Class 8

You can Download Force Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Basic Science Solutions Chapter 10 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 10 Force

When an object is pushed or pulled, a force is being applied on it. The unit of force is newton. It is indicated by the letter N.

Force is that which changes or tends to change the shape, size, volume, state of rest or state of motion of a body.

There are various types of forces such as muscular force, magnetic force, electric force, gravitational force, frictional force etc.

Contact force and non contact force

The force applied by the contact between objects is contact force. The force applied on an object without contact on it is non contact force.
Ex:pushing a trolley- contact force, Magnet attracting a nail-non contact force

Kerala Syllabus 8th Standard Physics Notes Frictional force:

When a ball is rolling on the floor after some time it stops. This is because of the frictional force between the ball and floor. On the surface of the objects which are on contact with each other there are many mounts and pits get interlocked.

When we try to move a body by applying a force parallel to their surfaces an opposing force is experienced.
When a body moves or tends to move on the surface of another body, a force is experienced parallel to the surface which opposes the relative motion between them. This is friction.

Rolling friction and sliding friction

When a body rolls over the surface of another body, the force of friction which originates is rolling friction. When a body slides on the surface of another body the force of friction which originates is sliding friction.

Rolling friction is less than sliding friction. Striking a match stick on the side of a match box, wear and tear of machines, ability to hold objects, walking etc are advantages of the friction and treading of tyres of vehicles, wearing out of tyres, fuel loss are disadvantages of friction.

The materials like oil, grease which are used to reduce friction are the lubricants. Graphite is a solid lubricant.

Force Class 8 Kerala Syllabus Thrust and Pressure

The total normal force experienced by a ‘ surface is thrust. Thrust per unit area is pressure.
Pressure = \(\frac{Thrust}{area}\)

Unit of pressure is N/m2. This is known as pascal. The force acting on a larger area exerts a smaller pressure, while a force acting on a smaller area exerts a large pressure.

Hsslive Guru 8th Class Physics Kerala Syllabus Liquid Pressure

The pressure is experienced in liquids also. As the height of liquid column increases the pressure exerted by it increases.

The pressure exerted by a liquid column increases with increase in height. The thrust acting per unit area by a liquid is liquid pressure.

Liquids exert force on all sides of the container in which they are taken.

If the height of the liquid column is Ti’, density of liquid ‘d’ and acceleration due to gravity ‘g’, then liquid pressure is P = hdg

Class 8 Physics Kerala Syllabus Atmospheric pressure

Atmospheric air also can exert pressure. An envelope of air surrounds the earth. This is earth’s atmosphere. The density of atmospheric air near the surface of earth is greater and it decreases when we go up. The weight of air column over the unit area of earth’s surface is atmospheric pressure. One atmospheric pressure is the weight of mercury column of 0.76 m height and unit area (1m²). This is standard atmospheric pressure .

The unit of atmospheric pressure is* bar. The instrument used to measure atmospheric pressure is Barometre

Force Textbook Questions and Answers

Kerala Syllabus 8th Standard Physics Notes Pdf Question 1.
Classify the following situations into contact and noncontact forces.
a. Applying break in a bicycle.
b. A mango falling from a mango tree.
c. The earth revolving around the sun.
d. The speed of a hall rolling on ground is reduced.
Answer:
Contact force
a. Applying break in a bicycle.
d. The speed of a ball rolling on ground is reduced
Non contact force
b. A mango falling from a mango tree.
c. The earth revolving around the sun.

Class 8 Physics Notes Kerala Syllabus Question 2.
State reason
a. We are able to walk on the ground without slipping.
b. It is easy to cut vegetables using a sharp knife.
c. The number of tyres is more for goods vehicles.
d. The moving parts of machines experience wear and tear.
Answer:
a. Because of the friction between ground and feet
b. When the area decreases pressure increases. As the area of the edge of the knife is veiy less, it is easy to cut
c. When the area of surface increases pressure decreases, so when more wheels are used it won’t depressed in soil.
d. Because of the friction between the area of contacts.

Hss Live Guru 8th Physics Kerala Syllabus Question 3.
Match the following

Answer:

Hsslive Guru Physics 8th Standard Kerala Syllabus Question 4.
Bubbles rising from the bottom of the water filled in a bottle are depicted in the figure. Which is the correct figure? Justify your answer.

