physics assignment 1 class 9

WorkSheets Buddy

Download Math, Science, English and Many More WorkSheets

CBSE Worksheets for Class 9 Physics

CBSE Worksheets for Class 9 Physics: One of the best teaching strategies employed in most classrooms today is Worksheets. CBSE Class 9 Physics Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and problem-solving capabilities. So in order to help you with that, we at WorksheetsBuddy have come up with Kendriya Vidyalaya Class 9 Physics Worksheets for the students of Class 9. All our CBSE NCERT Class 9 Physics practice worksheets are designed for helping students to understand various topics, practice skills and improve their subject knowledge which in turn helps students to improve their academic performance. These chapter wise test papers for Class 9 Physics will be useful to test your conceptual understanding.

Board: Central Board of Secondary Education(www.cbse.nic.in) Subject: Class 9 Physics Number of Worksheets: 25

CBSE Class 9 Physics Worksheets PDF

All the CBSE Worksheets for Class 9 Physics provided in this page are provided for free which can be downloaded by students, teachers as well as by parents. We have covered all the Class 9 Physics important questions and answers in the worksheets which are included in CBSE NCERT Syllabus. Just click on the following link and download the CBSE Class 9 Physics Worksheet. CBSE Worksheets for Class 9 Physics can also use like assignments for Class 9 Physics students.

• CBSE Worksheets for Class 9 Physics All Chapters Assignments 
• CBSE Worksheets for Class 9 Physics Force and Laws of Motion Assignment 1
• CBSE Worksheets for Class 9 Physics Force and Laws of Motion Assignment 2
• CBSE Worksheets for Class 9 Physics Force and Laws of Motion Assignment 3
• CBSE Worksheets for Class 9 Physics Gravitation Assignment 1
• CBSE Worksheets for Class 9 Physics Gravitation Assignment 2
• CBSE Worksheets for Class 9 Physics Gravitation Assignment 3
• CBSE Worksheets for Class 9 Physics Motion Assignment 1
• CBSE Worksheets for Class 9 Physics Motion Assignment 2
• CBSE Worksheets for Class 9 Physics Sound Assignment
• CBSE Worksheets for Class 9 Physics Work & Energy Assignment 1
• CBSE Worksheets for Class 9 Physics Work & Energy Assignment 2
• CBSE Worksheets for Class 9 Physics Assignment 1
• CBSE Worksheets for Class 9 Physics Assignment 2
• CBSE Worksheets for Class 9 Physics Assignment 3
• CBSE Worksheets for Class 9 Physics Assignment 4
• CBSE Worksheets for Class 9 Physics Assignment 5
• CBSE Worksheets for Class 9 Physics Assignment 6
• CBSE Worksheets for Class 9 Physics Assignment 7
• CBSE Worksheets for Class 9 Physics Assignment 8
• CBSE Worksheets for Class 9 Physics Assignment 9
• CBSE Worksheets for Class 9 Physics Assignment 10
• CBSE Worksheets for Class 9 Physics Assignment 11
• CBSE Worksheets for Class 9 Physics Assignment 12
• CBSE Worksheets for Class 9 Physics Assignment 13

Advantages of CBSE Class 9 Physics Worksheets

• By practising NCERT CBSE Class 9 Physics Worksheet , students can improve their problem solving skills.
• Helps to develop the subject knowledge in a simple, fun and interactive way.
• No need for tuition or attend extra classes if students practise on worksheets daily.
• Working on CBSE worksheets are time-saving.
• Helps students to promote hands-on learning.
• One of the helpful resources used in classroom revision.
• CBSE Class 9 Physics Workbook Helps to improve subject-knowledge.
• CBSE Class 9 Physics Worksheets encourages classroom activities.

Worksheets of CBSE Class 9 Physics are devised by experts of WorksheetsBuddy experts who have great experience and expertise in teaching Maths. So practising these worksheets will promote students problem-solving skills and subject knowledge in an interactive method. Students can also download CBSE Class 9 Physics Chapter wise question bank pdf and access it anytime, anywhere for free. Browse further to download free CBSE Class 9 Physics Worksheets PDF .

Now that you are provided all the necessary information regarding CBSE Class 9 Physics Worksheet and we hope this detailed article is helpful. So Students who are preparing for the exams must need to have great solving skills. And in order to have these skills, one must practice enough of Class 9 Physics revision worksheets . And more importantly, students should need to follow through the worksheets after completing their syllabus.  Working on CBSE Class 9 Physics Worksheets will be a great help to secure good marks in the examination. So start working on Class 9 Physics Worksheets to secure good score.

