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Cbse, ncert and icse solution online, class 8 science case study question, case study question class 8 science (cbse / ncert board).

Class 8 Science Case Study Question and Answer: CBSE / NCERT Board Class 8 Science Case Study Question prepared by expert Science Teacher. Students can learn Case Based Question / Paragraph Type Question for NCERT Class 8 Science.

There are total 18 chapter Crop Production and Management, Microorganisms: Friend and Foe

, Synthetic Fibres and Plastics, Materials: Metals and Non-Metals, Coal and Petroleum, Combustion and Flame, Conservation of Plants and Animals, Cell – Structure and Functions, Reproduction in Animals, Reaching the Age of Adolescence, Force and Pressure, Friction, Sound, Chemical Effects of Electric Current, Some Natural Phenomena, Light, Stars and the Solar System, Pollution of Air and Water

For any problem during learning any Case or any doubts please comment us. We are always ready to help You.

CBSE Class 8 Science Case Study Question

• Chapter 1 Crop Production and Management Case Study Question
• Chapter 2 Microorganisms: Friend and Foe Case Study Question
• Chapter 3 Synthetic Fibres and Plastics Case Study Question
• Chapter 4 Materials: Metals and Non-Metals Case Study Question
• Chapter 5 Coal and Petroleum Case Study Question
• Chapter 6 Combustion and Flame Case Study Question
• Chapter 7 Conservation of Plants and Animals Case Study Question
• Chapter 8 Cell – Structure and Functions Case Study Question
• Chapter 9 Reproduction in Animals Case Study Question
• Chapter 10 Reaching the Age of Adolescence Case Study Question
• Chapter 11 Force and Pressure Case Study Question
• Chapter 12 Friction Case Study Question
• Chapter 13 Sound Case Study Question
• Chapter 14 Chemical Effects of Electric Current Case Study Question
• Chapter 15 Some Natural Phenomena Case Study Question
• Chapter 16 Light Case Study Question
• Chapter 17 Stars and the Solar System Case Study Question
• Chapter 18 Pollution of Air and Water Case Study Question

What is Case Study Question?

Ans. At case Study there will one paragraph and on the basis of that concept some question will made. Students have to solve that question.

How many marks will have at case based question?

Most of time 5 questions will made from each case. There will 1 or 2 marks for each question.

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CBSE 8th Standard CBSE all question papers, important notes , study materials , Previuous Year questions, Syllabus and exam patterns. Free 8th Standard CBSE all books and syllabus online. Practice Online test for free in QB365 Study Material. Important keywords, Case Study Questions and Solutions. Updates about latest education news and Scholorships in one place.

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• Important Questions For Class 8 Science

CBSE Important Questions for Class 8 Science

CBSE Important Questions for Class 8 Science help students face the exams confidently. As the exam nears, they should complete the syllabus and start with the revision. The CBSE Important Questions for Class 8 Science will help them revise the subject quickly. Here, we have compiled the important questions for all the chapters of the CBSE Class 8 Science subject. By solving them, students will understand the types of questions asked in the exam. All types of questions are provided for students’ practice from basic to high level. CBSE Class 8 Science is a fascinating subject and includes interesting topics, thus providing basic knowledge about light, current, force, human beings, etc. Solving these CBSE Class 8 Science Important Questions is the best way to revise the major concepts of the subject.

The subject experts created the questions based exclusively from the exam perspective. Students must start solving them at least 20 days before the annual exam. By doing so, they will be able to analyse their weak areas and work on them. Practising the CBSE Important Questions will improve their performance and overall score in CBSE Class 8 Science exam.

There is a total of 18 chapters in NCERT Class 8 Science book. In the table below, we have provided the Important Questions for CBSE Class 8 Science for all the chapters. We have also included all types of questions, i.e. very short, short and long answers. These questions will cover all the crucial topics and the CBSE Class 8 Science syllabus.

How to Practise the CBSE Class 8 Science Important Questions

• Practise the question of a particular chapter in one go.
• Refrain from looking for the answer while solving the questions.
• Allocate a time duration for solving the questions and try to finish it accordingly.
• After completing all the questions, do a self-evaluation.
• Go through those questions again that could not be solved.

We, at BYJU’S, also provide other study materials such as CBSE Class 8 Science sample papers , notes, exemplars, solutions, etc.

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Some Natural Phenomena Class 8 Science Extra Questions and Answers

Some Natural Phenomena Class 8 Science Chapter 15 Extra Questions and Answers are provided here. We prepared these extra questions based on the latest NCERT Class 8 Science Book. CBSE Class 8 Science Chapter 15 Some Natural Phenomena Extra Questions will help you to properly understand a particular concept of the chapter.

Class 8 Science Chapter 15 Some Natural Phenomena Extra Questions

Very short answer type question.

Question 1: List three states in India where earthquakes are more likely to strike.

Answer: Gujarat, Kashmir, Rajasthan

Question 2: How will you charge a glass rod by the method of friction?

Answer: A glass rod can be charged by rubbing it with silk cloth.

Question 3: What are the two kinds of electric charges?

Answer: There are two kinds of charges — positive charge and negative charge.

Question 4: What is a flash of light which occurs in the sky during the rainy season called?

Answer: Lightning

Question 5: How a ballpoint pen refill can be charged by the method of friction?

Answer: A ballpoint pen refill can be charged by rubbing it against the woolen cloth.

Question 6: What are the three layers of earth?

Answer: Earth can be divided into three main layers: the core, the mantle and the crust.

Question 7: What happens when two like charges are brought together?

Answer: When two like charges are brought together then they will repel each other.

Question 8: What is the purpose of the Richter scale?

Answer: The Richter scale is used to measure the magnitude (or intensity) of an earthquake.

Question 9: What is directly below the epicenter of an earthquake?

Answer: Focus of an earthquake is located below the epicenter of an earthquake.

Question 10: What kind of electric charge is acquired by a plastic comb rubbed with dry hair?

Answer: The plastic comb acquires negative charge when it is rubbed with dry hair.

Question 11: How an inflated rubber balloon can be charged by the method of friction?

Answer: An inflated balloon can be charged by rubbing it against the woolen cloth.

Question 12: What is focus of an earthquake?

Answer: The place inside the earth’s crust where the earthquake begins is called focus of an earthquake.

Question 13: What is the instrument used to measure and record an earthquake?

Answer: Seismograph is the instrument which is used to measure and record an earthquake.

Question 14: What is epicentre of an earthquake?

Answer: The point on the earth’s surface directly above the focus is called epicentre of an earthquake.

Question 15: What type of electric charge is acquired by a rubber balloon when rubbed with a woolen cloth?

Answer: When balloon is rubbed with a woolen cloth, it acquires negative electric charge.

Question 16: What are seismic waves?

Answer: The tremors produce waves on the surface of the earth. These are called seismic waves.

Question 17: Which device is used to protect buildings from the effect of lightning?

Answer: Lightning Conductor is a device used to protect buildings from the effect of lightning.

Question 18: What is an earthquake?

Answer: An earthquake is a sudden shaking or trembling of the earth lasting for a very short time.

Question 19: What causes them to happen?

Answer: Earthquake is caused by a disturbance deep inside the earth’s crust.

Question 20: Name the scientist who showed that lightning and the spark from your clothes are essentially the same phenomena.