Answer:
Fig.c. There will be more pressure in the lower side of the liquid. The pressure decreases when it comes upward.

Hss Live Guru 8 Physics Kerala Syllabus Question 5.
A toy car of about 50 g placed on a polished table with threads attached to it carrying two pans passed through pulleys is depicted.

(a) What do you observe if too g each is placed on both pans?
(b) What do you observe if too g is placed on pan A and 200 g in pan B
(c) Justify your answers.
Answer:
a. The car will be stationary because the force excreted from both sides are equal.
b. The pan of weight 200 will pull with more force and it moves that direction.
c. In the first phase same force is excerted. But in second phase force is exerted to one direction.

Force Additional Questions and Answers

Force And Pressure Class 8 Worksheets With Answers Pdf Kerala Syllabus Question 1.
Tabulate the following as contact force and non contact force.
1. pushing a trolley
2. falling coconut from coconut tree
3. magnet attract iron nail
4. Drowing water from the well
5. pushing the car
6. earth is moving around the sun
Answer:
contact force – 1,4,5
non contact force – 2,3,6

8th Standard Physics Notes Kerala Syllabus Question 2.
Ramu tried to push a round stone and rectangle stone of same weight. Which is easy to push.
Answer:
It is easy to push round stone because rolling friction is less than sliding friction.
3. Which are the following occation friction is beneficial and non beneficial
1. striking a match stick on the side of a match box
2. wear and tear of machines.
3. ability to hold objects
4. wearing out of tyres
5. walking 6. loss of fuel
Answer:
Beneficial : 1, 3, 5
non beneficial : 2, 4, 6

8th Std Physics Notes Kerala Syllabus Question 4.
Why does the design of ship and aeroplane are streamlining?
Answer:
To reduce friction while moving through air and water.

Kerala Syllabus 8th Standard Physics Question 5.
Write one solid lubricant and liquid lubricant
Answer:
Solid-graphite
Liquid-oil

Kerala Syllabus 8th Standard Science Notes Question 6.
Fill up suitably.
Force – newton
Pressure – ………..
Answer:
Pressure – pascal

8th Standard Physics Notes Pdf Kerala Syllabus Question 7.
Write the reason
1. Constructing a knife with sharp edge
2. Constructing the building with wide basement
Answer:
1. When the area of contact decreases the pressure increases
2. When area increases the chance to depress the base in soil is very less.

Class 8 Basic Science Solutions Kerala Syllabus Question 8.
Given figures, one can filled with water and other can filled with kerosine.

Which balloon is filled with water
Answer:
fig 2. When density of liquid increases the pressure increases, water is denser than kerosene.

Physics Class 8 Kerala Syllabus Question 9.
What are the factors effecting liquid pressure? Write the mathematical equation.
Answer:
The height of the liquid column is ‘h’, density of liquid ‘d’ and acceleration due to gravity ‘g’ are the factors. The equation is liquid pressure P = hdg

Basic Science Class 8 Kerala Syllabus Question 10.
As mountaineers climb higher altitude there is a possibility of nasal bleeding. Why?
Answer:
Going higher altitude the atmospheric pressure decreases. So blood veins are broken and blood is oozing out.

Question 11.
What is standard atmospheric pressure
Answer:
Atmospheric pressure at the sea level is equal to the pressure exerted by 0.76m of mercury column. This is standard atmospheric pressure.

Question 12.
When a hole is formed at the bottom of the water bottle, Anu closed the lid of the bottle tightly. The flow of water stops. Explain
Answer:
When the lid is closed tightly, the effect of atmospheric pressure is not experienced on the surface of the water in the bottle. Hence water cannot flow out from the bottle.

Question 13.
Mercury remains at a height of 76cm in a barometer. Raju made a hole at the top end of the tube. What will you observe? How can it be explained?
Answer:
Mercury will flow downwards when a hole is made at the top part of the tube. The atmospheric pressure exerted on the mercury in the vessels supports the mercury level in the tube. When a hole is made at the top, air enters the tube and the pressure is distributed throughout uniformly and hence the mercury level comes down.

Question 14.
A sharp nail pierces more easily through wood than a blunt nail. What is the reason?
Answer:
The surface area of the sharp tip of a nail is very small. Hence the pressure that it can exert on the wood is very great. That is why it pierces easily. For a blunt nail, the surface area of the tip is greater. Hence it can exert only a small amount of pressure.