CBSE Worksheets For Class 9

Share this:.

• Click to share on Twitter (Opens in new window)
• Click to share on Facebook (Opens in new window)

Leave a Comment Cancel reply

Notify me of follow-up comments by email.

Notify me of new posts by email.

NCERT Solutions for Class 9 Physics Free PDF Download

Ncert solutions for class 9 physics.

NCERT Solutions for Class 9 Physics will help you to ace the unsolved problems in the Class 9 Science book prescribed by the NCERT for all the schools of CBSE. A thorough understanding of the NCERT Solutions for Class 9 Physics helps you in understanding Physics concepts to the point. The NCERT Solutions for Class 9 Physics cover all the 5 chapters of the prescribed Physics syllabus and are the best alternative.

NCERT Solutions for Class 9 Physics breaks down the solutions into detailed steps and explains the answer thoroughly, which helps you understand the pattern of questioning and a way to increase your score in exams.

NCERT Solutions for Class 9 Physics are prepared by our team of highly professional, qualified and experienced faculties. These NCERT Solutions for Class 9 Physics helps you to quickly grasp all the basic concepts of physics. In case you have a doubt while you are studying NCERT Solutions for Class 9 Physics, we have a team of teachers who are just a click away to solve your doubt any time.

Download NCERT Solutions for Class 9 of all subjects here .

Toppr provides free video lectures on each and every topic of physics, chemistry, and maths, free study materials and last 10 years of question papers. Download Toppr app for Android and iOS or sign up for free.

NCERT Solutions for Class 9 Physics Chapterwise

Class 9 Physics Chapter 1 – Motion

Class 9 Physics Chapter 2 – Force And Laws Of Motion

Class 9 Physics Chapter 3 – Gravitation

Class 9 Physics Chapter 4 – Work and Energy

Class 9 Physics Chapter 5 – Sound

Class 9 Chapterwise NCERT Solutions for Physics

Ncert solutions for class 9 physics chapter 1 motion.

Here, you will learn about motion including motion along a straight line, types of motion, the difference between Vector & Scalar, Speed & Velocity, Distance & Displacement, Acceleration – Rate of change of velocity and average speed and velocity, Graphical representation of motion and derivation of three equation of motion. Download NCERT Solutions for Motion here .

NCERT Solutions for Class 9 Physics Chapter 2 Force and Laws of Motion

In this chapter, you will learn the concept of balance and unbalance forces. Starting with the First law of motion, the Galileo’s concept, the law of inertia. You will also learn the Second law of motion and Third law of motion, momentum, rate of change of momentum. And applications on first, second and third laws of motion. Download NCERT Solutions for Force And Laws Of Motion here .

NCERT Solutions for Class 9 Physics Chapter 3 Gravitation

This chapter gravitation takes you through the depths of motion of objects under the influence of gravitational force on the earth. Gravitational force and Newton’s universal law of gravitation. Download NCERT Solutions for Gravitation here .

NCERT Solutions for Class 9 Physics Chapter 4 Work and Energy

In this chapter, you will learn about the relationship between work and energy, scientific conception of work and also different forms of energy such as Kinetic energy, Potential energy, application of kinetic and potential energy, and energy of an object at a certain height. Download NCERT Solutions for Work and Energy here .

NCERT Solutions for Class 9 Physics Chapter 5 Sound

This chapter is a very interesting one as you will get to learn about the Reflection of Sound i.e ECHO, reverberation, and applications of multiple reflections of sound. All these concepts are taught by implementing various activities needed to be done in the Physics laboratory that makes the learning process more effective and interactive. Download NCERT Solutions for Sound here .

Why choose Toppr?

With Toppr App, you get free online classes with conceptual videos. You will also have access to all the free pdf downloads of study materials and solutions along with absolutely free online tests to enhance your problem-solving speed. So, Download Toppr App for Android and iOS or sign up for free.

Solved Questions For You:

Question 1.  What is the acceleration of free fall?

Answer:  Acceleration of free fall is the acceleration experienced by the freely falling body the effect of gravitation of earth alone. It is also called acceleration due to gravity.

Answer:  The gravitation force between the earth and object is called weight.

Answer:  By crumpling the paper into a ball, the volume of the object decreases but the mass remains the same. Hence its density increases.

