve all the money bills or other financial bills of the state after they are approved by the state legislature.
  • He may reduce the salaries and allowances of all or any person serving in the states.
  • The President can reduce the salaries allowances of the persons working at the union level including the judges of the Supreme Court and the State High Courts. But so far the financial emergency has not been yet imposed in the country.

Question 3.
Discuss the powers and functions of the Prime Minister of India. [Mar. 18, 16]
Answer:
The Prime Minister is the real executive head of the Union Government. He occupies an important position in the administration of our country. Since India has a Parliamentary form of Government the real power rests with him. He is the ‘uncrowned king’ and “the keystone of the Cabinet arch in the Union Government”.

Qualifications :

  1. He should be citizen of India.
  2. He should have completed the age of 25 ytears.
  3. He should be qualified for election as a member of the Lok Sabha.
  4. He should not hold any office of profit under the union or state or local governments.

Appointment:
The President appoints the Prime Minister. Generally the President has to summon the leader of the majority party in the Lok Sabha to form the Ministry. If no party gets an absolute majority, the President can use his discretion and summon the leader of the party, who in his opinion can manage to form a ministry. Afterwards the Prime Minister will be asked to prove his majority in the Lok Sabha.

Oath of Office :
The President of India will administers the oath of office of the Prime Minister.

Term of Office :
The Prime Minister shall remain in office during the pleasure of the President. But actually he assumes his powers as long as he retains the confidence of the majority members in the Lok Sabha. He resigns when the Lok Sabha accepts a no-confidence motion against his ministry.

Salary and Allowances:
The salary and allowances of Prime Minister are decided by the Parliament from time to time. He gets his salary and allowances that are payable to a member of Parliament. At present the Prime Minister gets a salary and allowances of ₹ 1,60,000/- per month.

Powers and Functions :
The Prime Minister is the head of the union government. He is the real executive. The Council of Ministers cannot exist without the Prime Minister. His powers are explained here under.

1) Leader of the Union Cabinet:
The Prime Minister is the leader of the Union Cabinet and Union Council of Ministers. He selects some eminent members of his party in parliament and sees that they are appointed as ministers by the President. He has a free choice of both allocating portfolios and reshuffling the ministry. All the ministers are personally and politically loyal to the Prime Minister. He decides the agenda, of the cabinet meetings. Further, he presides over the cabinet meetings.

2) Leader of the Union Government:
The Prime Minister acts as the leader of the union government. The union executive (Union Council of Ministers) initiates its business after the swearing in ceremony of the Prime Minister. All the ministers in the union ministry assume their office, owe their position and exercise their powers along with the Prime Minister. Infact, the Prime Minister influences the nature and working of the union government. He not only has a clear understanding but holds complete control over the affairs of the union government. All the high-level officers and the entire ministry in the union government behave and act according to the wishes of the Prime Minister.

3) Leader of the parliament:
The Prime Minister acts the leader of the Parliament in India. He is primarily a member of Parliament. He extends co-operation to the presiding officers in the smooth conduct of the two Houses. He wields complete control over his party members in the Parliament. He ensures that his party members maintain discipline during the sessions of the Parliament. He informs out the cabinet decisions to the Parliament. He communicates the major domestic and foreign policies of the union government to the members of Parliament. He maintains rapport with the opposition leaders and discusses the major issues confronted by the nation with them.

4) Link between the President and the Council of Ministers :
The Prime Minister acts as the main link between the President and the Union Council of Ministers. It is his duty to communicate to the President about the decisions of the Union Council of Ministers. He furnishes the every information required by the President concerning the affairs of union government. All the ministers shall formally meet the President only with the consent of the Prime Minister.

5) Leader of the Majority Party:
The Prime Minister acts as the leader of the majority party or group in the lower House of Parliament. He participates in the meetings of the party and acquaints his party members on various issues and steps taken by his ministry in implementing the party promises. He utilizes the services of the senior party leaders in running the government. He acts as the main link between the part and the government.

6) Leader of the Nation :
The Prime Minister is the leader of the nation. He takes initiative in finding solutions to several problems in the internal matters of the country. He plays an important role in the development of the nation.

7) Maker of Foreign Policy :
The Prime Minister plays a dominant role in shaping the foreign policy of the nation. He keeps in touch with the developments in all countries. He meets Heads of various countries and maintains friendly relations with them.

8) Chairman of NITI Aayog :
The Prime Minister heads the NITI Aayog (National Institution for Transforming India) NITI Aayog means policy commission. It is a policy think tank of government of India that replaces planning commission which aims to involve the states in economic policy making in India. It will provide strategic and technical advice to the central and state governments. It will have a governing council comprising Chief Ministers of all the states and it governors of Union Territories. Union government set up the NITI Aayog on January 1,-2015.

Question 4.
Explain the composition, powers and functions of the Union Council of Ministers. .
Answer:
Article 74 (1) of the Constitution provides for a Council of Ministers at the Centre. It’s main function is to aid and advice the President in the performance of his duties. It consists of Prime Minister and other Ministers. It is this body which runs the entire administration of our country. It is thefreal Executive authority of the country. It functions on the ‘Principle of Collective Responsibility1. It holds office till it continues to enjoy the confidence and support of the Lok Sabha.

Formation of Council of Ministers :
The formation of the Council of Ministers starts with the appointment of the Prime Minister. The President appoints the Prime Minister and on the advice of the Prime Minister, the other Ministers are appointed by the President.

Composition of Council of Ministers :
Our Constitution did not mention the exact size of the Union Council of Ministers. But there are three kinds of Ministers:

  1. Cabinet Ministers
  2. Ministers of State
  3. Deputy Ministers.

The Cabinet Ministers are entrusted with the maintenance of some important ministries. They enjoy independence and decision making powers.

The Ministers of State act as the heads of some important sections in the ministry. They are directly responsible to the Prime Minister for their activities.

The Deputy Ministers have no independent and discretionary powers. They assist the Cabinet Ministers and perform the functions assigned to them.

The Cabinet or the Council of Ministers is the pivot around which the entire administration of our country revolves. “It is the steering wheel of the ship of the State.” It is the hyphen that joins the Executive and Legislative organs of the Government.

Powers and Functions :
1. Executive Powers :
The Union Cabinet is a deliberative and policy formulating body. It discusses and decides all National and International policies of the country. The policies decided by the cabinet are carried out by the Ministers. It directs and leads the Parliament for action and gets its approval for all its policies. It coordinates and guides the activities of departments of the Government. It also plays an important role by suggesting persons for all major appointments. It considers the reports of various committees before they are presented to the Parliament.

2. Legislative Powers :
The Cabinet plans the legislative programme of the Government at the beginning of each session of Parliament. It drafts Bills on all important matters and introduces them in the Parliament. It also decides the time of summoning and prorogation of Parliament. The inaugural speech of the President to the Parliament is prepared by the Cabinet.

3. Financial Powers:
The Cabinet possess important financial powers. It has complete control over national finance. It prepares the Union Budget. It decides what taxes are to be imposed and how much of expenditure is to be incurred. Money Bills are always introduced in the Lok Sabha by the Finance Minister.

4. Foreign Relations :
In the field of foreign relations also the Cabinet plays an important role. It determines and formulates the foreign policy of the country and decides India’s relations with other countries. It considers and approves all international treaties and agreements.

Collective Responsibility :
Article 75(3) of Indian constitution stated that the union council of Ministers shall be collectively responsible to the Lok Sabha, for all their acts of commissions and commissions. They act as a team under the leadership of the Prime Minister. They sail together, they swim together and they sink together.

Conclusion:
It is thus clear that the Council of Minister or Cabinet enjoys far reaching powers both with regard to the internal and external policies of the country. Internally it maintains law and order within the country and externally protects the country from foreign aggression. The progress of the country largely depends upon the ability of the Cabinet.

Short Answer Questions

Question 1.
How is the President of India elected?
Answer:
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both Houses of Parliament. State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Each member of the Electoral College has one vote. But the value of the vote differes from State to State. The value of the vote of a Parliament member also differs from the value of the vote of a member of State Assembly.

The value of the vote of an M.L.A is worked out as follows. The total population of the State is divided by the total number of elected members of the Assembly. The quotient thus obtained is to be divided by 1,000. Fractions of half or more should be counted as one and added to the quotient. If it is less than half, it is ignored. This may be shown as follows.

  1. Value of vote of an M.L.A = Population of State / number of elected members of the Assembly / 1,000.
  2. Value of vote of M.P. = Total value of votes of all Assembly members / Total number of elected members of both Houses of Parliaments.

This method is followed to keep the election of the President above narrow political considerations.

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Question 2.
Write briefly about the procedure of impeachment of President
Answer:
The President may be removed from the office for violation of the constitution by a process of impeachment. Impeachment is a quasi-judicial procedure adopted by the Parliament. Either House of Parliament shall prefer the charge for removal of the President. The other House shall investigate into the charges itself or cause the charge to be investigated.

There are four stages in the impeachment of the President.

Firstly the impeachment resolution has to be moved with a 14 days prior notice in writing signed by not less than 1/4th of the total members of that House. Such a resolution has to be passed by a majority of not less than 2/3rds of the total members of the House.

Secondly, the resolution approved by the first House will be sent to the second House for consideration and approval.

Thirdly, the second House investigates into the charges directly or constitutes a committee to enquire into the charges. The President has the right to present his views directly or through a deputy during such enquiry.

Fourthly, if the charges against the President are established and adopted by the second House with 2/3rds majority of the total members, the President stands removed from the office. With regard to voting on the resolution for impeachment, only the elected members cast their vote. No president has so far been impeached in our country till today.

Question 3.
Mention any two Emergency powers of the Indian President.
Answer:
1) National Emergency : (Article 352)
The President exercises this power during the period of war, External aggression or armed rebellion. He declares emergency if he is satisfied that the sovereignty and security of India or any part thereof is threatened by external aggression.

When National Emergency is in force, the federal provisions of our constitution ceases to operate. So far, National Emergency was proclaimed on four occasions in our country. They are :

  1. Chinese Aggression (1962) ‘
  2. Indo – Pak war (1965) .
  3. Indo – Pak war in the context of Bangaldesh Liberation Movement (1971)
  4. Oppositions call for blocking Parliament (1975)

2) Constitutional Emergency : (Article 356)
Article 356 of Indian constitution Empowers the President to proclaim the constitutional emergency. If the President, on receipt of a report from the Governor or other wise, is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency, It is also called as the President’s rule. So far this type of emergency was proclaimed for over 100 times.

Question 4.
Explain the role and position of the President in Union Government.
Answer:
The President of India is the head of the union executive. He is the first citizen of India. He could exercise many powers as enstined in the constitution in the following manner.

Position of the President:
There are different opinions on the actual position of the president of India in the administration of our country. The farmers of the constitution wanted him to be a nominal Head of the state.

Dr. Ambedkar compared his position to that of the British King Sri. M. C. Setalved, the farmer Attorney General of India mentioned that the position of the President of India is like the king in England and the Governor General in a Dominion. Sri Alladi Krishna Swamya Ayyar also said that it was perfectly clear that our presidents position was similar to that of the constitutional Monarch in England.

Jawaharlal Nehru, the first Prime Minister of India, said that “We have not given our President any real power but we have made his position one of great authority and dignity.” These opinions make it clear that our President is only a nominal figure head and he does not have any real powers. This is confirmed by the 42nd Amendment of the constitution of India.

However, the President exercises independent powers under some conditions. He utilises these powers in regard to the appointment of the Prime Minister, dissolving the Lok Sabha and ordering midterm poll to the Lok Sabha. Some presidents like Sanjiva Reddy, Zail Singh, R.Venkata Raman, Dr. S.D. Sharma etc., utilised their discretionary powers where there was political instability or Hung Parliament after the general elections in the country. Like the Monarch of England, he still enjoys three rights the right to be ! consulted the right to encourage and the right to warn.

Question 5.
Write about any two powers of the Vice – President of India.
Answer:
The Vice President of India occupies the second highest position in the union government. He is accorded a rank next to the President.

Qualifications :
A person to be eligible as vice-president should possess the following qualifications.

  1. He should be a citizen of India.
  2. He should have completed 35 years of age.
  3. He should be qualified for election as a member of the eousil of states.
  4. He should not hold any office of profit under the union, state or local Governments in India.

Election :
The election of the vice – president like that of the President shall be indirect and in accordance with the system of proportional representation by means of the single transferable vote system. He is elected by the members of an electoral college consisting of the members of both the houses of Parliament.

Term of Office :
The Vice-President holds office for a term of 5 years from the date on which he enters his office.

Removal:
The Vice President may be removed from his office by a resolution of the council of states passed by a majority of all the members of the council and agreed to by the house of the people.

Salary and Allowances :
The Vice President of India receives a monthly salary of ₹ 1,25,000/-in addition, he is entitled to daily allowance, free furnished residence, medical, travel and other facilities.

Powers and Functions :
The Vice – President is the ex-officio Chairman of the Rajya Sabha. As. such he enjoys the same powers like the Speaker of Lok Sabha, such as (1) presiding over the meetings of Rajya Sabha, (2) maintaining discipline, decency and decorum in the House, (3) exercising casting vote in case of a tie, (4) admitting visitors, (5) protecting the privileges and rights of the members. He has no right with regard to money bills.

Acting as President of India :
He discharges the functions of the President during the temporary absertce of the President. He may take over the office of the President under 4 situations like (1) Death of the President, (2) Resignation of the President, (3) Removal of the President, (4) Inability of the President due to absence, illness or any other cause.

Question 6.
How is the Prime Minister appointed?
Answer:
Article 75 (1) of the Indian constitution deals with the appointment of the Prime Minister of India.

Appointment:
The constitution simply lays down that the Prime Minister shall be appointed by the President. After the conduct of General Elections to the Lok Sabha, the President has to invite the majority party leader of the Ldk Sabha to form the Government.

When no single party is able to secure majority seats in Lok Sabha, the President invites the leader of a coalition to form the Government. The president uses his discretionary powers in this regard. The President appoints the leader of the coalition as the Prime Minister on the condition that he has to prove his majority in the Lok Sabha within a specified period. Being the leader of the majority in Lok Sabha to be the Prime Minister, the person has to be a member of Parliament. If he is not a member at the time of appointment, he has to acquire it within six months from the date of his appointment as Prime Minister.

The powers of the President in choosing, inviting and appointing the Prime Minister cannot be questioned in any court of Law.

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Question 7.
Explain the role of the Prime Minister in Union Government.
Answer:
The prime Minister plays a predominant role in the affairs of the union government. He will have an indelible impression on every one in the administration of the union government. He is described as the Primus Interparus (first among equals). His role as the leader of the Union Council of Ministers, Union Cabinet, Party in power, Lok Sabha, Nation and as the link between the President and the Union Council of Ministers is unique. He wields tremendous political power and patronage. He enjoys enormous powers and fulfils innumerable tasks. It all depends on the image, influence, stature, and personality of the Prime Minister in the union government.

Jawaharlal Nehru, Dr. Ambedkar, and other eminent leaders of the National Movement and members of the Constituent Assembly described the Prime Minister as the linchpin of the union government. It is in this context that Sir william Harcourt remarked that every one expects from the Prime Minister dignity and authority, firmness to control, tact, practice and firmness, an impartial mind, a tolerant temper, a kind and prudent counsellorship, and accessibility to the people.

Question 8.
Describe the composition and powers of the Union Council of Ministers.
Answer:
Composition of Council of Ministers :
Our Constitution did not mention the exact size of the Union Council of Ministers. But there are three kinds of Ministers: 1) Cabinet Ministers 2) Ministers of State 3) Deputy Ministers.

The Cabinet Ministers are entrusted with the maintenance of some important ministries. They enjoy independence and decision making powers.

The Ministers of State act as the heads of some important sections in the ministry. They are directly responsible to the Prime Minister for their activities.

The Deputy Ministers have no independent and discretionary powers. They a&ist the Cabinet Ministers and perform the functions assigned to them.

The Cabinet Or the Council of Ministers is the pivot around which the entire administration of our country revolves. “It is the steering wheel of the ship of the State”. It is the hyphen that joins the Executive and Legislative organs of the Government.

Powers and Functions :
1) Executive Powers :
The Union Cabinet is a deliberative and policy formulating body. It discusses and decides all National and International policies of the country. The policies decided by the cabinet are carried out by the Ministers. It directs and leads the Parliament for action and gets its approval for all its policies. It co-ordinates and guides the activities of departments of the Government. It also plays an important role by suggesting persons for all major appointments. It considers the reports of various committees before they are presented to the Parliament.

2) Legislative Powers :
The Cabinet plans the legislative programme of the Government at the beginning of each session of. Parliament. It drafts Bills on all important matters and introduces them in the Parliament. It also decides the time of summoning and prorogation of Parliament. The inaugural speech of the President to the Parliament is prepared by the Cabinet.

3) Financial Powers :
The Cabinet possesses important financial powers. It has complete control over national finance. It prepares the Union Budget. It decides what taxes are to be imposed and how much of expenditure is to be incurred. Money Bills are always introduced in the Lok Sabha by the Finance Minister.

4) Foreign Relations :
In the field of foreign relations also the Cabinet plays an important role. It determines and formulates the foreign policy of the country and decides India’s relations with other countries. It considers and approves all international treaties and agreements.

Very Short answers

Question 1.
Composition of the Union Executive.
Answer:
The constitution of India provides for the Union Executive. Articles 52 to 78 in part V of the constitution deal with the union executive. The Union Executive consists.
i) The President
ii) The Vice-President
iii) The Prime Minister and
iv) The Union Council of Ministers

Question 2.
Qualifications required for contesting the Presidential elections.
Answer:
A person to be eligible to contest the office of the president shall possess the following qualifications.

  1. He should be a citizen of India.
  2. He should have completed the age of 35 years.
  3. He should be qualified for election as a member of the Lok Sabha.
  4. He should not hold any office of profit under the Union, State or Local Governments.
  5. Possess such other qualifications as prescribed by the Parliament.

Question 3.
Election of President.
Answer:
The President of India shall be elected indirectly by an electoral college consisting of the elected members of both houses of Parliament, State Legislative Assemblies and elected members of Delhi and Pondicheri. The election is held in accordance with the system of proportional representation by means of a single transferable vote system and secret ballot.

Question 4.
Important appointments of President.
Answer:
The president has the power to appoint the following high dignitaries :

  • The Prime Minister of India.
  • Members of the union council of Ministers.
  • The Affomey General of India.
  • The Comptroller and Auditor general of India.
  • The judges of the Supreme Court and the High Court.
  • State Governors etc.

Question 5.
Judicial Powers of the President.
Answer:

  1. The President can grant pardons, reprieves, respites or remission of punishments.
  2. He appoints the Judges of the Supreme Court and State High Courts.
  3. He can also remove them on an address by the Parliament.

Question 6.
Article 352. [Mar. 18, 16]
Answer:
Article 352 of the Indian constitution empowers the President to impose National Emergency during the period of war, External Aggression, Armed Rebellion or internal disturbance. So far National Emergency was proclaimed on Four occasions. They are :

  1. Chinese Aggression (1962)
  2. Indo – Pak War (1965)
  3. Indo – Pak war in the context of Bangladesh Liberation movement (1971) and
  4. Opposition’s call for blocking Parliament (1975).

Question 7.
Article 356.
Answer:
Article 356 of Indian Constitution empowers the President to proclaim the constitutional emergency. If the President on receipt of a report from the governor or otherwise is satisfied that a situation has arisen in which the government of a state cannot be carried on according to the constitutional provisions. He is empowered to proclaim this emergency. It is also called as the President’s Rule.

Question 8.
Financial Emergency.
Answer:
If the President is satisfied that a situation has arisen where by the financial stability or credit of India is threatened then he may proclaim financial emergency in the country as per Article 360 of the Indian Constitution.

Question 9.
National Emergency.
Answer:
Article 352 of the Indian constitution empowers the president to impose National Erpergency during the period of war, External Aggression, Armed Rebellion or Internal disturbances. When National Emergency is in force, the Federal provisions of our constitution cease to operate. So far, National Emergency was imposed four times in 1962, 1965, 1971 and 1975.

Question 10.
Qualifications required for contesting as Vice-President. [Mar. 16]
Answer:
A person to be eligible for election as Vice-president should possess the following qualifications : ,

  1. He should be a citizen of India.
  2. He should have completed 35 years of age.
  3. He should be qualified for election as a member of the council of states.
  4. He should not hold any office of profit under the Union, State or Local Governments in India.

Question 11.
Chairman of Rajya Sabha,
Answer:
The Vice President is the ex-officio chairman of the Rajya Sabha. As such he enjoys the same powers like the speaker of Lok Sabha such as

  1. Presiding over the meetings of Rajya Sabha.
  2. Maintaining discipline, decency and decorum in the House.
  3. Exercising casting vote in case of a tie.
  4. Protecting the privileges and rights of the members.

Question 12.
Appointment of Prime Minister.
Answer:
After the conduct of General elections of the Lok Sabha, the President has to invite the majority party leader of the Lok Sabha to form the Government. When no single party is able to secure majority seats in Lok Sabha, the President invites the Leader of a coalition to form the Government. The president uses his discretionary powers in this regard.

Question 13.
Categories of Union Council of Ministers.
Answer:
There are three kinds of ministers in the union council of ministers. They are

  1. Cabinet Minister.
  2. Ministers of State.
  3. Deputy Ministers.

1) The Cabinet Ministers are entrusted with the maintenance of some important ministries like Finance, Home, Defence etc.

2) The Ministers of state act as the heads of some important sections in the Ministry. They are directly responsible to the Prime Minister for their activities.

3) The Deputy Ministers have no independent and discretionary powers. They assist the Cabinet Ministers.

Question 14.
Any two functions of the Union Cabinet.
Answer:

  1. The Union Cabinet formulates the policies of the union government. It finalizes the domestic as well as foreign policies of the nation after having serious deliberations.
  2. It pilots several bills in the Parliament at various stages and strives to secure the approval of the latter.

AP Inter 2nd Year Civics Study Material Chapter 3 Union Executive

Question 15.
Collective Responsibility. [Mar. 18]
Answer:
Article 75 (3) of Indian constitution stated that the union council of Ministers shall be collectively responsible to the Lok Sabha, for all their acts of omissions and commissions. They act as a team tinder the leadership of the Prime Minister. They sail together, they swim together and they sink together.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 2nd Lesson Fundamental Rights and Directive Principles of State Policy Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 2nd Lesson Fundamental Rights and Directive Principles of State Policy

Long Answer Questions

Question 1.
Explain the characteristic Features of Fundamental Rights.
Answer:
Fundamental Rights :
Fundamental Rights are an important feature of Indian constitution. They are meant for Indian citizens realising the ideal of political democracy. These rights are assigned to the Indian citizens. They enable the citizens to realize their personality. Fundamental Rights will act as a means for leading a happy and honourable life by citizens their render strength and succor to the citizens. They serve as the main source for realising the ideals of political democracy in India.

The makers of Indian constitution have incorporated Fundamental Rights in Articles 12-35 in part III of the constitution.

Characteristics features of Fundamental Rights :
Fundamental Rights have the following characteristic features.

1) Some of the fundamental rights are granted to the ‘citizens’ alone for example, equality of opportunity in matters of public employment, protection against discrimination on any ground; freedom of speech, assembly, association, etc., and cultural and educational rights of the minorities. On the other hand, some of the fundamental rights are available to any person living in the country whether Indian of foreign. For example, equality before law and its equal protection, protection of life, freedom of religion etc.

2) Some of the fundamental rights are positive in nature. They provide scope for the citizens to enjoy some types of freedom. On the other hand, some of the fundamental rights are negative in nature. They impose some restrictions upon the activities of the state.

3) Fundamental Rights are not absolute. In this sense the state can impose reasonable restrictions on their utilisation and enjoyment in the interest of public order, morality, and friendly relations with foreign states.

4) State may impose some restrictions on all or some of the fundamental rights of the citizens during the emergency. The president of India can suspend all the fundamental rights except article 21 (Right to Life) during the national emergency. However, the freedoms guaranteed can’t be restricted by any body.

5) Fundamental Rights are component of the Indian constitution. So they can’t be altered through ordinary laws.

6) Fundamental Rights are comprehensive, integrative and detailed in nature. Some restraints have also been imposed against the utilization of these rights under specific conditions.

7) Fundamental Rights are protected by the judicial organizations in the country. Especially the Supreme court and state High courts play a crucial role in this regard. The ensure justice to those whose rights are infringed or confiscated by others including the state authorities. They issue several writes for the protection of fundamental rights.

8) Fundamental Rights serve as the main means for proper utilization of the capacities and intelligence hidden among the Indian citizens.

9) Though the constitution guarantees six categories of fundamental Rights, all are not of equal weight. That is three can be discovered a hierarchy of values. It becomes evident when justice M. Hidayatullah in the Golaknath case ruled that right to property is the Weakest of all rights’.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Question 2.
Explain the various types of Directive principles of state policy mentioned in Indian constitution. [Mar. 18, 17]
Answer:
Directive principles of state policy are enumerated in articles from 36 to 51 in past -IV of the Indian constitution. They are borrowed from the Irish constitution. They help in realizing the objectives mentioned in the preamble.

Types of Directive principles of state policy: Directive principles can be classified into three broad categories namely Socialistic, Liberal-intellectual and Gandhian principles.

Article 36 defines the term “State”.

Article 37 declares that the directive principles shall not be enforceable by any court.

1) Socialist Principles:
Articles 38,39,41,42,43 and 47 explains about the socialistic ideology of the directive principles of state policy.
1. Article 38 prescribes that the state shall strive to provide justice and promote welfare of the people by creating a proper economic, social and political atmosphere.

2. Article 39 directs the state to secure its citizens.

  • Adequate means of livelihood for all citizens.
  • Equitable distribution of wealth for sub-serving the common good.
  • Equal pay for equal work for all.
  • Protection of adult and child labour.
  • Decentralization of nation’s wealth.
  • Preserving the health and strength of workers, men and women.
  • Protecting childhood and youth against exploitation.

3. To secure right to work and education for all people* relief in the case of unemployment; old age; sickness and disablement and in dther cases of under served want. (Article 41).

4. To make provision for just and human conditions of work and maternity relief (Article 42).

5. To secure living wage and decent standard of life so as to ensure to the workers sufficient leisure and enjoyment of social and cultural Opportunities. (Article 43).

6. Raising the level of nutrition and standard of living of the people and the improvement of public health (Article 47).

2) Liberal-intellectual Principles:
The principles represent the ideology of liberalism and certain objective like provision of basic education, uniform civil code, independent judiciary and international peace. They are incorporated in Articles 44, 45, 50 and 51 of the Constitution.

  1. The State shall secure for the citizens uniform civil code throughout the country. (Article 44)
  2. The State shall provide free and compulsory education for all the children below 14 years of age. The Constitution (Eighty Sixth Amendment) Act, 2002 substituted, the following words in Article 45. “The State shall endeavor to provide early childhood care and education for all children until they complete the age of six years.” (Article 45)
  3. The state organize agriculture and animal husbandry on modem and scientific lines (Article 48)
  4. The state protect monuments which are declared to be of national importance (Article 49)
  5. The state protect and improve the environment and to safeguard forests and wild life. (Article 48 A)
  6. The State shall take steps to separate judiciary from executive in public services of the State. (Article 50)
  7. The State shall (a) promote international peace, justice and security, (b) Maintain just and honorable relations with other nations (c) protection of monuments and place of historical and cultural interest (d) respect for international laws and treaty obligations; and (e) encourage settlement of international dispute by arbitration. (Article 51)

3) Gandhian Principles :
These Principles are based on gandhian ideology. They represent the programme of reconstruction enunciated by Mahatma Gandhi during the national movement. These principles provide ideal rule in India. They are reflected in Articles 40, 43, 46 and 47. They may be enumerated as under.

  1. The State shall organize village Panchayats and endow them with adequate powers and authority so as to enable them to function as the units of self-government. (Article 40)
  2. The State shall strive for the promotion of cottage industries on individual or cooperative basis in rural areas. (Article 43)
  3. The State shall promote the educational and economic interests of the SCs, STs and BCs of society with special care, (Article 46)
  4. The State shall endeavour to bring about prohibition of intoxicating drinks and of drugs which are injurious to health (Article 47)

Other Principles :
The Constitution (Forty Second and Forty-Fourth Amendment) Acts of 1976 and 1978 added a few more subjects to the list of Directive Principles. While the Constitution (Forty Second Amendment) Act inserted Articles 39A, 43A, and 48 A, the Constitution (Forty Fourth Amendment) Act included Article 39 Clause (2). They cc the following provisions.

  1. Providing opportunities for healthy development of children.
  2. Promotion of equal justice and legal aid to the poor.
  3.  Securing participation of workers in the management of industries.
  4. Protecting the environment, forests and wild animals.

Question 3.
Describe the Fundamental Duties incorporated in Indian Constitution. [Mar. 16]
Answer:
Fundamental Duties are a significant feature of Indian Constitution. They are incorporated in our constitution by the constitution 42nd amendment act of 1976 in part – IV under article 51 A. They are borrowed from Russian Constitution. They are 11 in number as mentioned below :

  • To abide by the Constitution and respect the National Flag and the National Anthem.
  • To cherish and follow the noble ideals which inspired our national struggle for freedom.
  • To protect the sovereignty, unity and integrity of India.
  • To defend the country and render national service when called upon to do so.
  • To promote harmony and the spirit of common brotherhood amongst all the people of India and renounce practices derogatory to the dignity of women.
  • To value and preserve the rich heritage of our composite culture.
  • To protect and improve the natural environment including forests, rivers and wildlife and to have compassion for living creatures.
  • To develop the scientific temper, humanism and the spirit of inquiry and reform.
  • To safeguard public property and to abjure violence.
  • To strive towards excellence in all spheres of individual and collective activity.
  • To provide educational opportunities by the parent or guardian to his child or ward between the age of six and fourteen years.

