AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Chapter 8 అవకలన సమీకరణాలు Exercise 8(a) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Exercise 8(a)

అభ్యాసం – 8(ఎ)

I.

ప్రశ్న 1.
xy = c ex + b e-x + x2 నుంచి యాదృచ్ఛిక స్థిర సంఖ్యలు b, c లను తొలగిస్తే వచ్చే అవకలన సమీకరణం పరిమాణం కనుక్కోండి.
సాధన:
పరిమాణము = 2

ప్రశ్న 2.
మూలబిందువు కేంద్రంగా గల వృత్తాల కుటుంబపు అవకలన సమీకరణం పరిమాణం కనుక్కోండి. (Mar. ’11)
సాధన:
మూల బిందువు కేంద్రంగా గల వృత్తం x2 + y2 = r2
పరిమాణం = యాదృశ్చిక స్థిరాంకాల సంఖ్య = 1

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a)

II.

ప్రశ్న 1.
బ్రాకెట్లలో చూపిన పరామితులతో కింది ఇచ్చిన వక్రాల కుటుంబాల అవకలన సమీకరణాలను కనుక్కోండి. (T.S. Mar. ’16)
i) y = c(x – c)2; (c)
ii) xy = aex + be-x; (a, b)
iii) y = (a + bx)ekx; (a, b)
iv) y = a cos (nx + b); (a, b)
సాధన:
i) y = c(x – c)2 ——- (1)
x దృష్ట్యా అవకలనం చేయగా
(2) ను (1) తో భాగించగా
\(\frac{\mathrm{y}_1}{\mathrm{y}}\) = \(\frac{2 c(x-c)}{c(x-c)^2}\)
x – c = \(\frac{2 y}{y_1}\)
c = x – \(\frac{2 y}{y_1}\)
(1) లో ప్రతిక్షేపించగా
У = \(\left(x-\frac{2 y}{y_1}\right)\left(\frac{2 y}{y_1}\right)^2\)
= \(\frac{x y_1-2 y}{y_1} \cdot \frac{4 y^2}{y_1^2}\)
y.\(\mathrm{y}_1^3\) = 4y2 (xy1 -2y)
i.e., \(\mathrm{y}_1^3\) = 4y (xy1 – 2y)
= 4xyy1 – 8y2
\(\left(\frac{d y}{d x}\right)^3\) – 4xy \(\frac{d y}{d x}\) + 8y2 = 0

ii) xy = aex + be-x ; (a, b)
సాధన:
xy = aex + be-x
x దృష్ట్యా అవకలనం చేయగా
x. y1 + y = aex + b. e-x
x దృష్ట్యా మరల అవకలనం చేయగా
xy2 + y1 + y1 = aex + be-x
= xy
x\(\frac{d^2 y}{d x^2}\) + 2\(\frac{d y}{d x}\) – xy = 0

iii) y = (a + bx)ekx; (a, b)
సాధన:
y = (a + bx)ekx
x దృష్ట్యా అవకలనం చేయగా
y1 = (a + bx) ekx. k + ekx. b
= k. y + b.ekx
y1 – ky = b.ekx —– (1)
x దృష్ట్యా మరల అవకలనం చేయగా
y2 – ky1 = kb ekx
= k(y1 – ky) —– (2)
= ky1 – k2y
\(\frac{d^2 y}{d x^2}\) – 2k\(\frac{d y}{d x}\) + k2y = 0

iv) y = a cos (nx + b) ; (a, b)
సాధన:
y = a cos (nx + b)
y1 = -a sin (nx + b) n
y2 = – an. cos (nx + b) n
= – n2. y
\(\frac{d^2 y}{d x^2}\) + n2.y = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a)

ప్రశ్న 2.
కింది వక్రాల కుటుంబాలకి అనుగుణంగా ఉండే అవకలన సమీకరణాలను కనుక్కోండి.

i) నిరూపకాక్షాలు అనంత స్పర్శరేఖలుగా ఉన్న లంబ అతిపరావలయాలు.
సాధన:
లంబ అతి పరావలయ సమీకరణము
xy = c2, c యాదృశ్చిక స్థిరాంకం
x దృష్ట్యా అవకలనం చేయగా
x\(\frac{d y}{d x}\) + y = 0

ii) మూల బిందువు వద్ద కేంద్రం ఉండి నిరూపకాక్షాలు అక్షాలుగా ఉన్న దీర్ఘ వృత్తాలు.
సాధన:
దీర్ఘవృత్త సమీకరణము
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a) 1
(ii) ను X తో గుణించి (i) నుండి తీసివేయగా
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a) 2

III.

ప్రశ్న 1.
బ్రాకెట్లలో చూపిన పరామితులతో కింద ఇచ్చిన వక్రాల కుటుంబాల అవకలన సమీకరణాలను కనుక్కోండి.

i) ae3x + be4x; (a, b)
సాధన:
x దృష్ట్యా అవకలనం చేయగా
y1 = 3ae3x + 4be3x
y1 – 3a. e3x = 4b.e3x
= 4(y – a. e3x)
= 4y – 4a. e3x
y1 – 4y = -a.e3x —- (1)
x దృష్ట్యా మరల అవకలనం చేయగా
y2 – 4y1 = -3a. e3x
= 3 (y1 -4y) by (1)
= 3y1 – 12y
\(\frac{d^2 y}{d x^2}\) – 7\(\frac{d y}{d x}\) + 12y = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a)

ii) y = ax2 + bx; (a, b)
సాధన:
\(\frac{d y}{d x}\) = 2ax + b
\(\frac{d^2 y}{d x^2}\) = 2a
x2 \(\frac{d^2 y}{d x^2}\) = 2ax2 —- (i)
– 2x\(\frac{d y}{d x}\) = -4ax2 – 2bx —— (ii)
2y = 2ax2 + 2bx —— (iii)
ఈ మూడు సమీకరణాలు కూడగా
x2 \(\frac{d^2 y}{d x^2}\) – 2x \(\frac{d y}{d x}\) + 2y = 0

iii) ax2 + by2 = 1; (a, b)
సాధన:
ax2 + by2 = 1
by2 = 1 – ax2 — (1)
x దృష్ట్యా అవకలనం చేయగా
2by. y1 = – 2ax —- (2)
(2) ను (1) తో భాగించగా
\(\frac{\text { by. } y_1}{\text { by }^2}\) = \(\frac{-a x}{1-a x^2}\)
\(\frac{1-a x^2}{a x}\) = \(\frac{-y}{y_1}\)
y1 – ax2y1 = -axy1
y1 = ax2y1 – axy1
= ax(xy1 – y)
a = \(\frac{y_1}{\left(x^2 y_1-x y\right)}\)
x దృష్ట్యా అవకలన సమీకరణాలు.
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a) 3

iv) xy = ax2 + \(\frac{\mathbf{b}}{\mathbf{x}}\) ; (a, b)
సాధన:
xy = ax2 + \(\frac{b}{x}\)
x2y = ax3 + b
x దృష్ట్యా అవకలనం చేయగా
x2y1 + 2xy = 3ax2
x తో భాగించగా
xy1 + 2y = 3ax —— (1)
x దృష్ట్యా మరల అవకలనం చేయగా
xy2 + y1 + 2y1 = 3a
xy2 + 3y1 = 3a —- (2)
(1) ను (2) తో భాగించగా
\(\frac{x y_1+2 y}{x y_2+3 y_1}\) = \(\frac{3 a x}{3 a}\) = x
అడ్డగుణకారము చేయగా
xy1 + 2y = x2 y2 + 3xy
x2y2 + 2xy1 – 2y = 0
x2\(\frac{d^2 y}{d x^2}\) – 2x\(\frac{d y}{d x}\) – 2y = 0

ప్రశ్న 2.
కింది వక్రాల కుటుంబాలకు అనుగుణంగా ఉండే అవకలన సమీకరణాలను కనుక్కోండి.

i) మూల బిందువు వద్ద Y – అక్షాన్ని స్పృశించే వృత్తాలు.
సాధన:
కావలసిన వృత్త సమీకరణం
x2 + y2 + 2gx = 0
x2 + y2 = -2gx —- (1)
x దృష్ట్యా అవకలనం చేయగా
2x + 2yy1 = – 2g. —- (2)
(1) తో ప్రతిక్షేపించగా
(2) నుండి x2 + y2 = x(2x + 2yy1)
= 2x2 + 2xyy1
yy2 – 2xyy1 – 2x2 = 0
y2 – x2 = 2xy \(\frac{d y}{d x}\)

ii) ప్రతిదాని అక్షం x – అక్షానికి సమాంతరంగానూ, నాభి లంబం 4a గానూ ఉన్న పరావలయాలు.
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a) 5
కావలసిన పరావలయ సమీకరణం
(y – k)2 = 4a(x – h) —- (1)
x దృష్ట్యా అవకలనం చేయగా
2 (y – k) y1 = 4a — (2)
x దృష్ట్యా అవకలనం చేయగా
(y – k) y2 + \(\mathrm{y}_1^2\) = 0 — (3)
(2) నుండి y – k = \(\frac{2 a}{y_1}\)
(3) లో ప్రతిక్షేపిస్తే
\(\frac{2 a}{y_1}\)· y2 + \(\mathrm{y}_1^2\) = 0
2ay2 + \(y_1^3\) = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a)

iii) మూల బిందువు వద్ద నాభి, X – అక్షం గుండా అక్షం గల పరావలయాల కుటుంబం.
సాధన:
పరావలయ సమీకరణం y2 = 4a (x + a)
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a) 6
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(a) 7

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 \(\frac{1}{3}\)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. \(\frac{1}{3}\)rd are elected by the members of the State Assembly.
  2. \(\frac{1}{3}\)rd are elected by members of local bodies.
  3. \(\frac{1}{12}\)th are elected by teachers.
  4. \(\frac{1}{12}\)th are elected by graduates.
  5. The remaining \(\frac{1}{6}\)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 \(\frac{1}{3}\)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 \(\frac{1}{3}\)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. \(\frac{1}{3}\)rd members are elected by the Legislative Assembly.
  2. \(\frac{1}{3}\)rd members are elected by members of local bodies.
  3. \(\frac{1}{12}\)th members are elected by teachers.
  4. \(\frac{1}{12}\)th members are elected by graduates.
  5. The remaining \(\frac{1}{6}\)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.

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Differential Equations Solutions Exercise 8(c)

I.

Question 1.
Express x dy – ydx = \(\sqrt{x^2+y^2}\) dx in the form F (\(\frac{y}{x}\)) = (\(\frac{dy}{dx}\)).
Solution:
x. dy – y dx = \(\sqrt{x^2+y^2}\) dx
\(\frac{dy}{dx}\) – y = \(\sqrt{x^2+y^2}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 1

Question 2.
Express (x – y Tan-1 \(\frac{y}{x}\))dx + x Tan-1\(\frac{y}{x}\) dy = 0 in the form F(\(\frac{y}{x}\)) = \(\frac{dy}{dx}\).
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 2

Question 3.
Express x\(\frac{dy}{dx}\) = y(log y – log x + 1) in the from F (\(\frac{y}{x}\)) = \(\frac{dy}{dx}\)
Solution:
x . \(\frac{dy}{dx}\) = y(log y – log x + 1)
\(\frac{dy}{dx}\) = \(\frac{y}{x}\)(log\(\frac{y}{x}\) + 1)

II. Solve the following differential equations.

Question 1.
\(\frac{dy}{dx}=\frac{x-y}{x+y}\)
Solution:
\(\frac{dy}{dx}=\frac{x-y}{x+y}\)
Put y = vx
\(\frac{dy}{dx}\) = v + x\(\frac{dv}{dx}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 3

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 2.
(x² + y²) dy = 2xy dx
Solution:
(x² + y²) dy = 2xy dx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 4
1 + v² = A(1 – v²) + Bv(1 – v) + Cv(1 + v)
v = 0 ⇒ = A
v = 1 ⇒ 1 + 1 = C(2) ⇒ c = 1
v = -1 ⇒ 1 + 1 = B(-1) (2) ⇒ 2 = -2B
B = -1
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 5

Question 3.
\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)
Solution:
\(\frac{dy}{dx}=\frac{-(x^2+3y^2)}{3x^2+y^2}\)
Put y = vx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 6
Multiplying with (v + 1)³
3 + v² = A(v + 1)² + B(v + 1) + C
v = – 1 ⇒ 3 + 1 = C ⇒ C = 4
Equating the co-efficients of v²
A = 1
Equating the co-efficients of v
0 = 2A + B
B = -2A = -2
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 7

Question 4.
y²dx + (x² – xy)dy = 0
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 8
v – log v = log x + log k
v = log v + log x + log k
= log k (vx)
\(\frac{y}{x}\) = log ky
Solution is ky = ey/x

Question 5.
\(\frac{dy}{dx}=\frac{(x+y)^2}{2x^2}\)
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 9

Question 6.
(x² – y²)dx – xy dy = 0
Solution:
(x² – y²)dx – xy dy = 0
(x² – y²)dx = xy . dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 10
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 11

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 7.
(x²y – 2xy²)dx = (x³ – 3x²y)dy
Solution:
(x²y – 2xy²)dx = (x³ – 3x²y)dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 12
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 13

Question 8.
y²dx + (x² – xy + y²) dy = 0
Solution:
y²dx = – (x² – xy + y²)dy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 14
1 – v + v² = A(1 + v²) + (Bv + C)v
v  =  0 ⇒ 1 = A
Equating the co-efficients of v²
1 = A + .B ⇒ B = 0
Equating the co-efficients of v
-1 = C
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 15

Question 9.
(y² – 2xy)dx + (2xy – x²)dy = 0
Solution:
(y² – 2xy)dx + (2xy – x²)dy = 0
(2xy – x²)dy = – (y² – 2xy)dx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 16
2v – 1 = A(1 – v) + Bv
v = 0 ⇒ -1 = A ⇒ A = -1
v = 1 ⇒ 1 = B ⇒ B = 1
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 17

Question 10.
\(\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}\)
Solution:
\(\frac{dy}{dx}+\frac{y}{x}=\frac{y^2}{x^2}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 18
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 19

Question 11.
x dy – y dx = \(\sqrt{x^2+y^2}\)
Solution:
x dy – y dx = \(\sqrt{x^2+y^2}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 20

Question 12.
(2x – y)dy = (2y – x)dx
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 21
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 22
⇒ (y – x)² = c²(y + x)²(y² – x²)
⇒ y – x = c²(y + x)³
⇒ (x + y)³ = c(x – y) where c = \(\frac{-1}{c^2}\)
∴ (x + y)³ = c(x – y)

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 13.
(x² – y²)\(\frac{dy}{dx}\) = xy
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 23
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 24

Question 14.
2\(\frac{dy}{dx}=\frac{y}{x}+\frac{y^2}{x^2}\)
Solution:
Put y = vx
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 25
(y – x)² = y²c²x
Solution is y²x = c(x – y)²

III.

Question 1.
Solve : (1 + ex/y)dx + ex/y(1 – \(\frac{x}{y}\))dy = 0.
Solution:
Put x = vy
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 26

Question 2.
Solve : x sin \(\frac{y}{x}.\frac{dy}{dx}\) = y sin \(\frac{y}{x}\) – x
Solution:
Dividing with x, we have
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 27

Question 3.
Solve : x dy = (y + x cos² \(\frac{y}{x}\))dx
Solution:
x.\(\frac{dy}{dx}\) = y + xcos² \(\frac{y}{x}\)
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 28

Question 4.
Solve : (x – y log y + y log x)dx + x(log y – log x)dy = 0
Solution:
Dividing with x. dx we get
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 29
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 30

Question 5.
Solve : (y dx + x dy) x cos \(\frac{y}{x}\) = (x dy – y dx) y sin \(\frac{y}{x}\).
Solution:
The given equation can be written as
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 31
∴ This is a homogeneous equation
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 32
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 33

Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c)

Question 6.
Find the equation of a curve whose gradient is \(\frac{dy}{dx}=\frac{y}{x}-\cos^2\frac{y}{x}\), where x > 0, y > 0 and which passes through the point (1, \(\frac{\pi}{4}\)).
Solution:
Inter 2nd Year Maths 2B Differential Equations Solutions Ex 8(c) 34
tan v = – log |x| + c
This curve passes through (1, \(\frac{\pi}{4}\))
tan \(\frac{\pi}{4}\) = c – log 1
c = 1
∴ Equation of the curve is
tan v = 1 – log |x|
tan \(\frac{y}{x}\) = 1 – log |x|

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Integration Solutions Exercise 6(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Integration Solutions Exercise 6(c)

I. Evaluate the following integrals.

Question 1.
∫x sec² x dx on I ⊂ R\{\(\frac{(2n+1)\pi}{2}\) : n is integer}.
Solution:
∫uvdx = u∫vdx – ∫[\(\frac{d}{du}\)(U).∫vdx]dx
Let v = sec² x and u = x then
∫x sec² x dx = x(tan x) – ∫tan x dx
= x tan x – log|sec x| + C

Question 2.
∫ex(tan-1 x + \(\frac{1}{1+x^2}\))dx, x ∈ R.
Solution:
∫ex[f(x) + f'(x)] dx = ex. f(x) + C
Let f(x) = tan-1 x so that f'(x) = \(\frac{1}{1+x^2}\)
∴ ∫ex(tan-1 x + \(\frac{1}{1+x^2}\))dx = extan-1 x + C

Question 3.
∫\(\frac{log x}{x^2}\) dx on (0, ∞).
Solution:
∫\(\frac{log x}{x^2}\) dx = (log x)(-\(\frac{1}{x}\)) + ∫\(\frac{1}{x}\) . \(\frac{1}{x}\)dx
= – \(\frac{1}{x}\)log x – \(\frac{1}{x}\) + C

Question 4.
∫(log x)² dx (0, ∞).
Solution:
∫(log x)² dx = (log x)² x – ∫x . 2log x.\(\frac{1}{x}\) dx
= x (log x)² – 2 ∫log x dx
= x (log x)² – 2(x . log x – ∫x\(\frac{1}{x}\)dx)
= x(log x)² – 2x. log x + 2x + c

Question 5.
∫ex(sec x + sec x . tan x)dx on I ⊂ R\{(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
∫ex(sec x + sec x . tan x)dx = ex.sec x + C
[∫ex[f(x) + f'(x)]dx = ex f(x) + C]

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 6.
∫ex cos x dx on R.
Solution:
I = ∫ex cos x dx = ex sin x – ∫sin x . ex dx
= ex. sin x + ex. cos x – ∫ex . cos x dx
= ex(sin x + cos x) – I
2I = ex(sin x + cos x)
I = \(\frac{e^x}{2}\)(sin x + cos x) + C

Question 7.
∫ex(sin x + cos x) dx on R.
Solution:
∫ex(sin x + cos x) dx
f(x) = sin x ⇒ f'(x) = cos x
∴ ∫ex(sin x + cos x) dx = ex. sin x + C

Question 8.
∫(tan x + log sec x)ex dx on ((2n – \(\frac{1}{2}\))π, (2n + \(\frac{1}{2}\))π) n ∈ Z.
Solution:
t = log|sec| ⇒ dt = \(\frac{1}{\sec x}\). sec x .tan x dx
= tan x dx
∫(tan x + log sec x)ex dx = ex.log|sec x| + C

II. Evaluate the following integrals.

Question 1.
∫xnlog x dx on (0, ∞), n is a real nember and n ≠ -1.
Solution:
∫xnlog x dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 1

Question 2.
∫log (1 + x²) dx on R.
Solution:
∫log (1 + x²) dx on
= [log (1 + x²). x – ∫x\(\frac{1}{1+x^2}\)2x dx
= x log (1 + x²) – 2∫\(\frac{1+x^2-1}{1+x^2}\)dx
= x log (1 + x²) – 2∫dx + 2∫\(\frac{dx}{1+x^2}\)
= x log (1 + x²) – 2x + 2 tan-1 x + C

Question 3.
∫\(\sqrt{x}\) log x dx on (0, ∞).
Solution:
∫\(\sqrt{x}\) log x dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 2

Question 4.
∫e√x dx on (0, ∞).
Solution:
t = √x ⇒ x = t²
dx = 2t dt
∫e√x dx = 2∫t et dt
= 2 [t et – ∫et dt
= 2 (t et – et) + C
= 2√x e√x – 2e√x + C

Question 5.
∫x² cos x dx on R.
Solution:
∫x² cosx dx = x²(sin x) – ∫sin x(2x dx)
= x² sin x + 2∫x(-sin x)dx
= x². sin x + 2[x cos x – ∫cos x dx]
= x² sin x + 2x cos x – 2 sin + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 6.
∫x sin² x dx on R.
Solution:
∫x sin² x dx = \(\frac{1}{2}\)∫x(1 – cos 2x) dx
= \(\frac{1}{2}\)[∫x dx – ∫x cos 2x dx]
= \(\frac{1}{2}\)[\(\frac{x^2}{2}\) – {x . \(\frac{\sin 2x}{2}\) – \(\frac{1}{2}\)∫sin 2x dx}]
= \(\frac{x^2}{4}\) – \(\frac{x}{4}\)sin 2x + \(\frac{1}{4}\)∫sin 2x dx
= \(\frac{x^2}{4}\) – \(\frac{x}{4}\)sin 2x – \(\frac{1}{8}\)cos 2x + C

Question 7.
∫x cos² x dx on R.
Solution:
∫x cos² x dx = \(\frac{1}{2}\)∫x(1+ cos 2x)dx
= \(\frac{1}{2}\)[∫x dx + ∫x cos 2x dx]
= \(\frac{1}{2}\)[\(\frac{x^2}{2}\) + {x.\(\frac{\sin 2x}{2}\) – \(\frac{1}{2}\)∫sin 2x dx}]
= \(\frac{x^2}{4}\) + \(\frac{x}{4}\) sin 2x – \(\frac{1}{4}\) ∫sin 2x dx
= \(\frac{x^2}{4}\) + \(\frac{x}{4}\) sin 2x + \(\frac{1}{8}\)cos 2x + C

Question 8.
∫cos √x dx on R.
Solution:
t = x = t² ⇒ dx = 2t dt
I = 2∫t. cos t dt = 2 (t sin t – ∫sin t dt)
= 2(t sin t + cos t) + C
= 2√x sin √x + 2 cos√x + C

Question 9.
∫x sec² 2x dx on I ⊂ R\ {(2nπ + 1)\(\frac{\pi}{4}\) : n ∈ Z}.
Solution:
∫x sec² 2x dx = x\(\frac{\tan 2x}{2}\) – \(\frac{1}{2}\)∫tan 2x dx
= x\(\frac{\tan 2x}{2}\) – \(\frac{1}{2}\) . \(\frac{1}{2}\) log|sec 2x| + C
= x\(\frac{\tan 2x}{2}\) – \(\frac{1}{4}\) log|sec 2x| + C

Question 10.
∫x cot² x dx on I ⊂ R\{nπ : n ∈ Z}.
Solution:
∫x cot² x dx
= ∫x (cosec² x – 1)dx
= ∫x cosec²x dx – ∫x dx
= x (- cot x) + ∫cot x dx – \(\frac{x^2}{2}\)
= – x cot x + log|sin x| – \(\frac{x^2}{2}\) + C

Question 11.
∫ex(tan x + sec² x)dx on I ⊂ R\{(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z}.
Solution:
f(x) = tan x ⇒ f'(x) = sec² x dx
I = ∫ex[f(x) + f'(x)] dx = ex. f(x) + C
= ex. tan x + C

Question 12.
∫ex(\(\frac{1+log x}{x}\))dx on (0, ∞).
Solution:
∫ex(\(\frac{1+log x}{x}\))dx = ∫ex(log x + \(\frac{1}{x}\))dx
= ex.log x + C

Question 13.
∫eaxsin bx dx on R, a, b ∈ R.
Solution:
Let I = ∫eaxsin bx dx …………. (1)
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 3

Question 14.
∫\(\frac{x.e^2}{(x+1)^2}\) dx on I ⊂ R\{-1}
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 4

Question 15.
∫\(\frac{dx}{(x^2+a^2)^2}\), (a > 0) on R.
Solution:
Put x = tan t
Then dx = a sec² t dt
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 5

Question 16.
∫ex log (e2x + 5ex + 6) dx on R.
Solution:
∫ex log (e2x + 5ex + 6) dx
∵ e2x + 5ex + 6 = (ex + 2)(ex + 3)
= ∫ex. log((ex + 2)(ex + 3))dx
= ∫ex.{log(ex + 2) + log(ex + 3)}dx
= ∫exlog (ex + 2)dx + ∫exlog (ex + 3)dx
Put ex = t ⇒ exdx = dt
= ∫log(t + 2)dt + ∫log(t + 3) dt
= log(t + 2)∫1dt – ∫{\(\frac{d}{dt}\)log(t + 2). ∫1dt}dt + log(t + 3)∫1dt – ∫{\(\frac{d}{dt}\)log(t + 3). ∫1dt}dt
= t log(t + 2) – ∫\(\frac{1}{t + 2}\))t dt + t log(t+ 3) – ∫\(\frac{1}{t + 3}\)). t dt
= t {log(t + 2) + log (t + 3)} – ∫\(\frac{t}{t + 2}\)dt – ∫\(\frac{t}{t + 3}\)dt
= t log (t² + 5t + 6) – ∫(\(\frac{t+2-2}{t + 2}\))dt – ∫(\(\frac{t+3-3}{t + 3}\))dt
= t. log (t² + 5t + 6) – ∫{1 – \(\frac{2}{t + 2}\)}dt – ∫{1 – \(\frac{3}{t + 3}\)}dt
= t log (t² + 5t + 6) – t + 2 log|t + 2| – t + 3 log|t + 3| + C
= t log (t² + 5t + 6) – 2t + 2 log|t + 2| + 3log|t + 3| + C
= ex log(e2x + 5ex + 6) – 2ex + 2log ex + 2 + 3 log ex + 3 + C

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 17.
∫ex.\(\frac{x+2}{(x+3)^2}\) dx on I ⊂ R\ {-3}
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 6

Question 18.
∫cos (log x) dx on (0, ∞).
Solution:
Put log x = t
x = et
dx = et . dt
I = ∫et . cos t . dt
= et sin t – ∫sin.et dt
= et . sin t + cos t . et – ∫et . cos t dt
2I = et . (sin t + cos t)
I = \(\frac{e^t}{2}\)(sin t + cos t)
= \(\frac{x}{2}\)[sin(log x) + cos(log x)] + C

III. Evaluate the following integrals.

Question 1.
∫x tan-1 x dx, x ∈ R
Solution:
∫x tan-1 x dx = (tan-1 x)\(\frac{x^2}{2}\) – \(\frac{1}{2}\)∫x².\(\frac{1}{1+x^2}\)dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 7

Question 2.
∫x² tan-1 x dx, x ∈ R.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 8

Question 3.
∫\(\frac{tan^{-1}x}{x^2}\) dx, x ∈ I ⊂ R\{0}
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 9

Question 4.
∫x cos-1x dx, x ∈ (-1, 1).
Solution:
∫x cos-1x dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 10
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 11

Question 5.
∫x² sin-1x dx, x ∈ (-1, 1).
Solution:
∫x² sin-1x dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 12

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 6.
∫x log(1 + x) dx, x ∈ (-1, ∞).
Solution:
∫x log(1 + x) dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 13
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 14

Question 7.
∫sin √x dx, on (0, ∞).
Solution:
x = t² ⇒ dx = 2t dt
∫sin √x dx = 2∫t. sin t dt
= 2(t(-cos t) + ∫cos t dt)
= -2t cos t + 2 sin t
= -2√x cos √x + 2 sin √x + C

Question 8.
∫eax sin (bx + c)dx, (a, b, c ∈ R, b ≠ 0) on R.
Solution:
Let I = ∫eax sin (bx + c)dx
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 15
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 16

Question 9.
∫ax cos 2x dx on R (a > 0 and a ≠ 0).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 17

Question 10.
∫tan-1(\(\frac{3x-x^3}{1-3x^2}\)) dx on I ⊂ R\{-\(\frac{1}{\sqrt{3}}\), \(\frac{1}{\sqrt{3}}\)}.
Solution:
Put x = tan t ⇒ dx = sec² t dt
Then
∫tan-1(\(\frac{3x-x^3}{1-3x^2}\)) dx
= ∫tan-1(\(\frac{3 \tan t -\tan^3 t}{1-3\tan^2 t}\)) dx
= ∫tan-1(tan 3t).sec² t dt
= 3∫t sec² t dt
= 3[t∫sec² t dt – ∫{\(\frac{d}{dt}\)(t) ∫sec² t dt}dt]
= 3[t(tan t) – ∫(1) tan t dt]
= 3(x tan-1 x – log\(\sqrt{1+x^2}\)) + C
= 3x[tan-1 x – \(\frac{3}{2}\)log (1 + x²) + C
= 3x tan-1(x) – \(\frac{3}{2}\)log (1 + x²) + C

Question 11.
∫sinh-1 x dx on R.
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 18

Question 12.
∫cosh-1 x dx on (1, ∞).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 19
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 20

Inter 2nd Year Maths 2B Integration Solutions Ex 6(c)

Question 13.
∫tanh-1x dx on (-1, 1).
Solution:
Inter 2nd Year Maths 2B Integration Solutions Ex 6(c) 21

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d)

I.