Question 15.
What is meant by limiting friction.
Answer:
Limiting friction is the force of static friction acting just before the body starts moving.

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Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 4 Properties of Matter in Malayalam

Properties of Matter Text Book Questions and Answers

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Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 7 Metals in Malayalam

Metals Text Book Questions and Answers

You can Download Circles Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 5 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 5 Circles

Kerala Syllabus 9th Standard Maths Circles Text Book Questions and Answers

Textbook Page No. 68

Circles Class 9 Kerala Syllabus Question 1.
Prove that the line joining the centres of two intersecting circles is the perpendicular bisector of the line joining the points of intersection.

Answer:

AC = AD (Radii of the same circle)
AE = AE (Common side)
BC = BD (Radii of the same circle)
ΔABC = ΔABD (Three sides are equal)
In equal triangles, angles opposite to equal sides are equal.
So, ∠CAE = ∠DAE
Consider ΔCAE and ΔEAD.
∠CAE = ∠DAE
AC = AD (Radii of the same circle)
AE = AE (Common side)
ΔAEC = ΔAED (Two sides and the angle between them)
In equal triangles, sides opposite to equal angles are equal.
So, CE = DE (∠CAE = ∠DAE)
CE = DE ………(1)
In equal triangles, angles opposite to equal sides are equal. So, ∠AEC = ∠AED
∠AEC + ∠AED = 180° (Linear pair)
∠AEC = ∠AED = 90° ………(2)
From equation (1) and (2)
The line joining the centres of the circles is the perpendicular bisector of the chord.

Kerala Syllabus 9th Standard Maths Chapter 5 Question 2.
The picture on the right shows two circles centred on the same point and a line intersecting them. Prove that the parts of the line between the circles on either side are equal.

Answer:

AD and BC are chords.
OE bisects the chords perpendicularly AD and BC
BE = CE
AE = DE
AE – BE = DE – CE
AB = CD

Circles Class 9 State Syllabus Question 3.
The figure shows two chords drawn on either sides of a diameter: What is the length of the other chord?

Answer:

PO is the perpendicular bisector of AB, OQ is the perpendicular bisector of AC.
∠OAQ = ∠OAP = 30° (Given)
∠OQA = ∠OPA = 90° (Right angles)
∴ ∠AOQ = ∠AOP (Third angle also equal)
AO = AO (Common side)
ΔOQA = ΔOPA
In equal triangles, sides opposite to equal angles are equal. So AP = AQ.
AP = ½AB (Perpendicular from the centre of a circle to a chord bisects the chord)
AQ = ½AC
Since AP = AQ
½AB = ½AC
AB = AC
So length of AC = 3 cm

Which among MeX, RCH2X , R2CHX and R3CX is most reactive towards SN2 reaction Haloalkanes and Haloarenes.

Kerala Syllabus 9th Standard Maths Notes Question 4.
A chord and the diameter through one of its ends are drawn in a circle. A chord of the same inclination is drawn on the other side of the diameter.

Prove that the chords are of the same length.
Answer:

AB is the diameter
AC and AD are the chords
Given that ∠OAC = ∠OAD
In triangle OAC,
∠OAC = ∠OCA
In triangle OAD,
∠OAD = ∠ODA
Consider the ΔOAC and ΔOAD
∠OAC = ∠OAD; ∠OCA = ∠ODA
∠AOC = ∠AOD; AO = AO
Triangles are equal.
Sides opposite to equal angles are also equal.
∴ AC = AD

9th Class Maths Notes Kerala Syllabus Question 5.
The figure shows two chords drawn on either sides of a diameter. How much is the angle the other chord makes with the diameter?

Answer:

AD is the diameter and O is the centre of the circle.
∠OAB = 40°
Consider ΔOAB and ΔOAC
AB = AC = 3cm
OC = OB (Radius of the circle)
OA = OA (Common side)
Three sides ΔOAB of are equal to three sides of ΔOAC
In equal triangles, angle opposite to equal sides are equal.
∴ ∠OAB = ∠OAC
∴ ∠OAC = 40°

Hss Live Guru 9th Maths Kerala Syllabus Question 6.
Prove that the angle made by two equal chords drawn from a point on the circle is bisected by the diameter through that point.
Answer:
AB, AC are the chords of same length AD is the diameter of the circle.