Question 4.
Explain the differences between Fundamental Rights and Directive principles of State Policy.
Answer:
Fundamental Rights are incorporated in Part – III (Article 12 to 35) of our Constitution. Fundamental Rights are of great significance. They serve as the best means to safeguard the life, liberty and property of individuals. They act as the main instrument for releasing the inherent talents and capabilities of the individuals. Fundamental Rights are borrowed from the American Constitution. These are justiciable and can be enforced by the courts.

The Directive Principles of State Policy is a feature of Indian Constitution. They are included in Part -IV from Articles 36 to 51. The principles help to realize the objectives mentioned in the preamble of our Constitution. The makers of our Constitution drew them from the Irish Constitution. These are directives to different governments and agencies of our country. These principles aim at transforming our country into a Welfare State. The rulers have to respect them. They cannot be enforced through any Court of Law. The I governments have to implement them subject to availability of funds. They explain the I responsibilities of the State towards the people.

Differences between Fundamental Rights and Directive Principles of state policy :

Fundamental Rights Directive Principles of state policty
1) The concept was borrowed from the American Constitution. 1) The concept was borrowed from the Irish Constitution.
2) These are enumerated in Part – III of the Constitution covering articles from 12 to 35. 2) These are enumerated in Part – IV of the Constitution covering articles from 36 to 51.
3) These-are negative in character in the sense that they prohibit the Government from doing certain things. 3) These are positive in character in the sense that they direct the government to work for the attainment of certain objects.
4) These aim at establishing political democracy in the country. 4) These aim at establishing social and economic democracy in the country.
5) These have constitutional sanction and so their implementation needs no legislation. 5) These have no constitutional sanction and so their implementation needs legislation.
6) These are justicable and can be enforced by the courts. 6) These are non – justiciable and cannot be enforced by the courts.
7) A law violating a Fundamental Right can be declared by the court as unconstitutional. 7) A law violating a Directive Principle cannot be declared by the court as unconstitutional.
8) These are personal and individualistic in nature. 8) These are societarian and socialistic in nature.
9) These promote the welfare pf the individual. 9) These promote the welfare of the society.
10) These have been laid down in clear legal language. 10) These are stated in general terms.

Short Answer Questions

Question 1.
Write a note on the changing Relationship between Fundamental Rights and Directive Principles of State Policy.
Answer:
Fundamental Rights and Directive principles of state policy are the salient features of Indian Constitution.

Changing Relationship between Fundamental Rights and Directive Principles :
Although a distinction is mode, between fundamental rights and directive principles of sate policy by way of justiciable and non-justiciable nature. Yet over the years directive principles of state policy have become politically important and the relation between the two has undergone several changes.

The Supreme court consistently held the opinion that the directive principles of state policy should be subsidiary to the fundamental rights. Judgements in various disputes like Sajjan Singh Vs state of Rajasthan and Golaknath Vs state of Punjab, the Supreme court confirmed its stand and reiterated that it is the duty of Parliament to enforce the Directive principles without tampering the Fundamental Rights.

As a result of the invalidation of certain laws like the Nationalization of Banks, Abolition of privy purses, the Parliament enacted the (Twenty Fifth Amendment) constitution Act in 1971 which declated that the enforcement of the directive principles of state policy shall not be invalidated by any court on the grand that it violates the fundamental rights in articles 14, 19 and 31 of the constitution.

Again the (Forty Second Amendment) constitution Act passed by Parliament in 1976 declared that no law, giving effect to any of all directive principles, shall be invalid on the ground that it infringes on Fundamental Rights. However, in the Minerava Mills case, the Supreme court restricted the original supremacy and sanctity of the fundamental rights over the directive principles of state policy. Thus, the fundamental rights have primary over the directive principles.

In the case of Keshavananda Bharati Vs state of Kerala. The supreme court held that the Parliament cannot among the basic structure of the constitution. By implication the Supreme court considered fundamental rights as a part of the basic structure of the constitution.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Question 2.
Explain any three fundamental rights of a citizen. [Mar. 18]
Answer:
1) Right to freedom of Religion :
This right denotes the secular nature of Indian political system. It aims at transforming India into a secular state. Both the citizens and aliens of India enjoy this right.

Article 25 empowers every persons to profess, practice and propagate a religion of this liking.

Article 26 guarantees the following rights to every person.

  1. To establish and maintain religious and charitable institutions.
  2. To mange his their religious affairs.
  3. To own and acquire moveable and immovable properties and
  4. To maintain such properties in accordance with the provision of the law.

Article 27 prohibits the state to impose or collect taxes from individuals purely on religious grounds. It also prohibits the state to’impose and collect taxes for the benefit and maintenance of any particular religion or religious denominations.

Article 28 bans religious instructions in educational institutions wholly or partly maintained by the state funds.

2) Educational and Cultural Rights :
Indian constitution provided several cultural and educational opportunities for Indian citizens through this right. Article 29 enables every citizen to preserve and protect his own language and culture irrespective of one’s religion, language or region.

Article 30 prohibits special treatment to any citizen in the admission into educational institutions either wholly or partly funded by the state on the grounds of caste, religion, region, colour, language or sect. However, it allowed the minotities some special facilities for preserving and promoting their language and culture. The state can grant financial assistance to them in this regard.

3) Right to Constitutional Remedies :
This right enables the individuals to approach a high court under article 226 or the supreme court under article 32 to get any of the fundamental rights restored in case of their violation. The Supreme court and the state High courts issue various writs for the implementation of Fundamental Rights. Dr. Ambedkar described this right as the Heart and Soul of the constitution.

Question 3.
Describe the six Freedoms of a citizen. [Mar. 17]
Answer:
Our constitution in chapter III under Article 19 (clause 1) guarantees certain fundamental rights subject to certain restrictions. They are also known as fundamental freedoms. They are :

  1. Freedom of speech and expression.
  2. Freedom of peaceful Assembly without Arms.
  3. Freedom of Associations and Unions.
  4. Freedom of movement throughout the territory of India.
  5. Freedom of residence and settlement in any part of the Territory of India.
  6. Freedom of profession, trade occupation or business.

These freedoms would facilitate the progress of Indian citizens in social, political and economic spheres. These freedoms are not absolute. The state may, if necessary, impose certain reasonable restrictions on the enjoyment of the above freedoms by the Indian citizens. These restrictions relate to the maintenance and safe guarding of the independence, sovereignty, integrity, law and order.

Question 4.
Write briefly on the right to Constitutional Remedies.
Answer:
This right is the most significant of all the fundamental rights. It extends protection and relief to those whose Fundamental Rights were abridged, confiscated or infringed by others including the public, authorities. As this right gives a citizen the right to approach a rights restored in case of their violation. The supreme court and the High court can issue orders and give directions to the governments for the enforcement of the Fundamental Rights. The courts can issue various writs like habeas corpus, Mandamus, prohibition, Quowarranto and certiorari Dr. Ambedkar rightly described this right as Heart and Soul of the constitution.

Question 5.
Explain any five differences between Fundamental rights and Directive principles of state policy. [Mar. 16]
Answer:

Fundamental Rights Directive Principles
1) The concept was borrowed from American constitution. 1) The concept was borrowed from Irish Constitution.
2) These aim at establishing a political democracy in the country. 2) These aim at establishing a social and economical democracy in the society.
3) These are justicable and then enforced by the courts. 3) These are non-justicable and cannot be enforced by the courts.
4) These are personal and Individualistic in nature. 4) These are societarian and socialistic in nature.
5) These promote the welfare of the Individual. 5) These promote the welfare of the society.

Question 6.
Explain the important characteristics of Directive principles of State Policy.
Answer:
The following are the characteristics of the Directive principles :

  1. These are in the form of directives or instructions issued to the successive governments.
  2. These are positive in nature as they extend the jurisdiction of the powers and functions of the governments at various levels in India.
  3. Enforcement of these principles depends upon availability of financial resources.
  4. They are popular in nature as they aim at the establishment of egalitarian society.
  5. They are supposed to be implemented by any party in power irrespective of it’s policies and Ideology.
  6. Failure to implement these principles is not considered as a breach of law.
  7. They are non-justicable in nature as no one can force the governments to implement them immediately. The governments have discretion in implementing these principles.

Question 7.
Examine the implementation of Directive principles of state policy.
Answer:
Keeping in view the fact, Directive Principles of State Policy are fundamental in the governance of the country. The Union as well as the State Governments, since 1950, have been continuously taking various steps to implement them from to time. These are mentioned below:

  1. Abolition of Zamindari, Jagirdari and Inamdari systems.
  2. Introduction of Land Ceiling Acts.
  3. Abolition of Privy Purses.
  4. Nationalization of 14 leading commercial banks.
  5. Establishment of Khadi and Village Industries Board etc. .
  6. Organization of Village Panchayats.
  7. Reservation of seats are reserved for SCs and STs in educational institutions and representative bodies.
  8. Enactment of Ancient and Historical Monuments and Archeological Sites Remains Act 1951.
  9. Separating criminal procedure code from the executive.
  10. Prohibition of cow slaughter, calves and bullocks in some States.
  11. Establishment of primary health centers and hospitals throughout the country.
  12. Implementation of Non-Alignment and Panchasheel Principles.
  13. Initiation of old age pension schemes.
  14. Introduction of unemployment schemes.
  15. Enactment of Minimum Wages Act.
  16. Enactment of Wild Life Act.

Question 8.
Explain the significance of Directive principles of State Policy.
Answer:
Directive principles of state policy are considered as the most significant feature of the Indian constitution. They have great significance in the administration of our country. Though they lack legal sanction, they enjoy popular support and possess public sanction. If the party in power fails to implement these principles, it has to face resistance from the electorate in the coming elections.

So, no government can afford to ignore the implementation of these principles. The opposition, the press and the electorate would exert pressure on the government for implementing these principles. Hence the union government considered these principles as the basis for the functioning of National Planning Commission (or) NITI Ayog. They have helped the cburts in exercising their power of Judicial Review.

To conclude, directive principles of state policy are considered as an authoritative declaration of the aims and aspirations of Indians formulated by their representatives after solemn and mature deliberations as they are considered” as the goal of social Revolution”.

Very Short Questions

Question 1.
Fundamental Rights.
Answer:
Fundamental Rights are an important feature of Indian constitution. They are incorporated in part – III Articles from 12 to 35 in our constitution. They enable the citizens to realise their personality. They are :

  • Right to Equality
  • Right to Freedom
  • Right against exploitation
  • Right to Freedom of Religion
  • Educational and cultural rights
  • Right to constitutional remedies

Question 2.
Types of Directive principles.
Answer:
On the basis of their content and direction directive principles of state policy can be classified into three broad categories namely socialistic, Liberal – intellectual and Gandhian principles.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Question 3.
Habeas corpus. [Mar. 16]
Answer:
Habeas corpus literally means “To produce the body of”. It is in the nature of calling upon a person who has detained another to produce the latter before it. The court wants to know on what grounds a person has been detained. This writ frees a person whose detention has no legal justification.

Question 4.
Mandamus.
Answer:
Mandamus literally means “A mandate or command”. It is issued by a competent court for directing any person, corporation or inferior courts requiring him, it or them to do some particular, thing specified there in which appertains to his of their and is in the nature of public duty.

Question 5.
Cultural and Educational Rights.
Answer:
Article 29 guarantees to every citizen to protect his own language, script or culture. Article 30 provides that all minorities whether based on religion or language shall have the right to establish and maintain educational institutions of their choice.

Question 6.
Gandhian Ideas of Directive principles of state policy.
Answer:

  1. The state shall organize village panchayats and endow them with adequate powers and authority so as to enable them to function as the units of self government (Article 40).
  2. The state shall strive for the promotion of cottage industries on individual or cooperative basis in rural areas (Article 43).
  3. The state shall promote the educational and economic interests of the SCs, STs and BCs of society with special care (Article 46).
  4. The state shall endeavour to. bring about prohibition of intoxicating drinks and of drugs which are injurious to health (Article 47).

Question 7.
Significance of Fundamental Duties. [Mar. 17]
Answer:
Fundamental duties are considered most significant from the following view points.

  1. The Fundamental Duties act as a reminder to the citizens that while enjoying their rights, they should also be conscious of duties they owe to their country, their society and to their fellow citizens.
  2. They serve as warning against antinational and anti social activities.
  3. They serve as the source of inspiration for the citizens and promote sense of discipline and commitment among them.
  4. They help the courts in examining and determining the constitutional validity of a law.

Question 8.
Mention any three liberal principles.
Answer:

  1. The state shall secure for the citizens uniform civil code throughout the country (Article 44).
  2. The state organize Agriculture and Animal husbandry on modem and scientific lines, (Article 48).
  3. The state shall take steps to separate judiciary from executive in public services of the state (Article 50)

Question 9.
Quo – Warranto
Answer:
Quo – Warranto literally means ‘What Warrant or Authority”. It enables the competent court to enquire into the legality of the claim which a party assets to a public office and to oust him from its enjoyment if the claim is not well founded.

Question 10.
Right to Religion.
Answer:
This right denotes the secular nature of Indian political system. It aims at transforming India into a secular state. Both the citizens and aliens of India enjoy this right. Articles from 25 to 28 in part – III of Indian constitution deals with right to freedom of Religion. This right enables the individuals to profers, practice and propagate any religion according to their conscience.

AP Inter 2nd Year Civics Study Material Chapter 2 Fundamental Rights and Directive Principles of State Policy

Question 11.
Right against exploitation.
Answer:
In our country, there are millions of people who are underprivileged and deprived. They are subjected to exploitation by their fellow human beings. One such form of exploitations is ‘begar’ or ‘forced labour’ without payment. Another closely related form of exploitation is trafficking of human beings and using them as slaves. Both are prohibited under Article 23 of our constitutions.

Article 24 of the Indian constitution forbids all forms of child labour below the age of 14 years in factories, mines and other hazardous industries.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Andhra Pradesh BIEAP AP Inter 2nd Year Civics Study Material 1st Lesson The Constitution of India Textbook Questions and Answers.

AP Inter 2nd Year Civics Study Material 1st Lesson The Constitution of India

Long Answer Questions

Question 1.
What is a constitution? Explain the historical background of the Indian Constitution.
Answer:
The term Constitution implies a written document embodying the provisions relating to the powers and functions of the government organs, the rights of the people, and their relations with the government. In simple,

Powers and Functions of the state + Rights and Duties of the Citizens = Constitution

Historical Background of the Indian constitution :
The constitution of India is the best constitution in the world.

It was formulated by Drafting Committee headed by Dr. B.R. Ambedkar and approved by the Constituent Assembly Chaired by Dr. Rajendra Prasad.

Earlier the Leaders of the Indian National Movement demanded several times for a separate constitution for Indians. They at first emphasized, during the second decade of the 20th century, that the British government shall grant self-rule or the right to formulate a constitution for Indians. It is in this context that leaders like Bala Gangadhara Tilak and Annie Besant launched the Home Rule movement. They requested the British government to apply the “right of self-determination”.

The Swaraj Party Leaders in 1922 made it clear that constitutional arrangements must be made for providing representation to Indians in Legislative Assembly.

Later Motilal Nehru moved a resolution in the Central Legislative Assembly requesting the British government for convencing a Round Table Conference for formulating a constitution for Indians.

The Three Round Table conferences held in London in 1930, 1931 and 1932 and their recommendations led to the passage of Government of India Act, 1935.

The Act provided for the introduction of Federal Polity and the establishment of provincial autonomy in the British India Provinces.

In the wake of Second World War, the Congress Ministers in the provincial legislative councils proposed a resolution meant for recognizing India as independent state and the power to make a constitution by the provincial legislative councils.

Viceroy Linlithgow through his August offer (1940) announced for the first time that Indians must Cooperate with Britain in the Second World War and the primary responsibility of making a constitution wholly vests with the Indians.

In 1942 Prime Minister Winston Churchill made efforts for resolving the Indian constitutional crisis by sending Stafford Cripps, a member of war cabinet to India for extensive discussions with Indian leaders.

The Cripps Mission (1942) was a failure. In this back drop, Indian National Congress under Gandhiji’s leadership adopted the famous “Quit India Revolution on 8th August, 1942 for the immediate ending of British rule in India.

After the Second World War, the labour party came to power in Britain under the leadership of element Attlee.

Then Viceroy Lord Wavell announced the latest policy of government in September, 1945. The British Government despatched a three member cabinet mission in February 1946 to India for making deliberations with Indian leaders on the issue of setting up of the Constituent Assembly and the provision of independence for Indians.

The Cabinet Mission made it clear that elections will be held to the Constituent Assembly and citizens having franchise will elect the members. Except Muslim League all the parties in India have agreed broadly the suggestions of the above team.

Lord Wavell formed on interion government with Jawaharlal Nehru as it’s head in 1946, with an increasing communal violence it seems there is no alternative to partition of the sub-continent.

The Mount Batten plan (1947) provided for the partition of the country. The Indian Independence Act of 1947 provided for setting up of a dominion of India and a dominion of Pakistan.

The Native states were given the choice of joining and becoming an integral part of either of the two dominions.

The vision and constructive statemanship of Sardar Vallabhai Patel the deputy Prime Minister and Home Minister facilitated the peaceful merger of most of the Native states into the Indian union.

Of course, the merger of Jammu and Kashmir, Junagadh and Hyderabad states into the Indian Union provided to be a difficult task.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Question 2.
Explain in brief the salient features of Indian Constitution. [Mar. 18, 16]
Answer:
Introduction :
The Indian constitution was prepared and adopted by the Constituent Assembly which was set up in 1946. The Constituent Assembly took nearly three years (From 9th December, 1946 to 25th November, 1949) to complete the framing of the constitution.

The Constituent Assembly approved the Indian Constitution on 26th November, 1949. The Indian constitution came into force on 26th January, 1950. Which we have been celebrating as “The Republic Day”. The following are the salient or basic features of the Indian constitution.

1) Preamble :
The Indian constitution begins with a preamble. The preamble clearly defines the objectives of our constitution. It declares India as a Sovereign, Socialist, Secular, democratic Republic. It provides liberty, equality, fraternity and justice. It states that the people of India are the chief source of the political authority.

2) A lengthy Written Document:
The Constitution of India is the most written, lengthy and detailed document in the world. In 1950, the Indian Constitution had 22 parts, 8 schedules and 395 articles. After subsequent amendments, it contains 12 schedules and 444 articles for the present.

3) A combination of rigidity and flexibility :
The constitution of India is a blend of rigidity and flexibility. Our founding Fathers were programatic enough to provide for rigidity and flexibility as the situation demands. Article 368 provides the details of the amendment procedure.

i) Some of the provisions like admission of New states (Ex : Telangana), provisions relating to citizenship, salaries and allowances of the members of the constitutional bodies like President, Vice-President, Supreme Court and High Court judges etc. Can be amended by simple majority. It is said to be flexible.

ii) Some provisions can be amended by a special majority i.e., not less than Two – Thirds of the members of the House present and voting.
Ex : Fundamental Rights, Directive Principles of state policy etc.

It is said to be half rigid and half flexible.

iii) Some provisions can be amended by two-thirds majority of the Parliament and with the concurrence of half of the states.
Ex: Election of the President, executive powers of the union and the states, distribution of legislative powers between the union and the states etc. It is said to be rigid.

4) Quasi-Federal polity:
India is a states according to the constitution. Our constitution contains both the features of unitary and Federal Governments. It prescribed unitary system in emergencies and Federal system on ordinary occassions. Provisions of unitary state such as Single Citizenship, Single Integrated Judiciary, Single Election Commission, Role of All India Services Personnel etc., are found in our constitution. At the same time certain federal features like written, rigid constitution, Dual government, Bicameralism etc., are profoundly seen in our constitution. Thus it is a Quasi-Federal. Polity like Canada.

5) Republican government :
Unlike the colonial Master, the UK, India preferred a Republican government. Here all public offices right from World member to the top president of India are open to all eligible citizens and there is no place for hereditary principle.

6) Parliamentary government :
The constitution of India provided Parliamentary government of the British type but with an elected President of Irish Model. Accordingly, the features of Parliamentary government such as two executive heads, Ministerial accountability to the lower house of the legislature, Prime Minister leadership etc., are prevalent in our political system.

7) Fundamental Rights and Fundamental Duties:
Part – III of the constitution, Articles from 12 to 35, provides for a set of basic human rights to ail. They are justiceable and ensure basic freedoms. They are six in number. 1) Right to equality 2) Right to freedom 3) Right against exploitation 4) Right to religion 5) Cultural and educational right and 6) Right to constitutional remedies.

The 42nd Amendment to the constitution in 1976 incorporated the fundamental duties in Article 51A under part – IVA. Though they are not justiciable. But they put an obligation on the citizens to render certain duties in return for the protection they have been enjoying through fundamental rights. Fundamental duties relate to respecting the constitution, the National Flag and National Anthem, safeguarding public property etc.

8) Single citizenship :
Our Constitution provides for single citizenship for all persons who are born in India and who resided in India for a specific period. It enables the citizens to possess and enjoy identical rights and privileges. It also promotes unity, integrity and fraternity among the people.

9) Universal and Adult Franchise :
The makers of the Indian constitution provided for the universal adult Franchise for all citizens without any discrimination based on caste, colour, creed, community, language, religion, region, sex, property etc. At the beginning, adult Franchise was given to all the citizens who attained the age of 21 years. Later voting age was reduced to 18 years through the 61st constitution, Amendment Act in 1988.

10) Secular state :
Our constitution stands for a secular state. It does not uphold any particular religion as the official religion of. the Indian state. It ensures complete religious freedom to the people. It abolishes discrimination between individuals on religious grounds in the matters of employments education and legislation. It prohibits religious instructions in state owned or state – aided educational institutions.

11) Independent judiciary :
The Judiciary performs its functions independently. The legislature or the executive shall not interfere in the working of the Judiciary. The Judiciary carries on its obligations according to the constitutional norms and democratic principles.

12) Directive principles of state policy:
Our constitution hinted out certain directive principles as the policy of the state in part IV from Articles from 36 to 51. These principles are drawn from Irish constitution. These principles reflect the welfare state concept. They are the directions to be followed by the various governments. Equal pay for equal work, provision of employment opportunities, fair distribution of wealth, old age pension, protection of illhealth, provision of education, protection of women and children etc., are the some examples of these principles. Though these principles are non-justiciable, No responsible government can afford to ignore them.

13) Bi-cameralism:
The constitution of India introduced Bi-cameralism at the National level. Accordingly, The Indian Parliament consists of two houses namely the Rajya Sabha (upper house) and the Lok Sabha (lower house). While the Rajya Sabha represents the states, the Lok Sabha represents the people.

14) Panchayati Raj and Nagar Palikas Acts :
The Panchayati Raj and Nagar Palikas Acts are recent features of our constitution. The 73rd and 74th Constitutional Amendment Acts gave constitutional recognition to the Rural and Urban local governments. Which came into force in 1993 and 1994 respectively had become operative all over the territory of India. The ideals of democratic decentralisation or the grass roof democracy are realised by these acts.

These acts provides for adequate representation for women, scheduled castes, scheduled tribes and other weaker sections in the policy making bodies of the local governments.

Short Answer Questions

Question 1.
Write briefly the elements of the constitution.
Answer:
The term constitution implies a written document embodying the provisions relating to the powers and functions of the government organs, the rights of the people and their relations with the government.
Elements of the constitution :
1) The constitution prescribes a set of basic rules that ensure coordination amongst the members and groups of a society. The constitution specifies the basic allocation of power in a society. It decides who frames the laws. For example, in a democratic country like, the people through their elected representatives make the laws i.e., the Parliament at national level and state legislatures at state level are empowered to make laws.

2) It specifies the structure of the government and its limitations of the modern govemrttents are made up of three organs, viz.
i) Legislature which frames the laws within the limits set by the constitution.
ii) The executive, the President or the Governor, the Council of Ministers headed by the Prime Minister or Chief Minister would be taking policy decisions as per the guidelines provided by the Parliament or Legislature.
iii) The Judiciary by interpreting the laws would set limits on the powers of both the legislature and executive.

Through the Judiciary, the constitution ensures limited and responsible government.
For example : Article 13 of the Indian constitution establishes the Supremacy of the constitution.

3) The constitution establishes the relationship between the rulers and the ruled. Fundamental rights and Fundamental Duties spell out in detail the mutual obligations of the citizens and the state. Part III and Part IVA of the Indian constitution bind the state and the citizens towards each other.

4) Every society has certain aspirations and goals. The state came into existence to fulfill the bare needs of the people and continues to exist for the good life of all. It is the constitution which directs the state i.e., government to make certain policies for welfare of the people.
Ex : Directive principles of state policy which incorporated in part – IV of the Indian
1 Constitution.

5) The constitution, as the Supreme document, serves as a shock absorber in limiting the fluctuations of present and future generations. It is a living document that connects the past with the present and assures a predictable future. Modem societies cannot survive without a just constitution.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Question 2.
Write about the making of Indian constitution.
Answer:
While Negotiations were going on about the modalities of transfer of power, a l Constituent Assembly was set up to draw the constitution for India. The Cabinet Mission and the major political parties reached an agreement over the constitution of Constituent Assembly in 1946.

Elections to the Constituent Assembly were held in July, 1946 in which 292 members from British provinces, 93 members from Native states and 4 members from central provinces were elected. The Constituent Assembly of India held its first meeting on December 9, 1946. It elected Sachchidanand Sinha as its protemporaiy chairman.

On December 11, 1946 it elected Dr. Babu Rajendra Prasad as its permanent chairman.

The making of the constitution really began at its third session held between April 22 and May 2, 1947.

The Fourth session of the Assembly was held on July 14 and continued till July 31, 1947. It held discussions on Model constitution, Adopted the National Flag. The Assembly met for the fifth time on the eve of the independence day.

On August 29, 1947, it set up a seven member Drafting committee with Dr. B.R. Ambedkar as its chairman.

The Drafting Committee presented the Draft constitution on February 21, 1948. The Constituent Assembly adopted the Draft constitution on November 26, 1949. The last session of the Assembly was held on January 24, 1950.

It elected Dr. Babu Rajendra Prasad as the first President of Indian Republic under the new constitution.

On January 26, 1950 the New constitution came into operation and India was declared as a Republic state.

Question 3.
Write a note on the sources of Indian constitution. [Mar. 18, 16]
Answer:
The constitution of India was formulated on the basis of various experiences. Almost all the noble features of the world constitutions have been incorporated in it. Similarly the peculiar political, social and cultural conditions present in India have been taken into account at the time of drafting the constitution. The makers of Indian constitution ignored those conditions prevalent in other countries which are contrary to the socio-economic and political background of India.

On the whole the following sources figure prominently in making the Indian constitution.

  1. Many provision of Indian constitution have been drawn on the basis of West Minister Model (british). These include parliamentary traditions, rule of law, cabinet government, legislative-executive relations, single citizenship’ nominal executive head etc.
  2. Some provisions like fundamental rights, judicial review, federal system, president’s election, impeaching the president etc., have been taken from the American constitution.
  3. Items relating to directive Principles of State Policy have been drawn from the constitution of Ireland.
  4. The emergency powers of the President have been taken basing on the German constitution.
  5. Matters such as Concurrent List, Business, Commerce, Inter State trade, Special privileges of legislators etc., have been added on the model of Australia.
  6. The Union-State relations mentioned in Indian constitution have been designed basing on the Canadian constitution.
  7. Matters of constitutional amendment procedure were drawn from South African constitution.
  8. The idea of republic and the ideals of liberty, equality and fraternity have been taken from the constitution of France.
  9. Most of provisions of Indian constitution were drawn from the Government of India Act, 1935.

Question 4.
Explain any three salient features of Indian constitution.
Answer:
1) A combination of rigidity and flexibility :
The constitution of India is a blend of rigidity and flexibility. Article 368 provides the details of the amendment procedure. Some Of the’provisions like Admission of New states (Ex : Telanganaj, provisions relating to citizenship, salaries and allowances of the President, Vice president, Supreme Court and High Court judges etc., can be amended by simple majority. It is said to be flexible.

Some provisions can be amended by a special majority i.e., not less than two-thirds of the members of the House present and voting. Ex : Fundamental Rights, Directive principles of state policy etc. It is said to be half rigid and half flexible. Some provisions can be amended by two-thirds majority of the Parliament and with the concurrence of half of the states.
Ex : Election of the President, Distribution of Legislative powers between the union and the states etc. It is said to be rigid.

2) Republican government:
Unlike the colonial master, the U.K., India preferred a Republican government. Here all public offices right from ward member to the top of India are open to all eligible citizens and there is no place for hereditary principle.

3) Single citizenship :
Our constitution provides for single citizenship for all persons who are born in India and who resided in India for a specific period. It enables the citizens to possess and enjoy identical rights and privileges. It also promotes unity, integrity and fraternity among the people.

Question 5.
“The preamble is the soul of the constitution”. Explain.
Answer:
Preamble is the most salient features of Indian constitution. It denotes the essential philosophy of Indian constitution. It reflects the aims, aspirations and objectives of the makers of Indian constitution. Jawaharlal Nehru described the Preamble as a declaration, a firm resolve, a pledge, an understanding and something more than a resolution. Preamble begins with the sentence “we, the people of India’ adopt, enact and give to ourselves this constitution”.

It declared India as a sovereign socialist, secular, democratic and republic. It announced that Indians enjoy liberty, equality, fraternity and justice. It clearly stated that sovereignty in India belongs the people of India. Justice Madholkar pronounced that Preamble is the essence of Indian constitution. Justice Hidayatullah praised the Preamble as “the soul of the constitution”.

The Preamble after the Constitution (Forty Second Amendment) Act, 19718 is as follows :

We, the people of India having solemnly resolved to constitute India into a Sovereign Socialist Secular Democratic Republic and to secure to all its citizens.

