Class 9

You can Download Forces in Fluids Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Physics Solutions Part 1 Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.

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

Forces in Fluids Textual Questions and Answers

Activity

Class 9 Physics Forces Of Fluids Kerala Scert Solutions Question 1.
Take water in a bucket. Place a tightly closed bottle on it. What do you observe?
Answer:
Bottle floats on water surface

Physics Class 9 Chapter 1 Forces In Fluids Notes Question 2.
Immerse the plastic bottle to the bottom of the bucket. Don’t you have to exert a force? Why is it so?
Answer:
It is because water exerts an upward force

Kerala Syllabus 9th Standard Physics Notes Chapter 1 Question 3.
Leave the bottle free. What do you observe?
Answer:
Bottle rises to the surface of water

Forces In Fluids Questions And Answers Kerala Syllabus 9th Question 4.
When an object in water is lifted, what happens? What can be the reason?
Answer:
When an object in water is lifted, its weight appears diminished, when compared to that in air.
It is because water exerts an upward force on a body placed in it.
Fluids: Liquids and gases together are generally called fluids.
Buoyancy: When a body is immersed completely or partially in a liquid, the liquid exerts an upward force on the body. This force is the buoyancy.

Class 9 Physics Forces Of Fluids Kerala Scert Worksheets Question 5.
Tabulate some situations in your daily life where buoyant forces are experienced in liquids and gases.
Answer:

• A hydrogen-filled balloon rises in the air.
• Ships float on the surface of water.
• Submarines work by controlling the upthrust of water.
• While drawing water from well, weightlessness is felt when the bucket is underwater level.
• Life jacket floats on the surface of water.

To Measure Buoyancy?

Experiment:
Aim: To find method to measure the buoyancy experienced by a body in a liquid.
Materials: Stone, piece of metal, water, beaker, spring balance.
Procedure: Take a piece of stone and a piece of metal of almost the same size. Find out the weight of each in air using a spring balance calibrated in newton. Then find the weight of each of them when immersed in water. Record the data obtained in table.

Observation: The stone and the metal piece lose weight in water.
Conclusion: Buoyancy is the same as the loss of weight experience in water. That is in order to calculate the buoyancy experienced by a substance immersed in a fluid, it is enough to find out the loss of weight of the substance in that fluid.
Buopyant force of liquid = loss of weight

Factors That Influence Buoyancy

Experiment: To find the factors that influence buoyancy
Materials: Water, Kerosene, Saline water, beakers, spring balance, stone, copper block and iron block of same weight etc.
Activity 1: Take water, kerosene and saline water in three separate beakers. Find out the buoyancy of these liquids which is exerted on a piece of stone and tabulate.

• The stone experiences maximum buoyancy in saline water.
• The stone experiences least buoyancy in kerosene. Density of Kerosene is lower than that of other liquids.

Analysis:

Saline water has greatest density and kerosene has least density. As density increases, buoyancy also increases.
Conclusion
Density of a liquid is a factor that influences the buoyancy of a body in that liquid. As density of the liquid increases buoyancy increases.
Activity: Calculate the buoyancy exerted by water on two blocks of equal weight, one of copper and the other of iron.

Observation:

• Buoyancy experienced by each block can be seen different.
• Buoyancy exerted by water on copper block is greater and on iron block is less.

Analysis: The mass and weight of the copper and the iron blocks are the same. But their volumes are different. Copper block has greater volume, and the buoyancy experienced by copper block is greater. Iron block has less volume and the buoyancy experienced by iron block is less.
Conclusion: The buoyancy acting on a body depends also on its volume.

Archimedes principle

Experiment:
Aim: To find whether there is any relation between the weight of the fluid displaced and its buoyancy when an object is in a fluid.
Materials: Store, water, spring balance, overflowing jar, beaker, iron block

Activity: Find out the buoyancy of a piece of stone and an iron block in water. Simultaneously, using an overflowing jar, find out the volume of water they dis¬placed. Find out the weight of water overflowed using a spring balance. Tabulate the results in table.
Observation:

• The loss of weight of the stone in water (buoyancy) is the same as the weight of water overflowed.
• The loss of weight of the iron block in water (buoyancy) is the same as the weight of water overflowed.

Conclusion: The buoyancy experienced by an object in a liquid will be equal to the weight of the liquid displaced by it.
Archimedes principle:
When an object is immersed partially or completely in a liquid, the buoyancy experienced by it will be equal to the weight of the liquid displaced by it.

Archimedes:
Archimedes was born in 273 BC at Syracuse, a port city in Italy. He lived during the rule of Hiero II. King Heiro commissioned a goldsmith to fabricate a crown. He commanded Archimedes to find out whether there was any impurity in the crown. This problem confounded him. He knew that the density of pure gold could be calculated by dividing the mass of the gold bar by its volume.

His confusion was how to calculate the volume without damaging the crown. When he stepped into his bath, he could see water overflowing. This made him realise that he could calculate the volume of an object, by finding the volume of the water displaced by the same. By finding the density and volume of the gold he could prove that the crown contained impurities. He was stabbed to death by a Roman soldier in 212 BC during the Punic War. This happened when he was engaged in the complicated mathematical activities related to circle.

Forces In Fluids Worksheet Answers Kerala Syllabus 9th Floatation

Activity: Make a small vessel using aluminium foil of length and breadth 10 cm each. Place this small vessel on the surface of water in a trough. Then unfold the vessel of aluminium foil. Fold it till it becomes a small piece. Then put it in water.

Observation:

• The aluminium foil vessel floats on water.
• The folded foil sink in water.

Activity 2: Put some pieces of stone, wood, rubber cork etc., one by one into water. Note down the items which float on water. Find out the buoyancy and the weight of the water displaced by each floating body and record them in table. Observation: In the case of the floating object the buoyancy and weight of the displaced water is same.
Conclusion: A body float on a liquid when the weight of body is equal to the weight of displaced water.
Law of floatation:
Weight of a floating body is equal to the weight of the liquid displaced by it.

Kerala Syllabus 9th Standard Physics Notes Pdf Question 6.
Find out the reasons for the following.
a) A piece of stone experience a loss of weight within water.
b) Though an egg sinks in pure water, it floats on salt water.
c) Kerosene floats on water
d) A ship floats on water
e) When a body was placed in a liquid it remained in the same position.
Answer:
a) It is because of the buoyancy exerted by water.
b) As pure water has lower density it has less buoyancy. Saline water has greater density and has greater density and has greater buoyancy. So an egg sinker in pure water and floats on saltwater.
c) Kerosene has density less than that of water.
d) It is because weight of the ship is equal to the weight of the water displaced by it.
e) Density of the body is same as that of the liquid.

Kerala Syllabus 9th Standard Physics Notes Pdf Malayalam Medium Question 7.
Does a ship that enters a freshwater lake from the ocean sink more or rise move? Justify your answer.
Answer:
The ship sinks more.
Seawater has more density and has higher buoyancy. Pure water has less density and less buoyancy. So the ship that enters a freshwater lake from the ocean sinks more.

Kerala Syllabus 9th Standard Physics Notes English Medium Question 8.
What are the factors affecting buoyancy?
Answer:
Density of the liquid and volume of the body

9th Standard Physics Notes Kerala Syllabus Question 9.
As density of a liquid increases buoyancy (increases/decrease)
Answer:
increases

Relative Density

Relative density of a substance is the ratio of the density of the substance to the density of water.
Relative density = \(\frac{\text { Density of substance }}{\text { Density of water }}\)
Since it is a ratio, relative density has no unit.

Forces In Fluids Notes Kerala Syllabus Question 10.
Density of water is =
Answer:
1000 kg/m3

Hsslive 9th Physics Kerala Syllabus Question 11.
How is relative density of kerosene calculated?
Answer:
Relative density of a liquid = \(=\frac{\text { Density of liquid }}{\text { Density of water }}\)

Hsslive Std 9 Physics Kerala Syllabus Question 12.
Write examples for liquids of density greater and lesser than that of water.
Answer:

Kerala Syllabus 9th Standard Physics Notes Malayalam Medium Question 13.
Which device is used to measure the relative density of a liquid?
Answer:
Hydrometer

Question 14.
What will be the reading when the hydrometer is dipped in water?
Answer:
1

Question 15.
Suppose the hydrometer is dipped in a liquid of density greater than that of water. Will the liquid surface be above or below the marking of 1?
Answer:
The liquid surface will be below the marking of 1.

Question 16.
In which case does the hydrometer sink more? In liquids of greater density or those with a lower density?
Answer:
In liquids of lower density.

Question 17.
Why do the markings on the hydrometer increase towards the bottom?
Answer:
When a hydrometer is dipped in a liquid of greater density, the liquid surface will be below the marking of 1. But in a liquid of lower density, the hydrometer sinks more and the liquid surface will be above the marking of 1. Relative density of water is 1, and that of liquids with higher density will be greater than 1 and that of liquids with lower density will be less than 1.

Question 18.
Name two instruments which work on the principle of floatation?
Answer:
Hydrometer:

Lactometer:

Question 19.
Which instrument is used to test the purity of milk?
Answer:
Lactometer

Pascal’S Law

Activity 1: Take a plastic bag without holes and fill it with water. Tie its mouth tightly so that no air gets inside. Make holes on the bag with a pin. Observe what happens?
Observation: There are identical streams of water from all the holes.
Conclusion: Pressure applied at any point of the bag is transmitted equally to entire bag.
Pascal’s Law:
The pressure applied at any point of a liquid at rest in a closed system will be experienced equally at all parts of the liquid.
Volume of a liquid can not be changed using pressure. This is the basis of Pascal’s law.
Activity 2: Fill two identical syringes with water. Connect them with a plastic tube. Then arrange them as shown in figure.

Question 20.
Press slightly on the end X. What do you observe.
Answer:
Y rises up

Question 21.
Bring the ends X and to the same level. Place a slotted weight of 1N (100 gwt) at the end of X. What happens at the end Y?
Answer:
Y rises

Question 22.
Place a weight of 1 N at the end Y as well. What do you observe now?
Answer:
X and Y balance
Then replace the syringe X with one of slightly smaller diameter and repeat the activity.

Question 23.
Are you able to balance with 1N at Y?
No
Now increase the weight at Y and find out the weight required to bring both the ends to the same level.

Question 24.
More weight was required at to attain equilibrium. Is it not due to the increased surface area of water at that end?
Answer:
Yes

It is the figure of a hydraulic jack. Two cylinders of different area of cross-section, filled with water, are connected using a pipe.
A₁ is the area of cross-section of the smaller piston and A₂, that of the bigger one.
When a force F₁ is applied on the piston X, a force F₂ will be experienced on the piston Y. Then according to Pascal’s Law, pressure applied on the side X will be equal to the pressure experienced on the side Y.
So P_x = P_y

If the area of cross-section of the 100 times that of the first piston
A₂ = 100A,
Therefore A₂/A₁ = 100
F₂ = 100F₁
That means the large piston will experience a force of 100N when a force of 1N is applied on the small piston. The amount of work done on the two dies of the tube is equal. Therefore if the section Y is to be raised by 1cm, the section X has to be lowered by 100cm.

Question 25.
Devices constructed on the basis of Pascal’s Law
Answer:

• Hydraulic brake of vehicles
• Hydraulic jack
• Hydraulic press
• Excavator, Hydraulic lift

Capillarity

Activity: Take two boiling tubes, one containing water, and the other mercury. Dip a capillary tube in each tumbler.

Question 26.
What happened to the water level in the capillary tube?
Answer:
Water rises up in the tube
The rising of water in a tube against gravitation or against the weight of water is known as capillary rise.

Question 27.
Does the capillary rise occur in mercury as well? What did you observe?
Answer:
No. depression occurs in the case of mercury.

Capillarity:
The rise or depression of a liquid in a narrow tube or a minute hole is capillarity.
Cohesive Force:
Cohesive force is the force of attraction between molecules of the same type.
Adhesive Force:
Adhesive force is the force of attraction between molecules of different types of substance.

Question 28.
What is the reason for capillary rise?
Answer:
Capillary rise occurs when the adhesive force is greater than the cohesive force.

Question 29.
What is the reason for capillary depression?
Answer:
Capillary depression occurs when the cohesive force is greater than the adhesive force.

Question 30.
Arrange capillary tubes of different diameter on a piece of thermocol. Dip the capillary tubes in water. Compare the capillary rises in the tubes.

a) Which has greater capillary rise?
In the tube of smaller / greater diameter
b) Which has lower capillary rise?
c) How is the diameter of the tube and capillary rise related?
Answer:
a) In the tube of smaller
b) Larger diameter
c) As the diameter of the tube decreases, capillary rise increases. As diameter increases, capillary rise decreases.

Question 31.
Why does the capillary rise increases when the diameter of the tube decreases?
Answer:
Weight of the liquid in the tube is supported by adhesive force. When the diameter of the tube decreases, the weight of the liquid it can contain also decreases. The adhesive force with the tube is greater than the cohesive force of the liquid. So the liquid is able to rise. Capillary rise increases with the decrease in the diameter.

Question 32.
Why does the capillary rise decreases when the diameter is increased?
Answer:
In bigger tube, as the liquid level rises, the volume increases and the weight of the liquid also increases. Capillary rise is opposed by the weight of the liquid in the tube.

Question 33.
How does the oil rise up along the wick made of cotton cloth in lamps?
Answer:
This is due to capillarity. Oil rises up through the narrow capillary tubes in the cotton wick.

Question 34.
Why is the land ploughed before the beginning of summer? Does it have any relation to the capillary rise?
Answer:
When the land is ploughed the separation between the soil particles increases. The diameter of the capillary tubes formed in the soil increases. So the capillary rise decreases. This helps to retain moisture in the soil.

Viscous Force

Place one drop each of kerosene, water, glycerine and honey at various places along a single line at one end of a glass plate. Hold the glass plate slightly tilted. Compare the speed of flow of each liquid and writedown.

• Water flows faster than honey
• Kerosene flows faster than water
• Water flows faster then glycerine.

Conclusion: Different liquids flow at different speed. There is frictional force between the layers of each. liquid. This frictional force is parallel to the layers which try to prevent the relative motion between the layers. This frictional force is viscous force. Every layer of liquid prevents the flow of the layer is contact with it. This is the reason for the variation in speed between the layer of liquid.

Question 35.
What is viscosity?
Answer:
Viscosity is the characteristic property of liquid to initiate a force that tries to prevent the relative motion between the layers of the liquid.

Question 36.
Find out liquids of viscosity greater and lower than that of water and tabulate them.