Question 1.
Find the area of the region enclosed by the given curves.
i) y = cos x, y = 1 – \(\frac{2x}{\pi}\)
Solution:
Equations of the given curves are
y = cos x ………….. (1)
y = 1 – \(\frac{2x}{\pi}\) ………….. (2)
Eliminating y from eq’s (1) and (2)
cos x = 1 – \(\frac{2x}{\pi}\)
When x = \(\frac{\pi}{2}\), cos x = cos\(\frac{\pi}{2}\) = 0
1 – \(\frac{2}{\pi}\), x = \(\frac{2}{\pi}\) . \(\frac{\pi}{2}\) = 1 – 1 = 0
When x = 0, cos x = cos 0 = 1
1 – \(\frac{2x}{\pi}\) = 1 – 0 = 1
∴ Point of intersection are A = (\(\frac{\pi}{2}\), 0) B = [π – 1]
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 1

Question 2.
y =cos x, y = sin 2x, x = 0, x = \(\frac{\pi}{2}\).
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 2
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 3

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

Question 3.
y = x³ + 3, y = 0, x = -1, x = 2.
Solution:
Required area PABQ
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 4

Question 4.
y = ex, y = x, x = 0, x = 1.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 5

Question 5.
y = sin x, y = cos x; x = 0, x = \(\frac{\pi}{2}\).
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 6
Between 0 and \(\frac{\pi}{4}\).
cos x > sin x
Between \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\)
cos x < sin x
Required area
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 7

Question 6.
x = 4 – y², x = 0.
Solution:
The given parabola x = 4 – y² meets, the x – axis at A(4, 0) and Y – axis at P(0, 2) and Q(0, -2).

The parabola is symmetrical about X – axis
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 8
Required area = 2 Area of OAP
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 9

Question 7.
Find the area enclosed with in the curve |x| + |y| = 1
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 10

II.

Question 1.
x = 2 – 5y – 3y², x = 0.
Solution:
Solving the equation of given curves
2 – 5y – 3y² = 0
3y² + 5y – 2 = 0
(y + 2)(3y – 1) = 0 y = -2 or \(\frac{1}{3}\)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 11

Question 2.
x² = 4y, x = 2, y = 0.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 12

Question 3.
y² = 3x, x = 3.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 13
The parabola is symmetrical about X – axis
Required area = 2\(\int_0^3\)√3. √x dx
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 14

Question 4.
y = x², y = 2x.
Solution:
Given equation are y = x² ………….. (1)
y = 2x …………. (2)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 15
Eliminating y, we get x² = 2x
x² – 2x = 0
x(x – 2) = 0
x = 0 or x = 2
y = 0, or y = 4

Point of intersection are O(0, 0), A(2, 4)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 16

Question 5.
y = sin 2x, y = √3 sin x, x = 0, x = \(\frac{\pi}{6}\).
Solution:
Given equation are y = sin 2x ………… (1)
y = √3 sin x …………. (2)
sin 2x = √3 sin x
2 sin x. cos x = √3 sin x
sin x = 0 or 2 cos x = \(\frac{\sqrt{3}}{2}\)
x = 0, cos x = \(\frac{\sqrt{3}}{2}\) ⇒ x = \(\frac{\pi}{6}\)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 17
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 18

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

Question 6.
y = x², y = x³.
Solution:
Given equations are y = x² ………….. (1)
y = x³ ………. (2)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 19
From equation (1) and (2) x² = x³
x³ – x² = 0
x²(x – 1) = 0
x = 0 or 1
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 20

Question 7.
y = 4x – x², y = 5 – 2x
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 21
y = 4x – x² ………… (i)
y = 5 – 2x …………..(ii)
y = -([x – 2]²) = 4
y – 4 = -(x – 2)²
Solving equations (i) and (ii) we get
4x – x² = 5 – 2x
x² – 6x + 5 = 0
(x – 5)(x – 1 ) = 0
x = 1, 5
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 22

Question 8.
Find the area in Sq. units bounded by the x – axis, part of the curve y = 1 + \(\frac{8}{x^2}\) and the ordinates x = 2 and x = 4.
Solution:
Given equations y = 1 + \(\frac{8}{x^2}\)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 23

Question 9.
Find the area of the region bounded by the parabolas y² = 4x and x² = 4y.
Solution:
Equations of the given curve are
y² = 4x
x² = 4y
(\(\frac{x^2}{4}\))² = 4x
\(\frac{x^4}{16}\) = 4x
x4 = 64x ⇒ x4 = 0 or x³ = 64, x = 4
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 24

Question 10.
Find the area bounded by the curve y = lnx the X – axis and the straight line x = e.
Solution:
Equation of the curve is y = lnx
x = 1 ⇒ y = 0
The curve y = lnx meets
X – axis at C(1, 0)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 25
Required area = \(\int_1^e\)lnx dx
= (x.lnx)e1 – \(\int_1^e\)x.\(\frac{1}{x}\) dx
= (e.ln e – 1.ln 1) – (x)e1
= e – (e – 1)
= e – e + 1 = 1 Sq.unit.

III.

Question 1.
y = x² + 1, y = 2x – 2, x = -1, x = 2.
Solution:
Equation of the curve are
y = x² + 1 …………. (1)
y = 2x – 2 ………….. (2)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 26
Area between the given curves
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 27

Question 2.
y² = 4x, y² = 4(4 – x).
Solution:
Equations of the curve are y² = 4x ………… (1)
y² = 4(4 – x) …………. (2)
Eliminating y, we get
4x = 4(4 – x)
2x = 4 ⇒ x = 2
Substituting in equation (1), y² = 8
y = ± 2√2
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 28
Points of intersection are
A(2, 2√2), B(2, -2√2)
Required area is symmetrical about X – axis
Area OACB
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 29

Question 3.
y = 2 – x², y = x².
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 30
y = 2 – x² …………. (1)
y = x² …………. (2)
x² = -(y – 2)
From equation (2)
2 – x² = x²
2 = 2x² or x² = 1
x = ±1
Area bounded by two curve be
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 31

Question 4.
Show that the area enclosed between the curve y² = 12(x + 3) and y² = 20(5 – x) is 64\(\sqrt{\frac{5}{3}}\).
Solution:
Equation of the curve are
y² = 12(x + 3) ……….. (1)
y² = 20(5 – x) ……….. (2)
Eliminating y
12(x + 3) = 20(5 – x)
3x + 9 = 25 – 5x
8x = 16
x = 2
y² = 12(2 + 3) = 60
y = √60 = ±2√15
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 32
Points of intersection are B’ (2, 2√15)
B’ (+2, -2√15)
The required area is symmetrical about X – axis
Area ABCB’
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 33

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

Question 5.
Find the area of the region {(x, y)/x² – x – 1 ≤ y ≤ -1}
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 34

Question 6.
The circle x² + y² = 8 is divided into two parts by the parabola 2y = x². Find the area of both the parts.
Solution:
Equations of the curves are
x² + y² = 8 ………… (1)
2y = x² ………… (2)
Eliminating y between equations (1) and (2)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 35
Let x² = t
4t + t² = 32
t² + 4t – 32 = 0
(t + 8)(t – 4) = 0
t = -8 (not possible) x² = 4 ⇒ x = ±2
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 36
As curve is symmetric about Y – axis, total area be
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 37

Question 7.
Show that the area of the region bounded \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 (ellipse) is π ab. also deduce the area of the cricle x² + y² = a².
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 38
The ellipse is symmetrical about X and Y axis
Area of the ellipse = 4 Area of CAB
= 4.\(\frac{\pi}{4}\) ab
Equation of elliple = \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 39
Substituting b = a, we get the circle
x² + y² = a²
Area of the circle = πa(a) = πa² sq. units.

Question 8.
Find the area of region enclosed by the curves y = sin πx, y = x² – x, x = 2.
Solution:
Required area
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 40
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 41

Question 9.
Let AOB be the positive quadrant of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 with OA = a, OB = b. Then show that the area bounded the chord AB and the arc AB of the elliple is \(\frac{(\pi-2) a b}{4}\).
Solution:
Let OA = a, OB = b
Equation of AB is \(\frac{x}{a}+\frac{y}{b}\) = 1
\(\frac{y}{b}\) = 1 – \(\frac{x}{a}\)
y = b(1 – \(\frac{x}{a}\))
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 42
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 43

Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)

Question 10.
Prove that curves y² = 4x and x² = 4y divide the area of the square bounded by the lines x = 0, x = 4, y = 4 and x = 0 into three equal parts.
Solution:
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 44
The given equations are y² = 4x …………. (1)
x² = 4y …………. (2)
The points of intersection are O(0, 0) A(4, 4)
Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d) 45

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Andhra Pradesh BIEAP AP Inter 2nd Year Economics Study Material 4th Lesson Agriculture Sector Textbook Questions and Answers.

AP Inter 2nd Year Economics Study Material 4th Lesson Agriculture Sector

Essay Questions

Question 1.
Explain the importance of agriculture sector in the Indian Economy.
Answer:
Agriculture plays a vital role in Indian economy and is the backbone of Indian Economy. It provides employment to around 55 percent of the total work force in the country. It provides raw materials for industries and market for industrial goods. It is the supplier of food for 121 crore people. India is the land of producing multiple goods (crops). In our country agriculture is not only an important occupation of people but also a way of life culture and customs.

Importance:
1) Share of Agriculture in the National Income: Agriculture sector, including forestry, fishing, mining, quarrying and allied activities like animal husbandry, horticulture, silk industry etc.’, is significantly contributing to the Gross Domestic Product in India.

The share of agriculture in National Income has been disclining gradually from 1950-51. This is mainly due to the development of non-agriculture sectors during the five years plans in the economy. The share of agriculture in national income declined gradually from 56.5 percent in 1950 – 51 to 13.9 percent by 2013 -14. The share of agriculture in national income in U.K. and U.S.A is 2 to 3 percent. The proportion is about 7 percent in France and 6 percent in Australia.

2) Employment Providing Sector: Agriculture dominates the economy to such an extent that a very high proportion of working population in India is engaged in agriculture.

Agriculture provided employment to 98 million peoples in 1951. Thenumber of people working on lands both the cultivators and agricultural laborers increased to’234 million in 2011.

In case of developed countries the population engaged in agriculture is very much less. Ex: In USA and UK only 2 percent, in Japan and France 4 percent.

3) Importance in International Trade: For a long time three agricultural based exports of India like cotton textiles, Jute and tea accounted more than 50 percent. If we add other agricultural commodities the share of agriculture in exports will rise to 70 to 75 percent.

At present we are exporting Sugar, Cotton, Tobacco, Rice, Cashew nuts, Spices, Oilcakes, Coffee, Tea, Fish, Meat, Fruits, Pulses etc., to foreign countries. By exporting all these products we are earning foreign exchange.

4) Effective Social Safety Net: Though agriculture presently contributes less than 15 percent of Indians GDP yet it continues to employ more than half of the work force. If crops, animal husbandry, fisheries, agro-forestry, silk industry and horticulture are developed at household level, hunger and poverty will be alleviated. Green Revolution, along with White Revolution and Blue Revolution, provides income and employment to the rural poor by eradicating poverty. Thus, agriculture acts as an effective social safety net.

5) Food Supplier of the Expanding Population: The Hunger Index shows that India’s rank is 55 out of 76 countries. According to FAO, India is a hunger affected country. Because of the heavy pressure of population the existing levels of food consumption in our country are very low. The per capita availability of food grains was 510.1 grams per day in 1991. But it has declined to 468.9 grams in 2013. Therefore, unless agriculture is able to increase continuously its marketed surplus of food grains a crisis will emerge.

6) Role of agriculture in Industrialization : Industries which depend on agriculture products for their raw material are called agro-based industry. Industries like Cotton, Jute, Textile, Sugar, Flour mills, Edible oil etc., directly depend on agriculture for raw material. Some other industries like handloom weaving, rice husking, food processing, oil crushing, horticulture etc., are depend on agriculture. The development of agriculture sector will expand .the demand for industrial goods. Thus the development of industrial sector facilities the development of agriculture sector by supplying inputs like machines, fertilizers, pesticides etc.

7) Market for Industrial Products: Two thirds of the populations of developing country like India lives in rural areas. The purchasing power of these people in rural areas is too low to purchase industrial products. If steps are taken to increase agriculture produce and productivity, the income of the rural sector will increase. As a result of increase in income of those who depend on agriculture in rural areas causes demand for industrial goods. The increased demand for industrial goods leads to the development of industrial sector.

8) Other Factors:
a) The development of agriculture sector directly promotes growth in transport sector.
b) Development of agriculture sector invites expansion of branches of banks into rural areas.
c) Agriculture development minimizes migration from rural to urban.
d) Farm tourism in rural areas can be developed through the development of agriculture.
e) Agriculture and its allied sectors play a key role in protecting biodiversity.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 2.
Explain the present conditions of agricultural labour and suggest the measures to improve the conditions of agricultural labour.
Answer:
The number of people living in agriculture sector in developing country like India is very high. Unfortunately agriculture labour belong to economically depressed and socially backward sectors of rural economy.
The people who work on agriculture sector for their survival can be treated as agricultural labour.

Conditions of Agricultural Labourers :
1) Low Social Status: Most of the agricultural labourers are belonged to the depressed classes. They have been neglected for ages. In case of such farm labourers, exploitation has become common and they have not fought for their rights.

2) Unorganised : Agricultural labourers are living in scattered villages. Moreover, they are illiterates. Hence, they cannot easily to be organized. As a results it is difficult for farm labourers to bargain with the land owners for good wages.

3) Seasonal Employment : Agricultural labourers have to face the problems of unemployment and underemployment. They are employed while sowing and harvesting season, of the year they remain unemployed.

4) Low Wages and Income : Agricultural wages and family incomes of agricultural labourer are very low. The wages of farm labourers vary from state to state. In case of attached labourers these wages are pathetic. However, as prices also increased considerably, the real wages rates did not increase much.

5) Rural Indebtedness : Because of the low level of income agricultural labour generally seek debts. In fact the debt of agricultural labourers passes from generation to generation and become bonded labour.

6) Femanisation of Agricultural labour : Female agricultural workers are generally forced to work harder. They are paid less wages comparatively to male workers.

7) High Incidence of Child labour: It is estimated that one-third of child labourers in Asia are in India. The largest number of child workers are engaged in agriculture. Wages paid to child labourers are too low which adversely affect the incomes of their households.

8) Lack of subsidaty professions : Another serious problem that is being faced by farm labourers is lack of non-agricultural occupations in the villages. As there is no work on the fields, agricultural labour has no other means to earn for subsistence. Thus, they are disguised and are burden on land.

Measures to Improves the Conditions of Agricultural Labour:
1) Minimum Wages Act: It was passed in 1948. According to this act every State Government was asked to fix minimum wages for agricultural labour within three years. According to this law while fixing the minimum wages, the total costs and standards of living in various states of the country or in different parts within the states are to be kept in view.

2) Providing Land to Landless Labourers : The Government has distributed land to landless labourers with a view to improve their economic position. The surplus land obtained by enforcing Land Ceiling Acts and those donated in Bhoodan and Gramdan were distributed among the landless labourers.

3) Provision of House sites and Houses: Mostly, agriculture labourers do not possess their own houses. Governments have taken several steps during the plans periods to provide free house live “Indira Avasa Yojana (IAY)”, “Minimum Needs Programme (MNP)” etc.

4) Organization of Labour co-operation : During the second five year plan efforts were made for the formation of labour co-operatives. They provide employment to farm labourers and also eliminate the exploitation.

5) Special schemes for Providing Employment: A number of schemes have been initiated in planning period to provide employment to agricultural workers. Some important schemes are Rural Works Programme (RWP), Crash Scheme for Rural Employment (CSRE), Employment Guarantee Scheme (EGS), Food for work Programme (FWP) etc.

6) Sanction of Loans and Subsidies: Loans should be provided to the farm labourers at low rate of interest in order to start their own business. Sometimes Governments resorts debt moratorium for immediate relief of debts of agricultural labour and marginal farmers.

7) Abloition of Bonded Labour: Both exploitation and slavery are inhuman activities and punishable offences. The Government of India passed a legislation known as the Bonded Labour System Abolition Act, 1976.

8) Development of Cottage Industries: In order to minimize the pressure of population on land and to improve the economic conditions of farm labour, the Government has been making efforts to start the cottage and smalLscale industries in the rural areas of the country.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 3.
What are the factors affecting cropping pattern in India ?. Suggest the measures to correct the cropping pattern.
Answer:
The economic development of any nation depends on the utilization pattern of natural resources like land, water, the fertility of soil, irrigation facilities etc. The cropping pattern determines the development of a country.

Factors affecting cropping pattern in India : The cropping pattern in India can be affected by various factors like physical, economical, technical and Government policies.
I. Physical Factors : Physical factors play a vital role in determining the cropping pattern. These factors are classified below.
1) Climate and Rain fall: Climatic conditions and rainfall determine cropping pattern. Some crops require cool climate while some other crops require hot climate. For instance, apples will be produced in cool climate. For instance, crops like Paddy, Sugarcane require abundance of water.

2) Nature of Soil and Fertility: Nature of soil and fertility determine the production of certain crops. For instance, wheat requires well drained silt and fertile loam soils but for cotton black soils are ideal.

3) Irrigation facility: It also determine the cropping pattern. For instance crops like Paddy, Sugarcane, Wheat etc., require assured irrigation facility. Some other crops like Jowar, Maize, Ragi etc., will grow in the areas where irrigation facilities are insufficient.

II. Economic Factors:
1) Price and Incphie Maximization: Generally, farmers try to maximize their income. Consequently, they produce those crops whose market prices are high. As a result of fixed procurement prices of wheat and rice and other Government controls, the farmers are induced to shift to cash crops like’sugarance and cotton etc., Proff. M. L. Dantwala opined that the area under commercial crops is increasing in India as the farmers fetching higher returns.

2) Farm Size: There is a relationship between farm size and cropping pattern. At first the farmers are interested in producing food grains for their requirements but large farmers are used to produce both food grains and commercial crops.

3) Available of Inputs : The availability of agricultural inputs like seeds, fertilizers, pesticides machines etc., affect the cropping pattern in our country. Infrastructural facilities such as transport, storage, marketing, water storage etc., influence the cropping pattern significantly.

4) Insurance Against Risk: Generally, the farmers resort diversified cropping pattern to minimize the risk of crop failure. If Government introduces crop insurance whichprotects the farmers against all risks in farming, the farmers will farm those insurable crops. Thus, insurance affects cropping pattern.

5) Tenancy System: Existing tenancy system in India influences the cropping pattern. Generally, the Landlord decides the cropping pattern to ensure maximum profit before he leases out his land.

6) Social Factors : Social Environment, customs, traditions etc., also influence crop pattern to some extent. These factors induce farmers to cultivate traditional crops by using traditional varieties of seeds and methods.

III. Government Policies : Policies of Government relating to different crops, exports, taxes, subsidies, supply of inputs, available of credit, fixing support prices etc., can affect the cropping pattern in a significant manner.

Measures to correct the cropping pattern: Among all economists there is a common opinion that the cropping pattern in India is not suitable to satisfy the requirements of growing population. In order to maintain an optimum cropping pattern. “The National Council of Applied Economic Research (NCAER) has made the following suggestions for better cropping pattern in India.

  1. Government should enact some legislation fixing the production of certain crops in certain suitable regions.
  2. Government should appoint officials at local level to encourage fanners to produce more of food grains to meet the requirements of the growing population.
  3. Government should encourage mechanization in agriculture sector by supplying required machines at a cheaper rate.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 4.
What are the causes for low productivity in agriculture in India ? Suggest some measures to improve it. [A.P. Mar. 17]
Answer:
Agriculture plays a predominant role in Indian economy. The productivity in Indian Agriculture is too low when we compared to the agricultural productivity of other countries in the world.

The causes for low level of agriculture productivity in India are manifold. They can be grouped into four broad categories.

  1. General causes
  2. Institutional causes
  3. Technical causes
  4. Environmental causes

1) General causes:
i) Pressure of Population on Agriculture : Pressure of population on agriculture is heavy as a result of high growth rate of population and slow growth rate of other sectors of the economy. In 2011, about 263 million workers out of 348 million rural working populations are employed in agriculture. Increasing pressure of population on agriculture results, in the subdivisions and fragmentation of holding. Consequently, the productivity in agriculture sector remains low in India.

ii) Social Environment: The social environment of villages is an obstacle in agricultural development. The farmers in rural are illiterate, superstitious, conservative and unresponsive to new agriculture techniques. The decline of joint family system and land hunger are also discouraging the rural atmosphere. Peasants are not able to take proper care of their agriculture. Unless this atmosphere is changed, it is too difficult to enhance the productivity of agriculture.

iii) Lack of Infrastructural Facilities: Infrastructural facilities like transport, storage, credit and marketing are inadequate in rural areas to the growing population, due to lack of these adequate infrastructure facilities, the agricultural productivity in rural areas is very low.

iv) Impact of the British Regime : During British rule in India, they have not shown any interest in developing agriculture sector but made our economy as colonial one. Moreover, their policies like land tenure system, collection of land cess gave a deadly blow to the Indian agriculture.

2) Institutional causes:
i) Uneconomic Land Holding: According to the National Sample Survey, 52 percent land holdings had a size of less than 2 hectares in 1961 – 62. In 2010 -11,85 percent of total land holdings are less than 2 hectares. As a result of laws of inheritance and other reasons there is a further divisions and fragmentation of land holdings. Hence, these small holdings are adversely affecting productivity of agriculture.

ii) Defects in Land Tenancy System : The Indian agriculture system was adversely affected before Independence because of defectives in Zamindari, Jagirdari, Mahalvari systems which exploited the farmers. In this system lack of certainty in rent, security of tenure and ownership right the tenants don’t show any attention to develop agriculture. Hence, India has become less productivity.

iii) Lack of credit and marketing facilities : The cultivators are not able to invest requisite sources in agriculture due to lack of marketing facilities and required credit at fair rate of interest. Even support price policy and subsidies to inputs of agriculture fired by the Government are unsatisfactory. Hence, peasants follow traditional methods which results in low productivity.

3) Technical causes:
i) Outmoded Agricultural Techniques : T. W Schultz of famous economist opined that the peasants in India are still using traditional or outmoded techniques. Indian farmers are still using wooden ploughs, bullock carts, sickles etc. Use of fertilizers and new high yielding varieties of seeds is also extremely limited. Hence, the productivity in agriculture is low.

ii) Inadequate Irrigation Facilities : Gross cropped area in India in 2010 – 11 was 198.97 million hectares but only 89.36 million hectares of land had irrigation facilities. It implies that 55 percent of the gross cropped area continues to depend on rains. Rainfall is often insufficient, uncertain and irregular. In such atmosphere it is difficult to extend the new agricultural technology all over the country.

iii) Scarcity of Agricultural Inputs : The supply of modem agricultural inputs like fertilizers, pesticides, hybrid seeds, farm machinery etc., are inadequate to meet the requirements of our country. In order to achieve high production in agriculture requisite supply of inputs is essential.

4) Environmental causes : Environment also plays a vital role in affecting the productivity of agriculture, increase in the temperature. Degradation of soil, changes in temperature, pollution of water and air etc., adversely affect the productivity of agriculture are

  1. Global warming.
  2. Soil Degradation.
  3. The intensive cultivation of high yielding variety crops.
  4. The reckless use of fertilizers.
  5. Shifting cultivation.
  6. Displacement of the traditional practices of crops.

Measures to increase agricultural productivity in India :

  1. The proportion of people depending upon agriculture must decrease. Development of non- farm activities in rural areas go a long way for decreasing the dependency load in agriculture.
  2. A favourable support price policy leads to increased yield levels in agriculture.
  3. Agriculture, like industry must be protected for favourable terms of trade.
  4. Public investment in agriculture must increase.
  5. Timely and adequate institutional credit automatically enhances the access to vital inputs in agriculture.
  6. Strict implementation of land reforms is necessary. Land reforms remove the structural dificiencies in agriculture.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 5.
Explain various sources of Irrigation and its importance.
Answer:
The sources of irrigation in India are mainly classified into four. They are :

  1. Canal Irrigation,
  2. Well Irrigation,
  3. Tank Irrigation and
  4. Other Sources.

1) Canal Irrigation : Canals are considered to be the important source of irrigation. Though construction and maintenance of canals are highly expensive, they can irrigate a wide extent of cultivated area. Canal irrigation is more prevalent in Uttar Pradesh, Punjab, Haryana, Rajasthan and West Bengal. Canals are classified into two types. They are :
A) Perennial Canals
B) Inundation Canals

2) Well Irrigation : Wells are important and dependable source of irrigation. Wells are classified into common wells and tube wells. These common wells irrigate less area comparatively with tube wells. Well irrigation is more prevalent in Uttar Pradesh, Punjab, Gujarat and Bihar. The area irrigated under tube wells is more in Uttar Pradesh.

3) Tank Irrigation : Tanks form another sources of irrigation. Tank irrigation is common in those areas where canal and well irrigation is not possible. Tank irrigation is popular in Southern States like Andhra Pradesh, Tamilnadu and Karnataka.

Importance of Irrigation:
1) Insufficient, uncertain and irregular rains : The period of rainfall is restricted to only four months during monsoons. Even during monsoons the rainfall is scanty. Sometimes monsoons are delayed while some other times they are prematured. This type of atmosphere results in drought conditions. Hence, with the help of proper development of irrigation, droughts and famines can be effectively controlled.

2) Higher productivity on irrigated land : Irrigation helps greatly in raising the productivity of land than unirrigated land. Because this enables the application of modem inputs like fertilizers, high yielding varieties of seeds etc.

3) Multiple cropping possible: India is a land of tropical and sub-tropical climate. It has potentialities to grow crops throughout the year but rainfall is restricted to less than four months. Provision of irrigation facilities can make possible the growing of multiple crops throughout the year.

4) Crucial role in new agricultural strategy: The successful implementation of high yielding varieties programmed depends on timely and adequate supply of water. These seeds require chemical fertilizers and substantial water at regular intervals of time. Therefore, irrigation facilities facilitates the expansion of new agricultural strategy to larger areas of land.

5) Bringing more land under cultivation: The total reporting area according to land utilization statistics was 305.56 million hectares in 2009-10. Thus, some portion of the waste land can be brought under cultivation if there is availability of irrigation facilities.

6) Prosperity : Irrigation helps in stabilizing the output and yield level. Irrigation plays.a protective role during drought years. Thus, irrigation facilities prevent fall in output during drought and achieve stability in income and output which in turn result prosperity.

7) Indirect benefits of Irrigation: Expansion of irrigation facilities all over the country avoid disparities in the production of food grains. Irrigation promotes the growth of agriculture and allied sectors. Thus, increased production in agriculture stabilizes the prices of agricultural products.

Question 6.
What are the causes for small size.of land holdings in India ? Mention the problems of small holdings.
Answer:
Small size land holdings adversely affect the productivity of agriculture. The following are the causes for low productivity in agriculture.
1) The law of inheritance : The system of inheritance’is causing for division and fragmentation of land holdings. According to the law of inheritance children have an equal share in father’s property. Both the sons and daughters have equal share in father’s property either in Hindu Law or in Mohammedan Law. Consequently the division and fragmentation of land holdings are taken place.

2) Pressure of population: The population of India is growing at a faster rate of 1.64 percent per annum. While land under agriculture has increased only marginally. Because of th6 unsatisfactory expansion of the non-agricultural sector and its inability to absorb the growing population led to subdivision of land holdings.

3) Decline of joint family system : All the family members live together under the joint family system and this kept the land holding intact. As a result, subdivision and fragmentation of land holdings are taken place.

4) Farmers indebtedness: Most of the farmers in India are neck – deep in indebtedness. Frequently they are forced to sell off parts of their land to pay off the debts. It is observed that many money lenders in villages adopt unfair methods of fending and trap the illiterate peasants with a view keeping m mind to grab their lands. Thus the land continues to get subdivided and fragmented.

5) Land hunger In India of rural area are fond of land because of psychological, sentimental, social and economical attachments to land. Generally people in rural areas feel having l^nd as a matter of prestige and social symbol. This sort of attachment with land is called “Land Hunger”. Because of this land hunger peasants are reluctant to leave their land even though it is uneconomical.