“Sovereign” – which means its authority within the country is undisputed and externally it is free from others contort. “Socialist” means a state that doesn’t, allow any kind of exploitation – social, economic and political.

“Secular” – It gives equal freedom to all religions.

“Democratic” – Stipulates that the sovereign power vests with the people. They exercise it periodically through universal Adult Franchise.

A “Republican” state assures that public offices are open to every citizen without any discrimination. There is no place for hereditary principle.

  1. Justice – social, economic and political.
  2. Liberty of-thought, expression, belief, faith and worship.
  3. Equality of status and of opportunity; and to promote among them all.
  4. Fraternity assuring the dignity of the individual and the unity and integrity of the nation.

In our Constituent Assembly this twenty-sixth day of November, 1949 do hereby adopt, enact and give to ourselves this Constitution”.

Very Short Answer Questions

Question 1.
Written constitutibn
Answer:
The constitution of India is the most written, lengthy and detailed document in the world. In 1950 the Indian constitution had 22 parts, 8 schedules and 395 articles. After subsequent amendments, it contains 12 schedules and 444 articles for the present.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Question 2.
Rigid constitution
Answer:
Rigid constitution is one whose provisions cannot be charged easily. In this system the constitutional amendment methods are different from those of ordinary laws. There will be a special procedure for amending the provisions of the rigid constitution. The rigid constitution will have firmness due to its special procedures of amendment.
Ex : United states and India.

Question 3.
Parliamentary form of government
Answer:
The constitution of India provided Parliamentary Government of the british type but with an elected president of Irish model. According to the features of Parliamentary government such as two executive heads, ministerial accountability to the lower house, of the legislature, Prime Minister, leadership etc., are prevalent in our political system.

Question 4.
Fundamental Rights and Duties
Answer:
Fundamental Rights and Fundamental Duties are the salient features of Indian constitution.

Six Fundamental Rights were incorporated in Part – III articles from 12 to 35. They are justiciable and ensures basic freedoms to all Indians.

Eleven fundamental duties were incorporated in Part IV A under article 51 A. They put an Obligation on the citizens to render certain duties in return for the protection they have been enjoying through Fundamental Rights.

Question 5.
Secular state
Answer:
Our constitution stands for a secular state. It does not uphold any particular religion as the official religion of the Indian state. It ensures complete religious freedom to the people it abolishes discrimination between individuals on religious grounds in the matters of employment,’ education and legislation. It prohibits religious instructions’in state-owned or state-aided educational institutions.

Question 6.
Universal Adult Franchise
Answer:
The makers of the Indian constitution provided for the universal adult Franchise for all citizens without any discrimination based on caste, colour, creed, community, language, religion, region, sex, property etc. At the beginning, adult Franchise was given to all .the citizens who attained the age of 21 years. Later voting age was reduced to 18 years through the 61st constitution Amendment Act in 1988.

Question 7.
Bicameralism
Answer:
The constitution of India introduced Bi-cameralism at the national level. Accordingly, the Indian Parliament consist of two houses namely the Rajya Sabha (upper house) and the Lok Sabha (Lower house). While the Rajya Sabha represents the states, the Lok Sabha represents the people.

Question 8.
Directive principles of state policy
Answer:
Our constitution hinted out certain directive principles as the policy of the state in part IV from Articles from 36 to 51. These principles are drawn from Irish constitution. These principles reflect the welfare state concepts. They are the directions to be followed by the various governments. Though these principles are non-justiciable, No responsible government can afford to ignore them.

Question 9.
Independence of Judiciary
Answer:
The Judiciary performs its functions independently. The legislature of the executive shall not interfere in the working of the judiciary. The Judiciary carries on its obligations according to the constitutional norms and democratic principles.

AP Inter 2nd Year Civics Study Material Chapter 1 The Constitution of India

Question 10.
Preamble [Mar. 18, 17, 16]
Answer:
The Indian constitution begins with a preamble. The preamble clearly defines the objectives of our constitution. It declares India as a sovereign, as a socialist, Secular Democratic Republic. It provides liberty, equality, fraternity and justice. It states that the people of India are the chief sources of the political authority.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 14th Lesson Kinetic Theory

Very Short Answer Questions

Question 1.
Define mean free path. [A.P. Mar. 18, 17, 15; T.S. Mar. 17, 15]
Answer:
The average distance covered by a molecule between two successive collisions is called the mean free path.

Question 2.
How does kinetic theory justify Avogadro’s hypothesis and show that Avogadro Number in different gases is same ?
Answer:
For two different gases, we have [latex]\frac{P_1 V_1}{N_1 T_1}=\frac{P_2 V_2}{N_2 T_2}[/latex] = KB (constant)
If P, V, T are same, then N is also same for two gases.
N is called Avogadro’s number. According to Avogadro’s hypothesis, that the number of molecules per unit volume is same for all gases at a fixed temperature and pressure.
In this way kinetic theory justify Avogadro’s hypothesis.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 3.
When does a real gas behave like an ideal gas ? [T.S. Mar. 16; Mar. 14]
Answer:
At low pressures and high temperatures real gases behave like an ideal gas.

Question 4.
State Boyle’s Law and Charle’s Law. [A.P. Mar. 18; T.S. Mar. 15]
Answer:
Boyle’s law : At constant temperature, the pressure of a given mass of gas varies inversely with volume.
Charle’s law: At constant pressure, the volume of the gas is proportional to its absolute temperature.

Question 5.
State Dalton’s law of partial pressures. [T.S. Mar. 18; A.P. Mar. 16; Mar. 14]
Answer:
The total pressure exerted by a mixture of non-reacting gases occupying a vessel is equal to the sum of the individual pressures which each gas would exert if it is alone occupied the whole vessel.
i.e., P = P1 + P2 + ………….

Question 6.
Pressure of an ideal gas in container is independent of shape of the container – explain. [T.S. Mar. 17]
Answer:
The kinetic theory expression for the pressure of a given mass of an ideal gas in a container is 1 -2.
P = [latex]\frac{1}{3}[/latex] nm[latex]\bar{v}^2[/latex], where [latex]\overline{\mathrm{V}}[/latex] indicates mean-square-speed, n is number of molecules, m is mass of molecule. From the above equation, shape of the container is immeterial. Hence pressure of an ideal gas is independent of shape of the container.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 7.
Explain the concept of degrees of freedom for molecules of a gas.
Answer:
The degrees of freedom of a particle indicate the number of independent motions which the particle can undergo, or the number of independent methods of exchanging energy.

A monoatomic molecule (He) has three degrees of freedom, a diatomic (H2, O2) one has five, while a triatomic (H2O) one has six.

Question 8.
What is the expression between pressure and kinetic energy of a gas molecules ? [AP – Mar. ’17, ’16, ’15]
Answer:
By kinetic theory pressure, P = [latex]\frac{1}{3} m n \overline{V}^2[/latex] and kinetic energy = [latex]\frac{1}{2} m n \overline{V}^2[/latex] where m is the mass of the molecules, n is the number of moles per unit volume, v is the mean-square-speed.
∴ P = [latex]\frac{2}{3}\left(\frac{1}{2} m n \overline{V}^2\right)[/latex] = [latex]\frac{2}{3}[/latex] (kineticenergy)

Question 9.
The absolute temperature of a gas is increased 3 times. What will be the increase in rms velocity of the gas molecule ?
Answer:
Case (i) : The r.m.s velocity of gas molecule, [latex]\overline{V}_1=\sqrt{\frac{3 R T_1}{M}}[/latex]
Case (ii) : The r.m.s velocity of gas molecule, [latex]\overline{V}_2=\sqrt{\frac{3 R T_2}{M}}[/latex]
[latex]\frac{\overline{\mathrm{V}}_2}{\overline{\mathrm{V}}_1}=\sqrt{\frac{\mathrm{T}_2}{\mathrm{~T}_1}}=\sqrt{\frac{3 \mathrm{~T}_1}{\mathrm{~T}_1}}[/latex]  [∵ T2 = 3T1] ⇒ [latex]\overline{\mathrm{V}}_2=\sqrt{3} \overline{\mathrm{V}}_1=1.732 \overline{\mathrm{V}}_1[/latex]
∴ Increase in r.m.s velocity of gas molecule = [latex]\overline{\mathrm{V}}_2-\overline{\mathrm{V}}_1=1.732 \overline{\mathrm{V}}_1-\overline{\mathrm{V}}_1=0.732 \overline{\mathrm{V}}_1[/latex]

Short Answer Questions

Question 1.
Explain the kinetic interpretation of Temperature.
Answer:
Kinetic interpretation of Temperature : Since the pressure of the gas is given by
P = [latex]\frac{1}{3} \mathrm{mn} \overline{\mathrm{V}}^2[/latex], where m = mass of the gas, n = [latex]\frac{\mathrm{N}}{\mathrm{V}}[/latex] = Number of molecules per unit volume
[latex]\overline{V}[/latex] = r.m.s velocity of gas, P = [latex]\frac{1}{3} m \frac{N}{V} \overline{V}^2[/latex] ⇒ PV = [latex]\frac{1}{3} \mathrm{mN} \overline{\mathrm{V}}^2[/latex]
Also for a gram molecule of the gas, PV = RT
Hence RT = [latex]\frac{1}{3} m N \overline{V}^2 \Rightarrow 3 \frac{R T}{N}=m \overline{V}^2 \text { or } \frac{1}{2} m \overline{V}^2=\frac{3}{2} K_B T[/latex] [∵ [latex]\frac{R}{N}[/latex] = KB]
Here KB is Boltzman constant. So mean kinetic energy of a molecule is [latex]\frac{3}{2}[/latex] KB T, which depends upon the temperature. As temperature increases mean kinetic energy of the molecules also increases.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 2.
How specific heat capacity of mono atomic, diatomic and poly atomic gases can be explained on the basis of Law of equipartition of Energy ? [Mar. 13]
Answer:
Specific heat capacity:
1) Mono atomic gases : According to law of equipartition of energy, a molecule of mono atomic gas has only 3 (translational) degrees of freedom, i.e., f = 3.
The molar specific heat of the gas at constant volume is given by Cv = [latex]\frac{f}{2}[/latex] R where f is degree of freedom.
∴ Cv = [latex]\frac{3}{2}[/latex]R = 3 cal/mole k [∵ R = 2 cal/mole – k]
The molar specific heat at constant pressure is given by
Cp = ([latex]\frac{f}{2}[/latex] + 1)R = ([latex]\frac{3}{2}[/latex] + 1) R = [latex]\frac{5}{2}[/latex]R = 5 cal/ m0le – k

2) Diatomic gases : A molecule of diatomic gas has 5 degrees of freedom, 3 translational and 2 rotational i.e., f = 5
Therefore Cv = [latex]\frac{f}{2}[/latex] R = [latex]\frac{5}{2}[/latex]R = 5 cal/mole – k, Cp = ([latex]\frac{f}{2}[/latex] + 1) R = [latex]\frac{7}{2}[/latex] R = 7 cal/mole – k

3) Polyatomic gases : Polyatomic molecule has 3 translational, 3 rotational degrees of freedom, i. e., f = 6
Therefore Cv = [latex]\frac{f}{2}[/latex]R = 3R = 6 cal / mole – k, Cp = ([latex]\frac{f}{2}[/latex] + 1) = 4R = 8 cal / mole – k.

Question 3.
Explain the concept of absolute zero of temperature on the basis of kinetic theory.
Answer:
Concept of absolute zero on the basis of kinetic theory :
Since the pressure of the gas is given by
P = [latex]\frac{1}{3} \mathrm{mn} \overline{\mathrm{V}}^2[/latex], where m = mass of the gas, n = [latex]\frac{\mathrm{N}}{\mathrm{V}}[/latex] = number of molecules per unit volume.
V = r.m.s velocity of gas, P = [latex]\frac{1}{3} \frac{\mathrm{mN}}{\overline{\mathrm{V}}} \overline{\mathrm{V}}^2[/latex] ⇒ PV = [latex]\frac{1}{3} m N \overline{V}^2[/latex]
Also for a gram molecule of the gas, PV = RT
Hence RT = [latex]\frac{1}{3} M \overline{V}^2[/latex] [∵ M = mN = molecular weight]
[latex]\overline{\mathrm{V}}^2=\frac{3 R T}{M}[/latex]
∴ [latex]\overline{v}=\sqrt{\frac{3 R T}{M}}[/latex]
Putting T = 0 in the above equation [latex]\overline{\mathrm{V}}[/latex] = 0. Hence pressure becomes zero. Then the gas converts into liquids. Thus, this temperature is called absolute zero.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 4.
Prove that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.
Answer:
Since the pressure of the gas is given by
P = [latex]\frac{1}{3} \mathrm{mn} \overline{\mathrm{V}}^2[/latex], where m = mass of the gas, n = [latex]\frac{\mathrm{N}}{\mathrm{V}}[/latex] = number of molecules per unit volume
[latex][/latex] = r.m.s velocity of gas, P = [latex]\frac{1}{3} m \frac{N}{V} \overline{V}^2[/latex] ⇒ PV = [latex]\frac{1}{3} \mathrm{mNV}^2[/latex]
Also for a gram molecule of the gas, PV = RT
Hence RT = [latex]\frac{1}{3}[/latex] mNV2
3[latex]\frac{\mathrm{R}}{\mathrm{V}}[/latex]T = [latex]\mathrm{m} \overline{\mathrm{V}}^2[/latex] or [latex]\frac{1}{2}[/latex] mV2 = [latex]\frac{3}{2}[/latex]KBT [∵ [latex]\frac{\mathrm{R}}{\mathrm{V}}[/latex] = KB]
Where KB is the Boltzmann constant and T is the absolute temperature. ∴ E = [latex]\frac{3}{2}[/latex]KBT
Hence the average kinetic energy of a molecule is proportional to the absolute temperature of the gas.

Question 5.
Two thermally insulated vessels 1 and 2 of volumes V1 and V2 are joined with a valve and filled with air at temperatures (T1, T2) and pressures (P1, P2) respectively. If the valve joining the two vessels is opened, what will be the temperature inside the vessels at equilibrium.
Answer:
During adiabatic process, there is no loss of energy, i.e. K.ET of molecules before mixing = K.ET of molecules after mixing.
AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 1

Question 6.
What is the ratio of r.m.s speed of Oxygen and Hydrogen molecules at the same temperature ?
Answer:
The r.m.s speed of gas is [latex]\overline{\mathrm{V}}=\sqrt{\frac{3 R T}{M}} \Rightarrow \frac{\overline{\mathrm{V}}_0}{\mathrm{~V}_{\mathrm{H}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{H}}}{\mathrm{M}_{\mathrm{O}}}}[/latex], Given MH = 2 and M0 = 32
[latex]\frac{\overline{\mathrm{V}}_0}{\overline{\mathrm{V}}_{\mathrm{H}}}=\sqrt{\frac{2}{32}}=\frac{1}{4}[/latex]
∴ V0 : VH = 1 : 4

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 7.
Four molecules of a gas have speeds 1, 2, 3 and 4 km/s. Find the rms speed of the gas molecule.
Answer:
Given V1 = 1 km/s; V2= 2 km/s; V3 = 3 km/s; V4 = 4 km/s; = ?
Vrms = [latex]\sqrt{\frac{V_1^2+V_2^2+V_3^2+V_4^2}{n}}=\sqrt{\frac{1^2+2^2+3^2+4^2}{4}}[/latex]
= [latex]\sqrt{\frac{1+4+9+16}{4}}=\sqrt{\frac{30}{4}}=\sqrt{7.5}[/latex]
= 2735 kms-1

Question 8.
If a gas has ‘f’ degrees of freedom, find the ratio of CP and CV.
Answer:
If a gas has ‘f’ degrees of freedom, then
CV = [latex]\frac{f}{2}[/latex]R and CP = CV + R = [latex]\frac{f}{2}[/latex] R + R = ([latex]\frac{f}{2}[/latex] + 1)R
Hence the ratio of the two specific heats of a gas is, r = [latex]\frac{C_p}{C_V}=\left[\frac{\frac{f}{2}+1}{\frac{f}{2} R}\right]=1+\frac{2}{f}[/latex]

Question 9.
Calculate the molecular K.E of 1 gram of Helium (Molecular weight 4) at 127°C. Given R = 8.31 J mol-1 K-1.
Answer:
Given, t = 127°C, T = 273 + 127 = 400; R = 8.31 J mol-1 K-1
K.E. = [latex]\frac{3}{2}[/latex]KBT = [latex]\frac{3}{2}[/latex] × 1.38 × 10-23 × 400 = 8.28 × 10-21 J.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 10.
When pressure increases by 2%, what is the percentage decrease in the volume of a gas, assuming Boyle’s law is obeyed ?
Answer:
The gas obeys the Boyles law PV = constant, on differentiating, PdV + VdP = 0
PdV = – V dP ⇒ [latex]\frac{\mathrm{dV}}{\mathrm{V}}=\frac{-\mathrm{dP}}{\mathrm{P}}[/latex] ⇒ [latex]\frac{\mathrm{dV}}{\mathrm{V}}[/latex] × 100% = -[latex]\frac{-\mathrm{dP}}{\mathrm{P}}[/latex] × 100%
% change in volume = – percentage change in pressure
∴ % change in volume = – 2%
Here negative sign indicates decrease in volume.

Long Answer Questions

Question 1.
Derive an expression for the Pressure of an ideal gas in a container from Kinetic Theory and hence give Kinetic Interpretation of Temperature.
Answer:
Let us consider cubical vessel of side T with perfectly elastic wall containing gas molecules. Let the three sides of the cube be taken as co-ordinate axis. Consider a molecule moving with velocity V, in any direction at any instant. The components of V1 along the three sides are Vx, Vy and Vz respectively. Then
[latex]\overline{V}^2=V_x^2+V_y^2+V_z^2[/latex] ……………………. (1)
If ‘m is the mass of this molecule, it transfers a momentum mV when it strikes the face ABCD of the cube. Since the wall is perfectly elastic, this molecule is reflected back with a velocity – Vx and momentum – mVx. So the change in momentum = mVx – (- mVx)
= 2mVx
This molecule then travels towards the opposite face, collide with it, rebounds and travels again towards the face ABCD. The distance travelled between two successive collisions is 2l. Time taken between two successive collisions is [latex]\frac{2 l}{\mathrm{~V}_{\mathrm{x}}}[/latex]
AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 2
∴ Number of collisions per second = [latex]\frac{V_x}{2 l}[/latex]
Change in momentum per second = (2mVx) × [latex]\left(\frac{\mathrm{V}_{\mathrm{x}}}{2 l}\right)=\frac{\mathrm{mV}_{\mathrm{x}}^2}{l}[/latex]
Thus force exerted by this molecule = [latex]\frac{\mathrm{mV}_{\mathrm{x}}^2}{l}[/latex] [∵ F = [latex]\frac{\mathrm{dP}}{\mathrm{dt}}[/latex]]
Hence, force exerted by N such molecule in the X-direction.
f1 = [latex]\mathrm{N} \frac{\mathrm{mV}_{\mathrm{x}}^2}{l}=\frac{\mathrm{mNV}_{\mathrm{x}}^2}{l}[/latex]
∴ Pressure exerted by the molecules in the x-direction is Px = [latex]\frac{f_1}{l^2}=\frac{m N \overline{V}_x^2}{l^3}[/latex]
Similarly, pressure exerted by the molecules in the y and z-directions are
Py = [latex]\frac{\mathrm{m}}{l^3} \mathrm{~N} \overline{\mathrm{V}}_{\mathrm{y}}^2[/latex] and Pz = [latex]\frac{\mathrm{m}}{l^3} \mathrm{~N} \overline{\mathrm{V}}_{\mathrm{z}}^2[/latex]
Since the pressure exerted by a gas in all the directions is same, the average pressure
P = [latex]\frac{P_x+P_y+P_z}{3}=\frac{m N}{3 l^3}\left[V_x^2+v_y^2+V_z^2\right]=\frac{m N}{3 V} \overline{V}^2[/latex] [ from (1) and V = l3]
Here V2 is the mean square velocity of the molecule, V is the volume of the vessel.
If M is the mass of the gas then M = mN
∴ P = [latex]\frac{1}{3} \frac{M^2}{V}=\frac{1}{3} m n \overline{V}^2[/latex] [∵ n = [latex]\frac{N}{V}[/latex]]
This pressure is actually the pressure exerted by an ideal gas.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

b) Kinetic interpretation of temperature :
Since the pressure of the gas is given by
P = [latex]\frac{1}{3} m n \overline{V}^2[/latex], where m = mass of the gas, n = [latex]\frac{N}{V}[/latex] = Number of molecules per unit volume
[latex]\overline{V}[/latex] = r.m.s velocity of gas
AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 3
Here KB is Boltzmann constant. So mean kinetic energy of a molecule is [latex]\frac{3}{2}[/latex] KBT, which depends upon the temperature. As temperature increases means kinetic energy of the molecules also increases.

Additional Problems

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Solution:
Here, diameter,
d = 3A°, r = [latex]\frac{\mathrm{d}}{2}=\frac{3}{2}[/latex] A° = [latex]\frac{3}{2}[/latex] × 10-8 cm.
Molecular volume,
v = [latex]\frac{4}{3}[/latex] πr3N, where N is Avagadros No.
= [latex]\frac{4}{3}[/latex] × [latex]\frac{22}{7}[/latex] (1.5 × 10-8)3 × (6.023 × 1023)
= 8.52 cc.
Actual volume occupied by 1 mole of oxygen at STP v = 22400 cc
= [latex]\frac{v}{v^{\prime}}=\frac{8.52}{22400}[/latex] = 3.8 × 10-4 = 4 × 10-4.

Question 2.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C.) Show that it is 22.4 litres.
Solution:
For one mole of an ideal gas, ρv = RT
∴ v = [latex]\frac{\mathrm{RT}}{\rho}[/latex]
Put R = 8.31 J mole-1 k-1, T = 273k, ρ = 1 atm = 1.013 × 105 Nm-2
v = [latex]\frac{8.31 \times 273}{1.013 \times 10^5}[/latex] = o.0224m3 = 0.0224 × 106
cc = 22400 cc = 22.4 lit.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 3.
Figure 14.8 shows plot of PV/T versus P for 1.00 × 10-3 kg of oxygen gas at two different temperatures.
AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 4
a) What does the dotted plot signify ?
b) Which is true: T1 > T2 or T1 < T2 ?
c) What is the value of PV/T where the curves meet on they y-axis ?
d) If we obtained similar plots for 1.00 × 10-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis ? If not, what mass of hydrogen yields the same value of PV/T (for low pressurehigh temperature region of the plot) ? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mole-1 K-1.)
Solution:
a) The dotted plot shows that [latex]\frac{\mathrm{PV}}{\mathrm{T}}[/latex] (= μR) is a constant quantity, independent of pressure p. This sign signifies the ideal gas behaviour.

b) The curve at temperature T1 is closer to dotted plot than the curve at temperature T2. As the behaviour of a real gas approaches the behaviour of a perfect gas when temperature is increased, therefore T1 > T2.

c) Where the two curves meet, the value of
[latex]\frac{\mathrm{PV}}{\mathrm{T}}[/latex] on y-axis is equal to μR.
As mass of oxygen gas = 1.00 × 10-3 kg = 1g
[latex]\frac{\mathrm{PV}}{\mathrm{T}}[/latex] = nR ([latex]\frac{1}{32}[/latex]) × 8.31 Jk-1.

d) If we obtained similar plots for 1.00 × 10-3 kg of hydrogen, we will not get the same value of [latex]\frac{\mathrm{PV}}{\mathrm{T}}[/latex] at the point, where the curve meets on the y-axis. This is because molecular mass of hydrogen is different from that of oxygen.
For same value of [latex]\frac{\mathrm{PV}}{\mathrm{T}}[/latex] mass of hydrogen required is obtained from.
[latex]\frac{\mathrm{PV}}{\mathrm{T}}[/latex] = nR = [latex]\frac{\mathrm{M}}{2.02}[/latex] × 8.31 = 0.26
M = [latex]\frac{2.02 \times 0.26}{8.31}[/latex] gm = 6.32 × 10-2 grams.

Question 4.
All oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol-1 K-1, molecular mass of O2 = 32 u).
Solution:
Initially in the oxygen cylinder,
v1 = 30 lit = 30 × 10-3 m3
P1 = 15 atm = 15 × 1.01 × 105 Pa;
T1 = 27 + 273 = 300 k.
If the cylinder contains n1 mole of oxygen gas, then
P1v1 = nRT1
(or) n1 = [latex]\frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{RT}_1}=\frac{\left(15 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 300}[/latex]
= 18.253 .
For oxygen, molecular weight, M = 32 g
Initial mass of cylinder,
M1 = n1 M = 18.253 × 32 = 584.1 g
Finally in the oxygen cylinder, let n2 moles of O2 be left, were,
v2 = 30 × 10-3 m3, P2 = 11 × 1.01 × 105 Pa, T2 = 17 + 273 = 29K
Now
n2 = [latex]\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{RT}_2}=\frac{\left(11 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 290}[/latex]
= 13.847
∴ Final mass of oxygen gas in the cylinder,
m2 = 13.847 × 32 = 453.1 g
∴ Mass of oxygen taken out = m1 – m2 = 631.0 g.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 5.
An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grown when it reaches the surface, which is at a temperature of 35 °C ?
Solution:
v1 = 1.0 cm3 = 1.0 × 10-6 m3,
T1 = 12 °C = 12 + 273 = 285 k
P1 = 1 atm + h1 ρg = 1.01 × 105 + 40 × 103 × 9.8
= 493000 Pa.
When the air bubble reaches at the surface of lake, then
v2 = 2, T2 = 35 °C = 35 + 273 = 308 K,
P2 = 10 atm = 1.01 × 105 pa.
Now [latex]\frac{P_1 v_1}{T_1}=\frac{P_2 v_2}{T_2}[/latex] (or) v2 = [latex]\frac{P_1 v_1 T_2}{T_1 P_2}[/latex]
∴ v2 = [latex]\frac{(493000) \times\left(1.0 \times 10^{-6}\right) \times 308}{285 \times 1.01 \times 10^5}[/latex]
= 5.27 × 10-6m3.

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure.
Solution:
Here, v = 25.0 m3, T = 27 + 273 = 300 k,
k = 1.38 × 10-23 Jk-1
Now Pv = nRT = n(NK) T = (nN) kT = N’kT
When nN = N’ = total no. of air molecules in the given gas.
N’ = [latex]\frac{\mathrm{Pv}}{\mathrm{kT}}=\frac{\left(1.01 \times 10^5\right) \times 25}{\left(1.38 \times 10^{-23}\right) \times 300}[/latex] = 6.10 × 1026

Question 7.
Estimate the average thermal energy of helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).
Solution:
i) Here, T = 27 °C = 27 + 273 = 300 k
Average thermal energy
= [latex]\frac{3}{2}[/latex] kT = [latex]\frac{3}{2}[/latex] × 1.38 × 10-23 × 300
= 6.2 × 10-21 J

ii) At T = 6000 k. Average thermal energy = [latex]\frac{3}{2}[/latex]kT = [latex]\frac{3}{2}[/latex] × 1.38 × 10-23 × 6000
= 1.24 × 10-19 J.

iii) At T = 10 million K = 107 K.
Average thermal energy
= [latex]\frac{3}{2}[/latex]kT = [latex]\frac{3}{2}[/latex] × 1.38 × 10-23 × 107
= 2.1 × 10-16 J.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (poly-atomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases ? If not, in which case is urms the largest ?
Solution:
All the three vessels (at the same temperature and pressure) have same volume. So in accordance with the avagadro’s law, the three vessels will contain equal number of respective molecules, being equal to
Avagadro’s number N = 6.023 × 1023
As vrms = [latex]\sqrt{\frac{3 k T}{m}}[/latex]
i.e., vrms ∝ [latex]\frac{1}{\sqrt{m}}[/latex] at a given temp, therefore, rms speed of molecules will not be same in the three cases.
As neon has the smallest mass, therefore, rms speed will be largest in case of neon.

Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Solution:
Let c and c’ be the rms velocity of argon and a helium gas atoms at temperature TK and TK’ respectively.
Here μ = 39.9, μ = 4.0,
T = 2, T’ = -20 + 273 = 253 K.
AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 5

Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).
Solution:
Here, λ = 2, f = 2
P = 2 atm = 2 × 1.013 × 105 Nm-2
T = 17 °C = (17 + 273) K = 290 k
σ = 2 × 1 = 2A° = 2 × 10-10 m
K = 1.38 × 10-23 J molecule-1 k-1,
μ = 28 × 10-3 kg
λ = [latex]\frac{\mathrm{KT}}{\sqrt{2} \pi \sigma^2 \rho}[/latex]
= [latex]\frac{1.38 \times 10^{-23} \times 290}{1.414 \times 3.14\left(2 \times 10^{-10}\right)^2 \times 2.026 \times 10^5}[/latex]
λ = [latex]\frac{1.38 \times 29 \times 10^{-7}}{1.44 \times 3.14 \times 2.06}[/latex] = 1.11 × 10-7 m.
vrms = [latex]\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 \times 8.31 \times 290}{28 \times 10^{-3}}}[/latex] = 508.24 m/s.
Collision frequency = no. of collisions per second = [latex]\frac{v_{r m s}}{\lambda}=\frac{508.24}{1.11 \times 10^{-7}}[/latex] = 4.58 × 109

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 11.
A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom ?
Solution:
When the tube is held horizontally, the mercury of length 76 cm traps length of air = 15 cm. A length of 9 cm of the tube will be left at the open end. Fig. (a). The pressure Of air enclosed in tube will be atmospheric pressure. Let area of cross section of the tube be 1 sq.cm.
∴ P1 = 76 cm and v1 = 15 cm3.
When the tube is held vertically, 15 cm air gets another 9 cm of air (filled in the right hand side position) and let h cm of the mercury flows out to balance the atmospheric pressure. Fig. (b). Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively.
The pressure of air = 76 – (76 – h) = h cm of mercury
AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 6
∴ v2 = (24 + h) cm3 and P2 = hcm.
If we assume that temperature remains constant, then
P1v1 = P2v1 or 76 × 15 = h × (24 + h) or h2 + 24h – 1140 = 0
(or) h = [latex]\frac{-24 \pm \sqrt{(24)^2+4 \times 1140}}{2}[/latex] = 23.8 cm
(or) -47.8 cm.
Since h cannot be negative, therefore h = 23.8 cm.
In vertical position it flows out (23.8) cm.

Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3s-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm3s-1. Identify the gas.
[Hint: Use Graham’s law of diffusion : R1/R2 = (M2/M1) where R1, R2 are diffusion rates of gases 1 and 2 and M1 and M2 their respective molecular masses. The law is a simple consequence of kinetic theory.]
Solution:
According to Graham’s law of diffusion
[latex]\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}[/latex]
Where
r1 = diffusion rate of hydrogen = 28.7 cm3s-1
r2 = diffusion rate of unknown gas = 7.2 cm3s-1
M1 = molecule wt. of hydrogen = 2u M2 = ?
M2 = ?
∴ [latex]\frac{28.7}{7.2}=\sqrt{\frac{\mu_2}{\mu_1}} \text { or } M_2=\left(\frac{28.7}{7.2}\right)^2 \times 2[/latex] = 31.78
≈ 32 µ

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so- called law of atmospheres.
n2 = n1 exp [-mg(h2 – h1)/KBT] Where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column :
n2 = n1 exp [-mg NA (ρ – P) (h2 – h1)/(ρRT]
Where ρ is the density of the suspended particle and ρ that of surrounding medium. [NA is Avogadro’s number and R the universal gas constant.]
[Hint: Use Archimedes principle to f ind the apparent weight of the suspended particle.]
Solution:
According to the law of atmospheres n2 = n1 exp. [-[latex]\frac{m g}{K_B T}[/latex] (h2 – h1)] ………… (1)
Where n2 n1 refer to number density of particles at heights h2 and h1 respectively. If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of suspended particles.
Let
v = average vol. of suspended particle
ρ = density of suspended particle
ρ’ = density of liquid
m = mass of one suspended particle
m’ = mass of equal vol. of liquid displaced
According to Archimedis principle, effective weight of one suspended particle.
= actual weight – weight of liquid displaced
= mg – m’ g
= mg – v P’g = mg – (m/P) P’g = mg [latex]\left(\frac{1-\rho^{\prime}}{\rho}\right)[/latex]
Also boltzman constant, KB = [latex]\frac{R}{N A}[/latex] ………….. (i)
Where R is gas constant and NA is Avagadros number.
Putting mg [latex]\left(\frac{1-\rho^{\prime}}{\rho}\right)[/latex] in place of mg and value of KB in (i).
We get
n2 = n1 exp [latex]\left[-\frac{\text { mgNA }}{R T} \cdot\left(1-\frac{\rho^{\prime}}{\rho}\right)\left(h_2-h_1\right)\right][/latex]
Which is the required relation.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 14.
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :
AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory 7
[Hint: Assume the atoms to be tightly packed1 in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of few Å].
Solution:
If r is the radius of the atom, then volume of each atom = [latex]\frac{4}{3}[/latex] πr3
Volume of all atoms in one mole of substance
= [latex]\frac{4}{3}[/latex] πr3 × N = μ/P
r = [latex]\left[\frac{3 \mu}{4 \pi \mathrm{PN}}\right]^{\frac{1}{3}}[/latex]
For carbon p = 12.01 × 10-3 kg,
P = 2.22 × 103 kg m-3
r = [latex]\frac{3 \times 12.01 \times 10^{-3}}{4 \times \frac{22}{7} \times\left(2.2210^3\right) \times\left(6.02310^{23}\right)}[/latex]
r = 1.29 × 10-10 m = 1.29A°
Similarly for gold
r = 1.59A°
For liquid nitrogen r = 1.77A°
For lithium r = 1.73A°
For liquid fluorine r = 1.88A°

Textual Examples

Question 1.
The density of water is 1000 kg m-3. The density of water vapour at 100° C and 1 atm pressure is 0.6 kg m-3. The volume of a molecule multiplied by the total number gives. What is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.
Answer:
For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is 1000/0.6 = /(6 × 10-4) times, large. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. 6 × 10-4.

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 2.
Estimate the volume of a water molecule using the data in Example 1.
Answer:
In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water = 1000 kgm-3. We know that 1 mole of water has a mass approximately equal to (2 + 16)g = 18 g = 0.018 kg.
Since 1 mole contains about 6 × 1023 molecules (Avogadro’s number).
The mass of a molecule of water is (0.018) / (6 × 1023) kg = 3 × 10-26 kg.
Volume of a water molecule = (3 × 10-26 kg) (1000 kg m-3)
= 3 × 10-29 m3
= (4/3)π (Radius)3
∴ Radius » 2 × 10-10 m = 2Å

Question 3.
What is the average distance between atoms (interatomic distance) in water ? Use the data given in Examples 1 and 2.
Answer:
A given mass of water in vapour state has 1.67 × 103 times the volume of the same mass of water inliquid state (e.g. 1). When volume increases by 103 times the radius increases by V1/3 or 10 times,
i.e., = 10 × 12 = 2Å.
So the average distanceis 2 × 10 40 Å.

Question 4.
A vessel contains two non-reactive gases : neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3 : 2. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of Ne = 20.2 u, molecular mass of O2 = 32.0 u.
Answer:
Each gas (assumed ideal) obeys the gas law. Since V and T are common to the two gases, we have P1V = μ1 RT and P2V = μ2 RT. i.e., (P1 / P2) = (μ1 / p2). Here 1 and 2 refer to neon and oxygen respectively.
Since (P1 / P2) = (3/2) (given), (μ1 / μ2) = 3/2.

i) By definition μ1 = (N1 /NA) and μ2 = (N2/NA) where N1 and N2 are the number of molecules of 1 and 2, and NA is the Avogadro’s number. Therefore,
(N1/N2) = (μ1/μ2) = 3/2.

ii) We can also write μ1 = (m1/ M1) and μ2 = (m2 / M2) where m1 and m2 are their molecular masses. (Both m1 and M1; as well as m2 and M2 should be expressed in the same units). If ρ1 and ρ2 are the mass densities of 1 and 2 respectively, we helve
[latex]\frac{\rho_1}{\rho_2}=\frac{m_1 / V}{m_2 / V}=\frac{m_1}{m_2}=\frac{\mu_1}{\mu_2} \times\left[\frac{M_1}{M_2}\right][/latex]
= [latex]\frac{3}{2} \times \frac{20.2}{30.2}[/latex] = 0.947

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 5.
A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the mixtures is 27° C. Obtain the ratio of (i) average kinetic energy per molecule and (ii) root mean square speed vrms of the molecules of the two gases. Atomtic mass of argon = 39.9 u. Molecular mass of chlorine = 70.9 u.
Answer:
The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to (3/2) kBT. It depends only on temperature and is independent of the nature of the gas.
i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1 : 1.
ii) Now 1/2 mvrms2 = average kinetic energy . per molecule = (3/2) KBT where m is the mass of a molecule of the gas. Therefore,
[latex]\frac{\left(\mathrm{V}_{\mathrm{rms}}^2\right)_{A \mathrm{C}}}{\left(\mathrm{V}_{\mathrm{ms}}^2\right)_{\mathrm{cl}}}=\frac{(\mathrm{m})_{\mathrm{C}}}{(\mathrm{m})_{\mathrm{Ar}}}=\frac{(\mathrm{M})_{\mathrm{Cl}}}{(\mathrm{M})_{\mathrm{Ar}}}=\frac{70.9}{39.9}[/latex] = 1.77
where M denotes the molecular mass of the gas.
[latex]\frac{\left(\mathrm{V}_{\text {rms }}\right)_{\mathrm{Ar}}}{\left(\mathrm{V}_{\mathrm{rms}}\right)_{\mathrm{cl}}}[/latex] = 1.33

Question 6.
a) When a molecule (or an elastic ball) hits a (massive) wall, it rebounds with the same speed. When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower ? (ch. 6 will refresh your memory on elastic collisions).
b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above.
c) What happens when a compressed gas pushes a piston out and expands. What would you observe ?
d) Sachin Tendulkar uses a heavy cricket bat while playing. Does it help him in anyway ?
Answer:
a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed V relative to the wicket, then the relative speed of the ball to bat is V + u towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is V + u moving away from the bat. So relative to the wicket the speed of the rebounding ball is V + (V + u) = 2V + u, moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than u if the bat is not massive. For a molecule this would imply an increase in temperature.
You should be able to answer (b) (c) and (d) based on the answer to (a).

AP Inter 1st Year Physics Study Material Chapter 14 Kinetic Theory

Question 7.
A cylinder of fixed capacity 44.8 litres contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0° C ? (R = 8.31 J mol-1 K-1).
Answer:
Using the gas law PV = μRT, you can easily show that 1 mol of any (ideal) gas at standard temperature (273 K) and pressure (1 atm = 1.01 × 105 Pa) occupies a volume of 22.4 litres. This universal volume is called molar volume. Thus the cylinder in this example contains 2 mol of helium. Further, since helium is monatomic, its predicted (and observed) molar specific heat at constant volume, Cv = (3/2) R, and molar specific heat at constant pressure, Cp = (3/2) R + R =(5/2) R. Since the volume of the cylinder is fixed, the heat required is determined by Cv.
Therefore,
Heat required = no. of moles × molar specific heat × rise in temperature
= 2 × 1.5 R × 15.0 = 45
R = 45 × 8.31 = 374 J.

Question 8.
Estimate the mean free path for water molecule in water vapour at 373 K. Use information from Exercises 1 and above. l = 2.9 × 10-7 m = 1500d.
Ans. The d for water vapour is same as that of air. The number density is inversely proportional to absolute temperature.
So n = 2.7 × 1025 × [latex]\frac{273}{373}[/latex] = 2 × 1025 m-3
Hence, mean free path l = 4 × 10-7 m.

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 13th Lesson Thermodynamics Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 13th Lesson Thermodynamics

Very Short Answer Questions

Question 1.
Define Thermal equilibrium. How does it lead to the Zeroth Law of Thermodynamics?
Answer:
If the temperatures of the two systems are equal, then they are said to be in thermal equilibrium.
Zeroth law of thermodynamics: “If two systems (A, B) are in thermal equilibrium with the third system (c) separately, then two systems (A, B) thermal equilibrium with each other.”

Question 2.
Define Calorie. What is the relation between calorie and mechanical equivalent of heat?
Answer:
Calorie: It is defined to be the amount of heat required to raise the temperature of 1g of water from 14.5°C to 15.5°C.
The relation between calorie and mechanical equivalent of heat is given by 1 cal = 4186 J Kg-1K-1 or 1 cal = 4.186 J g-1K-1.

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Qeustion 3.
What thermodynamic variables can be defined by
a) Zeroth Law
b) First Law ?
Answer:
a) Temperature
b) Internal energy.

Question 4.
Define specific heat capacity of the substance. On which factors does it depend ?
Answer:
Specific heat capacity :
The amount of heat required to raise the 1 gm of substance through 1°c or 1 k is called specific heat capacity (or) the heat capacity per unit mass is called specific heat.
S = [latex]\left(\frac{1}{m}\right) \frac{\Delta Q}{\Delta T}[/latex]
It depends on 1) natuture of the substance 2) temperature.

Question 5.
Define molar specific heat capacity.
Answer:
Molar specific heat capacity : The amount of heat required to raise the 1gm mole of substance through 1°C or 1°K is called molar specific heat capacity.

Question 6.
For a solid, what is the total energy of an oscillator ?
Answer:
For a mole of a solid, the total energy of an oscillator, U = 3KB T × NA = 3RT.

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 7.
Indicate the graph showing the variation of specific heat of water with temperature. What does it signify ?
Ans.
AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 1
Significance : the specific heat of water varies slightly with temperature in the range 0 to 100°C.

Question 8.
Define state variables and equation of state.
Answer:
State variables : The state of a system is described by pressure P, temparature T, density p (intensive variables) and internal energy U, Volume V, total mass M (Extensive variables). These are called state variables.

Equation of state : The connection between the state variables is called the equation of state.

Question 9.
Why a heat engine with 100% efficiency can never be realised in practise ?
Answer:
The efficiency of heat engine is η = 1 – [latex]\frac{Q_2}{Q_1}[/latex]
For Q2 = 0, η = 1, i.e., the engine will have 100% efficiency in converting heat into work.
The first law of thermodynamics does not rule out such an engine. But experience shows that such an ideal engine with η = 1 is never possible.

Question 10.
In summer, when the valve of a bicycle tube is opened, the escaping air appears cold.
Answer:
This happens due to adiabatic expansion of the air of the tube of the bicycle.

Question 11.
Why does the brake drum of an automobile get heated up while moving down at constant speed ?
Answer:
The work done by the brake drum on wheel is converted into heat due to friction.

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 12.
Can a room be cooled by leaving the door of an electric refrigerator open ?
Answer:
No, a room can not be cooled by leaving the door of a refrigerator open, rather it will get slightly heated.

Question 13.
Which of the’ two will increase the pressure more, an adiabatic or an isothermal process, in reducing the volume to 50% ?
Answer:
Isothermal process obeys , P1V1 = P2V2
But V2 = [latex]\frac{\mathrm{v_1}}{2}[/latex] ⇒ P1V1, = P2 [latex]\frac{\mathrm{v_1}}{2}[/latex]
∴ P2 = 2P1
Adiabatic process obeys, P1V1γ = P2V2γ
P2 = [latex]P_1\left[\frac{v_1}{v_2}\right]^\gamma=p_1\left[\frac{v_1}{\frac{v_1}{2}}\right][/latex]
∴ P2 = 2γ P1 Where r = [latex]\frac{C_p}{C_v}[/latex]
∴ pressure is more in adiabatic than isothermal.

Question 14.
A thermoflask containing a liquid is shaken vigorously. What happens to its temperature ? .
Answer:
Work is done by the liquid on the wall of flask, since it is vigorously shaken. Hence Internal energy and temperature of the liquid increases.

Question 15.
A sound wave is sent into a gas pipe. Does its internal energy change ?
Answer:
Yes, a sound wave is sent into a gas pipe internal energy increases.

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 16.
How much will be the internal energy change in

  1. isothermal process
  2. adiabatic process

Answer:

  1. Change in internal energy during isothermal process is dU = 0 [∵ U is constant)]
  2. a) Change in internal energy during adiabatic compression increases,
    b) Change in internal energy during adiabatic expansion decreases.

Question 17.
The coolant in a chemical or a nuclear plant should have.high specific heat. Why ?
Answer:
In chemical and nuclear plants more heat is liberated. To absorb this heat, the coolant should have a property of high specific heat for small rise in temperature.
Since S = [latex]\frac{1}{m} \frac{d Q}{d t}[/latex] ⇒ S ∝ [latex]\frac{1}{d t}[/latex]

Question 18.
Explain the following processes

  1. Isochoric process
  2. Isobaric process

Answer:

  1. Isochoric process : A process takes place at constant volume is called Isochoric process. During this process, no work is done on or by the gas, its internal energy and its temperature changes.
  2.  Isoboric process : A process takes place at constant pressure is called isobaric process During this process temperature, internal energy changes. The heat absorbed goes partly to increase internal energy and partly to do work.

Short Answer Questions

Question 1.
State and explain first law of thermodynamics.
Answer:
Statement: The amount of heat supplied to system is equal to the algebraic sum of the change in internal energy of the system and the amount of external work-done.

Explanation : When ∆Q is the quantity of heat is supplied to system, a part of it is used to increase its internal energy ∆U and the rest is used in doing external work ∆W. The mathematical expressions of this law is ∆Q = ∆U – ∆W. It is a special case of law of conservation of energy.

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 2.
Define two principal specific heats of a gas. Which is greater and why ?
Answer:
Two principal specific heats of a gas are

  1. Molar specific heat capacity at constant pressure
  2. molar specific heat capacity at constant volume.

1. Molar Specific heat capacity at constant pressure (Cp): The amount of heat required to raise the temperature of 1 gm – mole of a gas through 1°C at constant pressure is called molar specific heat at constant pressure.
i.e., Cp = [latex]\frac{1}{\mu} \frac{\Delta Q}{\Delta T}[/latex] where μ is no of moles.

2. Molar Specific heat capacity at constant volume (Cv): The amount of heat required to raise the temperature of 1gm – mole of a gas through 1°C at constant volume is called molar specific heat at constant volume i.e., Cv = [latex]\frac{1}{\mu} \frac{\Delta Q}{\Delta T}[/latex]

Explanation of Cp is greater that Cv: When a gas is heated at a constant pressure, it expands. The heat supplied to it is used partly in raising its temperature and partly in doing work against the external pressure. If, on the other hand, the gas is heated at constant volume, no work is done. Therefore, the heat supplied is to be used only in raising the temperature. Hence the amount of heat required to be supplied to a gas to raise its temperature by 1 °C (say) at constant pressure will be greater than the amount required at constant volume.

Question 3.
Derive a relation between the two specific heat capacities of gas on the basis of first law of thermodynamics
Answer:
The mathematical equation of first law of thermodynamics for 1 mole of gas is given by
∆Q = ∆U + P∆V
if ∆Q is absorbed heat at constant volume, ∆V = 0
Cv = [latex]\left[\frac{\Delta Q}{\Delta T}\right]_V=\left[\frac{\Delta U}{\Delta T}\right]_V=\left[\frac{\Delta U}{\Delta T}\right][/latex] ……………. (1)
Where the subscript V is dropped in the last step, since U of an ideal gas depends only on temperature. (The subscript denotes the quantity kept fixed.) If, on the other hand, ∆Q is absorbed heat at constant pressure,
Cp = [latex]\left[\frac{\Delta Q}{\Delta T}\right]_P=\left[\frac{\Delta U}{\Delta T}\right]_P+P\left[\frac{\Delta V}{\Delta T}\right]_P \Rightarrow\left[\frac{\Delta Q}{\Delta T}\right]_P=\left[\frac{\Delta U}{\Delta T}\right]+P\left[\frac{\Delta V}{\Delta T}\right]_P[/latex] ………………… (2)
The subscript p.can be dropped from the first term since U of an ideal gas depends only on T. Now, for a mole of an ideal gas.
PV = RT
Which gives P[latex]\left[\frac{\Delta V}{\Delta T}\right]_P[/latex] = R
Putting the equations (1) and (3) in (2), we get
Cp = Cv + R
∴ Cp – Cv = R

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 4.
Obtain an expression for the work done by an ideal gas during isothermal change.
Answer:
Expression for the work done by an ideal gas during isothermal change : Let a certain mass of gas expands from volume V1 to V2 isothermally at constanst temperature T. Let the pressure changes form P1 to P2.
The work done during the small change in volume dv at a pressure ‘p’ is dw = pdv
The total work done during the change of volume from V1 to V2 is W = [latex]\int d w=\int_{V_1}^{V_2} p d v[/latex] …………. (1)
The Isothermal change is represented by
PV = constant
PV = μRT ⇒ P =[latex]\frac{\mu \mathrm{RT}}{\mathrm{V}}[/latex] where μ = no. of moles
substituting this value in equation (1)
W = [latex]\int_{V_1}^{v_2} p d v=\int_{V_1}^{V_2} \frac{\mu R T}{V} d V=\mu R T \int_{V_1}^{V_2} \frac{d v}{V}[/latex]
W = μRT loge [latex]\frac{V_2}{V_1}[/latex]
∴ Work done during the isothermal change W = 2.3026 RT log10 [latex]\frac{V_2}{V_1}[/latex] .

Question 5.
Obtain an expression for the work done by an ideal gas during adiabatic change and explain.
Answer:
Expression for the work done by an ideal gas during adiabatic change :
During an adiabatic chnage the state of an ideal gas changes from (P1, V1, T1) to (P2, V2, T2)
The workdone during a small change in volume dV at a pressure p is dw = pdV
Total work done by gas from volume V1 to V2 is ,
W = [latex]\int[/latex] dw = [latex]\int_{v_1}^{v_2}[/latex] pdv ………………. (1)
The adiabatic relation between pressure and volume is given by PVr = constant K …………. (2)
P = [latex]\frac{k}{v^r}[/latex] …………… (3)
and P1V1r = P2V2r = K
Substituting eg. (3) in (1) becomes
AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 2
This is equal to the work done during the adiabatic change.

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 6.
Compare isothermal and an adiabatic process.
Answer:
Isothermal change

  1. Changes in volume and pressure of a gas taking place at constant temperature are called isothermal changes,
  2. Temperature of the gas remains constant.
  3. The gas remains in good thermal contact with the surroundings and heat is exchanged.
  4. Internal energy remains constant. Change in internal energy ∆U = 0
  5. This process takes place slowly.
  6. Boyle’s law PV = Constant holds good.
  7. Work done W = RT loge[latex]\frac{V_2}{V_1}[/latex] .

Adiabatic change

  1. Changes in volume and pressure of a gas taking place in a thermally isolated system are called adiabatic changes.
  2. Temperature of the gas changes.
  3. The gas is isolated from the surroundings and heat is not exchanged ∆Q = 0.
  4. Internal energy changes.
  5. This process takes place quickly.
  6. PVr = Constant
  7. Work done W = [latex]\frac{R}{(r-1)}[/latex](T1 – T2)

Question 7.
Explain the following processes
i) Cyclic process with example
ii) Non cyclic process with example
Answer:
i) Cyclic process: “A process in which the system after passing through various stages (of pressure, volume and temperature changes) returns to its initial state1′ is called as a cyclic process.
For cyclic process P.V graph is a closed curve. The area under P-V graph gives work done by the substance. In a cyclic process there will be no change in the internal energy i.e., ∆U = 0
The first law of thermodynamics states that
∆Q = ∆U + ∆W
∴ For cyclic process ∆Q = ∆W
In cyclic process, the total heat absorbed by the system equals the work done by the system. Example : Fleat engine is a device by which a system is made to undergo a cyclic process that results in conversion of heat to work.

ii) Non cyclic process : A process which is not cyclic is called Non-cyclic process. In Non-cydic process, the system does not attains its initial state. Irreversible process is a Non cyclic process. Workdone in non-cyclic process is given by the area between the curve and volume axis.

Example :

  1. Diffusion of liquids or gases
  2. Free expansion of a perfect gas.

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 8.
Write a short note on Quasi – static process.
Answer:
A quasi-static process is defined as “an infinitesimally slow process in which at each and every intermediate stage the system remains in thermal and mechanical (thermodynamic) equilibrium with the surroundings through out the entire process.”

In this process the pressure and temperature of the surroundings at each and every stage will differ from those of the system only in a very small amount.

Any process taking place sufficiently slowly, not involving acceleration motions and large temperature gradients can be considered as quasi -static process.
Example : isothermal process, adiabatic process.

Question 9.
Explain qualitatively the working of a heat engine.
Answer:
“A device used to convert heat energy into work (or mechanical energy)” is called heat engine. A heat engine converts heat into work.
A heat engine consists of the following essential parts :
AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 3

  1. A body at a higher temperature T1. Heat Q1 is extracted from this body and hence it is called a “hot reservior” (or source).
  2. The engine contains working substance. In a steam engine the working substance is the steam. In a diesel engine the working substance is a mixture of fuel vapour and air.
  3. A body at lower temperature T2. Heat Q2 is rejected by the working substance to this body and hence it is called a cold reservoir (or sink).

The difference in heat absorbed (Q1) and heat rejected (Q2) is equal to the workdone by the system (W). This is because the engine is operated in a cyclic process. W = Q1 – Q2.

The heat engine operates in a cyclic process and in each cycle, the work done (W) by the system is transferred to the environment via some arrangement (shaft) to drive vehicles.

The efficiency (η) of aheat engine is defined as the ratio of the workdone (W) by the engine to the amount of heat absorbed (Q1) by the engine.
Efficiency (η) = [latex]\frac{\mathrm{W}}{\mathrm{Q}_1}=\frac{\mathrm{Q}_1-\mathrm{Q}_2}{\mathrm{Q}_1}=1-\frac{\mathrm{Q}_2}{\mathrm{Q}_1}[/latex]

Long Answer Questions

Question 1.
Explain reversible and irreversible processes. Describe the working of Carnot engine. Obtain an expression for the effieciency. [AP – Mar. ’18, ’17, ’16, 15; TS – Mar. ’17, ’15, ’14]
Answer:
Reversible process : A process that can be retraced back in the opposite direction in such away that the system passes through the same states as in the direct process, and finally the system and the surroundings return in their original states, is called a reversible process.
A reversible process is only a purely idealised process.
Examples:

  1. Slow isothermal and slow adiabatic changes
  2. Peltier effect and seeback effect.
  3. Fusion of ice and vaporisation of water.

Irreversible process : “A process that cannot be retraced back in the opposite direction” is called as irreversible process.

In this process the system does not pass through the same intermediate states as in the reversible process.
All natural processes such as conduction, radiation, radioactivity etc, are irreversible.
Example:

  1. Work done against friction.
  2. Heat produced in conductors by passing a current through it. (Joule heating)
  3. Diffusion of gases.

Carnot Engine : A reversible heat engine operating between two temperatures is called a carnot engine. The cycle operating it is known as carnot’s cycle. In this cycle the working substance(say an ideal gas) is taken through a cycle by means of two isothermals and two adiabatics. The four operations are shown in P-V (Indicator) diagram.
AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 4

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Step 1 → 2 : Isothermal expansion of the gas taking its state from (P1, V1, T1) to (P2, V2, T2). It is shown in curve (a).
The heat absorbed by the gas (Q1) from the reservoir at temperature
T1 equals the workdone by the gas. i.e.
W1 → 2 = Q1 = μRT1 loge [latex]\frac{V_2}{V_1}[/latex]

Step 2 →3 : Adiabatic expansion of the gas from (P2, V2, T1) to (P3, V3, T2). It is shown in curve (b). The work done by the gas is gÑen by
W2 → 3 = [latex]\frac{\mu R\left(T_1-T_2\right)}{(r-1)}[/latex]

Step 3 → 4: Isothermal compression of the gas from (P3, V3, T2) to (P4, V4, T2). It is shown in curve (c).
Heat released (Q2) by the gas to the reservoir at temperature T2 equals the work done on the goes.
i.e., W3 → 4;= μRT2 loge [latex]\frac{V_4}{V_3}[/latex] = -μRT1 loge [latex]\frac{V_3}{V_4}[/latex]

Step 4 → 1: Adiabatic compression of the gas from (P4, V4, T2) to (P1, V1, T1). It is shown in curve (d) Work done on the gas is given by.
W4 → 1 = μR [latex]\frac{\left(T_2-T_1\right)}{(r-1)}=-\mu R \frac{\left(T_1-T_2\right)}{(r-1)}[/latex]
Total work done by the gas in one complete cycle is
W = W1 → 2 + W2 → 3 + W3 → 4 + W4 → 1
= [latex]\mu R T_1 \log _e \frac{V_2}{V_1}+\frac{\mu R\left(T_1-T_2\right)}{(r-1)}-\mu R T_2 \log _e \frac{V_3}{V_4}-\frac{\mu R\left(T_1-T_2\right)}{(r-1)}[/latex]
∴ W = μRT1 loge [latex]\frac{V_2}{V_1}[/latex] – μRT2 loge [latex]\frac{3}{V_4}[/latex] = Q1 – Q2 …………… (2)
The efficiency of the cannot engine is
η = [latex]\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}[/latex] ………… (3)
Now since step 2 → 3 is an adiabatic process,
T1V2r – 1 ≈ T2V3r – 1
Similarly, since step 4 → 1 is an adiabatic process,
T2V2r – 1 = T1V1r – 1 ⇒ T1V1r – 1 = T2V2r – 1
AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 5

Question 2.
State second law of thermodynamics. How is heat engine different form a refrigerator. [TS – Mar. ’18, ’16; AP – Mar. ’15, ’13]
Answer:
Second law of thermodynamics gives the direction of flow of heat. Second law consists two statements.

  1. Kelvin – Planck statement: “No process is possible whose sole result is the absorption of heat from a reservoir and the complete conversion of the heat into work”.
    (or)
    “It is impossible to derive a continuous supply of engines in cooling a body below the coldest of its surroundings1′.
  2. Clausius statement : “No process is possible whose sole result is the transfer of heat from a colder object to a hotter object”.
    (or)
    “Heat can not itself flow from cold body to hot body”.

Heat engine : A device which converts heat energy into work is called heat engine.
A heat engine consists of the essential parts.
AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 6

  1. Source or hot reservoir: It is maintained at a very high temperature T1. Heat is extracted from this body.
  2. Working substance : In a steam engine working substance is steam. In a diesel engine working substance is mixture of fuel vapour and air.
  3. Sink or cold reservoir: It is maintained at a very low temperature T2. It absorbs heat energy, rejected by working substance.

Work done: The difference of heat absorbed from source and heat rejected to sink is equal to work done by the engine, i.e., W = Q1 – Q2.
Efficiency : The ratio of workdone (W) by the engine to the amount of heat absorbed (Q1) by the engine is called efficiency.
i.e., η = [latex]\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}[/latex]
Refrigerator : Refrigerator is a heat pump which is the reverse of a heat engine.
AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 7
In refrigerator the working substance extracts (Q2) an amount of heat from sink of lower temperature and a external work W is done on the working substance and finally amount of heat (Q1) is transfered to source.
The coefficient of performance (α) of a refrigerator is given by
α = [latex]\frac{\mathrm{Q}_2}{\mathrm{~W}}=\frac{\mathrm{Q}_2}{\mathrm{Q}_1-\mathrm{Q}_2}[/latex] (∴ heat released Q1 = Q2 + W)
For heat engine η can never exceed 1. For refregirator a can be greater than 1.
Therefore working of refrigerator is opposite to that of heat engine.