Answer: The Importance of Universal law of gravitation lies in the fact, that it was successful in explaining many phenomena such as.

Customize your course in 30 seconds

Which class are you in.

NCERT Solutions for Class 9

• NCERT Solutions for Class 9 History Chapter 3 Free PDF Download
• NCERT Solutions for Class 9 History Chapter 6 Free PDF Download
• NCERT Solutions for Class 9 Maths Chapter 15 Free PDF Download
• NCERT Solutions for Class 9 Maths Chapter 7 Free PDF Download
• NCERT Solutions for Class 9 Maths Chapter 14 Free PDF Download
• NCERT Solutions for Class 9 Maths Chapter 12 Free PDF Download
• NCERT Solutions for Class 9 Maths Chapter 1 Free PDF Download
• NCERT Solutions for Class 9 Maths Chapter 9 Free PDF Download
• NCERT Solutions for Class 9 Maths Chapter 4 Free PDF Download
• NCERT Solutions for Class 9 Maths Chapter 11 Free PDF Download

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Download the App

AssignmentsBag.com

Assignments For Class 9 Physics

Assignments for Class 9 Physics have been developed for Standard 9 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 9 Physics from our website as we have provided all topic wise assignments free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 9 Physics. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Physics book and get good marks in class 9 exams.

Assignments for Class 9 Physics as per CBSE NCERT pattern

All students studying in Grade 9 Physics should download the assignments provided here and use them for their daily routine practice. This will help them to get better grades in Physics exam for standard 9. We have made sure that all topics given in your textbook for Physics which is suggested in Class 9 have been covered ad we have made assignments and test papers for all topics which your teacher has been teaching in your class. All chapter wise assignments have been made by our teachers after full research of each important topic in the textbooks so that you have enough questions and their solutions to help them practice so that they are able to get full practice and understanding of all important topics. Our teachers at https://www.assignmentsbag.com have made sure that all test papers have been designed as per CBSE, NCERT and KVS syllabus and examination pattern. These question banks have been recommended in various schools and have supported many students to practice and further enhance their scores in school and have also assisted them to appear in other school level tests and examinations. Its easy to take print of thee assignments as all are available in PDF format.

Some advantages of Free Assignments for Class 9 Physics

• Solving Assignments for Physics Class 9 helps to further enhance understanding of the topics given in your text book which will help you to get better marks
• By solving one assignments given in your class by Physics teacher for class 9 will help you to keep in touch with the topic thus reducing dependence on last minute studies
• You will be able to understand the type of questions which are expected in your Physics class test
• You will be able to revise all topics given in the ebook for Class 9 Physics as all questions have been provided in the question banks
• NCERT Class 9 Physics Workbooks will surely help you to make your concepts stronger and better than anyone else in your class.
• Parents will be able to take print out of the assignments and give to their child easily.

All free Printable practice assignments are in PDF single lick download format and have been prepared by Class 9 Physics teachers after full study of all topics which have been given in each chapter so that the students are able to take complete benefit from the worksheets. The Chapter wise question bank and revision assignments can be accessed free and anywhere. Go ahead and click on the links above to download free CBSE Class 9 Physics Assignments PDF.

You can download free assignments for class 9 Physics from https://www.assignmentsbag.com

You can get free PDF downloadable assignments for Grade 9 Physics from our website which has been developed by teachers after doing extensive research in each topic.

On our website we have provided assignments for all subjects in Grade 9, all topic wise test sheets have been provided in a logical manner so that you can scroll through the topics and download the worksheet that you want.

You can easily get question banks, topic wise notes and questions and other useful study material from https://www.assignmentsbag.com without any charge

Yes all test papers for Physics Class 9 are available for free, no charge has been put so that the students can benefit from it. And offcourse all is available for download in PDF format and with a single click you can download all assignments.

https://www.assignmentsbag.com is the best portal to download all assignments for all classes without any charges.

Related Posts

Assignments For Class 11 Mathematics Straight Lines

Assignments For Class 12 Physics

Assignments For Class 12 Mathematics Matrices and Determinants

CBSE NCERT Solutions

NCERT and CBSE Solutions for free

Class 9 Science Assignments

We have provided below free printable Class 9 Science Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 9 Science . These Assignments for Grade 9 Science cover all important topics which can come in your standard 9 tests and examinations. Free printable Assignments for CBSE Class 9 Science , school and class assignments, and practice test papers have been designed by our highly experienced class 9 faculty. You can free download CBSE NCERT printable Assignments for Science Class 9 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Science Class 9. Students can click on the links below and download all Pdf Assignments for Science class 9 for free. All latest Kendriya Vidyalaya Class 9 Science Assignments with Answers and test papers are given below.