6) The decline of handicrafts : Decline of village handicrafts is an another important historical factor for the small sized holdings in our country. Those handicrafts had provided employment and a source of livelihood to the artisans. As a result of competition from machine made goods these artisans left their ancestral occupations and they were forced to depend on agriculture. Hence, this further increased subdivision and fragmentation.

Problems of fragmentation of Land holdings : Small and fragmented land holdings impede agricultural progress. The following are the main disadvantages of subdivision and fragmentation.
1) Wastage of land : Because of subdivision and fragmentation, the size of plots becomes so small. In Ratnagiri district of Maharashtra land holdings were often as small as 0.006 acres. It is not possible to cultivate on them. In addition to such wastage, another 3 to 5 percent of land is wasted in drawing boundaries, hedges, pathways etc.

2) Supervisory problems: In general, agricultural activities are seasonal in nature. It is too difficult to supervise simultaneously the agricultural activities on scattered land holdings in various parts of the village. Hence, the productive efficiency will decline and the productivity will also be affected adversely.

3) Difficulties in modernisation : Small size land holdings are not suitable for mechanization. The small peasants are not able to invest on costly equipments like tractors, electric motors, sprayers, drillers, harvesting machines etc. There is no possibility to implant new technology to improve agricultural productivity in small holdings.

4) Disputes over boundaries : Disputes over boundaries, hedges, pathways, theft of crops, grazing by animals belonging to other villages etc., are very common in villages. These litigations lead to court causing wastage of valuable time and money. All these incidents disturb the rural atmosphere.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 7.
Explain the advantages and disadvantages of co-operative farming.
Answer:
Meaning of Co-operative Fanning: Co-operative farming indicates that all farmers of a village from themselves voluntarily a society. After forming the society entire land holdings of the farmers will be pooled into one unit. Later they will handover their land, cattle, implements etc., to the society. The co-operative society will cultivate all these holdings as one farm.

Advantages:
1) Increase in Production: Co-operative farming will bring more land into cultivation either consolidation of land holdings or bring into use waste land. Consequently, both the production and marketable surplus will increases.

2) Large Scale Economies: Large scale economies will occur in co-operative fanning such as technical, marketing, financial economies etc., As a result, the cost of production will decrease and the returns will be maximized.

3) Land Development Activities: Land development activities, like land conservation, land reclamation, construction of water sheds etc., can easily be made by co-operative farming societies. These activities will improve the production of agriculture.

4) New Farm Technology : Co-operative farming enables the farmers to adopt new technical implements in cultivation. New agricultural practices can be made in the farm sector. This results in agriculture development.

5) Effective Farm Management : Co-operative farm societies can easily ensure effective farm management by appointing experts in agriculture science. It is possible to implement division of labour and specialization in farm activities.

6) Higher Demand for Labour: The demand for labour increases because of intensive and extensive cultivation in farm sector. As a result, seasonal and disguised unemployment can be evicted from farm sector.

7) Social Equality: Co-operative farming inculcates the spirit of co-operation among the members of the society. Such a spirit of co-operation results in confidence, collective action, joint thinking and feeling of fraternity among the members of the society.

Problems of co-operative Farming:
1) Opposition from Farmers : The farmers feared that they may lose their right of ownership of the land and their position will decline to the level of agricultural labour. Hence, peasants reluctant to join in co-operative farming.

2) Management problems : Generally, Indian farmers are used to operate the small land holdings but it is difficult to them to operate the large size holdings effectively. Due to lack of efficient managerial persons in villages the co-operative farming was discouraged.

3) Danger of Unemployment: There is a lot of scope for implantation of mechanization in co-operative farms which are large in extent. Usually mechanization generates unemployment of displacing labour with machines.

4) Domination of Landlords : The landlords in the system of co-operative farming never treat the small and marginal farmers equal to them. The interests of share croppers, marginal and small farmers cannot be protected because of the domination of landlords in the co-operative farming.

5) Lack of trained Employees: Professional skill is necessary to operate and manage co-operative farms which are large in extent on contrary to the small individual holdings. Lack of such trained staff in our country, these co-operative farming societies are inactive. .

6) Other Problems:

  1. It is a voluntary programme without any motivation to the peasants.
  2. The principles of distribution of profit between farmers and workers are not clearly defined.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 8.
Explain the tenancy reforms in India.
Answer:
Tenancy system is quite common under the Zamindari and Ryotwari Systems. The farmers who take the land from the landlords on leased basis for cultivation called tenants. Tenants are classified into three categories. They are occupancy tenants, subtenants and tenants at will.

1) Occupancy Tenants : Occupancy tenants are called permanent tenants because the rights of occupancy tenants are permanent and inheritable. These tenants enjoy a fixity and security of tenure. No landlord can evict them until they pay the rent. The difference between occupancy tenant and landlord is that the occupancy tenant pays rent to the landlord while landlord pays rent to the Government.

2) Sub Tenants: Sub tenants are called as temporary tenants. Permanent tenant will lease out a part of land under their control to some other farmers who are called subtenants. These tenants do not have any ownership rights on the land which they cultivate on temporary basis. These tenants are ruthlessly exploited. They can be evicted from land on minor pretexts. The rent fixed under this system is oral.

3) Tenants at will: The position of tenants at will is precarious and pitiable. They are subjective to exploitation in the forms of enhancement of rent, eviction from the land without any reason.

Because of ruthless exploitation, Government implemented some tenancy reforms. They are regulation of rent, security of tenure and ownership right for tenants.
1) Regulation of Rent: During the pre-independence period rents were very high as a result of custom, inelastic supply of land and increasing population. Hence, in order to regulate rent various State Governments have enacted various legislations.

2) Security of tenure : Tenants take much care in agriculture if they are provided security of tenure. Then only tenants can invest on lands for the purpose of development of land, like wells or tube wells, permanent fence, preserving soil fertility etc. Due to the fear of loss of tenancy rights, the tenants do not show much personal interest on land.

3) Ownership rights for Tenants: So far as the right of ownership is concerned, tenants have been declared as the owners of land they cultivate. The main aim of tenancy legislation is to provide “Land to the tiller”. They were allowed to purchase their holdings at a fair price determined by tribunals.

Question 9.
Briefly explain various land reforms in India.
Answer:
The Government of India has implemented various land reforms in the country after independence. The following land reforms are introduced for the welfare of depressed groups of farming community. They are

  1. Abolition of Intermediaries
  2. Tenancy Reforms
  3. Ceiling on Land Holdings.

1) Abolition of Intermediaries: Soon after independence the abolition of Zamindari system and its alien forms like Jagirdary and Inamadari systems and Mahalwari system were abolished. The defectives pertain to the Ryotwari system were also regulated. The first act to abolish intermediaries was passed in Madras in 1948. As a result 30 lakh tenants and share croppers acquire ownership rights over a total cultivated area of 62 lakh acres. Compensation was paid to all the intermediaries on installment basis in our country unlike communist countries.
Advantages:

  1. As a result of abolition of intermediaries, tenants became owners of the land and the exploitation came to an end.
  2. The ownership right to the tillers of the land enabled them to have a direct contact with Government which resulted in agricultural development in India.
  3. The Government provided infrastructure facilities for the development of farm sector as the land revenue increased substantially.
  4. A considerable area of cultivable waste land.

2) Tenancy Reforms : Tenancy system is quite common under the Zamindari and Ryotwari systems. The farmers who take the land from the landlords on leased basis for cultivation called tenants. Tenants are classified into three categories. They are Occupancy tenants. Subtenant and Tenants at will.

Because of ruthless exploitation, Government implemented some tenancy reforms. They are regulation of rent, security of tenure and ownership right for tenants.

3) Ceiling on land holdings : Ceiling on land is an important land reform. This is more essential not only to build rural economy on the basis of socialistic pattern of society but also imperative to save the rural masses from the exploitation. The ceiling on land holdings means statutory absolute limit on the amount of land which an individual cultivator or a household may possess. The supply of land is limited and the claimants for it possession are numerous. Therefore, it is unjust to allow exploitation by single individual over a large surface of land. Hence, the Government will take over the land beyond the limit of ceiling and redistributes it among landless labour and marginal farmers.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 10.
Explain the factors responsible for “Green Revolution? in India and its impact on Indian Economy.
Answer:
The Government of India has announced the New Agricultural Strategy in 1965 to ensure rapid agricultural progress. Prof. Norman Borlog is the father of green revolution. The new strategy of agriculture which resulted in revolutionary progress in the farm sector during the period 1960-70 is termed as Green Revolution. Willian S. Gand is the first economist who used the term green revolution.

Achieving high produce and productivity in farm sector by implementing hybrid seeds, fertilizers, pesticides, machines etc., and also by inducing farmers is called green revolution.

Factors responsible for Green Revolution :
1. Intensive Agriculture District Programme (IADP) : In 1964 Government of India has introduced this programme on the basis of recommendation of Ford Foundation Team. This programme was introduced in.Seven Districts like West Godavari in Andhra Pradesh, Shahabad in Bihar, Raipur in Madhya Pradesh, Thanjavur in Tamilnadu, Ludhiana in Punjab, Aligarh in Uttar Pradesh and Pali in Rajasthan where assured irrigation facilities, high fertility, proper rainfall and less hazards (like floods, drainage problem soil conservation problem etc) are available.

2. Intensive Agricultural Area Programme (IAAP) : In 1967, Government of India has introduced this programme with a view to extend the area under intensive cultivation. Under this programme intensive cultivation has extended to other districts of various states. Both the programmes IADP and IAAP are restricted to the intensive cultivation of selected regions but IAAP is limited to some crops only. As a result of intensive cultivation in the selected regions, the neighboring districts will also develop which is treated as spread effect. This programme had extended to 164 Districts in the country.

3. High Yielding Variety Programme (HYVP) : In the new farm technology high yielding variety of seeds programme is crucial because it enables the Indian Agriculture to ensure high productivity Hence, Government of India initiated the programme in 1965.
Various types of hybrid, seeds are innovated with collective efforts of Indian Council of Agricultural Research (ICAR), International Crops Research Institute for Semi-Arid Tropic, (ICRISAT), Universities in Punjab and other research institutes, As a result of HYVP remarkable rise in the production of paddy, wheat, sugarcane, cotton etc., has taken place.

4. Introduction of crops with short gestation period: Various crops with short gestation period have been developed in India with the continuous collective research of various organizations like ICAR, ICRISAT. Considerable decrease has taken place in the gestation period of paddy, wheat, maize etc. Some kinds of paddy like IR 8, IR 20,1001,1010, Masuri, Basmati, Jaya, Padma etc., and mexican type of wheat known as miracle wheat which have short gestation period are being produced by Indian farmers.

5. Expansion of Irrigation Facilities: Agriculture research and experiments take place only in those areas where proper irrigation facilities are available. So far all the high yielding variety seeds require more irrigation facilities. The area under the irrigation facilities increased from 21 million hectares in 1951 to 63 million hectares by 2010.

6. Farm Mechanization : Mechanization is an integral part of green revolution. Mechanization in the farm of sector in the form of tractors, oil engines, electric motors, crushers, drillers, harvesting machines etc. enriches the operations in agriculture. As agricultural operations are being seasonal, mechanization is essential for timely operations in agriculture.

7. Consumption of fertilizers and pesticides: High yielding variety seeds are highly responsive to the use of fertilizers and pesticides. In order to achieve high production by using fertilizers and pesticides the Government of India has been providing subsidies favorably affects productivity of agriculture.

8. Other Factors:
a) Agricultural Extension Services centers like “Rural Knowledge Centers”, “Agricultural Technology Management Agency” (ATMA) etc, are established by the Government of India.
b) Adult education centers have been established by the Government to literate the rural people.
c) Credit facilities are indispensible for the successful accomplishment of timely operations in agriculture. The Government of India has been providing credit to the peasants through Commercial, Regional Rural banks, Primary Agricultural Co-operative Credit Societies etc.

Impact of Green Revolution on Indian Economy:
1. Increase in Foodgrain production : The rice production from 35.0 mts in 1960-61 increased to 99.37 mts in 2008-09. The production of wheat was 11.0 mts and rose to 77.63 mts. The production of pulses was only 12.7 mts in 1960-61 which has increased to 14.2 mts. Totally production of food grains was 82.0 mts in 1960-61 and that has increased 229.9 mts by the year 2008-09.

2. Boost to employment generation : Green revolution is the small farm revolution. Labour intensive crops like rice, sugarcane, potato, vegetables, fruits have increased the employment opportunities in agriculture sector. The entiy of corporate houses has generated more employment opportunities in retailing business of fruits and vegetables.

3. Improvement in incomes: The impact of green revolution revealed.that the farmers in Kerala, Madhya Pradesh, Andhra Pradesh, Tamilnadu, Gujarat, Punjab and Himachal Pradesh had good chances of improving their incomes. It facilitates the farmers to follow simple but scientific and technical ways like grading the produce in the field itself selling directly to the corporate retail companies by avoiding middlemen. Organised retailers have provided better remunerations to the farmers. So the consumers were benefited in the form of quality produce at lower prices.

Therefore the green revolution has also widened the inequalities between big and small farmers, in rural areas.

4. Forward and backward linkages strengthened: Agriculture supplies raw material to industries which is known as forward linkage. The new technology in agriculture has strengthened the backward linkage. This way the linkage between agriculture and industry has got strengthened.

5. Decrease in Poverty : The result of green revolution surplus in the production of food grains is actived.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 11.
What are the various sources of rural credit in India?
Answer:
Credit plays a significant role in the farm activities. The credit needs of the peasants can be satisified either by institutional or non-institutional sources. Credit plays on important role to meet the requirements of the farmes. But at present the importance of institutional credit has been steadly increasing in our Country.

I) Institutional Credit:
1. Government : The Government had played a key role in providing credit for agricultural operations before emerging institutional credit organizations. Generally, the volume of credit supplied by the Government directly to the farmer for farm operations is very low. The Government will sanction direct loans to the farmers whenever natural calamities like drought, famine etc., are happened. The Government will sanction these direct loans at low rate of interest which are called as “Takkavi loans”. These loans can be repaid by the farmers in easy installments. These loans are insignificant because of rigid rules in lending and are about 2.3 percent in the total requisite credit of the farmers.

2. Role of Reserve Bank of India: Reserve Bank of India was established in 1935 and was nationalized in 1949. It has started Agriculture Credit Department and two separate funds in 1956 to supply to the agriculture sector. They are i) National Agricultural Credit fund, ii) National Agricultural Credit Stabilization fund. RBI provides credit to the peasants through State Co-operative Banks.

3. Co-operative Credit Societies : Co-operative credit system was successfully implemented in Germany by providing cheap credit. Keeping it in view this co-operative movement was started in India in 1904. These societies were organized to relieve the indebtedness of rural pqople and to promote thrift.

4. Commercial Banks: Commercial Banks are defined “Those institutions which take up all types of banking activities with a view of profit”. Commercial Banks are extending credit facilities to agricultural allied activities like dairying, poultry farming, piggery, fisheries etc.

5. Regional Rural Banks : Establish Regional Rural Banks in our country on October 2,1975. The Main objective of RRBs is to provide credit and other facilities to the small and marginal farmers, agricultural labourers, artisans and small entrepreneurs so as to develop commerce, agriculture, industry and other productive activities.

6. National bank for agriculture and rural development (NABARD) : It was established in 1982. NABARD provides short term credits, medium term and long term credits to State co-operative Banks, Regional Rural Banks, Land Development Banks-and other financial institutions engaged in rural development and approved by R.B.I.

Non-Institutional sources:
1. Money lenders : Money lenders are the major source of agricultural credit for long period of time. Money lenders are two types, i) Agricultural money lenders ii) Professional money lenders. Money lenders lend money to the farmers at exorbitant rate of interest unproductive purposes. They are exploiting farmers by using unfair methods like manipulating accounts. The share of money lenders in the total farm credit was around 70% in 1951. But it declined to 19.6% in 2002.

2. Land lords : Mostly small farmers and tenants depend on land lords in order to meet their financial requirement. Land lords lend money both fob productive and un productive purposes at exorbitant rate of interest. They lend money to the farmers at high rates of interest keeping in view of grabbing their lands. The share of landlords in total farm credit was 15% in 1951-52 and declined 1% in 2002.

3. Traders and Commission agents : They advance loans to agriculturalists for i productive purposes against the crops without any legal agreement. They force the borrowers to sell their products at low prices to them and charge heavy commission on the produce of the farmers. The share of traders and commission agents in the total farm credit was 55% in 1951 and declined 2,6% in 2002.

4. Relatives and friends : Farmers often borrow from their relatives and friends in order to meet their temporary credit needs. They do not involve in any sort of exploitation. The share of these loans in total farm credit was 14.2% in 1951 and declined 7.1 in 2002.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 12.
What are the causes for rural indebtedness ? Suggest some remedial measures to reduce it. [A.P. Mar. 16]
Answer:
A majority of rural households are in indebtedness. Among the rural households, in-debtedness is highly prevalent in cultivators household, especially in the case of marginal and small farmers. Generally, these households are borrowing from non-institutional sources at a high rate of interest which results in rural indebtedness. The following are the main causes for increasing rural indebtedness in India.

1. Ancestral debt: The most important cause of the existing rural indebtedness is the ancestral debt which is inherited from ancestors. Infact, the volume of inherited debt should be limited to the extent of inherited property. Hence, many of the rural poor in our country are starting their career with a heavy burden of ancestral debt.

2. Poverty : The basic cause of the indebtedness of the farmers is their poverty. The farmers have to borrow for various purposes, as he has no past savings of his own. Poverty either forces the peasants to borrow or prevents them to pay off debts.

3. Natural calamities : Another cause of rural indebtedness is that indian agriculture is still a gamble in monsoons. Frequent failure of monsoons results is drought. On the other hand, excessive rains cause havoc in the form of floods which damage crops. All these problems force the farmers to borrow funds.

4. Extravagance of the farmers : It has been observed that Indian farmers spend more on social and religious functions like marriages, festivals, births, funerals, dinners, ornaments etc, beyond their capacity. All these funds borrowed for such unproductive purposes cannot be paid off easily.

5. Money lenders : Money lenders are the main source for the provision of credit facilities in the rural areas. Money lenders are least interested in the well being of farmers. Hence, they tempt the farmers to borrow funds for unproductive purposes at high rates of interest keeping in grab of their valuable assets. They adopt many unfair methods to exploit the rural people.

6. Small size land holdings: The average size of land holdings in India is very small due to sub-division and fragmentation of land holdings. When the holdings are small, modernization of agriculture becomes impossible.

7. Litigations: Litigation either civil or criminal is another cause of rural indebtedness. People in rural areas generally indulge in various kinds of disputes like disputes over boundaries, pathways, fencing etc. Hence, their valuable time and money is being wasted and this adversely affect farm production. All these lead to increase in their debts.

8. Passion for land : The peasants have a tremendous passion for land and desire to make improvements on land. These improvements on land should be done through saving and not through borrowing. But farmers mostly borrow for these purposes.

9. Other causes : In addition to the above causes, purchase of household luxuries, spending on bad habits, increasing cost of cultivation, dependence On non-institutional sources, expenditure on medicines, lack of support prices for the crops etc, are becoming responsible factors for increasing rural indebtedness.

Remedial measures:
1. Expansion of Institutional credit: In order to reduce the dependence of the rural people cm money lenders, the Government should provide timely and adequate credit through commercial banks, RRBs, co-operative credit societies etc. Consequently, the rural people get relief of their debts.

2. Regulation of Money lenders: The Government should enact legislations to control money lenders. These legislations should consist of licensing and registration of money lenders, maintenance of accounts in prescribed form, fixing maximum rate of interest, furnishing receipts to debtors after payment etc.

3. Debt Moratorium : Central Government and State Governments are resorting the policy of debt moratorium for redemption of debt of the marginal farmers, small farmers and landless laborers as it was introduced during the period of emergency.

4. Educating the farmers : All the farmers in rural areas should be educated. Then only they can understand various legislations pertaining to ancestral debt and fixed rate of interest.

5. Supply of Inputs : The institutional credit sources should sanction loans to the rural poor not in the form of cash but in the form of inputs in order to avoid unproductive expenditure. Such mode of sanction definitely enhances the repaying capacity of the debtor and minimizes the problem of indebtedness.

6. Others: In addition to the above the Government has to frame various schemes to eradicate poverty and employment which in turn increase the income and the repaying capacity of rural poor. Coverage of women farmers under micro-finance methodology must be increased.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 13.
Explain the role and progress of NABARD in the filed of agriculture and rural credit
Answer:
National Bank of Agriculture and Rural Development (NABARD) is an apex bank for rural credit and does not deal with rural people directly. The NABARD set up on July 12, 1982. It provides grants assistance through the co-operative banks, commercial banks, Regional Rural banks and self help groups etc.

Disbursement of Refinance : NABARD has initiated several innovative projects like Rural Infrastructure Development Fund (RIDF), Farm Innovation and Promotion Fund (FIPF), Kissan Credit Cards. It is playing an energetic role in strengthenning the rural credit structure in the country. The following are the features of these schemes.

1. R.I.D.F. : Rural Infrastructure Development Fund (RIDF) was created in 1995-96 from out of short fall of commercial banks lending to priority sectors and agriculture. Since the NABARD has partnered State Government in the creation of rural infrastructure. The annual allocation of funds under the RIDF has gradually increased from ? 2000 Crore in 1995-96 to 20000 crore in 2012-13.

2. Kisan Credit Card System: KCC scheme was introduced in 1998-99 with a view to facilitate short term timely credit to farmers. Commercial banks, Regional Rural Banks and Co-operative Banks which are refinanced by NABARD are implementing this scheme. Since its inception till the end of August, 2012 ? 91676 crore have been issued to nearly 9.54 crore kisan credit card holders.

3. Micro Finance: The aim of this programme is to bring banking service to the door steps of the poor. Micro Finance is a novel approach to banking with the poor. Under this scheme bank credit is extended to the poor through Self Held Groups (SHGs) which were incepted by NABARD in 1986-87 as non-govt, organisations. Micro credit attempts to combine lower transactions cost and high degree of repayments. NABARD disbursed ? 3916.64 crore in 2012-13 to SHGs.

4. Swarnajayanthi Gram Swarozgar Yojana (SGSY): It was launched by the Govt, of India on April 1,1999 by combining various programmes like IRDP, TRYSEM, DWCRA and allied schemes. Under this scheme up to 2011,168.46 lakh Swarojgaries have been assisted with a total out lay of ₹ 42.168 crore. NABARD disbursed an amount of ₹ 111.72 crore to SGSY in its annual budget 2012-13.

Progress of NABARD: The performance of NABARD and the amount to various sectors for the development of rural poor is presented in the following table.
Sector-wise Disbursement of Refinance (Crore)
AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector 1
AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector 2
In the table the share of non-farm sector is high i.e., ₹ 5150.88 crore in 2012-13. Minor irrigation and development, mechanization, plantation, dairy development and self help groups derived much attention Of NABARD in its disbursement of refinance. The total amount disbursed by NABARD in 2012-13 is ₹ 17674.29 crore.

Question 14.
Explain the defects in agricultural marketing and suggest some remedial measures. [A.P. Mar. 18]
Answer:
No doubt, agricultural markets are operating efficiency in India, being influenced by the reforms and the process of globalization. However, the position of agricultural markets is deplorable in our country. The farmer is poor illiterate and ignorant. He does not have facilities to store his produce. He has caught in the evil hands of the middlemen. He has been suffering from the following defects of Agricultural marketing.

1. Inadequate storage facilities: There is lade of proper and sufficient storage facilities for the produce in rural areas. The available facilities are so bad and unscientific that more than 20 percent of the produce is eaten away by the rats.

2. Absence of proper Grading: The farmers are not getting a relevant price according to the equality of their products. They are not properly graded. They are sold in one common tot in heap of all quantities of produce consequently. The farmer producing products of better quality in not assured of a better price.

3. Inadequate transport facilities: Even today the farmers are using traditional means of transportation to move his produce to the markets. Railways and roadways are not properly connecting the village. Bullock carts are used to move the products. As a result, the farmer is forced to domp his produce at nearby weekly or daily markets. This is true particularly in the case of perishable commodities.

4. Existance of middlemen : More number of middlemen are operating in the agriculture marketing. As a result, the share of farmers is reduced in the prices realised for products sold in the market. Many studies observed that middlemen are grabbing almost 60 to 70 percent of the market price.

5. Malpractices in the markets: In the markets, the farmer is defective by the use of wrong weights and measures. He is also cheated by brokers and traders. The farmer has to pay weighing charge unloading charges, separation of impurities in the produce charges and many other miscellaneous undefined and unspecified charges.

6. Lack of adequate market information : The farmers are not having proper information about the prices existing in the markets. They are deprived of getting reasonable price for their products due to the inadequate information.

7. The farmers are unorganised : The farmers in India belong to different places, different languages and dialects and unorganized. The middlemen are strong and well organized. As a result, the farmers are unable to get proper price for their products.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Remedial measures: After independence the Government of India adopted a number of measures to improve the system of agricultural marketing.

  1. Regulated markets : Regulated markets have been started will view to ensure remunerative price to the products of the farmers. These markets check all unfair practices prevalent in the marketing of farm produce.
  2. Co-operative marketing: In this system all the farmers of a village form together into a co-operative marketing society to sell their products at a fair price rate.
  3. Contract farming: It is another good remedy for solving the problems Of agricultural marketing in our country.
  4. Rythu Bazars : The Andhra Pradesh Government has introduced the concept of Rythu Bazars as 26th January 1999. In these markets the farmers will sell their products directly to the consumers without the interference of middlemen.
  5. Facilities of grading and Standardisation : The Government of India has done much to grade and standardise many agricultural goods. For this purpose Government of India enacted legislation in 1937 and established grading centres at Jaipur, Bhopal and Nagpur etc.
  6. Warehousing facilities : Many warehouses have been constructed by the Central Government throughout the country. As a result, the farmers will store their products until they get fair prices.
  7. Transport facilities: Transportation plays an important role in marketing system. Cheap and easy means of transportation encourage the farmers to carry their products to the market and create confidence among the farmers.
  8. Credit facilities : The Government of India established institutions like Primary agricultural co-operative credit societes, Commercial banks, Regional rural banks etc.
  9. Market information : The farmers should have perfect knowledge of prevailing market prices. Thus, market information relating to the prices can be provided to the farmers through Radio, T.V.s and News papers etc.

Short Answer Questions

Question 1.
Explain the features of Indian agriculture.
Answer:
1) Uncertainty in crop output: Monsoon and climatic conditions play a significant role in Indian agriculture which often affects agriculture productivity adversely. Moreover, floods and drought are quite common. Hence, Indian agriculture is rightly to be called a “gamble on monsoon”. As a result of these unforeseen contigencies production and productivity are uncertain.

2) Feudal relation of Agriculture : After independence Zamindari and Mahalwari systems were abolished and Ryotwari system came into existence. Under this system tenants do not possess security of tenure, regulation of rent and ownership rights.

3) Rural Indebtedness: After independence the Government has initiated co-operative credit societies and banks for providing rural credit. However, the small and marginal farmers continue to depend on money lenders for their requirements. Hence, rural indebtedness prevailing in agriculture sector has become common. Because of indebtedness farmers are unable to invest more on agriculture to reap much harvest.

4) Dualism in Labour market: Because of pressure of population on land, wages in agricultural sector are considerably lower comparatively to the industrial and service sectors of the economy. The cheapness of labour in the traditional agricultural sector causes that labour in agriculture sector to be used extensively rather than other sectors of the economy where wages are high.

5) Diversities in Agricultural sector : The nature of soil, magnitude of rainfall, irrigation facilities etc, vary from one region to another region. Similarly, drought conditions, floods, problems of water salinity etc., vary among various regions of our country. The size of land holdings are differ in various regions. These variations affect cropping pattern.

6) Outmoded Farming techniques : Most of the Indian farmers continue to use outmoded farming techniques. The traditional agriculture depends on human and animal labour, rains and dung manure. Thus, this results in subsistence farming.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 2.
Explain the present conditions of agricultural labourers. [A.P. Mar. 18]
Answer:
Present conditions of Agricultural Labourers :
1) Low Social Status: Most of the agricultural labourers are belonged to the depressed classes. They have been neglected for ages. In case of such farm labourers, exploitation has become common and they have not fought for their rights.

2) Unorganised : Agricultural labourers are living in scattered villages. Moreover, they are illiterates. Hence, they cannot easily to be organized. As a results it is difficult for farm labourers to bargain with the land owners for good wages.

3) Seasonal Employment : Agricultural labourers have to face the problems of unemployment and underemployment. They are employed while sowing and harvesting season, of the year they remain unemployed.

4) Low Wages and Income : Agricultural wages and family incomes of agricultural labourer are very low. The wages of farm labourers vary from state to state. In case of attached labourers these wages are pathetic. However, as prices also increased considerably, the real wages rates did not increase much.