Problems

Question 1.
If a monoatomic ideal gas of volume 1 litre at N.T.P. is compressed (i) adia- batically to half of its volume, find the work done on the gas. Also find (ii) the work done if the cdmpression is iso- thermal, (γ = 5/3)
Solution:
i) During an adiabatic process
T1V1r – 1 = T2V2r – 1
Here T1 = 273 K
V2 = [latex]\frac{V_1}{2}[/latex]
r = [latex]\frac{5}{3}[/latex]
T2 = [latex]\left[\frac{V_1}{V_2}\right]^{\frac{5}{3}-1}=273\left[\frac{V_1}{V_1}\right]^{\frac{2}{3}}[/latex]
T2 = (2)[latex]\frac{2}{3}[/latex] = 273 = 431.6 K
Number of moles = [latex]\frac{1 \text { litre }}{22.4 \text { litre }}=\frac{1}{22.4}[/latex]
Work done = [latex]\frac{\mu R}{(r-1)}[/latex] [T1 – T2]
= [latex]\frac{8.314}{22.4\left[\frac{5}{3}-1\right]}[/latex] × (273 – 431.6)
= [latex]\frac{8.314 \times 3}{22.4 \times 2}[/latex] (-158.6)
= -89 J

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

ii) Work done during isothermal compression is
w = 2.3026 μRT log10[latex]\frac{V_2}{V_1}[/latex]
μ = Number of moles = [latex]\frac{1}{22.4}[/latex]
T = 273K
R = 8.314 Jmol-1K-2
[latex]\frac{V_2}{V_1}=\frac{1}{2}[/latex] = 0.5
∴ W = [latex]\frac{2.3026 \times 8.314 \times 273 \log _{10}(0.5)}{22.4}[/latex]
Or w = -70J

Question 2.
Five moles of hydrogren when heated through 20 K expand by an amount of 8.3 × 10-3m3 under a constant pressure of 105 N/m2. If Cv = 20.1/mole K, find Cp.
Solution:
Mayer’s relation Cp – Cv = R
Multiplying throughout by μ∆T
μCp∆T – μCv∆T = μ R∆T
μ ∆T(Cp – Cv) = P∆T [∴ μ R∆T = P∆V]
5 × 20 (Cp – 20) = 105 × 8.3 × (10 – 3)
[∴ μ = 5, ∆T = 20 K, P = 1 × 105 N/m2 Cv = 20 J/mole K and ∆V = 8.3 × 103 M3]
Cp – 20 = 8.3
∴ Cp = 28.3 J/mole-K

Additional Problems

Question 1.
A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operate on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g ?
Solution:
Here volume of water heated = 3.0 lit/min.
Mass of water heated m = 3000 g/min
Rise in temperature ∆T = 77 – 27 = 50°C
Sp. heat of water C = 4.2 Jg-1C-1
Amount of heat used
∆Q = mc∆T = 3000 × 4.2 × 50
= 63 × 104 J/min
Heat of combustion = 4 × 104 J/g
Rate of combustion of fuel = [latex]\frac{63 \times 10^4}{4 \times 10^4}[/latex]
= 15.75 g/min

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 2.
What amount of heat must be supplied t 2.0 × 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.)
Solution:
Here mass of gas m = 2 × 10-2kg = 20g
Rise in temperature ∆T = 45°C
Heat required ∆Q = ?
Molecular mass M = 28
Number of moles n = [latex]\frac{m}{M}=\frac{20}{28}[/latex] = 0.714
As nitrogen is a diatomic gas, molecular specific heat of constant pressure is
Cp = [latex]\frac{7}{2}[/latex] R = [latex]\frac{7}{2}[/latex] × 8.3 J mole-1 K-1
As ∆Q = nCp∆T
∆Q = 0.714 × [latex]\frac{7}{2}[/latex] × 8.3 × 45 J = 933.4 J

Question 3.
Explain why
a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2)/2.
b) The coolant in a chemical or nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
c) Air pressure in a car tyre increases during driving.
d) The climate of a harbour town is more temperate then that of a town in a desert at the same latitude.
Solution:
a) In thermal contact heat flows from the body at higher temp to the body at lower temperature till temperatures becomes equal. The final temperature can be the mean temp (T1 + T2)/2 only when thermal capacities of the two bodies are equal.

b) This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.

c) During, driving the temperature of air inside the tyre increases due to motion. According to Charles law, P α T. Therefore, air pressure inside the tyre increases.

d) This is because in a harbour town, the relative humidity is more than in a desert town. Hence the climate of a harbour town is without extremes of hot and cold.

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ?
Solution:
As no heat is allowed to be exchanged, the process is adiabatic.
∴ P2V2r = P1V1r (or) [latex]\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^r[/latex]
As V2 = [latex]\frac{V_1}{2}[/latex]
∴ P2/P1 = [latex]\left(\frac{V_1}{\frac{1}{2} V_1}\right)^{1 .4}[/latex]
= 21.4 = 2.64

Question 5.
In changing the state of a gas adiabatically form an equilibrium state A to another equilibrium state B, and amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal. how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J)
Solution:
Here, when the change is adiabatic, ∆Q = 0, ∆w = -22.3 J
If ∆u is change in internal energy of the system, then as
∆Q = ∆u + ∆w
O = ∆u – 22.3 (or) ∆u = 22.3 J
In the second case ∆Q = 9.35 cal
= 9.35 × 4.2 J = 39.3 J
∆w = ?
As ∆u + ∆w = ∆Q
∆w = ∆Q – ∆u
= 39.3 – 22.3 = 17.0 J

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :
a) What is the final pressure of the gas in A and B ?
b) What is the change in internal energy of the gas ?
c) What is the change in the temperature of the gas ?
d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ?
Solution:
a) When the stopcock is suddenly opened, the volume available to the gas at 1 atm pressure will become two times. Therefore pressure will decrease to one-half i.e., 0.5 atm.

b) There will be no change in the internal energy of the gas as no work is done on/ by the gas.

c) Also there will be no change in temp of the gas as gas does no work in expansion.

d) No, because the process called free expansion is rapid and cannot be controlled. The intermediate states are non equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.

Question 7.
A steam engine delivers 5.4 × 108J . of work per minute and services 3.6 × 109J of heat per minute form its boiler, what is the efficiency of the engine ? How much heat is wasted per minute ? .
Solution:
Output i.e., useful workdone per min
= 5.4 × 108 J
Input i.e., heat absorbed per min
= 3.6 × 109 J
Efficiency = [latex]\frac{\text { Output }}{\text { Input }}=\frac{5.4 \times 10^8}{3.6 \times 10^9}[/latex] = 0.15
= 0.15 × 100% = 15%
heat energy wasted / minute = Heat absorbed / minute – useful work done / minute
= 3.6 × 109 – 5.4 × 108 = 109 (3.6 – 0.54)
= 3.06 × 109

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 8.
An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing ?
Solution:
Heat supplied, ∆Q = 100 w = 100 J/s
Useful workdone ∆w = 75 J/s
Increase in internal energy/ sec, ∆u = ?
As ∆Q = ∆u + ∆w
∴ ∆u = ∆Q – ∆w
= 100 – 75 = 25 J/S

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig.
AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics 8
Its volume is then reduced to the original value form E to F by an isobaric process. Calculate the total work done by the gas from D to E to F
Solution:
A is clear from fig.
Change in pressure dp = EF = 5.0 – 2.0
= 3.0 atm = 3.0 × 1015 Nm-2
Change in volume dv = DF = 600 – 300
= 300 cc = 300 × 10-6 m3
Workdone by the gas from D to E to F = area of ∆DEF
w = [latex]\frac{1}{2}[/latex] × DF × EF
= [latex]\frac{1}{2}[/latex] × (300 × 10-6 ) × (3.0 × 105 ) = 45 J

AP Inter 1st Year Physics Study Material Chapter 13 Thermodynamics

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C. calculate the coefficient of performance.
Solution:
Here T1 = 36°C = 36 + 273 = 309 K
T2 = 10°C = 10 + 273 = 283 K
COP = [latex]\frac{T_2}{T_1-T_2}=\frac{283}{309-283}[/latex]
= [latex]\frac{283}{26}[/latex] = 10.9

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter

Very Short Answer Questions

Question 1.
Distinguish between heat and temperature. [T.S. Mar. 15]
Answer:
Heat

  1. It is a form of energy.
  2. It is the cause on the body.
  3. It is determined with calorimeter.
  4. It’s unit is Joule and Calories.

Temperature

  1. It is a degree of hotness or coldness of a body.
  2. It is an effect on a body.
  3. It is measured with thermometer.
  4. It’s unit is degree’Celsius or Kelvin or degree Fahrenheit.

Question 2.
What are the lower and upper fixing points in Celsius and Fahrenheit scales ? [T.S. Mar. 16]
Answer:
In Celsius scale, lower fixed point is ice point or 0°C and upper fixed point is steam point or 100°C. In Fahrenheit scale, lower fixed point is 32°F and upper fixed point is 212°F.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 3.
Do the values of coefficients of expansion differ, when the temperatures are measured on Centigrade scale or on Fahrenheit scale ?
Answer:
Yes. The coefficients of expansion depends on scale of temperature because
[latex]\frac{\alpha}{{ }^{\circ} \mathrm{C}}=\frac{9}{5} \alpha /{ }^{\circ} \mathrm{F}[/latex]

Question 4.
Can a substance contract on heating ? Give an example.
Answer:
Yes. Rubber, type metal, cast iron contract on heating.

Question 5.
Why gaps are left between rails on a railway track ? [A.P. Mar. 17, 16]
Answer:
The length of the rails increases in summer due to high temperature. Therefore a gap is left to allow this expansion.

Question 6.
Why do liquids have no linear and areal expansions ?
Answer:
Liquids occupy the same shape of vessel. They do not have individual length and area. Hence, liquids have no linear and areal expansions.

Question 7.
What is latent heat of fusion ?
Answer:
The amount of heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure is called the latent heat of fusion (Lf).

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 8.
What is latent heat of vapourisation ? [Mar. 13]
Answer:
The amount of heat per unit mass required to change a substance from the liquid to the vapour state at the same temperature and pressure is called the latent heat of vaporisation (Lv).

Question 9.
What is specific gas constant ? Is it same for all gases ?
Answer:
Universal gas constant per molecular mass is called specific gas constant.
i.e., r = [latex]\frac{\mathrm{R}}{\mathrm{m}}[/latex]. No, it is different for different gases.

Question 10.
What are the units and dimensions of specific gas constant ?
Answer:
The units of specific gas constant is J/Kg/K. Dimensional formula is (m0.L2t-2K1).

Question 11.
Why utensils are coated black ? Why the bottom of the utensils are made of copper ?
Answer:

  1. Utensils are coated black, because it is a good absorber of heat.
  2. Copper is a good conductor of heat. So, copper is used at the bottom of cooking utensils.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 12.
State Weins displacement law. [A.P. Mar. 17]
Answer:
The wavelength (λm) corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature
i.e., λm ∝ [latex]\frac{1}{T}[/latex]

Question 13.
Ventilators are provided in rooms just below the roof. Why ? [Mar. 14]
Answer:
Ventilators are provided in rooms just below the roof, because the hot air escapes out and fresh air enters into the rooms due to convection.

Question 14.
Does a body radiate heat at 0 K ? Does it radiate heat at 0°C ?
Answer:

  1. No. A body does not radiate heat at 0k.
  2. Yes. A body radiate heat at 0°C.

Question 15.
State the different modes of transmission of heat. Which of these modes require medium ?
Answer:
The different modes of transmission are : 1) conduction 2) convection and 3) radiation.
Among three modes conduction and convection require medium.

Question 16.
Define coefficient of thermal conductivity and temperature gradient.
Answer:
Coefficient of thermal conductivity : It is defined as the quantity of heat energy flowing per second between the opposite faces of cube of unit side, which are maintained at unit temperature difference.
Temperature gradient: The change of temperature per unit length of conductor is called temperature gradient.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 17.
What is thermal resistance of a conductor ? On what factors does it depend ?
Answer:
The resistance offered by the conductor for the flow of heat is called thermal resistance (R) of the conductor. R = [latex]\frac{\mathrm{d}}{\mathrm{kA}}[/latex]. It depends on

  1. The nature of the material, i.e., thermal conductivity (K).
  2. d; length of conductor along which heat flows and
  3. A; area of cross section of the conductor.

Question 18.
State the units and dimensions of coefficient of convection.
Answer:
The unit of coefficient of convection is, wm-2k-1.
Dimensional formula (m.L0T3θ-1).

Question 19.
Define emissive power and emissivity ?
Answer:
Emissive power : The energy flux emitted by unit surface area of a radiating body is known as emissive power (eλ).

Missivity (e): It is defined as the ratio of the emissive power of the body to that of black body at the same temperature.

Question 20.
What is greenhouse effect ? Explain global warming. [A.P. Mar. 15, 13]
Answer:
Greenhouse effect: When the earth receives sun light, it gets heated up and emits infrared radiation into air. CO2, CH4, N2O, O3, Chlorofluoro carbon (green house gases) present in the air absorbs the heat content of infrared radiation and keeps the earth warm. This is called green house effect.

Global warming : As CO2 content increases, more heat is retained in the atmosphere and the temperatures all over the world increases. This is called global warming.

Effects : a) Polar ice caps melt and fed more water in rivers and seas results in severe floods, b) In some areas, water resources dry up leading to drought conditions.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 21.
Define absorptive power of a body. What is the absorptive power of a perfect black body ?
Answer:
Absorptive power : At a given temperature and wavelength, the ratio of the amount of radiant energy absorbed to the amount of radiant energy incident in a wavelength range is called the absorptive power at that temperature and wavelength. Absorptive power of a perfect black body is 1.

Question 22.
State Newton’s law of cooling. [A.P. Mar. 18, 16; T.S. Mar. 18]
Answer:
The rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings provided the temperature difference is small. i.e., – [latex]\frac{\mathrm{dQ}}{\mathrm{dt}} \alpha\left(\mathrm{T}_B-\mathrm{T}_{\mathrm{S}}\right)[/latex]

Question 23.
State the conditions under which Newton’s law of cooling is applicable.
Answer:

  1. Loss of heat is negligible by conduction and only when it is due to convection.
  2. Loss of heat occurs in a streamlined flow of air i.e., forced convection.
  3. Temperature of the body is uniformly distributed over it.

Question 24.
The roof of buildings are often painted white during summer. Why ? [T.S. Mar. 17, 15]
Answer:
The roof of buildings are painted white during summer, because it reflects more heat radiations and absorbs less heat radiations. Hence we feel comfort inside the house.

Short Answer Questions

Question 1.
Explain Celsius and Fahrenheit scales of temperature. Obtain the relation between Celsius and Fahrenheit scales of temperature.
Answer:
Centigrade (Celsius) scale of temperature: In the Celsius scale of temperature, the lower fixed point is called the ice point and is assigned the value 0°C. The upper fixed point is called the steam point and is assigned the value 100°C. The interval between these two points (i.e., 100°C – 0 = 100°C) is subdivided into 100 equal parts each one corresponding to 1°C.
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 1
Fahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32°F and the upper fixed point is the staempoint and is assigned a value 212°F. The interval between these two points (i.e., 212°F – 32°F = 180°F) is subdivided into 180 equal parts, each one corresponding to 1°F.

Fahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32°F and the upper fixed point is the staempoint and is assigned a value 212°F. The interval between these two points (i.e., 212°F – 32°F = 180°F) is subdivided into 180 equal parts, each one corresponding to 1°F.
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 2
Relation between Celsius and Fahreinheit scales of temperature:
Difference of 100 Celsius degrees = Difference of 180 Fahrenheit degress
When the temperature of a body is measured on both the Celsius and Fahrenheit scales, let the readings be tC and tF respectively. Then
[latex]\frac{t_c-0}{t^{100}}=\frac{t_F-32}{180} \Rightarrow \frac{t_C}{5}=\frac{t_F-32}{9}[/latex]
C = [latex]\frac{5}{9}[/latex] (F – 32)

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 2.
Two identical rectangular strips, one of copper and the other of steel, are riveted together to form a compound bar. What will happen on heating ?
Answer:
Bimetallic strip:
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 3
Two similar strips of copper and steel are revitted side by side as a compound strip called bimetallic strip. It is straight at ordinary temperatures. When the bimetallic strip is heated, copper expands more than steel. So, it bends with copper on convex side. When the bimetallic strip is cooled below room temperature, copper contracts more than steel and it bends with copper on concave side. Bimetallic strip is used in refrigerators, automatic iron, fire alarm etc.

Question 3.
Pendulum clocks generally go fast in winter and slow in summer. Why ?
Answer:
The time period of a pendulum at t1°C is T1 = 2π [latex]\sqrt{\frac{L_1}{g}}[/latex] ………….. (1)
Where L1 is length of pendulum at t1°C.
If L2 is length of pendulum at t2°C,
then T2 = 2π [latex]\sqrt{\frac{L_2}{g}}[/latex] ………………. (2)
[latex]\frac{(2)}{(1)}[/latex] gives [latex]\frac{T_2}{T_1}=\sqrt{\frac{L_2}{L_1}}=\sqrt{\frac{L_1(1+\alpha t)}{L_1}}[/latex]
Where a is the linear coefficient of expansion of pendulum clock. Where t = t2 – t1
[latex]\frac{T_2}{T_1}=\sqrt{(1+\alpha t)}=(1+\alpha t)^{\frac{1}{2}}[/latex]
[latex]\frac{T_2}{T_1}=1+\frac{\alpha t}{2}[/latex]
[latex]\frac{T_2}{T_1}-1=\frac{\alpha t}{2}[/latex]
[latex]\frac{T_2-T_1}{T_1}=\frac{\alpha t}{2}[/latex]
[latex]\frac{T_2-T_1}{T_1}[/latex] = time lost by pendulum per second.
Due to expansion in length, during summer, time period increases or the clock looses time in summer. In winter due to fall in temperature, the length contracts, time period decreases, hence clock gains time.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 4.
In what way is the anomalous behaviour of water advantageous to acquatic animals ? [A.P. Mar. 18, Mar. 14]
Answer:
In cold countries, as atmospheric temperature decreases, the upper layers of the lakes, rivers etc., cool, contract and sink to the bottom (fig). This goes on until the whole of the water reaches the temperature of 4°C. When the top layers cool further temperature falls below 4°C, it expands and becomes lighter. It does not sink downwards and remains at the top. With further cooling the top layer gradually form ice at the top (fig). Ice and water are bad conductors of heat. So the lower layers are protected against freezing by the layers of ice and cold water at 1°C, 2°C and 3°C. This results in water remaining at the bottom at 4°C. So that aquatic animals survive in those layers of water.
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 4

Question 5.
Explain conduction, convection and radiation with examples. [TS – Mar. ’18, ’16, ’15; AP – Mar. ’15]
Answer:
The heat is transmitted in three types. They are 1) Conduction 2) Convection 3) Radiation.

  1. Conduction : The process of transmission of heat from one place to other without actual movement of the particles of the medium is called conduction.
    Ex : When long iron rod is heated at one end, heat transmits to the other end.
  2. Convection : The process of transmission of heat from one place to another by the actual
    movement of the particles is called convection.
    Ex. : If water in a beaker is heated, the particles of water at the bottom receive the heat first. These particles expand, become lighter and rise up. At the same time colder and denser particles reach the bottom. They get in their turn heated and move up. This process is known as convection.
  3. Radiation : The process of transmission of heat from one place to another without any intervening medium is called radiation.
    Ex. : Earth receives heat radiations from the sun.

Long Answer Questions

Question 1.
State Boyle’s law and charle’s law. Hence, derive ideal gas equation which of two laws is better for the purpose of Thermometry and why ?
Answer:
P ∝ [latex]\frac{1}{\mathrm{V}}[/latex]
⇒ PV = constants ⇒ P1V1 – P2V2
Charles law at constant volume : At constant volume the pressure of given mass of a gas is directly proportional to absolute temperature of gas.
P ∝ T
⇒ [latex]\frac{P}{T}[/latex] = constant ⇒ [latex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/latex]
Charles law at constant pressure : At constant pressure the volume of given mass of gas is directly proportional to absolute temperature.
V ∝ T
⇒ [latex]\frac{V}{T}[/latex] = constant ⇒ [latex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/latex]
Ideal gas equation : A gas which obeys all the gas laws at all temperature and all pressures is called an ideal gas.
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 5
Consider are mole of a gas initially at a pressure P1 volume V1 and temperature T1 and Let P2, V2 and T2 be the final pressure, volume and absolute temperature T2.
From Boyle’s law, P1V1 = P2V2
⇒ V = [latex]\frac{P_1 V_1}{P_2}[/latex] ………….. (1)
Now the gas is heated at constant pressure P2, then its volume changes from V to V2 and temperature changes from T1 and T2.
From charles law, [latex]\frac{V}{T_1}=\frac{V_2}{T_2} \Rightarrow V=\frac{V_2 T_1}{T_2}[/latex] ………….. (2)
From (1) & (2) [latex]\frac{P_1 V_1}{P_2}=\frac{V_2 T_1}{T_2} \Rightarrow \frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}[/latex]
⇒ [latex]\frac{P V}{T}[/latex] = constant ⇒ [latex]\frac{P V}{T}[/latex] = R
Where ‘R’ is coniversal gas constant.
⇒ PV = RT
From ‘n’ gram moles PV = nRT. This law is called ideal gas equation.
Out of Boyle’s law, charles law is better for the purpose of thermometry because with increasing temperature, pressure and volume of gas also increase.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 2.
Explain thermal conductivity and coefficient of thermal conductivity. A copper bar of thermal conductivity 401 W/(mK) has one end at 104°C and the other end at 24°C. The length of the bar is 0.10 m and the cross – sectional area is 1.0 × 10-6 m-2. What is the rate of heat conduction, along the bar ?
Answer:
Thermal conductivity : Thermal conductivity of a solid body is ability to conduct heat in the body. Coefficient of thermal conductivity: The coefficient of thermal conductivity of a material may be defined as the quantity of heat energy flowing per second between the opposite faces of the cube of unit side, which are maintained at unit temperature difference.

Explanation of conduction : In steady state, conduction between the opposite faces which are at temperatures θ1°C and θ2°C separated by distance d is

  1. Directly proportional to area of cross section of the rod.
    Directly proportional to temperature difference (θ2 – θ1) between the opposite faces.
  2. Time of flow of heat, t
  3. Inversely proportional to the separation of faces ‘d’.
    AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 6
    ∴ Q ∝ [latex]\frac{\mathrm{A}\left(\theta_2-\theta_1\right) t}{\mathrm{~d}}[/latex]
    or Q = [latex]\frac{K A\left(\theta_2-\theta_1\right) t}{d}[/latex]
    K is called coefficient of thermal conductivity of the material.
    When A = 1; (θ2 – θ1) = 1; t = 1; d = 1
    ∴ K = Q

Problem:
Kc = 401 W/mK, θ2 = 104°C, θ2 = 24°C, d = 0.10 m, A = 1.6 × 10-6 m-2
Rate of heat conduction = [latex]\frac{Q}{t}=K_c A \frac{\left(\theta_2-\theta_1\right)}{d}[/latex]
= 401 × 1 × 10-6 × (104-24) = 0.32 W

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 3.
State and explain Newton’s law of cooling. State the conditions under which Newton’s law of cooling is applicable. A body cools down from 60°C to 50°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surroundings.
Answer:
Expression for Newton’s law of cooling : Consider a hot body at temperature T. Let T0 be the temperature of its surroundings. According to Newton’s law of cooling,
Rate of loss of heat ∝ Temperature difference between the body and its surroundings.
[latex]\frac{-\mathrm{dQ}}{\mathrm{dt}}[/latex] ∝ (T – T0)
[latex]\frac{-\mathrm{dQ}}{\mathrm{dt}}[/latex] = k (T – T0) ⇒ (1) where k is proportionality constant. Let m be the mass and C the specific heat of the body at temperature T. If the temperature of the body falls by small amount dT in time dt, then the amount of heat lost is
dQ = mc dT
∴ Rate of loss of heat is given by
[latex]\frac{\mathrm{dQ}}{\mathrm{dt}}[/latex] = mc [latex]\frac{\mathrm{dT}}{\mathrm{dt}}[/latex] ………….. (2)
putting (2) in (1) ⇒
-mc [latex]\frac{\mathrm{dT}}{\mathrm{dt}}[/latex] = k(T – T0)
[latex]\frac{\mathrm{dT}}{\mathrm{dt}}=\frac{\mathrm{k}}{\mathrm{mc}}[/latex] (T – T0) = -K(T – T0)
where K = [latex]\frac{\mathrm{k}}{\mathrm{mc}}[/latex] is another constant.
⇒ [latex]\frac{d T}{\left(T-T_0\right)}[/latex] = – K dt
on integrating both sides, we get
[latex]\int \frac{d T}{\left(T-T_0\right)}=-K \int d t[/latex]
loge (T – T0) = – Kt + c …………… (3)
T – T0 = e-Kt + c
T = T0 + ec e-Kt
T = T0 + Ce-Kt ……………… (4)
Where c is integration constant and C = ec
equations (1), (2), (3) and (4) are the different expressions for Newton’s law of cooling.

Explanation of Graphs:
1) If we plot a graph by taking equation (4) different values of temperature difference ∆T = T – T0 along y – axis and the corresponding values of t along X – axis, we get a curve of the form shown in figure. It clearly shows that the rate of cooling is higher initially and then decreases as the temperature of the body falls.
Curve showing cooling of hot water with time.
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 7
2) More over, the equation (3) is of the form y = mx + c. So if we plot a graph, by taking loge(T – T0) along Y – axis and time t along x – axis, we must get a straight line, as shown in figure. It has a negative slope equal to – K and intercept on Y – axis equal to C.
In both of the above situations, Newton’s law of cooling stands verified.
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 8
Newton’s law of cooling is applicable :

  1. Loss of heat is negligible by conduction and only when it is due to convection.
  2. Loss of heat occurs in a stream lined flow of air i.e., forced convection.
  3. Temperature of the body is uniformly distributed over it.
  4. Temperature differences are moderate i.e., upto 30 K, however if heat body is due to forced convection the law is valid for large differences of temperature also.
    AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 9
    ∴ Surrounding temperature, θ0 = [latex]\frac{85^{\circ} \mathrm{C}}{3}[/latex] = 28.33°C

Problems

Question 1.
What is the temperature for which the readings on Kelvin and Fahrenheit scales are same?
Answer:
The relation between Kelvin and Fahrenheit scales is given by
[latex]\frac{K-273.15}{100}=\frac{F-32}{180}[/latex]
but K = F
[latex]\frac{F-273.15}{100}=\frac{F-32}{180}[/latex]
F – 273.15 = [latex]\frac{5}{9}[/latex] F – [latex]\frac{160}{9}[/latex]
F – [latex]\frac{5}{9}[/latex] F = 273.15 – [latex]\frac{160}{9}[/latex]
= 273.15 – 17.77
[latex]\frac{4F}{9}[/latex] = 255.38
∴ F = [latex]\frac{9}{4}[/latex] (255.38)
= 574.6°F

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 2.
Find the increase in temperature of aluminium rod if its length ¡s to be increased by 1%. (a for aluminium = 25 × 10-6/0 C). [A.P (Mar. ‘15)]
Answer:
Percentage increase in length = [latex]\frac{l_2-l_1}{l_1}[/latex] × 100
= α(t2 – t1) × 100 [∵ [latex]\frac{l_2-l_1}{l_1}[/latex] = α(t2 – t1)]
Here, percentage increase in length = 1,
α = 25 × 10-6/ °C
1 = 25 × 10-6(t2 – t1) × 100
t2 – t1 = [latex]\frac{1}{25 \times 10^{-4}}=\frac{10^4}{25}[/latex] = 400°C

Question 3.
How much steam at 100°C is to be passed into water of mass 100 g at 20°C to raise its temperature by 5°C ? (Latent heat of steam is 540 cal/g and specific heat of water is 1 cal/g°C).
Answer:
In the method of mixtures,
Heat lost by steam = Heat gained by water
mSLS + mSS(100 – 1) = mWS (t – 20)
Where mS is the mass of steam, LS is the latent heat of steam, S is the specific heat of steam and mW is the mass of water.
Here, LS = 540 cal/g
S = 1 cal/g°C;
mW = 100 g
t = 20 + 5 = 25°C
mS × 540 + mS × 1 × (100 – 25) = 100 × 1 × (25 – 20)
615 mS = 500
mS = [latex]\frac{500}{615}[/latex]
= 0.813 g.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 4.
2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313 K. If the specific heat of air at constant volume is 0.718 kJ/kgK, find the amount of heat absorbed in kJ and kcal.
(J = 4.2 kJ/kcal).
Answer:
Here, M = 2 kg
dT = 313- 293 = 20 K
CV = 0.718 × 103 J/Kg – K
J = 4.2 KJ/Kcal
CV = [latex]\frac{1}{\mathrm{~m}} \frac{\mathrm{d} \theta}{\mathrm{dt}}[/latex]
dθ = mCVdT
= 0.718 × 103 × 2 × 20
= 28720 J
dθ = 28.72 KJ
dθ = [latex]\frac{28720}{4.2 \times 10^3}[/latex]
= 6.838 Kcal.