Answer:

Viscous liquids:
Liquids of greater viscosity are the viscous liquids, eg. honey, glycerine
Mobile liquids:
Liquids of very low viscosity are the mobile liquids eg. kerosene, diesel, petrol

Question 37.
What is the relation between temperature of a liquid and its viscosity?
Answer:
As temperature of a liquid increases, its viscosity decreases

Question 38.
Why is it said that the body of a person who gets an electric shock should be massaged well?
Answer:
The body temperature of a person who gets an electric shock falls suddenly. As a result, the viscosity of the blood increases, causing hindrance to the flow of blood, resulting in a heart attack. When massaged, the body becomes warm and the viscosity of the blood attains normal level. The person thus overcomes the dangerous situation.

Let Us Assess

Question 1.
The weight of a piece of stone in air is 120 N and its weight in water is 100 N. Calculate the buoyancy, experienced by the stone.
Answer:
Buoyancy = weight of stone in air – weight in water = 120 N – 100 N = 20 N

Question 2.
A body which floated in water sank when put in kerosene. Why did it happen?
Answer:
Kerosene has lower density than water. So kerosene can not exert more buoyancy when compared with water.

Question 3.
Observe the figures of an object placed in different liquids.

a) Compare the gravitational force and the buoyancy acting on the body when it is in the liquids A and B.
b) Among A and B, which is the liquid whose density is greater than that of the object? Why?
Answer:
a) In the liquid A, buoyancy is greater than the gravitational force. In B, buoyancy is less than the gravitational force,
b) Liquid A. As the body floats, the density of the liquid is greater.

Question 4.
A body of weight 1000 N sinks in water. The weight of the liquid overflowed is 250 N.
a) What will be the weight of the body in water?
b) A body of the same weight as above floats in water. What is its weight in water? What will be the weight of the water displaced?
Answer;
a) Weight of the body in water = 1000 N – 250 N = 750 N
b) Zero, 1000N

Question 5.
The area of one end of a U-tube is 0.01 m² and that of the other end the force 1 m², When a force was applied on the liquid at the first end?
Answer:
A₁ = 0.01 m²
A₂ = 1 m²
F₁ = ?
F2 = 20000 N

Question 6.
Write down the reason for the following:
a) Ink can be blotted with chalk.
b) Sweat can be blotted with tissue paper.
Answer:
a) Due to capillarity
b) Due to capillarity

Question 7.
Which is the correct figure? Why?

Answer:
Fig. A
Reason: The centre portion of the liquid surface is seen higher than the edge portion. This is due to the reason that in this liquids, the cohesive force greater than the adhesive force. Insides liquids undergo capillary depression instead of capillary rise.

Question 8.
A hydraulic lift has been constructed to lift vehicles of maximum 3000 Kg. weight. The area of cross-section of the piston’s platform on which the vehicles are lifted is 580 Sq.cm. Calculate the maximum pressure experienced on the small piston.
Answer:
Area = 580cm² = 0.058m²
Force = 3000 × 10 = 30000N
Pressure = \(\frac { 3000 }{ 0.058 }\) = 517241.36Pa

Forces in Fluids More Questions and Answers

Question 1.
What is meant by fluids?
Answer:
Liquids and gases together are called fluids

Question 2.
What are the force experienced by an object inside a fluid?
Answer:
a) Weight of the object (gravitational force)
b) Buoyancy applied by the fluid

Question 3.
What is buoyancy? How is buoyancy measured?
Answer:
When a body is immersed completely or partially in a liquid, the liquid exerts an upward force on the body. This force is called buoyancy.
Buoyancy will be equal to the loss of weight experienced by the body inside the liquid.
Buoyancy = Weight of the body in air – weight of the body in the liquid.

Question 4.
What are the factors affecting buoyancy?
Answer;

• Density of liquid
• Volume of the object

Question 5.
State Achiemes principle.
Answer:
When an object is immersed partially or completely in a liquid, the buoyancy experienced by it will be equal to the weight of the liquid displaced by it.

Question 6.
The given below datas are tabulated by Gokul while conducting an experiment to prove Archimedes principles. Fill the missing parts.

Answer:

Question 7.
Kerosene and saline water are taken in two beakers. The same stone is immersed in both liquids.
a) In which liquid, more weight difference is experienced?
b) In which liquid, the stone experiences more buoy¬ancy?
c) Write whether the buoyancy of that liquid is more or less than that of the other liquid?
d) What is the relation between the density of a liquid and buoyancy?
Answer:
a) In saline water
b) In saline water
c) More
d) Buoyancy increases with increase in density of a liquid

Question 8.
X represents kerosene and Y represents water

Among A and B, which is the right figure? Why?
Answer:
B is the right figure.
Reason: Density of water is greater than that of kerosene. When taken in the same container due to the difference in density, water comes to the lower part and kerosene to the upper part.

Question 9.
A body of weight 10 N sinks in water. Weight of the water displaced is 2N.
a) What is the weight of the object in water?
b) Which principle is related to this?
Answer;
a) 8 N
b) Archimedes principle

Question 10.
State principle of floatation
Answer:
Weight of a floating body is equal to the weight of the liquid displaced by it.

Question 11.
Does a ship enters a freshwater lake from a sea sink more or rise more? Justify your answer.
Answer:
Sinks more
Seawater has more density. Seawater exerts more buoyancy on the ship. Freshwater has less density. It exerts less buoyancy.

Question 12.
An object of weight 500 N sinks in water. Weight of water overflowed is 50 N.
a) What is the weight of the object in water?
b) If another object of the same weight floats on water, what is the weight in water? What is the buoyancy? What is the loss of weight in water?
Answer:
a) Weight in water = Weight in air – weight of water overflowed = 500 N – 50 N = 450 N
b) When floats,
Weight in water = 0 N
buoyancy = 500 N
Loss of weight in water = 500 N
(When a body floats on a liquid, weight of the body in air = loss of weight = buoyancy = weight of water displaced)

Question 13.
Density of water is = ………….
Answer:
1000 kg/m³

Question 14.
Relative density of water = ………
Answer:
1

Question 15.
Classify the following liquids into liquids having density greater than that of water and liquids having density less than that of water.
Honey, kerosene, glycerine, mercury, saline water, petrol, diesel
Answer:

Question 16.
What is the relation between the diameter of capillary tube and capillary rise? Why?
Answer:
As the diameter of the tube increases, capillary rise decreases.
Weight of the liquid inside the tube opposes capillary rise.

Question 17.
On which law, hydraulic break works? Name other three devices that works on the basis of this law. State the law.
Answer:
Hydraulic break works on the basis of Pascal’s law. Other devices are hydraulic jack, hydraulic press and excavator.
Pascal law: The pressure applied at any point of a liquid at rest in a closed system, will be experienced equally at all parts of the liquid.

Question 18.
Honey : Viscous liquid
Petrol : ……………
Answer:
Mobile liquid

Question 19.
P, Q and R are three capillary tubes. If these are dipped in water.

a) Which has greater capillary rise?
b) Which factor influenced capillary rise?
c) Why is the land ploughed before the beginning of summer?
Answer:
a) In R
b) Diameter of the tube
c) When the land is ploughed the separation between the soil particles increases. The diameter of the capillary tubes formed in the soil increases. So the capillary rise decreases. This help to retain moisture in the soil.

Question 20.
In the given figure, the area of cross-section of B is 10 times greater than that of A

a) What is the working principle behind this device?
b) If a force of 20 N is applied on A, what will be the force experienced on B?
c) If the section A is lowered by 1 cm, how much will B rise up?
Answer:
a) Pascal’s law

c) Rises 10 cm

Question 21.
An iron block of weight 2N is immersed in three dif-ferent liquids A, B, C weight of the iron block experi-enced inside the liquids are labelled in the figure.

a) In which liquid, the iron block experiences more buoyancy?
b) Among these three liquids, which has less density?
c) What is the relation between density of a liquid and buoyancy?
Answer:
a) C
b) B
c) As density of the liquid increases, buoyancy also increases.

Question 22.
When a hydrometer is dipped n a liquid, the reading is found to be 1.25. What is the density of the liquid?
Answer:
Relative density = \(\frac{\text { density of liquid }}{\text { density of water }}\)
∴ Density of liquid = Relative density x density of water
= 1.25 × 1000 = 1250 kg/m³

Question 23.
What is the relation between viscosity and temperature?
Answer:
Viscosity decreases with increase in temperature.

Question 24.
Whan a midrib of a coconut leaf, pencil etc. are dipped in water, and taken out, water is seen sticking to them. Why?
Answer:
Due to the adhesive force between the midrib of a coconut leaf and water.

Question 25.
When a piece of chalk is used on a blackboard, particles of the chalk stick to the board. Why?
Answer:
Due to the adhesive force between particles of chalk and particles of board.

Question 26.
Fingers are wetted at intervals when currency notes are being counted. Why?
Answer:
To make use of the adhesive force between notes and water. Otherwise, notes stick together.

Question 27.
Which attractive force between the molecules is responsible for the surface tension?
Answer:
Force of cohesion.

Students can Download Kerala Padavali Unit 3 Chapter 2 Arbhatattilninn Lalityattilekk Questions and Answers, Summary, Notes Pdf, Activity, Kerala Padavali Malayalam Standard 9 Solutions helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Padavali Malayalam Standard 9 Guide Unit 3 Chapter 2 Arbhatattilninn Lalityattilekk

Arbhatattilninn Lalityattilekk Questions and Answers, Summary, Notes

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Kerala Padavali Malayalam Standard 9 Guide Unit 2 Chapter 1 Amma

Amma Questions and Answers, Summary, Notes

You can Download Food Through Digestive Tract Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Biology Solutions Part 1 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Biology Solutions Chapter 2 Food Through Digestive Tract

Food Through Digestive Tract Textual Questions and Answers

Kerala Syllabus 9th Standard Biology Notes Chapter 2 Question 1.
What are the nutrients we get from food?
Answer:
Carbohydrate, Protein, Fat, Minerals, vitamins & water.

Food Through Digestive Tract Class 9 Notes Kerala Syllabus Question 2.
Complete the table given below.

Answer:

Kerala Syllabus 9th Standard Biology Notes Chapter 2 English Medium Question 3.
Illustrate & identify the different parts of the digestive system.

Answer:

Food Through Digestive Tract Class 9 Kerala Syllabus Question 4.
What are the functions performed by the nutrients in our body – Prepare a table?

Answer:

Food Inside Mouth

Kerala Syllabus 9th Standard Biology Notes Question 5.
State true or false
a) Vitamin A is required to the healthy eyesight
b) Incisor helps to bite and cut the food
c) Molar helps to tear the food
Answer:
a) True
b) True
c) False

Kerala Syllabus 9th Standard Biology Notes Malayalam Medium Chapter 2 Question 6.
Complete the Illustration

Answer:
Canine: Helps to tear the food
Premolar: Helps to chew at the food
Molar: Helps to chew the food

9th Class Biology Chapter 2 Notes Kerala Syllabus Question 7.
What do you mean by digestion?
Answer:
Digestion is the process of conversion of complex food materials into simple absorbable forms.

Kerala Syllabus 9th Standard Biology Notes English Medium Question 8.
…………. helps to tear the food?
Answer:
Canine

Kerala Syllabus Biology 9th Standard Question 9.
Molar helps to the food
Answer:
Chew

9th Biology Notes Kerala Syllabus Question 10.
The process of digestion begins from the.
Answer:
Mouth

9th Class Biology Book Chapter 2 Kerala Syllabus Question 11.
……….. helps to chew the food
Answer:
Premolar

Kerala Syllabus 9th Standard Biology Notes Pdf Question 12.
……… helps to bite and cut the food
Answer:
Incisor

Question 13.
Define enamel.
Answer:
Enamel is the hardest part in our body. It is white in color. It is a dead tissue seen in the out layer of our teeth.

Question 14.
What is the role of the tongue in the digestive system?
Answer:
The tongue mixes food items with saliva and helps the teeth to masticate it. It taste buds in the tongue also help us to sense taste.

Question 15.
………. is the living tissue which forms the tooth.
Answer:
Dentine.

Question 16.
……… is a calcium-containing connective tissue that holds the tooth in the socket of the gum.
Answer:
Cementum

Question 17.
………….. is the soft connective tissue seen in the pulp cavity.
Answer:
Pulp

Question 18.
Complete the word relation
Enamel: dead tissue
……….: living tissue
Answer:
Dentine

Saliva And Digestion

Question 19.
Where is saliva produced?
Answer:
Saliva is produced in the Salivary glands.

Question 20.
Write a note on the significance of salivary glands?
Answer:
There are three pairs of salivary glands in the mouth. The saliva secreted from the salivary glands contains mucus and enzymes like salivar amylase & lysozyme.

Question 21.
……… makes the food slimy, so that it can be swallowed.
Answer:
Mucus

Question 22.
………….. helps to destroy the germs that enter the body through food.
Answer:
Lysozyme

Question 23.
……….. partially converts starch to maltose?
Answer:
Salivary amylase

Question 24.
What are the components of saliva?
Answer:
Mucus and enzymes like salivary amylase and lysozyme

Question 25.
Complete the word relation.
Salivary amylase: ………….
Lysozyme: destroy germs
Answer:
Maltose

Question 26.
When iodine added to starch color appears.
Answer:
Blue

Food Through Oesophagus

Question 27.
Complete the illustration?

Answer:
a) Uvula
b) Pharynx
c) Epiglottis
d) Oesophagus

Question 28.
Trachea begins from the
Answer:
Pharynx

Question 29.
From the mouth, food enters the………. through the pharynx
Answer:
Oesophagus

Question 30.
Prepare a flow chart regarding the swallowing of food?
Answer:
Mouth → buccal cavity → pharynx → Oesophagus → food into the stomach.

Question 31.
Observe the illustration and write down the questions below.

1. What is the role of uvula?
2 compresses the food into balls with the help of plate?
3. State whether true or false?
The trachea tilts up and it is closed by the epiglottis
Answer:

1. Uvula doses the nasal cavity that opens to the pharynx
2. Tongue
3. True

Question 32.
How does the food we swallowed enter the esophagus properly without entering the trachea? Explain?
Answer:
The Tongue compresses the food into balls with the help of the palate. That time Ovula closes the nasal cavity that opens to the pharynx and the trachea tilts up and it is closed by the epiglottis.

Food Inside The Stomach

Question 33.
Food reaches the stomach by
Answer:
Peristalsis

Question 34.
What do you mean by peristalsis?
Answer:
Peristalsis is the wave-like movement of the esophagus, stomach, and intestines.

Question 35.
What is the role of peristalsis in the process of digestion?
Answer:
Peristalsis in the stomach converts foods into a pastelike form.

Question 36.
What are the changes that happen to food while it is in the stomach?
Answer:
Peristalsis in the stomach converts food into a paste-like form. Special kind of circular muscles present in the posterior part of stomach retains food for a specific period. Different components in the gastric juice, secreted by the glands in the stomach wall also play a very important role in the process of digestion.

