Inter 2nd Year

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Telugu Study Material Textbook Solutions Guide Questions and Answers PDF Free Download, TS AP Inter 2nd Year Telugu Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Telugu Textbook Pdf Download are part of AP Inter 2nd Year Study Material Pdf.

Students can also read the AP Inter 2nd Year Telugu Syllabus & AP Inter 2nd Year Telugu Important Questions for exam preparation.

AP Intermediate 2nd Year Telugu Study Material Pdf Download | Sr Inter 2nd Year Telugu Textbook Solutions

Inter 2nd Year Telugu Poems పద్య భాగం

• Poem 1 సత్య ప్రాశస్త్యము
• Poem 2 శాంతి కాంక్ష
• Poem 3 హనుమత్సందేశము
• Poem 4 కన్యక
• Poem 5 కర్మభూమిలో పూసిన ఓ పువ్వా….!
• Poem 6 ఈ దారి ఎక్కడికి పోతుంది?

Inter 2nd Year Telugu Textbook Lessons గద్య భాగం

• Chapter 1 మన ఆటలు
• Chapter 2 అర్థ విపరిణామం
• Chapter 3 చాటువులు
• Chapter 4 వేమన కవిత్వము
• Chapter 5 నా జీవిత యాత్ర
• Chapter 6 జాతీయోద్యమ కవిత్వం

Inter 2nd Year Telugu Non-Detailed ఉపవాచకం నందిని

• Chapter 1 గవేషణ
• Chapter 2 తెరచిన కళ్ళు
• Chapter 3 ఆశ ఖరీదు అణా

Inter 2nd Year Telugu Grammar

Intermediate 2nd Year Telugu Grammar Pdf

• భాషాభాగాలు
• వాక్య విజ్ఞానం
• ఛందస్సు
• అలంకారాలు
• సంక్షిప్తీకరణ

Intermediate 2nd Year Telugu Syllabus

AP Inter 2nd Year Telugu Syllabus

We hope that this Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Telugu Study Material Textbook PDF Download 2022-2023 helps the student to come out successful with flying colors in this examination. This Sr Inter 2nd Year Telugu Study Material will help students to gain the right knowledge to tackle any type of questions that can be asked during the exams.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers

Very Short Answer Questions

Question 1.
Explain why propanol has a higher boiling point than that hydrocarbon-butane.
Answer:
Propanol has a higher boiling point (391 K) than that hydrocarbon butane (309K).

Reason: In propanol, strong intermolecular hydrogen bonding is present between the molecules. But in the case of butane weak van der Waals force of attractions are present.

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Explanation: .

• Alcohols and water are both polar solvents. Alcohol is dissolves in water, due to formation of hydrogen bonding with water molecules.
• Hydro carbons are non polar and these dorv.t form hydrogen bonds with water molecules. So alcohols are soluble in water where as hydrocarbons are not soluble in water.

Question 3.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:
Given molecular formula of monnhydric phenols is C₇H₈O. The no. of possible isomers with molecular formula C₇H₈O are three.

Question 4.
Give the reagents used for the preparation of phenol from chiorobenzene.
Answer:
Phenol is prepared from chiorobenzene as follows. Reagents required are

1. NaOH, 623K, 300 atm
2. HCl.
   Chemical reaction :
   

Question 5.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:
Only 1° – alcohols form Ethers on acid dehydration. But not 2° or 3°-alcohols.

Reason : In case of 2° or 3° alcohols steric hindrance arises. Due to this steric hindrance alkenes are formed but not Ethers.

Question 6.
Write the mechanism of the reaction of HI with methoxymethane.
Answer:
Case – I: When methoxy methane reacts with cold.dil. HI then methyl alcohol and methyl iodide are formed.
Mechanisms:

Case – II : When methoxy methane reacts with hot.conc.HI then only methyl iodide is formed.
Mechanisms:

Question 7.
Name the reagents used in the following reactions.

1. Oxidation of primary alcohol to carboxylic acid
2. Oxidation of primary alcohol to aldehyde.

Answer:

1. The reagents used for the oxidation of 1° – alcohols to carboxylicacid are acidified K₂Cr₂O₇7 (or) Acidic/alkaline KMnO₄ (or) Neutral KMnO₄
2. The reagents used for the oxidation of 1°- alcohols to aldehyde are pyridine chloro chromate (PCC) in CH₂Cl₂.

Question 8.
Write the equations for the following reactions.
i) Bromination of phenol to 2,4, 6-tribromophenol
ii) Benzyl alcohol to benzoic acid.
Answer:
i) Bromination of phenól to 2, 4, 6 tribromophenol.

Question 9.
IdentIfy the reactant needed to form t-.butylalcohol from acetone.
Answer:
When acetone reacts with methyl magnesium bromide followed by the hydrolysis forms t-butyl alcohol.

Question 10.
Write the structures for the following compounds

1. Ethoxyethane
2. Ethoxybutane
3. Phenoxyethane

Answer:

1. Ethoxyethane → CH₃ – CH₂ – O – CH₂ – CH₃
2. Ethoxybutane → CH₃ – CH₂ – O – CH₂ – CH₂ – CH₂ – CH₃
3. Phenoxyethane → C₆H₅ – O – CH₂ – CH₃

Short Answer Questions

Question 1.
Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O₂ and give their IUFAC names and classify them as primary, secondary and tertiary alcohols.
Answer:

• Given molecular formula of compound is C₅H₁₂O.
• It has eight isomeric alcohols.
  
  

In the above isomeric alcohols (i), (ii), (iii), (iv) and 1°-alcohols; (v), (vi) and (viii) are 2°- alcohols, (vii) is 3°-alcohol.

Question 2.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give rèason.
Answer:
While separating a mixture of ortho and para nitrophenols by steam distillation, ortho nitrophenol is steam volatile.

Reason: In ortho nitrophenol intra molecular hydrogen bonding is present and in case of para nitrophenol inter molecular hydrogen bonding is present. So O-nitrophenol is steam volatile.

Question 3.
Give the equations for the preparation of phenol from Cumene. [Mar. 14]
Answer:
Phenol.is prepared from Cumene as follows.

1. Oxidation of Cumene to Cumene hydroperoxide.
2. Cumene hydroperoxide on acidic hydrolysis to form phenol.
   

Question 4.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Hydration of Ethene to yield ethanol involves 3—step mechanism.
Step – 1: In step – 1 formation of carbocation takes place by the protonation of ethene.

Step – 2 : In step – 2 carbocation formed in the above step attacked by water.

Step – 3 : Ethyl alcohol is formed by he deprotonation in step -3

Question 5.
Explain the acidic nature of phenols and compare with that of alcohols.
Answer:
The reaction of phenol with sodium metal and with aq.NaOH indicates the acidic nature of phenol.
i) Phenol reacts with sodium metal to form sodium phenoxide.

•  In phenol hydroxyl group is attached to the Sp² hydridised carbon of benzene ring which acts as electron with drawing group. The formed phenoxide ion from phenol is more stabilised due to delocalisation of negative charge.

Comparison of acidic character of Phenol and Ethanol:

• The reaction of phenol with aq. NaOH indicates that phenols are stronger acids than alcohols.
• The hydroxyl group attached to an aromatic ring is more acidic than in hydroxyl group is attached to an alkyl group.
• Phenol forms stable phenoxide ion stabilised by resonance but ethoxide ion is not.
  

Question 6.
Write the products formed by the reduction and oxidation of phenol.
Answer:
a) Reduction of phenol: Phenol undergo reduction in presence of zinc dust to form benzene.

b) Oxidation of phenol : Phenol undergo oxidation with chromicacid and forms a conjugated diketone known as benzoquinone.

Question 7.
Ethanol with H₂SO₄, at 443K forms ethene while at 413 K it forms ethoxy ethane. Explain the mechanism.
Answer:
Case – 1: Ethanol reacts with Cone. H₂SO₄ at 443K forms ethene

Mechanism:
Step – 1: Formation of protonated alcohol

Step 2 : Formation of carbo cation

Step 3 : Formation of ethene by elimination of a proton

Case – II : Ethanol reacts with Cone. H₂SO₄ at 413 K to form ethoxy ethane.

Mechanism:
In the above reaction ether formation is a SN reaction. This involve attack of alcohol molecule on a protonated alcohol.

Question 8.
Account for the statement: Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.
Answer:
Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.

Explanation : Consider ethanol, propane and methoxy methane which are having comparable molecular masses. The boiling points, molecular masses and structures of the above compounds mentioned below.

The higher boiling points of alcohols are due to the presence of intermolecular hydrogen bonding in them which is lacking in ethers and hydrocarbons.

Question 9.
Explain why in anisole electrophilic substitution takes place at ortho and para positions and not at meta position.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH₃ influences + R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophihic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m—position. So 0-and P-products are mainly formed during electrophillic substitution reactions.

Question 10.
Write the products of the following reactions :

Answer:

Long Answer Questions

Question 1.
Write the IUPAC name of the following compounds :

Answer:

Question 2.
Write structures of the compounds whose IUPAC names are as follows:
i) 2, Methyl butan—ol
ii) 1-Phenylprpan-2-ol
iii) 3, 5-Dhuethylhexane-1, 3, 5-triol
iv) 2, 3-Diethylphenol
v) 1-Ethoxypropane
vi) 2-Ethoxy-3-methylpentane
vii) Cyclohexylmethanol
viii) 3-Chloromethylpentan-1-ol
Answer:

Question 3.
Write the equations for the preparation of phenol using benzene, conc. H₂SO₄ and NaOH. [Mar. 14]
Answer:
The equations for the preparation of phenol using conc.H₂SO₄ and NaoH as follows

Question 4.
Illustrate hydroboration-oxidation reaction with a suitable example.
Answer:
When alkenes undergo addition reaction with diborane to form tri alkyl boranes. These followed by the oxidation by alkaline H₂O₂ to form alcohols. This reaction is called as hydroboration-oxidation reaction.

Question 5.
Write the IUPAC name of the following compounds:

Answer:

Question 6.
How will you synthesise:
i) 1 – Phenylethanol from a suitable alkene
ii) Cyclohexylmethanol using an alkyl halide by an SN² reaction.
iii) Pentan-1-ol using a suitable alkyl halide.
Answer:
i) Synthesis of 1-phenylethanol from a suitable alkene : When styrene undergo hydrolysis in presence of dil.H₂SO₄ to form 1-phenyl ethanol. It is an example of Marknowni koff’s rule.

ii) Synthesis of cyclohexyl methanol using an alkyl halide by SN² reaction : When cyclohexyl methyl bromide reacts with aq. NaOH to form cyclohexyl methanol.

iii) Synthesis of 1-pentanol using a suitable alkyl halide: When 1-Bromo pentane reacts with aq.KOH to form 1-pentanol.

