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Preservation of Earned Leave to the Staff who have attended SSC Advanced Supplementary Examination works during summer vacation in the Year 2025-26

Preservation of Earned Leave to the Staff who have attended SSC Advanced Supplementary Examination works during summer vacation in the Year 2025-26

PROCEEDINGS OF THE DIRECTOR OF SCHOOL EDUCATION ANDHRA PRADESH, AMARAVATI.

Present : Smt Thameem Ansariya.A, I.A.S.

Rc.No.ESE02- 30027/14/2024-A&I, Dated:09-09-2026

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Sub: SE – Academics and Inspections - Preservation of Earned Leave to the Staff who have attended SSC Advanced Supplementary Examination works during summer vacation in the Year 2025-26 – Reg.

Read:1.This office Procs.Rc.No.362/E1-1/2013, Dated:16.11.2013.

2. Certain representations from Teachers Associations.

*****

The attention of all the District Educational Officers in the State is invited to the reference 2nd cited, wherein certain representations have been received from teachers’ associations requesting for preservation of Earned Leave to the Head Masters/Teachers/Non-Teaching staff under the control of the School Education Department,who worked during the summer vacation for the conduct of SSC Advanced Supplementary Examinations held during the academic year 2025–26.

2. In this regard, all the District Educational Officers in the State are informed that, clear instructions have already been issued regarding the preservation of Earned Leave for teachers and non- teaching employees working in schools.

3. Therefore, all the District Educational Officers in the State are requested to follow the instructions issued vide reference 1st cited scrupulously, without any deviation, duly ensuring the actual days of duties done by the concerned by preventing from availing summer vacation, as per the schedule, and report compliance.


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AP EAPCET - 2026 MPC, Bipc 3rd and Final Phase Web Counselling Notification

AP EAPCET - 2026 MPC, Bipc  3rd and Final Phase Web Counselling Notification

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Web Options from 9th to 13th Sep

APEAPCET-2026 ADMISSIONS (M.P.C. & Bi.P.C. STREAMS) NOTIFICATION

CERTIFICATE VERIFICATION & OPTION EXERCISING FOR WEB BASED COUNSELLING

M.P.C. STREAM : 

M.P.C. STREAM (3rd & Final Phase): The qualified candidates of APEAPCET-2026 (M.P.C. stream) for admission into B.E / B.Tech Courses are informed to attend the web based counselling for 3rd & Final Phase from 09-09-2026 to 13-09-2026. For details visit website: https://cap.apcfss.in


Bi.P.C. STREAM:

Final Phase: The qualified candidates of APEAPCET-2026 (Bi.P.C. stream) for admission into B.E / B.Tech / B.Pharmacy /Pharm-D Courses are informed to attend the web based counselling for final phase from 09-09-2026 to 13-09-2026. For details refer the website: https://cap.apcfss.in

THIRD & FINAL PHASE COUNSELLING SCHEDULE 

APEAPCET-2026 ADMISSIONS [ M P C & BI P C STREAM]


SNO ACTIVITY DATES
1

Online Certificates Verification &  Fee Payment (Candidate Registration)

09.09.2026 To
 11.09.2026
2

Verification of Uploaded Certificates at HLCs by Online

09.09.2026 To 12.09.2026

3 Option Entry 09.09.2026 To 12.09.2026
4 change of options 13.09.2026
5 Allotment of Seats 16.09.2026
6

Self Reporting & Reporting at  college

16.09.2026 To 19.09.2026


For more Details Click here 



10th Class Physics 10th Lesson The Human Eye and the Colourful World Questions and Answers

10th Class Physics 10th Lesson The Human Eye and the Colourful World Questions and Answers

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10th Class Physics 10th Lesson Questions and Answers (Exercise)

Question 1.

The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to

a) presbyopia.

b) accommodation.

c) near-sightedness.

d) far-sightedness.

Answer:

b) accommodation.


Question 2.

The human eye forms the image of an object at its

a) cornea.

b) iris.

c) pupil.

d) retina.

