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AP 9th Class Physical Science 2nd Lesson Questions and Answers Is Matter Around Us Pure

AP 9th Class Physical Science 2nd Lesson Questions and Answers Is Matter Around Us Pure

9th Class Physics 2nd Lesson Is Matter Around Us Pure Questions and Answers (Exercise)


Question 1.

Which separation techniques will you apply for the separation of the following ?

a) Sodium chloride from its solution in water.

b) Ammonium chloride from a mixture containing sodium chloride and ammonium chortic.

c) Small pieces of metal in the engine oil of a car.

d) Different pigments from an extract of flower petals.

e) Butter from curd.

f) Oil from water.

g) Tea leaves from tea.

h) Iron pins from sand.

i) Wheat grains from husk.

j) Fine mud particles suspended in water.

Answer:

Separation techniques :

a) Evaporation

b) Sublimation

c) Filtration

d) Chromatography

e) Churning (or) Centrifugation

f) Separating funnel

g) Filtration

h) Magnetic separation

i) Winnowing

j) Sedementation / decantation.


Question 2.

Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

Answer:

Steps for making tea :

Take water as solvent and boil it a few minutes.

Add solutes, i.e., milk, tea leaves, sugar. Now, again boil the solution for few minutes.

Sugar is soluble in water thus it will be dissolve.

Colour of tea leaves goes into solution as filtrate. The remaining tea leaves being insoluble remains as residue.

Now, filter the solution. Collect the filtrate in cup.

The insoluble tea leaves will be left behind as residue.

Question 3.

Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K ?

Answer:

Solubility of potassium nitrate at 313 K = 62/100

100 g of water contains potassium nitrate =62 g

50 g of water contains potassium = 62100 × 50 = 31g


b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

Answer:

As solution cools, potassium chloride gets crystallized or precipitated. Because at 353 K solubility of potassium chloride is 54 g, 100 g of water. Where at room temperature it is 35 g. So, excess of potassium chloride dissolved gets precipitated.


AP 9th Class Physics 2nd Lesson Questions and Answers Is Matter Around Us Pure


c) Find the solubility of each salt at 293 K . Which salt has the highest solubility at this temperature?

Answer:

Potassium nitrate = 32 g

Sodium chloride = 36 g

Potassium chloride = 35 g

Ammonium chloride = 37 g

Ammonium chloride has the highest solubility of 37 g at 293 k


d) What is the effect of change of temperature on the solubility of a salt ?

Answer:

The rate of solubility increases with the increase of temperature.

Question 4.

Explain the following giving examples.

a) Saturated solution

b) Pure substance

c) Colloid

d) Suspension

Answer:

a) Saturated solution : A solution in which the maximum possible amount of a solute is dissolved at a given temperature.

Ex:



Take 50 ml of water in a cup.

Add one spoon of sugar to the cup and stir still it dissolves.

Keep on adding sugar to the water in the cup and stir till no more sugar can be dissolved.

The solution so formed is a saturated solution.

b) Pure substance : A single form of a substance (or) matter is called pure substance.

Ex : Elements & compounds.


c) Colloid : It is a heterogeneous mixture which has intermediate properties between solution & suspension. Ex : Smoke, starch solution, ink, butter, cheese.


d) Suspension : A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout. Ex : Muddy water, Syrups, Chalk powder, mixed with water.

Question 5.

Classify each of the following as a homogeneous or heterogeneous mixture.

soda water, wood, air, soil, vinegar, fitered tea

Answer:


Soda water :

Heterogeneous : When bottle if it opened.

Homogeneous : When a bottle if it is not opened.

Wood : Heterogeneous

Air : Homogeneous

Soll : Heterogeneous

Vinegar : Homogeneous

Filtered tea : Homogeneous

Question 6.

How would you confirm that a colourless liquid given to you is pure water ?

Answer:

Every liquid hasia characteristic boiling point at 1 atmospheric pressure. If the given odourless liquid boils at exactly 373 K at 1 atmospheric pressure, then it confirms that the given liquid is pure water. Otherwise, it is contaminated.


Question 7.

Which of the following materials fall in the category of a “Pure substance”?

a) Ice

b) Milk

c) Iron

d) Hydrochloric acid

e) Calcium oxide

f) Mercury

g) Brick

h) Wood

i) Air

Answer:

Pure substance :

a) Ice

c) Iron

d) Hydrochloric Acid

e) Calcium oxide

f) Mercury.



Question 8.

Identify the solutions among the following mixtures.

a) Soil

b) Sea water

c) Air

d) Coal

e) Soda water

Answer:

b) Sea water

c) Air

e) Soda water are solutions.



Question 9.

Which of the following will show “Tyndall effect”?

a) Salt solution

b) Milk

c) Copper sulphate solution

d) Starch solution

Answer:

b) Milk

d) Starch solution show tyndall effect as they are colloids.

L

Question 10.

Classify the following into elements, compounds and mixtures.

a) Sodium

b) Soil

c) Sugar solution

d) Silver

e) Calcium carbonate

f) Tin

g) Silicon

h) Coal

i) Air

j) Soap

k) Methane

l) Carbon dioxide

m) Blood

Answer:

Elements

a) Sodium

d) Silver

f) Tin

g) Silicon


Compounds :

e) Calcium carbonate

k) Methane

l) Carbondioxide



Mixtures

b) Soil

c) Sugar solution

h) Coal

i) Air

j) Soap

m) Blood



Question 11.

