Kerala Syllabus 9th Standard Physics Solutions Chapter 1 Forces in Fluids in Malayalam

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Kerala Syllabus 9th Standard Physics Solutions Chapter 6 Current Electricity in Malayalam

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Kerala Syllabus 9th Standard Social Science Solutions Part 2 Chapter 2 The Signature of Time

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The Signature of Time Textual Questions and Answers

The Signature Of Time Class 9 Kerala Syllabus Question 1.
You have recognized from the map that the earthquake zones, volcanoes and mountain ranges overlap. Why is this so?
Answer:

  • The crust, which is the outermost layer of the earth is solid.
  • The crust, together with the upper part of the mantle, is the lithosphere.
  • The lithosphere exists as several fragments just like the broken shell of an egg.
  • Compared to the thickness of the portion from the crust to the inner core, the lithosphere is very thin.
  • The portions of the lithosphere which are several thousand kilometers wide and roughly, 100 kilometers thick are called lithospheric plates.
  • Whether big or small, each plate contains oceanic as well as continental areas.

The Signature Of Time Class 9 Notes Kerala Syllabus Question 2.
Observe the following figures.
The Signature Of Time Class 9 Kerala Syllabus
Zones of severe earthquakes:
The Signature Of Time Class 9 Notes Kerala Syllabus
Volcanic Zones:
Hss Live Guru 9th Geography Kerala Syllabus
Can you mark the information in each of the above maps in a single map? Don’t forget to use different colors or symbols for each type of feature.
Answer:
Hss Live Class 9 Social Science Kerala Syllabus

Hss Live Guru 9th Geography Kerala Syllabus Question 3.
Look at the following diagrams and find out different movements of the lithospheric plates.
Hsslive Class 9 Social Science Kerala Syllabus
Answer:
a) Divergent margin: Plates move apart from each other.
b) Convergent margin: Plates move towards each other.
c) Transform margin (sheer margins): Plates slide past each other

Hss Live Class 9 Social Science Kerala Syllabus Question 4.
What is the average thickness of the lithosphere?
Answer:
The average thickness of the lithosphere is 100 kilometers.

Hsslive Class 9 Social Science Kerala Syllabus Question 5.
The following are the pictures of some landforms formed due to plate movements.
Hsslive Social Science 9th Kerala Syllabus
Identify their respective plate margins.
Answer:
a) Convergent margins. Fold mountains are formed along the convergent margins, eg. Himalaya.
b) Divergent margins: Magma comes out through the gap formed due to the divergence of plates and solidifies to form mountains.
Eg. Mid Atlantic Ridge.

Hsslive Social Science 9th Kerala Syllabus Question 6.
Identify the Plate margins where the World’s major fold mountains are formed?
Answer:
Fold mountains are formed along the convergent margins. The Himalayas, the Alps, the Andes, the Atlas, etc are all folded mountains. The Himalaya fold mountain is formed between the Indian Plate and the Eurasian plate.

Hsslive Guru Class 9 Social Science Kerala Syllabus Question 7.
Identify and mark the focus and epicenters in the given diagram
Hsslive Guru Class 9 Social Science Kerala Syllabus
Answer:
A – shows the epicenter B – shows the focus

Social Science Class 9 Notes State Syllabus Question 8.
Collect the details of earthquakes that have occurred since 2005.
Answer:

Year Earthquake zone Intensity
2005 Indonesia 8.6
2006 Curil Islands 8.3
2009 Samoa Islands 8.1
2011 Japan 9.0
2012 Sumatra 8.6
2014 Chile 8.2
2015 Nepal 7.9

Let Us Assess

Kerala Syllabus 9th Standard Social Science Notes Question 9.
Identify the different plate margins. Which are the associated landforms:
ANswer:

  • There are three types of plate margins. They are Divergent margin, Convergent margin and Trans-form margin.
  • Divergent margin: Plates move apart from each other. Oceanic ridges are formed.
  • Convergent margin: Plates move towards each other. Fold mountains and ocean trenches are formed.

Kerala Syllabus 9th Standard Social Science Notes Malayalam Medium Question 10.
Answer the following questions based on earthquakes.
a) How are earthquakes formed?
b) Which are the different types of seismic waves?
c) Which seismic wave causes maximum destruction on the earth’s surface?
d) On what scale are earthquakes measured?
Answer:
a) The processes that take place in the interior of the earth cause in the storage of energy in some portions in the Earth. Earthquakes occur on those instances. When this energy is bring released.
b) i. Primary waves
ii. Secondary waves
iii. Surface waves
c) The surface waves are the most destructive.
d) In Richter scale

9th Geography Notes Kerala Syllabus Question 11.
What does the term’ the Pacific Ring of Fire’ mean?
Answer:
Nearly 80 percent of the world’s volcanoes are situated around the Pacific Ocean. This zone containing more than 452 volcanoes is therefore known as ‘the Pacific Ring of Fire’.

Hss Live Guru 9th Social Science Kerala Syllabus Question 12.
Explain the various instances where volcanoes are useful to man.
Answer:

  • Though the volcanic eruptions are threats to human life. The regions where it occurs are useful to man.
  • These areas are rich in minerals. The Black soil that formed as a result of volcanic eruption is very suitable for cotton cultivation.
  • The hot springs that are found on volcanic mountains are supposed to have healing properties.
  • Volcanic ash is a good manure.
  • Geysers are formed in many volcanic regions.
  • These places have been developed as tourist spots. They are known as spas.

The Signature of Time Model Questions and Answers

Hsslive Guru 9th Social Science Kerala Syllabus Question 13.
Analyze the maps given in textbook and make a list of the conclusions that you have reached.
Answer:

  • Earthquakes are comparatively frequent in certain parts of the earth.
  • Volcanoes are more common in certain specific places.
  • There are some peculiarities in the distribution of mountains.
  • Earthquake zones and distribution of mountains on the earth’s surface more or less coincide.

Hsslive Guru Social Science 9th Kerala Syllabus Question 14.
Define Lithosphere. What is its important feature?
Answer:
The crust, together with the upper part of the mantle is known as the Lithosphere. The portions of the Lithosphere which are several thousand kilometers wide and roughly 100 kilometers thick enable to exist the life on the earth. Lithosphere exists like a broken eggshell. It consists of 7 major plates and about a dozen of minor plates.

Hss Live Guru Social Science 9th Kerala Syllabus Question 15.
Define Lithospheric plates?
Answer:
The portions of the Lithosphere, which are several thousand Kilometres wide and roughly 100 Kilometres thick are called Lithospheric plates.

Hss Live Guru Class 9 Social Science Kerala Syllabus Question 16.
Identify and list the different lithospheric plates from the following map.
Social Science Class 9 Notes State Syllabus
Answer:

  1. Pacific plate
  2. Australian plate
  3. North American plate
  4. South American plate
  5. Eurasian plate
  6. African plate
  7. Antarctic plate

9th Standard Social Science Map Kerala Syllabus Question 17.
Complete the following
Kerala Syllabus 9th Standard Social Science Notes
Answer:

  1. Major plates
  2. Minor plates

Question 18.
Write the names of the landforms that are created along the plate margins by the movements of plates?
Answer:

  • Fold Mountains
  • Seafloor
  • Ocean Trenches
  • Oceanic Mountain Ranges

Question 19.
Classify the following into major and minor plates.
1. Pacific
2. Cocos
3. Arabian
4. Eurasian
5. Scotia
6. African
Answer:
Major plates:
1. Pacific
4. Eurasian
6. African
Minor plates:
2. Cocos
3. Arabian
5. Scotia

Question 20.
The plates move at a speed of cms a year.
Answer:
2 to 12 cms

Question 21.
Analyze the figure given below summarise your findings
Kerala Syllabus 9th Standard Social Science Notes Malayalam Medium
Answer:

  • The lithospheric plates are situated above the asthenosphere which is in a semi-plastic state.
  • Magma, which is a part of the mantle, remains molten due to the high temperature at the earth’s interior and undergoes continuous convection.
  • The plates move due to convection. As a result of this, new ocean floors and subduction zones are formed.

Question 22.
What do you mean by subduction zones?
Answer:
Due to difference in density between the plates along a convergent margin, the denser plate will submerge under the lighter one. These zones are called subduction zones.

Question 23.
Who put forward the idea of continental drift.
Answer:
Alfred Wegener

Question 24.
Prepare a flow chart based on plate margins.
Answer:
9th Geography Notes Kerala Syllabus

Question 25.
Complete the following flow chart.
Hss Live Guru 9th Social Science Kerala Syllabus
Answer:
Hsslive Guru 9th Social Science Kerala Syllabus

Question 26.
Which are the Earth Movements that cause a lot of destruction to the mankind?
Answer:

  • Volcanic eruptions
  • Earthquakes

Question 27
Give examples of fold mountains.
Answer:

  • The Himalayas
  • The Alps
  • TheAndes
  • TheAtlas

Question 28.
How does the Earthquake occur?
Answer:
Rocks in the deeper interior of the earth undergo displacement and faults due to the plate movements and other causes. Under such situations, severe pressure is exerted on the earth’s lithosphere and seismic waves are generated. These waves create tremors on the earth’s surface: These are experienced by as Earthquakes.