Answer:

d) retina.



Question 3.

The least distance of distinct vision for a young adult with normal vision is about

a) 25 m.

b) 2.5 cm.

c) 25 cm.

d) 2.5 m.

Answer:

c) 25 cm.


Question 4.

The change in focal length of an eye lens is caused by’the action of the

a) pupil.

b) retina.

c) ciliary muscles.

d) iris.

Answer:

c) ciliary muscles.


Question 5.

A person needs a lens of power -5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting

 (i) distant vision, and (ii) near vision?

Answer:

i) Focal length of the lens for distant vision = 1 Power  = 100−5.5 = cm = -18 cm (approx)

ii) Focal length of the lens for near vision = 1001.5 cm = 66.66 cm



Question 6.

The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem ?

Answer:

The far point of a normal eye is infinity. Since the far point of the defective eye is given as 80 cm, the eye is short-sighted. To correct it, the lens should be such that an object at infinity must form its image at the far point of defective eye.

∴ u = -∝, v = -80 cm.

Using lens formula 1f=1v−1u

∴ 1f = 1−80 – 1(−∞) = 1−80

∴ Focal length of lens is – 80 cm

The correction is done by using a concave lens of focal length 80 cm.

Power of the lens = 100f( in cm) = –10080 = -1.25 D is needed.


Question 7.

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is 25 cm.

Answer:



AP 10th Class Physics 10th Lesson Questions and Answers The Human Eye and the Colourful World 1

N = Near point of a hypermetropic eye

N’ = Near point of a normal eye

To correct the defect, the image of an object at 25 cm should be brought at 100 cm.

∴  = 1−100 – 1−25

1f = −1100 + 125 = −1+4100 = 3100

∴ f = + 1003 = +33.3 cm

So, a convex lens of focal length 33.3 cm is required power, P = 10033.3 = 3.0 D


Question 8.

Why is the normal eye not able to see clearly the objects placed closer than 25 cm?

Answer:

The focal length of the eye lens cannot be reduced below a certain limit.


Question 9.

What happens to the image distance in the eye when we increase the distance of an object from the eye ?

Answer:

In eye, the image is always formed on the retina. The image distance is the distance between the eye lens and the retina. When we increase the distance of the object from the eye, the focal length of the eye lens increases due to the action of ciliary muscless so that the image of object is formed on the retina and therefore, the image distance remains the same.


Question 10.

Why stars are Twinkle?

Answer

Stars appear to twinkle due to atmospheric refraction.

The light of stars passes through the Earth’s atmosphere, which contains air layers of varying temperatures and densities.

These variations cause the star’s light to refract and create small, rapidly changing differences in brightness and position, leading to the twinkling effect.

Question 11.

Explain why the planets do not twinkle.

Answer:

The planets are much closer to the Earth, and are thus seen as extended sources. If we consider a planet as a collection of a large number of point- sized sources of light, the total variation in the amount of light entering our eye from’all the individual, point – sized sources will average out to zero, thereby nullifying the twinkling effect.


Question 12.

Why does the sky appear dark instead of blue is an astronaut ?

Answer:

At such huge heights due to absence of atmosphere, no scattering out the light takes place. Therefore, sky appears dark.

AP TET Paper 1A & 2A Psychology Concept of Childhood and it's importance

AP TET Paper 1A & 2A Psychology Concept of Childhood and it's importance

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AP TET Paper 2A – Child Development & Pedagogy

बालक को समझना: बाल्यावस्था की अवधारणा एवं महत्व

AP TET Paper 1A & 2A Psychology Concept of Childhood and it's importance

AP TET Paper 1A & 2A Psychology Concept of Childhood and it's importance

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AP TET Paper 2A – Child Development & Pedagogy

Understanding a Child: Concept and Importance of Childhood

AP TET Paper 1A & 2A Psychology Concept of Childhood and it's importance

AP TET Paper 1A & 2A Psychology Concept of Childhood and it's importance

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AP TET Paper 1A & 2A – Child Development & Pedagogy

బాలుడిని అర్థం చేసుకోవడం – బాల్య భావన

Mission March 2026-27 Corbon compounds Level 1 & Level 2 Answers

Mission March 2026-27 Corbon And it's Compounds Level 1 & Level 2 Answers



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1. Fill the table with suitable answers related to functional groups, structural formulae, ex-amples and suffixes.