Question 7.
Explain why-
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
ii) OH group attached to benzene ring activates it towards electrophilic substitution.
Answer:
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
Explanation: .

• -NO₂ is an electron withdrawing group and -OCH₃ is electron releasing group.
• By the presence of electron withdrawing group the phenoxide ion is more stabilized. By the presence of electron releasing group the phenoxide ion is less stabilized.
• Due to high stability of phenoxide ion, acidic nature increases.
  

ii) The -OH group attached to benzene ring activates it towards electrophilic substitution.

Explanation : When an electrophile is attacked, the – OH group exerts +R effect on the benzene ring. So electrodensity in the ring increases at ortho and para positions. When an electrophile attacks, substitution takes place at O and p-positions. So benzene ring activates towards electrophilic substitution.

Question 8.
Wth a suitable example write equations for the following:. [T.S. MAr. 19, 18; A.P. Mar. 18, 16] [A.P. Mar. 16]
i) Kolbe’s reaction
ii) Reimer-Tiemann reaction
iii) Williamsons ether synthesis
Answer:
i) Kolbes reaction: Phenol reacts with sodium hydroxide to form sodium phenoxide. This undergoes electrophilhic substitution with CO₂to form salicylic acid.

ii) Relmer-Tlemann reactIon : Phenol reacts with chloroform in presence of NaOH to form salicylaldehyde (O-Hydroxy benzaldehyde). This reaction is known as Reimer-Tiemann reaction. .

iii) Wilhiamsons ether synthesis:

• This method is used for the preparation of symmetrical and unsymmetrical ethers.
• The reaction of an alkyl halide with sodium alkoxide to form ethers is known as Williamsons Synthesis.
  

Question 9.
How are the following conversions carried out?
i) Benzyl chloride to Benzyl alcohol
ii) Ethyl magnesium bromide to Propan-1-ol
iii) 2-Butanone to 2-Butanol
Answer:
i) Conversion of Benzyl chloride to Benzyl alcohol : Ben.zyl chloride reacts with aq. NaOH to form benzyl alcohol.

ii) Conversion of Ethyl magnesium bromide to Propan-1-ol : Ethyl magnesium bromide reacts with form aldehyde followed by hydrolysis to form 1-propanol.

iii) Conversion of 2-Butanone to 2-Butanol: 2-Butanone undergo reduction in presence of LiA/H₄ to form 2-Butanol.

Question 10.
Write the names of the reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
i) 1-Propoxypropane
ii) Ethoxybenzene
iii) 2-Methoxy-2-methylpropane
iv) 1 -Methoxyethane
Answer:
i) Preparation of 1-propoxy propane :

Question 11.
How is 1-propoxypropane synthesized from propan-1-ol ? Write mechanism of this reaction.
Answer:
1 – Proponal reacts with Conc. H₂SO₄ at 413 K to form 1-propoxy propane.

Question 12.
Explain the fact that in aryl alkyl ethers the alkoxy group activates the benzene ring towards electrophilic substitution.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH₃ influences +R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophillic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m-position. So O-and P-products are mainly formed during electrophillic substitution reactions.

Question 13.
Write equations of the below given reactions:
i) Alkylation of anisole
ii) Nitration of anisole
iii) Friedel-Crafts acetylation of anisole
Answer:
i) Friedel crafts Alkylation of anisole :

ii) Nitration of anisole

Question 14.
Show how you would synthesize the following alcohols from appropriate alkenes?

Answer:

Question 15.
Explain why phenol with bromine water forms 2,4,6-tribromophenol while on reaction with bromine in CS₂ at low temperatures forms para-bromophenol as the major product.
Answer:
a) Phenol under goes Bromination in presence of CS₂ to form p-bromophenol as major product.

b) Phenol undergoes bromination in aqueous medium form 2,4,6 -tribromo phenol (white ppt).

Explanation: In bromination of phenol, the polarisation of Br₂ molecule takes place even in the absence of Lewis acid. This is due to the highly activating effect of -OH group attached to the benzene ring.

Textual Examples

Question 1.
Give IUPAC names of the following compounds:

Solution:
i) 4-Chloro-2, 3-dimethylpentan-1-ol
ii) 2-Ethoxypropane
iii) 2, 6-Dimethyiphenol
iv) 1-Ethoxy-2-nitrocyclohexane

Question 2.
Give the structures and IUPAC names of the products expected from the following reactions :
a) Catalytic reduction of butanal.
b) Hydration of propene in the presence of dilute sulphuric acid.
c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis.
Solution:

Question 3.
Arrange the following sets of compounds in order of their increasing boiling points :
a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
b) Pentan-1-ol, n-butane, pentanal, ethoxyethane.
Solution:
a) Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
b) n-Butane, ethoxyethane, pentanal and pentan-1-ol.

Question 4.
Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3, 5-dinitrophenol, phenol, 4-methylphenol.
Solution:
Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2, 4, 6-trinitrophenol.

Question 5.
Write the structures of the major products expected from the following reactions:
a) Mononitratlon of 3-methylphenol .
b) Dinitratlon of 3methylphenol
c) Mononitration of phenyl methanoate.
Solution:
The combined influence of -OH and -CH₃ groups determine the position of the incoming group.

Question 6.
The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

i) What would be the major product of this reaction?
ii) Write a suitable reaction for the preparation of t-butylethyl ether.
Solution:
i) The major product of the given reaction is 2-methylprop-1-ene.
It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

Question 7.
Give the major products that are formed by heating each of the following ethers with HI.

Solution:

Intext Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols:

Answer:
Primary alcohols (i), (ii), (iii)
Secondary alcohols (iv) and (y)
Tertiary alcohols (vi)

Question 2.
Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols are (ii) and (vi)

Question 3.
Name the following compounds according to IUPAC system.


Answer:
i) 3-Chioremethyl 2-isopropylpentan-1-ol
ii) 2, 5-Dimethylhexane-1, 3-dial
iii) 3-Bromocyclohexanol
iv) Hex-1-en-3-ol
v) 2-Bromo-3-methylbut-2-en-1-ol.

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent of methanol?

Answer:

Question 5.
Write structures of the products of the following reactions :


Answer:

Question 6.
Predict the major product of acid catalysed dehydration of

1. 1-methylcyclohexanol and
2. butan-1-ol.

Answer:

1. 1-Methylcyclohexene
2. A mixture of but-1-ene and but-2-ene. But-1-ene is the major product formed due to rearrangement to give,secondary carbocation.

Question 7.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer:

Question 8.
Predict the products of the following reactions:

Answer:

Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 2 System of Circles to solve questions creatively.

Intermediate 2nd Year Maths 2B System of Circles Formulas

Definition:
→ The angle between two intersecting circles is defined as the angle between the tangents at the point of intersection of the two circle’s. If θ is the angle between the circles, then
cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
Here d = distance between the centres, r₁, r₂ be their radii.

→ If θ is angle between the two circles.
x² + y² + 2gx + 2fy + c = 0 and x² + y² + 2g’x + 2f’y + c’ = 0, then
cos θ = \(\frac{-2 g g^{\prime}-2 f f^{\prime}+c+c^{\prime}}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

→ Circles cut orthogonally if
2g’g + 2f’f = c + c’ [∵ cos 90° = 0]
(or) d² = r²₁ + r²₂ then also two circles cut orthogonally.

Theorem:
If d is the distance between the centers of two intersecting circles with radii r₁, r₂ and θ is the angle between the circles then cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\).

Proof:
Let C₁, C₂ be the centre s of the two circles S = 0, S’ = 0 with radii r₁, r₂ respectively. Thus C₁C₂ = d. Let P be a point of intersection of the two circles. Let PB, PA be the tangents of the circles S = 0, S’ = 0 respectively at P.

Now PC₁ = r₁, PC₂ = r₂, ∠APB = θ
Since PB is a tangent to the circle S = 0, ∠C₁PB = π/2
Since PA is a tangent to the circle S’ = 0, ∠C₂PA = π/2
Now ∠C₁PC₂ = ∠C₁PB + ∠C₂PA – ∠APB = π/2 + π/2 – θ = π – θ
From ∆C₁PC₂, by cosine rule,
C₁²C₂² = PC₁² + PC₂² – 2PC₁ . PC₂ cos ∠C₁PC₂ ⇒ d² = r₁² + r₂² – 2r₁r₂ cos(π – θ) ⇒ d² = r₁² + r₂² + 2r₁r₂ cos θ
⇒ 2r₁r₂ cos θ = d² – r₁² – r₂² ⇒ cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\)

Corollary:
If θ is the angle between the circles x² + y² + 2gx + 2fy + c = 0, x² + y² + 2g’x + 2f’y + c’= 0 then cos θ = \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)
proof:
Let C₁, C₂ be the centre s and r₁, r₂ be the radii of the circles S = 0, S’ = 0 respectively and C₁C₂ = d.
∴ C₁ = (- g, – f), C₂ = (- g’, – f’),
r₁ = \(\sqrt{g^{2}+f^{2}-c}\), r₂ = \(\sqrt{g^{2}+f^{2}-c}\)
Now cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\) = \(\frac{\left(g-g^{\prime}\right)^{2}+\left(f-f^{\prime}\right)^{2}-\left(g^{2}+f^{2}-c\right)-\left(g^{\prime 2}+f^{\prime 2}-c^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)
= \(\frac{\mathrm{g}^{2}+\mathrm{g}^{\prime 2}-2 \mathrm{gg}^{\prime}+\mathrm{f}^{2}+\mathrm{f}^{\prime 2}-2 \mathrm{ff} \mathrm{f}^{\prime}-\mathrm{g}^{2}-\mathrm{f}^{2}+\mathrm{c}-\mathrm{g}^{2}-\mathrm{f}^{\prime 2}+\mathrm{c}^{\prime}}{2 \sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}} \sqrt{\mathrm{g}^{2}+\mathrm{f}^{\prime 2}-\mathrm{c}^{\prime}}}\)
= \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

Note: Let d be the distance between the centers of two intersecting circles with radii r₁, r₂. The two circles cut orthogonally if d² = r₁² + r₂².
Note: The condition that the two circles

S = x² + y² + 2gx + 2fy + c = 0, S’ = x² + y² + 2g’x + 2f’y + c’ = 0 may cut each other orthogonally is 2gg’ + 2ff’ = c + c’.
Proof: Let C₁, C₂ be the centers and r₁, r₂ be the radii of the circles S = 0, S’ = 0 respectively.
∴ C₁ = (- g, – f), C₂ = (- g’, – f’)
r₁ = \(\sqrt{g^{2}+f^{2}-c}\), r₂ = \(\sqrt{g^{2}+f^{2}-c^{\prime}}\)
Let P be point of intersection of the circles.
The two circles cut orthogonally at P
⇔ ∠C₁PC₂ = 90°.
⇒ C₁C₂² = C₁P² + C₂P² ⇔ (g – g’)² + (f – f’)² = r₁² + r₂²;
⇔ g² + g’² – 2gg’ + f² + f’² – 2ff’ = g² + f² – c + g’² + f’² + c’
⇔ – (2gg’ + 2ff’) = – (c + c’) ⇒ 2gg’+ 2ff’ = c + c’

Note:

• The equation of the common chord of the intersecting circles s = 0 and s¹ = 0 is s – s¹ = 0.
• The equation of the common tangent of the touching circles s = 0 and s¹ = 0 is s – s¹ = 0
• If the circle s = 0 and the line L = 0 are intersecting then the equation of the circle passing through the points of intersection of s = 0 and L = 0 is S + λL = 0.
• The equation of the circle passing through the point of intersection of S = 0 and S’ = 0 is s + λS’ = 0.