Answer: In 1752 Benjamin Franklin, an American scientist, showed that lightning and the spark from your clothes are essentially the same phenomena.

Short Answer Type Questions

Question 1: If scientists know so much about earthquakes, can they also predict when and where the next one will strike?

Answer: Although, we know for sure what causes an earthquake, it is not yet possible to predict when and where the next earthquake might occur.

Question 2: Other than earthquakes what causes tremors on the earth?

Answer: Tremors on the earth can also be caused when a volcano erupts, or a meteor hits the earth, or an underground nuclear explosion is carried out.

Question 3: A positively charged object attracts another charged object placed near it. What is the nature of charge on the other object?

Answer: The nature of charge on the other object will be negative. This is because like charges repel each other and unlike charges attract each other.

Question 4: Explain why a charged body loses its charge if we touch it with our hand.

Answer: When we touch a charged body with our hand, then the electric charge present on it flows to the earth through our hand and body. That is why a charged body loses its charge if we touch it with our hand.

Question 5: What was the magnitude of Bhuj and Kashmir earthquakes on the Richter scale?

Answer: Really destructive earthquakes have magnitudes higher than 7 on the Richter scale. Both Bhuj and Kashmir earthquakes had magnitudes greater than 7.5.

Question 6: Why does a plastic comb rubbed with dry hair attract tiny pieces of paper?

Answer: When a plastic comb is rubbed with dry hair, it acquires a small charge due to friction. The electrically charged comb then exerts an electric force on the tiny pieces of paper and attracts them.

Question 7:  Why is it not safe to stay under a tree during a thunderstorm?

Answer: During a thunderstorm, there is danger of lightning striking the tree and burning it up. This lightning can also pass through the body of the person standing under the tree and may kill him. Therefore, it is not wise to stand under a tree during a thunderstorm.

Question 8: When a charged refill is touched with the metal top of an electroscope, its aluminium leaves diverge. Give reason.

Answer: The aluminium foil strips receive the same charge from the charged refill through the paper clip. The strips carrying similar charges repel each other and they become wide open.

Question 9: What happens when plates brush past one another, or a plate goes under another due to collision?

Answer: When plates brush past one another, or a plate goes under another due to collision, they cause disturbance in the earth’s crust. It is this disturbance that shows up as an earthquake on the surface of the earth.

Question 10: The weather department has predicted that a thunderstorm is likely to occur on a certain day. Suppose you have to go out on that day. Would you carry an umbrella? Explain.

Answer: Carrying umbrella is not a good idea at all during thunderstorms because lightning may strike the top end of the metal rod of umbrella and harm us.

Question 11: Sometime, a crackling sound is heard while taking off sweater during winters. Explain.

Answer: When we take off a woolen (or synthetic) sweater, it rubs against our shirt. Due to rubbing, opposite electric charges develop on them, which attract each other. The discharge of these electric charges produces tiny sparks of light as well as crackling sound.

Question 12:   What will happen when the metal top of an electroscope is touched with a glass rod which has been rubbed with silk cloth? Give reason for your answer.

Answer: The aluminium foil strips receive the same charge from the charged glass rod through the paper clip. The strips carrying similar charges repel each other and they become wide open.

Question 13: What happens when you touch the paper clip of a self-made Electroscope? Why does it happen? What is this process called?

Answer: Every time the foil strips collapse as soon as we touch the paperclip with hand. The reason is that the foil strips lose charge to the earth through our body. The process of transferring of charge from a charged object to the earth is called earthing.

Long Answer Type Questions

Question 1:  What damage is caused by lightning?

Answer: When lightning strikes a tree, it can burn up the tree. And when a person is hit by lightning during a thunderstorm, then electric energy passes through the body of the person due to which the person get severe burns and get killed. Thus, it can cause a lot of destruction by damaging property, trees and killing people.

Question 2:  How can we determine if an object is charged or not?

Answer: An electroscope may be used to detect whether a body is charged or not. When the object is touched with the metal cap of an electroscope, both the metal cap and the leaves acquire the same charge from the charged object through the paper clip. It will cause the leaves to diverge showing that the object was charged.

Question 3:  What is lightning conductor? How does it work? Or How does a lightning conductor protect a tall building?

Answer: Lightning Conductor is a device used to protect buildings from the effect of lightning. A metallic rod, taller than the building, is installed in the walls of the building during its construction. One end of the rod is kept out in the air and the other is buried deep in the ground. The rod provides easy route for the transfer of electric charge to the ground.

Question 4:  What are the various effects of an earthquake?

Answer: The various effects of an earthquake are:

• They can cause immense damage to buildings, bridges, dams and people. There can be a great loss to life and property.
• The earthquakes can cause floods, landslides and tsunamis.
• A major tsunami occurred in the Indian Ocean on 26th December 2004. All the coastal areas around the ocean suffered huge losses.

Question 5: What will you do to protect yourself from lightning if you are in an open field?

Answer: If no shelter is available and we are in an open field, we should stay far away from all trees. We should stay away from poles or other metal objects. We should not lie on the ground instead we should squat low on the ground. We should place our hands on our knees with our head between the hands. This position will make us the smallest target to be struck.

Question 6: Explain why a charged balloon is repelled by another charged balloon whereas an uncharged balloon is attracted by another charged balloon?

Answer: A charged balloon is repelled by another charged balloon because both the balloons have same type of charges and like charges repel each other. Whereas an uncharged balloon is attracted by another charged balloon because charged balloon induces opposite charges in the nearer end of the uncharged balloon by electric induction and unlike charges attract each other.

Question 7: Suppose you are outside your home and an earthquake strikes. What precaution would you take to protect yourself?

Answer: If we are outside our home and an earthquake strikes, then

• We should try to find a clear spot, away from buildings, trees and overhead power lines. Drop to the ground.
• If we are in a car or a bus, we should not come out. We should ask the driver to drive slowly to a clear spot. We should not come out till the tremors stop.

Question 8: Name the scale on which the destructive energy of an earthquake is measured. An earthquake measures 3 on this scale. Would it be recorded by a seismograph? Is it likely to cause much damage?

Answer: The destructive energy of an earthquake is measured by Richter scale. An earthquake of 3 Richter can be recorded by a seismograph. An earthquake of magnitude 3 on Richter scale is often felt but not likely to cause much damage. Really destructive earthquakes have magnitudes higher than 7 on the Richter scale.

Question 9: What precautions would you take to protect yourself during an earthquake if you are at home?

Answer: In order to protect ourselves during an earthquake we must take following precautions:

• We should take shelter under a table and stay there till shaking stops.
• We should stay away from tall and heavy objects that may fall on us.
• If we are in bed then we should not get up and should protect our head with a pillow.

Question 10: Suggest three measures to protect ourselves from lightning.

Answer: Measures to protect ourselves from lightning

• A house or a building is a safe place. If we are travelling by car or by bus, we are safe inside with windows and doors of the vehicle shut.
• Lightning can strike telephone cords, electrical wires and metal pipes.      During a thunderstorm when lightning is taking place, contact with these should be avoided.
• Carrying umbrella is not a good idea at all during lighting because lightning may strike the top end of the metal rod of umbrella and harm us.

Question 11: What causes an earthquake?