Structural Formula Functional Group Name Suffix Example
R-OH Alcohol -ol C2H5OH
R-CHO Aldehyde -al CH3CHO
R-COOH Carboxylic acid -oic acid CH3COOH
R-CO-R Ketone -one CH3COCH3

2. Complete the table of alkanes, alkenes and alkynes based on the number of carbon atoms.

No. of CorbonsAlkanesAlkenesAlkynes
2EthaneEthene
Ethyne
3PropanePripenePropyne
4ButaneButeneButyne
5PentanePentenePentyne

3. Explain the following

a) Addition reaction of unsaturated hydrocarbons

An addition reaction is a reaction in which atoms or groups of atoms are added to an unsaturated hydrocarbon at the double or triple bond to form a saturated compound.

Example:

CH₂=CH₂ + H₂ → CH₃–CH₃

Ethene + Hydrogen → Ethane

This reaction takes place in the presence of a catalyst such as Ni/Pt/Pd.

b) Substitution reaction of saturated hydrocarbons

A substitution reaction is a reaction in which one or more hydrogen atoms of a saturated hydrocarbon are replaced by another atom or group of atoms.

Example:

CH₄ + Cl₂ → CH₃Cl + HCl

Methane + Chlorine → Chloromethane + Hydrogen chloride

The reaction occurs in the presence of sunlight (UV light).

4. Explain the cleansing action of soap with a well-labelled diagram

A soap molecule has two parts:

Hydrophilic end – water-loving ionic end.

Hydrophobic end – water-repelling hydrocarbon chain that attracts oil and grease.

When soap is added to water containing oily dirt, the hydrophobic ends attach themselves to the grease, while the hydrophilic ends remain in water. Many soap molecules surround the grease particle and form a micelle.

On rubbing and washing with water, the grease-containing micelles are removed.

Diagram:


                 Water

       ○  ○  ○  ○  ○  ○

        \ |  \ |  \ | /

         \|   \|   \|/

       ┌─────────────────┐

       │   OIL / GREASE  │

       └─────────────────┘

         /|   /|   /|\

        / |  / |  / | \

       ○  ○  ○  ○  ○  ○


○ = Hydrophilic end (water-loving)

Lines = Hydrophobic hydrocarbon chains

                 ↓

        MICELLE FORMATION

                 ↓

       Dirt/grease is washed away

5. Distinguish between soap and detergent

Soap

SoapDetergent
Soaps are sodium or potassium salts of long Chain Fatty AcidsDetergent s are salts of long chain alkyle benzene sulfonates or alkyl sulfonates
They are Generally less effective in hard waterThey are  effective in hard water
They form scumwith Ca2+ and Mg2+ ionsThey Generally don't form insoluble scum with Ca2+ and Mg2+ ions
Example: Sodium StearateExample Sodium alkyle benzene sulphonate


2 Marks Questions

1. What are the two properties of carbon that lead to the large number of carbon compounds?

The two important properties are:

Catenation: Carbon atoms can bond with one another to form long chains, branched chains and rings.

Tetravalency: Carbon has a valency of four and can form four covalent bonds with carbon and other elements.

2. Explain catenation with an example.

Catenation is the property of carbon by which carbon atoms form bonds with other carbon atoms to produce long chains, branched chains and rings.

Example:

CH₃–CH₂–CH₂–CH₂–CH₃

This is a chain of five carbon atoms (pentane).

3. What is allotropy? Write the allotropes of carbon.

Allotropy is the property of an element to exist in two or more different forms in the same physical state.