Theorem: The equation of the radical axis of the circles S = 0, S’ = 0 is S – S’ = 0.

Theorem: The radical axis of two circles is perpendicular to their line of centers.
Proof:
Let S = x² + y² + 2gx + 2fy + c = 0, S’ = x² + y² + 2g’x + 2f’y + c’= 0 be the given circles.

The equation of the radical axis is S – S’ = 0
⇒ 2(g – g’)x + 2(f – f’)y + (c – c’) = 0
⇒ a₁x + b₁y + c₁ = 0 where
a₁ = 2(g – g’), b₁ = 2(f – f’), c₁ = e – e’
The centers of the circles are (- g, – f), (- g’, – f’)
The equation to the line of centers is:
(x + g) (f – f’) = (y + f) (g – g’)
⇒ (f – f’)x – (g – g’)y – gf’ + fg’= 0
⇒ a₂x + b₂y + c₂ = 0 where
a₂ = f – f’, b₂ = – (g – g’), c₂ = fg’ – gf’
Now a₁a₂ + b₁b₂ = 2(g – g’) (f – f’) – 2(f – f’) (g – g’) = 0.

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Physics Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 2nd Year Physics Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Physics Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also read AP Inter 2nd Year Physics Syllabus & AP Inter 2nd Year Physics Important Questions for exam preparation. Students can also go through AP Inter 2nd Year Physics Notes to understand and remember the concepts easily.

AP Intermediate 2nd Year Physics Study Material Pdf Download | Sr Inter 2nd Year Physics Textbook Solutions

AP Inter 2nd Year Physics Solutions in English Medium

• Chapter 1 Waves
• Chapter 2 Ray Optics and Optical Instruments
• Chapter 3 Wave Optics
• Chapter 4 Electric Charges and Fields
• Chapter 5 Electrostatic Potential and Capacitance
• Chapter 6 Current Electricity
• Chapter 7 Moving Charges and Magnetism
• Chapter 8 Magnetism and Matter
• Chapter 9 Electromagnetic Induction
• Chapter 10 Alternating Current
• Chapter 11 Electromagnetic Waves
• Chapter 12 Dual Nature of Radiation and Matter
• Chapter 13 Atoms
• Chapter 14 Nuclei
• Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits
• Chapter 16 Communication Systems

AP Inter 2nd Year Physics Solutions in Telugu Medium

• Chapter 1 తరంగాలు
• Chapter 2 కిరణ దృశాశాస్త్రం, దృగ్ సాధనాలు
• Chapter 3 తరంగ దృశాశాస్త్రం
• Chapter 4 విద్యుత్ ఆవేశాలు, క్షేత్రాలు
• Chapter 5 స్థిర విద్యుత్ పోటెన్షియల్ – కెపాసిటెన్స్
• Chapter 6 ప్రవాహ విద్యుత్తు
• Chapter 7 చలించే ఆవేశాలు-అయస్కాంతత్వం
• Chapter 8 అయస్కాంతత్వం-ద్రవ్యం
• Chapter 9 విద్యుదయస్కాంత ప్రేరణ
• Chapter 10 ఏకాంతర విద్యుత్ ప్రవాహం
• Chapter 11 విద్యుదయస్కాంత తరంగాలు
• Chapter 12 వికిరణం, ద్రవ్యాల ద్వంద్వ స్వభావం
• Chapter 13 పరమాణువులు
• Chapter 14 కేంద్రకాలు
• Chapter 15 అర్ధవాహక ఎలక్ట్రానిక్స్, పదార్థాలు, పరికారాలు, సరళవలయాలు
• Chapter 16 సంసర్గ వ్యవస్థలు

TS AP Inter 2nd Year Physics Weightage Blue Print 2022-2023

TS AP Inter 2nd Year Physics Weightage 2022-2023 | TS AP Inter 2nd Year Physics Blue Print 2022

Intermediate 2nd Year Physics Syllabus

TS AP Inter 2nd Year Physics Syllabus

Chapter 1 Waves

• Introduction
• Transverse and longitudinal waves
• Displacement relation in a progressive wave
• The speed of a traveling wave
• The principle of superposition of waves
• Reflection of waves
• Beats
• Doppler effect

Chapter 2 Ray Optics and Optical Instruments

• Introduction
• Reflection of Light by Spherical Mirrors
• Refraction
• Total Internal Reflection
• Refraction at Spherical Surfaces and by Lenses
• Refraction through a Prism
• Dispersion by a Prism
• Some Natural Phenomena due to Sunlight
• Optical Instruments

Chapter 3 Wave Optics

• Introduction
• Huygens Principle
• Refraction and reflection of plane waves using the Huygens Principle
• Coherent and Incoherent Addition of Waves
• Interference of Light Waves and Young’s Experiment
• Diffraction
• Polarisation

Chapter 4 Electric Charges and Fields

• Introduction
• Electric Charges
• Conductors and Insulators
• Charging by Induction
• Basic Properties of Electric Charge
• Coulomb’s Law
• Forces between Multiple Charges
• Electric Field
• Electric Field Lines
• Electric Flux
• Electric Dipole
• Dipole in a Uniform External Field
• Continuous Charge Distribution
• Gauss’s Law
• Application of Gauss’s Law

Chapter 5 Electrostatic Potential and Capacitance

• Introduction
• Electrostatic Potential
• Potential due to a Point Charge
• Potential due to an Electric Dipole
• Potential due to a System of Charges
• Equipotential Surfaces
• Potential Energy of a System of Charges
• Potential Energy in an External Field
• Electrostatics of Conductors
• Dielectrics and Polarisation
• Capacitors and Capacitance
• The Parallel Plate Capacitor
• Effect of Dielectric on Capacitance
• Combination of Capacitors
• Energy Stored in a Capacitor
• Van de Graaff Generator

Chapter 6 Current Electricity

• Introduction
• Electric Current
• Electric Currents in Conductors
• Ohm’s law
• The drift of Electrons and the Origin of Resistivity
• Limitations of Ohm’s Law
• The resistivity of various Materials
• Temperature Dependence of Resistivity
• Electrical Energy, Power
• Combination of Resistors – Series and Parallel
• Cells, emf, Internal Resistance
• Cells in Series and in Parallel
• Kirchhoff’s Laws
• Wheatstone Bridge
• Meter Bridge
• Potentiometer

Chapter 7 Moving Charges and Magnetism

• Introduction
• Magnetic Force
• Motion in a Magnetic Field
• Motion in Combined Electric and Magnetic Fields
• Magnetic Field due to a Current Element, Biot-SavartLaw
• Magnetic Field on the Axis of a Circular Current Loop
• Ampere’s Circuital Law
• The Solenoid and the Toroid
• The force between Two Parallel Currents, the Ampere
• Torque on Current Loop, Magnetic Dipole
• The Moving Coil Galvanometer

Chapter 8 Magnetism and Matter

• Introduction
• The Bar Magnet
• Magnetism and Gauss’s Law
• The Earth’s Magnetism
• Magnetization and Magnetic Intensity
• Magnetic Properties of Materials
• Permanent Magnets and Electromagnets

Chapter 9 Electromagnetic Induction

• Introduction
• The Experiments of Faraday and Henry
• Magnetic Flux
• Faraday’s Law of Induction
• Lenz’s Law and Conservation of Energy
• Motional Electromotive Force
• Energy Consideration: A Quantitative Study
• Eddy Currents
• Inductance
• AC Generator

Chapter 10 Alternating Current

• Introduction
• AC Voltage Applied to a Resistor
• Representation of AC Current and Voltage by Rotating Vectors – Phasors
• AC Voltage Applied to an Inductor
• AC Voltage Applied to a Capacitor
• AC Voltage Applied to a Series LCR Circuit
• Power in AC Circuit: The Power Factor
• LC Oscillations
• Transformers

Chapter 11 Electromagnetic Waves

• Introduction
• Displacement Current
• Electromagnetic Waves
• Electromagnetic Spectrum

Chapter 12 Dual Nature of Radiation and Matter

• Introduction
• Electron Emission Photoelectric Effect
• Experimental Study of Photoelectric Effect
• Photoelectric Effect and Wave Theory of Light
• Einstein’s Photoelectric Equation: Energy Quantum of Radiation
• Particle Nature of Light: The Photon
• Wave Nature of Matter
• Davisson and Germer Experiment

Chapter 13 Atoms

• Introduction
• Alpha-particle Scattering and Rutherford’s Nuclear Model of Atom
• Atomic Spectra
• Bohr Model of the Hydrogen Atom
• The Line Spectra of the Hydrogen Atom
• DE Broglie’s Explanation of Bohr’s Second Postulate of Quantisation

Chapter 14 Nuclei

• Introduction
• Atomic Masses and Composition of Nucleus
• Size of the Nucleus
• Mass-Energy and Nuclear Binding Energy
• Nuclear Force
• Radioactivity
• Nuclear Energy

Chapter 15 Semiconductor Electronics: Materials, Devices and Simple Circuits

• Introduction
• Classification of Metals, Conductors and Semiconductors
• Intrinsic Semiconductor
• Extrinsic Semiconductor
• p-n Junction
• Semiconductor diode
• Application of Junction Diode as a Rectifier
• Special Purpose p-n Junction Diodes
• Junction Transistor
• Digital Electronics and Logic Gates
• Integrated Circuits

Chapter 16 Communication Systems

• Introduction
• Elements of a Communication System
• Basic Terminology Used in Electronic Communication Systems
• Bandwidth of Signals
• The bandwidth of Transmission Medium
• Propagation of Electromagnetic Waves
• Modulation and its Necessity
• Amplitude Modulation
• Production of Amplitude Modulated Wave
• Detection of Amplitude Modulated Wave

We hope that this Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Physics Study Material Textbook Solutions Guide PDF Free Download 2022-2023 in English Medium and Telugu Medium helps the student to come out successful with flying colors in this examination. This Sr Inter 2nd Year Physics Study Material will help students to gain the right knowledge to tackle any type of questions that can be asked during the exams.