Answer: Causes of an earthquake

• The outermost layer of the earth is not in one piece. It is fragmented. Each fragment is called a plate. These plates are in continual motion. When they brush past one another, or a plate goes under another due to collision, they cause disturbance in the earth’s crust. It is this disturbance that shows up as an earthquake on the surface of the earth.
• Tremors on the earth can also be caused when a volcano erupts, or a meteor hits the earth, or an underground nuclear explosion is carried out.

Question 12: What causes lightning in the sky?

Answer: During the development of a thunderstorm, the air currents move upward while the water droplets move downward. These vigorous movements cause separation of charges. The positive charges collect near the upper edges of the clouds and the negative charges accumulate near the lower edges. There is accumulation of positive charges near the ground also. When the magnitude of the accumulated charges becomes very large, the air which is normally a poor conductor of electricity is no longer able to resist their flow. Negative and positive charges meet, producing streaks of bright light and sound. We see streaks as lightning. The process is called an electric discharge.

Question 13: State some precautions that people should take to protect themselves against earthquakes if they live in seismic zones of the country.

Answer: People living in seismic zones should take the following measures for protection against earthquakes:

• All the buildings in these zones should be designed so that they can withstand major tremors. Modern building technology can make it possible.
• In highly seismic areas, the use of mud or timber is better than the heavy construction material. Keep roofs as light as possible. In case the structure falls, the damage will not be heavy.
• It is better if the cupboards and shelves are fixed to the walls, so that they do not fall easily.
• Wall clocks, photo-frames, water heaters etc., should be mounted securely on the walls, so that in the event of an earthquake, they do not fall on people.
• Since some buildings may catch fire due to an earthquake, it is necessary that all buildings, especially tall buildings, have fire fighting equipment in working order.

Question 14: Draw a labelled diagram of a seismograph. Explain its working.

Answer: The tremors produce waves on the surface of the earth. These are called seismic waves. The waves are recorded by an instrument called the seismograph. The instrument is simply a vibrating rod, or a pendulum, which starts vibrating when tremors occur. A pen is attached to the vibrating system. The pen records the seismic waves on a paper which moves under it. By studying these waves, scientists can construct a complete map of the earthquake. They can also estimate its power to cause destruction.

At Study Path, you can also learn more about Class 8 Science Chapter 15 Some Natural Phenomena by accessing the free exhaustive list of study materials and resources related to the chapter such as NCERT Solutions, Notes, Important Questions, and MCQ.

Case Study Questions Class 8 Science Reaching the age of adolescence

Case study questions class 8 science chapter 10 reaching the age of adolescence.

CBSE Class 8 Case Study Questions Science Reaching the age of adolescence. Important Case Study Questions for Class 8 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Reaching the age of adolescence.

CBSE Case Study Questions Class 8 Science Reaching the age of adolescence

Case study 1.

Que. 1) The protruding part of the throat usually seen in boys that is responsible for deep voice is known as?

(b) Swollen lymph

(a) Broader shoulder

Que. 2) (d) Region below waist wider

Que. 3) (a) Ovaries

Case study 2

Que. 1) Sex hormones are under the control of hormones from the ………………………………………………………………………………….. gland.

(c) Thyroid

(a) Superficial glands

(b) Doubtful glands

Case study 3

Que. 1) The first menstrual flow begins at puberty and is termed as ………………………………………………………………………………………. .

(d) Menarche

Que. 4) Explain in detail the process of menstruation.

Que. 4)Answer: In about 28-30 days, one ovum maturesand is released from the ovary. Walls of the uterus are thickened to receive the ovum. If fertilisation of that ovum does not occur, the released ovum along with the thickened lining and its blood vessels are shed off causing bleeding.

Que. 3) In a person suffering from Diabetes, the pancreas does not produce …………………………………………………………………………………….. hormone in sufficient quantity.

(d) Melatonin

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NCERT Solutions for Class 8 Science

Ncert solutions for class 8 science - free pdf download.

NCERT Solutions for Class 8 Science Chapterwise PDF

• NCERT Solutions for Chapter 1 - Crop Production and Management
• NCERT Solutions for Chapter 2 - Microorganisms: Friend and Foe
• NCERT Solutions for Chapter 3 - Synthetic Fibres and Plastics
• NCERT Solutions for Chapter 4 - Materials: Metals and Non-Metals
• NCERT Solutions for Chapter 5 - Coal and Petroleum
• NCERT Solutions for Chapter 6 - Combustion and Flame
• NCERT Solutions for Chapter 7 - Conservation of Plants and Animals
• NCERT Solutions for Chapter 8 - Cell: Structure and Functions
• NCERT Solutions for Chapter 9 - Reproduction in Animals
• NCERT Solutions for Chapter 10 - Reaching the Age of Adolescence
• NCERT Solutions for Chapter 11 - Force and Pressure
• NCERT Solutions for Chapter 12 - Friction
• NCERT Solutions for Chapter 13 - Sound
• NCERT Solutions for Chapter 14 - Chemical Effects of Electric Current
• NCERT Solutions for Chapter 15 - Some Natural Phenomena
• NCERT Solutions for Chapter 16 - Light
• NCERT Solutions for Chapter 17 - Stars and The Solar System
• NCERT Solutions for Chapter 18 - Pollution of Air and Water

Why Class 8 NCERT Solutions of Science are important

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Motion Class 9 Notes CBSE Science Chapter 8 (Free PDF Download)

CBSE Class 9 Science Chapter 8 - Motion Revision Notes - Free PDF Download

The theories and principles taught in NCERT chapter 8 of motion for class 9 are explained in motion chapter class 9 notes. When the position of a body changes in reference to a stationary object, it is said to be in motion. Understanding higher-level physics subjects requires a comprehension of motion. There are several types of motions, and the phenomenon of motion is regulated by definite principles. As part of the Class 9 science chapter 8 notes, we will go over all of this in great detail.

Vedantu also provides free NCERT Solutions to all the students. You can download Class 9 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations. 

Download free NCERT Solutions Class 9 Maths to amp up your preparations and to score well in your examinations.

Download CBSE Class 9 Science Revision Notes 2024-25 PDF

Also, check CBSE Class 9 Science revision notes for All chapters:

Access CBSE Class 9 Science Chapter 8 - Motion Notes

Introduction: 

One of the most common phenomena in the physical world is motion. Mechanics is the branch of Physics that deals with the behavior of moving objects.

Mechanics is divided further into two sections: Kinematics and Dynamics.

Kinematics is the study of motion without regard for the cause of motion.

Dynamics is concerned with the source of motion, which is force. 

Motion and Rest: 

An object is said to be in motion if its position in relation to its surroundings changes in a given time.

An object is said to be at rest if its position in relation to its surroundings does not change.

A frame of reference is another object or scene against which we compare the position of an object.

Take a look at the numbers. Figure 1 shows the car to the right of the tree. Figure 2 shows the car to the left of the tree after 2 seconds. The car must have moved from one location to another because the tree does not move. As a result, the tree serves as the frame of reference in this case. 

Types of Motion: 

There are three types of motion: 

Translatory motion 

Rotatory motion 

Vibratory motion 

Translatory Motion: 

A particle in translatory motion moves from one point in space to another. This movement may be in a straight line or in a curved path.

Rectilinear motion is defined as motion along a straight line.

Curvilinear motion is defined as movement along a curved path.

As an example, consider a car driving down a straight road.

Rectilinear Motion: 

Example: A car negotiating a curve 

Curvilinear Motion: 

Rotatory Motion: 

The particles of the body describe concentric circles around the axis of motion in rotatory motion.