Question 29.
Write down the various reasons for the Earthquake.
Answer:

  • Plates movements and faulting
  • Collapse of roofs of mines
  • Pressure in reservoirs
  • Volcanic eruptions.

Question 30.
What is developed in subduction zones?
Answer:
Ocean trenches are developed in subduction zones.
Eg. The Challenger Deep in the Pacific Ocean.

Question 31.
Prepare a note on the Continental Drift Hypothesis.
Answer:
Alfred Wegener, a German meteorologist, put forward the idea of continental drift in 1912. He argued that millions of years ago, all the present-day continents were a single unit forming supercontinent named Pangea which was encircled by an ocean called Panthalassa. Wegener believed that over millions of years, the continental portions drifted over the ocean floor forming the present continents.

Some recent studies indicate that mostly all the continents on earth come together once in every 500 million years. It is believed that the most recent for¬mation of Pangea was about 200 million years back, which means that we will have to wait for another 300 years for the next Pangea

Question 32.
What is seafloor spreading?
Answer:
New ocean floor is continuously being created as a result of magma that comes out through the divergent margins and solidification along the edges of the plates. This results in the phenomenon called seafloor spreading.

Question 33.
Prepare a table showing earth’s major fold mountains? Where have they formed?
Answer:
Fold mountains → Plates
Himalayas → Between the Indo-Australian Plate and the Eurasian plate.
The Andes → Between South American and the Nasca plates.
The Alps → Between Eurasia and African plates.
The Atlas → Between Eurasia and African plates.

Question 34.
Give an example of fault zone.
Answer:
The San Andreas Fault Zone in North America.

Question 35.
The plate margins are generally vulnerable. Why?
Answer:
The plate margins are generally vulnerable to earthquakes, volcanoes, and faults. This is because the plate margins are weaker than other areas.

Question 36.
Prepare a table showing the features of earthquakes having different intensities.
Answer:
The features of earthquakes having different intensities

Richter Scale Results
From 1 to 3 We don’t experience the tremors of the quake, but it is recorded in the seismograph.
From 3 to 4 Tremors are felt by us.
from 4 to 5 Small objects fall to the ground. Doors and windows shake.
From 5 to 6 Weak buildings are damaged.
From 6 to 7 Strong tremors are experienced in a large area. Weak buildings collapse.
From 7 to 8 Comparatively strong earthquake – large scale destruction takes place over a large area. Even strong building may collapse. May cause a Tsunami.
From 8 to 9 Strong earthquake causes dev­astation at a radius of 100 km from the epicenter.
Above 9 Extremely strong earthquake. Such quakes have been rare.

Question 37.
Give two examples of sudden movements of the earth.
Answer:

  • Earthquakes
  • Volcanic eruption

Question 38.
How are volcanoes formed?
Answer:
The plate margins are active with volcanoes. The hot molten rock that comes out through the fissures on the crust. Volcanoes are formed by such molten rock material coming out through the fissures along the plate margins.

Question 39.
How are earthquakes formed?
Answer:
Rocks in the deeper interior of the earth undergo displacement and faults due to plate movements and other causes. Under such situations, severe pressure is exerted on the earth’s lithosphere and seismic waves are generated just like waves in a pond spreading, in all directions when a heavy object falls into it. These waves create tremors on the earth’s surface. These tremors are experienced by us as an earthquake.

Question 40.
Name the most destructive type of seismic wave.
Answer:
The surface waves

Question 41.
Point out the relief features on the surface of the earth due to the movement of plates.
Answer:

  • Volcanoes
  • Plateaus
  • Fold mountains

Question 42.
Give reasons for the occurrence of earthquakes.
Answer:

  • Plate movements
  • Faulting
  • Collapse of roofs of mines
  • Pressure in reservoirs
  • Volcanic eruptions.

Question 43.
Distinguish between focus and epicenter.
Answer:
The deep points inside the earth where the earth-quake occurs are known as focus and the point vertically above it on the earth’s surface is known as epicenter.

Question 44.
The seismic waves are recorded by
Answer:
Seismograph

Question 45.
Which was the most severe earthquake occurred? What was its intensity?
Answer:

  • The earthquake that occurred in Chile.
  • It recorded an intensity of 9.5 in the Richter Scale.

Question 46.
Find out and prepare a note on different types of Vol-canoes?
Answer:

  • Active Volcanoes: The frequently erupted Volcanoes. Eg. Mount Fujiyama in Japan
  • Extinct Volcanoes: Active at a time. But will not erupt again. Eg. Kilimanjaro in Africa
  • Dormant Volcanoes: Volcanoes that have erupted earlier and not active in the recent past. But can become active. Eg. Vesuvius in Italy

Question 46.
What are the three types of waves produced from Focus during the Earthquake? Which is the most destructive among them?
Answer:

  • Primary waves
  • Secondary waves
  • Surface waves
  • The surface waves are the most destructive.

Question 47.
What is Seismograph?
Answer:
It is an instrument used to record the seismic waves.

Question 48.
The word ‘Tsunami’ means.
Answer:
Harbourwaves

Kerala Syllabus 9th Standard Social Science Solutions Chapter 3 National Income in Malayalam

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Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 4 फूलों का शो

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9th Hindi Notes Kerala Syllabus प्रश्ना 1.
किसी एक त्योहार की ऐतिहासिक या सांस्कृतिक पृष्ठभूमि पर लेख लिखें।
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उत्तर:
ओणम : फूलों का त्योहार
ओणम केरल का त्योहार है। केरल के लोग बड़े उत्साह से ओणम’ मनाते हैं। ‘ओणम’ के साथ राजा महाबली और वामन-अवतार की पौराणिक कथा जुड़ी हुई है। राजा महाबली बहुत ही न्यायप्रिय और परम दानवीर था। उसके शासन में प्रजा बहुत सुखी थी। ऐसा विश्वास है कि ‘ओणम’ के दिन महाबली अपनी प्रजा को देखने आते हैं। प्रजा उनका स्वागत-सत्कार करती है। वैसे भी ‘ओणम’ को ‘फूलों का पर्व’ कहा जाता है। क्योंकि ओणम के दस दिन पहले से ही घरों के आँगन में फूलों से रंगोली बनाते हैं।

 

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Kerala Syllabus 9th Standard Physics Solutions Chapter 4 Gravitation

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Kerala State Syllabus 9th Standard Physics Solutions Chapter 4 Gravitation

Gravitation Textual Questions and Answers

Activity – 1

Lifting a small stone to a certain height and then dropping it downwards.

Kerala Syllabus 9th Standard Physics Notes Chapter 4 Question 1.
What do you observe?
Answer:
The stone falls down

Hss Live Guru 9th Physics Kerala Syllabus Chapter 4 Question 2.
What could be the reason for the falling of the stone?
Answer:
Earth’s attraction is the reason

Gravitation Class 9 Kerala Syllabus Chapter 4 Question 3.
What change takes place in the speed of the stone as it is thrown up?
Answer:
Speed decreases

Gravitation Class 9 State Syllabus Chapter 4 Question 4.
What about the speed when it falls down?
Answer:
Speed increases

9th Class Physics 4th Lesson Questions And Answers Question 5.
Did you apply any force on the stone to bring it down?
Answer:
No

9th Class Physics Chapter 4 Kerala Syllabus Question 6.
From where did the stone get the force for the acceleration?
Answer:
The force required for acceleration got from earth’s attraction

Activity – 2

The stone tied to a thread is suspending from a spring balance.

Hsslive Guru Std 9 Physics Kerala Syllabus Chapter 4 Question 7.
What do you observe?
Answer:
Spring is stretched downwards

9th Class Physics Notes Kerala Syllabus Chapter 4 Question 8.
The spring stretched down when the stone was suspended from it. Why?
Answer:
Earth’s attraction is the reason

The earth attracts all objects towards its centre. This force of attraction is the force of gravity.

Conclusion:

  • Attraction between two bodies is gravitational force.
  • Attraction between bodies and earth is force of gravity.

Kerala Syllabus 9th Standard Physics Notes Pdf Chapter 4 Question 9.
Write down instances where the force of gravity is felt.
Answer:

  • Mango falling naturally form a mango tree.
  • Stone which is thrown upwards reaches the ground.
  • Raindrops falling downwards.
  • When we open the bottle filled with water and keeps it upside down water flows out.
  • An object in our hands falls down when it is released.