Important allotropes of carbon are:

Diamond

Graphite

Fullerenes (such as C₆₀)

4. Name the following functional groups

(i) –CHO → Aldehyde group

(ii) >C=O → Ketone (carbonyl) group

5. What are hydrocarbons? How are they classified?

Hydrocarbons are compounds containing only carbon and hydrogen.

They are mainly classified as:

Saturated hydrocarbons – contain only single C–C bonds.

Example: Ethane (C₂H₆)

Unsaturated hydrocarbons – contain double or triple bonds.

Alkenes – contain C=C double bond. Example: Ethene (C₂H₄)

Alkynes – contain C≡C triple bond. Example: Ethyne (C₂H₂)


6. Give two examples each of saturated and unsaturated hydrocarbons.

Saturated hydrocarbons:

Methane – CH₄

Ethane – C₂H₆

Unsaturated hydrocarbons:

Ethene – C₂H₄

Ethyne – C₂H₂

7. Give one example each of a carbon compound containing:

1. Alcohol functional group:

Ethanol – C₂H₅OH

2. Carboxylic acid functional group:

Ethanoic acid – CH₃COOH

8. Why are cooking oils hydrogenated before making vanaspati ghee?

Cooking oils contain unsaturated fatty acids. Hydrogenation adds hydrogen to the double bonds in these oils in the presence of a nickel catalyst, converting them into more saturated and solid fats.

Example:

Unsaturated oil + H₂ → Saturated fat

(Ni catalyst)

Thus, hydrogenation helps convert liquid vegetable oils into solid/semi-solid vanaspati ghee.

9. A student says ethanol and ethanoic acid have the same properties. Is the statement correct?

No, the statement is incorrect.

Ethanol and ethanoic acid have different functional groups and therefore different physical and chemical properties.

Ethanol: C₂H₅OH — alcohol

Ethanoic acid: CH₃COOH — carboxylic acid

Ethanoic acid turns blue litmus red, whereas ethanol does not.

10. Suggest a way to identify whether a carbon compound is saturated or unsaturated.

Treat the compound with bromine water.

Unsaturated compound: Decolourises bromine water.

Saturated compound: Does not decolourise bromine water under ordinary conditions.

Another test is the alkaline KMnO₄ test.


1 Mark Questions

1. A student proposes to make a sweet-smelling compound from ethanol and acetic acid. Which reaction should be carried out?

Answer: Esterification reaction.

Ethanol reacts with ethanoic acid in the presence of concentrated H₂SO₄ to form ethyl ethanoate, which has a sweet/fruity smell.

C₂H₅OH + CH₃COOH → CH₃COOC₂H₅ + H₂O

2. Write the structural formula of any hydrocarbon with eight carbon atoms.

Octane:

CH₃–CH₂–CH₂–CH₂–CH₂–CH₂–CH₂–CH₃

3. Identify the functional group present in propanal.

Answer: Aldehyde group (–CHO)

4. Which of the following undergoes substitution reaction?

A) CH₄

B) C₃H₆

C) C₂H₂

D) C₇H₁₄

Answer: A) CH₄ (Methane)

5. Which of the following hydrocarbons undergoes addition reaction?

A) C₂H₆

B) C₃H₈

C) C₃H₆

D) CH₄

Answer: C) C₃H₆ (Propene)


6. Identify the alkene among the following hydrocarbons.

A) C₂H₆

B) C₃H₈

C) C₃H₆

D) CH₄

Answer: C) C₃H₆ (Propene)

7. The general formula of alkanes is CₙH₂ₙ₊₂. Write the first member of alkanes.

Answer: Methane (CH₄)

8. A hydrocarbon has four carbon atoms and ten hydrogen atoms. Write its name

Formula = C₄H₁₀

Answer: Butane

9. Which compound is formed when two carbon atoms are linked by a single bond and six hydrogen atoms are added?

Answer: Ethane (C₂H₆)

Structural formula:

CH₃–CH₃

10. How can you convert ethene into ethane?

By hydrogenation of ethene in the presence of a nickel catalyst.