Vibrational Motion:

Particles in vibratory motion move back and forth around a fixed point.

Distance and Displacement: 

The distance between termini A and B is 150 kilometers. A bus connects Terminus A and Terminus B. The bus travels a distance of 150 kilometers. The bus returns from terminus B to terminus A along the same route. As a result, the total distance traveled by the bus from A to B and then from B to A is 150 km + 150 km = 300 km. 

A bus traveling from point A to point B and back again.

The distance traveled by a moving object is the length of the path the object takes.

The measure of distance is a scalar quantity. The meter is the SI unit of distance.

The bus's position changed when it moved from Terminus A to Terminus B. The distance between A and B is 150 kilometers. The distance traveled on the return trip is also 150 kilometers.

Displacement is the shortest path covered by a moving object in a specified direction from the point of reference (the initial position of the body).

Note:  

However, the displacement when the bus moves from A B to B B is zero. The meter is the SI unit of displacement.

Displacement is a vector, which means that it is represented by a number with appropriate units and direction.

To emphasize the distinction between displacement and distance, consider a few more examples.

Assume a person moves 3 meters from point A to point B and 4 meters from point B to point C, as shown in the figure. He has travelled a total distance of 7 meters. But is he really 7 meters away from his starting point? No, he is only 5 meters away from his initial position, implying that he is displaced by the shortest distance between his initial and final positions.

Distance and Displacement  

To determine the displacement in this example, we can use Pythagoras' theorem. Consider an object that is changing its position with respect to a fixed point known as the origin 0. 

$\mathrm{x}_{\mathrm{i}}$ and $\mathrm{x}_{\mathrm{f}}$ are the initial and final positions of the object. Then the displacement of the

object $=x_{f}-x_{i}$

Suppose the object is travelling from $+1$ to $+4$, then displacement

$=\mathrm{x}_{\mathrm{f}}-\mathrm{x}_{\mathrm{i}}$

Displacement: Case 1 

Case 2:  

If the object is travelling from -3 to -1, then displacement 

If the object is travelling from $-3$ to $-1$, then displacement

$=x_{f}-x_{i}$

Displacement: Case 2 

Case 3:  

If the object is travelling from +4 to +2, then displacement 

Displacement: Case 3

Case 4:  

If the object follows the path depicted in the figure, the final and initial positions are the same, implying that the displacement is zero.

Displacement: Case 4 

We can conclude from the preceding examples that a body's displacement is positive if its final position is on the right side of its initial position and negative if its final position is on the left side of its initial position. The displacement of a moving object is said to be zero when it returns to its original position. Consider an athlete running in a clockwise direction along a circular track with radius r, beginning at A.

A Circular Track of Radius r 

What is the athlete's total distance traveled when he arrives at point B?

The athlete's total distance traveled when he arrives at point B equals to half of the circumference of the circular track, that is, $\frac{2 \pi r}{2}=\pi r$.

Displacement $=A B=2 r=$ Diameter of circle (the shortest distance between the initial and final positions).

If the athlete arrives at the starting point $A$, the distance covered is equal to the circumference of the circular track, i.e., $2 \pi r .$ However, the displacement is zero because the athlete's initial and final positions are the same.

Difference between Distance and Displacement

Motion: 

Uniform Motion and Non-uniform Motion: 

The distances covered by car A and car B with respect to time is given below:

Car A: 

Car B: 

The car A travels equal distances in equal time intervals, whereas the car B does not travel equal distances in equal time intervals. That is, car A's motion is an example of uniform motion, whereas car B's motion is an example of non-uniform motion.

A body is said to describe uniform motion when it covers equal distances in equal intervals of time.

When a body moves unequal distances in equal time intervals, or vice versa, this is referred to as non-uniform motion.

Speed: 

Ram and Krishna compete in various races over varying distances. Ram covers $1000 \mathrm{~m}$ in 20 minutes and Krishna covers $700 \mathrm{~m}$ in 10 minutes. Who is the fastest?

To determine who is faster, we will calculate the distance they cover in one minute.

Distance covered by Ram in one minute $=\frac{1000 \mathrm{~m}}{20 \mathrm{~min}}=500 \mathrm{~m} / \mathrm{min}$

Distance covered by Krishna in one minute $=\frac{700 \mathrm{~m}}{10 \mathrm{~min}}=70 \mathrm{~m} / \mathrm{min}$

Krishna covered more ground in the same amount of time. We conclude that Krishna is the faster of the two. 

Speed is defined as the distance traveled by a moving object in one unit of time.

$\text { speed }=\frac{\text { distance }}{\text { time }}=\frac{\mathrm{S}}{\mathrm{t}}$

Where S denotes the distance traveled and t denotes the time spent.

The SI unit of speed is millimeters per second (m/s). Speed is defined as a scalar quantity.

Uniform Speed:

The graph depicts the distance traveled by a ball every 2 seconds.

Every 2 seconds, the ball travels 10 meters. At any point between A and E, the ball moves at a speed of 5 m/s. The object is moving at a constant speed.

If an object travels the same distance in the same amount of time, it is said to be moving at a uniform speed.

Surface friction or resistance is ignored in this case.

Variable Speed or Non-Uniform Speed: 

The distance covered varies with time.

Variable Speed:  

For example, when a rubber ball is dropped from a certain height (h 1 ), it bounces up to a height less than the initial one (h 2 ). It keeps bouncing, but the height to which it rises keeps decreasing (h 3 , h 4 ). The ball's distance traveled per unit time decreases. The ball's speed varies from point to point. This type of speed is known as variable speed.

Average Speed and Instantaneous Speed:

When we travel by car, the speed varies depending on the road conditions at the time. The speed is calculated in this case by dividing the total distance traveled by the vehicle by the total time required for the journey. This is known as the average speed. 

The average speed of an object traveling S 1 in time t 1 , S 2 in time t 2 , and S n in time t n is given by,

$\text { Average speed }=\frac{\mathrm{S}_{1}+\mathrm{S}_{2}+\mathrm{S}_{3}+\ldots+\mathrm{S}_{\mathrm{n}}}{\mathrm{t}_{1}+\mathrm{t}_{2}+\mathrm{t}_{3}+\ldots+\mathrm{t}_{\mathrm{n}}}$

When we say that the car travels at an average speed of 60 km/h, we do not mean that it will travel at that speed for the duration of the journey. The actual speed of the car may be less than or greater than the average speed at a given location.

The speed of a moving body at any given point in time is referred to as instantaneous speed.

The diagram below depicts the various routes Shyam can take from his house to school. 

Shyam drives himself to school every day, averaging 60 km/h. Is it possible to find out how long it will take to get to the destination? Yes, you can use the relation to determine the time.

$\text { speed }=\frac{\text { distance }}{\text { time }}$

But you don't know what path he would have taken. As a result, simply providing the speed of a moving object does not allow one to determine the exact position of the object at any given time. As a result, there is a need to define a quantity that has both magnitude and direction.

Starting with A, consider two objects P and Q. Allow them to travel equal distances in equal time intervals, i.e. at the same speed. Can you guess where each of them will be in 20 seconds? P and Q are free to move in any direction. To determine the exact position of P and Q, we must also know their direction of motion.