5) Rural Indebtedness : Because of the low level of income agricultural labour generally seek debts. In fact the debt of agricultural labourers passes from generation to generation and become bonded labour.

6) Femanisation of Agricultural labour : Female agricultural workers are generally forced to work harder. They are paid less wages comparatively to male workers.

7) High incidence of Child labour: It is estimated that one-third of child labourers in Asia are in India. The largest number of child workers are engaged in agriculture. Wages paid to child labourers are too low which adversely affect the incomes of their households.

8) Lack of subsidary professions : Another serious problem that is being faced by farm labourers is lack of non-agricultural occupations in. the villages. As there is no work on the fields, agricultural labour has no other means to earn for subsistence. Thus, they are disguised and are burden on land.

Question 3.
Explain the factors affecting cropping pattern.
Answer:
Factors affecting cropping pattern in India : The cropping pattern in India can be affected by various factors like physical, economical, technical and Government policies.
Physical Factors: Physical factors play a vital role in determining the cropping pattern. These factors are classified below.

  1. Climate and Rain fall: Climatic conditions and rainfall determine cropping pattern. Some crops require cool climate while some other crops require hot climate. For instance, apples will be produced in cool climate. For instance, crops like Paddy, Sugarcane require abundance of water.
  2. Nature of Soil and Fertility: Nature of soil and fertility determine the production of certain crops. For instance, wheat requires well drained silt and fertile loam soils but for cotton black soils are ideal.
  3. Irrigation facility: It also determine the cropping pattern. For instance crops, like Paddy, Sugarcane, Wheat etc., require assured irrigation facility. Some other crops like Jowar, Maize, Ragi etc., will grow in the areas where irrigation facilities are insufficient.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 4.
Explain the importance of irrigation.
Answer:
Importance of Irrigation :
1) Insufficient, uncertain and irregular rains : The period of rainfall is restricted to only four months during monsoons. Even during monsoons the rainfall is scanty. Sometimes monsoons are delayed while some other times they are prematured. This type of atmosphere results in drought conditions. Hence, with the help of proper development of irrigation, droughts and famines can be effectively controlled.

2) Higher productivity on irrigated land : Irrigation helps greatly in raising the productivity of land than unirrigated land. Because this enables the application of modern inputs like fertilizers, high yielding varieties of seeds etc.

3) Multiple cropping possible : India is a land of tropical and sub-tropical climate. It has potentialities to grow crops throughout the year but rainfall is restricted to less than four months. Provision of irrigation facilities can make possible the growing of multiple crops throughout the year.

4) Crucial role in new agricultural strategy: The successful implementation of high yielding varieties programmed depends on timely and adequate supply of water. These seeds require chemical fertilizers and substantial water at regular intervals of time. Therefore, irrigation facilities facilitates the expansion of new agricultural strategy to larger areas of land.

5) Bringing more land under cultivation: The total reporting area according to land utilization statistics was 305.56 million hectares in 2009-10. Thus, some portion of the waste land can be brought under cultivation if there is availability of irrigation facilities.

6) Prosperity : Irrigation helps in stabilizing the output and yield level. Irrigation plays a protective role during drought years. Thus, irrigation facilities prevent fall in output during drought and achieve stability in income and output which in turn result prosperity.

7) Indirect benefits of Irrigation: Expansion of irrigation facilities all over the country avoid disparities in the production of food grains. Irrigation promotes the growth of agriculture and allied sectors. Thus, increased production in agriculture stabilizes the prices of agricultural products.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 5.
What are the causes for low productivity ?
Answer:
The causes for low level of agriculture productivity in India are manifold. They can be grouped into four broad categories.

  1. General causes
  2. Institutional causes
  3. Technical causes
  4. Environmental causes

1) General causes:
i) Pressure of Population on Agriculture : Pressure of population on agriculture is heavy as a result of high growth rate of population and slow growth rate of other sectors of the economy. In 2011, about 263 million workers out of 348 million rural working populations are employed in agriculture. Increasing pressure of population on agriculture results, in the subdivisions and fragmentation of holding. Consequently, the productivity in agriculture sector remains low in India.

ii) Social Environment: The social environment of villages is an obstacle in agricultural development. The farmers in rural are illiterate, superstitious, conservative and unresponsive to new agriculture techniques. The decline of joint family system and land hunger are also discouraging the rural atmosphere. Peasants are not able to take proper care of their agriculture. Unless this atmosphere is changed, it is too difficult to enhance the productivity of agriculture.

iii) Lack of Infrastructural Facilities: Infrastructural facilities like transport, storage, credit and marketing are inadequate in rural areas to the growing population, due to lack of these adequate infrastructure facilities, the agricultural productivity in rural areas is very low.

iv) Impact of the British Regime : During British rule in India, they have not shown any interest in developing agriculture sector but made our economy as colonial one. Moreover, their policies like land tenure system, collection of land cess gave a deadly blow to the Indian agriculture.

2. Institutional causes:
i) Uneconomic Land Holding: According to the National Sample Survey, 52 percent land holdings had a size of less than 2 hectares in 1961 – 62. In 2010 -11,85 percent of total land holdings are less than 2 hectares. As a result of laws of inheritance and other reasons is a further divisions and fragmentation of land holdings. Hence, these small holdings are adversely affecting productivity of agriculture.

ii) Defects in Land Tenancy System : The Indian agriculture system was adversely affected before Independence because of defectives in Zamindari, Jagirdari, Mahalvari systems which exploited the farmers. In this system lack of certainty in rent, security of tenure and ownership right the tenants don’t show any attention to develop agriculture. Hence, India has become less productivity.

iii) Lack of credit and marketing facilities : The cultivators are not able to invest requisite sources in agriculture due to lack of marketing facilities and required credit at fair rate of interest. Even support price policy and subsidies to inputs of agriculture fired by the Government are unsatisfactory. Hence, peasants follow traditional methods which results in low productivity.

3. Technical causes:
i) Outmoded Agricultural Techniques : T. W. Schultz of famous economist opined that the peasants in India are still using traditional or outmoded techniques. Indian farmers are still using wooden ploughs, bullock carts, sickles etc. Use of fertilizers and new high yielding varieties of seeds is also extremely limited. Hence, the productivity in agriculture is low.

ii) Inadequate Irrigation Facilities : Gross cropped area in India in 2010 – 11 was 198.97 million hectares but only 89.36 million hectares of land had irrigation facilities. It implies that’55 percent of the gross cropped area continues to depend on rains. Rainfall is often insufficient, uncertain and irregular. In such atmosphere it is difficult to extend the new agricultural technology all over the country.

iii) Scarcity of Agricultural Inputs : The supply of modem agricultural inputs like fertilizers, pesticides, hybrid seeds, farm machinery etc., are inadequate to meet the requirements of our country. In order to achieve high production in agriculture requisite supply of inputs is essential.

4. Environmental causes : Environment also plays a vital role in affecting the productivity of agriculture, increase in the temperature. Degradation of soil, changes in temperature, pollution of water and air etc., adversely affect the productivity of agriculture are

  1. Global warming.
  2. Soil Degradation.
  3. The intensive cultivation of high yielding variety crops.
  4. The reckless use of fertilizers.
  5. Shifting cultivation.
  6. Displacement of the traditional practies of crops.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 6.
Examine the present pattern of land utilization.
Answer:
Land is the most important resource of natural resources of any country. Land has the characteristic of inelasticity in supply. The economic development of a nation is directly depends on the supply of land. Hence, certain modifications can occur in the existing pattern of land utilization. The total area land is about 306 million hectares. The total cropped area is about 192 million hectares. The total area under fallow lands is about 26 million hectares and the area under forests is 70 million hectares. Asa result of increased irrigation facilities culturable waste land has been declined to 12.85 million hectares. The Government has provided loans and subsidies to the farmers for land reclamation.

Question 7.
Consolidation of land holding.
Answer:
Consolidation of holdings is a proper solution to the problems of scattered holdings. Consolidation means all the holdings of the village are pooled together into one compact block. The owners of scattered land holders may voluntarily exchange their pieces of land into one compact block. The State Governments have adopted coercive methods to the cosolidation of holdings. It was started in 1951 – 52. The States are Punjab, Haryana, M.P, U.P implemented this programme. But the States of A.P, Tamilnadu, Rajasthan, West Bengal, Assam have not enacted the laws of consolidation. According to annual report of Ministry of Rural Development 2004, total area brought under consolidation was 163.3 million hectares.

Question 8.
Creation of Economic land holding.
Answer:
Creation of economic holdings is another important method to solve the problems of small size land holdings. To develop and strengthen Indian agriculture sector is inevitable to create economic holdings. The economic holdings also termed as ‘Family holding1 or ‘Optimum holding’.

According to Dr. Mann the economic holding as “one which will provide for an average family the minimum standard of life”.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 9.
Need for land reforms. [A.P. Mar. 17, 16]
Answer:

  1. Agriculture development : Agricultural development takes places when land reforms are entrusted in agrarian sector to avoid the hindrances to agricultural development. In such atmosphere technical reforms will be fruitful in agricultural sector.
  2. Economic. Development : Agrarian sector influences largely the economic development of our country. In order to attain sustainable growth rate in Indian agriculture, it is inevitable to implement land reforms.
  3. Social Justice: Land reforms are aimed at alleviating rural poverty by distributing land among the landless, providing security to tenant, protecting the interests of tribals. Land reforms aim at achieving social justice in the economy by eradicating poverty and disparities in income.
  4. Increase in Agricultural productivity: Land reforms are essential to increase the production and productivity in Agriculture.

Question 10.
Abolition of intermediaries.
Answer:
The first act to abolish intermediaries was passed in Madras in 1948. Later state after state enacted legislations abolishing intermediaries. As a result 30 lakh tenants and share croppers acquire ownership rights over a total cultivated area of 62 lakh acres. Compensation was paid to all the intermediaries on installment basis in our country unlike communist countries.

The Effects of Abolition of Intermediaries :
The following are the advantages of the abolition of Intermediaries.

  1. As a result of abolition of intermediaries, tenants became owners of the land and the exploitation came to an end. This helped to secure social justice in agraian structure.
  2. The ownership right to the tillers of the land enabled them to have a direct contact with Government which resulted in agriculture development in India.
  3. The Government provided infrastructure facilities for the development of farm sector as the land revenue increased substantially.
  4. A considerable area of cultivable waste land and private forests belonging to the intermediaries have been brought under cultivation.

Question 11.
Ceiling on land holdings.
Answer:
Ceiling on land is an important land reform. This is more essential not only to build rural economy on the basis of socialistic pattern of society but also imperative to save the rural masses from the exploitation. The ceiling on land holdings means statutory absolute limit on the amount of land which and individual cultivator or a household may possess. The supply of land is limited and the claimants for it possession are numerous. Therefore, it is unjust to allow exploitation by single individual over a large surface of land. Hence, the Government will take over the land beyond the limit of ceiling and redistributes it among landless labour and marginal farmers.

Objectives of celling of land:

  1. To reduce inequalities in agrarian sector.
  2. To enlarge the sphere of self employment.
  3. To eliminate exploitation and to provide equal opportunities to all..
  4. To meet the desire that land must belong to the tiller.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 12.
Reasons for poor performance of land reforms.
Answer:
Land reforms were implemented with good objectives, aiming at the development and empowerment of rural poor. However, in practice they were internally rejected. The following reasons can be mentioned for the poor performance of land reforms.

  1. Lack of political will.
  2. The rural poor are unorganised.
  3. Absence of updation of land records.
  4. Legal hurdles in the way of implementation.
  5. Weak administrative set up.
  6. Very little surplus lands were talan into possession.
  7. Provision of exemptions were used for evading the ceiling on land holdings.
  8. Laws were challenged on number of other grounds like rates of compensation, calculation of standard holdings etc.
  9. Reforms were given a low priority in development strategy.

Question 13.
Describe the impact of Green Revolution on Indian Economy. [A.P. Mar. 18, 17]
Answer:
Impact of Green Revolution on Indian Economy :
1. Increase in Foodgrain production: The rice production from 35.0 mts in 1960-61 increased to 99.37 mts in 2008-09. The production of wheat was 11.0 mts and rose to 77.63 mts. The production of pulses was only 12.7 mts in 1960-61 which has increased to 14.2 mts. Totally production of food grains was 82.0 mts in 1960-61 and that has increased 229.9 mts by the year 2008-09.

2. Boost to employment generation : Green revolution is the small farm revolution. Labour intensive crops like rice, sugarcane, potato, vegetables, fruits have increased the employment opportunities in agriculture sector. The entry of corporate houses has generated more employment opportunities in retailing business of fruits and vegetables.

3. Improvement in incomes: The impact of green revolution revealed that the farmers in Kerala, Madhya Pradesh, Andhra Pradesh, Tamilnadu, Gujarat, Punjab and Himachal Pradesh had good chances of improving their increases. It facilitates the farmers to follow simple but scientific and technical ways like grading the produce in the field itself selling directly to the corporate retail companies by avoiding middlemen. Organised retailers have provided better remunerations to the farmers. So the consumers were benefited in the form of quality produce at lower prices.

Therefore the green revolution has also widened the inequalities between big and small farmers in rural areas.

4. Forward and backward linkages strengthened: Agriculture supplies raw material to industries which is known as forward linkage. The new technology in agriculture has strengthened the backward linkage. This way the linkage between agriculture and industry has got strengthened.

5. Decrease in Poverty : The result of green revolution surplus in the production of food grains is actived.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 14.
Role of Regional Rural Bank in rural credit.
Answer:
Regional Rural Banks : According to the recommendations of the working group on Rural Banks headed by prof. M.Narasimham, the Government of India issued an ordinance country establish Regional Rural Banks in our country on October 2, 1975. Five Regional Rural Banks were started. Later the number rose to 196. The Government of India amalgamated Regional Rural Banks in order to consolidating and strengthening them. As on March, 2013, the total member of branches of RRBs were 17,856 across 635 districts in 26 states and one Union Territory.

The Nationalized commercial banks are sponsoring the RRBs. Each Regional Rural Bank had an authorized capital of ₹ 1 crore and paid up capital ₹ 25 lakhs. The share capital has subscribed by the Central Govt. 50% the State Govt, concerned 15% and sponsoring commercial banks 35%.

The main objective of RRBs is to provide credit and other facilities to the small and marginal farmers, agricultural labourers, so as to develop commerce, agriculture and other productive activities.

Question 15.
Primary Agricultural Co – operative Credit Societies.
Answer:
Co-operative credit societies : Co-operative credit system was successfully implemented in Germany by providing cheap credit. Keeping it in view this co-operative movement was started in India in 1904. These societies were organized to relieve the indebtedness of rural people and to promote shift.

The co-operative credit institutions in India have been organized into short term and long term structures.
AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector 3
PACSs are organized at village level. They can be formed by any ten or more than ten persons. In order to strengthen PACSs financially the Reserve Bank of India, in collaboration with State Governments had been taking a series of steps. Commercial banks in India introduced a scheme of financing through PACSs for disbursing agricultural loans. The StCB advances loans to DCCBs in order to augment their capacity to advance loans to the village PACSs. PACSs are organized at village level. The management of the society is under an elected body consisting of a president, secretary and treasurer.

The total volume of credit supplied by these co-operative banks for agricultural sector has reached to ₹ 87,963 crore by 2012. The co-operative credit system is organizationally and financially weak to meet the credit needs of agricultural sector. Hence, co-operation has failed, but co-operation must succeed.

Question 16.
Commercial banks and Rural credit.
Answer:
The commercial banks started a vital role after nationalization of 14 banks in 1969 and 6 banks in 1980. As far as rural credit is concerned, the role of commercial banks is highly appreciable. These banks are supplying credit to the rural areas in the following manner. –
1) Commercial banks are supplying credit for short term and long term requirements of the needy farmers in rural areas. They are providing short term crop loans which accounted 42 to 45 percent of total loans disbursed by commercial banks. Long term loans are extended by commercial banks for purchasing pump sets, tractors and other agricultural machinery which accounted 35 to 37 % of total loans disbursed by commercial banks.

2) Commercial banks are extending credit facilities to agricultural allied activities like dairying, poultry, farming, piggery, fisheries etc.

3) Commercial banks working for the implementation of various rural development programmes like IRDP ; JRY etc by sanctioning and disbursing credit to the beneficiaries.

4) Commercial banks are indirectly helping the rural farmers by extending credit for fertilisers and pesticides companies. Central warehousing Corporation and Regional Rural banks.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 17.
Role of Reserve Bank of India in Rural Credit.
Answer:
RBI was established in 1935 and was nationalized in 1949. It has been rendering viable services for the development of rural areas. It has started agriculture credit department and two separate funds in 1956 to supply to the agriculture sector. They are 1. National Agriculture Credit Fund. 2. National Agricultural Credit Stabilization Fund.

R.B.I provides credit to the farmers through State co-operative banks in the following ways.
a) Short Term Credit: RBI provides short term credit facilities to State co-operative banks at a lower interest rate for a period of 15 months by providing rediscounting facilities on Govt, securities and debentures of land development bank.

b) Medium Term Credit: R.B.I has been providing medium term credit to the State co-operative banks for a period of 15 months to 5 years.

c) Long term Credit: R.B.I provides long term credit to the State Govt, for agricultural development activities. The term credit varies from 5 years to 20 years.
Other services:

  1. It provides credit to all the institutions which are engaged in rural credit.
  2. It provides loans for small farmers development agency and marginal farmers development agency.

Question 18.
Defects in the Agricultural marketing in India.
Answer:
No doubt, agricultural markets are operating efficiency in India, being influenced by the reforms and the process of globalization. However, the position of agricultural markets is deplorable in our country. The farmer is poor illiterate and ignorant. He does not have facilities to store his produce. He has caught in the evil hands of the middlemen. He has been suffering from the following defects of Agricultural marketing.

1. Inadequate storage facilities: There is lack of proper and sufficient storage facilities for the produce in rural areas. The available facilities are so bad and unscientific that more than 20 percent of the produce is eaten away by the rats.

2. Absence of proper Grading: The farmers are not getting a relevant price according to the equality of their products. They are not properly graded. They are sold in one common tot in heap of all quantities of produce consequently. The farmer producing products of better quality in not assured of a better price.

3. Inadequate transport facilities: Even today the farmers are using traditional means of transportation to move his produce to the markets. Railways and roadways are not properly connecting the village. Bullock carts are used to move the products. As a result, the farmer is forced to domp his produce at nearby weekly or daily markets. This is true particularly in the case of perishable commodities.

4. Existance of middlemen : More number of middlemen are operating in the agriculture marketing. As a result, the share of farmers is reduced in the prices realised for products sold in the market. Many studies observed that middlemen are grabbing almost 60 to 70 percent of the market price.

5. Malpractices in the markets : In the markets, the farmer is defective by the use of wrong weights and measures. He is also cheated by brokers and traders. The farmer has to pay weighing charge unloading charges, separation of impurities in the produce charges and many other miscellaneous undefined and unspecified charges.

6. Lack of adequate market information : The farmers are not having proper information about the prices existing in the markets. They are deprived of getting reasonable price for their products due to the inadequate information.

7. The farmers are unorganised : The farmers in India belong to different places, different languages and dialects and unorganized. The middlemen are strong and well organized. As a result, the farmers are unable to get proper price for their products.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 19.
Various stages of agricultural marketing.
Answer:
Farms cannot sell away all the their products instaniously after harvesting. These products must go through a series of stages before they are actually marketed. Such stages in agricultural marketing.

  1. Assembling: The total produce of various farmers from various places should be brought to one place. The process of pooling up of small surpluses of individual farmers in the market of the producing area is called assembling.
  2. Transportation : All the farm products must be transported from the producer markets to consumer markets. Transportation facilitates the availability Of goods in the market.
  3. Grading: All the assembled products should be graded and standardized according to their quality and durability.
  4. Processing : All the agricultural products assembled and graded cannot be used directly. So those products should be made useful for consumption.
  5. Sampling: Samples are to be made from the graded, standardized and processed produce. This sampling process enables the consumer to choose the best goods from the market.
  6. Packing: All the farms products whether processed or not must be packed to ensure better quality.
  7. Storing: All the farms products processed and packed cannot be sold immediately. They have to be preserved in store houses until they are sold.

Question 20.
Regulated markets. [A.P. Mar. 16]
Answer:
Regulated markets have been started with view to ensure remunerative price .to the products of farmers reduce the price spread between the producers and consumers. It check all unfair practices in the market. Most of the States implemented Act of agriculture marketing. In India 1951, 200 regulated markets were started. In 2005 the number of regulated markets increased 7521.
Functions :

  1. Fixation of charges for weighing and brokerage.
  2. Enforcing the use of standardized weights.
  3. Providing up to date information of market prices.
  4. Prevention of unauthorised deductions and under hand dealings.
  5. Market licenses are given to the middle men and their number is to be minimized.
  6. Setting of disputes among the parties arising in market operations.
  7. Warehouses and cold storages should be provided to the farmers wherever necessary.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 21.
Co-operative marketing.
Answer:
The first co-operative marketing society was started in India in 1915 which was implemented in Denmark. It was started with the purpose of giving credit to the farmers and marketing their surplus products. According to this system all the farmers of a village form together into a co-operative marketing society to sell their products at a fair price. The members of the society sell their products to the society. These societies are maintained by efficient paid staff. These societies sell their products at high prices in order to provide remunerative prices to their members.
Advantages: .

  1. Prices of farm produces will be remunerative.
  2. The co-operative marketing has its own storage and warehousing facilities. Thus it can avoid damage to agriculture products through rain, rats etc.
  3. These markets provide grading and standardizing facilities.
  4. Co-operative marketing control the flow of supplies and influences the price.
  5. These markets advance loans to the farmers and ensure them wait for better price.
  6. Co-operative marketing reduces cost of transportation.

Question 22.
Contract farming.
Answer:
Contract farming means that farming in which sale contract is made between fanners and users of farm products. Industries sugar, cotton, jute etc., have sale contracts with the farmers. The. following are the advantages of contract farming.

  1. Collective sale contracts of farmers provide better result than individual contracts.
  2. Fluctuations in the prices of farm products can be minimized and income of farmers can be maximized.
  3. Farmers will receive credit and technological support from the industries with whom they have sale contract.
  4. Prices of these products are fixed well in advance the farmers and put to effort to maintain quality of the products which have sale contracts.

Very Short Answer Questions

Question 1.
Agriculture sector.
Answer:
It is the sector which includes forestry, fishing, mining and quarrying and allied activities like animal husbandry etc., along with agriculture.

Question 2.
Agro based industries.
Answer:
Industries which depend an agriculture products for their raw materials are called agro based industries. Ex : Cotton, textile, flour mills, sugar etc., directly depend on agriculture for raw materials.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 3.
Food security.
Answer:
Food security is such a security which enables the people to have all time enough food for an active and healthy life.

Question 4.
Land reclamation.
Answer:
It means regaining the ownership on land after abolition of Zamindari system to make the land useful.

Question 5.
Cropping pattern.
Answer:
It is the pattern of utilisation of total farm land for producing different crops in a country at a point of time.

Question 6.
Perennial canals.
Answer:
These canals supply water through the year to irrigate lands as they are connected to the dams or rivers.

Question 7.
Drip irrigation.
Answer:
It is the system where water is deliver at or near root zones of plants, drop by drop.

Question 8.
Sprinkler irrigation.
Answer:
This is also called overhead irrigation system, water is piped to one or more central locations within the held and distributed by overhead high pressure sprinklers or guns.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 9.
Land reforms.
Answer:
Land reforms are the introduction of economic and non-economic changes relating to land in order to achieve social justice and agriculture development.

Question 10.
Organic farming. [A.P. Mar. 16]
Answer:
It is the farming which uses natural fertilizers and pesticides.

Question 11.
Economic holdings.
Answer:
Economic holdings is the size of holdings which provides a decant standard of giving of the members of the family.

Question 12.
Farm mechanization.
Answer:
It is the implementation of machines like tractors, pumpsets, harvesting, machines etc, in agriculture operations.

Question 13.
Consolidation of holdings.
Answer:
Consolidation of holdings means all the holdings of the village are pooled together into one unit. .

Question 14.
Co-operative farming. [A.P. Mar. 18]
Answer:
Co-operative farming means where the total land of village pooled into one unit and farmed together.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 15.
Objectives of land reforms.
Answer:
Land reforms facilitates the redistribution of land with a view to safe guard the interests of small and marginal farmers.

  1. Establishment of social justice.
  2. To provide security for the filler of soil.
  3. Assurance of equality Of status and opportunity for all sections of the rural population.

Question 16.
Zamindari system.
Answer:
Lord Comwalls introduced this system in Bengal in 1793. In this .system, the land was held by a few zamindars. They were entrusted with the responsibility of collection of revenue fixed by the British Government.

Question 17.
Ryotwari system. [A.P. Mar. 16]
Answer:
Thomas Manro introduced this system in Tamil Nadu in 1792. In this system had full rights regarding sale, transfer and leasing of land. Under this system no middle man exists between farmer and Government.

Question 18.
Occupancy tenant.
Answer:
Occupancy tenants are those tenants whom cannot be evicted by land lords until they pay rent.

Question 19.
Green Revolution.
Answer:
William S. Gand is the first economist who used the term green revolution. It was also called a “new strategy of agriculture development”. It is the result of the technological break through composed of improved irrigation facilities, better agricultural practices and mechanisation of agricultural operations.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 20.
IADP.
Answer:
Intensive Agriculture District Programme. This was introduced oh the basis of recommendations of Ford team. The Govt, provide fertilizers, pesticides, hybrid seeds etc., to all the farmers of the selected districts at a subsidize rate. Ex: West Godavari in A.P.

Question 21.
IAAP.
Answer:
Intensive Agriculture Area Programme. This was introduced in the year 1967. Under this system intensive cultivation has extended to other districts in various states.

Question 22.
HYVP.
Answer:
High Yielding Variety Programme. This was introduced in the year 1965.

Question 23.
RIDF.
Answer:
Rural Infrastructure Development Fund was created in 1995 – 96 from out short form of commercial banks landing priority sectors and agriculture.

Question 24.
Kisan Credit Cards. [A.P. Mar. 17]
Answer:
It was introduced in 1998 to facilitate the flow for crop loan by providing adequate, timely cost, effective short term loans. It also enables the farmers to purchase agriculture inputs and to draw cash for their production needs.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 25.
SGSY.
Answer:
Swama Jayanti Gram Swarozgar Yojana started on 1999. It is sponsored scheme. SGSY is a credit – cum – subsidy programme.

Question 26.
Micro Finance.
Answer:
It is the provision of finance on a small scale to the rural and urban poor.

Question 27.
Assembling.
Answer:
The process of pooling up of small surplus of individual farmers in the market of the producing area is called assembling.

Question 28.
Processing.
Answer:
Processing is the conversion of agriculture products into consumption.

Question 29.
AGMARK. [A.P. Mar. 18]
Answer:
It is simply an abbreviation for agriculture marketing which is the symbol of quality of produce.

AP Inter 2nd Year Economics Study Material Chapter 4 Agriculture Sector

Question 30.
Marketable suplus. [A.P. Mar. 18, 17]
Answer:
Marketable surplus is the available surplus for marketing after meeting all the requirements of the farmers.

Question 31.
Rythu Bazar. [A.P. Mar. 17, 16]
Answer:
It is a market where there is no existence of middle men between farmer Vs buyer.

AP Inter 2nd Year Botany Study Material Chapter 11 Biotechnology: Principles and Processes

Andhra Pradesh BIEAP AP Inter 2nd Year Botany Study Material 11th Lesson Biotechnology: Principles and Processes Textbook Questions and Answers.

AP Inter 2nd Year Botany Study Material 11th Lesson Biotechnology: Principles and Processes

Very Short Answer Questions

Question 1.
Define Biotechnology.
Answer:
It is a science which utilizes properties and uses of micro-organisms or exploits cells and the cell constituents at the industrial level for generating useful products essential to life and human welfare.

Question 2.
What are molecular scissors? Where are they obtained from?
Answer:
Molecular scissors are the restriction enzymes which cut the DNA at specific locations. Usually, they are obtained from Bacteria.

Question 3.
Name any two artificially restructured plasmids.
Answer:
PBR322, PUC19 101.

Question 4.
What is E coRI? How does it function?
Answer:
E.coRI is a restriction enzyme obtained from Escherichia coli. It specifically recognises GAA sites on the DNA and cuts it between G and A.

AP Inter 2nd Year Botany Study Material Chapter 11 Biotechnology: Principles and Processes

Question 5.
What are cloning vectors? Give an example.
Answer:
Vectors used for multiplying the foreign DNA sequences are called cloning vectors.
Ex: Plasmids, Bacteriophages, Cosmids.

