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

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.

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

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 1

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

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.

AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and 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:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 3
Case – II : When methoxy methane reacts with hot.conc.HI then only methyl iodide is formed.
Mechanisms:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 4

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 5

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 6

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

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.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 7
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 8

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 9

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

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.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 10

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 11
Step – 2 : In step – 2 carbocation formed in the above step attacked by water.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 12
Step – 3 : Ethyl alcohol is formed by he deprotonation in step -3
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 13

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 14

  •  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.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 15

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

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 16
b) Oxidation of phenol : Phenol undergo oxidation with chromicacid and forms a conjugated diketone known as benzoquinone.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 17

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
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 18
Mechanism:
Step – 1: Formation of protonated alcohol
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 19
Step 2 : Formation of carbo cation
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 20
Step 3 : Formation of ethene by elimination of a proton
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 21
Case – II : Ethanol reacts with Cone. H2SO4 at 413 K to form ethoxy ethane.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 22
Mechanism:
In the above reaction ether formation is a SN reaction. This involve attack of alcohol molecule on a protonated alcohol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 23

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 24
The higher boiling points of alcohols are due to the presence of intermolecular hydrogen bonding in them which is lacking in ethers and hydrocarbons.

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

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 25

Question 10.
Write the products of the following reactions :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 26
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 27
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 28

Long Answer Questions

Question 1.
Write the IUPAC name of the following compounds :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 29
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 30

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:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 31
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 32

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

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
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 33

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 34

Question 5.
Write the IUPAC name of the following compounds:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 35
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 36

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

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 37
ii) Synthesis of cyclohexyl methanol using an alkyl halide by SN2 reaction : When cyclohexyl methyl bromide reacts with aq. NaOH to form cyclohexyl methanol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 38
iii) Synthesis of 1-pentanol using a suitable alkyl halide: When 1-Bromo pentane reacts with aq.KOH to form 1-pentanol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 39

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: .
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 40

  • -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.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 41

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 42

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

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 43

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. .
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 44

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.
    AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 45

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 46

ii) Conversion of Ethyl magnesium bromide to Propan-1-ol : Ethyl magnesium bromide reacts with form aldehyde followed by hydrolysis to form 1-propanol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 47

iii) Conversion of 2-Butanone to 2-Butanol: 2-Butanone undergo reduction in presence of LiA/H4 to form 2-Butanol.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 48

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 :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 49
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 50

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

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 51

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 52

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 :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 53
ii) Nitration of anisole
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 54

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

Question 14.
Show how you would synthesize the following alcohols from appropriate alkenes?
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 55
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 56
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 57

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 58
b) Phenol undergoes bromination in aqueous medium form 2,4,6 -tribromo phenol (white ppt).
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 59
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:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 60
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:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 61

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

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 62

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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 63

Question 6.
The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 64
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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 65

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

Question 7.
Give the major products that are formed by heating each of the following ethers with HI.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 66
Solution:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 67

Intext Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 68
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.
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 69
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 70
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.

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

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent of methanol?
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 71
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 72

Question 5.
Write structures of the products of the following reactions :
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 73
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 74
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 75

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:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 76

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

Question 8.
Predict the products of the following reactions:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 77
Answer:
AP Inter 2nd Year Chemistry Study Material Chapter 12(a) Alcohols, Phenols, and Ethers 78

Inter 2nd Year Maths 2B System of Circles Formulas

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.

Inter 2nd Year Maths 2B System of Circles Formulas

→ 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.
Inter 2nd Year Maths 2B System of Circles Formulas 1
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 = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\)

Inter 2nd Year Maths 2B System of Circles Formulas

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.

Inter 2nd Year Maths 2B System of Circles Formulas

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.
Inter 2nd Year Maths 2B System of Circles Formulas 2
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.

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

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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.

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

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

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

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°

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

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
Inter 2nd Year Maths 2B System of Circles Important Questions 1

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

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

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)

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

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
Inter 2nd Year Maths 2B System of Circles Important Questions 2
\(\frac{1}{\sqrt{2}}=\frac{6 k}{4 \cdot \sqrt{\frac{k^{2}}{4}+68}}\)
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.

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

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.

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

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

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

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)
Inter 2nd Year Maths 2B System of Circles Important Questions 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).

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

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.

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

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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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Inter 2nd Year Maths 2B Important Questions with Solutions | Sr Inter Maths 2B Important Questions PDF 2022-2023

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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

Inter 2nd Year Maths 2B Blue Print Weightage

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Inter 2nd Year Maths 2A Important Questions 2022-2023 | Sr Inter Maths 2A Important Questions

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  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

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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

Inter 2nd Year Maths 2B System of Circles Solutions

Inter 2nd Year Maths 2B Parabola Solutions

Inter 2nd Year Maths 2B Ellipse Solutions

Inter 2nd Year Maths 2B Hyperbola Solutions

Inter 2nd Year Maths 2B Integration Solutions

Inter 2nd Year Maths 2B Definite Integrals Solutions

Inter 2nd Year Maths 2B Differential Equations Solutions

Inter 2nd Year Maths 2B Solutions in Telugu Medium

Inter 2nd Year Maths 2B వృత్తం Solutions

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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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AP Inter 2nd Year Maths 2A Formulas Chapter 10 యాదృచ్ఛిక చలరాశలు, సంభావ్యత విభాజనాలు

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 10 యాదృచ్ఛిక చలరాశలు, సంభావ్యత విభాజనాలు to solve questions creatively.

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(Xr) ≥ 0.

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

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

AP Inter 2nd Year Maths 2A Formulas Chapter 10 యాదృచ్ఛిక చలరాశలు, సంభావ్యత విభాజనాలు

→ 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) = nCr . pr qn-r; r = 0, 1, 2, ….. n మరియు q = 1 – p అవుతుంది.

→ ద్విపద విభాజనాన్ని X ~ B(n, p) గా లేదా P(X = r) = nCr pr qn-r . pr, 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

AP Inter 2nd Year Maths 2A Formulas Chapter 9 సంభావ్యత

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

AP Inter 2nd Year Maths 2A Formulas Chapter 9 సంభావ్యత

→ 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 EC తో సూచిస్తాం. EC = 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, ……., En లు i ≠ j లకు 1 ≤ i, j ≤ n లకు Ei ∩ Ej = Φ అయ్యేటట్లుంటే, వాటిని పరస్పర వివర్జిత ఘటనలు అంటాం.

→ ఘటనలు E1, E2, …… Ek లు E1 ∪ E2 ∪ E3 ∪ Ek = 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) = \(\frac{n(B \cap A)}{n(A)}\)

AP Inter 2nd Year Maths 2A Formulas Chapter 9 సంభావ్యత

→ 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, ……., En లు పరస్పర వివర్జిత, పూర్ణ ఘటనలవుతూ, P(Ei) > 0, i = 1, 2, …. n అయినప్పుడు k = 1, 2, 3, ….. n లకు
\(P\left(\frac{E_k}{A}\right)=\frac{P\left(E_k\right) \cdot P\left(\frac{A}{E_k}\right)}{\sum_{i=1}^n P\left(E_i\right) \cdot P\left(\frac{A}{E_i}\right)}\)

AP Inter 2nd Year Maths 2A Formulas Chapter 8 విస్తరణ కొలతలు

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 = Σfi, మరియు \(\bar{x}\) మధ్యమం
  • మధ్యగతం నుంచి మధ్యమ విచలనం = \(\frac{1}{N} \sum f_i \mid x_i\) – మధ్యగతం|, N = Σfi

→ 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

AP Inter 2nd Year Maths 2A Formulas Chapter 8 విస్తరణ కొలతలు

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