Question 37.
Complete the illustration

Answer:
a) Pepsin – Converts protein to peptones partially
b) Hydrochloric acid
c) Mucus: Protects the stomach wall from the action of digestive Juices

Question 38.
What is the role of Hydrochloric acid in the gastric juice?
Answer:
Hydrochloric acid in the gastric juice kills germs in the food. It also regulates pH for the digestion in the stomach.

Question 39.
………….. protects the stomach wall from the action of digestive juices.
Answer:
Mucus

Question 40.
………….. converts protein to peptones partially
Answer:
Pepsin

Question 41.
……………. regulates pH for the digestion in stomach.
Answer:
Hydrochloric acid

Food In The Small Intestine

Question 42.
Define Liver and its main function?
Answer:
Liver is the largest solid organ in the body situated in the upper part of the abdomen on the right side. The main function of liver is to filter the blood coming from the digestive tract, before passing it to the rest of the body.

Question 43.
What is the role of gall bladder?
Answer:
The excess bile secreted by the Liver is stored in the gall bladder.

Question 44.
………… secretes pancreatic juice
Answer:
Pancreas

Question 45.
Complete the table given below.

Answer:

Question 46.
Complete the word relation.
Liver: …………….
Pancreas: Pancreatic juice
Answer:
Bile

Question 47.
Pancreatic juice contains the enzymes …………. & ……………
Answer:
Amylase, lipase, trypsin

Question 48.
Prepare a flow chart showing the role of pancreas in the process of digestion?
Answer:

Question 49.
Complete the word relation.
Starch: Maltose
Protein: ……….
Answer:
Peptides

Question 50.
Peptidase converts peptides to …………..
Answer:
Amino acids.

Question 51.
………. converts lactose to glucose & galactose.
Answer:
Lactase

Question 52.
Prepare a flow chart showing the digestion within the small intestine?
Answer:
Small Intestine → Intestinal glands →
Intenstinal juice → Peptidase → Peptides → AminoAcids

Question 53.
Explain the action of disaccharidases?
Answer:
Disaccharidases like Maltose converts maltose to glucose, Lactase converts lactose to glucose and galactose and sucrase converts sucrose to glucose and fructose.

Question 54.
Which are the end products formed after digestion? Write in table.
Answer:

Question 55.
Which are the nutrients that do not undergo digestion?
Answer:

• Water
• Minerals
• Vitamins

Absorption Begins

Question 56.
Which are the simple components formed by the digestion of protein, fat, and carbohydrate?
Answer:
Protein: amino acids and glycerol
Fat: fatty acid and glycerol
Carbohydrate: Monosasacchrides (glucose, fructose, galactose).

Question 57.
The small intestine of humans is about ………… long.
Answer:
5 to 6 meters.

Question 58.
illustrate the structure & position of villus?
Answer:

Question 59.
State whether true or false.
a) Movement of food through the small intestine is very fast.
b) Many foldings are seen in the inner wall of small intestine.
Answer:
a) False
b) True

Question 60.
Small intestine is seen in the
Answer:
Abdominal cavity.

Question 61.
What are Villi?
Answer:
Villas are small finger-like projections seen in the walls of the intestine.

Question 62.
Fatly acids and glycerol are absorbed to the ………… of villus.
Answer:
Lacteal

Question 63.
……… are small finger-like projections seen on the walls of the intestine.
Answer:
Villi

Question 64.
What is the function of villi? Write down your interference?
Answer:
Villi increase the area of absorption of nutrients to a great extent within the small intestine. They also contain blood capillaries & lacteals, the lymph capillaries. Absorption of nutrients & 90% of water absorption take place in the villi.

Question 65.
……….. are the lymph capillaries?
Answer:
Lacteals

Question 66.
Villi are covered with ……………
Answer:
Single-layer of cells

Behind Absorption

Question 67.
‘Diffusion does not require energy’ Kalyani said to Ishan. Do you agree with Kalyani’s opinion? Why? )
Answer:
Yes. Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration. By diffusion, molecules move across the cell through the cell membrane. This process continues till the concentration becomes equal in both sides of the membrane. So this process does not require any energy.

Question 68.
Absorption of fatty acids and glycerol into lacteals is the example of
Answer:
Diffusion

Question 69.
What do you mean by facilitated diffusion?
Answer:
Diffusion of certain molecules with the help of protein molecules in the cell membrane is called facilitated diffusion.

Question 70.
Asha stated that “Absorption of water in the small intestine and the large intestine is an example of Osmosis” Do you agree? Substantiate your answer by explaining the term osmosis?
Answer:
Osmosis is the movement of water molecules from a region of higher concentration to a region of lower concentration across a semi-permeable mebrane. This process continues till the concentration becomes equal. It can be stated that in small intestine and large intestine water molecules move from a region of higher concentration to a region on lower concentration.

Question 71.
Explain the process of Active Transport?
Answer:
The process by which molecules are absorbed with the help of carrier Proteins against the concentration gradient is called active transport.

Question 72.
Absorption of glucose and salt are examples of
Answer:
Active Transport

Question 73.
Complete the word relation:
Osmosis: absorption of water
…………: absorption of amino acid.
Answer:
Facilitated diffusion

Question 74.
Which are the process that help in the transport and substances in cells?
Answer:
Diffusion & Osmosis

Question 75.
Differences and similarities between Diffusion and Osmosis – Prepare a table.
Answer:

Question 76.
How does diffusion differ from facilitated diffusion?
Answer:
Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration but facilitated diffusion is the diffusion of certain molecules with the help of protein molecules in the cell membrane.

Question 77.
What is the difference noticed in active transport when compared to other processes?
Answer:
In Active transport energy is needed by other process energy is not needed.

In And Out Of Large Intestine

Question 78.
The digestive wastes left after the absorption of nutrients move towards the …………
Answer:
Large intestine

Question 79.
Bacteria residing in the large intestine produced ………….
Answer:
Vitamin K

Question 80.
Absorption of vitamin K takes place in the ………….
Answer:
Large intestine

Question 81.
What are the main ideas that can include in the seminar relationship between food and health?
Answer:

• Importance of roughage for a healthy digestive system.
• Health issues created by junk food and fast food.
• Harmful chemical substances added to food to enhance its taste and color.

Let Us Assess

Question 82.
Identify the correct statement with regard to bile.
a) Secreted in liver
b) Enzymes are seen
c) Secreted into the stomach
d) Converts fat into tiny particles
Answer;
a) Secreted in liver

Question 83.
Complete the table related to the process of digestion in humans.

Answer:

Food Through Digestive Tract Additional Question and Answers

Question 84.
Illustrate the process of Osmosis
Answer:

Question 85.
How much time is required for the completion of digestion?
Answer:
4 to 5 hours

Question 86.
“You should take food only after an interval of 4 hours” – said Arun to Archa. What is your opinion?
Answer:
The process of digestion is completed through many complex actions. It requires 4 to 5 hours. Hence it is necessary to arrange the intake of food accordingly keeping an interval of four hours between the intake.

Question 87.
Doctor advised Minnu who had a complaint of stomach pain to avoid cool drinks like cola &junk food. What can be the reason for this advice? Point out your opinion?
Answer:
The nutrients required for our health are derived from our food. If we choose food only on the basis of taste, we may not get all the required nutrients. Regular use of colo drinks & junk foods are not good for our health. Therefore try to take all types of food instead of giving priority to cola & junk food.

Question 88.
Prepare a poster on ‘Good food habits’
Answer:

• Eat plenty of fruits & vegetables every day
• Eat less unhealthy food
• Balance your food choices over time. (Draw a suitable picture using these ideas).

You can Download Redox Reactions and Rate of Chemical Reactions Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Chemistry Solutions Part 1 Chapter 3 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Chemistry Solutions Chapter 3 Redox Reactions and Rate of Chemical Reactions

Redox Reactions and Rate of Chemical Reactions Textual Questions and Answers

Kerala Syllabus 9th Standard Chemistry Notes Chapter 3 Activity -1

Given below is a table showing the mass of the reactants and the products when hydrogen combines with chlorine to form hydrogen chloride in two situations.

HSSlive chemistry redox reactions Question 1.
Write down the total mass of the reactants and also the total mass of product in the above experiment.
a) Situation 1: …………………
b) Situation 2: ……………….
Answer:
a) Total mass of reactants 73g products 73g
b) Reactants 146g products 146 g

Class 9 Chemistry Chapter 3 Notes Kerala Syllabus Activity-2

Given below is a table showing the mass of the reactants and the products when hydrogen combines with oxygen to form water, in two situations.

Kerala Syllabus 9th Standard Chemistry Chapter 3 Question 2.
Write down the total mass of the reactants and the total mass of product in the above experiment.
a) Situation 1:
b) Situation 2:
Answer:
a) Reactants 18 g, Product 18 g
b) Reactants 36 g, Products 36 g

Redox Reaction And Rate Of Chemical Reaction Kerala Syllabus 9th Question 3.
What is the relation between the total mass of the reactants and the total mass of products?
Answer:
Both are equal.

State The Law Of Conservation Of Mass

In a chemical reaction mass is neither created nor destroyed. This is the law of conservation of mass.
This law was proposed by the scientist Antoine Lavoisier. That is, the total mass of the reactants will be equal to the total mass of the products.

Activity-3

Kerala Syllabus 9th Standard Chemistry Chapter 3 Notes Question 4.
A piece of magnesium is burned in air. What do you observe?
Answer:
It burns with sparkling light

Redox Reactions And Rate Of Chemical Reactions Kerala Syllabus 9th Question 5.
What is the white powder formed?
Answer:
Magnesium oxide (MgO)

Hss Live Guru 9th Chemistry Kerala Syllabus  Question 6.
Which are the reactants here?
Answer:
Mg and O₂

Chemistry 9th Class Notes Kerala Syllabus Question 7.
Which is the product?
Answer:
MgO

9th Class Chemistry Chapter 3 Kerala Syllabus Question 8.
Note down the number of atoms in the reactant side and the number of atoms in the product side in the table below.

Answer:

9th Class Chemistry Chapter 3 Notes Kerala Syllabus Question 9.
Is the number of atoms of each element equal on both sides?
Answer:
No

Kerala Syllabus 9th Standard Chemistry Notes Question 10.
The number of which atom is not equal on both sides?
Answer:
O

Chemistry Class 9 Kerala Syllabus Question 11.
How many product molecules are needed to equalize the number of oxygen atoms on both sides?
Answer:
2

Kerala Syllabus 9th Standard Chemistry Notes Pdf Question 12.
How will you represent two molecules of magnesium oxide?
Answer:
2 MgO

Hss Live Class 9 Chemistry Kerala Syllabus Question 13.
Now, is the number of magnesium atoms equal on both sides?
Answer:
No

Chemistry Notes For Class 9 Kerala Syllabus Question 14.
How many magnesium atoms are needed in the reactant side to equalize the number of magnesium atoms on both sides?
Answer:
2

Chemistry 9th Class Chapter 3 Kerala Syllabus Question 15.
Then how can you rewrite the above equation?
Answer:
2 Mg + O₂ → 2MgO

Question 16.
Is the total number of atoms of each element in the molecules present in the reactant side and that in the product side equal in this equation?
Answer:
Yes

Question 17.
What is balancing of equations?
Answer:
Equalizing the number of atoms of each element in molecules in the reactant side and that in the product side is called balancing of equation.
Zn + HCl → ZnCl₂ + H₂
Tabulate the total number of atoms of each elements in the reactant side and that in the product side of the above reaction.

Question 18.
How many Zn atoms are there in the reactant side and in the product side?
Answer:
1 each

Question 19.
Which are the atoms showing a difference in their number?
Answer:
H and Cl

Question 20.
In order to make them equal on both sides how many molecules of HCI should be taken as reactant.
Answer:
2
Now rewrite the equation
Zn + HCl → ZnCl₂ + H₂

Question 21.
Balance the chemical equation H₂ + O₂ → H₂O
Answer:
Step I: H₂+ O₂ → H₂O
Step II: 2H₂ + O₂ → 2H₂O

Question 22.
Balance the chemical equation Al + O₂ → Al₂O₂
Answer:
Step I: 2Al + O₂ → Al₂O₃
Step II: 2Al+ 3O₂ → 2Al₂O₃

Question 23.
Balance the chemical equation H₂ + O₂ → H₂O
Answer:
Step I: H₂ + O₂ → 2H₂O
Step II: 2H₂ + O₂ → 2H₂O

Question 24.
Some chemical equations are given below. Note down the number of reactant atoms and that of product atoms in the table given below.
1. C + O₂ → CO₂
2. N₂ + H₂ → NH₃
3. 2H₂ O₂ → 2H₂O + O₂
4. SO₂ + O₂ → SO₃
5. BaCl₂ + H₂SO₄ → BaSO+2HCl

Balance the equations which are unbalanced
Answer:
1. N₂ + 3H₂ → 2NH₃
2. 2SO₂ + O₂ → 2SO₃

Oxidation And Reduction

Question 25.
The electronic configuration of magnesium and chlorine are 2, 8, 2 and 2, 8, 7 respectively. How many electrons does a magnesium atom donate? What charge will it get?
Answer:
2, two positive (2+)

Question 26.
Let us complete the equation for this process,
Mg → Mg²⁺ + ……………
Answer:
Mg → Mg²⁺ + 2e^–

Question 27.
How many electrons are accepted by each chlorine atom? What will be the charge acquired by each atom?
Answer:
One 1^– Cl^1-

Question 28.
Complete the equation of this process.
Cl + 1e^– →
Answer:
Cl + 1e^– → Cl^1-

Question 29.
What are oxidation and reduction?
Answer:
Oxidation is the process of loss of electrons. Reduction is the process of gain of electrons. The atom which loss electron is called the reducing agent and the atom which gains electron is called the oxidizing agent.

Question 30.
In the above chemical reaction, which atom is oxidized?
Answer:
Sodium

Question 31.
Which atom is reduced?
Answer:
Chlorine

Question 32.
Which is the oxidizing agent in this chemical reaction?
Answer:
Chlorine (Cl)

Question 33.
Which is the reducing agent in this chemical reaction?
Answer:
Sodium (Na)

Question 34.
Analyze the following equations and list the oxidized atom, reduced atom, oxidizing agent and reducing agent.
a) Mg + F₂ → MgF₂
b) 2Na + Cl → NaCl
Answer:
a) Oxidized atom — Mg
Equation of oxidation — Mg → Mg²⁺ + 2e^–
Reduced atom — F
Equation of reduction — F + e^– → F^–
Oxidizing agent — F
Reducing agent — Mg

b) Oxidized atom — Na
Equation of oxidation — Na → Na + 1e^–
Reduced atom — Cl
Equation of reduction — Cl+ 1e^– → Cl^–
Oxidizing agent — Na
Reducing agent — Cl

Question 35.
Analyse the following equations and complete the Table given below.
1. Mg → Mg²⁺ + 2e^–
F+ le^– → F^–
2. Na → Na⁺+1e^–
Cl + 1e^– → Cl^–
3. Fe → Fe²⁺+ 2e^–
O + 2e^– → O^2–

Answer:

Oxidation Number

Oxidation number of an atom in a molecule is the formal charge assigned on the atom if all the bonds in the substance arc considered to be ionic.