Which of the following are chemical changes?

a) Growth of a plant

b) Rusting of iron

c) Mixing of iron filings and sand

d) Cooking of food

e) Digestion of food

f) Freezing of water

g) Burning of a candle

Answer:

Chemical changes :

a) Growth of a plant

b) Rusting of iron

d) Cooking of food

e) Digestion of food

g) Burning of a candle


12. A startup company launches three new products:

1. A room freshener spray

2. A medicinal antacid suspension (like milk of magnesia)

3.A coloured decorative glass panel

Which of the following correctly identifies the type of colloid in each product?

A) Spray - Foam; Antacid - Emulsion; Glass - Gel

B) Spray - Aerosol; Antacid - Sol; Glass - Solid sol

C) Spray - Emulsion; Antacid - Gel; Glass - Aerosol

D) Spray - Sol; Antacid - Foam; Glass - Emulsion

✅ Correct Answer: B) Spray – Aerosol; Antacid – Sol; Glass – Solid sol

Explanation:

1. Room freshener spray → Aerosol

Liquid droplets are dispersed in a gas (air).

2. Milk of magnesia (antacid suspension) → Sol

Solid particles (Mg(OH)₂) are dispersed in a liquid.

3. Coloured decorative glass panel → Solid sol

Solid particles are dispersed in a solid medium (glass).

๐Ÿ‘‰ Answer: B) Spray – Aerosol; Antacid – Sol; Glass – Solid sol

School Management Committees Elections for the Academic Year 2026-27 Schedule Guidelines

School Management Committees Elections for the Academic Year 2026-27 Schedule Guidelines

PROCEEDINGS OF THE STATE PROJECT DIRECTOR, SAMAGRA SHIKSHA, ANDHRA PRADESH. Present: Sri B.Srinivasa Rao, I.A.S.,

File.No.-SS-16021/54/2021-CMO SEC, Dated: 15-09-2026

Sub: APSS- CMO Section Conduct of Elections for reconstitution of School Management Committees for the Academic Year 2026-27 Schedule Guidelines-Issued. Election

Read: 1 D.O. Letter No.1-1/2025-SS.17- Part (1) Dt.02.05.2026 of Joint Secretary (Coord & Media), DoSE&L, MoE, Gol.

2D. O. No. 1-1/2025-SS.17, 11th May 2026, Launching Guidelines 06.05.2026. DOSE&L, MoE, Gol

3D. O. No. 1-1/2025-SS.17, DoSE&L, MoE, Gol, 15th May 2026

4 SMC Guidelines English and Telugu Versions issued by Gol

5 File.No.-SS-16021/54/2021-CMO SEC, Dated:28-05-2026, SS, AP

6 Lr.Rc.No.SSA-16021/54/2023-SEC-SSA, dated: 16-07-2026.

7 Memo No.977203/Prog.II/A2/2019-11, Dated:11.09.2026

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Order:

All the District Educational Officers and Additional Project Coordinators in the state are hereby informed that the Department of School Education and Literacy (DoSE&L), Ministry of Education, Government of India have issued "Guidelines for SMCs - 2026" guidelines relating to the constitution and functioning of School Management Committees, including the composition of SMCs.

Accordingly, the Government of Andhra Pradesh have issued directions for Reconstitution of School Management Committees in the state duly conducting elections on 29.09.2026 as per the schedule.

Vide ref 7th cited, the Government have issued a MEMO for reconstitution of School Management Committees along with the following schedule.


SI. No. Date Activity to be conducted Time
1 15.09.2026 (Tuesday) Issue of Notification for conducting elections for reconstitution of School Management Committee members, Chairperson and Vice-Chairperson 10:00 AM
2 18.09.2026 ( Friday) Board Display of Voter List for conduct of elections on the Notice  2:00 PM
3 25.09.2026 (Friday) Calling for objections on the Voter List and redressal grievances, if any 9:00 AM to 1:00 PM


Finalization of Voter List for conduct of elections and PM to display of the finalized list on the Notice Board 3:00 PM to 4:00 PM
4 29.09.2026 (Tuesday) Conduct of elections and finalization of reconstitution of AM to School Management Committee Members 7:00 AM to
1:00 PM




Conduct of election of Chairperson & Vice-Chairperson by the School Management Committee Members

1:30 PM  



Oath-taking by School Management Committee Members, Chairperson & Vice-Chairperson 2:00 PM

Conduct of the First School Management Committee PM to Meeting 3:00 PM to 3:30 PM

All the District Educational Officers and Additional Project Coordinators in the state are directed to take necessary action in conducting Elections in all schools to reconstitute the School Management Committees in their respective


Click here to Download Guidelines 

Click here to Download FAQ on SMC Ekections

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.


Click here to Download Complete Proceedings 

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

เคฌाเคฒเค• เค•ो เคธเคฎเคเคจा: เคฌाเคฒ्เคฏाเคตเคธ्เคฅा เค•ी เค…เคตเคงाเคฐเคฃा เคเคตं เคฎเคนเคค्เคต

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

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