When we consider ΔAOB, ΔAOC
AB = AC (Given)
OB = OC (Radius)
OA = OA (Common side)
Three sides ΔOAB of are equal to three sides of ΔOAC.
In equal triangles, angle opposite to equal sides are equal.
∠BAO = ∠CAO
∴ the diameter AD bisects ∠A.

Kerala Syllabus 9th Standard Maths Notes Malayalam Medium Question 7.
Draw a square and a circle through all four vertices. Draw diameters parallel to the sides of the square and draw a polygon joining the end points of these diameters and the vertices of the square.

Prove that this polygon is a regular octagon.
Answer:
The diameters are parallel to the sides

∠ADC = ∠BDC = 90°
Consider ΔADC & ΔBDC
AD = BD (Perpendicular from the centre of a circle to a chord bisects the chord)
DC = DC (Common side)
∠ADC = ∠BDC (90° each)
Two sides and the angle between them of ΔADC are equal to two sides and the angle between them of ΔBDC.
So ΔADC & ΔBDC are equal.
∴ AC = BC
In the same way we can see that other sides of the octagon are also equal.
So it is a regular octagon.

Textbook Page No. 72

Scert Class 9 Maths Solutions Kerala Syllabus Question 1.
Prove that chords of the same length in a circle are at the same distance from the centre.
Answer:

AB, CD are the chords of same length.
AB = CD
AP = ½AB (Perpendicular from the centre of a circle to a chord bisects the chord)
Similarly CQ = ½CD
AP = CQ [Since AB = CD]
Consider the right angled triangle ΔAOP and ΔCOQ
OP² = OA² – AP²
OP² = OB² – CQ² [Since OA = OB, AP = CQ]
OP² = OQ²
∴ OP = OQ
So, the chords of the same length in a circle are at the same distance from the centre.

Kerala Syllabus 9th Standard Maths Guide Question 2.
Two chords intersect at a point on a circle and the diameter through this point bisects the angle between the chords. Prove that the chords have the same length.
Answer:
OA = OC (radius of the same circle)
OB = OB (common side)

∠OBA = ∠OBC (given)
∠BAO = ∠BCO [Base angle of isosceless triangle ΔOCB & ΔOCA]
∠AOB = ∠BOC;
∴ ΔAOB = ΔBOC
So the sides AB and BC opposite to equal angles are also equal.

Kerala Syllabus 9th Standard Notes Maths Question 3.
In the picture on the right, the angles between the radii and the chords are equal. Prove that the chords are of the same length.

Answer:
Perpendiculars are drawn from the centre of the circle to the chords.

Consider ΔAOM, ΔCON;
OM = ON
OA = OC (radii)
∠AMO = ∠CNO = 90°
∠A = ∠C (given)
ΔAOM ≅ ΔCON (A.A.S)
In equal triangles, angle opposite to equal sides are equal.
AM = CN
½AB = ½CD
∴ AB = CD

Textbook Page No. 73

Kerala Syllabus 9th Standard Maths Guide In Malayalam Question 1.
In a circle, a chord I cm away from the centre is 6 cm long. What is the length of a chord 2 cm away from the centre?
Answer:

Radius of the circle =
\(\sqrt{3^{2} + 1^{2}} = \sqrt{9 + 1} = \sqrt{10}\)
AB = \(\sqrt{\sqrt{10}^{2} – 2^{2}} = \sqrt{10 – 4} = \sqrt{6}\)
Length of the chord \(\sqrt 6 + \sqrt 6 = 2\sqrt 6\)

Kerala Syllabus 9th Std Maths Notes Question 2.
In a circle of radius 5cm, two parallel chords of lengths 6cm and 8cm are drawn on either side of a diameter. What is the distance between them? If parallel chords of these lengths are drawn on the same side of a diameter, what would be the distance between them?
Answer:

OM = \(\sqrt{5^{2} – 3^{2}}\)
= \(\sqrt{25 – 9}\)
=\(\sqrt {16}\) = 4 cm
ON = \(\sqrt{5^{2} – 4^{2}}\)
= \(\sqrt{25 – 16}\)
=\(\sqrt 9\) = 3 cm
The distance between the chords = 4 + 3 = 7cm
If the chords are on same side = 4 – 3 = 1cm

Kerala Syllabus Maths Class 9 Question 3.
The bottom side of the quadrilateral in the picture is a diameter of the circle and the top side is a chord parallel to it. Calculate the area of the quadrilateral.