Activity – 3

Weighing a stone of lower mass and another of higher mass by using a spring balance.
Mass of a body is the amount of matter contained in it

Hsslive Guru 9th Physics Kerala Syllabus Chapter 4 Question 10.
In which case was the reading higher?
Answer:
Reading of the spring balance is higher when the stone having greater mass is weighed.

Hss Live Guru Class 9 Physics Kerala Syllabus Chapter 4 Question 11.
Which of the stones experienced greater force of attraction of the earth?
Answer:
Greater force of attraction of the earth is experienced in stone having greater mass.

Physics Chapter 4 Class 9 Notes Kerala Syllabus Question 12.
On the basis of these observations, find out the factor that influences the force of attraction from the earth.
Answer:
The factor is mass

  • As the distance between the bodies increases gravitational force decreases.
  • Another factor which influences the gravitational force is distance between the objects.

Universal Law of Gravitation:
All bodies in the universe attract each other. The force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. If two bodies of masses m1 and m2 are separated by a distance d, then
Kerala Syllabus 9th Standard Physics Notes Chapter 4
F ∝ m1, m2 → (1)
F ∝ 1/d2 → (2)
Combining the two, we get
Kerala Syllabus 9th Standard Physics Solutions Chapter 4 Gravitation 2
G is the gravitational constant.
The value of G is 6.67 x 10-11 Nm2/kg2. The scientist Henry Cavendish determined the value of G for the first time through experiments.

Hss Guru 9 Physics Kerala Syllabus Chapter 4 Question 13.
Complete the table given below, based on Newton’s law of gravitation?
Hss Live Guru 9th Physics Kerala Syllabus Chapter 4

Answer:
Gravitation Class 9 Kerala Syllabus Chapter 4
Observe the table and find out the answers to the following questions.

Physics Class 9 Chapter 4 Kerala Syllabus Question 14.
Two bodies are at a specific distance so as to attract each other. How many times will the mutual force of attraction be if the mass of one of them is doubled?
Answer:
Mutual force of attraction doubled

Hss Live 9th Physics Kerala Syllabus Chapter 4 Question 15.
What if the mass of both the bodies are doubled?
Answer:
Force of attraction becomes 4 times.

Hsslive Guru Class 9 Physics Kerala Syllabus Chapter 4 Question 16.
What if the distance between the bodies is doubled?
Answer:
When distance is doubled, attractive force reduced as to one fourth (1/4th).

Kerala Syllabus 9th Standard Physics Notes Chapter 4 Question 17.
What happens when the distance between the bodies is halved?
Answer:
When the distance is halved, force of attraction becomes four times.

9th Physics Notes Kerala Syllabus Chapter 4 Question 18.
Why two children sitting close to each other do not come closer due to mutual force of attraction?
Answer:
The attractive force between two children sitting close to each other are mutual. The force felt by each of them are equal and very short in magnitude. So they do not come closer to each other.

9th Std Physics Notes Kerala Syllabus Chapter 4 Question 19.
A body of mass 50 kg and another body of mass 60 kg are separated by a distance of 2 m. What is the force of attraction between them?
Answer:
m1 = 50 kg, m2 = 60 kg
d = 2m, G = 6.67 × 10-11 Nm2/kg2
Gravitation Class 9 State Syllabus Chapter 4

Force Of Gravity

9th Class Physics 4th Lesson Questions And Answers
Force of attraction by a body on earth GMm
F = \(\frac { GMm }{ R2 }\)
G – Gravitational constant
M – Mass of earth
m – Mass of body
R – Radius of earth

Class 9 Physics Chapter 4 Notes Kerala Syllabus Question 20.
Is the earth really spherical in shape?
Answer:
No

Question 21.
Is the radius of the earth the same everywhere?
Answer:
No

Question 22.
Where on the surface of the earth is the radius maximum?
Answer:
At the equator

Question 23.
Where is it minimum?
Answer:
At the poles

Question 24.
At which part of the earth must a body be placed so that it will experience the maximum force of attraction? Where the radius is large/where the radius is small?
Answer:
Where the radius is small. Force of attraction is maximum at the poles.

Question 25.
What change occurs in the force of attraction, if a body is being continuously raised from the surface of the earth?
Answer:
If a body is being continuously raised from the surface of the earth, force of attraction decreases.

Question 26.
Suppose if the body is moved from the surface of the earth to the centre. What happens?
Answer:
When it reaches the centre attractive force becomes zero.

Acceleration Due To Gravity

According to II Law of motion.
F = ma
Acceleration due to gravity a = g
i. e, a = g
According to Newton’s law of graviation F = \(\frac { GMm }{ R2 }\)
By Newton’s second law of motion, F = ma = mg
9th Class Physics Chapter 4 Kerala Syllabus

Question 27.
Find the factors affecting the value of g?
Answer:

  • Mass of the earth
  • Radius of the earth

Question 28.
The earth is not a perfect sphere, its radius is not the same everywhere If so, will the value of g be the same everywhere on earth?
Answer:
Value of g be different everywhere on earth

Question 29.
Where will the value of g be the maximum on the earth’s surface?
Answer:
At the poles

Question 30.
Where will it be the minimum?
Answer:
At the equator

Question 31.
What will be the value of g at the centre of the earth?
Answer:
Zero

Consolidation :

  • The value of g on the surface of the earth is 9.8 m/s2,
    value of g at pole = 9.78 m/s2
    value of g at equator = 9.83 m/s2
  • Value of g at moon is 1/6 of the value of g on the earth.

Question 32.
When a stone of mass 50 kg and another of mass 5 kg fall down simultaneously form the top of a five-storey building, which one will reach the ground first?
Answer:
They reaches the ground at the same time.
Reason: Acceleration due to gravity does not depend on the mass of the body. It will be the same for all
bodies falling to the earth, g = GM/R2

Question 33.
A stone and a sheet of paper are dropped together. Which of the following statements regarding their fall is true?
1. Both of them reach simultaneously
2. The paper reaches first
3. The stone reaches first
Answer:
The paper reaches slowly
Reason: Resistance of air
Feather and Coin Experiment:
The scientist who discovered that freely falling bodies offers air resistance. This was proved by Issac Newton.
Objects like paper fall slowly Galileo was the first person to; argue that this is due to air resistance. Fie wasn’t able to prove it then because at that time: there were no facilities to create a vacuum. Sir Isaac, Newton could prove this later through the ‘feather and coin’ experiment.

Newton placed a feather and a coin in a long transparent tube with closed ends. The tube was: kept vertical at first and then; suddenly turned upside down.

The coin reached the bottom first followed feather a short while later. The experiment was repeated after removing the air inside the tube and it was found that the feather and the coin reached the bottom simultaneously. The conclusion was that the feather took more time to reach the bottom due to air resistance. Thus Galileo’s argument was proved right.
Hsslive Guru Std 9 Physics Kerala Syllabus Chapter 4

Question 34.
When a stone falls, it attract the earth just as the earth attracts the stone. But it is only the stone that falls, the earth does not rise up. Give reason?
Answer:
Mass of the earth (M) is far greater than that of stone (m). So the acceleration acquired by earth is lesser and that by the stone is greater.

Question 35.
A stone falls down from the top of a wall in 1 s to the ground (g=10 m/s2)
a) What is the speed of the stone just before it touches the ground?
b) Calculate the average speed when the stone is falling down
c) How much is the height of the wall?
Answer:
9th Class Physics Notes Kerala Syllabus Chapter 4

Question 36.
A ball thrown vertically upward reached a maximum height of 20m
i) What was the velocity of the stone at the instant of throwing up?
ii) How much time did the ball take to reach the height 20 m?
Answer:
Kerala Syllabus 9th Standard Physics Notes Pdf Chapter 4

Mass and Weight

Hsslive Guru 9th Physics Kerala Syllabus Chapter 4
Mass is measured using common balance. Spring balance is used to measure weight.

  • Mass is measured using common balance.
  • The quantity of matter present in an object is called mass.
  • Spring balance is used to measure weight.
  • Weight is the force of attraction exerted by the earth on an object.
    F = mg

Question 37.
Where does a body experience maximum weight on the earth? What is the reason?
Answer:
Maximum weight experiences at the poles. Because value of g is greater at the poles.

Question 38.
Where on the earth does a body experience minimum weight? What is the reason?
Answer:
Minium weight experiences at the equator. Value of g is lesser at the equator.

Question 39.
What is the weight of the body when it is at the center of the earth? Give reasons.
Answer:
At the center, weight is zero. Because value of g is zero

Question 40.
Find out the weight of a body of mass 20 kg. Express the value in newton.
Answer:
F = mg
= 20 × 9.8
= 196 N

Free Fall

When a body is allowed to fall from a height, it falls to the earth due to the force of gravity. This is free fall.