Pictorial Representation of the Position of the Objects P and Q:

As a result, another physical quantity known as velocity is introduced to provide us with an idea of both speed and direction.

Velocity is defined as the distance traveled in a given direction by a moving object in a given time or speed in a given direction.

$\operatorname{velocity}(v)=\frac{\text { distance travelled in a specified direction }(s)}{\text { time taken }(t)}$

Note: 

Velocity is defined as the distance traveled in a given direction in a given amount of time. Displacement is the distance traveled in a specified direction.

As a result, velocity can be defined as the rate at which displacement changes.

Uniform Velocity and Non-Uniform Velocity:

Assume that two athletes, Ram and Shyam, are running at a constant speed of 5 m/s. Ram moves in a straight line, while Shyam follows a circular path. For a layperson, both Ram and Shyam are moving with uniform velocity, but for a physicist, only Ram is running with uniform velocity because his speed and direction of motion do not change.

In the case of Shyam, who is running on a circular track, the direction of motion changes at every instant because a circle is a polygon with infinite sides, and Shyam must change his direction at every instant.

A body is said to be moving with uniform velocity if it travels the same distance in the same amount of time in the same direction.

A body is said to be moving with variable velocity if it covers unequal distances in equal intervals of time and vice versa in a specified direction, or if its direction of motion changes.

Acceleration:

We are all aware that a car moving down the road does not have a uniform velocity. Either the speed or the direction of travel shifts. We say that a vehicle is accelerating when it is speeding up, i.e. when the speed increases.

Let us look at the change in velocity of a train traveling from Bangalore to Mysore to get an idea of acceleration. The train, which was initially at rest, begins to move; its velocity gradually increases until it reaches a constant velocity after a certain time interval. As the train approaches the next station, its speed gradually decreases until it comes to a halt.

When a train starts from a stop, its speed increases from zero, and we say it is accelerating. After a while, the speed becomes uniform, and we say that the train is moving at a uniform speed, which means that it is not accelerating. However, as the train approaches Mysore, it slows down, indicating that the train is accelerating in the opposite direction. When the train comes to a halt in Mysore, it stops accelerating once more.

As a result, it is clear that the term "acceleration" does not always imply that the speed of a moving body increases; it can also decrease, remain constant, or become zero.

In general, acceleration is defined as the rate at which the velocity of a moving body changes over time.

This change could be a change in the object's speed, direction of motion, or both.

Let us now look up a mathematical formula for calculating acceleration.

If an object moves with an initial velocity 'u' and reaches a final velocity 'v' in time 't,' then the acceleration 'a' produced by the object is

Acceleration = Rate at which velocity changes over time.

$a=\frac{v-u}{t}$

Unit of Acceleration:

The SI unit of acceleration is m/s 2 and it is a vector quantity.

Different Types of Acceleration:  

It is clear from the preceding example that acceleration takes various forms depending on the change in velocity.

Positive acceleration:

When an object's velocity increases, it is said to be moving with positive acceleration.

Positive Acceleration 

Example: a ball rolling downhill on an inclined plane.

Negative acceleration:

When an object's velocity decreases, it is said to be moving with negative acceleration. Negative acceleration can also be referred to as retardation or deceleration.

(1) A ball moving up an inclined plane.

(2) A vertically thrown upwards ball has a negative acceleration as its velocity decreases over time.

Zero Acceleration:  

If the change in velocity is zero, indicating that the object is either at rest or moving at uniform velocity, the object is said to have zero acceleration.

A parked car, for example, or a train moving at a constant speed of 90 km/hr.

Uniform Acceleration:  

The object is said to be moving with uniform acceleration if the change in velocity at equal intervals of time is always the same.

As an example, consider a body falling from a great height towards the earth's surface.

Non-uniform or Variable Acceleration:  

If the change in velocity over equal time intervals is not the same, the object is said to be moving with variable acceleration.

Motion:  

Distance-Time Table and Distance-Time Graph: 

Mr. X is taking a bus from New Delhi to Agra and recording his observations.

According to the table above, the bus travels equal distances at equal times. The bus is moving at a constant speed. In such a case, we can compute the distance traveled by the bus at any given point in time.

Consider an object moving from its initial position x i to its final position x f in time t at a uniform speed v. 

$\text { uniform speed }=\frac{\text { total distance }}{\text { time taken }}$

$v=\frac{x_{f}-x_{i}}{t}$

$x_{f}-x_{i}=v t \cdots \cdots(1)$

The relationship between distance, time, and average speed is given by equation (1). This relationship can be used to generate distance-time tables as well as to determine the position of any moving object at any given time. However, it is a time-consuming and tedious process, especially when we need to determine the position after a long period of time or compare the motion of two objects. In such cases, graphs such as the distance-time graph can be useful. A distance-time graph is a line graph that shows how distance changes over time. A distance-time graph plots time along the x-axis and distance along the y-axis. 

Distance-Time Graph for Non - Uniform Motion  

Let us now look at the nature of a distance-time graph for a non-uniform motion. The distance traveled by a bus every 15 minutes is shown in the table below.

We can deduce from the above table that the motion is non-uniform, i.e. it covers unequal distances in equal time intervals.

Measure time along the x-axis and distance along the y-axis.

Analyze the provided data and select the appropriate scale for time and distance.

Plot the points.

Join the points.

Consider any two points (A, B) on the graph.

Draw perpendicular from A to B to x and y axes.

Join A to C to get a right angled ACB.

The slope of the graph is shown below.

$\mathrm{AB} =\frac{\mathrm{BC}}{\mathrm{AC}}$

$=\frac{\mathrm{S}}{\mathrm{t}}$

$=\mathrm{speed}$

Write the title and scale chosen for the graph.

$\text { speed }=\frac{15-5}{30-15}$

$=\frac{10}{15}$

$=\frac{2}{3}$

$=0.666 \mathrm{~km} / \mathrm{min}$

Consider another two points on the graph, P and Q, and draw a right-angled triangle PRQ.

$\text { slope }=\text { speed }$

$\mathrm{PQ}=\frac{\mathrm{QR}}{\mathrm{PR}}$

$=\frac{35-30}{90-75}$

$=\frac{5}{15}$

$=0.333 \mathrm{Km} / \mathrm{min}$

Complete Graph

Nature of S- t Graph for Non- Uniform Motion and Uses of Graphs

Let us now see the nature of S-t graph for non-uniform motion. 

Nature of s-t Graph for Non-Uniform Motion: 

Figure (a) depicts the S-t graph as the speed of a moving object increases, while Figure (b) depicts the S-t graph as the speed of a moving object decreases. The nature of the S-t graph allows us to determine whether the object is moving at a constant or variable speed.

Uses of Graphical Representation:  

Because it provides a visual representation of two quantities, graphical representation is more informative than tables (e.g., distance vs. time)

A graph provides more information than a table at a glance. Both of the graphs shown here depict increasing speed.

Figure (1) depicts the nature of the variation in speed, indicating that the increase is greater in the beginning up to time t 1 and relatively lower after t 2 .

Similarly, fig (2) depicts how the increase in speed becomes greater after t1. A similar explanation applies to the decreasing speed.

Graphs are simple to read at a glance.

Graph plotting takes less time and is more convenient.

Graphs can be used to determine the position of any moving object at any point in time.

Two moving objects' motions can be easily compared.

Graphs reveal information about the nature of motion.