Question 36.
Consider the equation H₂ +Cl₂ → 2HCl What is the oxidation number of hydrogen in H₂?
Answer:
O

Question 37.
What is the oxidation number of chlorine in Cl₂?
Answer:
O

Question 38.
In this reaction, does the oxidation number of hydrogen increase or decrease?
Answer:
Increases

Question 39.
What change takes place in the oxidation number of chlorine?
Answer:
Decreases

Question 40.
What are oxidation and reduction on the basis of change in oxidation number?
Answer:
The process in which the oxidation number increases is called oxidation.
The process in which the oxidation number decreases is called reduction.

Question 41.
During the formation of hydrogen chloride, which atom was oxidized?
Answer:
H

Question 42.
Which is the reducing agent?
Answer:
Cl

Question 43.
Which atom was reduced, during this reaction?
Answer:
Cl

Question 44.
Which is the oxidizing agent?
Answer:
H

Question 45.
Analyze oxidation numbers in the given equation and list the oxidizing agent and reducing agent in the table given below.

Answer:
Oxidizing agent – O
Reducing agent-C

Question 46.
Analyze the oxidation number and note down the oxidizing agent and reducing agent in the following reaction.

a) Oxidation number of which atom is increased?
b) The oxidized atom is
c) Oxidation number of which atom is decreased?
d) The reduced atom is
e) Oxidizing agent
f) Reducing agent
Answer:
a) Na
b) Na
c) Cl
d) Cl
e) Na
f) Cl

Question 47.
Analyze the following equation fill-up the blanks.

Question 48.
The oxidation number of zinc increases/decreases from … to…
Answer:
0 to +2

Question 49.
The oxidized atom is…
Answer:
Zn

Question 50.
The oxidation number of hydrogen increases/decreases from …. to
Answer:
+1 to 0

Question 51.
The reduced atom is …
Answer:
H
Here HCI is the oxidizing agent and Zn is the reducing agent. „

Question 52.
How do you determine the oxidation number of sulfur in H₂ SO₄?
Answer:
Oxidation state of hydrogen = +1
Oxidation state of oxygen = -2
Let the oxidation state of sulphur be ‘x’ We know the sum of oxidation states of all atoms in a compound is zero. Therefore,
[2×(+1)] + x + (4x – 2) = 0
(+2) + x + (-8) = 0
x – 6 = 0
x = +6

Question 53.
Find the oxidation number of Mn in KMnO₄ (oxidation number of K is +1, oxidation number of O is -2)
ANswer:
Oxidation state of potassium = +1
Oxidation state of oxygen = -2
Let the oxidation state of Mn be ‘x’.
1 x (+1) + x + 4(-2) = 0
(+1) + x + (-8) = 0
x – 7 =0
x = +7

Question 54.
Find the oxidation number of Mn in MnO₂, Mn₂O₃ and Mn₂O₇
Answer:
MnO2
Mn + 2x -2 = 0
Mn + 4 = 0
Mn = +4

Mn2O3
2Mn + 2x -2 = 0
2Mn + -4 = 0
2 Mn = +4
Mn = +2

Mn2O7
2Mn + -2×7 = 0
2Mn + -14 = 0
2Mn = +14
Mn = +7

Question 55.
What is Redox reaction?
Answer:
The process of oxidation and reduction take place simultaneously. Hence these two reactions together are known as redox reaction.
Examples:
H₂ + Cl₂ → 2HCl
Mg + F2 → MgF₂
2Na + Cl₂ → 2NaCl

Question 56.
What are the methods usually adopted to make firewood burn faster? ‘
Answer:
1. Provide more air
2. Split up into small pieces
3. Make firewood dry

Question 57.
Describe an experiment to prove that nature of the reactants affects the rate of chemical reaction.
Answer:
Materials required for the experiment.
Zn, Mg, dil. HCl arid test tubes.
Procedure:
Take equal volume of dil. HCl in two test tubes. Add Zn to one and Mg of same mass to the other. Hydrogen gas is produced in both the test tubes. Rate of reaction is faster in the test tube containing Mg.

Question 58.
Write an experiment to prove that concentration of reactants affect the rate of reactions.
Answer:
Materials required: Mg, dil. HCl, Con. HCl and test tubes.
Procedure: Take magnesium ribbons of equal mass in two test tubes. Add concentrated HCl to one test tube and dilute HCl to the other in equal volume

Question 59.

Write your observation
Test tube 1: ……………
Test tube 2: …………..
Test tube 1: Reaction is faster in con. HCl
Test tube 2: Reaction rate is slow

Question 60.
Why rate of reaction increases when concentration increases?
Answer:
As the concentration of reactants increases, the number of molecules per unit volume and the number of effective collisions increase. Consequently the rate of reaction increases.

Question 61.
What is the relation between rate of reaction and surface area? Write an experiment to prove it.
Answer:
Take equal volume of dil. HCI in two beakers. Add a small piece of marble into one and marble powder of equal mass into the other. Reaction rate is greater when marble powder is used. Rate of reaction increases when surface area increases

Question 62.

Is there any difference in the rate of reaction in the two beakers?
Answer:
Reaction rate is greater when powdered marble is used.

Question 63.
What about the concentration of acid in both the reactions?
Answer:
Same

Question 64.
Is there any difference in the mass of the marble?
Answer:
No

Question 65.
What about the surface area of marble?
Answer:
Different

Question 66.
In which case is the chance for greater number of acid molecules to get in contact with marble, in a given time?
Answer:
If marble is powdered

Question 67.
What is the change in the rate of collision when surface area increases?
Answer:
When surface area increases rate of reaction also increases.

Question 68.
What will happen to the rate of this reaction if marble is further crushed or powdered?
Answer:
Will increase

Question 69.
Why rate of reaction increases when surface area increases?
Answer:
When solids are made into small pieces or powder, their surface area increases. As a result the number of molecules undergoing effective collision also increases. Hence the rate of reaction increases.

Question 70.
Write an experiment to prove the relation between temperature and rate of reaction.
Answer:
Materials required: Sodium thiosulphate, hydrochloric acid, water, boiling tube, spirit lamp. Procedure: Prepare dilute solution of sodium thiosulphate in a beaker. Take equal volumes of this solution in two boiling tubes. Heat one boiling tube for some time. Add dilute hydrochloric acid in equal amounts in both the boiling tubes.
Observation: Reaction is faster in heated test tube.

Question 71.
In which of the boiling tubes is the precipitate formed faster?
Answer:
In heated test tube

Question 72.
What is the color of the precipitate formed?
Answer:
Pale yellow

Question 73.
What is Threshold Energy?
Answer:
To take part in a chemical reaction, molecules should attain certain minimum kinetic energy. This energy is called threshold energy.

Question 74.
Write an experiment to prove the influence of catalysts in a chemical reaction.
Answer:
Take some hydrogen peroxide solution in a test tube. Show a glowing incense stick into the test tube.
What is the observation? Is there any difference occurring in the way in which the incense stick bums? No . ‘
Now add some manganese dioxide (MnO₂) into the test tube. Again show the glowing incense stick. Observation:
Glowing incense stick flares up now.

It indicates that when manganese dioxide is added, the rate of reaction increases and oxygen is formed faster. Filter the solution using a filter paper when the reaction is completed.

The substance remaining in the filter paper is manganese dioxide itself When examined carefully it becomes clear that there is no change in its amount or property. The presence of manganese dioxide has increased the rate of the reaction. Manganese dioxide acts as a catalyst in this reaction.

Question 75.
What are catalysts?
Answer:
Catalysts are substances which alter the rate of chemical reactions without themselves undergoing any permanent chemical change.

Redox Reactions and Rate of Chemical Reactions Additional Questions and Answers

Magnesium combines with chlorine to form magnesium chloride. The equation is given below
Mg + Cl₂ → MgCl₂

Question 1.
Write down the electronic configuration of magnesium and chlorine
Answer:
Mg – 2, 8, 2
Cl – 2, 8,7

Question 2.
How many electrons are donated by magnesium? Write the chemical equation for this process
Answer:
2, Mg → Mg²⁺ +2e^–

Question 3.
How many electrons are accepted by chlorine?
Answer:
1, charge – ve

Question 4.
Write the chemical equation.
Answer:
Cl + 1e^– → cl^–

Question 5.
Explain oxidation, reduction, oxidizing agent and reducing agent?
Answer:

• Oxidation is the process by which the removal of electrons.
• Reduction is the process by which the addition of electrons.
• In the above process, magnesium reduces chlorine by donating electrons to it hence magnesium is the reducing agent. In this process since chlorine oxidizes magnesium by accepting electrons it is considered as the oxidizing agent.

Question 6.
From the given chemical equation identify the oxidizing agent and the reducing agent by writing the chemical equation of oxidation and reduction
(i) Mg + F₂ → MgF₂
(ii) 2Na + Cl₂ → 2NaCl
Answer:
(i) Mg → Mg²⁺ + 2e^– (Oxidation)
F + 1e^– → F^– (Reduction)
Here magnesium is the reducing agent and fluorine is the oxidising agent.
(ii) Na → Na⁺ + 1e^– (Oxidation)
Cl + 1e → Cl^– (Reduction)
Here sodium is the reducing agent and chlorine is the oxidizing agent.

Question 7.

In the formation of hydrogen chloride does the oxidation number of hydrogen increase or decrease?
Answer:
The oxidation number of hydrogen increases from 0 to +1

Question 8.
Whether the oxidation number of chlorine decrease or increase?
Answer:
The oxidation number of chlorine decreases from 0 to — 1.

Question 9.
Based on oxidation number which is the oxidizing agent and reducing agent?
Answer:
H₂ is the reducing agent and Cl₂ is the oxidizing agent.

Question 10.
What is meant by Redox reaction?
Answer:
A process in which the increase in oxidation number is the oxidation and decrease in oxidation number is the reduction. A process in which both oxidation and reduction takes place is the redox reaction. The molecule in which the atom whose oxidation number decreases during the process is the oxidizing agent and that in which oxidation number of an atom increase is the reducing agent.

Question 11.
In the following chemical reactions which is the oxidizing agent which is the reducing agent? Analyze the oxidation number and find out?

Reducing agent – C
Oxidising agent – Cl₂

Question 12.
Given below is the equation of the reaction between zinc and hydrochloric acid also indicating the oxidation number.

Analyze and find the following:
Answer:
In this process, oxidation number of zinc increases from 0 to +2.
Oxidation number of hydrogen decrease from +1 to 0 The reducing agent is zinc and oxidising agent is HCl.

Question 13.
Calculate the oxidation number of‘S’ in H₂ SO₄?
Answer:
Oxidation number of hydrogen = +1
Oxidation number of oxygen = -2
Let the oxidation number of ‘S’ be x
The sum of the oxidation numbers of all the Elements in a compound is zero,
Hence
[2x (+1] + x + 4 x (-2] = 0
2 + x – 8 = 0
x = +8 – 2
x = +6
ie., oxidation number of ‘S’ in H₂ SO₄ is +6.

Question 14.
Find the oxidation number of the ions in ionic compound?
The charge of ions in ionic compound is the oxidation number. Eg. In FeCl₂ the oxidation number of Fe is +2 and FeCl3 is +3.

Let Us Assess

Some chemical equations are given below.
C + O₂ → CO₂
CH₄ + 2O₂ → CO₂ + 2H₂O
N₂ + O₂ → NO
CaCO₂ → CaO + CO₂
H₂ + l₂ → Hl
Fe + HCl → FeCl₂ + H₂
CO₂ + C → CO
a) Which of these are balanced equations?
b) Balance the unbalanced equations.
c) Which among these are redox reactions?
Answer:
a) C + O₂ → CO₂
CH₄ + 2O₂ → CO₂ +2 H₂O
CaCO₃ → CaO +CO₂
b) N₂ + O₂ → 2NO
H₂ + l₂ → 2Hl
Fe + 2HCl → FeCl₂ + H₂
CO₂ + C₂CO

Question 2.
The chemical reaction between marble and dilute HCl is given.
CaCO₃ + 2HCl → CaCl₂ + H₂O +CO₂
a) Which gas is formed here? How can you identify this gas?
b) Suggest any two ways you would choose to increase the rate of this chemical reaction. Explain the reason.
Answer:
a) Carbon dioxide (CO₂). Show a burning splinter in the gas. It will get extinguished.
or
Show a glass rod dipped in clear lime water into the gas. The lime water will turn milky
b) 1) Use powdered marble
2) Use concentrated HCl
3) Heat the mixture.

Question 3.
Sulfur pieces do not react with cold concentrated nitric acid. But sulfur powder reacts.
a) Explain the reason why the rate of chemical reaction is increased here?
b) Suppose you want to increase the rate of reaction again. Which way you would choose? Give reason.
Answer:
a) When surface area increases rate of effective collision increases. So the rate of reaction increases.
b) Heat the mixture.
When temperature increases more molecules will get threshold energy. So rate of effective collision and rate of reaction increases.