Question 41.
Suppose a spring balance with a body suspended from it is allowed to fall. What will be the reading shown by the balance?
Answer:
Reading will be zero.

Question 46.
During the rotation of a giant wheel, a person experiences loss of weight on the descent. Explain why.
Answer:
Because the giant wheel is in free fall

Question 43.
Why does a freely falling body experience weightlessness? Note it down in your science diary.
Answer:
Due to free fall, the balance cannot exert a reaction force. Also, the force of gravity is utilized to give acceleration for the body.

Question 44.
What is the weight of a body of mass 10 kg?
Answer:
Weight F = mg = 10 × 9.8 = 98 N

Question 45.
If this body is allowed to fall freely, will there be any change in the force experienced by the body?
Answer:
Weight lessens occurs

Question 46.
Drill a hole at the bottom of an open bottle and fill it with water. Water goes out through the hole. Then allow the bottle to fall freely. What do you observe?
Answer:
During free fall, water and the bottle possess some acceleration. So there is no reaction force experiences in water. As a result water does not go out.

Let Us Assess

Question 1.
If the distance between two bodies that attract each other is trebled, how many times will their mutual force of attraction be? (9 time, 3 times, 1/3, 1/9)
Answer:
1/9

Question 2.
A body, the mass and the weight of which were already determined at the Equator, is now placed at the Pole. In this context, choose the correct statement from the following:
a) Mass does not change, weight is maximum
b) Mass does not change, weight is minimum
c) Both mass and weight are maximum
d) Both mass and weight are minimum
Answer:
a) Mass does not change, weight is maximum

Question 3.
Mass of the zeroth is 6 × 1024 kg and that of the Moon is 7.4 × 1022 kg. The distance between Earth and Moon is 3.84 × 105 km. Calculate the force of attraction of Earth of Moon. (G=6.7 × 10-11 Nm2 kg-2)
Answer:
F = \(\frac { GMm }{ d2 }\)
Hss Live Guru Class 9 Physics Kerala Syllabus Chapter 4

Question 4.
a) What is meant by the terms mass and weight?
b) Are they vector of scalar quantities? Why?
c) The mass of a body is 30 kg. What is its weight on earth? (g = 9.8 m/s2)
d) What is its weight on the moon? (g = 1.62 m/s2) )
Answer:
a) 1. The quantity of matter present in a body is called mass.
2. The force of attraction exerted by earth on a body is referred as weight.
b) Mass is a scalar quantity (No direction)
Weight is a vector quantity (Possess both magnitude and direction)
c) Weight = mg = 30 × 9 = 294 N
d) Weight on the moon = 30 × 1.62 = 48N

Question 5.
If a body of mass 40 kg is kept at a distance of 0.5 m from a body of mass 60 kg, what is the mutual force of attraction between them?
Answer:
Physics Chapter 4 Class 9 Notes Kerala Syllabus

Question 6.
Observe the figure and complete the table
Hss Guru 9 Physics Kerala Syllabus Chapter 4
Physics Class 9 Chapter 4 Kerala Syllabus
Answer:
Hss Live 9th Physics Kerala Syllabus Chapter 4

Gravitation More Questions and Answers

Question 1.
Complete the table
Hsslive Guru Class 9 Physics Kerala Syllabus Chapter 4
Answer:

SI. No. Mass of he bodies Distance between the bodies d(m) Gravitational force F(N)
m1kg) m2(kg)
1 12 5 2 G x 15
2 50 20 5 G x 40
3 10 15 0.5 G x 600

Question 2.
How the mass and distance between objects affect gravitational force?
Answer:
Among the two bodies if the mass of one body is doubled gravitational force becomes two times. When the mass of both the bodies are doubled. Gravitational force becomes four times. When the distance increases gravitational force decreases. Gravitational force varies in reciprocal of the square of the distance between the bodies.

Question 3.
Mass and weight of a body is determined at the pole and at the equator
a) Is there any difference in the mass?
b) Is there any change in the weight?
c) Justify the answer
Answer:
a) No change in mass
b) Change will occur
c) Mass is the quantity of matter present in a matter. Value of g is greater at the poles and lesser at the equator. So weight varies. Value of g does not affect the mass

Question 4.
Fill in the blanks.
a) Acceleration due to gravity does not affect the ……………. of the object
b) 1 kg wt = ………….. N
Answer:
a) Mass
b) 9.8

Question 5.
a) What is mean by free fall?
b) Weight of an object in free fall is zero? Why.
c) If a bottle having a hole at the bottom filled with water falling down freely water will not flow out. Why?
Answer:
a) Falling of a body towards earth due to force of gravity is said to be free fall.
b) In free fall the object cannot be able to give reaction force. Also, the gravitational force is utilized to give acceleration to the object.
c) During free fall, the acceleration of water and bottle remains the same. So no reaction force feel in water.

Question 6.
Correct the following.
a) Force of attraction increases when an object is raised from earth’s surface
b) Value of g remains the same on all regions of earth.
Answer:
a) Force of attraction decreases when an object is raised from earth’s surface.
b) Value of g is different on all regions of earth.

Question 7.
a) Calculate the weight of a body in the moon if it weights 150 kg in earth?
b) What will be the weight of a body at the center of earth.
Answer:
a) Weight in the moon = 150 × 1/6 = 25 kg
b) Weight at the center of earth is zero.

Question 8.
a) State Universal Law of gravitation?
b) Represent it mathematically.
c) Mention each letter indicates to what?
d) How mass and distance affect gravitational force?
Answer:
a) Every bodies in the universe attract each other. This attractive force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Kerala Syllabus 9th Standard Physics Solutions Chapter 4 Gravitation 18
c) m1 m2 – Mass of the bodies
d – Distance between the objects
F – Gravitational force
G – Gravitational constant
d) Mass increases gravitational force increases Distance increases gravitational force decreases.

Question 9.
List the situations in which gravitational force experiences.
Answer:
A mango falls naturally from the mango tree

  • Stone reaches the ground when it throws upwards.
  • Raindrops falling downwards
  • While releasing the hands the object holding in the hand falls to the ground
  • Airplane fall to earth when the engine fails.

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Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms in Malayalam 16
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms in Malayalam 17

Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms in Malayalam 18
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms in Malayalam 19
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms in Malayalam 20

Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms in Malayalam 21
Kerala Syllabus 9th Standard Maths Solutions Chapter 2 Decimal Forms in Malayalam 22

Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers

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Kerala State Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers

Kerala Syllabus 9th Standard Maths New Numbers Text Book Questions and Answers

Textbook Page No. 49

New Numbers Class 9 Kerala Syllabus Chapter 4 Question 1.
In the picture, the square on the hypotenuse of the top most right triangle is drawn. Calculate the area and the length of a side of the square.
New Numbers Class 9 Kerala Syllabus Chapter 4
Answer:
Hypotenuse of first right triangle
\(\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { {1} + {1} } = \sqrt {2} \)
Hypotenuse of second right triangle
\(\sqrt {\sqrt {{2}}^{ 2 } + 1^{2}} = \sqrt {2 + 1} = \sqrt {3} \)
Hypotenuse of third right triangle
\(\sqrt {\sqrt { { 3 } }^{ 2 } + 1^{ 2 }} = \sqrt {3 + 1} =\sqrt {4} \) = 2
Hypotenuse of fourth right triangle
\(\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { { 4 } + { 1 } } =\sqrt { 5 } \)
i. e. the length of one side of square is \(\sqrt 5 \)
Area \(\sqrt 5 \) × \(\sqrt 5 \) = 5 sq. m

Class 9 Maths Chapter 4 Kerala Syllabus Question 2.
A square is drawn on the altitude of an equilateral triangle of side 2m
Class 9 Maths Chapter 4 Kerala Syllabus
(i) What is the area of the square?
(ii) What is the altitude of the triangle?
iii) What are the lengths of the other two sides of the triangle shown below?
Kerala Syllabus 9th Standard Maths Chapter 4
Answer:
(i)
Class 9 Maths Chapter 4 New Numbers Kerala SyllabusBD = 1 m
AD = \(\sqrt { { 2 }^{ 2 }-{ 1 }^{ 2 } } = \sqrt { { 4 }-{ 1 } } = \sqrt { 3 } \)
Area of square
= \(\sqrt 3 \) × \(\sqrt 3 \) = 3 sq. m

(ii) Height of triangle \(\sqrt 3\)cm

(iii)
9th Standard Maths Chapter 4 Kerala Syllabus
Sides are 1 m and \(\sqrt 3\) m
Sides opposite to 30° angle = 1 m
Sides opposite to 60° = \(\sqrt 3\) m