Velocity-Time Graph:  

The variation of velocity with time can be graphically represented to calculate acceleration in the same way that we calculated speed from the distance-time graph.

Let us now create a velocity-time(v-t) graph using the data below.

Draw time on the x-axis and velocity on the y-axis. 

Analyze the provided data and select the appropriate scale for the x and y axes.

Plot the Given Points.

Join the Points

Consider Any Two Points A and B on the Straight-Line Graph.

Draw Perpendiculars from A and B to x and y-axes.

Join A to C, ACB forms a Right-Angled Triangle.

Slope of the Graph

$\mathrm{AB}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\text { Change in velocity }}{\text { time }}=\text { acceleration }$

Calculations:

$\text { Acceleration }=\frac{30-20}{6-4}$

$=\frac{10}{2}$

$=5 \mathrm{~m} / \mathrm{s}^{2}$

Write the title for the graph.

V - T Graph:

Let us now examine the nature of the v - t graph for various types of motion.

a) Increasing Acceleration:

Uniform Acceleration

Non-Uniform Acceleration

(b) Decreasing Acceleration:

Non-uniform Retardation

Uniform Retardation

Zero Acceleration

Uses of Velocity-time Graphs  

The velocity-time graph can be used to derive the following results. 

The acceleration produced in a body. 

The distance traveled by a moving object. 

The equations of motion.

Speed - Time Graph

To compute the distance traveled by a moving object using a speed-time graph. 

The graph below depicts the speed-time graph of a car traveling at a constant speed of 60 km/h for 5 hours .

Speed-Time Graph of a Car Moving with Uniform Speed 

Distance travelled by the car,

Distance travelled by the car

$(S)=v \times t$

$=60 \times 5$

$=300 \mathrm{~km}$

But $60 \mathrm{~km} / \mathrm{h}=\mathrm{OC}=$ breadth of the rectangle $\mathrm{OABC}$

$5 h=O A=$ length of the rectangle $O A B C$

i.e., the distance covered by the car = length $\times$ breadth $=300 \mathrm{~km}$.

To calculate the distance traveled by a moving object using a speed-time graph, find the area enclosed by the speed-time graph and the time axis. In the case of non-uniform motion, the distance covered by the object increases in steps as the object's speed increases. During the time intervals $0-t_{1}, t_{1}-t_{2}, t_{2}-t_{3}, \ldots \ldots$, the speed remains constant.

The motion of an object moving at a variable speed is depicted in the figure below.

Speed - Time Graph for an Object Moving with Variable Speed

Calculation of Distance:

The object's total distance traveled during the time interval. 

0- t 6 = Area of rectangle 1 + area of rectangle 2 + …… + area of rectangle 6. 

Motion 

Equations of Motion  

Time, speed, distance covered, and acceleration are the variables in a uniformly accelerated rectilinear motion. These quantities have simple relationships. These relationships are expressed using equations known as equations of motion. 

The equations of motion are: 

(1) $v=u+a t$

(2) $S=u t+\dfrac{1}{2} a t^{2}$

(3) $v^{2}-u^{2}=2 a S$

Derivation of the First Equation of Motion

Consider a particle moving in a straight line with a constant acceleration ‘a’. Let the particle be at A at t=0, and u be its initial velocity, and v be its final velocity at t=t.

$a=\dfrac{v-u}{t}$ 

$v-u=a t$ 

I Equation of Motion

Second Equation of Motion

$\text { Average velocity } =\dfrac{\text { total distance travelled }}{\text { total time taken }}$

$=\dfrac{S}{t} \ldots \ldots(1)$

Average velocity can alsobe written as

$\frac{u+v}{2} \cdots \cdots(2)$

From equation (1) and (2),

$\frac{S}{t}=\frac{u+v}{2} \cdots \cdots(3)$

The first equation of motion is $v=u+a t$. Substituting the value of $v$ in equation (3), we get

$\frac{S}{t}=\frac{u+u+a t}{2}=\frac{(u+u+a t) t}{2}$

$\quad=\frac{(2 u+a t) t}{2}$

$S=u t+\frac{1}{2} a t^{2}$

II equation of motion

Third Equation of Motion

The first equation of motion is

$v-u=a t \ldots(1)$

Average velocity $=\frac{S}{t} \cdots \cdots(2)$

Average velocity $=\frac{u+v}{t} \cdots \cdot(3)$

From equation (2) and equation (3) we get,

$\frac{u+v}{t}=\frac{S}{t} \cdots \ldots(4)$

Multiplying equation (1) and equation (4) we get,

$(v-u)(v+u)=a t \times \frac{2 S}{t}$

$(v-u)(v+u)=2 a S$

$\left(v^{2}-u^{2}\right)=2 a S$

III equation of motion

Derivations of Equations of Motion (Graphically) 

First Equation of Motion

Consider an object moving in a straight line with a uniform velocity u. When its initial velocity is u, give it a uniform acceleration an at time t = 0. The object's velocity increases as a result of the acceleration to v (final velocity) in time t, and S is the distance covered by the object in time t.

The graph depicts the velocity-time graph of the object's motion.

The acceleration of a moving object is given by the slope of the v - t graph.

Thus, acceleration = slope

$\mathrm{AB}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{v-u}{t-0}$

I equation of motion

Second Equation of Motion  

Let u be an object's initial velocity and 'a' be the acceleration produced in the body. The area enclosed by the velocity-time graph for the time interval 0 to t gives the distance travelled S in time t.

Graphical Derivation of Second Equation

Distance travelled S = area of the trapezium ABDO

Distance travelled $S=$ area of the trapezium ABDO

$=$ area of reactangle $A C D O+$ area of $\triangle A B C$

$=\mathrm{OD} \times \mathrm{OA}+\frac{1}{2} \mathrm{BC} \times \mathrm{AC}$

$=\mathrm{t} \times \mathrm{u}+\frac{1}{2}(\mathrm{v}-\mathrm{u}) \times \mathrm{t}$

$=\mathrm{ut}+\frac{1}{2}(v-u) \times t$

$v=u+$ at $\mid$ equation of motion; $v-u=a t$

$S=u t+\frac{1}{2} a t \times t$

$=u t+\frac{1}{2} a t^{2}$

Third Equation of Motion  

Let 'u' be an object's initial velocity and a be the acceleration produced in the body. The area enclosed by the v - t graph gives the distance travelled 'S' in time 't'.

Graphical Derivation of Third Equation

$\mathrm{S}=$ area of trapezium $\mathrm{OABD}$

$=\frac{1}{2}\left(b_{1}+b_{2}\right) h$

$=\frac{1}{2}(\mathrm{OA}+\mathrm{BD}) \mathrm{AC}$

$=\frac{1}{2}(u+v) t \cdots \cdots(1)$

But we know that $a=\frac{v-u}{t}$ or $t=\frac{v-u}{a}$

Substituting the value of $t$ in equation (1) we get,

$S=\frac{1}{2} \frac{(u+v)(v-u)}{a}$

$=\frac{1}{2} \frac{(v+u)(v-u)}{a}$

$2 a s=(v+u)(v-u)$

$(v+u)(v-u)=2 a s$

$v^{2}-u^{2}=2 a s$

III Equation of Motion 

Circular Motion 

We classified motion along a circular track as non-uniform motion in the example discussed under the topic uniform and non-uniform motion. Let's take a look at why circular motion is considered non-uniform motion. The figure depicts an athlete running at a constant speed on a hexagonal track.