Question 4.
Small amounts of phosphoric acid is usually added to hydrogen peroxide to prevent its decomposition?
a) What is the function of phosphoric acid here?
b) By which name are these type of substances known?
c) Which substance would you add to increase the rate of decomposition of hydrogen peroxide?
Answer:
a) Slow down the rate of decomposition of hydrogen peroxide
b) Negative catalyst
c) Manganese dioxide

Extended Activities

Question 1.
Find the oxidation number of the elements which are underlined in the compounds given below. Among these find out the elements which show variable oxidation numbers.
MnO₂, Mn₂O₇, K₂Cr₂O₇, KCrO₃, MnCl₂, MgO, MgCl₂, Al₂O₃,AlCl₃
(Hint. Oxidation number 0 = -2, Cl = -1, K = +1)
Answer:
MnO₂ Let x be the oxidation number of Mn

Question 2.
Some apparatus and chemicals are given.
Zn, Mg, dilute HCl, CaCO₃, test tube, water.
a) Design an experiment to prove that the nature of reactants can influence the rate of reaction.
b) Write the equations for the chemical reactions.
c) Write the expression for the rate of the reaction.
a) Take equal volume of dil HCl in two test tubes. Add a piece of Mg m to one of them. Add a piece of Zn of equal mass to the other.
Observation:- Mg reacts faster with dil. HCl than Zn. This is due to the difference in the chemical nature of Mg and Zn.
b) Mg + 2HCl → MgCl₂ +H₂
Zn + 2HCl → ZnCl₂ + H₂
c) Rate of a chemical reaction \(=\frac{\text { amount of reactants used up }}{\text { time }}\)
= \(\frac{\text { amount of product formed }}{\text { time }}\)

Question 3.
The experiments conducted by two students are given below.
Experiment 1: 2 mL of sodium thiosulphate solution is taken in a test tube, heated and to it 2 mL of HG solution is added.
Experiment 2:2 mL of sodium thiosulphate solution is taken in a test tube and to it 2 mL of HC1 solution is added.
a) In which experiment is the precipitate formed quickly? Justify your answer.
b) Write the balanced equation for the reaction.
Answer:
a) ln heated test tube.
When temperature increases more molecules get threshold energy. So rate of effective collision increases and rate of reaction increases.
b) Na₂ S₂O₃ + 2HCl → 2NaCl + SO₂ + S

Question 4.
Some materials available in the laboratory are given below. Magnesium ribbon, marble powder, marble pieces, dilute HCl, concentrated HCl.
a) Which materials will you choose for the preparation of more CO₂ in less time?
b) Write the balanced chemical equation of the reaction.
Answer:
a) Marble powder and concentrated HCl CaCO₃ + 2 HCl → CaCl₂ + H₂O + CO₂

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Kerala State State Syllabus 9th Standard Physics Solutions Chapter 5 Work, Energy and Power

A The Trio (Story) Textual Questions and Answers

Hss Live Guru 9th Physics Kerala Syllabus Chapter 5 Question 1.
Observe figure try to write down the activities shown in them.
Answer:

1. A man pushes a trolly.
2. Batting of a cricket ball.
3. Pushing a wall.

Kerala Syllabus 9th Standard Physics Notes Chapter 5 Question 2.
Write down more activities familiar to you
Answer:

1. A man carrying a load
2. Throwing a ball
3. Lifting the bag on to the shoulder
4. Pushing a car into motion

Hsslive Guru 9th Physics Kerala Syllabus Chapter 5 Question 3.
You have understood that a force is to be applied on a body to do an activity. Find out the source of applied force for every activity and note them down in the table.


Answer:

Objects undergo displacement only when the force is applied on it them.

Work :

Work is said to be done only when a body under¬goes displacement in the direction of the applied force.

work energy and power Question 4. Observe figure and write down the situation where work is said to be done
Answer:
The person loaded on head and goes upward.

Kerala Syllabus 9th Physics Notes Chapter 5 Question 5.
A boy pushed an object of mass 30Kg horizontally across the floor through 50m. Another boy pushed an object of mass 50kg across the same floor through 50m. Both of them gave the same speed for moving the objects.
a) Who applied greater force here?
b) In which case was the work greater?
c) Write down a factor influencing work.
Answer:
a) Second boy applied greater force
b) More work is done in the second situation
c) The Factor influencing work is force.

Learn Mass Transfer MCQ questions & answers are available for a Chemical Engineering students.

Hss Live Guru Physics 9th Kerala Syllabus Chapter 5 Question 6.
A boy pushed an object of mass 30kg across a horizontal floor through 20m. Another boy pushed the same body through 30m on the same floor with the same speed.
a) Who pushed a greater distance here?
b) What about the force applied
c) Who did the greater work?
d) Which is the factor influencing work here?
Answer:
a) Second boy
b) Force is same
c) Second boy done greater work
d) The Factor influencing work is displacement

The factor affecting work done are force (F) and displacement (S)
Equation for calculating work done Work W = Fs
Unit of work is Joule (J), Kilojoule (KJ)
1KJ = 1000 J

If a force of F newton is applied continuously on a body and the body undergoes a displacement of s metre in the direction of the force, then the work done by the applied force is W = Fs

Get the free “Nuclear Equation Calculator” widget for your website, blog, WordPress, Blogger, or iGoogle.

Physics Notes For Class 9 Kerala Syllabus Chapter 5 Question 7.
When a force of 10N is applied continuously on a body it undergoes a displacement of 2m find out the magnitude of the work done?
Answer:
F = 10N, s = 2m
Work W= Fs = 10 × 2 = 20 Nm
Look at the Figure

Hsslive Guru Physics Class 9 Kerala Syllabus Chapter 5 Question 8.
A body of mass m kg is placed on a table. What are the forces experienced by this body?
Answer:
Weight of the body applies downwards and the table applies an equal force upwards.

Hss Live 9th Physics Kerala Syllabus Chapter 5 Question 9
In which directions do these forces act?
Answer:
Forces acting on both directions, upward direction and downward direction.

9th Physics Notes Kerala Syllabus Chapter 5 Question 10.
A book of mass 100g is raised to the top of a table of height 1 m. Find the magnitude of the work done by the force applied against the gravitational force(g = 10m/s²)
Answer:
m = 100g = 0.1kg
g = 10m/s²
h = 1m
W = mgh
= 0.1 × 10 × 1 = 1J
1 J is the amount of work done to raise a body of mass 100 g through a height of 1m.

Kerala Syllabus 9th Standard Physics Solutions Chapter 5 Question 11.
If a force of 50N is applied on a body and it under¬goes a displacement of 2m in the direction of the force, calculate the amount of work done.
Answer:
F = 50N
s = 2m
W = Fs = 50 x 2 = 100J

Hss Live Guru 9 Physics Kerala Syllabus Chapter 5 Question 12.
a) If a force of 200N is applied on a table of mass 50kg, it undergoes a displacement of 0.5m in the direction of force. Calculate the amount of work done.
b) If the same table is raised by 3 m, what would be the work done against the gravitational force?
Answer:
a) F= 200N, s = 0.5m
W = Fs = 200 x 0.5 = 100J
b) m = 50 kg, g = 10m/s²
h = 3 m
W = mgh = 50 x 10 x 3 = 1500J
Observe Figure

Let a body mass m be pulled by a force F. If the body has a displacement s in the direction of the force, then the work done by the force F, Wf = Fs Here the displacement produced is in the direction of the force itself.

Kerala Syllabus Class 9 Physics Solutions Chapter 5 Question 13.
Write whether this work is negative or positive.
Answer:
Work is positive

9th Class Physics Notes Kerala Syllabus Chapter 5 Question 14.
The displacement is opposite to the frictional force, is the work done by the frictional force positive or negative
Answer:
If the displacement is in opposite direction work done by frictional force is negative.

Class 9 Physics Kerala Syllabus Chapter 5 Question 15.
In which direction is the force of gravity on the body?
Answer:
Direction of the gravitational force will be in the downward direction

9th Standard Physics Notes Kerala Syllabus Chapter 5 Question 16.
Is there a displacement for the body in the direction of the gravitational force?
Answer:
No displacement occurs in the direction of the gravitational force.
When a body on a floor is pulled and if it is displaced in the direction of the applied force, the work done by the applied force will be positive and the work done by the frictional force exerted by the floor will be negative.

Energy

Question 17.
What is the work to be done to raise a body of mass m kg through h meter?
Physics chapter 5 work and energy Answer:
w = mgh
Energy is the capacity to do work. Unit is Joule (J)

work energy book Question 18. In daily life, we use different forms of energy for various activities. List the forms of energy familiar to you
Answer:

1. Mechanical energy
2. Heat energy
3. Electrical energy
4. Chemical energy
5. Light energy
6. Nuclear energy

There are two type of Mechanical energy Kinetic energy and potential energy

Kinetic Energy

Pulling the toy car backwards a little and allow it to hit the plastic ball

Question 19.
What happened to the ball when the moving car hit it?
Answer:
Ball moves forward

10 to the 9th power Question 20.
How did the car get the energy to move the ball forward?
Answer:
The energy is obtained from the motion of toy car
Conclusion:

• Moving objects possess energy
• The energy possessed by a body by virtue of its motion is the Kinetic energy

Experiment:
Let’s do another activity. Allow a powder tin to slide down a polished, inclined plane as shown in the figure and let it hit a toy car. Try to measure the displacement of the toy car. Repeat the experiment by increasing the height of the inclined plane and filling the tin with sand.

Observation:

• The displacement of the toy car is greater When it is hit by the powder filled with sand
• Also the displacement is greater when it is dropped from a greater height

Conclusion: Kinetic energy depends on mass (m) and velocity (v)
Derivation of the equation for calculating K.E
Work W = Fs
As per II law of motion
F = ma
∴ W = mas
According to third equation of motion
v² = u² + 2as = 0 + 2as (u = 0) = 2as
∴ as = \(\frac{v^{2}}{2}\)
W = mas , if we put \(\frac{v^{2}}{2}\) instead of as

Work is equal to the magnitude of kinetic energy
i.e, Kinetic energy K = \(\frac { 1 }{ 2 }\) mv2,
M – mass, V – Velocity
When a body of mass m moves with a velocity v, its kinetic energy will be K = \(\frac { 1 }{ 2 }\) mv2

Question 21.
A man having a mass of 70kg is riding a scooter of mass 80kg. What is the total kinetic energy if the velocity of the scooter is 10m/s?
Answer:
m = 70kg + 80kg = 150kg
v = 10m/s

Question 22.
A car of mass 1500 kg is moving at a velocity of 20m/s. What is its kinetic energy?
Answer:
m = 1500 kg
v = 20m/s
K = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 1500 × 20²
= 300000J = 300kJ

Question 23.
A boy of mass 50kg is riding a bicycle with a speed 2m/s. The bicycle has a mass of 10kg. Calculate the total kinetic energy?
Answer:
m = 50kg + 10kg = 60kg
v = 2m/s
K = \(\frac { 1 }{ 2 }\) mv²
1/2 × 60 × 2² = 120J

Potential Energy

When we lift a body in perpendicular direction, the work is said to be done against gravitational force. According to figure, maximum work is done when it reaches at a height fb from the group, That means as the height increases work done also increases.

• The energy received by the body is equal to the work done on it.
• The body attains more energy when the height from the ground level increase

The energy possessed by a body due to its position is the potential energy.
ie, Potential energy
U = mgh
m – mass, g – acceleration due to gravity
h = height from the ground

Question 24.
Identify more situation in which potential energy is acquired by virtue of position
Answer:

• Coconut in a coconut tree
• Water stores in huge reservoirs
• Objects placed above the buildings,

Inference:
Height increases, potential energy increases

Question 25.
Write down situations in which potential energy varies.
Answer:

• Falling of a coconut from a coconut tree
• Pumping water to a tank at a height
• Climbing a ladder

Question 26.
Calculate the potential energy of a body of mass 1 kg at a height of 6m from the ground?
Answer:
Mass m = 1 kg, Acceleration due to gravity
g = 10m/s²
h = 6 m,
U = mgh = 1 × 10 × 6 = 60J

Question 27.
A bird of mass 0.5 kg is flying at the same speed at the same height of 5m. In this state, if its Kinetic energy and potential energy are equal.
a) What is the potential energy of the bird?
b) What is the velocity of the bird?
Answer:
a) m = 0.5kg
g = 10m/s²
h = 5m
u = mgh
= 0.5 × 10 × 5 = 25J
b) Kinetic energy K = 25J (∴U = K)

Question 28.
Write other examples for getting potential energy due to strain
Answer:

• A Stretched bow
• An elongated rubber band
• Compressed spring
• Compressed spring in a toy car

Conclusion: The factor which gives potential energy are position and strain

Law Of Conservation Of Energy

Question 29.
What form of energy does the flower pot have when it is on the sunshade of a building?
Answer:
Potential energy

Question 30.
While the flower pot is falling down, what forms of energy does it possess?
Answer:
potential energy and kinetic energy

Question 31.
Answer:
Does its potential energy increase/decrease when the pot falls down.
Answer:
Potential energy decreases

Question 31.
Will the kinetic energy increase/decrease at that time?
Answer:
Kinetic energy increases

Question 33.
What energy transformation takes place just before the flower pot reaches the ground?
Answer:
Potential energy is converted completely into kinetic energy

Question 34.
Let the mass of the flower pot 15kg and the height of the sunshade 4m.
a) When the flower pot is on the sunshade, what is its potential energy? (g = 10m/s2).
b) When it is on the sunshade, what is its kinetic energy?
c) If so, what is its total energy?
Answer:
U = mgh
= 15 × 10 × 4
= 600J
b) Kinetic energy will be zero
c) Total energy = PE + KE
= 600J + 0 = 600J

Question 35.
While falling, when the flower pot is at a height of 2m from the ground, what will be its kinetic energy?
Answer:
K = \(\frac { 1 }{ 2 }\) mv²
u = 0, g = 10m/s² ,
s = 4 – 2 = 2m
v² = u2² + 2as
=0 + 2 × 10 × 2 = 40
K = 1/2 × 15 × 40
= 300J

Question 36.
What is the potential energy when it is at a height of 2m from the ground? What is the total energy now?
Answer:
Potential energy at 2m height
U = mgh = 15 × 10 × 2 = 300J
∴ Total energy = 300 + 300 = 600J

Question 37.
What is the kinetic energy of the flower pot just before it touches the ground?
Answer:
K = \(\frac { 1 }{ 2 }\) mv²
v² = u² + 2as
= 0 + 2 x 10 x 4 = 80
Kinetic energy K = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 15 × 80 = 600J

Question 38.
The potential energy
U = mgh
= 15 x 10 x 10 x = 0.
What will be the total energy?
Answer:
Total energy = 600 + 0 = 600J

Question 39.
To sum up the amount of energy at each situation:
a) When on the sunshade
b) When at a height of 2m from the ground
c) Just before hitting the ground
Answer:
a) When on the sunshade = 600J
b) At a height of 2m from the ground = 600J
c) Just before hitting the ground = 600J
Energy can neither be created nor be destroyed. It can be transformed from one from to another

Sun is the major source of energy. We utilize solar energy in different ways. It is renewable source of energy Plants prepare food by using sunlight. From these fossil fuels, firewood are formed. Windmills works by utilizing kinetic energy of wind energy from infra¬red rays. Tidal energy is obtained due to gravitation. When we consider any energy sources, they were originated from solar energy.

Power

Question 40.
Given below is the information regarding the working of pumps in three neighboring houses. Complete the table (g = 10 m/s²).

Answer:
b) 150000J
c) 150000J

Question 41.
Is the amount of work done by the pump to fill water in the three tanks equal?
Answer:
Amount of work done is equal

Question 42.
Find the amount of work done per second by each pump.