Kerala Syllabus 9th Standard Maths Chapter 4 Question 3.
We have seen in Class 8 that any odd number can be written as the difference of two perfect squares. (The lesson, Identities). Using this, draw squares of areas 7 and 11 square centimetres.
Answer:
3² – 2² = 5
4² – 3² = 7
5² – 4² = 9
n² – (n – 1)² = 2n – 1
so 4² – 3² = 7
6² – 5² = 11
Draw a right angled triangle with hypotenuse 4cm and one of its side as 3 cm. It’s one side is \(\sqrt 7\) cm
Kerala Syllabus 9th Standard Maths Notes Chapter 4
Draw square BCDE with BC as the side.
New Numbers Class 9 Questions And Answers Kerala Syllabus Chapter 4
Area of square BCDE = \(\sqrt 7\) × \(\sqrt 7\) = 7 sq.cm
Draw a right angled triangle with hypotenuse 6 cm and one of its side as 5 cm. It’s one side is \(\sqrt 11\) cm
9th Class Maths Notes Chapter 4 Kerala Syllabus Chapter 4

Draw square BCDE with BC as the side.
Hsslive Guru 9th Maths Kerala Syllabus Chapter 4
Area of square QRST = \(\sqrt 11\) × \(\sqrt 11\) = 11 sq.m

Class 9 Maths Chapter 4 New Numbers Kerala Syllabus Question 4.
Explain two different methods of drawing a square of area 13 square centimetres.
Answer:
Method 1
72 – 62 = 13 Draw a right angled triangle with one side 6 cm and hypotenuse 7 cm.
Hsslive Guru Maths 9th Kerala Syllabus Chapter 4

Draw square QRST with QR as the side.
Hsslive Guru Class 9 Maths Kerala Syllabus Chapter 4

Method 2.
Hsslive Class 9 Maths Kerala Syllabus Chapter 4
Draw a right angled triangle with 1, 3 as perpendicular sides. Hypotenuse is \(\sqrt 10\)
Draw a right angled triangle with \(\sqrt 10\), 1, as perpendicular sides. Hypotenuse is \(\sqrt 11\)
Draw a right angled triangle with \(\sqrt 11\), 1 as perpendicular sides. Hypotenuse is \(\sqrt 12\)
Draw a right angled triangle with \(\sqrt 12\), 1 as perpendicular sides. Hypotenuse is \(\sqrt 13\). If AB = \(\sqrt 13\) cm.
Draw a square ABCD with AB as side.
9th Maths Notes Kerala Syllabus Chapter 4

9th Standard Maths Chapter 4 Kerala Syllabus Question 5.
Find three fractions larger than \(\sqrt 2\) and less than \(\sqrt 3\)
Answer:
\(\sqrt 2\) = 1.41; \(\sqrt 3\) = 1.73
Numbers in between \(\sqrt 2\) and \(\sqrt 3\) are 1.5, 1.6, 1.65 So fractions are \(\frac {15}{10}\), \(\frac {16}{10}\), \(\frac {165}{100}\)

Textbook Page No. 52

Kerala Syllabus 9th Standard Maths Notes Chapter 4 Question 1.
The hypotenuse of a right triangle is 1\(\frac {1}{2}\) m and another side is \(\frac {1}{2}\) m. Calculate its perimeter correct to a centimetre.
Answer:
Square of the third side is =
\((1\frac {1}{2})^{2} – (\frac {1}{2})^{2} = (1\frac {1}{2} + \frac {1}{2}) (1\frac {1}{2} – \frac {1}{2}) \) = 2 × 1 = 2
∴ Third side is \(\sqrt 2\)
Perimeter = 1\(\frac {1}{2}\) + \(\frac {1}{2}\) + \(\sqrt 2\) = 2 + \(\sqrt 2\)
= 2 + 1.41 = 3.412 m = 341.2 cm

New Numbers Class 9 Questions And Answers Kerala Syllabus Chapter 4 Question 2.
The picture shows an equilateral triangle cut into halves by a line through a vertex.
i. What is the perimeter of a part? (See the second problem at the end of the previous section)
ii. How much less than the perimeter of the whole triangle is this?
Kerala Syllabus 9th Standard Maths Notes Pdf Chapter 4
Answer:
9th Standard Maths Notes Kerala Syllabus Chapter 4
The sides of the new triangle is 2 cm,
1 m, \(\sqrt {2}^{2}-{1}^{2}\) m
(i) Perimeter of one of the triangle =
2 + 1 +\(\sqrt 3\) = 3 + 1.73 = 4.73 m

(ii) Perimeter of the whole triangle
= 2 + 2 + 2 = 6m
Less in the perimeter
= 6 – 4.73 = 1.27m

9th Class Maths Notes Chapter 4 Kerala Syllabus Chapter 4 Question 3.
Calculate the perimeter of the triangle shown below
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 16
Answer:
Draw BD perpendicular to AC
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 17
AB = 2;
BD= 1;
AD= \(\sqrt 3\)
BD = 1, DC = 1:
BC = \(\sqrt 2\)
AC \(\sqrt 3\) + 1; AB = 2: BC = \(\sqrt 2\)
Perimeter = \(\sqrt 3\) + 1 + 2 + \(\sqrt 2\)
= 3 + \(\sqrt 3\) + \(\sqrt 2\)
= 3 + 1.73 + 1.41 = 6.14 m

Hsslive Guru 9th Maths Kerala Syllabus Chapter 4 Question 4.
We have seen how we can draw a series of right triangles as in the picture.
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 18
(i) What are the lengths of the sides of the tenth triangle drawn like this?
(ii) How much more is the perimeter of the tenth triangle than the perimeter of the ninth triangle?
(iii) How do we write in algebra, the difference in perimeter of the n‘h triangle and that of the triangle just before it?
Answer:
i. Sides of the first triangle = 1 m, 1 m, \(\sqrt 2\) m
Sides of the second triangle = 1 m, \(\sqrt 2\) m, \(\sqrt 3\) m
Sides of the third triangle = 1 m, \(\sqrt 3\) m, \(\sqrt 4\) m
…………………………
…………………………
Sides of the tenth triangle = 1 m, \(\sqrt 10\) m, \(\sqrt 11\) m
Hypotenuse of the 10th triangle is \(\sqrt 11\)
Perpendicular sides are \(\sqrt 10\), 1

(ii) Perimeter of tenth triangle
= 1 m + \(\sqrt 10\) m + \(\sqrt 11\) m
Perimeter of ninth triangle = 1 m + \(\sqrt 9\) m + \(\sqrt 10\) m
More in the perimeter
= (\(\sqrt 10\) + \(\sqrt 11\) + 1) – (\(\sqrt 9\) + \(\sqrt 10\) + 1)
= \(\sqrt 11\) – \(\sqrt 9\) = (\(\sqrt 11\) – 3)m

iii. Sides of the nth triangle are
\(\sqrt n\), \(\sqrt n + 1\), 1
The sides of the (n – 1)th triangle are
\(\sqrt n – 1\), \(\sqrt n\), 1
The difference in perimeter
= (\(\sqrt n + 1\)) – (\(\sqrt n – 1\))

Hsslive Guru Maths 9th Kerala Syllabus Chapter 4 Question 5.
What is the hypotenuse of the right triangle with perpendicular sides \(\sqrt 2\) centimetres and \(\sqrt 3\) centimetres? How much larger than the hypotenuse is the sum of the perpendicular sides?
Answer:
Hypotenuse = \(\sqrt {\sqrt 3^{2}} + {\sqrt 2^{2}}\)
= \(\sqrt 3 + 2\) = \(\sqrt 5\) cm
Sum of perpendicular sides = \(\sqrt 3\) + \(\sqrt 2\) Difference with the hypotenuse
= (\(\sqrt 3\) + \(\sqrt 2\)) – \(\sqrt 5\)
= (1.73 + 1.41) – 2.24 = 3.14 – 2.24 cm = 0.9

Textbook Page No. 57

Hsslive Guru Class 9 Maths Kerala Syllabus Chapter 4 Question 1.
Of four equal equilateral triangles, two cut vertically into halves and two whole are put together to make a rectangle.
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 19
If a side of the triangle is 1 m, what is the area and perimeter of the rectangle?
Answer:
Sides of the triangle are \(\frac {1}{2}, \frac{\sqrt 3}{2}\), 1
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 20
Perimeter = 1 + \(\frac{\sqrt 3}{2}\) + \(\frac{\sqrt 3}{2}\) + 1 + \(\frac{\sqrt 3}{2}\) + \(\frac{\sqrt 3}{2}\)
= 2 + 2\(\sqrt 3\) m
Area of the rectangle = 1 × \(\sqrt 3\) = \(\sqrt 3\) m²