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B System of Circles Important Questions which are most likely to be asked in the exam.

Intermediate 2nd Year Maths 2B System of Circles Important Questions

Question 1.
x² + y² + 4x + 8 = 0, x² + y² – 16y + k = 0 [A.P. & T.S. Mar. 16]
Solution:
g₁ = 2; f₁ = 0; c₁ = 8
g₂ = 0; f₂ = – 4; c₂ = k
Two circles are said to be orthogonal
2g₁g₂ + 2f₁f₂ = c₁ + c₂
2(2) (0) + 2(0) (-8) = 8 + k
0 + 0 = 8 + k
⇒ k = -8

Question 2.
(x – a)² + (y – b)² = c², (x – b)² + (y – a)² = c² (a ≠ b) [A.P. Mar. 15]
Solution:
(x² + y² – 2xa – 2yb – c²)
– (x² + y² – 2xb – 2ya – c²) = 0
– 2x (a – b) – 2y(b – a) = 0
(or) x – y = 0

Question 3.
Find the angle between the circles given by the equations. [T.S. Mar.17]
i) x² + y² – 12x – 6y + 41 = 0, x² + y² + 4x + 6y – 59 = 0.
Solution:
C₁ = (6, 3)
r₁ =(36 + 9 – 41)^1/2
r₁ = 2

C₂ = (-2, -3)
r₂ =(4 + 9 + 59)^1/2
r₂ = (72)^1/2 = 6\(\sqrt{2}\)

C₁C₂ = d = \(\sqrt{(6+2)^{2}+(3+3)^{2}}\)
= \(\sqrt{64+36}\) = 10
cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
= \(\frac{100-4-72}{2 \times 2 \cdot \sqrt{72}}=\frac{24}{4 \times 6 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
θ = 45°

Question 4.
Find the equation of the circle which cuts the circles x² + y² – 4x – 6y + 11 = 0 and x² + y² – 10x – 4y + 21 = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7. [A.P. Mar. 16; May 07]
Solution:
Let circle be x² + y² + 2gx + 2fy + c = 0 ……………….. (i)
Orthogonal to circle
2g (-2) +2f(-3) = 11 + c ……………………. (ii)
2g (-5) + 2f(-2) = 21 + c ……………………. (iii)
Subtracting it we get
-6g + 2f = 10 ……………………… (iv)
Circles centre is on 2x + 3y = 7
∴ -2g – 3f = 7 ……………………. (v)
Solving (iv) and (y)
f = -1, g = -2, c = 3
Equation of circle be x² + y² – 4x – 2y + 3 = 0

Question 5.
Show that the angle between the circles x² + y² = a², x² + y² = ax + ay is \(\frac{3\pi}{4}\) [Mar. 14]
Solution:
Equations of the circles are
S ≡ x² + y² – a² = 0
S’ ≡ x² + y² – ax – ay = 0
C₁ (0, 0), C₂ (\(\frac{a}{2}\), \(\frac{a}{2}\))
∴ C₁C₂² = (0 – \(\frac{a}{2}\))² + (0 – \(\frac{a}{2}\))²

Question 6.
If x + y = 3 ¡s the equation of the chord AB of the circle x² + y² – 2x + 4y – 8 = 0, find the equation of the circle having \(\overline{\mathrm{AB}}\) as diameter. [A.P. Mar. 15]
Solution:
Required equation of circle passing through
intersection S = 0 and L = 0 is S + λL = 0
(x² + y² – 2x + 4y – 8) + λ(x + y – 3) = 0
(x² + y² + x(-2 + λ)+ y(4 + λ) – 8 – 3λ = 0 ………………. (i)
x² + y² + 2gx + 2fy + c = 0 ……………………. (ii)
Comparing (i) and (ii) we get
g = \(\frac{(-2+\lambda)}{2}\), f = \(\frac{(4+\lambda)}{2}\)
Centre lies on x + y = 3
∴ \(-\left(\frac{-2+\lambda}{2}\right)-\left(\frac{4+\lambda}{2}\right)\) = 3
2 – λ – 4 – λ = 6
-2λ = 8 ⇒ λ = – 4
Required equation of circle be
(x² + y² – 2x + 4y – 8) – 4(x + y – 3) = 0
x² + y² – 6x + 4 = 0

Question 7.
If two circles x² + y² + 2gx + 2fy = 0 and x² + y² + 2g’x + 2f’y = 0 touch each other then show that f’g = fg’. [T.S. Mar. 16]
Solution:
C₁ = (-g, -f)
r₁ = \(\sqrt{g^{2}+f^{2}}\)
C₂ = (-g^-1, -f^-1)
r₂ = \(\sqrt{g^{\prime 2}+r^{\prime 2}}\)
C₁C₂ = r₁ + r₂
(C₁C₂)² = (r₁r₂)²
(g’ – g)² + (f’ – f)² = g² + f² + g’² + f’² + \(2 \sqrt{g^{2}+f^{2}} \sqrt{g^{2}+f^{1^{2}}}\)
-2(gg’ + ff’) = 2{g²g’² + g²f’² + f²g’²}^1/2
Squaring again
(gg’ + ff’)² = g²g’² + f²f’² + g²f’² + g’²f²
g²g’² + f²f’² + 2gg’ff’ = g²g’² + f²f’² + g²f’² + g’²f’²
2gg’ff’ = g²f’² + f²g’²
⇒ g²f’² + g’²f’² – 2gg’ff’ = 0
(or) (gf’ – fg’)² = 0 (or) gf’ = fg’

Question 8.
Show that the circles
S ≡ x² + y² – 2x – 4y – 20 = 0 …………………… (1)
and S ≡ x² + y² + 6x + 2y – 90 = 0 …………………(2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C₁, C₂ be the centres and r₁, r₂ be the radii of the given circles (1) and (2). Then
C₁ = (1, 2); C₂ = (-3, -1); r₁ = 5; r₂ = 10.
C₁C₂ = distance between the centres = 5
|r₁ – r₂| = |5 – 10| = 5 = C₁C₂
∴ The given two circles touch internally. In
this case, the common tangent is nothing but the radical axis. Therefore its equation is S – S’ = 0.
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides \(\overline{\mathrm{C}_{1} \mathrm{C}_{2}}\) in the ratio 5 : 10
i.e., 1: 2 (externally)
∴ Point of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5)

Question 9.
Find the angle between the circles x² + y² + 4x – 14y + 28 = 0 and x² + y² + 4x – 5 = 0
Solution:
Equations of the given circles are .
x² + y² + 4x – 14y + 28 = 0
x² + y² + 4x – 5 = 0
Centres are C₁ (-2, 7), C₂ (-2, 0)
C₁C₂ = \(\sqrt{(-2+2)^{2}+(7-0)^{2}}\)
= \(\sqrt{0+49}\) = 7
r₁ = \(\sqrt{4+49-28}\) = \(\sqrt{25}\) = 5
r₂ = \(\sqrt{4+5}\) = \(\sqrt{9}\) = 3
If θ is the angle between the given circles,
then cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)
cos θ = \(\frac{49-25-9}{2(5)(3)}\) = \(\frac{15}{2.5 .3}\) = \(\frac{1}{2}\) = cos 60°
Angle between the circles = θ = \(\frac{\pi}{3}\)

Question 10.
If the angle between the circles x² + y² – 12x – 6y + 41 = 0 and x² + y² + kx + 6y – 59 = 0 is 45° find k.
Solution:
Suppose θ is the angle between the circles
x² + y² – 12x – 6y + 41 = 0
and x² + y² + kx + 6y – 59 = 0
g₁ = -6, f₁ = -3, c₁ = 41,
g₂ = \(\frac{k}{2}\), f₂ = 3, c₂ = -59

\(\frac{1}{\sqrt{2}}=\frac{6 k}{4 \cdot \sqrt{\frac{k^{2}}{4}+68}}\)
Squaring and cross – multiplying
4(\(\frac{k^{2}}{4}\) + 68) = 18k²
\(\frac{2\left[k^{2}+272\right]}{4}\) = 9k²
k² + 272 = 18 k²
17k² = 272
k² = \(\frac{272}{17}\)
k² = 16
k = ±4.

Question 11.
Find the equation of the circle which passes through (1, 1) and cuts orthogonally each of the circles.
x² + y² – 8x – 2y + 16 = 0 and …………….. (1)
x² + y² – 4x – 4y – 1 = 0. ………………. (2)
Solution:
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0 ……………. (3)
Then the circle (3) is orthogonal to (1) and (2).
∴ By applying the condition of orthogonality give in x² + y² + 2gx + 2fy + c = 0 we get
2g(-4) + 2f(-1) = c + 16 and …………………. (4)
2g (-2) + 2f(-2) = c – 1 ……………….. (5).
Given that the circle (3) is passing through (1, 1)
∴ 1² + 1² + 2g(1) + 2f(1) + c = 0
2g + 2f + c + 2 = 0 ………………….. (6)
Solving (4), (5) and (6) for g, f and c, we get
g = –\(\frac{7}{3}\), f = \(\frac{23}{6}\), c = -5
Thus the equation of the required circle is
3(x² + y²) – 14x + 23y – 15 = 0.

Question 12.
Find the equation of the circle which is orthogonal to each of the following three circles.
x² + y² + 2x + 17y + 4 = 0 ………………… (1)
x² + y² + 7x + 6y + 11 = 0 ………………. (2)
and x² + y² – x + 22y + 3 = 0 ………………… (3)
Solution:
Let the equation of the required circle be
x² + y² + 2gx + 2fy + c = 0 ……………… (4)
Since this circle is orthogonal to (1), (2) and (3). by applying the condition of orthogonality given in x² + y² + 2gx + 2fy + c = 0
we have
2(g)(1) + 2(f) (\(\frac{17}{2}\)) = c + 4 ……………….. (5)
2(g) (\(\frac{7}{2}\)) + 2(f)(3) = c + 11 ………………. (6)
and 2(g) (-\(\frac{1}{2}\)) + 2(f)(11) = c + 3 ………………….. (7)
Solving (5), (6) and (7) for g, f, c we get
g = -3, f = -2 and c = -44
Thus the equation of the required circle is
x² + y² – 6x – 4y – 44 = 0.

Question 13.
If the straight line is represented by
x cos α + y sin α = p ………………….. (1)
intersects the circle
x² + y² = a² ……………… (2)
at the points A and B, then show that the equation of the circle with \(\overline{\mathrm{AB}}\) as diameter is(x² + y² – a²) – 2p(x cos α + y sin α – p) = 0.
Solution:
The equation of the circle passing through the points A and B is S ≡ x² + y² + 2gx + 2fy + c = 0
(x² + y² – a²) + λ(x cos α + y sin α – p) = 0 ……………… (3)
The centre of this circle is
\(\left(-\frac{\lambda \cos \alpha}{2},-\frac{\lambda \sin \alpha}{2}\right)\)
If the circle given by (3) has \(\overline{\mathrm{AB}}\) as diameter then the centre of it must lie on (1)
∴ \(-\frac{\lambda \cos \alpha}{2}\) (cos α) – \(\frac{\lambda \sin \alpha}{2}\) (sin α) = p
i.e., –\(\frac{\lambda}{2}\) (cos² α + sin² α) p
i.e., λ = -2p
Hence the equation of the required circle is
(x² + y² – a²) – 2p(x cos α + y sin α – p) = 0.