Answer:

Amount of work done per second is referred as power of the pump.
Work done per unit time or rate of doing work is power
Power = \(\frac{\text { work }}{\text { time }}, P=\frac{w}{t}\)
Units of power
watt (W), kilowatt (KW), Horse power (H.P)
1 kW= 1000W
1 HP = 746W

Question 43.
If a man of mass 70kg climbs up a mountain of height 30m in 5 minutes, what is his power?
Answer:
m = 70kg, g = 10m/s²
h = 30m
Work W = mgh = 70 × 10 × 30 = 21000J
Time t = 5mt = 5 × 60 = 300s
Power p = \(\frac { w }{ t }\) = \(\frac { 21000 }{ 300 }\) = 70W

Question 44.
If a man of mass 50kg takes 60s to climb up 20 steps, each 15cm high, calculate his power.
Answer:
m = 50kg, g = 10m/s²
h = 15cm × 20 = 300cm = 8m
t = 60s
work W = mgh = 50 × 10 × 3 = 1500J
Power p = \(\frac { w }{ t }\) = \(\frac { 1500 }{ 60 }\) = 25W

Let Us Assess

Question 1.
A boy is trying to push the concrete pillar of the building using a force of 300N. Calculate the amount of work done by the boy.
Answer:
F= 300N, s = 0
w = Fs = 300 × 0 = 0

Question 2.
From what you have learnt of potential energy and kinetic energy write down the from of energy possessed by the bodies given below.
a) water in a dam
b) Stretched rubber band
c) Mango falling from a tree
Answer:
a) Potential energy
b) Potential energy
c) Potential energy lesser. Kinetic energy greater

Question 3.
Calculate the kinetic energy of an athlete of mass 60kg running with a velocity 10 m/s
m = 60kg
v= 10m/s

Question 4.
A stone of mass 2kg is thrown upwards from the ground with a velocity of 3m/s. When it reaches maximum height, calculate its potential energy.
Answer:
u = 3m/s v = 0
a = g = -10m/s² (Velocity of a body moving upwards is decreasing so acceleration is negative)
v² = u² + 2as
0 = 32 + 2 × -10 × s
0 = 9 – 20s
20s = 9, s = \(\frac { 9 }{ 20 }\)
U = mgh = 20 × 10 × \(\frac { 9 }{ 20 }\) = 90J

Question 5.
The heart of a healthy person beats 72 times per minute and each beat uses up about 1J of energy. Calculate the power of the heart.
Answer:
W= 1 x 72 = 72J
t = 1mt = 60s
P = \(\frac { W }{ t }\) = \(\frac { 72 }{ 60 }\) = 1.2w

Question 6.
Which among the following is a vector quantity?
a) work
b) momentum
c) power
d) energy
Answer:
momentum

Question 7.
If the velocity of an object is doubled, its kinetic energy becomes
a) 2 times
b ) 1/2
c) 4 times
e) 1/4
Answer:
4 times

Question 8.
An object of mass 1 kg is falling from a height 10 m. What be the work done while falling?
a) 10 J
b) 1J
c) 100 J
d)1000 J
Answer:
w = mgh = 1 × 10 × 10 = 100J

Question 9.
Which one among the following is correct?

Answer:
W = p × t

Question 10.
A roller weighing 1 tonne is being dragged along a road. What is the work done against gravity? Why?
Answer:
Zero, No displacement in the direction of force.

Question 11.
Is it possible for an object to possess energy without momentum? Give one example for such a situation.
Answer:
Yes, possible. A coconut on a coconut tree has potential energy, but it has no momentum.

Question 12.
Say whether the following are positive work or negative work.
1) The work done by a person drawing water from a well using a rope without a pulley.
2) Work done by gravitational force in this situation
3) Work done by the frictional force while and object is sliding down along an inclined plane.
4) Work done by the force is moving along a plane surface.
Answer:

1. Positive work
2. Negative work
3. Negative work
4. Positive work

Question 13.
How much joule is 1 kWh?
Answer:
1 kWh = 1000 × 60 × 60 = 3600000J

Question 14.
Find out the work done against the gravitational force in the situations given below.
1. A child is standing still with a bundle of books of mass 5 kg
2. With the same bundle of books, she travels 1m along a plane surface with a speed 5 m/s
3. The bundle of books is lifted onto the top of a cupboard having 1m height (g=10m/s2).
Answer:
1. 0
2. 0
3. W = mg = 5 × 10 × 1 = 50 J

Question 15.
A ball of mass 0.4 kg is thrown vertically upward with a velocity 14 m/s. Calculate its kinetic energy and potential energy after 1 s.
(Hint: v + at, s = ut + 1/2at²)
Answer:
v = u + at
= 14 + 10 × 1
= 24 m/s
KE = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 0.4 × 24
= 115.2 J
S = ut + 1/2 at²
= 14 × 1 + 1/2 × 10 × 1²
= 14 + 5 = 19 m
E = mgh = 0.4 × 10 × 19
= 76 J

Question 16.
An object of mass 1000 kg is travelling with a velocity 72 km/h. Calculate the work done to bring it rest.
Answer:
m = 1000 kg
v = 72 km/h = 20 m/s
Work = difference of kinetic energy
= \(\frac { 1 }{ 2 }\) mv²
= \(\frac { 1 }{ 2 }\) × 1000 × 20 × 20 = 400000J

Question 17.
Estimate the work done on a object of mass 80 kg to change its velocity from 5 m/s to 10 m/s
Answer:
m = 80 kg
Work = difference of kinetic energy

Work, Energy and Power More Questions

Question 1.
Fill in the blanks
a) Work done when a body of mass 100g is lifted up to a height of 1 meter is ……….
b) The two factors related to potential energy to body are………….
c) 1HP = ……… Watt
Answer:
1 Joule
b) Position, Strain
c) 746W

Question 2.
Classify the following into work done and work is not done
1. A mango falling from a mango tree
2. Pushing a table while sitting on it
3. Pushing a wall
4. Kicking a football
5. A trolley is moving forward
6. Standing with a load above the head
Answer:
Work done:
1. A mango falling from a mango tree
2. Kicking a football
3. A trolling is moving forward
Work is not done:
1. Pushing a table while sitting on it
2. Pushing a wall
3. Standing with a load on the head

Question 3.
When Lekshmi is applied a force of 50N on an object it undergoes a displacement of 2m. When Vinitha applied the same force on itundergoesadisplacementof3m if so
a) Which person done more work?
b) Give reason
c) Calculate the work done by each person?
Answer:
a) Vinitha
b) Displacement is greater
c) Work done by Lekshmi
W = Fs = 50N × 2m = 100J
Work done by Vinitha
W = Fs = 50N × 3m = 150J

Question 4.
Calculate work done when a boy of mass 40kg climb¬ing a staircase of height 50cm, g = 10m/s2
Answer:
W= Fs
F = mg = 40 × 10 = 400N
S = h = 0.5m
W= 400 × 0.5 = 200J

Question 5.
a) What is mean by kinetic energy?
b) How mass and velocity affect kinetic energy?
c) A body of mass 20kg is moving with a velocity of 5m/s. Calculate the kinetic energy?
Answer:
a) The energy possessed by a body by virtue of its motion is called kinetic energy
b) Mass increases, kinetic energy increases mass . is doubled kinetic energy also doubled. Velocity increases, kinetic energy increases, Velocity is doubled kinetic energy becomes four times.
c) Kinetic energy KE = \(\frac { 1 }{ 2 }\) mv²
= 1/2 × 20 × 5² = 1/2 × 20 × 25 = 250J

Question 6.
a) Write the equation for finding potential energy?
b) indicate the representation of each letter
Answer:
a) U = mgh
b) M – mass,
g – acceleration
h – height from the ground

Question 7.
State law of conservation of energy?
Answer:
Energy can neither be created nor be destroyed. It can be transformed from one form to another.

Question 8.
A store of mass 5kg was raised from the ground to the second floor of height 7m and from there to the third floor of height 3m from the second floor. Calcu¬late the potential energy of the store with respect to the ground floor and the second floor?
Answer:
Mass m = 5kg
Height from ground to third floor
h = 7m + 3m
g = 10m/s²
Potential energy with respect to the ground
U = mgh = 5kg × 10m/s² × 10 = 500J
Height from second floor to third floor h = 3m
Potential energy with respect to the second Floor
U = mgh = 5kg × 10m/s × 3 = 150J

Question 9.
a) What is power?
b) The time takes to move an object at a distance of 5m is 10s is the force applied is 30N, what is the power?
Answer:
a) Rate of doing work is power
b) Power P = E
W = Fs = 30 × 5 = 150J
t = 10sec.
Power P = \(\frac { 150 }{ 10s }\) = 15W

You can Download Division for Growth and Reproduction Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Biology Solutions Part 2 Chapter 7 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Biology Solutions Chapter 8 Division for Growth and Reproduction

Division for Growth and Reproduction Textual Questions and Answers

Division For Growth And Reproduction Class 9 Question 1.
The phase at which a cell prepares for division is called
Answer:
Interphase

9th Standard Biology Notes Kerala Syllabus Question 2.
………. takes place after karyokinesis
Answer:
Cytokinesis

Kerala Syllabus 9th Standard Biology Notes Question 3.
………… begins after interphase
Answer:
Cell division

Hss Live Guru 9th Biology Kerala Syllabus Question 4.
What are changes that take place during interphase?
Answer:
a) Division of nucleus (Karyokinesis)
b) Division of Cytoplasm (Cytokinesis)

Hss Live Biology Class 9 Kerala Syllabus Question 5.
State whether true or false
Cytokinesis takes place after karyokinesis
Answer:
True

Kerala Syllabus 9th Standard Biology Notes Pdf Question 6.
Main stages of cell division
Answer:
Interphase, Division of nucleus, Division of cytoplasm

Hss Live Guru Biology 9 Kerala Syllabus Question 7.
What are the important changes that take place during interphase?
Answer:

• Number of cell organelles increase
• Quantity of cytoplasm increases
• Cell size increases
• Genetic material duplicate

Cell cycle

A cell attains its complete growth during interphase. The fully grown cell undergoes division and becomes daughter cells. As the interphase and the division phase get repeated in a cyclic manner, they together constitute cell cycle.

UML full form, stands for, meaning, what is, description, example, explanation, acronym for, abbreviation, definitions, full name.

9th Biology Notes Kerala Syllabus Question 8.
…………..is brought about by cell division and cell growth
Answer:
Growth of the body

Kerala Syllabus Biology 9th Standard Question 9.
What are the two types of cell division?
Answer:
Mitosis and meiosis.

Biology Class 9 Kerala Syllabus Question 10.
What do you mean by mitosis?
Answer:
A parent cell divides to form two daughter cells are called mitosis.

Karyokinesis

Question 11.
Point out the phases taken place in the changes in nucleus?
Answer:
Prophase, metaphase, anaphase and telophase

Question 12.
In which phase does the chromatin reticulum become chromosomes?
Answer:
Prophase

Question 13.
What changes occurs in telophase?
Answer:
In telophase chromosomes that moved to the poles become chromatin reticulum and daughter nuclei are formed.

Question 14.
State whether true or false in prophase chromosomes become chromatin reticulum.
Answer:
False

Question 15.
Complete the table of stages of nuclear division
Answer:

Cytokinesis (Division Of Cytoplasm)

Question 16.
The division of the cytoplasm is taken place in plant is entirely different. Give reason?
Answer:
Because it is due to the presence of the cell wall in plant cell.

Question 17.
What is the significance of mitosis?
Answer:
The significance of mitosis is that there is no change in the number of chromosomes.

Question 18.
Mitosis helps ………… & ……….
Answer:
For the repair of tissues and growth.

Question 19.
Which condition leads to cancer?
Answer:
Mitosis is a controlled process. A disruption in this controlled process leads to the excessive division of a cell and its proliferation. This condition leads to cancer.

Different Stages Of Growth

Question 20.
List out the different stages in the growth of human beings.
Answer:

• Zygote
• Infancy
• Old age
• Embryo
• Childhood
• Fetus
• Adolescence
• Youth

Question 21.
What are the physical peculiarities of old age?
Answer:
Rate of cell division decreases, Availability of oxygen to the cells decreases, Deterioration of cells increase Muscles shrink, Production of energy decrease

Question 22.
The elders should be cared. Do you agree with this statement? Why?
Answer:
Old age is inevitable in life. The aged who worked for the welfare of their family and society during their younger age deserve special consideration.

Question 23.
What are the differences between the growth in plants and animals? Draw a comparison and complete table
Answer:

Question 24.
Plants grow due to the rapid division and differentiation of cells.
Answer:
Meristematic cells

Question 25.
What do you mean by meristematic cells?
Answer:
Meristematic cells are special types of cells that have the capacity for continuous division.

Question 26.
Plants can grow throughout their life due to the presence of
Answer:
Meristematic cells

Question 27.
……… helps to increase the length of root and stem.
Answer:
Apical meristem

Question 28.
……… helps to increase the girth of the stem.
Answer:
Lateral meristem

Question 29.
Name the meristematic cells which help to increase the length of the stem.
Answer:
Intercalary meristem

Question 30.
Where do you find intercalary meristem?
Answer:
It seen above the nodes of monocot plants.

Question 31.
The stem of monocots increases in length faster than dicots. Why?
Answer;
Because the intercalary meristem is visible only in the monocot plants.

Question 32.
Dicot plants: Lateral meristem
……………..: Intercalary meristem
Answer:
monocot plants

Question 33.
The stem of monocots does not increase its girth beyond an extent. Why?
Answer:
Because lateral meristem is absent in monocot plants.

Growth In Unicellular Organisms

Question 34.
Does cell division in unicellular organisms lead to growth or reproduction?

Answer:
Mitosis leads to reproduction in unicellular organisms.

Meiosis

Question 35.
What do you mean by meiosis? Explain.
Answer:
Meiosis is the mode of cell division in which gametes are formed. Meiosis occurs in the germinal cells of the reproductive organs. Human beings have 46 chromosomes. Germinal cells with 46 chromosomes divide continuously two times. These divisions in meiosis are known as meiosis I and meiosis II. Two daughter cells “with half the number of chromosomes (23 chromosomes) are formed in meiosis I. Each daughter cell again divides in meiosis II. There is no change in the chromosome number in this division. Hence meiosis II is similar to. mitosis. As a result of meiosis, four daughter cells, each with 23 chromosomes, are formed from a germinal cell.

Question 36.
What do you mean by polar body?
Answer:
When meiosis occurs in the female germinal cell, a large ovum and three small cells are formed. The smaller cells are the polar bodies. These sterile cells get destroyed.

Question 37.
Complete the illustration

Answer:
a = 46
b = 23
c= 23
d = 23
e = 23
f = 23
g = 23

Question 38.
What is the number of chromosomes in germinal cells?
Answer:
46

Question 39.
What is the number of chromosomes in the daughter cells formed after meiosis I?
Answer:
23

Question 40.
What is the peculiarity of meiosis II?
Answer:
Each daughter cell again divides in meiosis II. There is no change in the chromosome number in this division. Hence meiosis II is similar to mitosis.

Question 41.
What are the kinds of cell division occur in sexually reproducing organisms?
Answer:
There are two kinds of cell division occur in sexually reproducing organisms that are mitosis and meiosis.

Question 42.
What are the different stages in the growth of human being?
Answer:
Infancy, childhood, adolescence, youth and old age.