Answer:

AB = \(\sqrt{2.5^{2} – 1.5^{2}}\)
= \(\sqrt{6.25 – 2.25}\)
=\(\sqrt 4\) = 2 cm
The quadrilateral is a trapezium.
The distance between the parallel sides = 2 cm
Area = \(\frac{1}{2}\) × 2 × (5 + 3) = 8cm²

Circles Questions And Answers Kerala Syllabus 9th Question 4.
In a circle, two parallel chords of lengths 4 and 6 centimetres are 5 centimetres apart. What is the radius of the circle?
Answer:

MN = 5
ON = x
OM = 5 – x
x² + 3² = (5 – x)² + 2²
x² + 9 = 25 – 10x + x² + 4
9 = 25 – 10x + 4
10x = 25 + 4 – 9
10x = 20
x = 20/10 = 2
Radius = \(\sqrt{2^{2} + 3^{2}}\) = \(\sqrt {4 + 9}\) = \(\sqrt {13}\)cm

Textbook Page No. 78

9th Standard Maths Notes Kerala Syllabus Question 1.
Draw three triangles with lengths of two sides 4 cm and 5 cm and angle between them 60°, 90° and 120°. Draw the circumcircle of each . (Note how the position of the circumcentre changes).
Answer:

In this triangle all the angles are less than 90°. The circum centre ‘O’ is inside the triangle.

In the triangle with one angle is 90°. The circum centre is the midpoint of the hypotenuse.

In this triangle with an angle greater than 90°. The circumcentre ‘O’ is outside the triangle.

Question 2.
The equal sides of an isosceles triangle are 8 cm long and the radius of its circumcircle is 5 cm. Calculate the length of its third side.
Answer:
ΔABC is an isosceles triangle The bisector of ∠A bisects BC
OM = x; BM = \(\sqrt{5^{2} – x^{2}}\)
When we consider ΔAMB,
AB² = AM² + BM²

82 = (5 + r)² + \((\sqrt {5^{2} – x^{2}})^{2}\)
64 = 25 + 10x + x² + 25 – x²
64 = 10x + 50
14 = 10x
x = 14/10
= 1.4
BM = \((\sqrt {5^{2} – 1.4^{2}}\)
BC = \(2(\sqrt {5^{2} – 1.4^{2}} = 2\sqrt {23.04} = \sqrt {92.16}\)
= 9.6 cm

Question 3.
Find the relation between the length of a side and the circumradius of an equilateral triangle.
Answer:

In ΔABC,
AB = BC + AC (sides of an equilateral triangle)
∠DAO = 30°, ∠ADO = 90°
∴ ∠AOD = 60°
By using the properties of angles 30°, 60°, 90°.
If OA = r
OD = \(\frac{r}{2}\)
AD = \(\frac{\sqrt 3}{2}\)r
AB = 2 × \(\frac{\sqrt 3}{2}\)r = \(\sqrt 3\)r
One side of an equilateral triangle is \(\sqrt 3\) times its circumradius.

Kerala Syllabus 9th Standard Maths Circles Exam Oriented Text Book Questions and Answers

Question 1.
Draw a circle which passes through the points A, B and radius 5 cm.
Answer:

The 5 cm line drawn from A meet the perpendicular bisector of AB at point O. The a circle drawn with O as centre and OA as radius will pass through A and B.

Question 2.
The distance between the points A and B is 3 cm. Find out the radius of the smallest circle which passes through these points? What is AB about this circle?
Answer:
1.5 cm. Diameter.

Question 3.
Draw circles which passes through the points A and B and radius 3 cm, 4 cm and 5 cm.
Answer:
Centres of the circle lie on the same straight line.

Question 4.
Two circles in the diagram have same radius. Prove that AC = BD.

Answer:
OP is drawn perpendicular to the chord. OP bisect CD and AB.

Question 5.
A and B are the centres of two circles in the diagram. Circles meet at points O and P. MN || AB. Then prove that MN = 2 × AB.

Answer:

Draw perpendicular lines AX, BY
∴ XM = XO similarly YN = YO
MN = MX + XY + YN
= OX + XY + OY = XY + XY
= 2XY = 2AB

Question 6.
AB is he diameter of the circle with centre C. PQ || AB, AB = 50cm, PQ = 14cm. Find BQ.