Question 6.
What is recombinant DNA?
Answer:
The hybrid DNA formed by the fusion of DNA with desirable genes with the vector DNA by DNA ligase is called r-DNA.

Question 7.
What is palindromic sequence?
Answer:
The specific nucleotide sequence of DNA which is recognised by restriction enzyme is called palindrome, which is 4-6 base pairs in length.

Question 8.
What is the full form of PCR? How it is useful in Biotechnology?
Answer:
PCR stands for Polymerase chain reaction. It can be used for the diagnosis of diseases like AIDS, middle ear infection, and tuberculosis.

Question 9.
What is down stream processing?
Answer:
Separation and purification of products before they are ready for marketing is called
downstream processing.

Question 10.
How does one visualize DNA on an agar – gel?
Answer:
To make the DNA visible in the gel, Ethidium bromide is added to the gel solution and the buffer. This positively charged polycyclic compound binds to .DNA inserting itself between the base pairs. Southern blotting may also be used as visualization technique for agarose gels.

AP Inter 2nd Year Botany Study Material Chapter 11 Biotechnology: Principles and Processes

Question 11.
How can you differentiate between exonucleases and endonucleases?
Answer:

ExonucleaseEndonuclease
1. They remove nucleotides from the ends of the DNA1. They make cuts at specific locations with in the DNA

Short Answer Questions

Question 1.
Write short notes on restriction enzymes.
Answer:
Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coli were isolated in the year 1963. One of these added methyl groups to DNA and the other cut DNA. The latter was called restriction endonuclease. The first restriction endonuclease – Hind II which cut DNA molecules at a particular point by recognising a specific Sequence of six base pairs, called recognition sequence for Hind II. Today, more than 900 restriction enzymes were isolated from over 200 strains of Bacteria, each of which recognises a different recognition sequence.

E CORI is a restriction enzyme in which, the first letter comes from the Genus (Escherichia), and the second two letters from the species of the Prokaryotic cell [coli], the letter ‘R’ is derived from the name of strain. Roman numbers indicate the order in which the enzymes were isolated from that strain of Bacteria. Restriction enzymes belong to a larger class of enzymes called nucleases. They are of two types.
a) Exonucleases which remove nucleotides from the ends of the DNA.
b) Endonucleases which make cut’s at specific location with in the DNA-

Question 2.
Give an account of amplification of gene of interest using PCR.
Answer:
PCR stands for Polymerase chain reaction. In this reaction, multiple copies of the gene of interest are synthesized in vitro using two sets of primers(oligonucleotides) and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the. genomic DNA as template. As the process of replication of DNA is repeated many times, the segments of DNA can be amplified to approximately billion times.

Such repeated amplification is achieved by the use of a thermostable DNA polymerase such as Taq polymerase (isolated from thermus aquaticus) which remain active even during high temperature induced denaturation of ds DNA. The amplified fragment, if desired can: now be used to ligate with a vector for further cloning.

Question 3.
What is a bio-reactor? Describe briefly the stirring type of bio-reactor.
Answer:
Bio-reactor is a large vessel which is used for biological conversion of raw material into specific product.

A stirred-tank bio-reactor is usually cylindrical or with a curved base to facilitate the mixing of the reactor contents.

The stirrer facilitates even mixing and oxygen availability throughout the bio-reactor. Alternatively air can be bubbled through the reactor. It has an agitator system, an oxygen delivery system, a foam control system, a temperature .control system, pH control system and sampling ports, so that small volumes of the culture can be with drawn periodically.

AP Inter 2nd Year Botany Study Material Chapter 11 Biotechnology: Principles and Processes

Question 4.
What are the different methods of insertion of recombinant DNA into the host cell?
Answer:
There are several methods of introducing the ligated DNA into recipient cells. Recipient cells after making them competent to receive, take up the DNA present in their surrounding. R-DNA can be forced into such cells by incubating the cells with r-DNA on ice followed by placing them briefly at 42eC -(heat shock) and then putting them back on ice. This enables the bacteria to fake up the r-DNA.

In Micro-injection method, r-DNA is directly injected into the nucleus of an animal cell.

In Bio listic of gene gun method – cells are bombarded with high velocity micro particles of gold or tungsten coated with DNA.

In another method, ‘Disarmed pathogen’ vectors are used which when allowed to infect the cell, transfer the r-DNA into the host.

Long Answer Questions

Question 1.
Explain briefly the various processes of recombinant DNA technology.
Answer:
The important methods in recombinant DNA technology are performed through genetic engineering. They are :
i) Isolation of a desired gene,
ii) Insertion of isolated gene into a suitable vector,
iii) Introduction of recombinant vector into a host and
iv) Selection of the transformed host cells.

I) Isolation of a desired gene :

  1. The desired gene is isolated from the donor cell. Normally bacteria are the source of desired genes.
  2. The cell walls of bacteria are degraded with the help of enzymes.
  3. The cell membranes are lysed with the help of detergents.
  4. By treating the cellular constituents with phenols and suitable nucleases and by subjecting to gradient centrifugation, pure DNA is isolated.
  5. The purified DNA is cut into a number of fragments by restriction endonucleases.
  6. The restriction enzymes cleave DNA molecules in two ways.

i) In one way they cut both strands of DNA at exactly opposite points to each other. This results in DNA fragments with blunt ends or flush ends, where two strands end at the same point. Such cut is generally termed as even cut.
AP Inter 2nd Year Botany Study Material Chapter 11 Biotechnology Principles and Processes 1

ii) But commonly, most enzymes cut the two strands of DNA double helix at different locations. Such a cleavage is generally termed as staggered cut. This generates protruding ends i.e., one strand of DNA double helix extends some bases beyond the other. Since the target site is palindromic in nature, the protruding ends generated by such a cleavage have complimentary base sequence.

As a result, they readily pair with each other and such ends are called cohesive or sticky ends. This stickyness of the ends facilitates the action of the enzyme DNA ligase. When cut by the same restriction enzyme, the resultant DNA fragments have the same kind of ’sticky ends’ and these can be joined together readily by using DNA ligases. E.g.: The restriction enzyme E coRI.
E – The first letter, represents the name of genus Escherichia.
Co – The next two letters, represent the species Escherichia Coli.

The letter R is derived from the name of strain.

Roman numbers following the names indicate the order in which the enzymes were isolated from the strain of bacteria.

This enzyme specifically recognises GAA sites on the DNA and cuts it between G and A (G ↓ A)
AP Inter 2nd Year Botany Study Material Chapter 11 Biotechnology Principles and Processes 2

7) The resultant fragments are separated from each other by gel electrophoresis.
8) The desired fragments are selected by Southern blotting technique.

II) Insertion of isolated gene into a suitable vector :

  1. The selected fragments of DNA are inserted into a suitable vector to produce a large number of copies of genes. This is called gene cloning.
  2. There are two major types of vectors, namely plasmids and bacteriophages.
  3. Among the two types, plasmids are the ideal cloning vectors.
  4. To isolate a plasmid, the Bacterial cell is treated with EDTA (Ethylene diamine tetra acetic acid) along with lysozyme enzyme to digest the cell wall.
  5. Then the bacterial cell is subjected to centrifugation in sodium lauryl sulphate to separate the plasmid.
  6. The plasmid DNA is cut with the help of restriction endonuclease.
  7. The circular plasmid is converted into a linear molecule having sticky ends.
  8. The two sticky ends of linear plasmid are joined to the ends of desired gene by DNA ligase.
  9. The plasmid containing foreign DNA segments is called recombinant DNA (r DNA) or Chimeric DNA.

AP Inter 2nd Year Botany Study Material Chapter 11 Biotechnology Principles and Processes 3

III) Introduction of recombinant vector into a suitable host :

  1. The r DNA molecule is introduced into suitable bacterial host cell by transfor-mation.
  2. The cell containing r DNA is called transformed cell.
  3. Bacterial cell walls are not permeable to recombinant vectors, but keeping in dil. Calcium chloride renders the bacterial cell wall permeable to recombinant vectors.
  4. The r DNA replicates with in the host cell.
  5. The transformed cell grows on the culture medium. Each daughter cell contains r DNA.

IV) Selection of transformed host cells :
1) Selection of transformed cells depends on the nature of gene which is cloned. 2) It can be done in two ways. The are :
a) Without using probes,
b) By using probes.

a) Without using probes:

  1. If the gene is cloned for antibiotic resistance, the cells are first incubated on a medium without antibiotic for one hour, to allow the antibiotic resistance gene to be expressed.
  2. Then the cells are placed on a medium with an antibiotic for selection of colonies containing rDNA.
  3. The cells which have expressed the gene will survive and the others die.

b) By using probes:
When transformed cells are cultured on the nutrient medium, several cells are produced. To select the cells containing the desired gene colony hybridization method is used. In this gene specific probes are used. A probe is a small fragment of single stranded RNA or DNA which is tagged with radioactive, molecule. It can search out complimentary DNA sequences from an organism.

Question 2.
Give a brief account of the tools of recombinant DNA technology.
Answer:
Key tools are :
1) Restriction enzymes :
Two enzymes responsible for restricting the growth of Bacteriophage in Escherichia coli were isolated in the year 1963. One of these added methyl groups to DNA and the other cut DNA. The latter was called restriction endonuclease. The first restriction endonuclease – Hind II which cut.

DNA molecules at a particular point by recognising a specific sequence of six base pairs, called recognition sequence for Hind II. Today, more than 900 restriction enzymes were isolated from over 200 strains of Bacteria, each of which recognises a different recognition sequence.

E CORI is a restriction enzyme in which, the first letter comes from the genus (Escherichia), and the second two letters from the species of the Prokaryotic cell [coli], the letter ‘R’ is derived from the name of strain. Roman numbers indicate the order in which the enzymes were isolated from that strain of Bacteria. Restriction enzymes belong to a larger class of enzymes called nucleases. They are of two types.

a) ExOnucleases which remove nucleotides from the ends of the DNA.
b) Endonucleases which make cuts at specific location with in the DNA.

Most restriction enzymes cut the two strands of DNA double helix at different locations. Such a cleavage is know as staggered cut. E CoRI recognises 5′ GAATT 3′ sites on the DNA and cuts it between G and A results in the formation of sticky ends or cohesive end pieces. This stickyness of the ends facilitates the action of enzyme DNA ligase.

Cloning vectors :
The DNA used as a carrier for transferring a fragment of foreign DNA into a suitable host is called vector. Vectors used for multiplying the foreign DNA sequences are called cloning vectors. Commonly used cloning vectors are plasmids, bacteriophages, cosmids. Plasmids are extrachromosomal circular DNA molecules present in almost all bacterial species. They are inheritable and carry few genes are easy to isolate and reintroduce into the bacterium (Host).

Features required to facilitate cloning into a vector:
a) Origin of replication:
(ori) This is a sequence from where replication starts and any piece of DNA when linked to this sequence can be made to replicate within host cells. It is also responsible for controlling the copy number of the linked DNA.

b) Selectable marker :
In addition to ‘ori’, the vector requires a selectable marker, which helps in identifying and eliminating non-transformants and selectively permitting the growth of the any transformants normally, the genes encoding resistance to antibiotics such as ampicillin, chloramphenicol, tetracycline or kanamycin etc., are useful selectable markers for E.Coli.

c) Cloning sites :
In order to link the alien DNA, the vector needs to have very few, preferably single recognition sites for the restriction enzymes.

d) Molecular weight:
The cloning vector should have low molecular weight.

e) Vectors for cloning genes in plants and animals:
The tumour inducing (Ti) plasmid of Agrobacterium tumifaciens has now been modified into a cloning vector such that it is no more pathogenic to plants. Similarly retroviruses have also been disarmed and are now used to deliver desirable genes into animal cells.

Intext Questions

Question 1.
Do Eukaryotic cells have restriction Endonucleases? Justify your answer.
Answer:
No. only prokaryotic cells have restriction Endonucleases. Today we know 900 or more restriction enzymes have been Isolated from over 230 strains of Bacteria.

AP Inter 2nd Year Botany Study Material Chapter 11 Biotechnology: Principles and Processes

Question 2.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?
Answer:

  1. Small volumes of the culture can be withdrawn periodically.
  2. Air can be bubbled through the reactor.
  3. It has a control system that regulates the temperature and pH.

Question 3.
Can you recall Meiosis and indicate at what stage a recombinant DNA is made?
Answer:
Pachytene of Meiosis – I.

4. Describe briefly the following :
a) Origin of replication b) Bioreactors c) Down stream processing
Answer:
a) Origin of replication :
It is the specific DNA sequence which is responsible for initiating replication.

b) Bioreactors:
They are large vessels which are used for biological conversion of raw material into specific products.

c) Down stream processing :
Separation and purification of products is called down stream processing.

Question 5.
Explain briefly a) PCR b) Restriction enzymes and DNA c) Chitinase
Answer:
a) PCR:
Polymerase chain reaction: Is a biochemical technology in molecular biology to amplify a single or few copies of a piece of DNA generating thousands of copies of a particular DNA sequence.

b) Restriction enzymes and DNA :
Restriction enzymes are DNA cutting enzymes found in Bacteria because they cut with in the molecule.

c) Chitinase :
It is an enzyme obtained from the fungus, used in the digesting the bacterial cell walls.

Question 6.
Discuss with your teacher and find out how to distinguish between
a) Plasmid DNA and Chromosomal DNA
b) RNA and DNA
c) ExonuCleases and Endonucleases

a)

Plasmid DNAChromosomal DNA
1. Plasmid DNA is a small circular molecule found in Bacteria.1. It is main genetic material which is a circular, double stranded DNA
2. It carry a few genes.2. It carry several genes.

b)

RNADNA
1. It is made up of only polynucleotide chain.1. It is made up of 2 polynucleotide chains.
2. Purines or pyTamidines do not exist in 1 : 1 ratio.2. Purines and pyramidines exist in 1 : 1 ratio.
3. Nitrogen bases are Adenine, Guanine, uracil and cytosine.3. Nitrogen bases are Adenine, Guanine, Thymine and cytosine.

c)

ExonucleaseEndonuclease
1. They remove nucleotides grom the ends of the DNA.1. They make cuts at specific positions with in the DNA.

Question 7.
What does ‘H’ in’d’ and HI refer to in the enzyme Hind III?
Answer:
‘H’ – The genes from which it was isolated – Haemophilus
in – Influenza (the species name) .
d – Refers the bacterial strain from which enzyme was isolated.
III – Third isolated enzyme from the strain.

Question 8.
Restriction enzymes should not have more than one site of action in the cloning site of a vector. Comment.
Answer:
The presence of more than one recognition site for the same enzyme with in the vector will generate several fragments which will complicate the gene cloning. So vector needs to have very few, preferably single, recognition site for the commonly used restriction enzymes.

Question 9.
What does ‘competent’ refer to ‘incompetent cells’ used in transformation experiments?
Answer:
DNA is a hydrophilic molecule. It cannot pass through cell membranes. In order to force Bacteria to take up the plasmid, the bacterial cells must be made competent to take up DNA by treating them with a bivalent cation such as calcium which increaes the efficiency with which DNA enters the bacterium.

Question 10.
What is the significance of adding proteases at the time of Isolation of Genetic material (DNA)?
Answer:
Proteins can be removed by treatment with proteases.

Question 11.
While doing a PCR, ‘denaturation’ step is missed. What will be its effect on the process?
Answer:
The segment of DNA cannot be amplified.

Question 12.
What modification is done on the Ti plasmid of Agrobacterium tumefascians to convert it into a cloning vector?
Answer:
It is no more pathogenic to plants but is still able to use the mechanisms to deliver genes of our interest into a variety of plants.

Question 13.
What is meant by gene cloning?
Answer:
It is the replication of DNA fragments by the use of a self-replicating genetic material. It duplicates only individual genes of organism’s DNA.

AP Inter 2nd Year Botany Study Material Chapter 11 Biotechnology: Principles and Processes

Question 14.
Decide the ratio between ester bonds and hydrogen bonds that are broken in each palindromic sequence of DNA when treated with E coRI during the formation of sticky ends.
Answer:
1 : 4.

AP Inter 2nd Year Zoology Study Material Chapter 4(a) Endocrine System and Chemical Coordination

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 4(a) Endocrine System and Chemical Coordination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 4(a) Endocrine System and Chemical Coordination

Very Short Answer Questions

Question 1.
What is acromegaly? Name the hormone responsible for this disorder.
Answer:
Acromegaly is a hormonal disorder that results when the pituitary gland produces excess growth hormone (GH). This disease is characterized by enlargement of the bones of the jaw, hand, and feet, thickened nose, lips, eyelids, and wide fingertips, and gorilla-like appearance of the person affected.

Question 2.
Which hormone is called anti-diuretic hormone? Write the name of the gland that secretes it.
Answer:
Vasopressin is also called an anti-diuretic hormone that is secreted by the posterior pituitary.

Question 3.
Name the gland that increases in size during childhood and decreases in size during adulthood. What important role does it play in case of infection?
Answer:
The thymus is small at birth, it increases in size during childhood and reaches the maximum size at puberty. During adulthood, it shrinks to its size at birth.

In an old person thymus gland is degenerated, resulting in a decreased production of thymosin. Thymosin plays an important role in immune development. Thus, the immune response against infections of old people becomes weak.

Question 4.
Distinguish between diabetes insipidus and diabetes mellitus.
Answer:
Diabetes insipidus :
Deficiency of Vasopressin causes a disease called diabetes insipidus. It does not involve loss of sugar in urine.

Diabetes mellitus :
Under secretion of insulin by the pancreatic gland increases the level Of glucose in blood is called hyperglycemia. Prolonged hyperglycemia leads to a disease called diabetes mellitus, associated with loss of glucose through urine and formation of ketone bodies.

AP Inter 2nd Year Zoology Study Material Chapter 4(a) Endocrine System and Chemical Coordination

Question 5.
What are Islets of langerhans?
Answer:
The endocrine region of pancreas is called Islets of langerhans where it contain 1 to 2 millions Islets of langerhans. There are two main types of cells α – cells and β – cells.

α – cells produce the hormone glucagon, whereas β – cells produce insulin.

Question 6.
What is insulin shock?
Answer:
Hyper secretion of insulin leads to decreased level of glucose in blood (hypoglycemia) resulting in insulin shock.

Question 7.
Which hormone is commonly known as fight and flight hormone?
Answer:
Epinephrine and norepinephrine hormones are called fight and flight hormones because these hormones are secreted in response to stress and emergency situations.

Question 8.
What are androgens, which cells secrete them?
Answer:
Androgens are male sex hormones usually steroid hormones. E. g: testosterone.
Androgens are produced by the Leydig cells of the testes and to a minor extent by the adrenal glands in both sexes.

AP Inter 2nd Year Zoology Study Material Chapter 4(a) Endocrine System and Chemical Coordination

Question 9.
What is erythropoietin? What is its function?
Answer:
Erythropoietin is a hormone secreted the juxtaglomerular cells of the kidney. It plays an important role in the erythropoiesis i.e., in formation of RBC. Erythropoietin controls the formation of RBC by regulating the differentiation and proliferation of erythroid progenitor cells in the bone marrow.

Short Answer Questions

Question 1.
List out the names of the endocrine glands present in human beings and mention the hormone they secrete.
Answer:
1) Hypothalamus:
It secretes thyrotropin releasing hormone, corticotropin releasing hormone, gonadotropin releasing hormone, growth hormone releasing hormone, growth hormone release inhibiting hormone, prolactin release inhibiting hormone.

2) Pituitary glands :
Anatomically, it is divided into anterior and posterior pituitary.

Anterior Pituitary :
Produces Growth hormone, Prolactin, Thyroid stimulating hormone, Adreno corticotropic hormone, Follicular stimulating hormone, Luteinizing hormone.

Posterior Pituitary:
It releases two hormones namely Oxytocin and Vasopressin/ADH.

3) Pineal gland :
It secretes a hormone called Melatonin.

4) Thyroid gland :
It produces two hormones namely Thyroxine (T4) and Tri iodothyronine (T3).

5) Parathyroid gland :
Secretes a hormone called Parathyroid hormone.

6) Thymus gland :
It secretes peptide hormone called Thymosin.

7) Adrenal gland:
a) Adrenal cortex: GluCo corticoids, Mineralo corticoids, Androgens and Estrogens.
b) Adrenal medulla : Produces Epinephrine, norepinephrine.

8) Pancreas:
It secretes Glucagon and Insulin.

9) Testes :
Which secrete Androgens and Testosterone.

10) Ovaries :
Which produce Estrogen and Progesterone.

Question 2.
Describe the role of hypothalamus as a neuroendocrine organ. ‘
Answer:
Hypothalamus is located below the thalamus. It connects the neural and endocrine systems, as it closely tied to the pituitary gland. It responds to the sensory impulses received from different receptors by sending out appropriate neural or endocrine signals.

The hypothalamus is the master control centre of the endocrine system, as it contains several group of neuro secretary cells called nuclei, which produce hormones, called neuro hormones; These hormones directly control the pituitary gland which in turn secrete hormone that regulate the growth and functioning of other endocrine glands.

For example :
The two types of hormones produced by the hypothalamus are :
1) Releasing hormones :
Which stimulates the secretions of pituitary hormones.
Eg: 1) Thyrotropin releasing hormone – acts on anterior pituitary to release thyroid stimulating hormone.
2) Growth hormone releasing hormone – stimulate the release of growth hormone.

2) Inhibiting hormones :
Which inhibits the secretion of pituitary hormones.
Eg: 1) Growth hormone release inhibiting hormone – which inhibit the release of growth hormone from anterior pituitary.
2) Prolactin release inhibiting hormone – Inhibit the release of prolactin from anterior pituitary.

Question 3.
Give an account of the secretions of pituitary gland.
Answer:
The pituitary gland is also called hypophysis. Anatomically pituitary gland is divided into anterior and posterior pituitary.
I. Anterior Pituitary :
It produces six important peptides. They are ;
1) Growth hormone (GH) Somatotropin :
They promote growth of the entire body by increasing protein synthesis, cell division and cell differentiation.

2) Prolactin:
It causes enlargement of the mammary glands of the breast and initiate the maintenance of lactation in mammals. Prolactin also promote the growth of corpus luteum and stimulate the production of progesterone.

3) Thyroid stimulating hormone” (TSH) :
It stimulates the production of thyroid hormones from thyroid gland.

4) Adreno corticotropic hormone (ACTH) :
Controls the production of steroid hormones called gluco corticoids, by the adrenal cortex.

5) Follicle stimulating hormone (FSH) :
It stimulates growth the development of the ovarian follicles in females. In males FSH along with the androgens, regulates spermatogenesis.

6) Luteinizing hormone (LH) :
In males it stimulates production of androgens. In females it stimulates the ovaries to produce estrogens and progesterone and it maintains corpus luteum.

II. Posterior pituitary :
It stores and releases two hormones called oxytocin and vasopressin.

Oxytocin :
In females it stimulates contraction of pregnant uterus during child birth and ejection of milk from the mammary gland.

Vasopressin (ADH) :
Affects the kidney and stimulates reabsorption of water and electrolytes by the DCT and collecting duct.

AP Inter 2nd Year Zoology Study Material Chapter 4(a) Endocrine System and Chemical Coordination

Question 4.
Compare a pituitary dwarf and a thyroid dwarf in respect of similarities and dismilarities they posses.
Answer:

Pituitary dwarfThyroid dwarf
1. Hypersecretion of growth hormone from pituitary during childhood retards growth resulting in pituitary dwarf.1. Hyposecretion of thyroid hormones during pregnancy, defective development of baby i.e., physical and mental growth get severely stunted, resulting in thyroid dwarf.
2. Human growth hormone deficiency results in abnormally slow growth and short structure with normal proportion.2. Deficiency of thyroid hormones by birth results in enlarged head, short limbs, puff eyes, a thick and protudding tongue, dry skin, tow. I Qetc.
3. The pituitary dwarf is sexually and intellectually a normal individual.3. If the condition not treated the child will grow up dwarf, mentally retarded and sexually sterile.
4. Administration of purified HGH has been shown to induce skeletal growth in these patients.4. Early treatment can result in normal growth and development.

Question 5.
Explain how hypothyroidism and hyperthyroidism can affect the body.
Answer:
Hypothyroidism:
Inadequate supply of iodine or impairment in the function of thyroid glands leads to decrease in production of thyroid hormones (T3 & T4) results in hypothyroidism and enlargement of the thyroid gland called Simple goiter.

During pregnancy due to hypothyroidism, defective development of the growing body leads to a disorder called Cretinism. Physical and mental growth gets severely stunted due to untreated congenital hypothyroidism, stunted growth, mental retardation, low IQ, deafness, and mutism are some characteristics features of this disease.

In adult women it may cause irregular menstrual cycles. Hypothyroidism in adult causes Myxoedema characterized by bagginess under the eyes, puffiness of face, dry skin, slowness in physical and mental activities.

Hyperthyroidism:
Over activity of thyroid, cancer of the gland or development of nodule of thyroid lead to hyper thyroidism. In adults it causes an abnormal growth leads to a disease called Exophthalmic goiter with characteristically protruded eyeballs. Hyperthyroidism also affects the physiology of the body i.e., increased metabolic rate, nervousness, rapid heartbeat, sweating, increased appetite etc.

Question 6.
Write a note on Addison’s disease and Cushing’s Syndrome.
Answer:
Addison’s disease: It is caused due to hyposecretion of glucocorticoids by the adrenal cortex. This disease is characterised by loss of weight, muscle weakness, fatigue and reduced blood pressure. Sometimes darkening of the skin in both exposed and non-exposed parts of the body occurs in this disorder.

Cushing’s Syndrome :
It results due to over production of glucocorticoids. This condition is characterised by breakdown of muscle proteins and redistribution of body fat resulting in spindly arms and legs, a round moon-face, buffalo hump on the back and pendulous abdomen is also observed. Wound healing is poor. The elevated level of cortisols causes hyperglycemia, over deposition of glycogen in liver and rapid gain of weight.

Question 7.
Why does sugar appear in the urine of a diabetic?
Answer:
Hyposecretion of insulin of pancreatic gland increases the level of glucose in blood called hyperglycemia. Prolonged hyperglycemia leads to a disease called diabetes mellitus.

In diabetic patients glucose or sugar appears in urine because kidney plays a special role in the homeostasis of blood glucose.’Glucose is continuously filtered by the glomeruli, reabsorbed and returned to the blood. If the level of glucose in blood is above 160 -180 mg/dl. i.e., in hyperglycemia condition glucose in primary urine is not completely reabsorbed, and returned to the blood. Some of which is retained and excreted in urine.

Question 8.
Describe the male and female sex hormones and their actions.
Answer:
The hormones, which are responsible for the development of secondary sexual characters and changes in different stages of life are called sex hormones.

Male sex hormones :
Androgens :
Androgens are produced by the Leydig cells of the testes and to a minor extent by the adrenal glands in both sexes.

Functions :
→ Growth, development and maintenance of male reproductive organs.
→ Sexual differentiation and secondary sexual characteristics.
→ Spermatogenesis.
→ Male pattern of aggressive behaviour.
→ Increases the protein synthesis and increases the glycolysis.

Female sex hormones:
1) Estrogens :
Synthesized by the follicles and corpus luteum of ovary.
Functions :
→ Development and maintenance of female reproductive organs.
→ Maintenance of menstrual cycle.
→ Development of secondary sexual characters.
→ Estrogen promotes the protein synthesis and calcification and bone growth.

2) Progesterone :
It is synthesized and secreted by corpus luteum and placenta. Functions: required for implantation of fertilised ovum and maintenance of pregnancy.

3) Follicle stimulating and Lutenizing hormones :
Both these hormones produced from anterior pituitary gland in both sexes.

Functions:
Both these hormones play an important role in secondary sexual characters in both sexes.

AP Inter 2nd Year Zoology Study Material Chapter 4(a) Endocrine System and Chemical Coordination

Question 9.
Write a note on the mechanism of action of hormones.
Answer:
Hormones are primary messengers which interacting with receptors and they generate secondary messengers. These secondary messengers regulate cellular metabolism in the target cells.

Mechanism of action of lipid insoluble hydrophillic hormone :
→ The Hormone binds to a stimulatory membrane bound receptor, and stimulate ‘G’protein.
→ ‘G’ protein of the cell membrane binds to GTP and activates adenylate cyclase.
→ Adenylate Cyclase forms cAMP from ATP.
→ cAMP activates the protein kinase, which activates the enzyme phosphorylase.
→ Phosphorylase further phosphorylate the inactive enzyme and convert it to active form and involved in the metabolic process. Eg : Epinephrine.