Question 43.
Differentiate mitosis and Meiosis

Answer:

Let Us Assess

Question 1.
The stage of karyokinesis at which daughter nuclei are formed
A. Prophase
B. Metaphase
C. Anaphase
D. Telophase
Answer:
Telophase

Question 2.
List the meristems in various parts of the plant and list their functions
Answer:

Question 3.
In females, only a single ovum is formed from a germinal cell, whereas in males, more than one sperm is formed. Give reason.
Answer:
When meiosis occurs in the female germinal cell, a large ovum and three small cells are formed. The smaller cells are the polar bodies. These sterile cells get destroyed. So only a single ovum is formed from a germinal cell. But in males, after meiosis, four sperms having 23 chromosomes are formed form one germinal cell.

Question 4.
Observe the figures

a) Which stages of mitosis are indicated in the figures?
b) What are the changes that occur in the chromosomes during these stages?
Answer:
a) Metaphase
b) Chromosomes get aligned at the equator of the cell.

You can Download Proportion Questions and Answers, Activity, Notes, Kerala Syllabus 9th Standard Maths Solutions Chapter 12 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Maths Solutions Chapter 12 Proportion

Proportion Textual Questions and Answers

Textbook Page No. 182

Hss Live Guru 9th Maths Kerala Syllabus Question 1.
A person invests 10000 rupees and 15000 rupees in two different schemes. After one year, he got 900 rupees as interest for the first amount and 1200 rupees as interest for the second amount.
i. Are the interests proportional to the investments?
ii. What is the ratio of the interest to the amount invested in the first scheme? What about the second?.
iii. What is the annual rate of interest in the first scheme? And in the second?
Answer:
i. The ratio between the amounts invested = 10000:15000 = 2 : 3
The ratio between the interests = 900 : 1200 = 3 : 4
Since the ratios are different. So, interests will be not proportional to the investments.

ii. The ratio between the amount invested and interest in scheme 1
= 10000 : 900= 100 : 9
The ratio between the amount invested and interest in scheme 2
= 15000 : 1200 = 25 : 2

iii. Rate of interest in the first scheme
= 900/10000 × 100 = 9%
Rate of interest in the second scheme =
1200/15000 × 100 = 8%

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Kerala State Class 9 Maths Solutions Question 2.
The area of A0 paper is one square metre. Calculate the lengths of the sides of A4 paper correct to a millimetre, using a calculator.
Answer:
The area of A1 paper is half of the area of A0 paper. The area of A2 paper is half of the area of A1 paper. Let con-sider the area of the A0 paper be 1 square metre, the area of A1 paper is
1/2 square metre , the area of A2 paper is 1/4 square metre , the area of A3 paper is 1/8 square metre , and the area of A4 paper is 1/16 square metre ,
If the length of A4 paper is √2 x and breadth x
length of A4 paper : breadth = √2 : 1
Area of A4 paper = length x breadth
√2x × x = √2x²
This is 1/16 m², therefore √2 x2 = 1/16

Hsslive Guru 9th Maths Kerala Syllabus Question 3.
In calcium carbonate, the masses of calcium, carbon and oxygen are in the ratio 10 : 3: 12 . When 150 grams of a compound was analysed, it was found to contain 60 grams of calcium, 20 grams of carbon and 70 grams of oxygen. Is it calcium carbonate?
Answer:
The ratio between calcium, carbon and oxygen in calcium carbonate = 10 : 3: 12 The ratio of calcium, carbon and oxygen in the given compound = 60 : 20: 70 = 6 : 2: 7
Since this ratio is not equal to the ratio of calcium, carbon and oxygen in calcium carbonate, it is not calcium carbonate.

Textbook Page No. 185

Hss Live 9 Maths Kerala Syllabus Question 1.
For each pair of quantities given below, check whether the first is proportional to the second. For proportional quantities, calculate the constant of proportionality.
i. Perimeter and radius of circles.
ii. Area and radius of circles.
iii. The distance travelled and the number of rotations of a circular ring moving along a line.
iv. The interest got in a year and the amount deposited in a scheme in which interest is compounded annually.
v. The volume of water poured into a hollow prism and the height of the water level.
Answer:
i. The perimeter of a circle is n times its diameter. That is 2n times the radius.
∴Perimeter = 2πr
∴ The perimeter and radius are proportional.
The constant of proportionality is 2π

ii. The area of a circle is π times the square of the radius.
∴ Area = πr²
∴ Area and radius are not proportional.

iii. When the ring rotates once, the distance travelled is equal to its perimeter. When it rotates twice the distance travelled is twice its perimeter. When it rotates ‘n’ times, the distance travelled is ‘n ‘ times the perimeter of the ring.

iv. If the amount deposited is P and the rate of interest is R.
Annual interest = I = PNR,
I = P × R (N = 1)
The amount and interest are propor-tional.
Constant of proportionality is rate of interest, R.

v. Volume of a hollow prism
= base × height
Volume, of water and height of water level, are proportional. Constant of proportionality is the base area.

Kerala Syllabus 9th Standard Maths Notes Question 2.
During rainfall, the volume of water falling in each square metre may be considered equal.
i. Prove that the volume of water falling in a region is proportional to the area of the region.
ii. Explain why the heights of rainwater collected in different sized hollow prisms kept near one another are equal.
Answer:
i. The volume of water falling in each square metre are equal.
Let the volume of the rain falling on the 1 square metre be k The volume of the rain falling on the 2 square metre = 2k
The volume of the rain falling on the 3 square metre = 3k
The volume of the rain falling on the x square metre y = kx Here x and y are proportional.

ii. Let x be the base area of the vessel and h be the height of the water collected, then the volume of water y = xh.
y/x =h
The volume of water collected is different in hollow prisms having different base area.

But \(\frac { volume }{ Area }\) is always equal to the height of water level. So the height at which rainwater collected is same.

Hsslive Guru Maths 9th Kerala Syllabus Question 3.
When a weight is suspended by a spring, the extension is proportional to the weight. Explain how this can be used to mark weights on a spring balance.

Answer:
Mark the pointer when no weight is hanging on it. Then mark the point when a constant weight is hanging on it. For example when we hang 1 kg on it a 2 cm extension is made. So we mark 1, 2, 3, 4 from the points marked first, i.e.,
2, 4, 6 and 8.
These distances are divided into 10 equal parts, then we can mark the points 1.1 kg and 2.4 kg etc.

Hss Live Guru 9 Maths Kerala Syllabus Question 4.
In the angle shown below, for different points on the slanted line, as the distance from the vertex of the angle changes, the height from the horizontal line also changes.

i. In the picture, perpendicular to the horizontal lines are drawn from points on the slanted line.
Δ ABP, Δ ACQ, Δ ADRand Δ AES are similar triangles. (Because these are a right-angled triangle with is common to all)
∴ Ratio of their sides are equal. That is sides are proportional.

Let k be the constant of proportionality
PB = k × AB, QC = k × AC
RD = k × AD, SE = k × AE ,
i.e., the change in height is proportional to the distance.

An online geometric sequence calculator helps you to find geometric Sequence, first term, common ratio calculator, and the number of terms.

Textbook Page No. 189

Class 9 Maths Chapter 12 Kerala Syllabus Question 1.
i. Prove that for equilateral triangles, area is proportional to the square of the length of aside. What is the constant of proportionality?
ii For squares, is area proportional to square of the length of a side? If so, what is the constant of proportionality?
Answer:
i. Δ ABC is an equilateral triangle
Consider their sides are ‘a’,
Draw a perpendicular line CD to AB
from C. In right-angled triangle ADC, according to the Pythagoras theorem AD² + DC² = AC²
DC² = AC² – AD²

ii. If x be the one side of a square then its area y = x²
Area of the square is proportional to the square of their sides.
Constant of proportionality = 1

Hsslive Guru Class 9 Maths Kerala Syllabus Question 2.
In rectangles of area one square metre, as the length of one side changes, so does the length of the other side. Write the relation between the lengths as an algebraic equation. How do we say this in the language of proportions?
Answer:
Area of the rectangle is the sum product of length and breadth.
Let × be the length and y be the breadth, then
x × y = 1 m²

x = 1/y = m
Example when y = 3 m
x = 1/3 meter
when y = 4 m
x = 1/4 meter
The algebraic equation is x = 1/y
i.e., length of rectangle is proportional to the reciprocal of the breadth, i.e., length of rectangle is proportional to its breadth.

Chapter 12 Maths Class 9 Kerala Syllabus Question 3.
In triangles of the same area, how do we say the relation between the length of the longest side and the length of the perpendicular from the opposite vertex? What if we take the length of the shortest side instead?
Answer:
Let ‘a’ be the large side, h be the length of perpendicular from opposite vertices, ‘A’ be the area, then

Length of larger side is inversely proportional to the length of perpendicular from the opposite vertex.
i.e., The length of small side is inversely proportional to the length of perpendicular line from the vertex of small side.

Hss Live Guru Class 9 Maths Kerala Syllabus Question 4.
In regular polygons, what is the relation between the number of sides and the degree measure of an outer angle? Can it be stated in terms of proportion?
Answer:
The sum of the exterior angles of all polygon is 360°. If ‘n’ is the number of sides.
Measure of an exterior angle = \(\frac{\text { Sum of exterior angles }}{\text { No. of sides }}\)
One outer angle = 360/n
(n = number of sides)
If the measure of an exterior angle is ‘x’.
x= \(\frac{1}{n} \times 360^{0}\)
One outer angle and number of sides are inversely proportional. The constant of proportionality is 1/n.

Hsslive Guru Maths Kerala Syllabus 9th Question 5.
A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed by using different pipes. Write the relations between the following quantities as an algebraic equation and in terms of proportions.
i. The rate of water flow and the height of the water level.
ii The rate of water flow and the time taken to fill the tank.
Answer:
i. Let x be the rate of water flowing, y be the height of water in the tank and A be the base area of the tank, then
x = Ay
Height of the water level in the tank is proportional to the rate of the water flowing.

ii. If C is the volume of the tank, V be the volume of water flowing per second, the volume of water in ‘t’ second is given by
C = V × t
\(V=\frac{C}{t}=C \times \frac{1}{t}\)
That is the rate of water flow and the time taken for filling the tank are inversely proportional. C is the constant of proportionality.

Proportion Exam oriented Questions and Answers

Hsslive 9th Maths Kerala Syllabus Question 1.
the weight of 6 spheres of same size made of the same metal is 14 kg. When 9 more spheres are added the weight is 35 kg. Check whether the number of spheres and their weights are proportional.
Answer:
Weight of 6 spheres = 14kg
Ratio of number and weight = 6 : 14 = 3 : 7
Total number of sphers are added = 15
Total weight = 35kg
Ration of number and weight = 15 : 35 = 3 : 7
Since the ratios are equal, the number of sphers and their weights are propor-tional.

Question 2.
Raghu invested Rs. 60000 and Nazar Rs. 100000 and started a business. Within one month a profit of Rs. 4800 was obtained. Raghu took 1800 and Nazar took Rs. 3000 out of the profit obtained. What is the ratio of the investment? Is the investment and the profit divided proportionally?
Answer:
Ratio of investments = 60000 : 100000 = 6 : 10 = 3 : 5
Ratio of profit divided
1800 : 3000 = 18 : 30 = 3 : 5
Ratio of investments and Ratio of profit divided are equal. Hence they are proportional.

Question 3.
The two sides of a triangle having perimeter 10 m are 2\(\frac { 1 }{ 2 }\) m and 3 \(\frac { 1 }{ 2 }\) m.
What is the ratio of the length of the three sides of triangles?
Answer:
Perimeter of triangle = 10 m
First side = 3\(\frac { 1 }{ 2 }\) m = \(\frac { 5 }{ 2 }\) m
Second side = 3\(\frac { 1 }{ 2 }\) m = \(\frac { 7 }{ 2 }\) m
Third side = 4 m Ratio of three sides of a triangle
\(=\frac{5}{2}: \frac{7}{2}: 4 = 5: 7: 8\)

Question 4.
150 litres of water is flowing through a pipe in 6 minutes. If 200 litres of water flows through it in 8 minutes check whether the quantity of water and time of flow are proportional ?
Answer:
Quantity of water flowing in 6 minutes = 150 litres
Ratio between the quantity of water and timeofflow= 150: 6 = 25: 1
Quantity of water flowing in 8 minutes = 200 litres
Ratio between quantity of water and times of flow = 200 : 8 = 25: 1
Since the radios are equal, the amount of water flowing and the time of flow are proportional.

Question 5.
Unniyappam was made using 1 kg rice, 250 g plantain and 750g jaggery. Find the ratio between the ingredients.
Answer:
Rice = 1
kg= 1000g
Plantain = 250 g, Jaggery = 750 g.
Ratio between the ingredients
= 1000 : 250 : 750 = 4 : 1 : 3

Question 6.
Sathyan got Rs. 500 after working for 6 hours. Gopi got Rs. 400 after working for 4\(\frac { 1 }{ 2 }\) hours. Are the wages obtained proportional to the working time ?
Answer:
Ratio of working hours = 6 : 4 \(\frac { 1 }{ 2 }\)
= 12 : 9 = 4 : 3
Ratio of wages = 800 : 600 = 8 : 6 = 4 : 3
Since the ratios are equal, the working hours are proportional to the wages.

Question 7.
Are the length and breadth of a square having same perimeter inversely proportional?
Answer:
for a square the perimeter is 20 cm. so,
length + breadth = 10.
Let x be the length and y be the breadth then possible values of x and y are
x — y
9 — 1
8 — 2
7 — 3
Here xy is not a constant term i.e. the changes in the x and y is not in the firm of xy = kept
hence length and breadth are not in inversely proportional.

Question 8.
The weight of an object having mass 5 kg is 49 Newton. The weight of another object having mass 15 kg is 147 Newton. Check whether the mass and weight are proportional ? What is the constant of proportionality? What is the weight of an object having mass 8 kg ?
Answer:
Weight of 5 kg object = 49 N
Ratio between mass and weight = 5 : 49
Weight of 15 kg massed object = 147N
Ratio between mass and weight = 15:147 = 5:49
Mass and weight are proportional.
Weight of 8 kg massed object = 8 × 9.8 = 78.4N

Constant of proportionality = 9.8

Question 9.
The perimeter of a triangle is 60 cm. Sides are in the ratio 3: 4: 5. Then j find the length of the sides.
Answer:
Ratio of sides = 3 : 4 : 5
Therefore, the sides are the \(\frac { 3 }{ 12 }\), \(\frac { 4 }{ 12 }\) and \(\frac { 5 }{ 12 }\) part of the perimeter .
The perimeter is 60 cm, then the length of sides are

\(60 \times \frac{5}{12}=25 \mathrm{cm}\)

Question 10.
Will the dye which contains 10 L blue colour and 15 L white colour and another dye which contain 12L blue colour and 17 L white colour have j the same colour? Why?
Answer:
In the first dye , blue colour: white colour = 10 : 15 = 2 : 3
In the second dye, blue colour: white colour = 12: 17
The ratios are not same. Hence the both will not have same colour.