Answer:

Draw CM perpendicular PQ.
PQ = 14cm,
∴ MQ = 7 cm, CN = 7;
CB = 25 cm
CQ = 25 cm
NB = 25 – 7 = 18
CM = \(\sqrt{25^{2} – 7^{2}} = \sqrt{625 – 49}\)
=\(\sqrt{576}\) = 24
∴ NQ = 24
BQ = \(\sqrt{NQ^{2} – NB^{2}} = \sqrt{24^{2} + 18^{2}}\)
= \(\sqrt {576 + 524} = \sqrt {900}\) = 30 cm

Question 7.
AB, AC are two chords of a circle and the bisector of ∠BAC is a diameter of the circle. Prove that AB = AC.
Answer:
OA = OA (common side)
∠OPA = ∠OQA = 90°
(OP ⊥ AB, OQ ⊥ AC) ∠PAO
= ∠QAO (AE bisector)
∴ ΔOAP = ΔOAQ
OP = OQ therefore AB = AC (Equal chords of a circle are equidistant from the centre).

Question 8.
In the question above instead of assuming ∠OAB = ∠OCD assuming that AB = CD and then prove that ∠OAB = ∠OCD.

Answer:
∠OAB = ∠OCD (Given)
OA = OC (radius).
∠P = ∠Q = 90°
∴ (OP ⊥ AB, OQ ⊥ CD ) ∴ ΔOAP ≅ ΔOCQ;
∴ OP = OQ, Therefore AB = CD
[Equal chords of a circle are equidistant from die centre]

Question 9.
What is the distance from the centre of a circle of a circle of radius 5 cm to a chord of length 8 cm.
Answer:

Distance from the centre = cp = \(\sqrt{5^{2} – 4^{2}}\)
= \(\sqrt{25 – 16} = \sqrt{9}\) = 3cm

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Equal Triangles Text Book Questions and Answers

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You can Download ज्ञानमार्ग Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 8th Standard Hindi Solutions Unit 1 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 8th Standard Hindi Solutions Unit 1 Chapter 2 ज्ञानमार्ग (एकांकी)

ज्ञानमार्ग पाठ्यपुस्तक के प्रश्न और उत्तर

Hindi 8th Standard Kerala Syllabus प्रश्ना 1.
हर राजकुमार अपने को बड़ा ज्ञानी मानता है। असल में बड़ा ज्ञानी कौन है?

उत्तर:
सच्चे ज्ञानी के मन में अहंकार नहीं होता। वह हमेशा ज्ञान पाने की कोशिश करता रहता है। वह अपने ज्ञान को सदा अपूर्ण समझता है।

8th Standard Hindi Notes State Syllabus प्रश्ना 2.
‘गुरुजी, हम लोग संकट में पड़ गए हैं।’ राजकुमार क्यों संकट में पड़े? ।

उत्तर:
राजकुमार अपने ज्ञान पर मदांध हो गए। उन्होंने अपने ज्ञान का सदुपयोग नहीं किया। यह उनके संकट में पड़ने का कारण बन गया।

ज्ञानमार्ग Text book Activities

Hindi Class 8 Kerala Syllabus प्रश्ना 1.
‘ज्ञानमार्ग’ एकांकी का मंचन होनेवाला है। इसके लिए एक पोस्टर तैयार करें।

उत्तर:

ज्ञानमार्ग दोस्तों के प्रस्तुतीकरण में

उचित चौकोर में ✓लगाएँ।

रंगमंच:
उचित रंग सज्जा है।
सामग्रियों का उचित प्रयोग है।

पात्र:
वेषभूषा पात्रानुकूल है।
चाल-चलन उचित है।
हाव-भाव स्वाभाविक है।
अन्य पात्रों से और प्रसंग से तालमेल है।

कथोपकथन:
संवाद स्पष्ट व श्रव्य है।
संवाद प्रस्तुति स्वाभाविक है।

मंचीकरण Meaning In Malayalam Kerala Syllabus प्रश्ना 1.
वाक्य पढ़ें, रेखांकित शब्दों पर ध्यान दें

मैं मंत्र पढ़ता हूँ।
क्या, तुम इसमें प्राण भी डाल सकते हो?