Athlete Running on a Regular Hexagonal Track 

The athlete runs at a constant speed along the track segments AB, BC, CD, DE, EF, and FA, and at the turns, he quickly changes direction to stay on the track without changing his speed. Similarly, if the track had been a regular octagon, the athlete would have had to change directions eight times in order to stay on the track.

Athlete Running on a Regular Octagonal Track 

The athlete must turn more frequently as the number of sides of the track increases. If we increase the number of sides indefinitely, the track will take on the shape of a circle. As a result, because a circle is a polygon with infinite sides, motion along a circular path is classified as non-uniform motion.

Athlete Running on a Circular Track 

Thus, an object moving at uniform speed along a circular track is an example of non-uniform motion because the object's direction of motion changes at every instant of time. 

Examples of Uniform Circular Motion 

(1) A car negotiating a curve at a constant speed.

(2) An athlete spinning a hammer in a circle before throwing it.

(3) An aircraft looping the loop.

Expression for Linear Velocity

If an athlete takes t seconds to complete one circular path of radius r, then the velocity v is given by the relation,

$\mathrm{v}=\frac{\text { distance travelled }}{\text { time }}$

Distance travelled = circumference of the circle

$=2\pi\mathrm{r}$

Linear velocity $=\frac{\text { 2r }}{\mathrm{t}}$

Class 9 Science Chapter 8 Revision Notes - Free PDF Download

Contents of class 9 science motion notes.

The Class 9 science notes chapter 8 contains detailed discussions on the topic of motion. It tells you about the basic definition of motion and rest. Then the chapter discusses the different types of motion. Next comes distance and displacement. 

With all this knowledge in your hand, we thereby venture to find out the speed and velocities of different objects. We also look into acceleration and its types. 

The notes also contain worked-out examples of numerical problems and other questions to give us a clear idea of the type of questions we can expect in the examination.

Subtopics Covered in Class 9 Chapter 8 Science Notes

The subtopics covered in class 9 science ch 8 notes are:

Motion and Rest: The change in position of an object with time and with respect to a stationary object is moving. If a body undergoes no change in position with time and with respect to a stationary object then it is said to be at rest

Types of Motion: Motion is divided into the following types:

Translation Motion: Motion along a straight line or curved path

Vibrational Motion: Motion of a particle to and from fixed positions.

Rotatory Motion: The circular motion of an object around the axis of rotation.

Distance and Displacement: Distance is defined as the total length of the path traveled by an object. Displacement on the other hand is the net effective change in position. In this section, we shall also look into various examples of numerical problems that aim at finding the distance and displacement of various objects.

Uniform and Non-Uniform Motion: Motion is said to be uniform when the rate of change of position remains the same across intervals multiple times with respect to the same object. Non-uniform motion sees different rates of change of position with respect to the object of reference.

Speed: Uniform And Variable: Speed is defined as the rate of change of position. It can either be uniform or variable.

Velocity: The velocity of an object is defined as the rate of change of displacement.

Acceleration: Acceleration is the rate of change of velocity with respect to time.

Basics of Plotting a Graph: Graphs are basically illustrations that are used to show the rate of change of a character. A graph is made by the intersection of two mutually perpendicular lines. Each of these lines is called the axis. 

How to work out time vs distance or displacement

Laws of Motion: Motion is governed by laws of physics known as Newton’s law of motion or simply the laws of motion. 

Ch 8 science class 9 notes are very important from the point of view of examinations as the chapter of motion has a net weightage of 9. Apart from that this chapter is especially important for candidates who want to pursue science in higher classes.

Why Choose Notes of ch 8 Science Class 9 by Vedantu?

Vedantu's class 9th science chapter 8 notes are the greatest study companion available. Physics may be a difficult topic to master, especially on your own. Vedantu's Motion class 9 notes are designed just for students. Despite the fact that it is developed by specialists, the language used is relatively easy. The subject of motion is both wide and vital. It is critical for students to understand as much as they can about motion without becoming overburdened. This is why Vedantu's notes give the ideal blend of study material that is both easy and dense with knowledge. Some advantages of utilising Vedantu include:

It is easy to access

It is easy to practice from

It is easy to check your answers and compare them with the solved ones

It is free to download from the official website of Vedantu

• It has solved numerical problems in it to help improve your understanding

The chapter 'Motion' may provide pupils with a wealth of information. All of the major topics are discussed here. The students are required to understand the important coverage so that they can prepare accordingly, this can be done by studying these notes from the chapter 'Motion'.

The students can also download the free pdf which is mentioned in this article. 

FAQs on Motion Class 9 Notes CBSE Science Chapter 8 (Free PDF Download)

1. When can an object have a zero displacement according to what you have studied in Notes of the Chapter Motion?

If an item goes along a certain path and then returns to its original position, the associated displacement of that object is zero. The length of the path will be travelled, but because the beginning and ending points of the movement are the same, the displacement will be zero.

2. Explain the type of motions according to what you have studied in Notes of the Chapter Motion.

Motion is the term given to the change in position of an object with respect to time. There are three types of motions:

Rotatory Motion: This takes place in a circular path when the object moves around the axis of rotation.

Vibrational Motion: This occurs when the object moves to and from a fixed position.

Translation Motion: This is caused by an object moving along a curved path or a straight line.

3. What are the subtopics covered in Class 9 Science Chapter 8 notes?

The CBSE Class 9 Science Chapter 8 - Motion Revision Notes provide well-explained details on each topic that has been covered in the NCERT and is a part of the latest syllabus provided by CBSE for Class 9 Science. These subtopics include the following:

Motion and Rest

Types of Motion

Distance and Displacement

Uniform and Non-uniform Motion

Acceleration

Basics of Plotting a Graph

Newton’s Laws of Motion

4. What are the benefits of referring to Vedantu’s Class 9 Science Chapter 8 notes?

Referring to CBSE Class 9 Science Chapter 8 - Motion Revision Notes by Vedantu can provide students with several benefits like:

Providing one of the best study materials for preparation

Save time by providing notes that are ready for studying

Help in increasing clarity about the different concepts taught in chapter 8

Help in understanding and solving numerical that are an important part of this chapter

Free of cost PDF download for studying offline

5. What are the Laws of Motion discussed in Chapter 8 Class 9 Science?

Newton established three laws of motion. According to the first law of motion, an object at rest or in uniform motion will remain in that condition until an external force is applied to it. Newton's second law states that the acceleration of an item is proportional to the force applied to it and its mass. According to the third and final law, every action has an equal and opposite response. Students may learn more by visiting the Vedantu website or app.

STUDY MATERIALS FOR CLASS 9

• Algebraic Expressions and Identities Class 8 Case Study Questions Maths Chapter 8

Last Updated on August 25, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 8 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 8 maths. In this article, you will find case study questions for CBSE Class 8 Maths Chapter 8 Algebraic Expressions and Identities. It is a part of Case Study Questions for CBSE Class 8 Maths Series.

Table of Contents

Case Study Questions on Algebraic Expressions and Identities

A playground is in shape of a square. The area of the square PQRS is 256 m 2 with each side (x + 2) m. One day Suraj along with his two friends Ajay and Aman went to play there with bicycle. Someone stole Suraj bicycle, but Ajay and Aman helped him by contributing ₹(4a + 60) and ₹(6a + 10) respectively, to buy a new bicycle. The cost of bicycle was ₹4200.