Question 11.
The face perimeter of some vessels in the shape of square prisms are equal. 12 litres of water can be filled in the vessel having height 15 cm. 16 litres of water can be filled in the vessel having height 20 cm. Check whether the volumes of the vessel and height are proportional. What is the constant of proportionality?
Answer:
Since the face perimeters are equal,
Volume = face perimeter × height
Volume of the vessel having height 15 cm = 12 litres.
Ratio of height to volume =15 : 12 = 5 : 4 Volume of the vessel having height 20 cm = 16 litres
Ratio of height to volume =20 : 16 = 5 : 4 Since the ratios are equal, height and volume are proportional.
\(\frac { volume }{ height }\) = \(\frac { 4 }{ 5 }\) constant of proportionality
Volume of the vessel having height 35 cm
= 35 × \(\frac { 4 }{ 5 }\) = 28 liters.

Question 12.
The ratio of carbon, sulphur and potassium nitrate to make gun powder is 3 : 2: 1. How much quantity of each is required to make 1.2 kg of gun powder ?
Answer:

Question 13.
Raju got Rs. 400 after working for 8 hours. Damu worked for 6 hours and got Rs. 300. Are the wages obtained proportional to the work time?
Answer:
Ramu working hours = 8 hour
Wages Raju obtained = Rs. 400
Damu working hours = 6 hour
Wages obtained by Damu = Rs. 300

∴ Wages obtained are proportional to the work time.

Question 14.
150 L of water flows through a pipe for 6 minutes. 200 L of water flows for 8 minutes through the same pipe. Are the time and amount of water flowing proportional?
Answer:
Ratio of quantity of water
= 150 : 200 = 15 : 20 = 3 : 4
Ratio oftime =6 : 8 = 3 : 4
The ratios are same , hence they are proportional.

Question 15.
A car with 5L of petrol travels a distance of 75 km. What is the proportionality constant between the distance travelled and the quantity of petrol? How much petrol is needed for travelling 180 km?
Answer:
Taking the distance travelled as x and quantity of petrol as y, then the constant of proportionality

The quantity of petrol needed to travel 180 km= 12 litre

Question 16.
The angles of a triangle are in the ratio 1 : 3: 5. How much is each angle of the triangle?
Answer:
Sum of the angles of a triangle = 180°
The angles of the traingle are \(\frac { 1 }{ 9 }\), \(\frac { 3 }{ 9 }\) and \(\frac { 5 }{ 9 }\)
parts of 180° , hence the angles are

You can Download राग गौरी Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Hindi Solutions Unit 5 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

Kerala State Syllabus 9th Standard Hindi Solutions Unit 5 Chapter 2 राग गौरी (पद)

राग गौरी Text Book Activities

Hsslive Guru 9th Hindi Kerala Syllabus प्रश्ना 1.
समानार्थी शब्द कविता में ढूँढ़ें :
उत्तर:

Hss Live Guru 9th Hindi Kerala Syllabus प्रश्ना 2.
कविता में कृष्ण अपनी माँ से कुछ शिकायतें कर रहा है। कविता के प्रसंग में यशोदा और बालक कृष्ण के बीच का वार्तालाप लिखें।

उत्तर:
यशोदा : बेटा, क्या हुआ? बताओ न?
कृष्ण : क्या बताऊँ? जाके भैया से पूछ।
यशोदा : बोलो मेरे लाल,क्या हुआ?
कृष्ण : बलराम भैया बहुत तंग करता है मुझे। कहता है मुझे मोल में लिया है?
यशोदा : ऐसा कहा उसने!
कृष्ण : हाँ मैया, उसी के वास्ते मैं खेलने भी नहीं जाता। तेरी माँ कौन है, पिता कौन है कहकर बहुत चिढ़ाता है।
यशोदा : फिर?
कृष्ण : भैया कहता है माँ गोरी हैं, पिताजी गोरे हैं, केवल तू ही काला है।
यशोदा : सच? उसने ऐसा कहा?
कृष्ण : हाँ माँ, दूसरे लड़के भी मेरा मज़ाक करते हैं। भैया ने ही उन्हें सब कुछ सिखाया है।
यशोदा : ऐसा थोड़ा ही है?
कृष्ण : मुझे पता है, तू तो हमेशा भैया के साथ देती है। तू मुझे ही मारती है, भैया को कभी नहीं।
यशोदा : सुनो कान्ह, तेरा भैया तो जन्म से ही झूठा है। उसका कहना मत मान।
कृष्ण : माँ, तू मुझे मनाने के लिए ऐसा कहती है न?
यशोदा : नहीं..नहीं… मैं कसम खाती हूँ कि तू मेरा लाल है और मैं तेरी माँ।
कृष्ण : सच माँ?
यशोदा : सच।

राग गौरी Summary in Malayalam and Translation

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Kerala State Syllabus 9th Standard Hindi Solutions Unit 5 Chapter 1 मेरी ममतामई माँ (संस्मरण)

मेरी ममतामई माँ Textual Questions and Answers

मेरी ममतामई माँ विश्लेषणात्मक प्रश्न

Kerala Syllabus 9th Standard Hindi Solutions Unit 5 Chapter 1 प्रश्ना 1.
एकल परिवार और संयुक्त परिवार की अपनी – अपनी विशेषताएँ हैं। चर्च करें।

उत्तर:
कल परिवार का अर्थ है कि एक ऐसा परिवार जिसमें पति, पत्नी और उनके बच्चे ही रहते हैं। एक ऐसा परिवार जहाँ माता – पिता, बेटे – बहू, पोते – पोती, चाचा-चाची और ताऊ- ताई आदि एकसाथ रहते हैं उसे हम संयुक्त परिवार कहते हैं।
विशेषताएँ:
1. एकल परिवार छोटे होते हैं जबकि संयुक्त परिवार बड़े होते हैं।
2. एकल परिवार में खर्च कम होते हैं जबकि संयुक्त में अधिक खर्चे होते हैं।
3. एकल परिवार में लोगों को एकांत अधिक मिल पाता है जबकि संयुक्त परिवार में इस चीज़ की कमी होती है।
4. एकल परिवार में बच्चे अपने दादा-दादी, नाना-नानी के प्यार से वंचित रह जाते हैं जबकि संयुक्त परिवार में उन्हें अपने दादा – दादी, नाना – नानी आदि का भरपूर प्यार और संस्कार मिल पाते हैं ।
5. एकल परिवार में अच्छा-बुरा सलाह के लिए किसी बड़े अनुभवी का साथ नहीं मिल पाता जबकि संयुक्त में अनुभवी व्यक्ति की सलाह से कई समस्याएँ चुटकियो में सुलझ जाती है।
6. एकल परिवार में बच्चे संस्कृति और संस्कारो में पीछे रह जाते हैं जबकि संयुक्त परिवार में बच्चों को अच्छे संस्कार और संस्कृति को जानने का मौका मिल पाता है।
7. एकल परिवार में यदि कोई एक व्यक्ति बीमार हो जाता है तो देखभाल के लिए कोई भी नहीं होता है जबकि संयुक्त परिवार में यदि कोई बीमार हो जाता है तो सहायता और देखभाल के लिए कई लोग मिल जाते हैं और बड़ों का अनुभव भी मिलता है।

Kerala Syllabus 9th Standard Hindi Notes प्रश्ना 2.
‘शहर में सबसे पहले मैं ही लोगों तक समाचार-पत्र पहूँचाता था।’- इस प्रस्ताव से बालक कलाम का कौन-सा मनोभाव प्रकट होता है?


उत्तर:
कलाम एक ईमानदार बालक है। वह अपना दायित्व ठीक तरह से निभानेवाला है। समय की पाबंदी लगाने में वह सफल रहा। वह हमेशा मेहनती नज़र आता है।

9th Hindi Notes Kerala Syllabus प्रश्ना 3.
“उन्होंने अपने हिस्से की भी सारी रोटियाँ तुम्हें दे दी”-भाई की इस बात पर बालक कलाम की प्रतिक्रिया क्या होगी?

उत्तर:
बालक कलाम को अपनी माँ के प्रति ममता के मारे सिहरन आ गई। वह अपने आपको रोक नहीं पाया। दौड़कर माँ के पास गया और भावावेश में उनसे लिपट गया।

Std 9 Hindi Notes Kerala Syllabus प्रश्ना 4.
‘नारी ईश्वर की सुंदर रचना है।’ कलाम को ऐसा क्यों लगा?

उत्तर:
कलाम नारी को आदर करनेवाले है। कलाम अपनी माँ को ईश्वर समान मानते हैं। माँ ही उनके लिए सबकुछ है। माँ की ममता कलाम को सदा लिपटती थी। उनकी हर उन्नति पर माँ का हाथ रहा था। वे अपनी माँ के नाश्ते के बारे में अग्नि की उडान’ पर याद करते हैं। शायद इसीलिए उन्हें ऐसा लगा होगा।

मेरी ममतामई माँ Text Book Activities

Class 9 Hindi Notes Kerala Syllabus प्रश्ना 1.
मिलान करें:

उत्तर:

9th Class Hindi Notes Kerala Syllabus प्रश्ना 2.
बातचीत लिखें।
‘उस दिन पहली बार मुझे सिहरन की अनुभूति हुई। मैं अपने आपको रोक नहीं सका। दौड़कर अपनी माँ के पास गया और भावावेश में उनले लिपट गया।’ इस प्रसंग पर बालक कलाम और माँ के बीच क्या-क्या बातें हुई होंगी?

उत्तर:
कलाम : (दौडकर आता हुआ) माँ ………… ओ माँ ……..
माँ . : (लिपटती हुई) क्या हुआ बेटा?
कलाम : यह आपने क्या कर दिया? मुझे सिहरन हुई।
माँ : क्या बात है?
कलाम : आज आपने अपने हिस्से की रोटियाँ भी मुझे दे दी!
माँ : हाँ, तुम तो भूखा था न, इसलिए।
कलाम : तो आप स्वयं भूखी रहकर मुझे दे दिया?
माँ : तुम तो पढ़ाई और कमाई एकसाथ करते है न।
कलाम : इतनी कुरबानी मेरे लिए क्यों माँ?
माँ : क्योंकि मैं तेरी माँ हूँ और तुम मेरे लाडले हो।
कलाम : हमारे घर की हालत उतनी अच्छी नहीं है माँ।
माँ : मैं संभालूँगी बेटा। तुम फिकर मत कर।
कलाम : आप भूखों मत मरेगी। यह मेरी वादा है।
माँ : (प्यार से सहलाती हुई) जा जाकर कुछ पढ़ाई कर।

Kerala Syllabus 9th Standard Hindi प्रश्ना 3.
नमूने के अनुसार बदलकर लिखें।

उत्तर:

मेरी ममतामई माँ विधात्मक प्रश्न

Hindi Kerala Syllabus 9th Standard प्रश्ना 1.
“उन्होंने अपने हिस्से की भी सारी रोटियाँ तुम्हें दे दी। एक ज़िम्मेदार बेटा बनो और अपनी माँ को भूखों मत मारों।” बड़े भाई ने कलाम को डाँटा तो उसे क्या लगा होगा? उसकी डायरी कल्पना करके लिखें।

उत्तर:
25 जून 2019
बडे भाई का डाँट सुनकर आज पहली बार मुझे सिहरन की अनुभूति हुई। मैं अपने आपको रोक नहीं सका। मैं दोडकर अपनी माँ के पास गया। फिर भावावेश में उनसे लिपट गया। माँ की कुरबानी इतनी क्यों? घर की हालत उतनी अच्छी नहीं थी। युद्ध : के कारण सभी वस्तुओं की किल्लत हो गई थी। विशाल संयुक्त परिवार में रोटियों की भी कमी महसूस हुई थी। फिर भी माँ स्वयं भूखी रहकर उनकी सारी रोटियाँ मुझे दे दी। सिर्फ माँ ऐसा कर सकती है। वे कितने दयालू, स्नेहशील और धार्मिक प्रवृत्ति की महिला थीं। सदा मैं उनसे प्रेरित थी। मेरी माँ ही सबकुछ है। आगे मेरी दृष्टि उनके स्वास्थ्य पर भी होगा। अगले दिन की प्रतीक्षा में….

मेरी ममतामई माँ Additional Questions and Answers

9th Standard Hindi Notes Kerala Syllabus प्रश्ना 1.
‘कलाम के घर में खुशी और गम का आना जाना लगा रहता था’- इसका कारण क्या होगा?

उत्तर:
कलाम का परिवार संयुक्त परिवार है। संयुक्त परिवार में सदस्यों की संख्य ज़्यादा है। तो हमेशा कुछ खुशी या कुछ गम आते जाते रहेंगे।

9th Std Hindi Notes Kerala Syllabus प्रश्ना 2.
कलाम के गणित के शिक्षक की शिक्षण रीति कैसी थी?

उत्तर:
गणित के शिक्षक श्री स्वामियार निःशुल्क ट्यूशन पढ़ाते थे। वे एक साल में पाँच ही छात्रों को पढ़ाते थे। उनका एक ही शर्त था कि बच्चे स्नान करके पाँच बजे कक्षा में उपस्थित हो जाएँ।

Kerala Syllabus 9th Standard Notes Hindi प्रश्ना 3.
कलाम की चरित्रगत विशेषताओं पर टिप्पणी तैयार करें।

उत्तर:
सुबह पाँच बजे उठकर श्री स्वामियार के पास गणित पढ़ने जाते थे। साढ़े पाँच बजे घर वापस लौटता। बाद में नमाज़ उदा करने को और कुरान शरीफ़ सीखने के लिए पिता के साथ जाता। उसके बाद तीन किलोमीटर दूरी पर रामेश्वरम रोड़ रेलवेस्टेशन पैदल जाता था। प्रस्तुत पंक्तियों से बालक कलाम बहुत ही परिश्रमी जान पडता है और ईमानदार भी धनुष्कोडी मेल से समाचार पत्रों का बंडल लेता और तेज़ी से शहर के लोगों तक पहुँचाता था। कलाम की समय पर पाबंदी यहाँ बहुत ही स्पष्ट है। वह अपनी ज़िम्मेदारी ठीक तरह से निमाने वाला भी है। वह अपनी माँ को बहुत प्यार करता है। जब उसका भाई उसे डाँटा तो तुरंत ही दौडकर जाते हुए माँ को लिपट लेता है। वह नारी को ईश्वर की सुंदर रचना मानता है। स्त्रियों को आदर करने में वह सदा आगे है। इसप्रकार स्वाभिमानी, परिश्रमी, सेवाव्रती, ज़िम्मेदार एवं मानवता के व्यक्तित्व है बालक कलाम।

मेरी ममतामई माँ Summary in Malayalam and Translation

मेरी ममतामई माँ शब्दार्थ