Question 5.
A dock, with a brass pendulum, keeps correct time at 20°C, but loses 8.212 s per day, when the temperature rises to 30°C. Calculate the coefficient of linear expansion of brass.
Answer:
Here t1 = 20°C, t2 = 30°C
Loss of time per day = 8.212 sec
Loss of time per day = [latex]\frac{1}{2}[/latex] α (t2 – t1) × 86,400
8.212 = [latex]\frac{1}{2}[/latex] α (30 – 20) × 86,400
∴ α = [latex]\frac{16.424}{864 \times 10^3}[/latex]
= 19 × 10-6/C°

Question 6.
A body cools from 60°C to 40°C in 7 minutes. What will be its temperature after next 7 minutes if the temperature of its surroundings is 10°C ?
Answer:
Rate of cooling [latex]\frac{\mathrm{d} \theta}{\mathrm{dt}}[/latex] = K(θav – θ0)
Case (i) :
Given, .
θ1 = 60°C, θ2 = 40°C, θ0 = 10°C, t = 7 min
⇒ [latex]\frac{60^{\circ} \mathrm{C}-40^{\circ} \mathrm{C}}{7}=K\left[\frac{60^{\circ} \mathrm{C}+40^{\circ} \mathrm{C}}{2}-10^{\circ} \mathrm{C}\right][/latex]
[latex]\frac{20}{7}[/latex] = K[50 – 10]
⇒ [latex]\frac{20}{7}[/latex] = K × 40
∴ K = [latex]\frac{1}{14}[/latex] ……………. (1)

Case (ii) :
Given,
θ1 = 40°C, θ0 = 10°C, t = 7 min, θ2 = ?
⇒ [latex]\frac{40-\theta_2}{7}=K\left[\frac{40+\theta_2}{2}-10\right][/latex]
[latex]\frac{40-\theta_2}{7}=\frac{1}{14}\left[\frac{40+\theta_2}{2}-10\right][/latex]
80 – 2θ2 = [latex]\frac{40+\theta_2-20}{2}[/latex]
160 – 4θ2 = 20 + θ2
5θ2 = 140
∴ θ2 = [latex]\frac{140}{5}[/latex] = 28°C

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 7.
If the maximum intensity of radiation for a black body is found at 2.65 pm what is the temperature of the radiating body ? (Wien’s constant = 2.9 × 10-3 mK)
Answer:
λmax = 2.65 pm = 2.65 × 10-6m
Wien’s constant (b) = 2.90 × 10-3 mk
λmax T = b(constant)
T = [latex]\frac{b}{\lambda_{\max }}=\frac{2.9 \times 10^{-3}}{2.65 \times 10^{-6}}[/latex]
= 1094 K.

Additional Problems

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Answer:
Relation between kelvin scale and Celsius scale is TC = TK – 273.15
Where TC.TK = temperature on Celsius and kelvin scales respectively
For Neon TC = 24.57 – 273.15 = – 248.58°C
For CO2 TC = 216.55 – 273.15 = -56.60°C
Relation between Kelvin and Fahrenheit scales is
[latex]\frac{T_F-32}{180}=\frac{T_K-273.15}{100}[/latex]
TF = [latex]\frac{180}{100}[/latex] (TK – 273.15) + 32
FQ or Neon TF
= [latex]\frac{180}{100}[/latex](24.57 – 273.15) + 32 = – 415.44° F
FQ or CO2 TF = [latex]\frac{180}{100}[/latex] (216.55 – 273.15) + 32
= – 69.88° F

Question 2.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB ?
Answer:
Given triple point of water on scale A = 200
A triple point of water on scale B = 350 B
As per question 200 A = 350, B = 273.16 K
(Or)
IA = [latex]\frac{273.16}{200}[/latex] K and IB = [latex]\frac{273.16}{350}[/latex] K
If TA and TB represent the triple point of water on two scales A and B then
[latex]\frac{273.16}{200}[/latex] TA = [latex]\frac{273.16}{350}[/latex]TB (Or) [latex]\frac{T_A}{T_B}=\frac{200}{350}=\frac{4}{7}[/latex]
(Or) TA = [latex]\frac{4}{7}[/latex] TB

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = R0[1 + α(T – T0)]
The resistance is 101.6 Ω at the triple-point of water 273.16 K and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?
Answer:
Here, R0 = 101. 6 Ω, T0 = 273.16 K
Case (i) R1 = 165.5 Ω, T1 = 600.5 K
Case (ii) R2 = 123.4 Ω; T2 = ?
Using the relation R = R0(1 + α(T – T0))
Case (i):
165.5 = 101.6(1 + α(600.5-273.16))
α = [latex]\frac{165.5-101.6}{101.6 \times(600.5-273.16)}[/latex]
= [latex]\frac{63.9}{101.6 \times 327.34}[/latex]

Case (ii): 123.4 = 101.6(1 + α(T2 – 273.16))
(Or)
123.4 = 101.6
(1 + [latex]\frac{63.9}{101.6 \times 327.34}[/latex] (T2 – 273.16))
= 101.6 + [latex]\frac{63.9}{327.34}[/latex] (T2 – 273.16)
T2 = [latex]\frac{(123.4-101.6) \times 327.34}{63.9}[/latex] + 273.16
= 111.67 + 273.16 = 384.83 K

Question 4.
Answer the following :
a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?
b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?
c) The absolute temperature (Kelvin scale) T is related to the temperature tC on the Celsius scale by
tc = T – 273.15
Why do we have 273.15 in this relation and not 273.16 ?
Answer:
a) This is on account of the fact that the triple point of water has a unique value i.e., 273.16 K at a unique point, where exists unique values of pressure and volume. On the other hand, the melting point of ice and boiling point of water do not have unique set of value as they change in pressure and volume.

b) The other fixed point on the absolute scale is the absolute zero itself.

c) On Celsius 0°C corresponds to melting points of ice of normal pressure. The corresponding value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water.
From the given relation the corresponding value of triple point of water on Celsius scale
= 273.16 – 273. 15 = 0.01°C.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 5.
Two. ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 10
a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
b) What do you think is the reason behind the slight difference in answers of thermoineters A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?
Answer:
a). Let T be the melting point of sulphur, for water Ttr = 273.16 K
For thermometer A, T = P/Ptr × 273.16
= [latex]\frac{1.797 \times 10^5}{1.250 \times 10^5}[/latex] × 273.16 = 392.69 K
For thermometer B, T = P/Ptr × 273.16
= [latex]\frac{0.287 \times 10^5}{0.200 \times 10^5}[/latex] × 273.16 = 391.98 K

b) The cause of slight different answers is that the oxygen and hydrogen gases are not perfectly ideal. To reduce this discrepancy, the readings should be at lower and lower pressure as in that case, the gases approach to the ideal gas behaviour.

Question 6.
A steel tape 1 m long is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0°C ? Coefficient of linear expansion of steel = 1.20 × 10-5 K-1.
Answer:
Length of steell tape at 27°C is 100 cm i.e.
L = 100 cm and T = 27°C
The length of steel tape at 45°C is L1 = L + ∆L
= L + αL∆T
= 100 + (1.20 × 10-5) × 100 × (45° – 27)
= 100.0216 cm
Length of 1 cm mark at 27°C on this scale, at 45°C
= 100.0216/100 cm
Length of 63 cm measured’ by this tape at 45°C will be
= [latex]\frac{100.0216}{100}[/latex] × 63 = 63.0136 cm
Length of the same steel rod on a day when the temperature is 27°C = 63 × 1 = 63 cm.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 7.
A large steel wheel is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft ? Assume coefficient of linear expansion of the steel to be constant over the required temperature range :
αsteel = 1.20 × 10-5K-1.
Answer:
Here T1 = 27°C = 27 + 273 = 300 K
Length at temperature, T1K = LT1 = 8.70 cm
Length at temperature, T2K = LT2 – 8.69 cm
Change in length = LT2 – LT1 = LT2 α(T2 – T2)
Or 8.69 – 8.70 = 8.70 × (1.20 × 10-5) × (T2 – 300)
Or T2 – 300 = [latex]\frac{0.01}{8.70 \times 1.2 \times 10^{-5}}[/latex] = -95.8
Or T2 = 300 – 95.8 = 204.2 K = -68.8°C

Question 8.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C ? Coefficient of linear expansion of copper = 1.70 × 10-5K-1.
Answer:
In this problem superfical expansion of copper sheet will be involved on heating. Here, area of hole at 227°C, then area of the hole at 227°C,
S2 = [latex]\frac{\pi \mathrm{D}_2^2}{4}[/latex]cm2.
Coefficient of superficial expansion of copper is β = 2α = 2 × 1.70 × 10-5 = 3.4 × 10-5 c1
Increase in area = S2 – S1 = βS1 ∆T (or)
S2 = S1 + βS1 ∆T
= S1(1 + β∆T)
(Or)
[latex]\frac{\pi \mathrm{D}_2^2}{4}=\frac{\pi}{4}[/latex] (4.24)2 [1 + 3.4 × 10-5 (228 – 27)]
(Or)
D22 = (4.24)2 × 1.0068
D2 = 4.2544 cm
Change in diameter = D2 – D1
= 4.2544 – 4.24
= 0.0144 cm.

Question 9.
A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 × 10-5K-1; Young’s modulus of brass = 0.91 × 1011 Pa.
Answer:
Here L = 1.8m, T1 = 27°C, T2 = -39°C, r = 1
mm = 10-3m, F = ?
α = 2 × 10-5C-1, y = 0.91 × 1011 N/m2
From y = [latex]\frac{\mathrm{FL}}{a \Delta L}[/latex], ⇒ ∆L = [latex]\frac{\mathrm{FL}}{\mathrm{ay}}[/latex]
Also ∆L = αL∆T = ∴ [latex]\frac{\mathrm{FL}}{\mathrm{ay}}[/latex] = αL∆T
(Or) F = α∆Tay = α(T2 – T1) πr2y
= 2 × 10-5 × (-39 – 27) × (10-3)2 × 0.91 × 1011
= -3.77 × 10-2 N
Negative sign indicates that the force is inward due to contraction of the wire.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 10.
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C ? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2,0 × 10-5K-1, steel = 1.2 × 10-5K-1.
Answer:
∆L1 = L1α1∆T = 50 × (2.10 × 10-5) (250 – 40) = 0.2205 cm
∆L2 = L2α2∆T
= 50 × (1.2 × 10-5) (250 – 40) = 0.216 cm
∴ Change in length of combined rod
= ∆L1 + ∆L2
= 0.220 + 0.126 = 0.346 cm

Question 11.
The coefficient of volume expansion of glycerin is 49 × 10-5K-1. What is the fractional change in its density for a 30°C rise in temperature ?
Answer:
Here r = 49 × 10-5C-1, ∆T = 30°C .
As V1 = V + ∆V = V(1 + r∆T)
∴ V1 = V(1 + 49 + 10-5 × 30) = 1.0147 V
As P = [latex]\frac{m}{V}[/latex], p1 = [latex]\frac{m}{V^1}[/latex] = [latex]\frac{\mathrm{m}}{1.0147 \mathrm{~V}}[/latex]= 09855 P
Fractional change in density = [latex]\frac{\rho-\rho^{\prime}}{\rho}[/latex]
= [latex]\frac{\rho-0.9855 \rho}{\rho}[/latex]
= 0.0145

Question 12.
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 Jg-1K-1.
Answer:
Here p = 10 kw = 105w,
Mass m = 8.0 kg = 8 × 103 g
Rise in temp; ∆T = ?
time t = 2.5 min = 2.5 × 60 sec
Sp. heat C = 0.91 Jg-1°C-1
Total energy = p × t = 104 × 150
= 15 × 105 J
As 50% of energy is lost
∴ Energy available ∆Q = [latex]\frac{1}{2}[/latex] × 15 × 105
= 7.5 × 105 J
As ∆Q = mc ∆T
∴ ∆T = [latex]\frac{\Delta Q}{\mathrm{mc}}=\frac{7.5 \times 10^5}{8 \times 10^3 \times 0.91}[/latex] = 103°C

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 13.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt ? (Specific heat of copper = 0.39 Jg-1K-1 heat of fusion of water = 335 Jg-1.
Answer:
Here, mass of copper block m = 2.5 kg = 2500 kg
Fall in temperature ∆T = 500 – 0 = 500°C
Specific heat of copper C = 0.39 Jg-1°C-1
Latent heat of fusion L = 335 Jg-1
Let the mass of ice melted by m1
As heat gained by ice = Heat lost by copper
m-1L = Mc∆T
m-1 = [latex]\frac{\mathrm{mC} \Delta \mathrm{T}}{\mathrm{L}}[/latex]
m1 = [latex]\frac{2500 \times 0.39 \times 500}{335}[/latex]
= 1500 g = 1.5 kg

Question 14.
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?
Answer:
Here, mass of metal, m = 0.20 kg = 200 g
Fall in temperature of metal ∆T = 150 – 40
= 110°C
If L is specific heat of metal, then heat lost by the metal
∆Q = mc∆T = 200 × L × 110
Volume of water = 150 C.C
Mass of water m1 = 150 g
Water equivalent of calorimeter, w = 0.025 kg = 25 kg
Rise in temp of water and calorimeter
∆T1 = 40 – 27 = 13°C
Heat gained by water and calorimeter,
∆Q1 = (m1 + w)T1
= (150 + 25) × 13 = 175 × 13
As ∆Q = ∆Q1
∴ From (i) and (ii)
200 × C × 110 = 175 × 13
C = [latex]\frac{175 \times 13}{200 \times 110}[/latex] ≈ 0.1
(Or)
If some heat is lost to the sorroundings, value of L is so obtained will be less than the actual value of L.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 11
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?
Answer:
The gases which are listed in the above table are diatomic gases and not mono atomic gases. For diatomic gases, molar specific heat
= [latex]\frac{5}{2}[/latex] R = [latex]\frac{5}{2}[/latex] × 1.98 = 4.95, which agrees fairly well with all observations listed in the . table except for chlorine. A monoatomic gas molecules has only the transnational motion. A diatomic gas molecules, a part from translation motion. The vibrational as well as rotational motion is also possible. Therefore to raise the temperature of 1 mole of diatomic gas through 1°C heat is also to be supplied increase not only translation energy but also rotational and vibrational energies. Hence molar specific heat of diatomic gas is greater than that for monoatomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. Shows that for chlorine molecule, at room temperature vibrational motion. Also occurs along with translational and rotational motions. Where as other diatomic molecules at room temperature usually have rotational motion apart from their translational motion. This the reason that chlorine has some what larger value of molar specific heat.

Question 16.
Answer the following questions based on the P-T phase diagram of carbon dioxide :
a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium ?
b) What is the effect of decrease of pressure on the fusion and boiling point of CO2 ?
c) What are the critical temperature and pressure for CO2 ? What is their significance ?
d) Is CO2 solid, liquid or gas at (a) – 70°C under 1 atm, (b) -60°C under 10 atm, (c) 15°C under 56 atm ?
Answer:
a) The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the point i.e., temperature = -56.6°C and pressure = 5.11 atm.

b) With the decrease in pressure, both the fusion and boiling point of carbondioxide will decrease.

c) For carbondioxide, the critical tempera-ture is 31,1°C and critical pressure is 73.0 atm. If the temp of CO2 is more than 31.1°C. It cannot be liquified, however large pressure we may apply.

d) Carbondioxide will be (a) a vapour at – 70°C under 1 atm (b) a solid at 6°C under 10 atm (c) a liquid at 15°C under 56 atm.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 17.
Answer the following questions based on the P – T phase diagram of CO2 :
a) CO2 at 1 atm pressure and temperature – 60°C is compressed isothermally. Does it go through a liquid phase ?
b) What happens when CO2 at 4 atm pressure is cooled from room tem-perature at constant pressure ?
c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature – 65°C as it is heated up to room temperature at constant pressure.
d) CO2 is heated to a temperature 70°C and compressed isothermally. What changes in its properties, do you expect to observe ?
Answer:
a) Since the temp – 60°C lies to the left of 56.6°C on the curve i.e. lies in the region of vapour and solid phase. So CO2 will condense directly into solid with out becoming liquid.

b) Since the pressure 4 atm is less than 5.11 atm. The carbondioxde will condenses directly into solid without becoming liquid.

c) When a solid CO2 at 10 atm. Pressure and -65°C temp is heated, It is first converted into a liquid. A further increase in temperature brings it to the vapour phase. At P = 10 atm. If a horizontal line is drawn parallel to the T-axis. Then the points of intersection of this line with the fusion and vapourisation curve will give the fu-sion and boiling point of CO2 at 10 atm.

d) Since 70°C is higher than the critical temperature of CO2. So the CO2 gas can not be converted into liquid state on being compressed isothermally at 70°C. It will remain in the vapour state. Nowever the gas will depart more and more now its perfect gas behavious with the increase in pressure.

Question 18.
A child running a temperature of 101 °F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580 cal g-1.
Answer:
Here fall in temp = ∆T = 101 – 98 = 3°F
= 3 × [latex]\frac{5}{3}[/latex]°C = [latex]\frac{5}{3}[/latex]°C
Mass of child M = 30 kg
Sp. heat of human body = Sp heat of water
C = 1000 cal.kg-1C-1
∴ Heat last by the body of child, ∆Q = mC∆T
= 30 × 1000 × [latex]\frac{5}{3}[/latex] = 5000 calories
If m be the mass of water evapourated in 20 min then m1L = ∆Q
m1 = [latex]\frac{\Delta Q}{L}=\frac{5000}{580}[/latex] = 86.2 g
∴ Average rate of extra evapouration = [latex]\frac{86.2}{20}[/latex]
= 4.31 gmin-1

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 19.
A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45°C and co-efficient of thermal conductivity of thermacole is 0.01 Js-1m-1 K-1. [Heat of fusion of wafer = 335 × 103 J kg-1]
Answer:
Here length of each side l = 30 cm = 0.3 m
Thickness of each side ∆x = 5 cm = 0.05 m
Total surface area through which heat enters into the box
A = 6 l2 = 6 × 0.3 × 0.3 = 0 Jum2
Temp.diff ∆T = 45 – 0 = 45°C,
K = 0.01 JS-1m-1°C-1
Time ∆T = 6 hrs = 6 × 60 × 60 S
Latent heat of fusion L = 335 × 103 J/kg
Let m be the mass of ice melted in this time
∆Q = mL = KA [latex]\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)[/latex] ∆t
m = kA [latex]\left(\frac{\Delta T}{\Delta \mathrm{T}}\right) \frac{\Delta \mathrm{t}}{\mathrm{L}}[/latex] = 0.01 × 0.54 × [latex]\frac{45}{0.05}[/latex] × [latex]\frac{6 \times 60 \times 60}{335 \times 10^3}[/latex] = 0.313 kg
Mass of ice left = 4 – 0.313 = 3.687 kg

Question 20.
A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js-1m-1 K-1: Heat of vaporisation of water = 2256 × 103 Jkg-1.
Answer:
Were A = 0.15 m2 ∆x = 1.0 m = 10-2 m
[latex]\frac{\Delta Q}{\Delta t}=\frac{6 \times 10^3 \times 2256}{60}[/latex] JS-1 = 2256 × 102 JS-1
K = 609 JS-1m-1°C-1, ∆T = (t – 100)
From [latex]\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\mathrm{KA}\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right)[/latex]
2256 × 102 = 609 × 0.15 [latex]\frac{(t-100)}{10^{-2}}[/latex]
t – 100 = [latex]\frac{2256}{609 \times 0.15}[/latex] = 2470 t
t = 24.70

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 21.
Explain why :
a) a body with large reflectivity is a poor emitter.
b) a brass tumbler feels much colder than a wooden tray on a chilly day.
c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
d) the earth without its atmosphere would be inhospitably cold.
e) healting systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
Answer:
a) This is because a body with large refelectivity is a poor absorber of heat and poor absorbers of heat are poor imilters.

b) When we touch a brass tumbler on a chill day; heat flows from our body to the tumbler quickly cas thermal conductivity or brass is very high and as a result, it appears colder. On the other hand as the wood is bad conductor, heat does not flow to the wooden tray from our body, on touching it.

c) When the red hot iron pieces is in the oven, its temperature TK is the given by the relation.
E = αT4 But if the red hot iron piece is in the open having the surrounding temperature T0, its energy is radiated according to relation E1 = α(T4 – T40). As the working principle of optical pysometer is based on the fact that the brightness of – a glowing surface of a body depends , upon its temperature. Therefore, pysometes gives too low a value for the temperature of red iron in the open.

d) The lower layer’s of Earth’s atmosphere reflect infrared radiations from earth back to the surface of the earth. Thus the heat radiation recieved by the earth from the sun during the day are kept trapped by the atm. If atmosphere of earth were not there, its surface would become too cold to live.

e) Steam at 100°C possess more heat than the same mass of water at 100°C possess 540 calories of heat more than possessed by 1 gm of ware at 100°C. That is why heating systems based on circulation of stream are more efficient than those based on circulation of hot water.

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.
Answer:
[latex]\frac{\mathrm{d} \theta}{\mathrm{dt}}[/latex] = K [θav – θ0]
[latex]\frac{80-50}{5}[/latex] = K(65 – 20)
⇒ [latex]\frac{30}{5}[/latex] = K × 45 = K = [latex]\frac{2}{15}[/latex]
[latex]\frac{60-30}{t}=\frac{2}{15}[45-20][/latex]
[latex]\frac{30}{t}=\frac{2}{15} \times 15[/latex]
t = 5 min
Time of cooling is given by t = [latex]\frac{2.3026}{k}[/latex] log10
[latex]\frac{T_1-T_0}{T_2-T_0}[/latex]
As per question condition (i) T1 = 80°C, T2 = 50°C, T0 = 20°C, t = 5 min
5 × 60 = [latex]\frac{2.3026}{K} \log _{10} \frac{80-20}{50-20}[/latex]
= [latex]\frac{2.3026}{k} \log _{10^2}[/latex] ……………. (2)
As per question condition (i) T1 = 60°C, T2 = 30°C, T0 = 20°C, t = ?
t = [latex]\frac{2.3026}{K} \log _{10} \frac{60-20}{30-20}[/latex]
= [latex]\frac{2.3026}{k} \log _{10^4}[/latex] ……………. (3)
Div (3) by (2) we get
[latex]\frac{t}{5 \times 60}=\frac{\log _{10^4}}{\log _{10^2}}=\frac{0.6021}{0.3010}[/latex] = 2
(Or) t = 5 × 60 × 2 = 10 × 60 s
= 10 min

Textual Examples

Question 1.
Show that the coefficient of areal expansions. (∆A/A)/∆T, of rectangular sheet of the solid is twice its linear expansivity, α1.
Answer:
∆A3 = (∆a) (∆b)
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 12
When the temperature increases by ∆T, a increases by ∆a = α1a ∆T and b increases by ∆b = α1b ∆T. From fig. the increase in area
∆A = ∆A1 + ∆A2 + ∆A3
∆A = a ∆b + b ∆a + (∆a) (∆b)
= a α1b ∆T + b α1a ∆T + (α1)2 ab(∆T)2
= α1ab ∆T(2 + α1∆T)
= α1A ∆T(2 + α1∆T)
Since α1 = 10-5 K-1, from Table 12.1, the product α1∆T for fractional temperature is small in comparision with 2 and may be neglected.
Hence,
[latex]\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}\right) \frac{1}{\Delta \mathrm{T}}[/latex] ≈ 2α1

Question 2.
A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diametbr of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27°C. To what temperature should the ring be heated so as to fit the rim of the wheel ?
Answer:
Given,
T1 = 27°C
LT1 = 5.231 m
LT2 = 5.243 m
So,
LT2 = LT1 [1 + α1(T2 – T1)]
5.243 m = 5.231 m[1 + 1.20 × 10-5 K-1 (T2 – 27°C)]
or T2 = 218°C

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 3.
A sphere of aluminium of 0.047 kg is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100°C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg of water at 20°C. The temperature of water rises and attains a steady state at 23°C. Calculate the specific heat capacity of aluminium.
Answer:
Mass of aluminium sphere (m.,) = 0.047 kg
Initial temp, of aluminium sphere = 100°C
Final temp. = 23°C
Change in temp. (∆T) = (100°C – 23°C)
= 77°C
Let specific heat capacity of aluminium be sAl. The amount of heat lost by the aluminium sphere = m1sAl ∆T = 0.047 kg × sAl × 77°C s ……………. (i)
Mass of water (m2) = 0.25 kg
Mass of calorimeter (m3) = 0.14 kg
Initial temp, of water and calorimeter = 20°C
Final temp, of the mixture = 23°C
Change in temp. (∆T2) = 23°C – 20°C = 3°C
Specific heat capacity of water (sw) from table
12.3 = 4.18 × 103 J kg-1 K-1
Specific heat capacity of copper calorimeter = 0.386 × 103 J kg-1 K-1
The amount of heat gained by water and calorimeter = m2sw ∆T2 + m3scu ∆T2 = (m2sw + m3scu) (∆T2)
= (0.25 kg × 4.18 × 103 J kg-1 K-1 + 0.14 kg × 0.386 × 103 J kg-1 K-1) (23°C – 20°C) ………………… (ii)
In the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.
So, from (i) and (ii)
0.047 kg × sAl × 77°C = (0.25 kg × 4.18 × 103 J kg-1 K-1 + 0.14 kg × 0.386 × 103 J kg-1 K-1) (3°C)
sAl = 0.911 kJ kg-1 K-1

Question 4.
When 0.15 kg of ice of 0°C mixed with 0.30 kg of water at 50°C in a container, the resulting temperature is 6.7°C. Calculate the heat of fusion of ice.
(swater = 4186 J kg-1 K-1)
Answer:
Heat lost by water = msw (θf – θi)w
= (0.30 kg) (4186 kg-1 K-1) (50.0°C – 6.7°C)
= 54376.14 J
Heat required to melt ice = m2Lf = (0.15 kg) Lf
Heat required to raise temperature of ice water to final temperature
= mIsw (θf – θi)I
= (0.15 kg) (4186 J kg-1 K-1) (6.7°C – 0°C)
= 4206.93 J
Heat lost = heat gained
54376.14 J = (0.15 Kg)Lf + 4206.93 J
Lf = 3.34 × 105 J kg-1

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 5.
Calculate the heat required to convert 3 kg of ice at -12°C kept in a calorimeter-to steam at 100°C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg-1 K-1, specific heat capacity of water = 4186 J kg-1 K-1, latent heat of fusion of ice = 3.35 × 105 J kg-1 and latent heat of steam = 2.256 × 106 J kg-1.
Answer:
We have
Mass of the ice, m = 3 kg
Specific heat capacity of ice, sice
= 2100 J kg-1 K-1
Specific heat capacity of water, swater
= 4186 J kg-1 K-1
Latent heat of fusion ice, Lf ice
= 3.35 × 105 J kg-1
Latent heat of steam, Lsteam
= 2.256 × 106 J kg-1
Now,
Q = Heat required to convert 3 kg of ice at -12°C to steam at 100°C.
Q1 = Heat required to convert ice at – 12°C to ice at 0°C. .
msice ∆T1 = (3 kg) (2100 J kg-1 K-1) [0 – (-12)]°C = 75600 J
Q2 = Heat required to melt ice at – 0°C to water at 0°C.
mLf ice = (3 kg) (3.35 × 105 J kg-1) = 1005000 J
Q3 = Heat required to convert water at 0°C to water at 100°C.
msW ∆T2 = (3 kg) (4186 J kg-1 K-1) (100°C)
= 1255800 J
Q4 = Heat required to convert water at 100°C to steam at 100°C.
mLsteam = (3 kg) (2-256 × 106 J kg-1)
= 6768000 J
So,
Q = Q1 + Q2 + Q3 + Q4
= 75600 J + 1005000 J + 1255800 J + 6768000 J
= 9.1 × 106 J

Question 6.
What is the temperature of the steel- copper junction in the steady state of the system shown in fig. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300°C, temperature of the other end = 0°C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = 50.2 J s-1m-1K-1 and of copper = 385 J s-1m-1K-1].
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 13
Answer:
Let T be the temperature of the steel-copper junction in the steady state. Then,
[latex]\frac{K_1 A_1(300-T)}{L_2}=\frac{K_2 A_2(T-0)}{L_2}[/latex]
Where 1 and 2 refer to the steel and copper rod respectively. For A1 = 2, A2, L1 = 15.0 cm, L2 = 10.0 cm, K1 = 50.2 J s-1m-1K-1,
K2 = 385 J s-1m-1K-1, we have
[latex]\frac{50.2 \times 2(300-\mathrm{T})}{15}=\frac{385 \mathrm{~T}}{10}[/latex]
which gives T = 44.4°C

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 7.
An iron bar (L1 = 0.1 m, A1 = 0.02 m2, K1 = 79 W m-1 K-1) and a brass bar (L2 = 0.1 m, A2 = 0.02 m2, K2 = 109 W m-1K-1) are soldered end to end as shown in fig. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar and (iii) the heat current through the compound bar.
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 14
Answer:
Given,
L1 = L2 = L = 0.1 m, A1 = A2 = A = 0.02 m2, K1 = 79 W m-1 K-1, K2 = 109 W m-1K-1, T1 = 373 K and T2 = 273 K.
Under steady state condition, the heat current (H1) through iron bar is equal to the heat current (H2) through brass bar.
So, H = H1 = H2
= [latex]\frac{K_1 A_1\left(T_1-T_0\right)}{L_1}=\frac{K_2 A_2\left(T_0-T_2\right)}{L_2}[/latex]
For A1 = A2 = A and L1 = L2 = L this equation leads to
K1(T1 – T0) = K2(T0 – T2)
Thus the junction temperature T0 of the two bars is T0 = [latex]\frac{\left(K_1 T_1+K_2 T_2\right)}{\left(K_1+K_2\right)}[/latex] ………….. (a)
Using this equation, the heat current H through either bar is
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 15
Using these equations, the heat current H’ through the compound bar of length L1 + L2 = 2L and the equivalent thermal conductivity K, of the compound bar are given by
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 16
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 17

AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter

Question 8.
A pan filled with hot food cools from 94°C to 86°C in 2 minutes when the room temperature is at 20°C. How long will it take to cool from 71°C to 69°C ?
Answer:
The average temperature of 94°C and 86°C is 90°C, which is 70°C above the room temperature. Under these conditions the pan cools 8°C in 2 minutes.
Using [latex]\frac{\mathrm{dT}_2}{\mathrm{~T}_2-\mathrm{T}_1}=-\frac{\mathrm{k}}{\mathrm{ms}}[/latex] dt = – K dt we have
The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.
AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter 18
The average of 69°C and 71°C is 70°C, which is 50°C above room temperature. K is the same for this situation as for the original.
[latex]\frac{2^{\circ} \mathrm{C}}{\text { Time }}[/latex] = K(50°C)
When we divide above two equations, we have
[latex]\frac{8^{\circ} \mathrm{C} / 2 \min }{2^{\circ} \mathrm{C} / \text { time }}=\frac{K\left(70^{\circ} \mathrm{C}\right)}{\mathrm{K}\left(50^{\circ} \mathrm{C}\right)}[/latex]
Time = 0.7 min = 42 s

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Andhra Pradesh BIEAP AP Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids Textbook Questions and Answers.