Question 14.
Find the equation of the circle passing through the points of intersection of the circles.
x² + y² – 8x – 6y + 21 = 0 ……………….. (1)
x² + y² – 2x – 15 = 0 ……………….. (2)
and (1, 2).
Solution:
The equation of circle passing through the points of intersection of (1) arid (2) is
(x² + y² – 8x – 6y + 21) + λ (x² + y² – 2x – 15) = 0 ……………….. (3)
If it passes through (1, 2). we obtain
(1 + 4 – 8 – 12 + 21) + λ(1 + 4 – 2 – 15) = 0
i.e., 6 + λ(-12) = 0
i.e., λ = \(\frac{1}{2}\)
Hence the equation of the required circle is
(x² + y² – 8x – 6y + 21) + \(\frac{1}{2}\) (x² + y² – 2x – 15) = 0
i.e., 3(x² + y²) – 18x – 12y + 27 = 0.

Question 15.
Let us find the equation the radical axis of the circles S ≡ x² + y² – 5x + 6y + 12 = 0 and S’ ≡ x² + y² + 6x – 4y – 14 = 0
Solution:
The given equations of circles are in general form. Therefore their radical axis is (S – S’ = 0)
i.e., 11x – 10y – 26 = 0

Question 16.
Let us find the equation of the radical axis of the circles
2x² + 2y² + 3x + 6y – 5 = 0 ………………….. (1)
and 3x² + 3y² – 7x + 8y – 11 = 0 ……………….. (2)
Solution:
Hence the given equations are not in general form, we get :
x² + y² + \(\frac{3}{2}\)x + 3y – \(\frac{5}{2}\) = 0 and
x² + y² – \(\frac{7}{3}\)x + \(\frac{8}{3}\) y – \(\frac{11}{3}\) = 0
Now the radical axis equation of given circles is
(\(\frac{3}{2}\) + \(\frac{7}{3}\))x + (3 – \(\frac{8}{3}\))y + (-\(\frac{5}{2}\) + \(\frac{11}{3}\)) = 0
i.e., 23x + 2y + 7 = 0

Question 17.
Let us find the radical centre of the circles
x² + y² – 2x + 6y = 0 ………………… (1)
x² + y² – 4x – 2y + 6 = 0 …………………. (2)
and x² + y² – 12x + 2y + 3 = 0 ………………. (3)
Solution:
The radical axis of (1) and (2) and (3)
x + 4y – 3 = 0 ………………… (4)
8x – 4y + 3 = 0 …………………. (5)
10x + 4y – 3 = 0 ……………… (6)
Solving (4) and (5) for the point of intersection we get (0, \(\frac{3}{4}\)) which is the required radical centre. Observe that the co-ordinates of this point satisfies (6) also.

Question 18.
Find the equation and length of the common chord of the two circles
S ≡ x² + y² + 3x + 5y + 4 = 0
and S’ ≡ x² + y² + 5x + 3y + 4 = 0
Solution:
Equations of the given circles are
S ≡ x² + y² + 3x + 5y + 4 = 0 ………………… (1)
S’ ≡ x² + y² + 5x + 3y + 4 ………………. (2)
Equations of the common chord is S – S’ = 0
-2x + 2y = 0
L ≡ x – y = 0 …………….. (3)

Question 19.
Show that the circles
S ≡ x² + y² – 2x – 4y – 20 = 0 ……………….. (1)
and S’ ≡ x² + y² + 6x + 2y – 90 = 0 ……………. (2)
touch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
Solution:
Let C₁, C₂ be the centres and r₁, r₂ be the radii of the given circles (1) and (2). Then
C₁ = (1, 2); C₂ = (-3, -1); r₁ = 5; r₂ = 10
C₁C₂ = distance between the centres = 5
|r₁ – r₂| = |5 – 10| = 5 = C₁C₂
∴ The given two circles touch internally. In this case, the.common tangent is nothing but the radical axis. Therefore its equation is
S – S’ = 0
i.e., 4x + 3y – 35 = 0
Now we find the point of contact. The point of contact divides in the ratio 5 : 10 i.e., 1 : 2 (externally)
∴ Point, of contact
= \(\left(\frac{(1)(-3)-2(1)}{1-2}, \frac{(1)(-1)-2(2)}{1-2}\right)\)
= (5, 5).

Question 20.
Find the equation of the circle whose diameter is the common chord of the circles
S ≡ x² + y² + 2x + 3y + 1 = 0 ……………….. (1)
and S’ ≡ x² + y² + 4x + 3y + 2 = 0 ……………….. (2)
Solution:
Here the common chord is the radical axis of (1) and (2). The equation of the radical axis is S – S’ = 0.
i.e., 2x + 1 = 0 …………………… (3)
The equation of any circle passing through
the points of intersection of (1) and (3) is (S + λL = 0)
(x² + y² + 2x + 3y + 1) + λ(2x + 1) = 0
x² + y² + 2(λ + 1)x + 3y + (1 + λ) = 0 ……………………. (4)
The centre of this circle is (-(λ + 1), \(\frac{3}{2}\)).
For the circle (4), 2x + 1 = 0 is one chord. This chord will be a diameter of the circle (4) if the centre of (4) lies on (3).
∴ 2{-(λ + 1)} + 1 = 0
⇒ λ = –\(\frac{1}{2}\)
Thus equation of the circle whose diameter is the common chord (1) and (2)
(put λ = \(\frac{1}{2}\) in equation (4))
2(x² + y²) + 2x + 6y + 1 = 0

Question 21.
Let us find the equation of a circle which cuts each of the following circles orthogonally
S’ ≡ x² + y² + 3x + 2y + 1 = 0 ………………… (1)
S” ≡ x² + y² – x + 6y + 5 = 0 …………….. (2)
and S” ≡ x² + y² + 5x – 8y + 15 = 0 ……………………. (3)
Solution:
The centre of the required circle is radical centre of (1), (2) and (3) and the radius is the length of the tangent from this point to any one of the given three circles. First we shall find the radical centre. For, the radical axis of (1) and (2) is
x – y = 1 ………………… (4)
and the radical axis of (2) and (3) is
3x – 7y = -5 …………………… (5)
The point of intersection (3, 2) of (4) and (5) is the radical centre of the circles (1), (2) and (3). The length of tangent from (3. 2) to the circle (1)
= \(\sqrt{3^{2}+2^{2}+3(3)+2(2)+1}=3 \sqrt{3}\)
Thus the required circle is
(x – 3)² + (y – 2)² = (3\(\sqrt{3}\))²
x² + y² – 6x – 4y – 14 = 0.

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Economics Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 2nd Year Economics Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Economics Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also read AP Inter 2nd Year Economics Syllabus & AP Inter 2nd Year Economics Important Questions for exam preparation. Students can also go through AP Inter 2nd Year Economics Notes to understand and remember the concepts easily.

AP Intermediate 2nd Year Economics Study Material Pdf Download | Sr Inter 2nd Year Economics Textbook Solutions

AP Inter 2nd Year Economics Study Material in English Medium

• Chapter 1 Economic Growth and Development
• Chapter 2 Population and Human Resources Development
• Chapter 3 National Income
• Chapter 4 Agriculture Sector
• Chapter 5 Industrial Sector
• Chapter 6 Tertiary Sector
• Chapter 7 Planning and Economic Reforms
• Chapter 8 Environment and Sustainable Economic Development
• Chapter 9 Economy of Andhra Pradesh
• Chapter 10 Economic Statistics

AP Inter 2nd Year Economics Study Material in Telugu Medium

• Chapter 1 ఆర్థికవృద్ధి – అభివృద్ధి
• Chapter 2 జనాభా, మానవ వనరుల అభివృద్ధి
• Chapter 3 జాతీయ ఆదాయం
• Chapter 4 వ్యవసాయ రంగం
• Chapter 5 పారిశ్రామిక రంగం
• Chapter 6 తృతీయ రంగం
• Chapter 7 ప్రణాళికలు – ఆర్ధిక సంస్కరణలు
• Chapter 8 పర్యావరణం మరియు సుస్థిర ఆర్థికాభివృద్ధి
• Chapter 9 ఆంధ్రప్రదేశ్ ఆర్థిక వ్యవస్థ
• Chapter 10 అర్థగణాంక శాస్త్రం

TS AP Inter 2nd Year Economics Weightage Blue Print 2022-2023

TS AP Inter 2nd Year Economics Weightage 2022-2023 | TS AP Inter 2nd Year Economics Blue Print 2022

Intermediate 2nd Year Economics Syllabus

TS AP Inter 2nd Year Economics Syllabus

Chapter 1 Economic Growth and Development
Introduction, Economic Growth, Economic Development, Differences between Economic Growth and Economic Development, Classification of the World Countries, Indicators of Economic Development, Determinants of Economic Development, Characteristic Features of Developed Countries, Characteristic Features of Developing Countries with Special Reference to India

Chapter 2 Population and Human Resources Development
Introduction, Theory of Demographic Transition, World Population, Causes of rapid Growth of Population in India, Occupational Distribution of Population of India, Meaning of Human Resources Development, Role of Education and Health in Economic Development, Human Development Index

Chapter 3 National Income
Introduction, Trends in the growth of Indias National Income, Trends in the distribution of National Income by Industry Origin, Share of Public Sector and Private Sector in Gross Domestic Product, Share of Organised and Unorganised Sector in Net Domestic Product, Income Inequalities, Causes of Income Inequalities, Measures to Control Income Inequalities, Unemployment in India, Poverty, Micro Finance Eradication of Poverty

Chapter 4 Agriculture Sector
Introduction, Importance of Agriculture in India, Features of Indian Agriculture, Agriculture Labour in India, Land Utilization Pattern in India, Cropping Pattern in India, Organic Farming, Irrigation Facilities in India, Productivity of Agriculture, Landholdings in India, Land Reforms in India, Green Revolution in India, Rural Credit in India, Rural Indebtedness in India, Agricultural Marketing,

Chapter 5 Industrial Sector
Introduction, Significance of the Indian Industrial Sector in Post-Reform Period, Industrial Policy Resolution 1948, Industrial Policy Resolution 1958, Industrial Policy Resolution 1991, National Manufacturing Policy, Disinvestment, National Investment Fund, Foreign Direct Investment, Special Economic Zones, Causes of Industrial backwardness in India, Small Scale Enterprises, Industrial Estates, Industrial Finance in India, The Industrial Development under the Five Year Plans in India