Hsslive Class 9 Maths Kerala Syllabus Chapter 4 Question 2.
A square and an equilateral triangle of sides twice as long are cut and the pieces are rearranged to form a trapezium, as shown below:
If a side of the square is 2 cm, what are the perimeter and area of the trapezium?
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 21
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 22
Perimeter = 2 + 2\(\sqrt 3\) + 2 + 2\(\sqrt 2\) + 2\(\sqrt 3\) + 2\(\sqrt 2\)
= 4 + 4\(\sqrt 3\) + 4\(\sqrt 2\) = 4(1 + \(\sqrt 3\) + \(\sqrt 2\))
Area = 2² + \(\frac {1}{2}\) × 4 × 2\(\sqrt 3\) = 4 + 4\(\sqrt 3\)
= 4(1 + \(\sqrt 3\)) = 4(1 + 1.73)
= 4 × 2.73 = 10.92 cm

9th Maths Notes Kerala Syllabus Chapter 4 Question 3.
Calculate the perimeter and area of the triangle in the picture.
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 23
Answer:
Draw a perpendicular from B to AC
AB = 4cm
AD = 2
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 24
DB = 2\(\sqrt 3\)
DB = DC = 2\(\sqrt 3\)
BC = \(\sqrt {12} + {12}\) = \(\sqrt 24\)
Perimeter =
= 4 + \(\sqrt 24\) + 2\(\sqrt 3\) + 2
= 6 + \(\sqrt 24\) + 2\(\sqrt 3\) cm
Area = \(\frac {1}{2}\) × AC × DB
= \(\frac {1}{2}\) × (2 + 2\(\sqrt 3\)) × 2\(\sqrt 3\) = (2 + 2\(\sqrt 3\)) \(\sqrt 3\)
= 2\(\sqrt 3\) + 2\(\sqrt 3\) × \(\sqrt 3\) = 2\(\sqrt 3\) + 6
= 6 + 2\(\sqrt 3\)

Kerala Syllabus 9th Standard Maths Notes Pdf Chapter 4 Question 4.
From the pairs of numbers given below, pick out those whose product is a natural number or a fraction.
i. \(\sqrt 3\), \(\sqrt 12\)
ii. \(\sqrt 3\), \(\sqrt 1.2\)
iii. \(\sqrt 5\), \(\sqrt 8\)
iv. \(\sqrt 0.5\), \(\sqrt 8\)
v. \(\sqrt {7\frac {1}{2}}\), \(\sqrt 3{\frac{1}{3}}\)
Answer:
i. \(\sqrt 3\), \(\sqrt 12\)
\(\sqrt 3\) × \(\sqrt 12\) = \(\sqrt 3 × 12\) = \(\sqrt 36\) = 6
product is natural number

ii. \(\sqrt 3\), \(\sqrt 1.2\)
\(\sqrt 3\) × \(\sqrt 1.2\) = \(\sqrt 3 × 1.2\) = \(\sqrt 3.6\)
product is neither a natural number nor a fraction

iii. \(\sqrt 5\), \(\sqrt 8\)
\(\sqrt 5\) × \(\sqrt 8\) = \(\sqrt 5 × 8\) = \(\sqrt 40\)
product is neither a natural number nor a fraction

iv. \(\sqrt 0.5\), \(\sqrt 8\)
\(\sqrt 0.5\) × \(\sqrt 8\) = \(\sqrt 0.5 × 8\) = \(\sqrt 4\) = 2
product is natural number

v. \(\sqrt {7\frac {1}{2}}\), \(\sqrt 3{\frac{1}{3}}\)
\(\sqrt {7\frac {1}{2}}\) × \(\sqrt 3{\frac{1}{3}}\) = \(\sqrt {7\frac {15}{2}}\) × \(\sqrt 3{\frac{10}{3}}\)
= \(\sqrt {7\frac {150}{6}}\) = \(\sqrt 25\) = 5
product is natural number

Textbook Page No. 60

9th Standard Maths Notes Kerala Syllabus Chapter 4 Question 1.
Calculate the length of the sides of the equilateral triangle on the right, correct to a millimetre.
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 25
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 26
Let one side = 2x
\(\sqrt (2x)^{2} – x^{2}\) = 4
\(\sqrt 4x^{2} – x^{2}\) = 4;
\(\sqrt 3x^{2}\) = 4; \(\sqrt 3\) x = 4; x = \(\frac {4}{\sqrt 3}\) = 4;
x = \(\frac{4 \sqrt{3}}{\sqrt{3} \times \sqrt{3}} =\frac{4}{3} \sqrt{3} =\frac{4}{3} \times 1.73 = 2.309\)cm
= 2.31 cm
Length of sides = 2 × \(\frac {4}{\sqrt 3}\) = 2 × 2.309
= 4.62 cm = 46.2 cm

Question 2.
Prove that (\((\sqrt 2 + 1)(\sqrt 2 – 1 )\)) = 1. Use this to compute \(\frac {1}{\sqrt {2} – 1}\) correct to two decimal places.
Answer:
\((\sqrt{2} + 1)(\sqrt{2} – 1) = (\sqrt{2})^{2} – 1^{2}=2 – 1 = 1\)
\((\sqrt 2 + 1)(\sqrt 2 – 1 )\) = 1
\(\frac{1}{\sqrt{2} – 1} = \frac{1}{\sqrt{2} – 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{1} = \sqrt{2} + 1\)
\(\frac{1}{\sqrt{2} – 1} = \sqrt {2} + 1 \) = 1.41 + 1 = 2.41

Question 3.
Compute \(\frac{1}{\sqrt{2} + 1} \) corrects to two decimal places.
Answer:
\(\frac{1}{\sqrt{2} + 1} = \frac{1}{\sqrt{2} + 1} \times \frac{\sqrt{2} – 1}{\sqrt{2} – 1} = \frac{\sqrt{2} – 1}{1}\)
= \(\sqrt {2} \) – 1 = 1.41 – 1 = 0.41

Question 4.
Simplify (\(\sqrt 3 – \sqrt 2 \))(\(\sqrt 3 + \sqrt 2 \)). Use this to compute \(\frac{1}{\sqrt{3} – \sqrt {2}}\) and \(\frac{1}{\sqrt{3} + \sqrt {2}}\) correct to two decimal places.
Answer:
(a – b)(a+b) = a² – b²
\((\sqrt{3} – \sqrt{2})(\sqrt{3} + \sqrt{2}) = (\sqrt{3})^{2} – (\sqrt{2})^{2}\)
= 3 – 2 = 1
\(\frac{1}{\sqrt{3} – \sqrt{2}} = \frac{(\sqrt{3} -\sqrt{2})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} – \sqrt{2})} = \sqrt{3} + \sqrt{2}\)
≈ 1.732 + 1.414 = 3.146 = 3.15
\(\frac{1}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} – \sqrt{2})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} + \sqrt{2})}=\sqrt{3} – \sqrt{2}\)
≈ 1.732 – 1.414 = 0.318 = 0.32

Question 5.
Prove that \(\sqrt{2 \frac{2}{3}} = 2 \sqrt{\frac{2}{3}}\). Can you find other numbers like this?
Answer:
\(\sqrt{2 \frac{2}{3}} = \sqrt{\frac{8}{3}} = \sqrt{4 \times \frac{2}{3}} = 2 \sqrt{\frac{2}{3}}\)
\(\sqrt{3 \frac{3}{8}} = \sqrt{\frac{27}{8}} = \sqrt{9 \times \frac{3}{8}} = 3 \sqrt{\frac{3}{8}}\)
If \(\sqrt{x \frac{x}{n}} = x \sqrt{\frac{x}{n}}\) then
\(\sqrt{x + \frac{x}{n}} = \sqrt{x^{2} \frac{x}{n}} = \sqrt{\frac{x^{3}}{n}}\)
\(x + \frac{x}{n} = \frac{x^{3}}{n}\)
nx + x = x³
n + 1 = x²
n = x² – 1
similar numbers are
\(4 \frac{4}{4^{2} – 1} = 4 \frac{4}{15}\)
\(5 \frac{5}{5^{2} – 1} = 5 \frac{5}{24}\)
\(\sqrt{4 \frac{4}{15}} = 4 \sqrt{\frac{4}{15}}\)
\(\sqrt{5 \frac{5}{15}} = 5 \sqrt{\frac{5}{15}}\) etc.