Mechanism of action of lipid soluble hormone:
Lipid soluble hormones easily diffuse through the cell membrane.
→ It binds to a specific receptor in the cytoplasm forming hormone receptor complex molecule.
→ This complex molecule enters the nucleus and binds to the DNA and stimulate the production of specific m-RNA molecule.
→ The m-RNA passes into the cytoplasm, where it is involved in the translation process and synthesizes a protein. These proteins produced by the cell as a response of hormone and plays an important role in their respective metabolism.
Eg : Aldosterone

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Students get through AP Inter 2nd Year Physics Important Questions 8th Lesson Magnetism and Matter which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 8th Lesson Magnetism and Matter

Very Short Answer Questions

Question 1.
A magnetic dipole placed in a magnetic field experiences a net force. What can you say about the nature of the magnetic field?
Answer:
The nature of the magnetic field is uniform, magnetic dipole (bar magnet) experiences a net force (or torque).

Question 2.
Do you find two magnetic field lines intersecting? Why?
Answer:
Two magnetic field lines never intersect. If they intersect, at the point of intersection the field can have two directions. This is impossible. So, two field lines never intersect.

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 3.
What happens to the compass needles at the Earth’s poles? [T.S. Mar. 17; IPE 2014]
Answer:
At the Earth poles, the magnetic field lines are converging or diverging, vertically so that the horizontal component is negligible. Hence, the compass needle can point in any direction.

Question 4.
What do you understand by the ‘magnetisation’ of a sample ? Give its SI unit. [IPE 2016 (AP)]
Answer:
Magnetisation (M) of a sample is equal to its net magnetic moment per unit volume.
Magnetisation, M = \(\frac{m_{n e t}}{V}\), SI unit of magnetisation is Am-1.

Question 5.
What is the magnetic moment associated with a solenoid ? [IPE 2016 (TS)]
Answer:
The magnitude of magnetic moment of the solenoid is, M = n × 2l × i × πa2
where, ‘n’ is number of turns, ‘2l’ is length of the solenoid, ‘i’ is current passing through coil, and ‘a’ is area of cross section of solenoid.

Question 6.
What are the units of magnetic moment, magnetic induction and magnetic field ? [IPE 2016 (AP), (TS)]
Answer:
The unit of magnetic moment (M) is ampere-meter2 (Am2).
The unit of magnetic induction (B) is tesla (T) or N/A-m.
The unit of magnetic field (B) is tesla.

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 7.
Magnetic lines form continuous closed loops. Why ? [T.S. 2017; IPE 2016(AP)]
Answer:
Magnetic lines are imaginary lines. Within the magnet, they move from south pole to north pole and outside the magnet they move from north pole to south pole. Hence, magnetic lines form continuous closed loops.

Question 8.
Define magnetic declination. [IPE 2016 (TS)]
Answer:
Magnetic declination at a place is the angle between magnetic meridian and geographic meridian at that place.

Question 9.
Define magnetic inclination or angle of dip. [A.P. Mar. 17; A.P. & T.S. 2015 (TS), 14]
Answer:
Magnetic inclination at a place is the angle between direction of total strength of earth’s magnetic field and horizontal line in magnetic meridian.

Question 10.
Classify the following materials with regard to magnetism: Manganese, Cobalt, Nickel, Bismuth, Oxygen, Copper. [T.S. 2015, 2016 (TS); A.P. Mar. 17]
Answer:
Manganese …………. Paramagnetic
Cobalt ………….. Ferromagnetic
Nickel …………….. Ferromagnetic
Bismuth ……………. Diamagnetic
Oxygen ………………. Paramagnetic
Copper …………………. Diamagnetic

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 11.
The force between two magnet poles separated by a distance ‘d’ in air is ‘F’. At what distance between them does the force become doubled ?
Answer:
Force between two magnetic poles, F1 = F;
Distance between two magnetic poles, d1 = d
Force between two magnetic poles increased by double F2 = 2F
Distance between two magnetic poles, d2 = ?
From Coulombs law, F1d12 = F2 d22
Fd2 = 2 F d22
⇒ d22 = \(\frac{\mathrm{d}^2}{2}\)
d2 = \(\frac{\mathrm{d}}{\sqrt{2}}\)

Question 12.
If B is the magnetic field produced at the centre of a circular coil of one turn of length L carrying current I then what is the magnetic field at the centre of the same coil which is made into 10 turns ?
Answer:
For first circular coil; B1 = B, n1 = 1; I1 = I; a1 = \(\frac{\mathrm{L}}{2 \pi}\)
For second circular coil, B2 = ? n2 = 10; I2 = I; a2 = \(\frac{\mathrm{L}}{2 \pi}\)
As B = \(\frac{\mu_0 \mathrm{n} \mathrm{Ia}^2}{2 \mathrm{r}}\), B ∝ n.
\(\frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{n}_2}{\mathrm{n}_1}\)
\(\frac{\mathrm{B}_2}{\mathrm{~B}}=\frac{10}{1}\)
∴ B2 = 10 B

Question 13.
If the number of turns of a solenoid is doubled, keeping the other factors constant, how does the magnetic field at the axis of the solenoid change ?
Answer:
B1 = B (say); n1 = n; n2 = 2n; B2 = ?
Magnetic field at the centre of a solenoid is given by B = \(\frac{\mu_0 \mathrm{nI} \mathrm{a}^2(2 l)}{2 \mathrm{r}^3}\)
As I, a2, 2l and r are constants, B ∝ n
⇒ \(\frac{\mathrm{B}_2}{\mathrm{~B}_1}=\frac{\mathrm{n}_2}{\mathrm{n}_1} \Rightarrow \frac{\mathrm{B}_2}{\mathrm{~B}}=\frac{2 \mathrm{n}}{\mathrm{n}}\)
∴ B2 = 2B

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 14.
A closely wound solenoid of 800 turns and area of cross section 2.5 × 10-4 m2 carries a current of 3.0A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment ?
Answer:
Here n = 800, a = 2.5 × 10-4 m2, I = 3.0 A
A magnetic field develop along the axis of the solenoid. Therefore the current carrying solenoid behaves like a bar magnet
m = N IA = 800 × 3.0 × 2.5 × 10-4
= 0.6 Am2 along the axis of solenoid.

Short Answer Questions

Question 1.
Compare the properties of para, dia and ferromagnetic substances.
Answer:
Diamagnetic substances
a) When these materials placed in a magnetic field, they are magnetised feebly in the opposite direction to the applied external field.
b) When a rod of diamagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the perpe-ndicular direction to the magnetic field.
c) When they are kept in a non-uniform magnetic field, they move from the region of greater field strength to the region of less field strength.
d) The relative permeability is less than 1. μr < 1 and negative.
e) The susceptibility (χ) is low and negative.
E.g.: Copper, Silver, Water, Gold, Antimony, Bismuth, Mercury, Quartz Diamond etc.

Paramagnetic substances
a) When these materials placed in a magnetic field, they are magnetised feebly in the direction of the applied magnetic field.
b) When a rod of paramagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the direction of the applied magnetic field.
c) When they are kept in a non-uniform magnetic field„they move from the region of less field strength to the region of greater field strength.
d) The relative permeability is greater than 1. μr > 1 and positive.
e) The susceptibility (χ) is small and positive.
E.g.: Aluminium, Magnesium, Tungsten, Platinum, Mang-anese, liquid oxygen, Ferric chloride, Cupric chloride etc.

Ferromagnetic substances
a) When these materials placed in a magnetic field,they are magnetised strongly in the direction of the applied external field.
b) When a rod of ferromagnetic material is suspended freely in a uniform magnetic field, it comes to rest in the direction of the applied magnetic field.
c) When they are kept in a non-uniform magnetic field they move from the regions of lesser (magnetic field) strength to the regions of stronger (magnetic field) strength.
d) The relative permeability is much greater than 1. μr >> 1 and positive.
e) The susceptibility (χ) is high and positive.
E.g.: Iron, Cobalt, Nickel, Gadolinium and their alloys.

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 2.
Explain the elements of the Earth’s magnetic field and draw a sketch showing the relationship between the vertical component, horizontal component and angle of dip.
Answer:
The magnetic field of the earth at a point on its surface can be specified by the declination D, the angle of dip or the inclination I and the horizontal component of the earth’s field HE. These are known as the elements of the earth’s magnetic field.
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 1
Explanation:

  1. The total magnetic field at P can be resolved into a horizontal component HE and a vertical component ZE.
  2. The angle that BE makes with HE is the angle of dip, I.
  3. Representing the vertical component by ZE, we have
    ZE= BE Sin I
    HE = BE Cos I
    Which gives Tan I = \(\frac{\mathrm{Z}_{\mathrm{E}}}{\mathrm{H}_{\mathrm{E}}}\)

Question 3.
Define magnetic susceptibility of a material. Name two elements one having positive susceptibility and other having negative susceptibility.
Answer:

  1. Susceptibility: When a material is placed in a magnetic field, the ratio of the intensity of magnetization acquired by it to the intensity of the applied magnetic field is called its susceptibility.
    Suspectibility χ = \(\frac{\text { Intensity of magnetisation, } \mathrm{I}}{\text { Applied magnetic field, } \mathrm{H}}\)
  2. The susceptibility of a material represents its ability to get magnetism.
  3. Susceptibility is a dimension less quantity.
  4. Relation between μr and χ :
    a) Suppose that material is placed in a magnetic field of intensity H. Let I be the intensity of magnetisation acquired by it.
    b) Then the magnetic induction within the material is
    B = μ0H + μ0I ⇒ \(\frac{\mathrm{B}}{\mathrm{H}}\) = μ0[1 + \(\frac{\mathrm{I}}{\mathrm{H}}\)]
    ⇒ μ = μ0[1 + χ] ⇒ \(\frac{\mu}{\mu_0}\) = 1 + χ
    ∴ μr = 1 + χ [∵μr = \(\frac{\mu}{\mu_0}\)]
  5. Negative susceptibility (χ) of diamagnetic elements are Bismuth (-1.66 × 10-5) and copper (-9.8 × 10-6).
  6. Positive susceptibility of paramagnetic elements are Aluminium (2.3 × 10-5) and oxygen at STP (2.1 × 10-6).
  7. Large and positive susceptibility of Ferromagnetic elements are Cobalt and Nickel.

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 4.
Derive an expression for magnetic field induction on the equatorial line of a barmagnet. [Board Model Paper]
Answer:
At a point on equatorial line: Let us consider a point ‘P’ at a distance ‘d’ on the equatorial line from the centre of a bar magnet.
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 2
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 3

Qeustion 5.
What do you understand by “hysteresis” ? How does this propertry influence the choice of materials used in different appliances where electromagnets are used ?
Answer:

  1. Cycle of magnetisation : When a ferromagnetic specimen is slowly magnetised, the intensity of magnetisation varies with magnetic field through a cycle is called cycle of magnetisation.
  2. Hysterisis : The lagging of intensity of magnetisation (I) and magnetic induction (B) behind magnetic field intensity (H) when a magnetic specimen is subjected to a cycle of magnetisation is called hysterisis.
  3. Retentivity : The value of I for which H = 0 is called retentivity or residual magnetism.
  4. Coercivity: The value of magnetising force required to reduce I is zero in reverse direction of H is called coercive force or coercivity.
  5. Hysterisis curve : The curve represents the relation between B or I of a ferromagnetic material with magnetising force or magnetic intensity H is known as Hysterisis curve.
  6. Explanation of hysterisis loop or curve :
    a) In fig, a closed curve ABCDEFA in H – I plane, called hysteris loop is shown in fig.
    AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 4
    b) When ferromagnetic specimen is slowly magnetised, I increases with H.
    c) Part OA of the curve shows that I increases with H.
    d) At point A, the value I becomes constant is called saturation value.
    e) At B, I has some value while H is zero.
    f) In fig. BO represents retentivity. and OC represents coercivity.
  7. Uses : The properties of hysterisis curve, i.e., saturation, retentivity, coercivity and hysterisis loss help us to choose the material for specific purpose.
    1. Permanent magnets : A permanent magnet should have both large retentivity and large coercivity. Permanent magnets are used in galvanometers, voltmeres, ammeters, etc.
    2. An electromagnet core : The electromagnet core material should have maximum induction field B even with small fields H, low hysterisis loss and high initial permeability.
    3. Transformer cores, Dynamocore, Chokes, Telephone diaphragms: The core material should have high initial permeability, low hysterisis loss and high specific resistance to reduce eddy currents. Soft iron is the best suited material.

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 6.
Prove that a bar magnet and a solenoid produce similar fields.
Answer:
Bar magnet produce similar field of Solenoid :

  1. We know that the current loop acts as a magnetic dipole. According to Ampere’s all magnetic phenomena can be explained in terms of circulating currents.
  2. Cutting a bar magnet is like a solenoid. We get two similar solenoids with weaker magnetic properties.
  3. The magnetic field lines remain continuous, emerging from one face of solenoid and entering into other face of solenoid.
  4. If we were to move a small compass needle in the neighbourhood of a bar magnet and a current carrying solenoid, we would find that the deflections of the needle are similar in both cases as shown in diagrams.
    AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 5
    AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 6
    The axial field of a finite solenoid in order to demonstrate its similarity to that of a bar magnet
  5. The magnetic field at point P due to bar magnet in the form of solenoid is B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 m}{r^3}\)
  6. The total magnetic field, at a point P due to solenoid is given by
    B = \(\frac{\mu_0 \mathrm{n} \mathrm{I}}{2} \frac{\mathrm{a}^2}{\mathrm{r}^3}(2 l)=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{n}(2 l) \mathrm{I} \pi \mathrm{a}^2}{\mathrm{r}^3}\)
  7. The magnitude of the magnetic moment of the solenoid is, m = n(2l) I (πa2).
    ∴ B = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{~m}}{\mathrm{r}^3}\)
  8. Therefore, magnetic moment of a bar magnet is equal to magnetic moment of an equivalent solenoid that produces the same magnetic field.

Question 7.
A small magnetic needle is set into oscillations in a magnetic field B obtain an expression for the time period of oscillation.
Answer:
Expression for time period of oscillation :

  1. A small compass needle (magnetic dipole) of known magnetic moment m and moment of Inertia i is placing in uniform magnetic field B and allowing it to oscillate in the magnetic field.
  2. This arrangement is shown in Figure.
  3. The torque on the needle is τ = m × B
  4. In magnitude τ = mB sin θ.
    Here τ is restoring torque and θ is the angle between m and B.
  5. Therefore, in equilibrium i \(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) = – mB sinθ. Negative sign with mB sin0 implies that restoring torque is in opposition to deflecting torque.
    AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 7
  6. For small values of o in radians, we approximate sinθ ≃ θ and get \(i \frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) ≃ – mBθ
    \(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2} \approx \frac{-\mathrm{mB}}{\mathcal{j}} \theta\) …………….. (1)
    This represents a simple harmonic motion. .
  7. From defination of simple harmonic motion, we have \(\frac{\mathrm{d}^2 \theta}{\mathrm{dt}^2}\) = – ω2θ …………… (2)
    From equation (I) and (II), we get ⇒ ω2 = \(\frac{\mathrm{mB}}{\mathcal{J}}\)
    ∴ ω = \(\sqrt{\frac{\mathrm{mB}}{\mathcal{J}}}\)
  8. Therefore, the time period is T = \(=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathcal{J}}{\mathrm{mB}}}\)

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Qeustion 8.
A bar magnet, held horizontally, is set into angular oscillations in the Earth’s magnetic field. It has time periods T1 and T2 at two places, where the angles of dip are θ1 and θ2 respectively. Deduce an expression for the ratio of the resultant magnetic fields at the two places.
Answer:

  1. Suppose, the resultant magnetic fields is to be compared at two places A and B.
  2. A barmagnet, held horizontally at A and which is set into angular oscillations in the Earth’s magnetic field.
  3. Let time period of a bar magnet at place A’ is T1 and angular displacement or angle of dip is θ1.
  4. As the bar magnet is free to rotate horizontally, it does nqt remain vertical component (B1 sin θ1). It can have only horizontal component (B1 cosθ1).
  5. The time period of a bar magnet in uniform magnetic field is given by T = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{mB}_{\mathrm{H}}}}\)
    AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 8
  6. Now, in this case T = T1 and BH = B1Cosθ1
  7. Therefore time period of a bar magnet at place ‘A’ is given by
    T1 = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{mB}_1 \cos \theta_1}}\) …………… (1) Where I is moment of Inertia of a barmagnet and m is magnitude of magnetic moment.
  8. Similarly, the same bar magnet is placed at B and which is set into angular oscillations in the earth’s magnetic field.
  9. Let time period of a bar magnet at place B is T2 and angle of dip is θ2.
  10. Since horizontal component of earths field at B is BH = B2 cos θ2, time period,
    T2 = 2π \(\sqrt{\frac{\mathrm{I}}{\mathrm{mB}_2 \cos \theta_2}}\) ………………… (2)
  11. Dividing equation (1) by equation (2), we get \(\frac{T_1}{T_2}=\sqrt{\frac{\mathrm{mB}_2 \cos \theta_2}{\mathrm{mB}_1 \cos \theta_1}}\)
    Squaring on both sides, we have \(\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2}=\frac{\mathrm{B}_2 \cos \theta_2}{\mathrm{~B}_1 \cos \theta_1}\)
  12. But B1 = μ0H1, and B2 = μ0H2
    \(\frac{\mathrm{T}_1^2}{\mathrm{~T}_2^2}=\frac{\mu_0 \mathrm{H}_2 \cos \theta_2}{\mu_0 \mathrm{H}_1 \cos \theta_1} \)
  13. Therefore, \(\frac{\mathrm{H}_1}{\mathrm{H}_2}=\frac{\mathrm{T}_2^2 \cos \theta_2}{\mathrm{~T}_1^2 \cos \theta_1}\)
  14. By knowing T1, T2 and θ1, θ2 at different places A and B, we can find the ratio of resultant magnetic fields.

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 9.
Obtain Gauss’ Law for magnetism and explain it.
Answer:
Gauss law for Magnetism :

  1. According to Gauss’s law for magnetism, the net magnetic flux (ΦB) through any closed surface is always zero.
  2. The law implies that the no. of magnetic field lines leaving any closed surface is always equal to the number of magnetic field lines entering it.
  3. Suppose a closed surface S is held in a uniform magnetic field B. Consider a small vector area element ∆S of this surface as shown in figure.
  4. Magnetic flux through this area element is defined as ∆ΦB = B. ∆S. Then the net flux ΦB, is,
    ΦB = \(\sum_{\text {all }} \Delta \phi_B=\sum_{\text {all }} \text { B. } \Delta \mathrm{S}=0\)
  5. If the area elements are really small, we can rewrite this equation as
    ΦB = \(\oint\)B.ds = 0 …………………. (I)
    AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 9
  6. Comparing this equation with Gauss’s law of electrostatics i.e., electric flux through a closed surface S is given by
    ΦE = \(\oint \text { E. } \Delta S=\frac{q}{\varepsilon_0}\) …………….. (II) Where q is the electric charge enclosed by the surface.
  7. In an electric dipole were enclosed by the surface equal and opposite charges in the dipole add upto zero. Therefore, ΦE would be zero.
  8. The fact that ΦB = 0 indicates that the simplest magnetic element is a dipole or current loop.
  9. The isolated magnetic poles, called magnetic monopoles are not known to exist.
  10. All magnetic phenomena can be explained interms of an arrangement of magnetic dipoles and /or current loops.
  11. Thus corresponding to equation (II) of Gauss’s theorem in electrostatics, we can visualize equation (I) as
    ΦE = \(\int_S\) B . dS = μ0 (m) + μ0 (-m) = 0 where m is strength of N-pole and -m is strength of
    S – pole of same magnet.
  12. The net magnetic flux through any closed surface is zero.

Question 10.
What are ferromagnetic materials ? Give examples. What happens to a ferromagnetic material at curie temperature ? [IPE 2015 (TS)]
Answer:
Ferro magnetic substances : (a) These are strongly attracted by magnet, (b) Susceptibility is large, positive and temperature dependent, (c) Relative permeability, μr > > 1 (d) Atoms have permanent dipole moments which are organised in domains. Ex: Iron, Cobalt, Nickel

Curie temperature : The temperature above which a ferro magnetic substance changes in to para magnetic substance changes in to para magnetic substance is called curie temperature.

Problems

Question 1.
A coil of 20 turns has an area of 800 mm2 and carries a current of 0.5A. If it is placed in a magnetic field of intensity 0.3T with its plane parallel to the field, what is the torque that it experiences ? ,
Answer:
n = 20; A = 800 mm2 = 800 × 10-6 m2; i = 0.5A; B = 0.3T; θ = 0°.
When the plane parallel to the field,
T = Bin A cos θ = 0.3 × 0.5 × 20 × 800× 10-6 × cos 0°
∴ τ = 2.4 10-3 Nm

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 2.
In the Bohr atom model the electrons move around the nucleus in circular orbits. Obtain an expression the magnetic moment (p) of the electron in a Hydrogen atom in terms of its angular momentum L.
Answer:
Consider an electron of charge e, moves with constant speed v in a circular orbit of radius ‘r’ in Hydrogen atom as shown in fig.
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 10
The current constitute by revolving electron in circular motion around a nucleus, I = \(\).
Time period of orbiting electron, T = \(\frac{2 \pi \mathrm{r}}{\mathrm{v}}\) ⇒ I = \(\frac{e}{\frac{2 \pi r}{v}}=\frac{e v}{2 \pi r}\)
orbital magnetic moment, μ = IA = I (πr2)
⇒ μ = \(\frac{\mathrm{ev}}{2 \pi \mathrm{r}}\) (πr2) = \(\frac{\mathrm{evr}}{2}\)
μ = \(\frac{\mathrm{e}}{2 \mathrm{~m}}\) (mvr) [∵ Multiplying and dividing with ‘m’ on right side]
∴μ = \(\frac{\mathrm{e}}{2 \mathrm{~m}}\) L where L = mvr = angular momentum.

Qeustion 3.
A bar magnet of length 0.1m and with a magnetic moment of 5Am2 is placed in a uniform a magnetic field of intensity 0.4T, with its axis making an angle of 60° with the field. What is the torque on the magnet ?
Answer:
Given, 2l = 0.1m; m = 5A – m2; B = 0.4T; θ = 60°.
Torque, T = mB sin θ = 5 × 0.4 × sin 60° = 2 × \(\frac{\sqrt{3}}{2}\)
∴ T = 1.732 N – m

Question 4.
A solenoid of length 22.5 cm has a total of 900 turns and carries a current of 0.8 A. What is the magnetising field H near the centre and far away from the ends of the solenoid ?
Answer:
l = 22.5 cm = 22.5 × 10-2 m = \(\frac{45}{2}\) × 10-2m
N = 900; I = 0.8A; H = ?
H = \(\frac{\mathrm{NI}}{l}=\frac{900 \times 0.8}{\left(\frac{45}{2}\right) \times 10^{-2}}\)
H = \(\frac{900}{45}\) × 0.8 × 102 × 2
∴ H = 3200 Am-1

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Qeustion 5.
The horizontal component of the earth’s magnetic field at a certain place is 2.6 × 10-5T and the angle of dip is 60°. What is the magnetic field of the earth at this location ?
Answer:
Given HE = 2.6 × 10-5T;
D (or) δ = 60°
BE = \(\frac{\mathrm{H}_{\mathrm{E}}}{\cos \mathrm{D}}=\frac{2.6 \times 10^{-5}}{\cos 60^{\circ}}=\frac{2.6 \times 10^{-5}}{(1 / 2)}\) = 5.2 × 10-5 T
∴ BE = 5.2 × 10-5 T

Question 6.
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26 G and the dip angle is 60°. What is the magnetic field of the earth at this location ?
Solution:
It is given that HE = 0.26 G. From Fig., we have
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 11
The earth’s magnetic field, BE, its horizontal and vertical components. HE and ZE. Also shown are the declination, D and the inclination or angle of dip, I.
cos 60° = \(\frac{\mathrm{H}_{\mathrm{E}}}{\mathrm{B}_{\mathrm{E}}}\)
BE = \(\frac{\mathrm{H}_{\mathrm{E}}}{\cos 60^{\circ}}\)
= \(\frac{0.26}{(1 / 2)}\) = 0.52 G

Qeustion 7.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 8.0 cm at a distance of 50 cm from its mid-point ? The magnetic moment of the bar magnet is 0.40 A m2.
Solution:
From Eq.
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 12

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 8.
The earth’s magnetic field at the equator is approximately 0.4 G. Estimate the earth’s dipole moment. .
Solution:
The equatorial magnetic field is,
We are given that BE ~ 0.4 G = 4 × 10-5 T. For r, we take the radius of the earth 6.4 × 106 m.
Hence,
m = \(\frac{4 \times 10^{-5} \times\left(6.4 \times 10^6\right)^3}{\mu_0 / 4 \pi}\) = 4 × 102 × (6.4 × 106)30/4π = 10-7)
= 1.05 × 1023 A m2
This is close to the value 8 × 1022 A m2 quoted in geomagnetic texts.

Textual Examples

Question 1.
In Fig, the magnetic needle has magnetic moment 6.7 × 10-2 Am2 and moment of inertia i = 7.5 × 10-6 kg m2. It performs 10 complete oscillations in 6.70 s. What is the magnitude of the magnetic field ?
Solution:
The time period of oscillation is, :
T = \(\frac{6.70}{10}\) = 0.67 s
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 13
The axial field of a finite solenoid in order to demonstrate its similarity to that of a bar magnet.
From Eq. B = AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 14
= \(\frac{4 \times(3.14)^2 \times 7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times(.067)^2}\)
= 0.01 T

Question 2.
A short bar magnet placed with its axis at 30° with an external field of 800 G experiences a torque of 0.016 Nm.
(a) What is the magnetic moment of the magnet ?
(b) What is the work done in moving it from its most stable to most unstable position ?
(c) The bar magnet is replaced by a solenoid of cross-sectional area 2 × 10-4 m2 and 1000 turns, but of the same magnetic moment. Determine the current flowing through the solenoid.
Solution:
a) From Eq., τ = m B sin θ, θ = 30°, hence sin θ = 1/2.
Thus, 0.016 = m × (800 × 10-4 T) × (1/2)
m = 160 × 2/800 =0.40 Am2

b) From Eq . Um = -m. B, the most stable position is θ = 0° and the most unstable position is q = 180°. Work done is given by
W = Um (θ = 180°) – Um (θ = 0°)
= 2 m B = 2 × 0.40 × 800 × 10-4 = 0.064 J

c) From Eq., ms = NIA. From part (a), ms = 0.40 Am2
0.40 = 1000 × I × 2 × 10-4
I = 0.40 × 104/(1000 × 2) = 2A

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 3.
a) What happens if a bar magnet is cut into two pieces :
(i) transverse to its length,
(ii) along its length ?
b) A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why ?
c) Must every magnetic configuration have a north pole and a south pole ? What about the field due to a toroid ?
d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised ? If only one is magnetised how does one ascertain which one ? (Use nothing else but the bars A and B].
Solution:
a) In either case, one gets two magnets, each with a north and south pole.

b) No force if the field is uniform. The iron nail experiences a non-uniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. Then net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole.

c) Not necessarily. True only if the source of the field has a net nonzero magnetic moment. This is not so for a toroid or even for a straight infinite conductor.

d) Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see which one of the two bars is magnet, pick up one, (say, A) and lower one of its end first one of the ends of the other (say, B) and then on the middle of B. If you notice that in the middle of B, A experiences no force, then B is magnetised. If you do not notice any change from the end to the middle of B, then A is magnetised.

Question 4.
What is the magnitude of the equatorial and axial fields due to a bar magnet of length 8.0 cm at a distance of 50 cm from its mid-point ? The magnetic moment of the bar magnet is 0.40 A m2, the same as in Example – 2.
Solution:
From Eq.
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 12

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 5.
Figure shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q.
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 15
a) In which configuration the system is not in equilibrium ?
b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium ?
c) Which configuration corresponds to the lowest potential energy among all the configurations shown ?
Solution:
Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression. *
BP = –\(\frac{\mu_0}{4 \pi} \frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{r}^3}\) (on the normal bisector)
BP = \(\frac{\mu_0 2}{4 \pi} \frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{r}^3}\) (on axis)
where mp is the magnetic moment of the dipole P.
Equilibrium is stable when mQ is parallel to BP, and unstable when it is anti-parallel to BP. For instance for the configuration Q3 for which Q is along the perpendicular bisector of the dipole P, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence Q3 is stable. Thus,
a) PQ1 and PQ2
b) (i) PQ3, PQ6 (stable); (ii) PQ5, PQ4 (unstable)
c) PQ6.