Chapter 6 Tertiary Sector
Introduction, Importance of Service Sector, Indias Services Sector, State wise comparison of Services, Infrastructure Development, Tourism, Banking and Insurance, Communication, Science and Technology, Software Industry in India

Chapter 7 Planning and Economic Reforms
Meaning of Planning, NITI Ayog, Five Year Plans in India, XII Five Year Plan, Regional Imbalances, Role of Trade in Economic Development, Economic Reforms in India, GATT, W.T.O

Chapter 8 Environment and Sustainable Economic Development
Environment, Economic Development, Environment, and Economic Linkages, Harmony between Environment and Economy

Chapter 9 Economy of Andhra Pradesh
History of Andhra Pradesh, Characteristic Features of AP Economy, Demographic Features, Occupational Distribution of Labour, Health Sector, Education, Environment, Agriculture Sector, Industrial Sector, Service and Infrastructure Sector, Information and Technology, Tourism, Andhra Pradesh and Welfare Programmes/Schemes

Chapter 10 Economic Statistics
Measures of Dispersion, Definition of Dispersion, Importance of Measuring Variation, Properties of a good measure of Variation, Methods of Studying Variation, Measures of Dispersion for Average, Lorenz Curve, Correlation, Index Numbers, Weighted Aggregation method

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Maths 2B Important Questions Chapter Wise with Solutions Pdf 2022 | Intermediate 2nd Year Maths 2B Important Questions

To access the Sr Inter 2nd Year Maths 2B Important Questions Chapter Wise with Solutions Pdf 2022, click on the links below.

AP Inter 2nd Year Maths 2B Important Questions in English Medium

Inter 2nd Year Maths 2B Important Questions with Solutions

1. Inter 2nd Year Maths 2B Circle Important Questions
2. Inter 2nd Year Maths 2B System of Circles Important Questions
3. Inter 2nd Year Maths 2B Parabola Important Questions
4. Inter 2nd Year Maths 2B Ellipse Important Questions
5. Inter 2nd Year Maths 2B Hyperbola Important Questions
6. Inter 2nd Year Maths 2B Integration Important Questions
7. Inter 2nd Year Maths 2B Definite Integrals Important Questions
8. Inter 2nd Year Maths 2B Differential Equations Important Questions

AP Inter 2nd Year Maths 2B Important Questions in Telugu Medium

• Chapter 1 వృత్తం Important Questions
• Chapter 2 వృత్త సరణులు Important Questions
• Chapter 3 పరావలయం Important Questions
• Chapter 4 దీర్ఘవృత్తం Important Questions
• Chapter 5 అతిపరావలయం Important Questions
• Chapter 6 సమాకలనం Important Questions
• Chapter 7 నిశ్చిత సమాకలనులు Important Questions
• Chapter 8 అవకలన సమీకరణాలు Important Questions

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Sr Inter 2nd Year Maths 2A Important Questions with Solutions: Here we have created a list of Telangana & Andhra Pradesh BIEAP TS AP Intermediate Sr Inter 2nd Year Maths 2A Important Questions Chapter Wise with Solutions Pdf 2022-2023 Download just for you. Those who are preparing for Inter exams should practice Inter Maths 2A Important Questions and doing so will clear their doubts instantly. These Inter 2nd Year Maths 2A Important Questions Chapterwise Pdf enhances your conceptual knowledge and prepares you to solve different types of questions in the exam.

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Maths 2A Important Questions Chapter Wise with Solutions Pdf 2022 | Intermediate 2nd Year Maths 2A Important Questions

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AP Inter 2nd Year Maths 2A Important Questions in English Medium

Inter 2nd Year Maths 2A Important Questions with Solutions

1. Inter 2nd Year Maths 2A Complex Numbers Important Questions
2. Inter 2nd Year Maths 2A De Moivre’s Theorem Important Questions
3. Inter 2nd Year Maths 2A Quadratic Expressions Important Questions
4. Inter 2nd Year Maths 2A Theory of Equations Important Questions
5. Inter 2nd Year Maths 2A Permutations and Combinations Important Questions
6. Inter 2nd Year Maths 2A Binomial Theorem Important Questions
7. Inter 2nd Year Maths 2A Partial Fractions Important Questions
8. Inter 2nd Year Maths 2A Measures of Dispersion Important Questions
9. Inter 2nd Year Maths 2A Probability Important Questions
10. Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions

AP Inter 2nd Year Maths 2A Important Questions in Telugu Medium

• Chapter 1 సంకీర్ణ సంఖ్యలు Important Questions
• Chapter 2 డిమోయర్ సిద్ధాంతం Important Questions
• Chapter 3 వర్గసమాసాలు Important Questions
• Chapter 4 సమీకరణ వాదం Important Questions
• Chapter 5 ప్రస్తారాలు-సంయోగాలు Important Questions
• Chapter 6 ద్విపద సిద్ధాంతం Important Questions
• Chapter 7 పాక్షిక భిన్నాలు Important Questions
• Chapter 8 విస్తరణ కొలతలు Important Questions
• Chapter 9 సంభావ్యత Important Questions
• Chapter 10 యాదృచ్ఛిక చలరాశలు, సంభావ్యత విభాజనాలు Important Questions

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Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Maths 2B Textbook Solutions Study Material Guide PDF Free Download, Inter 2nd Year Maths 2B Blue Print Weightage 2022-2023, Maths 2B Chapter Wise with Solutions Pdf in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also go through Inter 2nd Year Maths 2B Formulas PDF to understand and remember the concepts easily. Students can also read Inter 2nd Year Maths 2B Syllabus & Inter 2nd Year Maths 2B Important Questions Chapter Wise with Solutions Pdf 2022-2023 for exam preparation.

Inter 2nd Year Maths 2B Textbook Solutions PDF | Intermediate 2nd Year Maths 2B Study Material

Inter 2nd Year Maths 2B Solutions in English Medium

Inter 2nd Year Maths 2B Circle Solutions

• Chapter 1 Circle Ex 1(a)
• Chapter 1 Circle Ex 1(b)
• Chapter 1 Circle Ex 1(c)
• Chapter 1 Circle Ex 1(d)
• Chapter 1 Circle Ex 1(e)

Inter 2nd Year Maths 2B System of Circles Solutions

• Chapter 2 System of Circles Ex 2(a)
• Chapter 2 System of Circles Ex 2(b)

Inter 2nd Year Maths 2B Parabola Solutions

• Chapter 3 Parabola Ex 3(a)
• Chapter 3 Parabola Ex 3(b)

Inter 2nd Year Maths 2B Ellipse Solutions

• Chapter 4 Ellipse Ex 4(a)
• Chapter 4 Ellipse Ex 4(b)

Inter 2nd Year Maths 2B Hyperbola Solutions

• Chapter 5 Hyperbola Ex 5(a)

Inter 2nd Year Maths 2B Integration Solutions

• Chapter 6 Integration Ex 6(a)
• Chapter 6 Integration Ex 6(b)
• Chapter 6 Integration Ex 6(c)
• Chapter 6 Integration Ex 6(d)
• Chapter 6 Integration Ex 6(e)
• Chapter 6 Integration Ex 6(f)

Inter 2nd Year Maths 2B Definite Integrals Solutions

• Chapter 7 Definite Integrals Ex 7(a)
• Chapter 7 Definite Integrals Ex 7(b)
• Chapter 7 Definite Integrals Ex 7(c)
• Chapter 7 Definite Integrals Ex 7(d)

Inter 2nd Year Maths 2B Differential Equations Solutions

• Chapter 8 Differential Equations Ex 8(a)
• Chapter 8 Differential Equations Ex 8(b)
• Chapter 8 Differential Equations Ex 8(c)
• Chapter 8 Differential Equations Ex 8(d)
• Chapter 8 Differential Equations Ex 8(e)

Inter 2nd Year Maths 2B Solutions in Telugu Medium

Inter 2nd Year Maths 2B వృత్తం Solutions

• Chapter 1 వృత్తం Ex 1(a)
• Chapter 1 వృత్తం Ex 1(b)
• Chapter 1 వృత్తం Ex 1(c)
• Chapter 1 వృత్తం Ex 1(d)
• Chapter 1 వృత్తం Ex 1(e)

Inter 2nd Year Maths 2B వృత్త సరణులు Solutions

• Chapter 2 వృత్త సరణులు Ex 2(a)
• Chapter 2 వృత్త సరణులు Ex 2(b)

Inter 2nd Year Maths 2B పరావలయం Solutions

• Chapter 3 పరావలయం Ex 3(a)
• Chapter 3 పరావలయం Ex 3(b)

Inter 2nd Year Maths 2B దీర్ఘవృత్తం Solutions

• Chapter 4 దీర్ఘవృత్తం Ex 4(a)
• Chapter 4 దీర్ఘవృత్తం Ex 4(b)

Inter 2nd Year Maths 2B అతిపరావలయం Solutions

• Chapter 5 అతిపరావలయం Ex 5(a)

Inter 2nd Year Maths 2B సమాకలనం Solutions

• Chapter 6 సమాకలనం Ex 6(a)
• Chapter 6 సమాకలనం Ex 6(b)
• Chapter 6 సమాకలనం Ex 6(c)
• Chapter 6 సమాకలనం Ex 6(d)
• Chapter 6 సమాకలనం Ex 6(e)
• Chapter 6 సమాకలనం Ex 6(f)

Inter 2nd Year Maths 2B నిశ్చిత సమాకలనులు Solutions

• Chapter 7 నిశ్చిత సమాకలనులు Ex 7(a)
• Chapter 7 నిశ్చిత సమాకలనులు Ex 7(b)
• Chapter 7 నిశ్చిత సమాకలనులు Ex 7(c)
• Chapter 7 నిశ్చిత సమాకలనులు Ex 7(d)

Inter 2nd Year Maths 2B అవకలన సమీకరణాలు Solutions

• Chapter 8 అవకలన సమీకరణాలు Ex 8(a)
• Chapter 8 అవకలన సమీకరణాలు Ex 8(b)
• Chapter 8 అవకలన సమీకరణాలు Ex 8(c)
• Chapter 8 అవకలన సమీకరణాలు Ex 8(d)
• Chapter 8 అవకలన సమీకరణాలు Ex 8(e)

Inter 2nd Year Maths 2B Blue Print Weightage 2022-2023 | Blueprint of Intermediate 2nd Year Maths 2B

AP Inter 2nd Year Maths 2B Blue Print | Inter 2nd Year Maths 2B Weightage 2022-2023

Inter 2nd Year Maths 2B Syllabus | Intermediate Second Year Maths 2B Syllabus

Inter 2nd Year Maths 2B Syllabus Coordinate Geometry

1. Circle

• 1.1 Equation of circle – standard form-center and radius of a circle with a given line segment as diameter & equation of a circle through three noncollinear points – parametric equations of a circle.
• 1.2 Position of a point in the plane of a circle – the power of a point-definition of tangent-length of tangent
• 1.3 Position of a straight line in the plane of a circle-conditions for a line to be tangent – chord joining two points on a circle – equation of the tangent at a point on the circle – point of contact-equation of normal.
• 1.4 Chord of contact – pole and polar-conjugate points and conjugate lines equation of chord with given middle point.
• 1.5 The relative position of two circles-circles touching each other externally, internally common tangents – centers of similitude- equation of pair of tangents from an external point.