AP Inter 1st Year Physics Study Material 11th Lesson Mechanical Properties of Fluids

Very Short Answer Questions

Question 1.
Define average pressure. Mention it’s unit and dimensional formula. Is it a scalar or a vector ? [A.P. Mar. 17]
Answer:
Average pressure (Pav) : Average power is defined as the normal force acting per unit area.
Pav = [latex]\frac{F}{A}[/latex]
units → N/m2 (or) pascal
Dimensional formula → [ML-1 T-2]
Pressure is a scalar quantity.

Question 2.
Define Viscosity. What are it’s units and dimensions ?
Answer:
Viscosity : The property of a liquid which opposes the relative motion between its layers is called viscosity. .
G.G.S unit poise
S.I unit → Nm-2s
Dimensional formula = [M1L-1T-1]

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 3.
What is the principle behind the carburetor of an automobile ? [AP – Mar. 15; TS – Mar. ’18, ’17]
Answer:
The carburetor of automobile has a venturi channel (nozzle) through which air flows with a large speed. The pressure is then lowered at the narrow neck and the petrol is sucked up in the chamber to provide the correct mixture of air to fuel necessary for combustion. ”

Question 4.
What is magnus effect ? [T.S. Mar. 16; A.P. Mar. 15]
Answer:
The difference in the velocities of air results in the pressure difference between the lower and upper faces and there is a net upward force on the ball. This dynamic lift due to spinning is called magnus effect.

Question 5.
Why are drops and bubbles spherical ? [A.P. Mar. 18, 17, 16, 14]
Answer:
The surface tension of a liquid tends to have minimum surface area. For a given volume, the surface area is minimum for a sphere. Hence rain drops are spherical shape.

Question 6.
Give the expression for the excess pressure in a liquid drop. [T.S. Mar. 17]
Answer:
Excess pressure in a liquid drop, pi – p0 = [latex]\frac{2 s}{r}[/latex]
where s = Surface tension; r = Radius of the liquid drop.

Question 7.
Give the expression for the excess pressure in an air bubble inside the liquid.
Answer:
Excess pressure in an air bubble inside the liquid, pi – p0 = [latex]\frac{2 s}{r}[/latex]
where s = Surface tension
r = Radius of the air bubble.
Air bubble forms inside the liquid, hence it has one liquid surface.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 8.
Give the expression for the excess pressure soap bubble in air. [T.S. Mar. 16]
Answer:
Soap bubble have two interfaces, hence excess pressure inside a soap bubble is pi – p0 = [latex]\frac{4 s}{r}[/latex]
where s = Surface tension
r = Radius of the soap bubble.

Question 9.
What are water proofing agents and water wetting agents ? What do they do ?
Answer:
Water proofing agents are added, to create a large angle of contact between water and fibres.
Soaps, detergents and dying substances are wetting agents. When they are added, the angle of contact becomes small. So that they may penetrate well and become effective.

Question 10.
What is angle of contact ? [A.P. Mar. 16]
Answer:
The angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid is termed as angle of contact (0).

Question 11.
Mention any two examples that obey Bernoullis theorem and justify them. [A.P. Mar. 18]
Answer:

  1. In heavy winds house roof’s are blown off. When the velocity of the wind is greater on the roof top than inside the house, then the pressure on the roof top becomes less than that inside the house. This pressure difference causes the dynamic lift.
  2. When a fan is rotating, papers are blown off from .the table top. The velocity of wind on the paper increases due to fan and hence pressure decreases. Due to this pressure difference papers are blown off.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 12.
When water flows through a pipe, which of the layers moves fastest and slowest ?
Answer:
Water flows through a pipe, the layers near the axis of the tube are fastest and at the walls of the tube are slowest
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 1

Question 13.
“Terminal velocity is more if surface area of the body is more”. Give reasons in support of your answer.
Answer:
Surface area (A) = 4πr2 and terminal velocity (υt) αr2
As surface area increases, r2 is also increases. Then terminal velocity is also increases.
∴ Terminal velocity is more, if surface area of the body is more.

Short Answer Questions

Question 1.
What is atmospheric pressure and how is it determined using Barometer ?
Answer:
Atmospheric pressure : Atmospheric pressure at any point is equal to the weight of a column of air of unit cross sectional area extending from that point to the top of the earth’s atmosphere.
1 atm = 1.013 × 105 pa
Determination of atmospheric pressure using barometer: A long glass tube closed at one end and filled with mercury is inverted into a trough of mercury. This device is known as mercury barometer.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 2
The space above the mercury column in the tube contains only mercury vapour whose pressure Pis so small, that it may be neglected. The pressure inside the column at point A must equal the pressure at B.
∴ Pressure at B = Atmospheric pressure = Pa
Pa = ρgh = Pressure at A …………………… (1)
Where ρ is density of mercury and h is the height of the mercury column in the tube. In the experiment it is found that the mercury column in the barometer has a height of about 76cm at the sea level equivalent to one atmosphere.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 2.
What is guage pressure and how is a manometer used for measuring pressure differences ?
Answer:
Guage pressure : The guage pressure is the difference of the actual pressure and the atmospheric pressure. Pg = P – Pa
measurement of pressure differences :

  1. The manometer consists of a U-shaped tube, which is filled with a low density liquid (oil) for measuring small pressure difference and high density liquid
    (mercury) for measuring large pressure difference.
  2. One end of the tube is connected to the vessel D whose pressure of air measure and the other end of the tube is open.
  3. If pressure of air in vessel D is more than the earth’s atmosphere, the level of liquid in arm I will go down up to point A and level of liquid in arm II rises up to C.
  4. Then the pressure of air in vessel is equal to pressure at point A.
    AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 3
  5. Note the difference of liquid levels in the two arms of U-tube.(Say h). p be density of the liquid. Pa is the atmospheric pressure.
  6. Pressure at point A (PA) = Pressure at point B = Pressure at point C + Pressure due to column of liquid.
    PA = Pc + hρg (or) PA – Pc = hρg
    Here Pc = Pa, PA = P, ∴ P – Pa = hρg
    P – Pa = Pg = guage pressure = hρg

Question 3.
State Pascal’s law and verify it with the help of an experiment.
Answer:
Pascal’s law : It states that if gravity effect is neglected, the pressure at every point of liquid in equilibrium of rest is same.
Proof :

  1. Imagine a circular cylinder of uniform cross-sectional area A, such that points C and D lie on flat faces of the cylinder.
  2. The liquid inside the cylinder is in equilibrium under the action of forces exerted by the liquid outside the cylinder.
  3. These forces are acting every where perpendicular to the surface of the cylinder.
  4. Thus the forces on the flat faces of the cylinder at C and D will be perpendicular to the forces on the curved surface of the cylinder.
  5. Since liquid is in equilibrium, the sum of the forces acting on the curved surface of the cylinder must be zero.
    AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 4
  6. If P1 and P2 are the pressures at points C and D respectively. F1 and F2 are the forces acting on the flat surfaces of the cylinder due to liquid, then
    F1 = P1 A and F2 = P2A Since liquid is in equilibrium, therefore
    F1 = F2
    P1 A = P2A
    (or) P1 = P2
    It means the pressures at C and D are the same. This proves the pascal’s law.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 4.
Explain hydraulic lift and hydraulic brakes.
Answer:
Hydraulic lift and hydraulic brakes are based on the Pascal’s law.

Hydraulic lift : Here C and D are two cylinders of different areas of cross section. They are connected to each other with a pipe E. Each cylinder is provided with airtight frictionless piston. Let a, A be the area’s of cross-sections of the piston at C and D (a < < A)
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 5
The cylinders are filled with an incompressible liquid.
Let f be the applied force at C. Pressure exerted on the liquid
P = [latex]\frac{f}{a}[/latex] ……………. (1)
According to pascal’s law, this pressure is transmitted to piston of cylinder D. Upward force at D will be
F = PA = [latex]\frac{f}{a}[/latex] A = f [latex]\frac{A}{a}[/latex] …………… (2)
As A > > a
∴ F > > f .
∴ Heavy load placed on the larger piston is easily lifted.

Hydraulic Brakes :
When the brake pedal is pressed, the piston (P) of the master cylinder is pushed inwards. There will be increased pressure on liquid at P, which is transmitted equally to P1 and P2 of wheel cylinder in accordance with pascal’s law. Due to which P1 and P2 move outwards. Breakshoes to move away from each other which in turn press against the inner rim of the wheel. The brake becomes operative.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 6

Question 5.
What is hydrostatic paradox ?
Answer:
Consider three vessels A, B and C of different shapes. They are connected at the bottom by a horizontal pipe. On filling with the level in the three vessels is the same, though they hold different amounts of water. This is so because water at the bottom has the same pressure below each section of the vessel. It means the liquid pressure at a point is independent of the quantity of liquid but depends upon the depth of point below the liquid surface. This is known as hydrostatic paradox.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 7

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 6.
Explain how pressure varies with depth.
Answer:
Consider a fluid at rest in a container. In figure point 1 is at a height h above a point 2. The pressure at points 1 and 2 are P1 and P2. As the fluid is at rest, the horizontal forces should be zero. The resultant vertical forces should balance the weight of the element. Pressure at top (P1A) acting downward, pressure at bottom (P2A) acting upward.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 8
(P2 – P1) A = mg ………………… (1)
Mass of the fluid (m) = ρv = ρhA
(P2 – P1) = ρgh ………………. (2)
Pressure difference depends on the vertical distance h.
If the point 1 under discussion is shifted to the top of the fluid (water), which is open to the atmosphere, P1 may be replaced by atmospheric pressure (Pa) and P2 by P. Then eq (2) gives
P – Pa = ρgh
P = Pa + ρgh ……………… (3)
Thus the pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρgh.

Question 7.
What is Torricelli’s law ? Explain how the speed of efflux is determined with an experiment.
Answer:
Torricelli’s theorem : The velocity of efflux i.e., the velocity with which the liquid flows out of an orifice (i.e., a narrow hole) is equal to that which is freely falling body would acquire in falling through a vertical distance equal to the depth of orifice below the free surface of liquid.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 9
Consider ideal liquid of density p contained in a tank provided with a narrow hole.
Let h = Height of free surface of liquid above O.
P = Atmospheric pressure
v = Velocity of efflux
Applying Bernoulli’s theorem at A and O
(P + ρgh + O)atA = [P + 0 + [latex]\frac{1}{2}[/latex]ρv2]atO
P + ρgh = P + [latex]\frac{1}{2}[/latex] ρv2 = ρgh = [latex]\frac{1}{2}[/latex]ρv2
v = [latex]\sqrt{2 g h}[/latex]

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 8.
What is Venturi-meter? Explain how it is used.
Answer:
Venturi meter: The Venturi-meter is a device to measure the flow speed of incompressible fluid.

  1. It consists of a tube with a broad diameter and a small constriction at the middle.
  2. A manometer in the form of a U-tube is also attached to it, with one of arm at the broad neck point of the tube and the other at constriction as shown in figure.
    AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 10
  3. The manometer contains a liquid of density ρm.
  4. The pressure difference causes the fluid in the U-tube connected at the narrow neck to rise in comparison to the other arm.
  5. Filter pumps, sprayers used for perfumes, carburetor of automobile has used on this principle.

Question 9.
What is Reynolds number ? What is it’s significance ?
Answer:
Reynolds number : “Reynold number is a pure number which determines the nature of flow of liquid through a pipe”.
Reynold number (Re) = [latex]\frac{\rho \mathrm{vd}}{\eta}[/latex]
Where ρ is density of the fluid
v is speed of the fluid
d is diameter of the pipe

  1. If the flow is stream line (or) laminor Re < 1000
  2. If the flow is turbulent, Re > 2000
  3. If the flow becomes unsteady, 1000 < Re < 2000

Physical significance of Reynolds number : Reynold’s number describes the ratio of the inertial force per unit area to the viscous force per unit area for a flowing fluid.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 11

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 10.
Explain dynamic lift with examples.
Answer:
Dynamic lift: Dynamic lift is the force that act on a body, by virtue of its motion through a fluid.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 12
e.g.1 : Fig (a) shows, ball moving without spin. Stream lines are equally distributed above and below the ball. The velocity above and below the ball is same resulting zero pressure difference. There no upward (or) downward force on the ball.

Fig (b) shows, ball moving with spin stream lines are more crowded above the ball than below. The velocity of air above the ball is large (v + vr) and below it is smaller (v – vr). As a result, there is a pressure difference between lower and upper faces. Pressure is less at top of the ball and pressure is morebelow the ball. There is a net upward force on the ball.
e.g. 2 : Dynamic lift also acts on the an aeroplane wing.

Question 11.
Explain Surface Tension and Surface energy. [Mar. 13]
Answer:
Surface tension (S): The force acting per unit length of an imaginary line drawn on the surface of a liquid, normal to it and parallel to the surface is called surface tension.
T = [latex]\frac{\mathrm{F}}{l}[/latex]
S.l unit → N/m
D.F → [MT-2]
Surface energy (E): The additional potential energy due to molecular forces per unit surface area is called surface tension.
Surface energy = [latex][/latex]
S.l Unit → J/m2
D.F → (MT-2 )
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 13
Consider a horizontal liquid film ending in bar free to slide over parallel guides. We move the bar by a small distance d. The area of the surface increases, the system now has more energy, this means – that some work has to be done against an internal force F.
Work done (W) = F.d
If the surface energy of the film is S per unit area, the extra area is 2dl. A film has two sides and the liquid in between them.
So there are two surfaces and the extra energy is
S (2dl) = Fd
S = [latex]\frac{\mathrm{F}}{2 l}[/latex]
Surface tension is equal to the surface energy and is also equal to the force per unit length exerted by the fluid on the movable bar.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 12.
Explain how surface tension can be measured experimentally.
Answer:
A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the otherside, with its horizontal edge just over water. The vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension. Weights are added till the plate just clears water. Suppose the additional weight requires is W.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 14
Surface tension of liquid air interface is
Sla = [latex]\frac{\mathrm{W}}{2 l}=\frac{\mathrm{mg}}{2 l}[/latex]
Where l is length of the plate edge, m is extra mass.

Long Answer Questions

Question 1.
State Bernoulli’s principle. From conservation of energy in a fluid flow through a tube, arrive at Bernoulli’s equation. Give an application of Bernoulli’s theorem.
Answer:
Bernoulli’s principle : Bernoulli principle states that in a stream line flow, the sum of the pressure, the K.E per unit volume and the P.E per unit volume remains a constant.
P + [latex]\frac{1}{2}[/latex] ρv2 + ρgh = constant

  1. Consider a non-viscous, incompressible fluid is flowing through the pipe in a steady flow.
  2. Consider the flow at two regions BC and DE. Initially the fluid lying between B and D.
    AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 15
  3. During short time interval At, this fluid would have moved. Suppose V1 is the speed at B and V2 is the speed at D.
  4. In time ∆t, distance moved from B to C is V1 ∆t in the same interval (∆t) distance moved from D to E is V2∆t.
  5. Let P1 and P2 be pressure act at area’s of cross¬sections A1 and A2 of the two regions.
  6. The work done on the fluid at left end (BC)
    = Force × displacement
    = Pressure × Area × displacement
    = P1A1 × V1∆t (∆V = A1V1∆t) = P1∆V ………………. (1)
  7. Similarly work done by the fluid at right end (DE)
    = P2A2 × V2∆t = P2∆V …………………. (2)
  8. Work done on the fluid is taken as positive and workdone by the fluid is taken as negative.
    ∴ Total work done (W) = (P1 – P1) ∆V ………………. (3)
    Part of this work goes into changing the K.E of the fluid and part goes into changing gravitational P.E.
  9. Mass of the fluid(∆m) passing through the pipe in time (∆t) is ∆m = ρA1V1∆t
    where ρ is the density of the fluid.
    ∆m = ρ∆V ………………….. (4)
  10. Gravitational P.E = ρg∆V (h2 – h1) ………………… (5)
    Change in K.E (∆K) = [latex]\frac{1}{2}[/latex] ρ∆V (V22 – V12) …………….. (6)
  11. According to law of conservation of energy
    (P1 – P2) ∆V = [latex]\frac{1}{2}[/latex] ρ ∆V (V22 – V12) + ρg∆V (h2 – h1)
    P1 – P2 = [latex]\frac{1}{2}[/latex] ρ (V22 – V12) + ρg (h2 – h1)
    P1 + [latex]\frac{1}{2}[/latex] ρ V12 + ρgh1 = P1 + [latex]\frac{1}{2}[/latex] ρ V22 + ρgh2
    P + [latex]\frac{1}{2}[/latex] ρ v2 + ρgh = constant ………………. (7)

∴ Sum of the pressure, K.E per unit volume and P.E per unit volume remains constant. Application of Bernoulli’s theorem :

  1. It explains the dynamic lift on the wings of aeroplanes.
  2. It explains the dynamic lift experienced by a spinning cricket ball.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 2.
Define coefficient of viscosity. Explain Stoke’s law and explain the conditions under which a rain drop attains terminal velocity, υt. Give the expression for υt.
Answer:
Coefficient of viscosity (η): The coefficient of viscosity is defined as the tangential force per unit area of the layer, required to maintain unit velocity gradient.
η = [latex]\frac{F}{A\left(\frac{\Delta V}{\Delta x}\right)}[/latex]
S.l. unit → Nm-2 s (or) PaS
C.G.S unit → Poise
Dimensional formula = [ML-1T-1]
Stoke’s law : According to this law the viscous force acting on a moving body which is spherical in shape is directly proportional to

  1. Coefficient of viscosity of fluid (η)
  2. Radius of the spherical body (r)
  3. Velocity of the body (v)

∴ F ∝ ηrv
F = Kηrv
Where K is a constant of proportionality. Experimentally it was found to be 6π.
∴ F = 6πηrv
When rain drops falling through air from the clouds reach the surface with almost constant speed through they are moving under gravitational force. This velocity is called terminal velocity. After attaining the terminal velocity, net force acting on the rain drop is zero.
According to stokes law, F ∝ ηrv
F = 6πηrv (∵ 6π = K = Proportionality constant)
Let ρ, r be the density and radius of the sphere.
The fluid density be σ.
The forces acting on the sphere are

  1. Weight of the sphere W = mg
    W = Vρg = [latex]\frac{4}{3}[/latex]πr3ρg …………….. (1)
  2. The force of buoyancy (B) = V σ g = [latex]\frac{4}{3}[/latex]πr3σg ……………. (2)
  3. Viscous force (f) = 6πηv ………………… (3)
    When the sphere attains terminal velocity (vt), the net force on the body becomes zero.
    AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 16
    ∴ At terminal velocity
    Net downward force = Net upward force
    W = B + f, W – B = f
    [latex]\frac{4}{3}[/latex]πr3ρg – [latex]\frac{4}{3}[/latex]πr3σg = 6πηrvt;
    ∴ [latex]\frac{4}{3}[/latex]πr3g (ρ – σ) = 6πηrvt
    Vt = [latex]\frac{2}{9} \frac{r^2 g(\rho-\sigma)}{\eta}[/latex] ………………. (4)

Problems

Question 1.
Calculate the work done in blowing a soap bubble of diameter 0.6 cm against the surface tension force. (Surface tension of soap solution = 2.5 × 102 Nm-1).
Solution:
D = 0.6 cm = 0.6 × 102 m
r = [latex]\frac{D}{2}=\frac{0.6 \times 10^{-2}}{2}[/latex] = 0.3 × 102 m
S = 2.5 × 102 N/m
W = 8πr2s
= 8 × 3.14 × (0.3 × 10-2)2 × 2.5 × 10-2
W = 5.652 × 10-6 J

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 2.
How high does methyl alcohol rise in a glass tube of diameter 0.06 cm ? (Surface tension of methyl alcohol = 0.023 Nm-1 and density = 0.8 gmcm-3. Assume that the angle of contact is zero)
Solution:
D = 0.06cm, θ = 0°
r = [latex]\frac{D}{2}=\frac{0.06}{2}[/latex] = 0.03 cm = 0.03 × 10-2 m
= 3 × 10-4 m
S = 0.023 N/m, Density = 0.8 gm/c.c.
= 0.8 × [latex]\frac{10^{-3}}{10^{-6}}[/latex]
ρ = 0.8 × 103 kg/m3
S = [latex]\frac{\mathrm{hr \rho g}}{2 \cos \theta}[/latex]
h = [latex]\frac{2 \cos \theta}{r \rho g}[/latex]
= [latex]\frac{2 \times 0.023}{3 \times 10^{-4} \times 0.8 \times 10^3 \times 9.8}[/latex] (∵ cos 0° = 1)
= 0.0019 × 10-1 m
≈ 0.002 m
h = 2 cm .

Question 3.
What should be the radius of a capillay tube if water has to rise to a height pf 6 cm in it ? (Surface tension of water = 7.2 × 10-2 Nm-1).
Solution:
h = 6 × 10-2 m
S = 7.2 × 10-2 N/m
Density of water (ρ) = 103 kg/m3
S = [latex]\frac{\mathrm{hr \rho g}}{2}[/latex]
r = [latex]\frac{2 \mathrm{~S}}{\mathrm{h \rho g}}=\frac{2 \times 7.2 \times 10^{-3}}{6 \times 10^{-2} \times 10^3 \times 9.8}[/latex]
r = [latex]\frac{14.4}{58.8} [/latex] × 10-3
r = 0.24 × 10-3 m
r = 0.24 mm

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 4.
Find the depression of the meniscus in the capillary tube of diameter 0.4 mm dipped in a beaker containing , mercury. (Density of mercury = 13.6 × 103 Kg m-3 and surface tension of mercury = 0.49 Nm-1 and angle of contact = 135°).
Solution:
D = 4 m.m
r = [latex]\frac{D}{2}=\frac{4}{2}[/latex] = 2m.m = 2 × 10-3 m
ρ = 13.6 × 103 kg/m3
θ = 135°, S = 0.49 Nm
cosθ = cos 135°
= – sin 45° = – [latex]\frac{1}{\sqrt{2}}[/latex]
S = [latex]\frac{\mathrm{hr \rho g}}{2 \cos \theta}[/latex]
h = [latex]\frac{2 \mathrm{~s} \cos \theta}{\mathrm{r \rho g}}[/latex]
= 2 × 0.49 [latex]\left(\frac{-1}{\sqrt{2}}\right) \frac{1}{2 \times 10^{-3} \times 13.6 \times 10^3 \times 9.8}[/latex]
h = [latex]\frac{-0.49}{13.6 \times 9.8 \times \sqrt{2}}[/latex]
h = -0.024m.

Question 5.
If the diameter of a soap bubble is 10 mm and its surface tension is 0.04 Nm-1, find the excess pressure inside the bubble. [Mar. 14]
Solution:
D = 10 m.m
r = [latex]\frac{D}{2}=\frac{10}{2}[/latex] = 5 m.m
= 5 × 10-3
S = 0.04 N/m
Pi – P0 = [latex]\frac{4 S}{r}=\frac{4 \times 0.04}{5 \times 10^{-3}}[/latex]
= 0.032 × 103
Pi – P0 = 32 N/m2 (or) Pascal.

Question 6.
If work done by an agent to form a bubble of radius R is W, then how much energy is required to increase its radius to 2R ?
Solution:
R1 = R, R2 = 2R
Initial work done (W) = 8πR2S
Final work done (W) = 8π[R22 – R12]S
= 8π [4R2 – R2]S
= 3 × 8π R2S
W = 3W.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 7.
If two soap bubbles of radii R1 and R2 (in vacuum) coalasce under isothermal conditions, what is the radius of the new bubble. Take T as the surface tension of soap solution.
Solution:
R1, R2 and R be the radii of first, second and resultant bubble. The soap bubbles coalesce in vacuum, so surface energy do not change
E = E1 + E2
8π R2T = 8π R12T + 8π R22 T
R2 = R12 + R22
R = [latex]\sqrt{R_1^2+R_2^2}[/latex]

Additional Problems

Question 1.
Explain why
a) The blood pressure in humans is greater at the feet than at the brain.
b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km.
c) Hydrostatic pressure is a scalar quantities even though pressure is force divided by area.
Answer:
a) The height of the blood column in the human body is more at feet than at the brain. That is why, the blood exerts more pressure at the feet than at the brain (∴ pressure = hρg)

b) We know that the density of air is maximum near the surface of earth and decreases rapidly with height and at a height of about 6 km, it decreases to near by half its value at the sea level. Beyond 6 km height the density of air decreases very slowly with height. Due to this reason, the atmospheric pressure at a height of about 6 km decreases to nearby half of its value at sea level.

c) Since due to applied force on liquid, the pressure is transmitted equally in all directions inside the liquid. That is why there is no fixed direction for the pressure due to liquid. Hence hydrostatic pressure is a scalar quantity.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 2.
Explain why
a) The angle of contact of mercurywith glass is obtuse, while that of water with glass is acute.
b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.) [T.S. Mar. 15]
c) Surface tension of a liquid is independent of the area of the surface.
d) Water with detergent disolved in it should have small angles of contact.
e) A drop of liquid under no external forces is always spherical in shape.
Answer:
a) When a small quantity of liquid is poured on solid, three interfaces, namely liquid- air, solid-air and solid-liquid are formed. The surface tensions corresponding to these three interfaces are SLA, SSA and SSL respectively. Let 0 be the angle of contact between the liquid and solid. The molecules in the region, where the three interfaces meet are in equilibrium. It means net force acting on them is zero. For the molecule at 0 to be in equilibrium, we have.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 17
In case of mercury glass, SSA < SSL, therefore cosθ is negative or θ > 90° i.e. obtuse. In case of water-glass, SSA > SSL, therefore cosθ is positive or θ < 90° i,e. acute.

b) For mercury glass, angle of contact is obtuse. In order to achieve this obtuse value of angle of contact, the mercury tends to form a drop. In case of water glass, the angle of contact is acute. To achieve this acute value of angle of contact, the water tends to spread.

c) Surface tension of liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. Since this force is independent of the area of liquid surface, therefore surface tension is also independent of the area of the liquid surface.

d) We know that the cloth has narrow spaces in the form of capillaries. The rise of liquid in a capillary tude is directly proportional to cos0. if 0 is small cos0 will be large. Due to which capillary rise will be more and so the detergent will penetrate more in cloth.

e) In the absence of external forces, the surface of the liquid drop tends to acquire the minimum surface area due to surface tension. Since for a given volume, the surface area of sphere is least, hence the liquid drop takes the spherical shape.

Question 3.
Fill in the blanks using the word (S) from the list appended with each statement ;
a) Surface tension of liquids generally … with temperatures (increases / decreases)
b) Viscosity of gases… with temperature, whereas viscosity of liquids … with temperature (increases / decreases)
c) For solids with elastic modulus of rigidity, the shearing force is proportional to…. while for fluids it is proportional to … (shear strain / rate of shear strain)
d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)
e) For the model of a plane in a wind tunnel, turbulence occurs at a… speed for turbulence for an actual plane (greater / smaller)
Answer:
a) Decreases
b) increases; decreases
c) Shear strain; rate of shear strain.
d) Conservation of mass; Bernoullis principle.
e) Greater.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 4.
Explain why
a) To keep a piece of paper horizontal, you should blow over, not under, it.
b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
e) A spinning cricket ball in air does not follow a parabolic trajectory.
Answer:
a) When we blow over the paper, the velocity of air blow increases and hence pressure of air on it decreases (according to Bernoullis theorem), whereas pressure of air below the paper is atmosphere. Hence the paper stays horizontal.

b) By doing so the area of the outlet of water jet is reduced, so velocity of water increases according to equation of continuity av = a constant.

c) When a fluid is flowing out of a small hole in a vessel it acquires a large velocity and hence possesses large momentum. Since no external force is acting on the system, a backward velocity must be attained by the vessel (according to law of conservation of momentum). As a result of it, impulse (backward thrust) is experienced by the vessel.

d) There, size of the needle controls velocity of flow and thumb pressure controls pressure. According to Bernoulli’s theorem, P + ρgh + [latex]\frac{1}{2}[/latex] ρV2 = a constant shows that P occurs with power one and V occurs with power two, hence the velocity has more influence. That is why the needle has a better control over flow.

e) If the ball is spinning well as moving linearly, the streamlines at the top of ball due to two types of motion are opposed to each other and those below are in the same direction. As a result of it, the velocity of air flow is greater below than above the ball. Now, according to Bernoullis principle, the pressure on the upper side of the ball becomes more than the pressure on the lower side of ball. Due to it, a resultant force F acts upon the ball at right angle to linear motion in the downward, direction, resulting the ball to move along a curved path.

Question 5.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor ?
Answer:
Here, m = 50kg; r =D/2 = 1/2 cm = [latex]\frac{1}{200}[/latex] m
Pressure = [latex]\frac{\text { Force }}{\text { area }}=\frac{m g}{\pi r^2}=\frac{50 \times 9.8}{(22 / 7)(1 / 200)^2}[/latex]
= 6.24 × 106 Nm-2.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 6.
Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m-3. Determine the height of the wine column for normal atmosphere pressure.
Answer:
P = 0.76 × (13.6 × 103) × 9.8
= h × 984 × 9.8 or
h = [latex]\frac{0.76 \times 13.6 \times 10^3 \times 9.8}{984 \times 9.8}[/latex]
= 10.5 m.