Question 6.
Many of the diagrams given in Fig. show magnetic field lines (thick lines in the figure) wrongly. Point out what is wrong with them. Some of them may describe electrostatic field lines correctly. Point out which ones.
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 16
Solution:
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 17
a) Wrong: Magnetic field lines can never emanate from a point, as shown in figure. Over any closed surface, the net flux of B must always be zero, i.e., pictorially as many field lines should seem to enter the surface as the number of lines leaving it. The field lines shown, in fact, represent electric field of a long positively charged wire. The correct magnetic field lines are circling the straight conductor.

b) Wrong: Magnetic field lines (like electric lines) can never cross each other, because otherwise the direction of field at the point of intersection is ambiguous. There is further error in the figure. Magnetostatic field lines can never form closed loops around empty space. A closed loop of static magnetic field line must enclose a region across which a current is passing. By contrast, electrostatic field lines can never form closed loops, neither in empty space, nor when the loop encloses charges.

c) Right: Magnetic lines are completely confined within a toroid: Nothing wrong here in field lines forming closed loops, since each loop encloses a region across which a current passes. Note, for clarity of figure, only a few field lines within the toroid have been shown. Actually, the entire region enclosed by the windings contains magnetic field.,

d) Wrong: Field lines due to a solenoid at its ends and outside cannot be so completely straight and confined; such a thing violates Ampere’s law. The lines should curve out at both ends, and meet eventually to form closed loops.

e) Right: These are field lines outside and inside a bar magnet. Note carefully the direction of field lines inside. Not all field lines emanate out of a north pole (or converge into a south pole). Around both the N-pole, and the S-pole, the next flux of the field is zero.

f) Wrong: These field lines cannot possibly represent a magnetic field. Look at the upper region. All the field lines seem to emanate out of the shaded plate. The net flux through a surface surrounding the shaded plate is not zero. This is impossible for a magnetic field. The given field lines, in fact, show the electrostatic field lines around a positively charged upper plate and a negatively charged lower plate. The difference between Fig. [(e) and (f)] should be carefully grasped.

g) Wrong: Magnetic field lines between two pole pieces cannot be precisely straight at the ends. Some fringing of lines is inevitable. Otherwise, Ampere’s law is violated. This is also true for electric field lines.

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 7.
a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field line’s also represent the lines of force on a moving charged particle at every point ?
b) Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why ?
c) If magnetic monopoles existed, how would the Gauss’s law of magnetism be modified ?
d) Does a bar magnet exert a torque on itself due to its own field ? Does one element of a current – carrying wire exert a force on another element of the same wire ?
e) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero ?
Solution:
a) No. The magnetic force is always normal to B (remember magnetic force = qv × B). It is misleading to call magnetic field lines as lines of force.

b) If field lines were entirely confined between two ends of a straight solenoid, the flux through the cross-section at each end would be non-zero. But the flux of field B through any closed surface must always be zero. For a toroid, this difficulty is absent because it has no ‘ends’.

c) Gauss’s law of magnetism states that the flux of,B thrugh any closed surface is always
zero \(\int_s B \cdot d s\) = o.
If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) qm enclosed by S. [Analogous to Gauss’s law of electrostatics, \(\int_s B \cdot d s\) = μ0qm
where qm is the (monopole) magnetic charge enclosed by S.]

d) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero).

e) Yes. The average of the cahrge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero.

Question 8.
The earth’s magnetic field at the equator is approximately 0.4 G. Estimate the earth’s dipole moment.
Solution:
The equatorial magnetic field is,
BE = \(\frac{\mu_0 \mathrm{~m}}{4 \pi \mathrm{r}^3}\)
We are given that BE ~ 0.4 G = 4 × 10-5 T. For r, we take the radius of the earth 6.4 × 106 m. Hence,
m = \(\frac{4 \times 10^{-5} \times\left(6.4 \times 10^6\right)^3}{\mu_0 / 4 \pi}\) = 4 × 102 × (6.4 × 106)30/4π = 10-7)
= 1.05 × 1023 A m2
This is close to the value 8 × 1022 A m2 quoted in geomagnetic texts.

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 9.
In the magnetic meridian of a certain place, the horizontal component of the earth’s magnetic field is 0.26 G and the dip angle is 60°. What is the magnetic field of the earth at this location ?
Solution:
It is given that HE = 0.26 G. From Fig., we have
AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter 11
The earth’s magnetic field, BE, its horizontal and vertical components. HE and ZE. Also shown are the declination, D and the inclination or angle of dip, I.
cos 60° = \(\frac{\mathrm{H}_{\mathrm{E}}}{\mathrm{B}_{\mathrm{E}}}\)
BE = \(\frac{\mathrm{H}_{\mathrm{E}}}{\cos 60^{\circ}}\)
= \(\frac{0.26}{(1 / 2)}\) = 0.52 G

Question 10.
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current Im.
Solution:
a) The field H is dependent of the material of the core, and is
H = nI = 1000 × 2.0 = 2 × 103 A/m

b) The magnetic field B is given by
B = μrμ0H
= 400 × 4π × 10-7 (N/A3) × 2 × 103 (A/m) = 1.0 T

c) Magnetisation is given by
M = (B – μ0 H)/μ0
= (μrμ0H – μ0H)/μ0 = (μr – 1) H = 399 × H ≃ 8 × 105 A/m

d) The magnetising current IM is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = μrn0 (I + IM). Using I = 2A, B = 1 T, we get IM = 794A.

AP Inter 2nd Year Physics Important Questions Chapter 8 Magnetism and Matter

Question 11.
A domain in ferromagnetic iron is in the form of a cube of side length 1 μm. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is 55 g/mole and its density is 7.9 g/cm3. Assume that each iron atom has a dipole moment of 9.27 × 10-24 A m3.
Solution:
The volume of the cubic domain is
V = (10-6 m)3 = 10-18 m3 = 10-12 cm3
Its mass is volume × density = 7.9 g cm-3 × 10-12 cm3 = 7.9 × 10-12 g
It is given that Afagadro number (6.023 × 1023) of iron atoms have a mass of 55g. Hence,the number of atoms in the domain is
N = \(\frac{7.9 \times 10^{-12} \times 6.023 \times 10^{23}}{55}\)
= 8.65 × 1010 atoms
The maximum possible dipole moment mmax is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned. Thus,
mmax = (8.65 × 1010) × (9-27 × 10-24)
= 8.0 × 10-13 Am2
The consequent magnetisation is
Mmax = mmax/DomainVolume :
= 8.0 × 10-13 Am2/10-18 m3
= 8.0 × 105 Am-1.

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Chapter 8 అవకలన సమీకరణాలు Exercise 8(c) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Exercise 8(c)

అభ్యాసం – 8(సి)

I.

ప్రశ్న 1.
x dy – ydx = \(\sqrt{x^2+y^2} d x\) ను \(\frac{d y}{d x}\) = F\(\left(\frac{y}{x}\right)\) రూపంలో రాయండి.
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 1

ప్రశ్న 2.
(x – y Tan-1\(\frac{y}{x}\))dx + x Tan-1\(\frac{y}{x}\)dy = 0 ని \(\frac{\mathbf{d y}}{\mathbf{d x}}\) = F\(\left(\frac{y}{x}\right)\) రూపంలో రాయండి.
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 2

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c)

ప్రశ్న 3.
x\(\frac{d y}{d x}\) = y (log y -log x + 1) ను \(\frac{d y}{d x}\) = F\(\left(\frac{y}{x}\right)\) రూపములో వ్రాయండి
సాధన:
x . \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y (log y – log x + 1)
\(\frac{d y}{d x}\) = \(\frac{y}{x}\left(\log \frac{y}{x}+1\right)\)

II. కింది అవకలన సమీకరణాలను సాధించండి.

ప్రశ్న 1.
\(\frac{d y}{d x}\) = \(\frac{x-y}{x+y}\)
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 3
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 4

ప్రశ్న 2.
(x2 + y2) dy = 2xy dx
సాధన:
(x2 + y2) dy = 2xy dx
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{2 x y}{x^2+y^2}\)
Put y = vx
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 5
1 + v2 = A(1 – v2) + BV(1 – V) + CV(1 + v)
v = 0 ⇒ 1 = A
v = 1 ⇒ 1 + 1 = C(2) ⇒ c = 1
y = -1 ⇒ 1 + 1 = B(-1) (2) ⇒ 2 = – 2B
B = -1
\(\int \frac{1+v^2}{v\left(1-v^2\right)} d v\) = \(\int \frac{d v}{v}\) – \(\int \frac{d v}{1+y}\) + \(\int \frac{d v}{1-v}\)
= log v – log (1 + v) – log (1 – v)
= log \(\frac{v}{1-v^2}\)
∴ log \(\frac{v}{1-v^2}\) = log x + log c = log cx
\(\frac{v}{1-v^2}\) = cx
v = cx(1 – v2)
v = cx\(\left(1-\frac{y^2}{x^2}\right)\)
\(\frac{y}{x}\) = cx\(c x \frac{\left(x^2-y^2\right)}{x^2}\) ⇒ సాధన y = c(x2 – y2)

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c)

ప్రశ్న 3.
\(\frac{d y}{d x}\) = \(\frac{-\left(x^2+3 y^2\right)}{3 x^2+y^2}\)
సాధన:
\(\frac{d y}{d x}\) = \(\frac{-\left(x^2+3 y^2\right)}{3 x^2+y^2}\)
put y = vx
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 6
(v + 1)3 గుణించగా
3 + v2 = A(v + 1)2 + B(v + 1) + C
v = -1 ⇒ 3 + 1 = C ⇒ C = 4
v2 గుణకాలను సమానం చేయగా
A = 1
v గుణకాలను సమానం చేయగా
0 = 2A + B
B = -2A = -2
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 7
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 8

ప్రశ్న 4.
y2 dx + (x2 – xy) dy = 0
సాధన:
y2 dx = -(x2 – xy) dy
= (xy – x2) dy
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 9
v – log y = log x + log k
v = log v + log x + log k
= log k (vx)
\(\frac{y}{x}\) = log ky
సాధన ky = ey/x

ప్రశ్న 5.
\(\frac{d y}{d x}\) = \(\frac{(x+y)^2}{2 x^2}\)
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 10

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c)

ప్రశ్న 6.
(x2 – y2) dx – xy dy = 0 (May 06)
సాధన:
(x2 – y2) dx – xy dy = 0
(x2 – y2) dx = xy . dy
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 11
\(-\frac{1}{4}\)log \(\left(\frac{x^2-2 y^2}{x^2}\right)\) = log x + log c
\(-\frac{1}{4}\)[log (x2 – 2y2) – log x2] = log x + log c
\(-\frac{1}{4}\)[log (x2 – 2y2) + \(\frac{1}{4}\). 2 log x = log x + log
\(-\frac{1}{4}\)(x2 – 2y2) = \(\frac{1}{2}\)logx + logc
– log (x2 – 2y2) = -2 log x – 4 log c
-log(x2 – 2y2) = -2 log x – 4 log c
log (x2 – 2y2) = – 2 log x + log k
ఇక్కడ k = \(\frac{1}{c^4}\) = log \(\frac{k}{x^2}\)
x2 – 2y2 = \(\frac{k}{x^2}\)
సాధన x2 (x2 – 2y2) = k

ప్రశ్న 7.
(x2y – 2xy2) dx = (x3 – 3x2y dy)
సాధన:
(x2y – 2xy2) dx = (x3 – 3x2y) dy
y = vx అనుకొంటే \(\frac{d y}{d x}\) = v + x . \(\frac{d v}{d x}\)
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 12
1 – 3v = A (1 + v) + Bv
v = 0 ⇒ 1 = A
v = -1 ⇒ 4 = -B ⇒ B = -4
\(\int\left(\frac{1}{v}-\frac{4}{1+v}\right) d\) = \(\int \frac{d x}{x}\)
log v – 4 log(1 + y) = log x + log c
log \(\frac{v}{(1+v)^4}\) = log cx
\(\frac{v}{(1+v)^4}\) = cx
v = cx (1 + v)4
\(\frac{y}{x}\) = cx\(\left(1+\frac{y}{x}\right)^4\)
\(\frac{y}{x}\) = cx \(\frac{(x+y)^4}{x^4}\)
x2y = e(x + y)4

ప్రశ్న 8.
y2dx + (x2 – xy + y2) dy = 0
సాధన:
y2dx = – (x2 – xy + y2) dy
\(\frac{d y}{d x}\) = \(\frac{-y^2}{x^2-x y+y^2}\)
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 13
1 – v2 + v2 = A(1 + v2) + (Bv + C)v
v = 0 ⇒ 1 = A
v2 గుణకాలను సమానం చేయగా
1 = A + B ⇒ B = 0
v గుణకాలను సమానం చేయగా
-1 = C
⇒ \(\frac{1-v+v^2}{v\left(1+v^2\right)}\) = \(\frac{1}{v}\) – \(\frac{1}{1+v^2}\)
\(\int \frac{1-v+v^2}{v\left(1+v^2\right)} d v\) = \(\int \frac{d v}{v}\) – \(\int \frac{d v}{1+v^2}\)
= log v – tan-1 v
(1) నుండి
log v – tan-1 v = – log x + log c
tan-1 v = log v + log x – log c
= log \(\frac{v x}{c}\)
= log \(\frac{y}{c}\)
\(\frac{y}{c}\) = \(\mathrm{e}^{\tan ^{-1} v}\) = \(e^{\tan ^{-1}(y / x)}\)
సాధన y = c. \(e^{\tan ^{-1}(y / x)}\)

ప్రశ్న 9.
(y2 – 2xy) dx + (2xy – x2) dy = 0
సాధన:
(y2 – 2xy)dx + (2xy – x2) dy = 0
(2xy – x2)dy = -(y2 – 2xy) dx
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 14
\(\int \frac{2 v-1}{v(1-v)} d v\) = \(3 \int \frac{d x}{x}\) —– (1)
\(\frac{2 v-1}{v(1-v)}\) = \(\frac{A}{v}\) + \(\frac{B}{1-v}\) అనుకుందాం
2v – 1 = A(1 – v) + Bv
v = 0 ⇒ -1 = A ⇒ A = -1
v = 1 ⇒ 1 = B ⇒ B = 1
\(\int\left(-\frac{1}{v}+\frac{1}{1-v}\right) d v\) = \(3 \int \frac{d x}{x}\)
-log v – log (1 – y) = 3 log x + log c
log \(\frac{1}{v(1-v)}\) = log cx3
\(\frac{1}{v(1-v)}\) = cx3
v(1 – v) = \(\frac{1}{c x^3}\)
\(\frac{y}{x}\left(1-\frac{y}{x}\right)\) = \(\frac{1}{c x^3}\)
\(\frac{y}{x}\left(\frac{x-y}{x}\right)\) = \(\frac{1}{\mathrm{cx}^3}\)
xy(x – y) = \(\frac{1}{c}\) = k
xy(y – x) = \(-\frac{1}{c}\) = k’

ప్రశ్న 10.
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = \(\frac{y^2}{x^2}\)
సాధన:
\(\frac{d y}{d x}\) + \(\frac{y}{x}\) = \(\frac{y^2}{x^2}\)
y = vx ⇒ \(\frac{d y}{d x}\) = v + x. \(\frac{d v}{d x}\)
v + x. \(\frac{d v}{d x}\) + v = \(\frac{v^2 x^2}{x^2}\)
x.\(\frac{d v}{d x}\) = v2 – 2v
\(\frac{d v}{v^2-2 v}\) = \(\frac{\mathrm{dy}}{\mathrm{x}}\)
\(\frac{1}{v^2-2 v}\) = \(\frac{A}{v}\) + \(\frac{B}{v-2}\) అనుకుందాం
1 = A(v – 2) + Bv
v = 0 ⇒ 1 = A(-2) ⇒ A = \(-\frac{1}{2}\)
v = 2 ⇒ 1 = 2B ⇒ B = \(\frac{1}{2}\)
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 15
సాధన
y – 2x = c2x2y
= kx2y, k = c2 అనుకుందాం

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c)

ప్రశ్న 11.
x dy – y dx = \(\sqrt{x^2+y^2}\) dx
సాధన:
x dy – y dx = \(\sqrt{x^2+y^2}\) dx
x \(\frac{\mathrm{dy}}{\mathrm{dx}}\) – y = \(\sqrt{x^2+y^2}\)
\(\frac{d y}{d x}\) – \(\frac{y}{x}\) = \(\frac{\sqrt{x^2+y^2}}{x}\)
y = vx ⇒ \(\frac{d y}{d x}\) = v + x . \(\frac{d v}{d x}\)
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 16

ప్రశ్న 12.
(2x – y)dy = (2y – x) dx.
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 17
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 18
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 19

ప్రశ్న 13.
(x2 – y2) \(\frac{d y}{d x}\) = xy (May ’11)
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 20
= log y + c
\(\frac{-x^2}{2 y^2}\) = (log y + c)
– x2 = 2y2 (c + log y)
⇒ సాధన x2 + 2y2 (c + log y) = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c)

ప్రశ్న 14.
2\(\frac{d y}{d x}\) = \(\frac{y}{x}\) + \(\frac{y^2}{x^2}\)
సాధన:
y = vx అనుకొందాం
\(\frac{d y}{d x}\) = v + x. \(\frac{d v}{d x}\)
2v + 2x. \(\frac{d v}{d x}\) = v + v2
2x. \(\frac{d v}{d x}\) = v2 – v
\(\frac{d v}{v(v-1)}\) = 2.\(\frac{d x}{x}\)
\(\int\left(\frac{1}{v-1}-\frac{1}{v}\right) d v\) = \(2 \int \frac{d x}{x}\)
log (v – 1) – log v = 2 log x + log c
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 21
సాధన (y – x) = cx2y

III.

ప్రశ్న 1.
(1 + \(e^{\frac{x}{y}}\))dx + \(e^{\frac{x}{y}}\)(1 – \(\frac{x}{y}\))dy = 0 ను సాధించండి.
సాధన:
x = vy అనుకొందాం
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 22
v + v.ev + y(1 + ev) + ev – v.ev = 0
y(1 + ev). dv = -(v + ev) dy
\(\int \frac{1+e^v}{v+e^v}\) = –\(\int \frac{d y}{y}\)
log (v + ev) = – log y + log c
v + ey = \(\frac{e}{y}\)
\(\frac{x}{y}\) + ex/y = \(\frac{e}{y}\) ⇒ x + y. e<sup.x/y = c

ప్రశ్న 2.
x sin \(\frac{y}{x}\). \(\frac{d y}{d x}\) = y sin \(\frac{y}{x}\) – x ను సాధించండి.
సాధన:
x తో భాగించగా
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 23

ప్రశ్న 3.
x dy = (y + x cos2 \(\frac{y}{x}\)) dx ను సాధించండి.
సాధన:
x. \(\frac{d y}{d x}\) = y + x. cos2 \(\frac{y}{x}\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{y}{x}\) + cos2\(\frac{y}{x}\)
put y = vx
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = v + x. \(\frac{\mathrm{dv}}{\mathrm{dx}}\)
v + x. \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = v + cos2 v
\(\frac{d v}{\cos ^2 v}\) = \(\frac{\mathrm{dx}}{\mathrm{x}}\)
\(\int \sec ^2 v \cdot d v\) = \(\int \frac{d x}{x}\)
tan v = log x + c
i.e., సాధన tan \(\left(\frac{y}{\dot{x}}\right)\) = log x + c.

ప్రశ్న 4.
(x – y log y + y log x)dx + x(log y – log x)dy = 0 ను సాధించండి
సాధన:
x. dx తో భాగించగా
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 24
y + cx = ylog\(\left(\frac{y}{x}\right)\) + x log x
= y log y – y log x + x log x
= (x – y) log x + y log y

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c)

ప్రశ్న 5.
(y dx + x dy) x cos \(\frac{y}{x}\) = (x dy – y dx) y sin \(\frac{y}{x}\) ను సాధించండి.
సాధన:
ఇచ్చిన సమీకరణమును క్రింది విధంగా వ్రాయగలరు.
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 25
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 26
∴ ఇది సమఘాత సమీకరణము
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 27
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 28

ప్రశ్న 6.
వాలు \(\frac{d y}{d x}\) = \(\frac{y}{x}\) – cos2 \(\frac{y}{x}\), (x > 0, y > 0) అవుతూ(1, \(\frac{\pi}{4}\)) బిందువు గుండా పోయే వక్రం సమీకరణం కనుక్కోండి.
సాధన:
\(\frac{d y}{d x}\) = \(\frac{y}{x}\) – cos2 \(\frac{y}{x}\)
ప్రతిక్షేపించగా y = vx
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(c) 29
వక్రం (1, \(\frac{\pi}{4}\)) నుంచి పోతాయి
tan (\(\frac{\pi}{4}\)) = x – log 1
c = 1
∴ వక్రం సమీకరణ
tan v = 1 – log |x|
tan (\(\frac{y}{x}\)) = 1 – log |x|

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Chapter 1 వృత్తం Exercise 1(c) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Exercise 1(c)

అభ్యాసం – 1(సి)

I.

ప్రశ్న 1.
కింద ఇచ్చిన ప్రతి S = 0 వృత్తానికి P వద్ద స్పర్శరేఖ సమీకరణాన్ని కనుక్కోండి.
i) S ≡ x2 + y2 – 6x + 4y – 12; P = (7, -5)
సాధన:
వృత్త సమీకరణము
S ≡ x2 + y2 – 6x + 4y – 12 = 0
P(7, -5) వద్ద స్పర్శరేఖ సమీకరణము
S1 = xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
⇒ x. 7 + y(-5) – 3(x + 7) + 2(y −5) – 12 = 0
⇒ 7x – 5y – 3x – 21 + 2y – 10 – 12 = 0
4x – 3y – 43 = 0

ii) S ≡ x2 + y2-6x + 4y – 12; P = (-1, 1)
సాధన:
P వద్ద స్పర్శరేఖ సమీకరణము
x(-1) + y. 1 – 3(x – 1) + 2(y + 1) – 12 = 0
⇒ -x + y – 3x + 3 + 2y + 2 – 12 = 0
⇒ – 4x + 3y – 7 = 0
4x – 3y + 7 = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

iii) S ≡ x2 + y2 + 4x + 6y – 39; P= (-6, -9)
సాధన:
P వద్ద స్పర్శరేఖ సమీకరణము S1 = 0
(i.e.,) x(-6) +y(-9) +2(x – 6) + 3(y – 9) – 39 = 0
⇒ -6x – 9y + 2x – 12 + 3y – 27 – 39 = 0
⇒ -4x – 6y – 78 = 0
⇒ 4x + 6y + 78 = 0
⇒ 2x + 3y+ 39 = 0

iv) S ≡ x2 + y2 – 4x – 6y + 11; P = (3, 4)
సాధన:
P వద్ద స్పర్శరేఖ సమీకరణము S1 = 0
⇒ x(3) + y(4) – 2(x + 3) – 3(y + 4) + 11 = 0
3x + 4y – 2x – 6 – 3y – 12 + 11 = 0
x + y – 7 = 0

ప్రశ్న 2.
కింద ఇచ్చిన ప్రతి S = 0 వృత్తానికి P వద్ద అభిలంబ రేఖ సమీకరణాన్ని కనుక్కోండి.
i) S ≡ x2 + y2 + x + y – 24; P = (3,-4)
సాధన:
అభిలంబరేఖ సమీకరణము
(x – x1) (y1 + f) – (y – y1) (x1 + g) = 0
(x -3) (- 4 + \(\frac{1}{2}\)) – (y + 4) (3 + \(\frac{1}{2}\)) = 0
–\(\frac{7}{2}\)(x – 3) – \(\frac{7}{2}\) (y + 4) = 0
⇒ (x – 3) + (y + 4)
x – 3 + y + 4 = 0
x + y + 1 = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ii) S ≡ x2 + y2 – 10x – 2y + 6; P = (3, 5)
సాధన:
అభిలంబ రేఖ సమీకరణము
(x – x1) (y1 + f) – (y – y1) (x1 + g) = 0
(x – 3) (5 – 1) – (y – 5) (3 – 5) = 0
4x – 12 + 2y – 10 = 0
4x + 2y – 22 = 0
లేదా
2x + y – 11 = 0

iii) S ≡ 3(x2 + y2) – 19x – 29y + 76; P = (1,3)
సాధన:
వృత్త సమీకరణము
x2 + y2 – \(\frac{19}{3}\)x – \(\frac{29}{3}\) y + \(\frac{76}{3}\) = 0
(x – 1) (3 – \(\frac{29}{3}\)) – (y – 3) (1 – \(\frac{19}{6}\)) = 0
– \(\frac{11}{6}\) (x – 1) + \(\frac{13}{6}\) (y – 3) = 0
11(x – 1) – 13(y – 3) = 0
11x – 11 – 13y + 39 = 0
11x – 13y + 28 = 0

iv) S ≡ x2 + y2 – 22x – 4y + 25; P = (1, 2)
సాధన:
P వద్ద అభిలంబరేఖ సమీకరణము
(x – 1) (2 – 2) – (y – 2) (1 – 11) = 0
10(y – 2) = 0 ⇒ y – 2 = 0
లేదా y = 2

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

II.

ప్రశ్న 1.
x2 + y2 −x + 3y – 22 = 0 y=x -3 రేఖపై ఏర్పరచే జ్యా పొడవును కనుక్కోంది. (Mar.’13, May ’11)
సాధన:
వృత్త సమీకరణము
S = x2 + y2 – x + 3y – 22 = 0
AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c) 1
= \(\sqrt{96}=4 \sqrt{6}\) యూనిట్లు.

ప్రశ్న 2.
x2+ y2 – 8x – 2y – 80 వృత్తం x + y + 1 = 0 రేఖపై ఏర్పరచే జ్యా పొడవును కనుక్కోండి. [T.S. Mar. ’16]
సాధన:
వృత్త సమీకరణము x + y – 8x – 2y – 8 = 0
కేంద్రం C(4, 1), r = \(\sqrt{16+1+8}\) = 5
రేఖ సమీకరణము x + y + 1 = 0
P = కేంద్రం నుండి లంబ దూరము = \(\frac{|4+1+1|}{\sqrt{1+1}}\)
= \(\frac{6}{\sqrt{2}}=3 \sqrt{2}\)
జ్యా పొడవు = 2 \(\sqrt{r^2-p^2}\)
= 2\(\sqrt{25-18}\)
= 2\(\sqrt{7}\) యూనిట్లు.

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 3.
x2 + y2 = a2 వృత్తం x cos α + y sin α = p రేఖపై ఏర్పరచే జ్యా పొడవును కనుక్కోండి.
సాధన:
వృత్త సమీకరణము x2 + y2 = a2
కేంద్రం C(0, 0), r = a
రేఖ సమీకరణము
x cos α + y sin α – P = 0
P = కేంద్రం నుండి లంబ దూరము
\(\frac{|0+0-p|}{\sqrt{\cos ^2 \alpha+\sin ^2 \alpha}}\) = p
జ్యా పొడవు = 2\(\sqrt{a^2-p^2}\)

ప్రశ్న 4.
(2, 3) కేంద్రంగా ఉంటూ 3x – 4y + 1 =0 రేఖను స్పృశించే వృత్త సమీకరణాన్ని కురుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c) 2
వృత్త సమీకరణము (x – h)2 + (y – k)2 = r2
(x – 2)2 + (y – 3)2 = 1
x2 + y2 – 4x – 6y + 12 = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 5.
(-3, 4) కేంద్రంగా ఉంటూ Y – అక్షాన్ని స్పృశించే వృత్త సమీకరణాన్ని కనుక్కోండి.
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c) 3
కేంద్రం C (-3, 4)
వృత్తం Y – అక్షాన్ని స్పృశిస్తుంది.
r యొక్క X నిరూపకం C = |-3 = 3
వృత్త సమీకరణము (x + 3)2 + (y – 4)2 = 9
x2 + 6x + 9 + y2 – 8y + 16 – 9 = 0
x2 + y2 + 6x – 8y + 16 = 0

ప్రశ్న 6.
x2 + y2 – + y − 8x − 2y + 12 = 0 వృత్తానికి y నిరూపకం 1 అయ్యే బిందువుల వద్ద స్పర్శరేఖ సమీకరణాలను కనుక్కోండి.
సాధన:
వృత్త సమీకరణము
x2 + y2 – 8x – 2y + 12 = 0
P నిరూపకాలు (X,, 1) అనుకొందాం.
P వృత్తం మీది బిందువు
x12 + 1 – 8x1 – 2 + 12 = 0
x12– 8x1 + 11 = 0
x1 = \(\frac{8 \pm \sqrt{64-44}}{2}=\frac{8 \pm 2 \sqrt{5}}{2}=4 \pm \sqrt{5}\)
x1 = 4 + \(\sqrt{5}\), x2 = 4 – \(\sqrt{5}\)
P నిరూపకాలు (4 + \(\sqrt{5}\), 1) మరియు
Q నిరూపకాలు (4 – \(\sqrt{5}\), 1)
P వద్ద స్పర్శరేఖ సమీకరణము (4 + \(\sqrt{5}\), 1)
x (4 + \(\sqrt{5}\)) + y . 1 – 4(x + 4 + \(\sqrt{5}\)) – (y + 1) + 12 = 0
⇒ 4x + \(\sqrt{5}\)x + y – 4x – 16 – 4\(\sqrt{5}\) – y – 1 + 12 = 0
⇒ x – 5 – 4\(\sqrt{5}\) = 0
⇒ \(\sqrt{5}\) (x – \(\sqrt{5}\) – 4) = 0
⇒ x – \(\sqrt{5}\) – 4 = 0

x = 4 + \(\sqrt{5}\)
Q వద్ద స్పర్శరేఖ సమీకరణము (4 –\(\sqrt{5}\) , 1)
⇒ x (4 – \(\sqrt{5}\) ) + y. 1 – 4 (x + 4 – \(\sqrt{5}\) ) – (y + 1) + 12 = 0
⇒ 4x – \(\sqrt{5}\)x + y – 4x – 16 + 4\(\sqrt{5}\) – y – 1 + 12 = 0
⇒ – \(\sqrt{5}\)x + 4\(\sqrt{5}\) – 5 = 0
⇒ –\(\sqrt{5}\)(x – 4 + \(\sqrt{5}\)) = 0
⇒ x – 4 + \(\sqrt{5}\) = 0
x = 4 – \(\sqrt{5}\)

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 7.
x2 + y2 – 10 = 0 వృత్తానికి x నిరూపకం 1 అయ్యే బిందువుల వద్ద స్పర్శరేఖ సమీకరణాన్ని కనుక్కోండి.
సాధన:
వృత్త సమీకరణము x2 + y2 = 10
P నిరూపకాలు (1, y1)
1+ y12 = 10 ⇒ y12 = 9 అనుకొందాం.
y1 = ±3
P నిరూపకాలు (1, 3) మరియు (1, -3) P(1, 3)
వద్ద స్పర్శరేఖ సమీకరణము
x. 1 + y. 3 = 10
x + 3y – 10 = 0
P(1, 3) వద్ద స్పర్శరేఖ సమీకరణము
x.1 + y(-3) = 10
⇒ x – 3y – 10 = 0

III.