Question 6.
All red triangles in the picture are equilateral. What is the ratio of the sides of the outer and inner squares?
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 27
Answer:
Let the sides of the shaded triangles are a units. The sides of the unshaded triangles are a, a, \(4 \frac{a}{2}\sqrt 3 \) units.
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 28
Side of the outer square = \(\sqrt{3}\frac{a}{2} + {\sqrt 3 \frac{a}{2}} + a\)
\(\sqrt{3}a + a = a(\sqrt {3} + 1)\)
Side of the inner square =
\(\sqrt{3}a + a – a(a + a) = \sqrt{3}a + a – a – a\)
\(\sqrt{3}a = a (\sqrt {3} – 1)\)
Ratio of the sides = \( a(\sqrt{3} + 1):a(\sqrt{3} – 1)\)
\( \sqrt{3} + 1 : \sqrt{3} – 1\)

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

Question 1.
Find the perimeter of the given triangle.
Hypotenuse = \(\sqrt{{….}^{2} + {….}^{2}}\) = \(\sqrt{….}\)
Perimeter = 2 + 3 + ………
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 29
Answer:
Hypotenuse = \(\sqrt{13}; \sqrt{2^{2} + 5^{2}}=\sqrt{4 + 9}\)
Perimeter = 2 + 3 +\(\sqrt{13} = 5 + \sqrt{13}\)

Question 2.
Find the perimeter of the rectangle with area 10 sq. centimetres.
Answer:
Let one side of the reactable be ‘a’
Area = a²
a² =10 sq. cm a = \(\sqrt{10}\)
Perimeter = 4 × a = 4 × \(\sqrt{10}\) cm

Question 3.
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 30
Answer:
\(\sqrt{x} \times \sqrt{y} = \sqrt{x \times y} = \sqrt{x y}\)
\(\sqrt{2} \times \sqrt{7} = \sqrt{2 \times 7} = \sqrt{14}\)
\(\sqrt{8} = \sqrt{4} \times \sqrt{2} =\sqrt{4} \times \sqrt{2} = 2\sqrt{2}\)
\(\sqrt{18} = \sqrt{9} \times \sqrt{2} =\sqrt{9} \times \sqrt{2} = 3\sqrt{2}\)
\(\sqrt{8} + \sqrt{18} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}\)

Question 4.
\(\frac{\sqrt{x}}{\sqrt {y}} = \frac{\sqrt{…..}}{\sqrt {…..}}\)
Answer:
\(\frac{\sqrt{x}}{\sqrt {y}} = \frac{\sqrt{x}}{\sqrt {y}}\)

Question 5.
\(\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{…….}\)
Answer:
\(\frac{\sqrt{1}}{\sqrt{2}} = \frac{\sqrt{1}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}\)

Question 6.
Find the value of \(\sqrt 50\) correct to 2 decimals.
\(\sqrt{50} = \sqrt{25 \times ……} = \sqrt{25} \times \sqrt{……} = 5\sqrt{2}\)
= 5 × 1.414 = ……
Answer:
\(\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}\)
= 5 × 1.414 = 7.070 = 7.07

Question 7.
Find the perimeter of the triangle correct to two, decimal places.
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 31
Answer:
Perimeter = \(\sqrt{2} + 2\sqrt{2} + 3\sqrt{2} = 6\sqrt{2}\)
= 6 × 1.414 = 8.484 = 8.48 cm

Question 8.
If x = \(\frac{1}{\sqrt{2}}\), find (x + \(\frac{1}{x})^{2}\)
Answer:
x = \(\frac{1}{\sqrt{2}}\)
\(\frac{1}{x} = \sqrt{2} ; \left(x + \frac{1}{x}\right)^{2} = \left(\sqrt{2} + \frac{1}{\sqrt{2}}\right)^{2}\)
2 + \(\frac{1}{2}\) + 2 = \(4\frac{1}{2}\) = 4.5

Question 9.
What is the total length of the line joining two lines of lengths \(\sqrt 2\)cm, \(\sqrt 3\). Find the length correct to 3 decimal places.
Answer:
Length of the line = \(\sqrt 2\)cm + \(\sqrt 3\)
= 1.414+ 1.732 = 3.146 cm

Question 10.
Which is greater \(\sqrt 3\) + \(\sqrt 2\), \(\sqrt 5\)cm?
Answer:
\(\sqrt 3\) + \(\sqrt 2\)
\(\sqrt 3\) + \(\sqrt 2\) = 3.146
\(\sqrt 5\) = 2.236;
\(\sqrt 3\) + \(\sqrt 2\) + \(\sqrt 3\) + \(\sqrt 2\) > \(\sqrt 5\).

Question 11.
A line of length \(\sqrt 27\)cm is cut from a line of length \(\sqrt 12\)cm. Find the length of the remaining part of the line?
Answer:
Remaining part = \(\sqrt 27\) – \(\sqrt 12\)
= \(3\sqrt 3\) – \(2\sqrt 3\) = \(\sqrt 3\) cm

Question 12.
Simplify the following.
(a) \(\sqrt 50\) × \(\sqrt 2\)
(b) \(\sqrt 27\) × \(\sqrt 3\)
(c) \(\sqrt 12\) × \(\sqrt 36\)
(d) \(\sqrt 24\) × \(\sqrt 6\)
(e) \(\sqrt 28\) × \(\sqrt 7\)
(f) \(\sqrt 32\) × \(\sqrt 2\)
(g) \(\sqrt 5\) × \(\sqrt 10\) × \(\sqrt 2\)
(h) \(\sqrt 8\) × \(\sqrt 5\) × \(\sqrt 10\)
Answer:
(a) \(\sqrt 25 × 2 \) x \(\sqrt 2\) = \(\sqrt 25\) × \(\sqrt 2\) × \(\sqrt 2\) = 5 × 2 = 10
(b) \(\sqrt 9 × 3\) × \(\sqrt 3\) = 3 × 3 = 9
(C) \(\sqrt 4 × 3\) × 6 = 2 × \(\sqrt 3\) × 6 = 12\(\sqrt 3\)
(d) \(\sqrt 6 × 4 \) × \(\sqrt 6\) = 2 × 6 = 12
(e) \(\sqrt{4 \times 7} \times \sqrt{7} = 2 \times 7 = 14\)
(f) \(\sqrt{16 \times 2} \times \sqrt{2} = 4 \times 2 = 8\)
(g) \(\sqrt{5} \times \sqrt{5} \times \sqrt{2} \times \sqrt{2} = 10\)
(h) \(\sqrt{4} \times \sqrt{2} \times \sqrt{5} \times \sqrt{2} \times \sqrt{5} = 2 \times 2 \times 5 = 20\)

Question 13.
Simplify the following.
(a) \(3\sqrt{2} + 5 \sqrt{2}\)
(b) \(7\sqrt{5} – 3 \sqrt{5}\)
(C) \(4\sqrt{7} + 5 \sqrt{7} – 3\sqrt{7}\)
(d) \(2\sqrt{3} + \sqrt{27}\)
(e) \(4\sqrt{3} – 3 \sqrt{12} + 3\sqrt{75}\)
(f) \(\sqrt{162} – \sqrt{72}\)
(g) \(4\sqrt{12 }- \sqrt{50} – 5\sqrt{48}\)
(h) \(\sqrt{8}0 + \sqrt{125}\)
(i) \(\frac{1}{\sqrt{5} – 1}\)
(j) \(\frac{3}{2 \sqrt{3} – \sqrt{2}}\)
(k) \(\frac{5}{2 \sqrt{7} – 3\sqrt{5}}\)
Answer:
(a) \(8\sqrt{2}\)

(b) \(4\sqrt{5}\)

(C) \(6\sqrt{7}\)

(d) \(\sqrt{3} + 3 \sqrt{3} = 5\sqrt{3}\)

(e) \(4\sqrt{3} – 3\sqrt{4 \times 3} + 3\sqrt{25 \times 3}\)
= 4 \(\sqrt{3}  – 6\sqrt{3} + 15 \sqrt{3}\)
= \(13\sqrt 13\)

(f) \(\sqrt{81 \times 2} – \sqrt{36 \times 2} = 9\sqrt{2} – 6\sqrt{2} = 3\sqrt{2}\)

(g) \(4\sqrt{4 \times 3} – \sqrt{25 \times 2} – 5\sqrt{12 \times 4}\)
\( = 8\sqrt{3} – 5\sqrt{2} -20 \sqrt{2}\)
\( = -12\sqrt{3} – 5\sqrt{2}\)

(h) \(\sqrt{16 \times 5}+\sqrt{25 \times 5} = 4\sqrt{5} + 5 \sqrt{5} = 9\sqrt{5}\)

Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 32
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 33

Question 14.
If \(\sqrt{75} + \sqrt{363} + x\sqrt{3} = 0\). Find the value of x.
Answer:
\(x \sqrt{3} = 0 – \sqrt{75} – \sqrt{363}\)
\(= – 1 \times(\sqrt{75} + \sqrt{363})\)
\(x = \frac{-1(5 \sqrt{3} + 11 \sqrt{3})}{\sqrt{3}} = \frac{-1 \times 16 \sqrt{3}}{\sqrt{3}} = -16\)

Question 15.
Calculate \(\sqrt{3}(\sqrt{48} + \sqrt{32} – \sqrt{18})\)
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 34

Question 16.
Write two fractions between \(\frac{1}{3}\) and \( \frac {1}{4}\)
Answer:
Fraction one = \(\frac{\frac{1}{3} + \frac{1}{4}}{2}=\frac{7}{24}\)
Another fraction =
\(\frac{\frac{1}{3} + \frac{7}{24}}{2} = \frac{8 + 7}{48} = \frac{15}{48}\)

Question 17.
In the figure given below, PQRS is a square of each side is 3 cm. Each sides of the square is divided into 3 equal parts. Joining these points to get an octagon, find its perimeter.
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 35
Answer:
ΔAPH is a right angled triangle.
AP = PH = 1 cm
∴ AH = \(\sqrt{1^{2}+1^{2}} = \sqrt{2}\)
AH = GF = ED = CB = \(\sqrt{2}\)
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 36

Perimeter of the octagon
= AB + BC + CD + DE + EF + FG + GH + HA
\(=1 + \sqrt{2} + 1 + \sqrt{2} + 1 + \sqrt{2} + 1 + \sqrt{2} = 4 + 4 \sqrt{2}\)
= 4 + 4 × 1.41 = 4 + 5.64 = 9.64cm.