Question 7.
A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to be roughly 3 km and ignore ocean currents.
Answer:
Here, maximum stress = 109Pa,
h = 3km = 3 × 103m;
ρ(water) = 103kg/m3 and g = 9.8 m/s2
The structure will be suitable for putting upon top of an oil well provided the pressure exerted by sea water is less than the maximum stress it can bear.
Pressure due to sea water, P = hρg
= 3 × 103 × 103 × 9.8
= 2.94 × 107 Pa
Since the pressure of sea water is less than the maximum stress of 109 Pa, the structure will be suitable for putting upon top of the oil well.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear ?
Answer:
The maximum force, which the bigger piston can bear,
F = 3000 kgf = 3000 × 9.8 N
∴ Area of piston, A = 425 cm2
= 425 × 10-4 m2
∴ maximum pressure on the bigger piston.
P = [latex]\frac{F}{A}=\frac{3000 \times 9.8}{425 \times 10^{-4}}[/latex] = 6.92 × 105 Pa
Since the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is 6.92 × 105 Pa.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 9.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit ?
Answer:
For water column in one arm of U-tube,
h1 = 10.0 cm; ρ1 (density) = 1g cm-3
For spirit column in other arm of U-tube, h2 = 12.5 cm; ρ1 = ?
As the mercury column in the two arms of U-tube are in level, therefore pressure exerted by each is equal. Hence h1ρ1g = h2ρ2g or
P2 = [latex]\frac{h_1 \rho_1}{h_2}=\frac{10 \times 1}{12.5}[/latex] = 0.8 g cm-3
Therefore, relative density of spirit = ρ2/ρ1
= [latex]\frac{0.8}{1}[/latex] = 0.8

Question 10.
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific, gravity of mercury = 13.6)
Answer:
On pouring 15.0 cm of water and spirit each into the respective arms of U-tube, the mercury level will rise in the arm containing spirit. Let h be the difference in the levels of mercury in two arms of U-tube and p be the density of mercury.
∴ The pressure exerted by h cm of mercury column = difference in pressure exerted by water and spirit.
∴ h1ρ1g = h2ρ2g ……………. (1)
Here h = ? ρ =13.6 g cm-3
ρ1 =1 cm-3
ρ2 = 0.8 g cm-3
h1 = 15 + 10 = 25 cm
h2 = 15 + 12.5 = 27.5 cm
Putting values in (i) we get h × 13.6 × g
= 25 × 1 × g – 27.5 × 0.8 × g = 3g
or h = 3/13.6 = 0.22cm

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain.
Answer:
No, Bernoulli’s theorem is used only for stream – line flow.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 12.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain.
Answer:
No, it does not matter if one uses gauge instead of absolute pressures in applying Bernoulli’s’ equation, provided the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10-3 kg s-1, what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 103 kg m-3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
Answer:
Here, l = 1.5 m, r = 1.0 cm = 10-2 m, ρ = 1; 1.3 × 10-3 kg/m3; η = 0.83 Nsm-2.
Mass of glycerine flouring per sec, M = 4 × 10-3 kg/s
Volume of glycerine flouring per sec, V = [latex]\frac{M}{\rho}[/latex]
V = [latex]\frac{4 \times 10^{-3}}{1.3 \times 10^3} \mathrm{~m}^3 \mathrm{~s}^{-1}[/latex] m3s-1
If ρ is the difference of pressure between two ends of the tube,then using poisevilles formula we have
V = [latex]\frac{\pi \rho r^4}{8 \eta l}[/latex] or P = [latex]\frac{V \times 8 \eta l}{\pi r^4}[/latex]
P = [latex]\left(\frac{4 \times 10^{-3}}{1.3 \times 10^3}\right) \times \frac{8 \times 0.83 \times 1.5}{3.142 \times\left(10^{-2}\right)^4}[/latex]
= 975.37 Pa

Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s-1 and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2 ? Take the density of air to be 1.3 kg m-3.
Answer:
Let V1, V2 be the speeds on the upper and lower surfaces of the wing of aeroplane and P1 and P2 be the pressures on upper and lower surfaces of the wing respectively.
Then V1 = 70ms-1; V2 = 63ms-1,
P = 1.3kg m-3.
This difference of pressure provides the lift to the aeroplane.
[latex]\frac{P_1}{\rho}[/latex] + gh + [latex]\frac{1}{2} V_1^2[/latex] = [latex]\frac{P_2}{\rho}[/latex] + gh + [latex]\frac{1}{2} V_2^2[/latex]
∴ [latex]\frac{P_1}{\rho}-\frac{P_2}{\rho}=\frac{1}{2}\left(V_2^2-V_1^2\right)[/latex]
or P1 – P2 = [latex]\frac{1}{2} \rho\left(V_2^2-V_1^2\right)[/latex]
= [latex]\frac{1}{2}[/latex] × 1.3 [(70)2 – (63)2]
= 605.15 Pa
This difference of pressure provides the lift to the aeroplane.
So, lift on the aeroplane = Pressure difference × area of wings
= 605.15 × 2.5
= 1512.875 N
= 1.51 × 103N.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 15.
Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 18
Answer:
Figure a is incorrect. According to equation of continuity i.e. av = a constant, where area of cross – section of tube is less, the velocity of liquid flow is more than the other portion of tube. According to Bernoulli’s theorem,
P + [latex]\frac{1}{2}[/latex] ρv2 = a constant i.e. where V is more P is less and avice versa.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min-1, what is the speed of ejection of the liquid through the holes ?
Answer:
Area of cross – section of tube, a1 = 8.0 cm2 = 8 × 10-4 m2
No. of holes = 40, Diameter of each hole, D = 1 mm = 10-3 m
∴ Radius of hole, r = [latex]\frac{D}{2}=\frac{1}{2}[/latex] × 10-3 m
= 5 × 10-4 m
Area of cross – section of each hole = πr2
= π(5 × 10-4)2m2
Total area of cross – section of 40 holes,
a2 = 40 × π (5 × 10-4)2m2
Speed of liquid inside the tube,
V1 = 1.5m/min
= [latex]\frac{1.5}{60}[/latex] ms-1
If V2 is the velocity of ejection of the liquid through the holes, then
a1V1 = a2V2 or V2 = [latex]\frac{a_1 V_1}{a_2}[/latex]
V2 = [latex]\frac{\left(8 \times 10^{-4}\right) \times 1.5}{60 \times 40 \times \pi \times\left(5 \times 10^4\right)^2}[/latex]
= 0.637 ms-1

Question 17.
A U-shaped wire is dipped in a soap solution and removed. The thin shaped film formed between the surfaces and the light slider supports a weight of 1.5 × 10-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film ?
Answer:
We know that soap film has two free surfaces, so total length of the film to be supported,
= 2l = 2 × 30 = 60 cm
= 0.6 m
Total force on the slider due to surface tension will be,
F = S × 2l
= S × 0.6 N
In equilibrium position, the force F on slider due to surface tension must balance the weight mg
(1.5 × 10-2)N i.e. F = 1.5 × 10-2
= S × 0.6 Or
S = [latex]\frac{1.5 \times 10^{-2}}{0.6}[/latex]
2.5 × 10-2Nm-1

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 18.
Figure (a) shows a thin liquid film supporting a small weight = 4.5 × 10-2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c) ? Explain your answer physically.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 19
Answer:
a) Here, length of film supporting the weight
= 40cm = 0.4 m
Total liquid film supported (or force)
= 4.5 × 10-2 N
film has two free surfaces, ∴ surface tension,
S = [latex]\frac{4.5 \times 10^{-2}}{2 \times 0.4}[/latex]
S = 5.625 × 10-2 Nm-1
Since the liquid is same for all the cases (a), (b), (c) and temperature is also same, therefore surface tension for cases (b) and (c) will also be the same = 5.625 × 10-2. In figure (b) and (c) the length of the film supporting the weight is also the same as that of(a), hence the total weight supported in each case is 4.5 × 10-2 N.

Question 19.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20°C) is 4.65 × 10-1 Nm-1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.
Answer:
Here, r = 3.0 mm = 3 × 10-3 m;
S = 4.65 × 10-1 Nm-1;
P = 1.01 × 105 Pa
Excess of pressure inside the drop of mercury is given by
P = [latex]\frac{2 S}{r}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}[/latex]
= 310 Pa
Total pressure inside the drop = P + ρ
= 1.01 × 105 + 310
= 1.01 31 × 105 Pa

Question 20.
What is the excess pressure inside a bubble of soap solution of radius 5.0 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 × 10-2 Nm-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 105 Pa).
Answer:
Here, S = 2.5 × 10-2 Nm-1, r = 5.00 mm = 5 × 10-3 m.
Density of soap solution, ρ = 1.2 × 103 kg m-3
Excess pressure inside the soap bubble,
P = [latex]\frac{4 s}{r}=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}[/latex] = 20 Pa
Excess pressure inside the air bubble, P’ = [latex]\frac{2S}{r}[/latex]
= [latex]\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}[/latex] = 10 Pa
Total pressure inside the air bubble at depth h in soap solution – ρ’ + atmospheric pressure + hρg
= 10 + 1.01 × 105 + 0.4 × 1.2 × 103 × 9.8
= 1.06 × 103 Pa

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 21.
A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.
Answer:
For compartment containing water,
h1 = 4m, ρ1 = 103 kg m-3
The pressure exerted by water at the door provided at bottom
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 20
P1 = h1ρ1g
= 4 × 103 × 9.8
= 3.92 × 104 Pa
For compartment containing acid,
h2 = 4m,
ρ2 = 1.7 × 1.03 kg/m3
The pressure exerted by acid at the door provided at bottom.
P2 = h2ρ2g
= 4 × 1.7 × 103 × 9.8
= 6.664 × 104 Pa
∴ Difference of pressure = P2P1
= 6.664 × 104 – 3.92 × 104
= 2.774 × 104Pa
Given, area of door, A = 20cm2
= 20 × 10-4m2
Force on the door = difference in pressure × area
= (P2 – P1) × A
= (2.774 × 104) × (20 × 104)
= 54.88N
≈ 55N
To keep the door closed, the force equal to 55 N should be applied horizontally on the door from compartment containing water to that containing acid.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in Fig. (a) when a pump removes some of the gas, the manometer reads as in Fig. (b) The liquid used in the manometers is mercury and the atmo-spheric pressure is 76 cm of mercury.
a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and b) in units of cm of mercury.
b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury, are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 21
Answer:
a) Here, atomospheric presure, p = 76 cm of mercury
In Fig (a) Pressure head
h = + 20 cm
∴ Absolute pressure = p + h
= 76 + 20
= 96 cm of mercury
Gauge pressure = h = 20 cm of mercury.
In fig (b) pressure head, h = -18 cm
Absolute pressure = p + h
= 76+ (-18)
= 58 cm of mercury
Gauge pressure = h = -18cm of mercury.

b) Here 13.6 cm of water added in right limb is equiralent to
[latex]\frac{13.6}{13.6}[/latex] = 1 cm of mercury column.
i.e., h1 = 1 cm of mercury column.
Now pressure at A, PA = P + h1 = 76 + 1
= 77 cm
Let the difference in mercury levels in the two lumbs be h1, then pressure at B,
PB = 58 + h1 or
As PA = PB = 77 = 58 + h1
h1 = 77 – 58 = 19 cm of mercury column.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular comon height. Is the force exerted by the water on the base of the vessel the same in the two cases ? If so, why do the vessels filled with water to that same height give different readings on a weighing scale ?
Answer:
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 22
Since the pressure depends upon the height of water column and the height of the water column in the two vessels of different shapes is the same.hence there will be same pressure due to water on the base of each vessel. As the base area of each vessel is same, hence there will be equal force acting on the two base areas due to water pressure. The water exerts force on the walls of the vessel also. In case, the walls of the vessel are not perpendicular to base, the force exerted by water on the walls has a net non-zero vertical component which is more in first vessel than that of second vessel. That is why, the two vessel:; filled with water to same vertical height show different readings on a weighing machine.

Question 24.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein ? Use the density of whole blood from Table 1.
Answer:
h = [latex]\frac{p}{p g}=\frac{200}{1.06 \times 10^3}[/latex] × 9.8 = 0.1925 m
The blood may just enter the vein if the height at which the blood container be kept must be slightly greater than 0.1925 m i.e. 0.2 m.

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.
a) What is the largest average velocity of blood flow in an artery of diameter 2 × 10-3 m if the flow must remain laminar ?
b) Do the dissipative forces become more important as the fluid velocity increases ? Discuss qualitatively.
Answer:
a) If dissipative forces are present, then some forces in liquid flow due to pressure different is spent against dissipative forces. Due to which the pressure drop becomes large.

b) The dissipative forces become more important with increasing flow velocity, because of tubulence.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 26.
a) What is the largest average velocity of blood flow in an artery of radius 2 × 10-3m if the flow must remain lanimar ?
Answer:
Here, r = 2 × 10-3m ;
D = 2r = 2 × 2 × 10-3 = 4 × 10-3m
η = 2.084 × 10-3 Pa s;
p = 1.06 × 103 kgm3
For flow tobe laminar, Nr = 2000
a) Now, Vc = [latex]\frac{N_r \eta}{\rho D}[/latex]
= [latex]\frac{2000 \times\left(2.084 \times 10^{-3}\right)}{\left(1.06 \times 10^3\right) \times\left(4 \times 10^{-3}\right)}[/latex] = 0.98m/s.

b) What is the corresponding flow rate ? (Take viscosity of blood to be 2.084 × 10-3 Pa s).
Answer:
Volume flowing per second = πr2Vc
= [latex]\frac{22}{7}[/latex] × (2 × 10-3)2 × 0.98
= 1.23 × 10-5 m3s-1

Question 27.
A plane is in level flight at constant speed and each of its two wings has an area of 25m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m-3)
Answer:
Here, V1 = 180 km/h = 50m/s, V2 = 234 km/ h = 65 m/s;
A = 2 × 25 = 50m2; P = 1kg/m3
P1 – P2 = [latex]\frac{1}{2}[/latex] p (V22 – V12)
= [latex]\frac{1}{2}[/latex] × 1 × (652 – 502)
Upward force = (P1 – P2) A = [latex]\frac{1}{2}[/latex] × (652 – 502) × 50N
As the plane is in level flight, so
mg = (P1 – P2)A
or m = [latex]\frac{\left(P_1-P_2\right) A}{g}=\frac{1 \times\left(65^2-50^2\right) \times 50}{2 \times 9.8}[/latex]
= 4.4 × 103N

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10-5 m and density 1.2 × 103 kg m-3. Take the viscosity of air at the temperature of the experiment to be 1.8 × 105 Pa s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air.
Answer:
Here, r = 2.0 × 10-5m; ρ = 1.2 × 103 kg m-3;
η = 1.8 × 105 Ns m-2
P0 = 0, V = ?, F = ?
Terminal velocity V = [latex]\frac{2 r^2\left(\rho-\rho_0\right) g}{9 \eta}[/latex]
= [latex]\frac{2 \times\left(2.0 \times 10^{-5}\right)^2 \times\left(1.2 \times 10^3-0\right) \times 9.8}{9 \times 1.8 \times 10^{-5}}[/latex]
= 5.8 × 10-2ms-1 = 5.8 cms-1
Viscous force on the drop, F = 6πηrv
= 6 × [latex]\frac{22}{7}[/latex] × (1.8 × 10-5) × (2.0 × 10-5) × (5.8 × 10-2)
= 3.93 × 10-10N.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 29.
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.456 N m-1. Density of mercury = 1.36 × 103 kg m-3.
Answer:
Here, θ = 140°, r = 1 × 10-3 m;
S = 0.465 Nm-1, ρ = 13.6 × 103 kg, h = ?]
Cos = 140° = – cos40° = -0.7660
Now h = [latex]\frac{2 S \cos \theta}{r \rho g}[/latex]
= [latex]\frac{2 \times 0.465 \times \cos 140^{\circ}}{10^{-3} \times 13.6 \times 10^3 \times 9.8}[/latex]
= [latex]\frac{2 \times 0.465 \times(-0.7660)}{10^{-3} \times 13.6 \times 10^3 \times 9.8}[/latex]
= -5.34 × 10-3m
= -5.34mm
Negative value of h shows that the mercury levels is depressed in the tube.

Question 30.
Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to forma a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube ? Surface tension of water at the temperature of the experiment is 7.3 × 10-2 N m-1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m-3 (g = 9.8 m-2).
Answer:
Here, S = 7.3 -2 Nm-1, ρ = 1.0 × 103 kg m-3; θ = 0°
For narrow tube, 2r1 = 3.00 mm = 3 × 10-3 m or r1 =1.5 × 10-3 m .
For wider tube, 2r2 = 6.00 mm = 6 × 10-3 m or r2 = 3 × 10-3 m
let h1, h2 be the heights to which water rises in narrow tube and wider tube respectively.
Then, h1 = [latex]\frac{2 s \cos \theta}{r_1 \rho g}[/latex] and h2 = [latex]\frac{2 s \cos \theta}{r_2 \rho g}[/latex]
∴ Difference in levels of water in two limbs of U-tube is
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 23

Question 31.
a) It is known that density of air decreases with height y as
ρ = ρ0e-y/y0
Where p0 = 1.25 kg m-3 is the density at sea level, and y0 is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.
Answer:
We know that the rate of decrease of density p of air is directly proportional to height y i.e.
[latex]\frac{-d \rho}{d y}[/latex] ∝ p or [latex]\frac{d \rho}{d y}[/latex] = – Kρ
Where K is a constant of proportionality. Here – ve sign shows that ρ decreases as y increases.
[latex]\frac{d y}{\rho}[/latex] = – Kρ
Integrating it with in the conditions, as y changes fromotoy density changes from ρ0 to ρ, we have
[latex]\left.\int_{P_0}^p \frac{d P}{P}=-\int_0^y k d y=\left[\log e^p\right)\right]_{\rho_0}^p=k y[/latex]
[latex]\frac{\rho}{\rho_0}=e^{-k y}=\rho=\rho_0 e^{-\mathrm{y} / y_0}[/latex]

b) A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assum that the balloon maintains constant radius as it rises. How high does it rise ?
(Take y0 = 8000 m and PHe = 0.18 kgm-3)
Answer:
The balloon will rise to aheight, where its density becomes equal to the air at that height.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 24
y = 8000 × 1 = 8000 m
= 8 km.

Textual Examples

Question 1.
The two thigh bones (femurs), each of cross-sectional area 10 cm2 mass 40 kg. Estimate the average pressure sustained by the femurs.
Answer:
Total cross-sectional area of the femurs is A = 2 × 10 cm2 20 × 10-4 m2. The force acting on them is F = 40 kg wt = 400 N (taking g 10 ms-2). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
Pav = [latex]\frac{F}{A}[/latex] = 2 × 105 N m-2

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 2.
What is the pressure on a swimmer 10m below the surface of a lake ?
Answer:
Here
h = 10 m and p = 1000 kg m-3.
Take g 10 m S-2
From Equation = P = Pa + ρgh
P = Pa + ρgh
= 1.01 × 105 Pa + 1000 kgm-3 × 10m S-2 × 10 m
= 2.10 × 105 Pa
≈ 2 atm
This is a 100% increase in pressure from surface level. At a depth of 1 km the increase in pressure is 100 atm! Submarines are designed to withstand such enormous pressures.

Question 3.
The density of the atmosphere at sea level is 1.29 kg/m3. Assume that it does not change with altitude. Then how high would the atmosphee extend ?
Answer:
We use P – Pa = ρgh
ρgh = 1.29 kg m-3 × 9.8 ms2 × hm
= 1.01 × 105 pa
∴ h = 7989 m ≈ 8 km
In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing press-ure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 4.
At a depth of 1000 m in an ocean
(a) What is the absolute pressure ?
b) What is the guage Pressure ?
c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sealevel atmospheric pressure. (The density of sea water is 1.03 × 103 kg m-3, g = 10ms-2.
Answer:
Here h = 1000 m and ρ = 1.03 × 103 kg m-3
a) From P2 – P1 = ρgh absolute pressure
P = Pa + ρgh
= 1.01 × 105 Pa + 1.03 × 103 kg m-3 × 10 m s-2 × 1000 m
= 104.01 × 105 Pa .
= 104 atm

b) Gauge pressure is P – Pa = ρgh = Pg
Pg = 1.03 × 103 kg m-3 × 10 m s2 × 1000 m
= 103 × 105 Pa
≈ 103 atm

c) The pressure outside the submarine is P = Pa + ρgh and the pressure inside it is Pa. Hence, the net pressure acting on the window is gauge pressure, Pg = ρgh. Since the area of the window is A = 0.04 m2, the force acting on it is
F = PgA = 103 × 105 Pa × 0.04m2
= 4.12 × 105 N
≈ 103 atm

Question 5.
Two syringes of different cross sections A1, A2 and lengths L1, L2 (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively.
a) Find the force exerted on the larger piston when a force of ION is applied to the smaller piston, b) if the smaller piston is pushed in through 6.0 cm,’how much does the larger piston move out ?
Answer:
a) Since pressure is transmitted undiminished throughout the fluid.
F2 = [latex]\frac{A_2}{A_1} F_1=\frac{\pi\left(3 / 2 \times 10^{-2} m\right)^2}{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2} \times 10 \mathrm{~N}[/latex]
= 90 N

b) Water is considered to be perfectly incompressible. Volume covered by the move-ment of smaller piston inwards is equal to volume moved outwards due to the larger piston.
L1A1 = L2A2
L2 = [latex]\frac{A_1}{A_2} L_1=\frac{\pi\left(1 / 2 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(3 / 2 \times 10^{-2} \mathrm{~m}\right)^2}[/latex]
= × 6 × 10-2m
≃ 0.67 × 10-2m = 0.67 cm.
Note, atmospheric pressure is common to both pistons and has been ignored.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 6.
In a car lift compressed air exerts a force F1 on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston or radius 15 cm (Fig). If the mass of the car to be lifted is 1350 kg, calculate F1. What is the pressure necessary to accomplish this task ? (g = 9.8 ms-2).
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 25
Answer:
Since pressure is transmitted undiminished throughout the fluid.
F1 = [latex]\frac{A_1}{A_2} F_2=\frac{\pi\left(5 \times 10^{-2} \mathrm{~m}\right)^2}{\pi\left(15 \times 10^{-2} \mathrm{~m}\right)^2}[/latex]
= (1350 N × 9.8m s-2) = 1470 N
= 1.5 × 103N
The air pressure that will produce this force is
P = [latex]\frac{F_1}{A_1}=\frac{1.5 \times 10^{-3} \mathrm{~N}}{p\left(5 \times 10^{-2}\right)^2 \mathrm{~m}}[/latex] = 1.9 × 105 Pa
This is almost double the atmospheric pressure.

Question 7.
Blooc velocity : The flow of blood in a large artery of an anesthetised dog is diverted through a Venturi meter. The wider part of the meter has a cross-sectional area equal to that of the artery. A = 8 mm2. The harrower part has an area = 4mn2. The pressure drop in the artery is 24 Pa. What is the speed of the blood in the artery ?
Answer:
We take the density of blood from whole blood to be 1.06 × 103 kg m3. The ratio of the areas is [latex]\left(\frac{\mathrm{A}}{\mathrm{a}}\right)[/latex] = 2
Using v1 = [latex]\left(\sqrt{\frac{2 \rho_m g h}{\rho}}\right)\left(\left(\frac{A}{2}\right)^2-1\right)^{-1 / 2}[/latex]
we obtain
v1 = [latex]\sqrt{\frac{2 \times 24 \mathrm{pA}}{1060 \mathrm{~kg} \mathrm{~m}^{-3} \times\left(2^2-1\right)}=0.125 \mathrm{~m} \mathrm{~s}^{-1}}[/latex]

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 8.
A fully loaded Boeing aircraft has a mass of 3.3 × 105 kg. Its total wing area is 500 m2. It is in level flight with a speed of 960 km/h. a) Estimate the pressure difference between the lower and upper surfaces of the wings (b) Estimate the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ = 1.2 kgm-3].
Answer:
a) The weight of’the Boeing aircraft is balanced by the upward force due to the pressure difference
∆P × A – 3.3 × 105 kg × 9.8 = mg.
∆P = (3.3 × 105 kg × 9.8 m s-2) / 500 m2
= 6.6 × 103 N m-2

b) We ignore the small height difference between the top and bottom sides in
P1 + ([latex]\frac{1}{2}[/latex]) ρv12 + ρgh1 = P2 + ([latex]\frac{1}{2}[/latex])ρv22 + ρgh2
The pressure difference between them is then
∆P = [latex]\frac{\rho}{2}\left(v_2^2-v_1^2\right)[/latex]
Where v2 is the speed of air over the upper surface and v1 is the speed under the bottom surface.
(v2 – v1) = [latex]\frac{2 \Delta P}{\rho\left(v_2+v_1\right)}[/latex]
Taking the average speed vav (v2 + v1)/2 – 960 km/h = 267 m s-1, we have
(v2 – v1)/vav = [latex]\frac{\Delta \mathrm{P}}{\rho v_{\mathrm{av}}^2}[/latex] ≈ 0.08
The speed of air above the wing needs to be only 8% higher than that below.

Question 9.
A metal block of area 0.10 m2 is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless and frictionless), as in Fig. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released, the block moves to the right with a constant speed of 0.085 ms-1. Find the co-efficient of viscosity of the liquid.
AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids 26
Answer:
The metal block moves to the right because of the tension in the string. The tension T is equal in magnitude to the weight of the suspended mass m. Thus the shear force F is
F = T = mg = 0.010 kg × 9.8 ms-2
= 9.8 × 10-2 N
Shear stress on the fluid = F/A = [latex]\frac{9.8 \times 10^{-2}}{0.10}[/latex]
Strain rate = [latex]\frac{v}{l}=\frac{0.085 \mathrm{~ms}^{-1}}{0.3 \times 10^{-3} \mathrm{~m}}[/latex]
η = [latex]\frac{\text { stress }}{\text { strain rate }}[/latex]
= [latex]\frac{\left(9.8 \times 10^{-2} \mathrm{~N}\right)\left(0.30 \times 10^{-3} \mathrm{~m}\right)}{\left(0.085 \mathrm{~ms}^{-1}\right)\left(0.10 \mathrm{~m}^2\right)}[/latex]
= 3.45 × 10-3 Pa s

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 10.
The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20°C is 6.5 cm s-1. Compute the viscosity of the oil at 20°C. Density of oil is 1.5 × 103 kg m-3, density of copper is 8.9 × 103 kg m-3.
Answer:
We have vt = 6.5 × 10-2 ms-1,
a = 2 × 10-3m,
g = 9.8 ms-2, ρ = 8.9 × 103 kg m-3,
σ = 1.5 × 103 kg m-3. From
vt = [latex]\frac{2 a^2(\rho-\sigma) g}{(9 \eta)}[/latex]
η = [latex]\frac{2}{9} \times \frac{\left(2 \times 10^{-3}\right) \mathrm{m}^2 \times 9.8 \mathrm{~ms}^{-2}}{6.5 \times 10^{-2} \mathrm{~ms}^{-1}}[/latex] × 7.4 × 103 kg m-3
= 9.9 × 10-1 kg m-1s-1

Question 11.
a) The flow rate of water from a tap of diameter 1.25 cm is L7min. The co-efficient of viscosity of water is 10-3 Pa s.
b) After sometime the flow rate is increased to 3L / min. Characterise the flow for both the flow rates.
Answer:
a) Let the speed of the flow be v and the diameter of the tap be d = 1.25 cm. The volume of the water flowing out per second is
Q = v × π d2/4
v = 4Q / d2 π
We then estimate the Reynolds number to be
Re = 4vQ / πdη
= 4 × 103 kg m-3 × Q / (3.14 × 1.25 × 10-2 m × 10-3 Pa S)
= 1.019 × 108 m-3 SQ
Since initially (a)
Q = 0.48 L/min = 8cm3/s
= 8 × 10-6 m3 s-1, we obtain,
Re = 815
Since this is below 1000, the flow is steady.
After some time

b) When Q = 3L/ min = 50 cm3, s = 5 × 10-5 m3 s-1 we obtain,
Re = 5095
The flow will be turbulent. You may carry out an experiment in your washbasin to determine the ransition from laminar to turbulent flow.

AP Inter 1st Year Physics Study Material Chapter 11 Mechanical Properties of Fluids

Question 12.
The lower end of a capillary tube of diameter 2.00 mm is dipped 8.00 cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water ? The surface tension of water at temperature of the experiments is 7.30 × 10-2 Nm1. 1 atmospheric pressure = 1.01 × 105 Pa, density of water = 1000 kg/m3, g = 9.8 × ms-2. Also calculate the excess pressure.
Answer:
The excess pressure in a bubble of gas in a liquid is given by 2S / r, where S is the surface tension of the liquid gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is 4S / r.) The radius of the bubble is r. Now the pressure outside the bubble, within water, P0 equals atmospheric pressure plus the pressure due to 8.00 cm of water column. That is
P0 = (1.01 × 105 Pa + 0.08 m × 1000 kg m-3 × 9.80 m s-2)
= 1.01784 × 105 Pa. .
Therefore, the pressure inside the bubble is
P1 = P0 + 2S / r (as r = 1 mm)
= 1.01784 × 105 Pa + (2 × 7.3 × 10-2 Pa m/10-3 m)
= (1.01784 + 0.00146) × 105 Pa
= 1.02 × 105 Pa
where the radius of the bubble is taken to be equal to the radius of the capillary tube, since the bubble is hemispherical ! (The answer has been rounded off to three significant figures). The excess pressure in the bubble is 146 Pa from (1.0178 + 0.00146) × 105 Pa.