ప్రశ్న 1.
వృత్తం x2 + y2 = c2, సరళరేఖ \(\frac{x}{a}+\frac{y}{b}\) = 1 లు A, B ల వద్ద ఖండించుకుంటే \(\overline{\mathrm{A B}}\) పొడవు కనుక్కొని, ఈ రేఖ వృత్తాన్ని స్పృశించడానికి నియమం కనుక్కోండి.
సాధన:
x2 + y2 = c2
వృత్త కేంద్రం C = (0, 0), వ్యాసార్ధం (r) = c
d = వృత్త కేంద్రం నుండి జ్యా \(\frac{x}{a}+\frac{y}{b}\) = 1 రేఖ మీదకు లంబదూరం.
= \(\frac{|0-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{a b}{\sqrt{a^2+b^2}}\)
ఇక జ్యా పొడవు = 2\(\sqrt{r^2-d^2}\)
= 2\(\sqrt{c^2-\left(\frac{a^2 b^2}{a^2+b^2}\right)}\)
జ్యా వృత్తాన్ని స్పృశిస్తే, జ్యా పొడవు సున్న కావలయును.
⇒ c2 = \(\frac{a^2 b^2}{a^2+b^2} \Rightarrow \frac{1}{c^2}=\frac{a^2+b^2}{a^2 b^2}\)
ఇదియే కావలసిన నియమం \(\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}\)

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 2.
x2 + y2 = a2 వృత్తాన్ని, y = mx + c రేఖ A, B ల వద్ద ఖండించుకుంటూ AB = 2λ అయితే c = (1 + m2) (a2 – λ2) అని చూపండి.
సాధన:
వృత్త కేంద్రం C (0, 0), వ్యాసార్ధం (r) = a
d = వృత్త కేంద్రం C = (0, 0) నుండి జ్యా y = mx + c
AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c) 4

ప్రశ్న 3.
(2, 3) కేంద్రంగా ఉంటూ 3x + 4y + 4 = 0 రేఖపై చేసే జ్యా పొడవు 2 అయ్యే వృత్త సమీకరణాన్ని (Mar. ’11) కనుక్కోండి.
సాధన:
కేంద్రం ( 2, 3) నుండి రేఖ మీదకు దూరం
d = \(\left|\frac{3(-2)+4(3)+4}{\sqrt{9+16}}\right|=\frac{10}{5}\) = 2
జ్యా AB పొడవు = 2 యూనిట్లు
AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c) 5
వృత్త వ్యాసార్ధం (r) అనుకొనిన
⇒ 2 = 2\(\sqrt{r^2-d^2}\)
⇒ r2 – d2 = 1
⇒ r2 – 4 = 1 ⇒ r2 = 5
వృత్త సమీకరణము
(x + 2)2 + (y – 3)2 = 5
x2 + y2 + 4x – 6y + 4 + 9 – 5 = 0
x2 + y2+ 4x – 6y + 8 = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 4.
(3, 2) బిందువు వద్ద x2 + y2 – x – 3y – 4 = 0 వృత్తానికి స్పర్శరేఖ, అభిలంబ రేఖ సమీకరణాలను కనుక్కోండి.
సాధన:
x2 + y2 – x − 3y – 4 = 0 వృత్తానికి (3, 2) బిందువు
వద్ద స్పర్శరేఖా సమీకరణం
xx1 + yy1 + g(x + x1) + f(y + y1) + c = 0
⇒ x(3) + y(2) + (-\(\frac{1}{2}\)) (x + 3) – \(\frac{3}{2}\) (y + 2) + (-4) = 0
⇒ 6x + 4y – x – 3 – 3y – 6 – 8 = 0
⇒ 5x + y – 17 = 0
అభిలంబరేఖ స్పర్శరేఖకు లంబంగా ఉంటుంది. కనుక P(3, 2) వద్ద అభిలంబ రేఖ x – 5y + k = 0 అనుకుందాం.
ఇది P(3, 2) గుండా పోతుంది కనుక
3 – 10 + k = 0 ⇒ k = 7
∴ P(3, 2) వద్ద అభిలంబ రేఖా సమీకరణం
x – 5y + 7 = 0

ప్రశ్న 5.
(1, 1) బిందువు వద్ద 2x2+ 2y2 – 2x – 5y + 3 = 0 వృత్తానికి స్పర్శరేఖ, అభిలంబ రేఖలను కనుక్కోండి.
సాధన:
వృత్త సమీకరణము 2x2 + 2y2 – 2x 5y + 3 = 0.
⇒ x2 + y2 – y2 – x – \(\frac{5}{2}\) y + \(\frac{3}{2}\) = 0
(1, 1) బిందువు వద్ద x2 + y2 – x – \(\frac{5}{2}\) y + \(\frac{3}{2}\) = 0
వృత్తానికి స్పర్శరేఖా సమీకరణం
x(1) + y(1) – \(\frac{1}{2}\) (x + 1) – \(\frac{5}{4}\) (y + 1) + \(\frac{3}{2}\) = 0
⇒ 4x + 4y – 2(x + 1) – 5(y + 1) + 6 = 0
⇒ 4x + 4y – 2x – 2 – 5y – 5 + 6 = 0
⇒ 2x – y – 1 = 0
అభిలంబరేఖ స్పర్శబిందువు P(1, 1) వద్ద స్పర్శరేఖకు లంబంగా ఉంటుంది. కనుక అభిలంబరేఖా సమీకరణం x + 2y + k = 0 అనుకుందాం. ఇది P(1, 1) గుండా పోతుందికనుక
1 + 2 + k = 0 ⇒ k = −3
∴ అభిలంబరేఖా సమీకరణం x + 2y – 3 = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 6.
x2 + y2 = 13 వృత్తానికి (3, 2) వద్ద గీసిన స్పర్శరేఖ, x2 + y2 + 2x – 10y – 26 = 0 వృత్తాన్ని స్పృశిస్తుందని చూపి, స్పర్శబిందువును కనుక్కోండి.
సాధన:
x2 + y2 = 13 వృత్తానికి (3, -2) వద్ద స్వరరేఖాసమీకరణం
x(3) + y (-2) = 13 ⇒ 3x – 2y – 13 = 0 ………………. (1)
వృత్త సమీకరణము x2 + y2 + 2x – 10y – 26 = 0
కేంద్రం C = (-1,5)
వ్యాసార్ధం (r) = \(\sqrt{1+25+26}=\sqrt{52}=2 \sqrt{13}\)
కేంద్రం C (-1, 5) నుండి 3x – 2y – 13 = 0 రేఖ మీదకు లంబదూరం
AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c) 6
∴ x1 + 1 = 6 ⇒ x1 = 5;
∴ y1 – 5 = -4 ⇒ y1 = 1
∴ స్పర్శబిందువు = (5, 1)

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 7.
x2 + y2 – 4x – 8y + 7 = 0 వృత్తానికి (-1, 2) వద్ద గీసిన స్పర్శరేఖ x2 + y2 + 4x + 6y = 0 వృత్తాన్ని స్పృశిస్తుందని చూపి, స్పర్శ బిందువును కనుక్కోండి.
సాధన:
x2 + y2 – 4x – 8y + 7 = 0 వృత్తానికి (-1, 2) వద్ద స్పర్శరేఖ S1 = 0
అంటే x(-1) + y(2) – 2(x – 1) – 4(y + 2) + 7 = 0
⇒ -3x – 2y + 1 = 0
⇒ 3x + 2y – 1 = 0
ఈ రేఖ x2 + y2 + 4x + 6y = 0 వృత్తానికి స్పర్శరేఖ అయిన r = d కావలయును.
ఇచ్చట వృత్త వ్యాసార్ధం (r) = \(\sqrt{4+9}=\sqrt{13}\)
d = వృత్త కేంద్రం C (-2, 3) నుండి రేఖ
3x + 2y – 1 = 0 మీదకు లంబదూరం
⇒ d = \(\frac{|3(-2)+2(-3)-1|}{\sqrt{13}}=\sqrt{13}\)
∴ r = d
⇒ రేఖ వృత్తానికి స్పర్శరేఖ అవుతుంది.
స్పర్శబిందువు P(x1, y1) అనుకుంటే, ఇది C (-2, -3) నుండి రేఖ మీదకు లంబపాదం అవుతుంది.
\(\frac{x_1+2}{3}=\frac{y_1+3}{2}=-\left(\frac{-6-6-1}{13}\right)\) = 1
⇒ x1 + 2 = 3 ⇒ x1 = 1
y1 + 3 = 2 ⇒ y1 = -1
∴ స్పర్శబిందువు = (1, −1)

ప్రశ్న 8.
x2 + y2 – 4x + 6y – 12 = 0 వృత్తానికి x + y − 8 = 0 రేఖకు సమాంతరంగా ఉండే స్పర్శరేఖ సమీకరణం(లు) కనుక్కోండి.
సాధన:
వృత్త సమీకరణము x2 + y2 – 4x + 6y – 12 = 0
కేంద్రం C (2, -3)
వ్యాసార్ధం (r) = \(\sqrt{4+9+12}=\sqrt{25}\) = 5
x + y – 8 = 0 రేఖకు సమాంతరంగా ఉండే రేఖా సమీకరణం x + y + k = 0 అనుకుందాం.
ఈ రేఖ x2 + y2 – 4x + 6y – 12 = 0 వృత్తాన్ని
r = d
⇒ 5 = కేంద్రం C (2, 3) నుండి x + y + k = 0 రేఖ మీదకు లంబదూరం
⇒ 5 = \(\frac{|2-3+k|}{\sqrt{1+1}}\)
⇒ 5\(\sqrt{2}\) = |k – 1|
∴ k – 1= = ±5\(\sqrt{2}\) ⇒ k = 1 ± 5\(\sqrt{2}\)
∴ కావలసిన స్పర్శరేఖా సమీకరణాలు
x + y + (1 ± 5 \(\sqrt{2}\)) = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 9.
x2 + y2 + 2 x – 2y – 3 = 0 వృత్తానికి 3x y + 4 = 0 రేఖకు లంబంగా ఉండే స్పర్శరేఖా సమీకరణాలను
సాధన:
3x – y + 4 = 0 కు లంబంగా ఉండే రేఖా సమీకరణం x + 3y + k = 0
ఇది వృత్తానికి స్పర్శరేఖ అయిన
r = d
(అంటే) వ్యాసార్ధం = కేంద్రం (−1, 1) నుండి
x + 3y + k = 0 కు లంబదూరం
⇒ \(\sqrt{1+1+3}=\left|\frac{-1+3(1)+k}{\sqrt{10}}\right|\)
⇒ \(\sqrt{50}\) = |k + 2 |
⇒ k + 2 = 5\(\sqrt{2}\)
k = -2 + 5\(\sqrt{2}\)
∴ స్పర్శరేఖా సమీకరణాలు x + 3y – 2 ± 5\(\sqrt{2}\) = 0

ప్రశ్న 10.
x2 + y2 – 4x – 6y + 3 = 0 వృత్తానికి గీసిన
స్పర్శరేఖ x
అక్షంతో 45° కోణం చేస్తే వాటి
సమీకరణాలను కనుక్కోండి.
సాధన. స్పర్శరేఖ వాలు = tan 45° = 1
కనుక రేఖా సమీకరణం y = x + k
(అంటే) x − y + k = 0 అనుకుందాం
ఇది వృత్తానికి స్పర్శరేఖ అయిన
వ్యాసార్ధం (r) = కేంద్రం C (2, 3) నుండి x – y + k = 0 కు గల లంబదూరం
⇒ \(\sqrt{4+9-3}=\left|\frac{2-3+k}{\sqrt{2}}\right|\)
⇒ \(\sqrt{20}\) = |k – 1|
⇒ k – 1 = ± 2\(\sqrt{5}\) ⇒ k = 1 ± 2\(\sqrt{5}\)
∴ సరళరేఖా సమీకరణాలు x – y + (1 ± 2\(\sqrt{5}\)) = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 11.
(-1, 0) గుండా పోతూ x + y – 7 = 0 రేఖను (3, 4) వద్ద స్పృశించే వృత్త సమీకరణాన్ని కనుక్కోండి.
సాధన:
వృత్త సమీకరణం
S ≡ x2 + y2+2gx + 2fy + c = 0 …………….. (i)
అనుకుందాం
ఇది (-1, 0) గుండా పోతుంది కనుక
1 + 0 – 2g(-1) + 2f(0) + c = 0
⇒ – 2g + c = – 1 …………….. (1)
S = 0 వృత్తం x + y – 7 = 0 రేఖను (3, 4) వద్ద స్పృశిస్తుంది. కనుక (3, 4) వృత్తంపై ఉంటుంది.
⇒ 9 + 16 + 2g(3) + 2f(4) + c = 0
⇒ 6g + 8f + c = 25 …………… (2)
(1) నుండి c = – 1 + 2g
(2) నుండి 6g + 8f + (-1 + 2g) = -25
⇒ 8g + 8f = -24
⇒ g + f = -3
⇒ f = -3 – g
x + y – 7 = 0 రేఖ వృత్తాన్ని స్పృశిస్తుంది కనుక
r = d
⇒ \(\sqrt{g^2+f^2-c}=\left|\frac{(-g)+(-f)-7}{\sqrt{1+1}}\right|\)
⇒ 2(g2 + f2 – c) = g + f+7)2
⇒ 2[g2 + (-3 – g)2 – (-1 2g)] = [g – 3 – g + 7]2
⇒ 2 [g2 + g2 + 9 + 6g + 1 – 2g] = 16
⇒ 2g2 + 4g + 10 = 8
⇒g2 + 2g + 1 = 0
(g + 1)2 = 0 g = − 1
∴ f = – 3 – g = -3(-1) = -2
c = 1 + 2g
⇒ c = (-1) + 2(-1) = -3
∴వృత్త సమీకరణం
x2 + y2 + 2(-1) x + 2(-2) y + (-3) = 0
⇒ x2 + y2 – 2x – 4y – 3 = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 12.
(1, -1) గుండా పోతూ, 4x + 3y + 5 = 0, 3x – 4y – 10 = 0 రేఖలను స్పృశించే వృత్త సమీకరణం కనుక్కోండి.
సాధన:
వృత్త సమీకరణం
x2 + y2 + 2gx + 2fy + c = 0 ……………. (1)
అనుకుందాం.
ఇది (1, -1) గుండా పోతుంది కనుక
1 + 1 + 2g(1) + 2f(-1) + c = 0
⇒ 2g – 2f + c = -2
∴ c = -2 + 2f – 2g
వృత్త లంబరేఖలు
4x + 3y + 5 = 0, 3x – 4y – 10 = 0 లను స్పృశిస్తుంది. కనుక
కేంద్రం (-g, -f) ⊥ నుండి లంబదూరాల సమూహాలు
\(\left|\frac{-4 g-3 f+5}{5}\right|=\left|\frac{-3 g+4 f+10}{5}\right|\)
-7g + f – 5 = 0 (లేదా) – g – 7f + 15 = 0
f = 7g + 5
ఇప్పుడు \(\left|\frac{-4 g-3 f+5}{5}\right|^2\) = (-g – 1)2 + (-f + 1)2
⇒ \(\frac{(-4 g-21 g-15+5)^2}{5}\) = (-g – 1)2 + (-7g – 5 + 1)2
⇒(5g + 2)2 = g2 + 1 + 2g + 16 +49 = g2 + 56g
సాధించగా
25g2 + 38g + 13 = 0
g = -1, \(\frac{-26}{50}\)
సందర్భం 1 : f = 7g + 5 = f = -2
వృత్తం (1, −1) గుండా పోతుంది.
∴ x2 + y2 + 2gx + 2fy + c = 0
1 + 1 + 2g – 2f + c = 0
2 – 2 + 4 + c = 0 (లేదా) c = -4
వృత్త సమీకరణము
x2 + y2 – 2x – 4y – 4 = 0

సందర్భం 2 : -g = \(\frac{-13}{25}\)
f = 7g + 5 = 7(\(\frac{-13}{25}\)) + 5
= \(\frac{-91+125}{25}\) = \(\frac{34}{25}\)
వృత్తం (1, – 1) గుండా పోతుంది.
x2 + y2 + 2gx + 2fy + c = 0.
x2 + y2+ \(\frac{26}{25}\)x + \(\frac{68}{25}\)y + c = 0
1 + 1 – \(\frac{26}{25}\) – \(\frac{68}{25}\) + c = 0
c = -2 + \(\frac{26}{25}\) + \(\frac{68}{25}\) = \(\frac{-50+26+68}{25}\) = \(\frac{44}{25}\)
∴ వృత్త సమీకరణము
x2 + y2 – \(\frac{26}{25}\)x + \(\frac{68}{25}\)y + \(\frac{44}{25}\) = 0
25 (x2 + y2) – 26x + 68y + 44 = 0

AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c)

ప్రశ్న 13.
x + y + 10 రేఖ x2+ y2 – 3x + 7y+ 14 = 0 వృత్తాన్ని స్పృశిస్తుందని చూపి, స్పర్శ బిందువును కనుక్కోండి.
సాధన:
వృత్త సమీకరణము x2 + y2 – 3x + 7y + 14 = 0
వృత్త కేంద్రము C = \(\left(\frac{3}{2}, \frac{-7}{2}\right)\)
వృత్త వ్యాసార్ధము (r) = \(\sqrt{\frac{9}{4}+\frac{49}{4}-14}\)
= \(\frac{\sqrt{58-56}}{4}=\frac{1}{\sqrt{2}}\)
కేంద్రం C నుండి రేఖ x + y + 1 = 0 కు లంబదూరం
d = \(\frac{\left|\frac{3}{2}-\frac{7}{2}+1\right|}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}\)
∴ r = d = \(\frac{1}{\sqrt{2}}\)
రేఖ వృత్తాన్ని స్పృశిస్తుంది.
రేఖ వృత్తాన్ని P(h, k) వద్ద స్పృశిస్తుంది అనుకొనుము.
AP Inter 2nd Year Maths 2B Solutions Chapter 1 వృత్తం Ex 1(c) 7
⇒ h = 2, k = -3
∴ స్పర్శ బిందువు (2, -3)

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b)

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Chapter 8 అవకలన సమీకరణాలు Exercise 8(b) will help students to clear their doubts quickly.

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Exercise 8(b)

అభ్యాసం – 8(బి)

I.

ప్రశ్న 1.
\(\sqrt{1-x^2}\) dy + \(\sqrt{1-y^2}\) dx = 0 సాధారణ సాధన కనుక్కోండి.
సాధన:
దత్త అవకలన సమీకరణం
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b) 1
sin-1 y = -sin-1 x + c
సాధన sin-1 x + sin-1 y = c, c స్థిరరాశి

ప్రశ్న 2.
\(\frac{d y}{d x}\) = \(\frac{2 y}{x}\)కు సాధారణ సాధన కనుక్కోండి.
సాధన:
\(\frac{d y}{d x}\) = \(\frac{2 y}{x}\)
\(\int \frac{d y}{y}\) = \(2 \int \frac{d x}{x}\)
log c + log y = 2 log x
log cy = log x2
సాధన cy = x2, స్థిరాంకము

II. క్రింది అవకలన సమీకరణాలను సాధించండి

ప్రశ్న 1.
\(\frac{d y}{d x}\) = \(\frac{1+y^2}{1+x^2}\)
సాధన:
\(\frac{d y}{d x}\) = \(\frac{1+y^2}{1+x^2}\)
\(\int \frac{d y}{1+y^2}\) = \(\int \frac{d x}{1+x^2}\)
tan-1 y = tan-1 x + tan-1 c, c స్థిరాంకం

ప్రశ్న 2.
\(\frac{d y}{d x}\) = ey-x
సాధన:
\(\frac{d y}{d x}\) = \(\frac{e^y}{e^x}\)
\(\frac{d y}{e^y}\) = \(\frac{d x}{e^x}\)
\(\int \frac{d y}{1+y^2}\) = \(\int \frac{d x}{1+x^2}\)
tan-1 y = tan-1 x + tan-1 c, c స్థిరాంకం

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b)

ప్రశ్న 3.
(ex + 1) y dy + (y + 1) dx = 0
సాధన:
(ex + 1)y. dy = -(y + 1) dx
\(\frac{y d y}{y+1}\) = –\(\frac{d x}{e^x+1}\)
\(\int\left(1-\frac{1}{y+1}\right) d y\) = \(\int-\frac{e^{-x} d x}{e^{-x}+1}\)
y – log (y + 1) = log(e-x + 1) + log c
⇒ y – log (y + 1) = log (e-x + 1)
⇒ y = log (y + 1) + log c (e-x + 1)
y = log c (y + 1) (e-x + 1)
సాధన
ey = c(y + 1) (e-x + 1)

ప్రశ్న 4.
\(\frac{d y}{d x}\) = ex-y + x2 e-y (Mar. ’06; May ’05)
సాధన:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = ex – y + x2 . e-y
= \(\frac{e^x}{e^y}+\frac{x^2}{e^y}\)
\(\int e^y \cdot d y\) = \(\int\left(e^x+x^2\right) d x\)
సాధన
ey = ex + \(\frac{x^3}{3}\) + c

ప్రశ్న 5.
tan y dx + tan x dy = 0
సాధన:
tan y dx = – tan x dy
\(\frac{d x}{\tan x}\) = \(\frac{-d y}{\tan y}\)
\(\frac{\cos x}{\sin x} d x\) = \(-\frac{\cos y}{\sin y} d y\)
log sin x = -log sin y + log c
log sin x + log sin y = log c
log (sin x. sin y) = log c
⇒sin x. sin y = c అనేది సాధన

ప్రశ్న 6.
\(\sqrt{1+x^2}\) dx + \(\sqrt{1+y^2}\) dy = 0
సాధన:
\(\sqrt{1+x^2}\) dx + \(\sqrt{1+y^2}\) dy
ఇరువైపుల సమాకలనం చేయగా
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b) 2

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b)

ప్రశ్న 7.
y – x\(\frac{d y}{d x}\) = 5(y2 + \(\frac{d y}{d x}\))
సాధన:
y – 5y2 = (x + 5)\(\frac{d y}{d x}\)
\(\frac{d x}{x+5}\) = \(\frac{d y}{y(1-5 y)}\)
‘ఇరువైపుల సమాకలనం చేయగా
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b) 3

ప్రశ్న 8.
\(\frac{d y}{d x}\) = \(\frac{x y+y}{x y+x}\)
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b) 4

III. కింది అవకలన సమీకరణములను సాధించండి.

ప్రశ్న 1.
\(\frac{d y}{d x}\) = \(\frac{1+y^2}{\left(1+x^2\right) x y}\)
సాధన:
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b) 5
log (1 + y2)= log x2 – log (1 + x2) + log c
log (1 + x2) + log (1 + y2) = log x2 + log c
సాధన (1 + x2) (1 + y2) = cx2

ప్రశ్న 2.
\(\frac{d y}{d x}\) + x2 = x2 . e3y
సాధన:
\(\frac{d y}{d x}\) + x2 = x2 . e3y
\(\frac{d y}{d x}\) = x2. e3y – x2
= x2 (e3y – 1)
\(\int \frac{d y}{e^{3 y}-1}\) = \(\int x^2 d x\)
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b) 6

ప్రశ్న 3.
(xy2 + x) dx + (yx2 + y) dy = 0 (Mar. ’07)
సాధన:
x(y2 + 1) dx + y (x2 + 1) dy = 0
x(y2 + 1) dx + y (x2 + 1 ) dy = 0
(1 + x2) (1 + y2) తో భాగించగా
\(\frac{x d x}{1+x^2}\) + \(\frac{y d y}{1+y^2}\) = 0
సమాకలనం చేయగా
\(\int \frac{x d x}{1+x^2}\) + \(\int \frac{y d y}{1+y^2}\) = 0
\(\frac{1}{2}\)[(log (1 + x2) + log (1 + y2)] = log c
log (1 + x2) (1 + y2) = 2 log c = log c2
సాధన (1 + x2) (1 + y2) = k. ఇక్కడ k = c2

ప్రశ్న 4.
\(\frac{d \mathbf{y}}{\mathbf{d x}}\) = 2y tanh x
సాధన:
\(\frac{d \mathbf{y}}{\mathbf{d x}}\) = 2y tanh x
\(\frac{\mathrm{dy}}{\mathrm{y}}\) = 2 tanh x dx
ఇరువైపులా సమాకలనం చేయగా
\(\int \frac{d y}{y}\) = 2\(\int \tanh x d x\)
log y = 2 log |cosh x | + log c
lny = 2ln cosh x + lnc
y = c cos2h x

ప్రశ్న 5.
sin-1 \(\left(\frac{\mathbf{d y}}{\mathbf{d x}}\right)\) = x + y (May ’07)
సాధన:
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = sin (x + y)
x + y = t
1 + \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{dt}}{\mathrm{dx}}\)
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) – 1 = sin t
\(\frac{\mathrm{dt}}{\mathrm{dx}}\) = 1 + sin t
\(\frac{d t}{1+\sin t}\) = dx
ఇరువైపులా సమాకలనం చేయగా
\(\int \frac{d t}{1+\sin t}[latex] = [latex]\int \mathrm{d} x[latex]
[latex]\int \frac{1-\sin t}{\cos ^2 t} d t\) = x + c
\(\int \sec ^2 t d t\) – \(\int \tan t \cdot \sec t d t\) = x + c
tan t – sec t = x + c
⇒ tan (x + y) – sec (x + y) = x + c

AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b)

ప్రశ్న 6.
\(\frac{d y}{d x}\) + \(\frac{y^2+y+1}{x^2+x+1}\) = 0
సాధన:
\(\frac{-d y}{y^2+y+1}\) = \(\frac{d x}{x^2+x+1}\)
ఇరువైపులా సమాకలనం చేయగా
AP Inter 2nd Year Maths 2B Solutions Chapter 8 అవకలన సమీకరణాలు Ex 8(b) 7

ప్రశ్న 7.
\(\frac{d y}{d x}\) = tan2 (x + y)
సాధన:
\(\frac{d y}{d x}\) = tan2 (x + y)
v = x + y అనుకుందాం
\(\frac{\mathrm{dv}}{\mathrm{dx}}\) = 1 + \(\frac{\mathrm{dy}}{\mathrm{dx}}\)
= 1 + tan2 v = sec2 v
\(\int \frac{d v}{\sec ^2 v}\) = \(\int d x\)
= \(\int \cos ^2 v \cdot d v\) = x + c
\(\int \frac{(1+\cos 2 v)}{2} d v\) = x + c
\(\int(1+\cos 2 v) d v\) = 2x + 2c
v + \(\frac{\sin 2 v}{2}\) = 2x + 2c
2v + sin 2v = 4x + c’
2(x + y) + sin 2(x + y) = 4x + c’
x – y – \(\frac{1}{2}\)[2(x + y)] = c