CH₂=CH₂ + H₂ → CH₃–CH₃

(Ni catalyst)


11. Give any example of an unsaturated hydrocarbon.

Answer: Ethene (C₂H₄)

12. How many hydrogen atoms are present in Butane?

Butane = C₄H₁₀

Answer: 10 hydrogen atoms

13. Which type of bond is present between carbon atoms in ethene?

Answer: Double covalent bond (C=C).


LEVEL – 2: SHINING STAR

8 Marks Questions

1. How can ethanol and ethanoic acid be differentiated based on their physical and chemical properties?

Property Ethanol Ethanoic acid
Formula C₂H₅OH CH₃COOH
Functional group –OH (Alcohol) –COOH (Carboxylic acid)
Smell Characteristic alcoholic smell Vinegar-like smell
Litmus test Does not change blue litmus Turns blue litmus red
Reaction with NaHCO₃ No brisk reaction Produces CO₂ gas with effervescence
Nature Neutral Acidic
Boiling point About 78°C About 118°C
Common use Solvent, fuel, sanitiser Vinegar, food preservative

Important test:


CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑


The evolution of CO₂ gas confirms ethanoic acid.


2. Explain the nature of the covalent bond using bond formation in CH₃Cl.

A covalent bond is formed by the sharing of electrons between atoms.

In CH₃Cl (chloromethane):

Carbon has 4 valence electrons and needs four more electrons to complete its octet.

Each hydrogen shares one electron with carbon, forming three C–H covalent bonds.

Chlorine has 7 valence electrons and shares one electron with carbon, forming one C–Cl covalent bond.

Thus carbon forms four covalent bonds.

Structure:


       H

       |

H — C — Cl

       |

       H

Therefore, CH₃Cl contains three C–H bonds and one C–Cl covalent bond.

3. Write the differences between saturated and unsaturated hydrocarbons.


saturated hydrocarbons unsaturated hydrocarbons
They Contain Only Single bonds between carbon atoms They Contain One or more Double or triple bonds between carbon atoms
They are generally less reactive  they are generally more reactive 
Example: Ethane
(C2H6)
Example: Ethene ( C2H4)

Saturated hydrocarbons Contain only single bonds between carbon atoms.Called alkanes.

Unsaturated hydrocarbons Contain double or triple bonds between carbon atoms. Include alkenes and alkynes.

General formula of open-chain alkanes: CₙH₂ₙ₊₂

Alkenes: CₙH₂ₙ; Alkynes: CₙH₂ₙ₋₂

Generally undergo substitution reactions.

Generally undergo addition reactions.

Do not decolourise bromine water.

Decolourise bromine water.

Example: Ethane (C₂H₆)

Example: Ethene (C₂H₄)

4. Differentiate between alkanes, alkenes and alkynes.


Alkanes Alkenes Alkynes 
SaturatedUnsaturated  Unsaturated
Single bond (C-C) Double bond
(C=C) 
Triple bond 
(C=-C)
CnH2n+2 CnH2n CnH2n-2
Methane(CH4) Ethene(C2H4) Ethyne (C2H2)
Substitution Addition Addition


Examples of Structure:

Alkanes: CH3-CH3

Alkenes: CH2=CH2

Alkynes: CH=-CH

2 Marks Questions

1. Differentiate between saturated and unsaturated hydrocarbons with one example each.

saturated hydrocarbonsunsaturated hydrocarbons
They Contain Only Single bonds between carbon atomsThey Contain One or more Double or triple bonds between carbon atoms
They are generally less reactive they are generally more reactive 
Example: Ethane
(C2H6)
Example: Ethene ( C2H4)


2. Draw the structural isomers of butane. Name them.

Butane has the molecular formula C₄H₁₀ and has two structural isomers.