Class 8 Hindi Chapter 1 Kerala Syllabus प्रश्ना 2.
चर्चा करें।
प्रत्येक वाक्य के रेखांकित शब्दों का आपसी संबंध क्या है?

पाठ भागों से ऐसे वाक्य चुनें और लिखें।


उदाहरण :
1. मैं बहुत चतुर नहीं हूँ।
2. मैं हर चीज़ को सीखना चाहता हूँ।
3. माफ़ी चाहता हूँ
4. मैं तो हमेशा ही सीखता हूँ।
5. पगड़ी तुम रख लो और खुद पढ़ो…।
6. मैं तुम दोनों से बड़ा हूँ।
7. तुम दोनों मेरे आगे मूर्ख हो।
8. मैं तुम सबसे बड़ा विद्वान हूँ।
9. मैं जानता हूँ।

ज्ञानमार्ग Meaning In Malayalam Kerala Syllabus प्रश्ना 3.
ज्ञान का मतलब :

i. हरेक में कुछ न कुछ ज्ञान है।

ii. ज्ञान बाहरी दिखावा नहीं है।

iii. ज्ञान सबकी भलाई के लिए है।

iv. अहंकार से ज्ञान असफल होता है।

“जिज्ञासा के बिना ज्ञान नहीं होता।”
— महात्मा गाँधी

“उस ज्ञान का कोई लाभ नहीं जिसे आप काम में नहीं लाते।”
— एंटन चेखोव

Hss Guru 8th Hindi Kerala Syllabus प्रश्ना 1.
ज्ञान से संबंधित उक्तियों का संकलन करें।

उत्तर:
1. केवल ज्ञान ही एक ऐसा अक्षय तत्व है, जो कहीं भी, किसी अवस्था और किसी काल में भी मनुष्य का साथ नहीं छोड़ता।
2. यदि कोई अपने धन को ज्ञान अर्जित करने में खर्च करता है तो उससे उस ज्ञान को कोई नहीं छीन सकता, ज्ञान के लिए किए गए निवेश से हमेशा अपना और दूसरों का भलाई होता है।
3. ज्ञान एक कामधेनु के समान है जो हर मौसम में अमृत प्रदान करती है। इसलिए ज्ञान को एक गुप्त धन कहा गया है। (आचार्य चाणक्य)
4. ज्ञान अगर बुद्धि में रहे तो बोझ बनता है और जब व्यवहार में आ जाए तो आचरण बनता है।
5. ज्ञान ज्ञान नहीं रह जाता जब वह इतना अभिमानी हो जाए कि रो भी ना सके, इतना गंभीर हो जाए कि हंस भी ना सके और इतना स्वार्थी हो जाए कि अपने सिवा किसी और का अनुसरण न कर सके।

ज्ञानमार्ग Summary in Malayalam and Translation

ज्ञानमार्ग शब्दार्थ Word meanings

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Kerala State Syllabus 8th Standard Maths Solutions Chapter 2 Equations in Malayalam

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Kerala State Syllabus 8th Standard Basic Science Solutions Chapter 5 Basic Constituents of Matter in Malayalam

Basic Constituents of Matter Text Book Questions and Answers

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Kerala State Syllabus 8th Standard Maths Solutions Chapter 3 Polygons in Malayalam

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Kerala State Syllabus 8th Standard Hindi Solutions Unit 2 Chapter 4 उजाला (कहानी)

उजाला Textbook Activities

Hss Live Guru 8th Hindi Kerala Syllabus प्रश्ना 1.
कहानी का कौन-सा प्रसंग शीर्षक को सार्थक बनाने में अधिक संगत है?

उत्तर:
कहानी का यह अंतिम प्रसंग शीर्षक को सार्थक बनाने में अधिक संगत है। यहाँ लेखक के पूछने पर अंधा व्यक्ति कहता है कि रात में लोग प्रकाश के बिना निकलते हैं। वे आकर हमसे टक्कर लेते हैं। उनसे बचने तथा उन्हें प्रकाश देने के लिए हम लालटेन लेकर चलते हैं। मतलब स्वयं अंधे होते हुए भी वे लालटेन से दूसरों को राह दिखाते हैं। यह प्रसंग ‘उजाला’ शीर्षक को सार्थक बनाने योग्य है।

उजाला Summary in Malayalam and Translation

उजाला शब्दार्थ Word meanings