2. System of circles

• 2.1 The angle between two intersecting circles.
• 2.2 Radical axis of two circles – properties – Common chord and common tangent of two circles – radical centre. The intersection of a line and a Circle.

3. Parabola

• 3.1 Conic sections – Parabola – equation of parabola in standard form-different forms of parabola – parametric equations.
• 3.2 Equations of tangent and normal at a point on the parabola (Cartesian and parametric) – conditions for a straight line to be a tangent.

4. Ellipse

• 4.1 Equation of ellipse in standard form- Parametric equations.
• 4.2 Equation of tangent and normal at a point on the ellipse (Cartesian and parametric) – condition for a straight line to be a tangent.

5. Hyperbola

• 5.1 Equation of hyperbola in standard form – Parametric equations.
• 5.2 Equations of tangent and normal at a point on the hyperbola (Cartesian and parametric) – conditions for a straight line to be a tangent – Asymptotes.

Intermediate 2nd Year Maths 2B Syllabus Calculus

6. Integration

• 6.1 Integration as the inverse process of differentiation- Standard forms – properties of integrals.
• 6.2 Method of substitution-integration of Algebraic, exponential, logarithmic, trigonometric, and inverse trigonometric functions. Integration by parts.
• 6.3 Integration – Partial fractions method.
• 6.4 Reduction formulae.

7. Definite Integrals

• 7.1 Definite Integral as the limit of a sum
• 7.2 Interpretation of Definite Integral as an area.
• 7.3 Fundamental theorem of Integral Calculus.
• 7.4 Properties
• 7.5 Reduction formulae
• 7.6 Application of Definite integral to areas

8. Differential Equations

• 8.1 Formation of differential equation-Degree and order of an ordinary differential equation.
• 8.2 Solving differential equations by
  • Variables separable method
  • Homogeneous differential equation
  • Non-Homogeneous differential equation
  • Linear differential equations

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Intermediate 2nd Year Maths 2B Textbook Solutions

• All textual problems are solved.
• Diagrams are drawn wherever necessary.
• To solve Questions, a formula relating to that problem is also given to understand the students easily.
• The newly introduced problems are solved in a better way.

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AP Intermediate 2nd Year Maths 2A యాదృచ్ఛిక చలరాశలు, సంభావ్యత విభాజనాలు Formulas

→ యాదృచ్ఛిక చలరాశి : ఒక యాదృచ్ఛిక ప్రయోగం శాంపిల్ ఆవరణం S అనుకుందాం. ఏదైనా ప్రమేయం X : S → R ను యాదృచ్ఛిక చలరాశి అంటాం.

→ సంభావ్యతా విభాజన ప్రమేయం : X ఒక యాదృచ్ఛిక చలరాశి అప్పుడు F : R → R ప్రతి X ∈ Rకు F(x) = P(X ≤ x) తో నిర్వచితమైన ప్రమేయాన్ని X కు సంభావ్యతా విభాజన ప్రమేయం అంటాం.

→ X : S → R ఒక యాదృచ్ఛిక చలరాశి. X వ్యాప్తి పరిమితం లేదా అపరిమిత గణ్యసమితి అయితే X ను విచ్ఛిన్న చలరాశి అని, అట్లా కాకపోతే అవిచ్ఛిన్న యాదృచ్ఛిక చలరాశి అని అంటాం.

→ X : S → R ఒక విచ్ఛిన్న యాదృచ్ఛిక చలరాశి. దాని వ్యాప్తి = {X₁, X₂, ….} అయిన \(\sum_{r=1}^n P\left(X_r\right)\) = 1, P(X_r) ≥ 0.

→ X ఒక విచ్ఛిన్న యాదృచ్ఛిక చలరాశి. దాని వ్యాప్తి {X₁, X₂, ……} అనుకుందాం. ప్రతి n కు P(X = x_n) తెలిసి, Σx_n P(X = x_n) అనే మొత్తం పరిమితమైతే, ఆ మొత్తాన్ని X కు మధ్యమం (లేదా సగటు) అంటాం. దీన్ని µ తో సూచిస్తాం. (i.e.,) µ = Σx_n P(X = x_n)
Σ(x_n – µ)² P(X = x_n) అనేది ఒక పరిమిత సంఖ్య అయితే ఆ మొత్తాన్ని X కు విస్తృతి అంటాం.

→ X విస్తృతిని σ² తో సూచిస్తే, σ ను X కు క్రమ విచలనం అంటారు.
∴ μ = Σx_n P(X = x_n); σ² = Σ(x_n – μ)² P(X = x_n) = \(\Sigma x_n^2 P\left(X=x_n\right)\) – μ²

→ ద్విపద విభాజనం: n ఒక ధన పూర్ణాంకం. p వాస్తవ సంఖ్య మరియు 0 ≤ p ≤ 1. యాదృచ్ఛిక చలరాశి వ్యాప్తి {0, 1, 2, 3, …. n}. X ద్విపద చలరాశి లేదా ద్విపద విభాజనాన్ని పాటిస్తూ, n, p లు పరామితులుగా గల్గి వుంటే, P(X = r) = ⁿC_r . p^r q^n-r; r = 0, 1, 2, ….. n మరియు q = 1 – p అవుతుంది.

→ ద్విపద విభాజనాన్ని X ~ B(n, p) గా లేదా P(X = r) = ⁿC_r p^r q^n-r . p^r, r = 0, 1, 2, 3, …. n లేదా (q + p)ⁿతో సూచిస్తాం.

→ ద్విపద విభాజనం యొక్క మధ్యమం = np. ద్విపద విభాజనం యొక్క విస్తృతి = npq, క్రమవిచలనం = \(\sqrt{n p q}\).

→ పాయిజాన్ విభాజనం : λ > 0 ఒక స్థిరరాశి. యాదృచ్ఛిక చలరాశి X యొక్క వ్యాప్తి {0, 1, 2, ….}
P(X = k) = \(\frac{\lambda^k}{k !} e^{-\lambda}\), (k = 0, 1, 2, …….) అనుకుంటే, λ పరామితిగా, X పాయిజాన్ విభాజనాన్ని అనుసరిస్తుందని అంటాం. X ను పాయిజాన్ యాదృచ్ఛిక చలరాశి అంటాం. పాయిజాన్ విభాజనానికి మధ్యమము = λ, విస్తృతి = λ, క్రమ విచలనం = √λ

→ ఈ క్రింది షరతులకు లోబడి పాయిజాన్ విభాజనాన్ని, ద్విపద విభాజనపు సమతాస్థితి (limiting case) గా ఉజ్జాయింపు చేయవచ్చు.

• యత్నాల సంఖ్య n అనిశ్చితమైనంత పెద్దది, అంటే n → ∞
• ప్రతి యత్నంలో గెలుపు సంభావ్యత (స్థిరం) అతిస్వల్పం, అంటే p → 0

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 9 సంభావ్యత to solve questions creatively.

AP Intermediate 2nd Year Maths 2A సంభావ్యత Formulas

→ యాదృచ్ఛిక ప్రయోగం : ఒక ప్రయోగంలో ఏ ఫలితం వస్తుందో ముందే చెప్పలేనిదై, ఆ ప్రయోగ ఫలితాల జాబితా ముందే తెలిసి ఉండి, ఒకే విధమైన పరిస్థితుల్లో ఆ ప్రయోగాన్ని ఎన్నిసార్లైనా చేయడానికి వీలుంటే ఆ ప్రయోగాన్ని యాదృచ్ఛిక ప్రయోగం అంటాం.

→ లఘుఘటన, ఘటన ఒక యాదృచ్ఛిక ప్రయోగంలోని ఫలితాన్ని లఘుఘటన అంటాం. కొన్ని లఘు ఘటనల సమూహాన్ని ఘటన అంటాం.

→ పరస్పర వివర్జిత ఘటనలు: రెండు లేదా అంతకంటే ఎక్కువ ఘటనల్లో ఏదైనా ఒక ఘటన సంభవించడం, మిగతా ఘటనల సంభవాన్ని నిరోధించేటట్టుంటే, అటువంటి ఘటనలను పరస్పర వివర్జిత ఘటనలు అంటారు.

→ సమ సంభవ ఘటనలు: రెండు లేదా అంతకంటే ఎక్కువ ఘటనల్లో ఏ ఘటన అయినా మిగతా ఘటనల కంటే ఎక్కువగా సంభవిస్తుందనడానికి కారణమేమి లేకపోతే, అటువంటి ఘటనలను సమసంభవ ఘటనలు అంటారు.

→ పూర్ణ ఘటనలు : ఒక యాదృచ్ఛిక ప్రయోగంలోని రెండు లేదా అంతకంటే ఎక్కువ ఘటనల్లో ఆ ప్రయోగం ఫలితం వాటిలో ఒక్కదానికైనా చెందేటట్లుంటే, అటువంటి ఘటనలను పూర్ణ ఘటనలు అంటాం.

→ సంభావ్యత సాంప్రదాయిక (లేదా గణితాత్మక) నిర్వచనం : ఒక యాదృచ్ఛిక ప్రయోగంలో n పూర్ణ, పరస్పర వివర్జిత, సమసంభవ ఘటనలుండి వాటిలో ఏదైనా ఘటన E జరగడానికి m అనుకూల ఫలితాలుంటే, ఆ ఘటన సంభావ్యతను P(E) తో సూచిస్తూ, P(E) = \(\frac{m}{n}\) గా నిర్వచిస్తాం. 0 ≤ P(E) ≤ 1

→ శాంపిల్ ఆవరణ : ఒక యాదృచ్ఛిక ప్రయోగపు ఫలితాన్ని లఘుఘటన అంటాం. ఒక యాదృచ్ఛిక ప్రయోగంలోని ఫలితాలన్నింటితో కూడిన సమితిని ఆ యాదృచ్ఛిక ప్రయోగానికి సంబంధించి శాంపిల్ ఆవరణ అంటాం. దీనిని S తో సూచిస్తాం. S లోని మూలకాలను శాంపిల్ బిందువులు అంటాం. S లో ప్రతిమూలకం, యాదృచ్ఛిక ప్రయోగం యొక్క ఒక ఫలితం S. ఒక ఉపసమితిని ఘటన అంటాం. అంటే, ఒక లఘుఘటనల సమితినే ఘటన అంటాం.