1. n-Butane:

CH₃ — CH₂ — CH₂ — CH₃

2. Isobutane (2-methylpropane):


       CH₃

        |

CH₃ — CH — CH₃


3. Why are carbon compounds poor conductors of electricity?

Carbon compounds are generally covalent compounds. They do not have free ions or free electrons that can carry electric current. Therefore, they are generally poor conductors of electricity.

4. Write two properties of ethane related to its use as a fuel.

Ethane is highly combustible and releases a large amount of heat on burning.

It burns in oxygen to produce carbon dioxide and water, releasing energy.

C₂H₆ + 7/2 O₂ → 2CO₂ + 3H₂O + heat

5. Explain why carbon forms a large number of compounds. Give any two reasons.

Carbon forms a large number of compounds mainly because of:

Catenation: Carbon atoms can bond with one another to form long chains, branched chains and rings.

Tetravalency: Carbon has valency 4 and can form four covalent bonds with carbon and other elements.


6. Why is a flame of unsaturated hydrocarbons smoky? How can we test for unsaturation?

Unsaturated hydrocarbons have a higher percentage of carbon. They often undergo incomplete combustion, producing tiny carbon particles (soot), which make the flame yellow and smoky.

Test for unsaturation: Add bromine water. An unsaturated hydrocarbon decolourises bromine water.

7. Write the chemical equation for esterification reaction. Name the ester formed from methanol and ethanoic acid.

Equation:

CH₃OH + CH₃COOH → CH₃COOCH₃ + H₂O

(conc. H₂SO₄, heat)

The ester formed is methyl ethanoate.

8. How is soap different from detergent? Give one point for each.

Soap: Forms scum with Ca²⁺ and Mg²⁺ ions in hard water and is therefore less effective.

Detergent: Does not form insoluble scum with Ca²⁺ and Mg²⁺ ions and works well in hard water.

9. What is meant by homologous series? Write two characteristics.

A homologous series is a group of organic compounds having the same functional group and the same general formula, where successive members differ by –CH₂–.

Characteristics:

Successive members differ by –CH₂– (14 u).

Members have similar chemical properties and show a gradual change in physical properties.


1 Marks Questions

1. The property of carbon to form chains, branched chains and rings is called:

Answer: B) Catenation

2. General formula of alkenes is:

Answer: CₙH₂ₙ

3. Which of the following is not an allotrope of carbon?

A) Diamond

B) Graphite

C) Fullerene

D) Methane

Answer: D) Methane

4. IUPAC name of CH₃CH₂OH is:

A) Methanol

B) Ethanol

C) Ethanal

D) Ethanoic acid

Answer: B) Ethanol

5. Which of the following belongs to the same homologous series?

Answer: B) C₂H₆ and C₃H₈

Both are alkanes and successive members differ by CH₂.


6. Functional group present in CH₃COOH is:

A) Alcohol

B) Aldehyde

C) Ketone

D) Carboxylic acid

Answer: D) Carboxylic acid

7. On adding sodium to ethanol, gas evolved is:

A) H₂

B) O₂

C) CO₂

D) CH₄

Answer: A) H₂ (Hydrogen)

2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑

8. Draw the electron dot structure of methane.

Methane is CH₄. Carbon shares one electron with each of four hydrogen atoms.


       H

       :

   H : C : H

       :

       H

Each ":" represents a shared pair of electrons (covalent bond).

9. Why does carbon form covalent bonds only?

Carbon has four valence electrons. It is difficult for carbon to either lose four electrons or gain four electrons. Therefore, carbon completes its octet by sharing electrons, forming covalent bonds.

10. Name the catalyst used in hydrogenation of vegetable oils.

Answer: Nickel (Ni)


11. Why is conversion of ethanol to ethanoic acid considered an oxidation reaction?

Conversion of ethanol to ethanoic acid involves the addition of oxygen.

CH₃CH₂OH + 2[O] → CH₃COOH + H₂O

Therefore, it is considered an oxidation reaction.

12. Write the IUPAC name of CH₃–O–CH₃.

Answer: Methoxymethane

Common name: Dimethyl ether.


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