On basis of this information given in passage answer following questions.

Q. 1. Find the value of x. (a) 16 (b) 18 (c) 14 (d) 12

Difficulty Level: Easy

Ans. Option (c) is correct. Explanation: Area of square = 256 m 2 (x + 2) 2 = 162 x + 2 = 16 x = 16 – 2 x = 14

Q. 2. Find the side of square quare-shaped ground? (a) 19 (b) 12 (c) 18 (d) 16

Ans. Option (d) is correct. Explanation: x = 14 Therefore, side of square = (x + 2) = 14 + 2 = 16

Q. 3. What is the value of a? (a) 410 (b) 403 (c) 413 (d) 423

Difficulty Level: Medium

Ans. Option (C) is correct. Explanation: (4a + 60) + (6a + 10) = 4200 4a + 60 + 6a + 10 = 4200 10a + 70 = 4200 10a = 4130 Hence, a = 413

Q. 4. What was the amount given by Ajay and Aman to Suraj?

Sol. Money contributed by Ajay = (4a + 60) Money contributed by Aman = (6a + 10) Total amount = ₹4200 4a + 60 + 6a + 10 = 4200 10a + 70 = 4200 a = 413 Money contributed by Ajay = 4 × 413 + 60 = 1652 + 60 = ₹1712 Money contributed by Aman = 6 × 413 + 10 = 2478 + 10 = ₹2488

Q. 5. What is the perimeter of the playground?

Sol. Perimeter of square shaped playground = 4 × (x + 2) x = 14 Substituting value of x inside, we get = 4 × (14 + 2) Perimeter = 4 × 16 = 64 m

• Comparing Quantities Class 8 Case Study Questions Maths Chapter 7
• Cube and Cube Roots Class 8 Case Study Questions Maths Chapter 6
• Square and Square Roots Class 8 Case Study Questions Maths Chapter 5
• Data Handling Class 8 Case Study Questions Maths Chapter 4
• Understanding Quadrilaterals Class 8 Case Study Questions Maths Chapter 3

Linear Equations in One Variable Class 8 Case Study Questions Maths Chapter 2

Rational numbers class 8 case study questions maths chapter 1, download ebooks for cbse class 8 maths.

• Rational Numbers Topicwise Worksheet for CBSE Class 8 Maths
• Linear Equations in One Variable Worksheet for CBSE Class 8 Maths
• Understanding Quadrilaterals Worksheet for CBSE Class 8 Maths
• Data Handling Worksheet for CBSE Class 8 Maths
• Squares and Square Roots Worksheet for CBSE Class 8 Maths
• Cube and Cube Roots Worksheet for CBSE Class 8 Maths
• Comparing Quantities Worksheet for CBSE Class 8 Maths
• Algebraic Expressions and Identities Worksheet for CBSE Class 8 Maths

Topics from which case study questions may be asked

• Multiplication of Algebraic Expressions
• Multiplying a Monomial by a Polynomial
• Multiplying a Polynomial by a Polynomial

Addition and Subtraction of Algebraic Expressions: While adding or subtracting polynomials, first look for like terms and then add or subtract these terms, then handle the unlike terms.

Multiplication of Algebraic Expressions (i) Multiplying two monomials (ii) Multiplying three or more monomials

Multiplying a monomial by a polynomial (i) Multiplying a monomial by a binomial (ii) Multiplying a monomial by a trinomial

Multiplying a polynomial by a polynomial (i) Multiplying a binomial by a binomial (ii) Multiplying a binomial by a trinomial.

Subtraction of a number is the same as addition of its additive inverse.

Case study questions from the above given topic may be asked.

Important Keywords

Like term: Terms having same literal factors are called like terms e.g., 7ab 2 , 2ab 2 , 3b 2 a are like terms. Monomial: An algebraic expression that contains only one term is called monomial. e.g., –6abc, 5xy etc Binomial: An algebraic expression that contains two terms are called Binomial. e.g., x + 5 etc Trinomial: An algebraic expression that contains three terms are called trinomial. e.g., 2x – 7y + 4 etc Polynomial: An algebraic expression that contains one or more terms are called Polynomial. e.g., 7xy + 2z + 3y + 4a

Frequently Asked Questions (FAQs) on Algebraic Expressions and Identities Case Study

Q1: what are algebraic expressions.

A1: Algebraic expressions are mathematical phrases that include numbers, variables (like x, y), and operations (like addition, subtraction, multiplication, and division). They represent a quantity and can include constants, coefficients, and variables. For example, 3x+4 is an algebraic expression.

Q2: What is the difference between a term and a coefficient in an algebraic expression?

A2: A term in an algebraic expression is a single part of the expression separated by addition or subtraction signs. A coefficient is the numerical factor of a term.

Q3: What are like and unlike terms?

A3: Like terms in an algebraic expression are those that have the same variables raised to the same powers. Unlike terms have different variables or different powers.

Q4: What is an identity in algebra, and how is it different from an equation?

A4: An identity is an equation that is true for all values of the variables involved. An equation, on the other hand, is true only for specific values of the variables.

Q5: How do you add and subtract algebraic expressions?

A5: To add or subtract algebraic expressions, first identify and group like terms. Then, add or subtract the coefficients of these like terms while keeping the variables and their powers the same.

Q6: How are algebraic identities useful in simplifying expressions?

A6: Algebraic identities are useful in simplifying complex algebraic expressions by allowing us to replace complicated expressions with simpler, equivalent forms. This makes it easier to perform calculations and solve problems efficiently.

Q7: What is factorization of algebraic expressions, and why is it important?

A7: Factorization is the process of breaking down an algebraic expression into a product of its factors. It is important because it simplifies expressions and is essential in solving equations, especially quadratic equations.

Q8: Are there any online resources or tools available for practicing comparing quantities case study questions?

A8: We provide case study questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

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Case Study Questions for Class 8 Science Chapter 13 Sound

• Last modified on: 1 year ago
• Reading Time: 4 Minutes

Here we are providing case study questions for CBSE Class 8 Science Chapter 13 Sound.

Case Study Questions

Question 1:

Sound plays an important role in our life. It is a form of energy which makes us hear . Vibrating objects produce sound. Vibration is the to and fro or back and forth movement of an object. Sound needs a medium to travel. Hence, it cannot travel in a vacuum.

Human beings have a voice box or larynx which is present in their throat on the upper side of the windpipe. The larynx has two vocal cords which have a narrow slit between them so that air can pass through it. As the lungs throw the air out of the windpipe, it passes through the slit and hence allows the production of sound as the vocal cords start vibrating. Since sound travels in the form of waves, it is important to study about characteristics of waves. The three main properties of waves are Amplitude, frequency, and time period.

The magnitude of disturbance in the medium on either side of the mean value is called as Amplitude(A). Larger the amplitude, louder the sound. The number of oscillations/vibrations per second is called frequency. which is expressed in Hertz (Hz). Time taken for one complete oscillation/vibration is the time period.

i) Which part of the human throat is responsible for the voice produced by a human?

ii) The form of energy which enable us to hear:

iii) The rapid to and fro or up and down movement of an object from its mean position:

iv) The maximum displacement of a vibrating object from its mean position:

iii) Vibration

iv) Amplitude

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