→ S శాంపిల్ ఆవరణ E ⊂ S.E లో ఒకే ఒక మూలకం ఉంటే E ని లఘుఘటన అంటాం. ఒక ప్రయోగ ఫలితం ఘటన E సంభవించింది లేదా జరిగింది అంటాం. అలాకాకపోతే ఘటన E సంభవించలేదు అంటాం.

→ Φ, S లు S కి ఉపసమితులు. వాటిని వరసగా అసంభవ ఘటన, నిశ్చిత ఘటన అంటాం.

→ ఘటన E కి పూరక ఘటనను E^C తో సూచిస్తాం. E^C = S – E శాంపిల్ ఆవరణ S కు E₁, E₂ లు రెండు ఘటనలు. అంటే E₁ ⊆ S, E₂ ⊆ S మరియు E₁ ∩ E₂ = Φ అయితే E₁, E₂ లను పరస్పర వివర్జిత ఘటనలు అంటాం.

→ ఘటనలు E₁, E₂, ……., E_n లు i ≠ j లకు 1 ≤ i, j ≤ n లకు E_i ∩ E_j = Φ అయ్యేటట్లుంటే, వాటిని పరస్పర వివర్జిత ఘటనలు అంటాం.

→ ఘటనలు E₁, E₂, …… E_k లు E₁ ∪ E₂ ∪ E₃ ∪ E_k = S అయితే వాటిని పూర్ణ ఘటనలు అంటాం. శాంపిల్ ఆవరణ S లో E₁, E₂లు రెండు ఘటనలు పూరక ఘటనలైన E₁ ∪ E₂ = S, E₁ ∩ E₂ = Φ.

→ ఒక యాదృచ్ఛిక ప్రయోగంకి శాంపిల్ ఆవరణ S. ఈ ప్రయోగపు అన్ని ఘటనల సమితి P(S) తో సూచిస్తాం. ఇచ్చట P(S), Sకు ఘాత సమితి.

→ సంభావ్యతా స్వీకృత నిర్వచనం: ఒక యాదృచ్ఛిక ప్రయోగపు శాంపిల్ ఆవరణ S. ప్రమేయం P : P(S) → R

→ క్రింది స్వీకృతాలను ధ్రువపరిస్తే Pని సంభావ్యతా ప్రమేయం అంటాం.

• P(E) ≥ 0 ∀ E ∈ P(S) (ధన స్వీకృతం)
• P(S) = 1 (పూరణ స్వీకృతం)
• E₁, E₂ ∈ P(S), E₁ ∩ E₂ = Φ అయితే P(E₁ ∪ E₂) = P(E₁) + P(E₂) (సమ్మేళును స్వీకృతం)
• ప్రతి E ∈ P(S) కు P(E) ను ఘటన E సంభావ్యత అంటాం.

→ శాంపిల్ ఆవరణ Sలో E ఒక ఘటన అయితే 0 ≤ P(E) ≤ 1. శాంపిల్ ఆవరణ Sలో E ఒక ఘటన అయితే

• P(E) : P(\(\bar{S}\)) ను E కు అనుకూలత అని
• P(E) : P(E) ను E కు ప్రతికూలత అంటాం.

→ సంభావ్యతపై సంకలన సిద్ధాంతం: ఒక యాదృచ్ఛిక ప్రయోగంలో E₁, E₂ లు రెండు ఘటనలైతే P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)

→ శాంపిల్ ఆవరణ Sకు E₁, E₂ లు రెండు ఘటనలు.
P(E₂ – E₁) = P(E₂) – P(E₁ ∩ E₂) మరియు P(E₁ – E₂) = P(E₁) – P(E₁ ∩ E₂)

→ E₁, E₂, E₃ లు శాంపిల్ ఆవరణ Sలో మూడు ఘటనలు అయిన P(E₁ ∪ E₂ ∪ E₃) = P(E₁) + P(E₂) + P(E₃) – P(E₁ ∩ E₂) – P(E₂ ∩ E₃) – P(E₃ ∩ E₁) + P(E₁ ∩ E₂ ∩ E₃)

→ ఒక యాదృచ్ఛిక ప్రయోగపు ఘటనలు A, Bలు అనుకుందాం. అప్పుడు “A జరిగిన తరువాత B జరగడం” అనే ఘటనను నియత ఘటన అంటాం. దీనిని \(\frac{B}{A}\) తో సూచిస్తాం. ఇట్లే \(\frac{A}{B}\) అనే ఘటన “B జరిగిన తరువాత A జరగడం” అనే ఘటనను సూచిస్తుంది.

→ నియత సంభావ్యత : ఘటన A జరిగిందని ఇస్తే, ఘటన B జరిగే సంభావ్యతను P(B/A) తో సూచిస్తాం. P(B/A) ని నియత సంఖ్య అంటాం.
దీనిని P(B/A) = \(\frac{P(B \cap A)}{P(A)}\), P(A) > 0 గా నిర్వచిస్తాం. ఇట్లే P(A/B) = \(\frac{P(B \cap A)}{P(B)}\), P(B) > 0.
సూచన : P(A/B) = \(\frac{n(A \cap B)}{n(B)}\); P(B/A) = \(\frac{n(B \cap A)}{n(A)}\)

→ నియత సంభావ్యతకు గణన సిద్ధాంతం ఒక శాంపిల్ ఆవరణం S లోని ఘటనలు A, Bలు ; P(A) > 0, P(B) > 0, అయితే P(A ∩ B) = P(A) . P(B/A) = P(B) . P(A/B).

→ స్వతంత్ర ఘటనలు : రెండు ఘటనలు A, Bలు P(A ∩ B) = P(A) . P(B) అయితే వాటిని స్వతంత్ర ఘటనలు అంటాం. అలాకాకపోతే A, B లను అస్వతంత్ర ఘటనలు అంటాం.

→ బేయీ సిద్ధాంతం: ఒక ప్రయోగంలో E₁, E₂, ……., E_n లు పరస్పర వివర్జిత, పూర్ణ ఘటనలవుతూ, P(E_i) > 0, i = 1, 2, …. n అయినప్పుడు k = 1, 2, 3, ….. n లకు
\(P\left(\frac{E_k}{A}\right)=\frac{P\left(E_k\right) \cdot P\left(\frac{A}{E_k}\right)}{\sum_{i=1}^n P\left(E_i\right) \cdot P\left(\frac{A}{E_i}\right)}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 8 విస్తరణ కొలతలు to solve questions creatively.

AP Intermediate 2nd Year Maths 2A విస్తరణ కొలతలు Formulas

→ అవర్గీకృత దత్తాంశానికి మధ్యమం = \(\frac{విచలనాల మొత్తం}{పరిశీలనల సంఖ్య}\) = \(\frac{\Sigma x_i}{n}\)

→ అవర్గీకృత దత్తాంశానికి మధ్యగతం: ముందుగా దత్త n పరిశీలనలను పరిమాణపరంగా అవరోహణ లేదా ఆరోహణ క్రమంలో వ్రాయవలెను.

→ n బేసి సంఖ్య అయితే \(\frac{n+1}{2}\) వ పరిశీలనల అవర్గీకృత దత్తాంశం యొక్క మధ్యగతం అగును.

→ n సరి సంఖ్య అయితే \(\frac{n}{2}\) మరియు \(\frac{n+2}{2}\) వ పరిశీలనల సరాసరిను అవర్గీకృత దత్తాంశం యొక్క మధ్యగతం అగును.

→ వర్గీకృత మరియు అవర్గీకృత దత్తాంశానికి వ్యాప్తి, మధ్యమ విచలనం, విస్తృతి మరియు ప్రామాణిక విచలనం కొన్ని విస్తరణ కొలతలు

→ వ్యాప్తిని దత్తాంశ గరిష్ట విలువకు, కనిష్ఠ విలువకు మధ్యగల భేదంగా నిర్వచిస్తారు.

→ అవర్గీకృత విభాజనానికి మధ్యమ విచలనం

• మధ్యమం నుంచి మధ్యమ విచలనం = \(\frac{\sum\left|x_i-\bar{x}\right|}{n}, \bar{x}\) మధ్యమం
• మధ్యగతం నుంచి మధ్యమ విచలనం = \(\frac{\sum \mid x_i-\text { మధ్యగతం } \mid}{n}\)

→ వర్గీకృత దత్తాంశానికి మధ్యమ విచలనం

• మధ్యమం నుంచి మధ్యమ విచలనం = \(\frac{1}{N} \sum f_i\left|x-\bar{x}_i\right|\); N = Σf_i, మరియు \(\bar{x}\) మధ్యమం
• మధ్యగతం నుంచి మధ్యమ విచలనం = \(\frac{1}{N} \sum f_i \mid x_i\) – మధ్యగతం|, N = Σf_i

→ అవర్గీకృత దత్తాంశానికి విస్తృతి, σ² = \(\frac{1}{n}\) = \(\sum\left(x_i-\bar{x}\right)^2\), ప్రామాణిక విచలనం σ = \(\sqrt{\frac{1}{n} \sum\left(x_1-\bar{x}\right)^2}\)

→ విచ్ఛిన్న పౌనఃపున్య విభాజనానికి విస్తృతి, σ² = \(\frac{1}{N}\) = \(\sum f_i\left(x_i-\bar{x}\right)^2\), \(\bar{x}\) మధ్యమం

→ ప్రామాణిక విచలనం σ = \(\sqrt{\frac{1}{N} \sum f_i\left(x_i-\bar{x}\right)^2}\)

→ అవిచ్ఛిన్న పౌనఃపున్య విభాజనానికి ప్రామాణిక విచలనం σ = \(\frac{1}{N} \sqrt{N \sum f_i x_i^2-\left(\sum f_i x_i\right)^2}\) (లేదా) σ = \(\frac{h}{N} \sqrt{N \Sigma f_i y_i^2-\left(\Sigma f_i y_i\right)^2}, y_i=\frac{x_i-A}{h}\)

→ విచలనాంకం = \(\frac{\sigma}{x} \times 100(\bar{x} \neq 0)\)

→ దత్తాంశంలోని ప్రతి పరిశీలనను ఒక స్థిరరాశి K చే గుణించినపుడు ఫలితంగా వచ్చే పరిశీలనల విస్తృతి, తొలిపరిశీలనల విస్తృతికి K² రెట్లు ఉంటుంది.

→ పరిశీలనలు x₁, x₂, …….., x_n లలో ప్రతిదానిని K కి పెంచితే లేదా కలిపితే (K ఒక ధనాత్మక లేదా ఋణాత్మక సంఖ్య), వచ్చే పరిశీలనల విస్తృతి మారదు.