Question 18.
If three points A, B,C. Such that AB = \(\sqrt{50}\)cm, BC = \(\sqrt{98}\)cm, AC = \(\sqrt{288}\)cm. Check whether the points A,B,C lie on a straight line?
Answer:
AB = \(\sqrt{50}\) = \(5\sqrt{2}\)cm
BC = \(\sqrt{98}\) = \(7\sqrt{2}\)cm
AC = \(\sqrt{288}\) = \(12\sqrt{2}\)cm
\(5\sqrt{2}\) + \(7\sqrt{2}\) = \(12\sqrt{2}\)
AB + BC = AC
The three points lie on a straight line.

Question 19.
Find the value of \(\sqrt{12} – \frac{1}{\sqrt{3}}\) correct to two decimal places.
Answer:
\(\sqrt{12} – \frac{1}{\sqrt{3}} = \frac{2 \sqrt{3}}{1} – \frac{1}{\sqrt{3}} = \frac{6}{\sqrt{3}} – \frac{1}{\sqrt{3}} = \frac{5}{\sqrt{3}}\)
= \(\frac {5}{1.73}\) = 2.89

Question 20.
Find the sum 745+7180+780
Answer:
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 37

Question 21.
A, B, C are three points such that AB = \(\sqrt {50}\) cm, BC = \(\sqrt {98}\) cm, AC =
\(\sqrt {288}\) cm. Do they lie on a straightline?
Answer:
If A, B, C are points on the same line then AB + BC = AC
Kerala Syllabus 9th Standard Maths Solutions Chapter 4 New Numbers 38
Here AB + BC = AC. So the three given points are on the same straight line.

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Gandhiji Gandhiji Kaise Bane Kerala Syllabus 9th प्रश्ना 1.
वे धीरे-धीरे गांधी बने’ इसका मतलब क्या है?
Gandhiji Gandhiji Kaise Bane Kerala Syllabus 9th
उत्तर:
अपने कर्मों से ही एक व्यक्ति महान बनता है। कोई भी एकाएक अपने आप महान नहीं बनते। गांधीजी पहले एक साधारण आदमी था। लेकिन उन्होंने अपने अनुभवों से अपने को गढ़ा और वे एक समाज सेवी बन गए। ऐसा परिवर्तन उनके जिंदगी में धीरे-धीरे आया था।

Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 प्रश्ना 2.
औरत की हालत देखकर गांधीजी ने एक ही धोती पहनने का फैसला कर लिया। ऐसा फैसला लेने का उद्देश्य क्या था?
Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2
उत्तर:
तत्कालीन समाज गरीबी से त्रस्त था। यह दृश्य तत्कालीन भारत की गरीबी का चित्र उनके सामने पेश किया। भारत की आम जनता गरीबी एवं अभावों से विवश थी.। यह समझकर गांधीजी ने उनके समान जीने का निश्चय किया। यह गांधीजी की ज़िंदगी का एक अहम मोड़ था।

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Gandhiji Gandhiji Kaise Bane Notes Kerala Syllabus 9th प्रश्ना 1.
भाषण तैयार करें। ‘मेरा जीवन ही मेरा संदेश है’ यह गांधीजी का कथन है। पाठभाग के आधार पर इसका विश्लेषण करके भाषण तैयार करें।
Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 गांधीजी गांधीजी कैसे बने 3Gandhiji Gandhiji Kaise Bane Notes Kerala Syllabus 9th
उत्तर:
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Gandhiji Gandhiji Kaise Bane Malayalam Kerala Syllabus 9th प्रश्ना 1.
गांधीजी की पहले की वेश-भूषा कैसी थी?
Gandhiji Gandhiji Kaise Bane Question Answer Kerala Syllabus 9th
उत्तर:
गांधीजी कुर्ता पाजामा पहन रखा था। पाँव में चप्पल थे। सर पर गांधी टोपी थी। बच्चों के स्कूल बैग की तरह गले में झोला टाँगता था।

Gandhiji Gandhiji Kaise Bane Question Answer Kerala Syllabus 9th प्रश्ना 2.
बचपन में गांधीजी के चरित्र की विशेषता क्या थी?
Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 गांधीजी गांधीजी कैसे बने 7
उत्तर:
बचपन में गांधीजी को अंधेरे से डर लगता था। उन्हें लगता था कोई भूत आकर उन्हें पकड़ लेगा।

9th Class Hindi Notes Kerala Syllabus प्रश्ना 3.
किस घटना गांधीजी की वेशभूषा में परिवर्तन लाया?
Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 गांधीजी गांधीजी कैसे बने 8
उत्तर:
एक बार गांधीजी भाषण देने मदुरै गए। वहाँ एक औरत को देखा जो तालाब में अपनी धोती धो रही थी। इसप्रकार धो रही थी कि आधी पहनती थी और बाकी आधी धोती थी। फिर धुली हुई पहन लेती थी और शेष को धोती थी। इस घटना ने गांधीजी को गरीबी पर सोचने को विवश किया। इससे उन्होंने एक धोती ही पहनने का फैसला ले लिया।

Kerala Syllabus 9th Standard Hindi Notes प्रश्ना 4.
गांधीजी सादगी से जीवन बिताते थे। पाठ भाग से एक उदाहरण पेश करें।
Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 गांधीजी गांधीजी कैसे बने 9
उत्तर:
गांधीजी का बढ़िया पेन एक दिन चोरी चला गया। तब से एक बच्चे से दी गई पेंसिल से लिखना शुरू किया। लिखते-लिखते पेंसिल छोटी हो गई। तो उन्होंने कागज़ की भोंगली लगाकर धागे से बाँधकर लिखते थे।

गांधीजी गांधीजी कैसे बने Grammar

गांधीजी गांधीजी कैसे बने व्याकरण के प्रश्न

9th Hindi Notes Kerala Syllabus प्रश्ना 1.
ये वाक्य पढ़ें :
Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 गांधीजी गांधीजी कैसे बने 22
प्रत्येक वाक्य में रेखांकित शब्दों का आपसी संबंध पहचानें।
Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 गांधीजी गांधीजी कैसे बने 10
i. औरत तालाब में अपनी धोती धो रही थी।
ii. परीक्षा में गांधीजी चौंतीस बच्चों में से बत्तीसवें स्थान पर रहे।
iii. ये लोग कविता करते है
iv. मेरे पड़ोस में एक गरीब और मेहरून्नीसा रहती है।
9th Class Hindi Notes Kerala Syllabus

Hss Live Guru Class 9 Hindi Kerala Syllabus 9th प्रश्ना 2.
तालिका के शब्दों से वाक्य बनाएँ:
Kerala Syllabus 9th Standard Hindi Notes
उत्तर:
1. धरती पानी के लिए तरस रही है।
2. रीता अभी -अभी घर आई है
3. गाती आती है
4. सलमान क्रिकेट खेलता है
5. गौरव पत्र लिखता हे.
6. सलमान परसो आएगा

गांधीजी गांधीजी कैसे बने Summary in Malayalam and Translation

9th Hindi Notes Kerala Syllabus
Hss Live Guru Class 9 Hindi Kerala Syllabus 9th
Class 9 Hindi Notes Kerala Syllabus
Hsslive Guru Class 9 Hindi Kerala Syllabus
Hindi Notes For Class 9 Kerala Syllabus
Kerala Syllabus 9th Standard Notes Hindi
Hindi Notes 9th Class Kerala Syllabus

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Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 गांधीजी गांधीजी कैसे बने 20
Kerala Syllabus 9th Standard Hindi Solutions Unit 2 Chapter 2 गांधीजी गांधीजी कैसे बने 21