Science Class 9 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 9 Science . Students and teachers can download and save all free Science assignments in Pdf for grade 9th. Our expert faculty have covered Class 9 important questions and answers for Science as per the latest syllabus for the current academic year. All test papers and question banks for Class 9 Science and CBSE Assignments for Science Class 9 will be really helpful for standard 9th students to prepare for the class tests and school examinations. Class 9th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 9 Science Download in Pdf

Advantages of Class 9 Science Assignments

• As we have the best and largest collection of Science assignments for Grade 9, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Science textbooks for Class 9 .
• All Science assignments for Class 9 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 9 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Science chapter wise worksheets and assignments for free in Pdf
• Class 9 Science question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 9 Science students

At https://www.cbsencertsolutions.com, we have provided the biggest database of free assignments for Science Class 9 which you can download in Pdf

We provide here Standard 9 Science chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Science in Grade 9, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

We have provided here topic-wise Science Grade 9 question banks, revision notes and questions for all difficult topics, and other study material.

We have provided the best collection of question bank and practice tests for Class 9 for all subjects. You can download them all and use them offline without the internet.

Related Posts

Class 9 Mathematics Triangles Assignments

Class 9 Malayalam Assignments

Class 9 French Assignments

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion

NCERT Solutions Class 9 Science Chapter 9 Force and Laws of Motion – Here are all the NCERT solutions for Class 9 Science Chapter 9. This solution contains questions, answers, images, step by step explanations of the complete Chapter 9 titled Force and Laws of Motion of Science taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Science, then you must come across Chapter 9 Force and Laws of Motion. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion in one place. For a better understanding of this chapter, you should also see Chapter 9 Force and Laws of Motion Class 9 notes , Science.

Topics and Sub Topics in Class 9 Science Chapter 9 Force and Laws of Motion:

• Force and Laws of Motion
• Balanced and Unbalanced Forces
• First Law of Motion
• Inertia and Mass
• Second Law of Motion
• Third Law of Motion
• Conservation of Momentum

These solutions are part of NCERT Solutions for Class 9 Science . Here we have given Class 9 NCERT Science Text book Solutions for Chapter 9 Force and Laws of Motion.

In-Text Questions Solved

NCERT Textbook for Class 9 Science – Page 118 Question 1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin? Answer: (a) A stone of the same size (b) a train (c) a five-rupees coin As the mass of an object is a measure of its inertia, objects with more mass have more inertia.

More Resources for CBSE Class 9

NCERT Solutions

• NCERT Solutions Class 9 Maths
• NCERT Solutions Class 9 Social Science
• NCERT Solutions Class 9 English
• NCERT Solutions Class 9 Hindi
• NCERT Solutions Class 9 Sanskrit
• NCERT Solutions Class 9 IT
• RD Sharma Class 9 Solutions

Formulae Handbook for Class 9 Maths and Science Educational Loans in India

Question 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Answer:  When the tree’s branch is shaken vigorously the branch attain motion but the leaves stay at rest. Due to the inertia of rest, the leaves tend to remain in its position and hence detaches from the tree to fall down.

Question 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest? Answer:  When a moving bus brakes-to a stop: When the bus is moving, our body is also in motion, but due to sudden brakes, the lower part of our body comes to rest as soon as the bus stops. But the upper part of our body continues to be in motion and hence we fall in forward direction due to inertia of motion. When the bus accelerates from rest we fall backwards: When the bus’ is stationary our body is at rest but when the bus accelerates, the lower part of our body being in contact with the floor of the bus comes in motion, but the upper part of our body remains at rest due to inertia of rest. Hence we fall in backward direction.

Class 9 Science NCERT Textbook – Page 126-127 Question 1. If action is always equal to the reaction, explain how a horse can pud a cart? Answer:  The third law of motion states that action is always equal to the reaction but they act on two different bodies. In this case the horse exerts a force on the ground with its feet while walking, the ground exerts an equal and opposite force on the feet of the horse, which enables the horse to move forward and the cart is pulled by the horse.

Question 2. Explain, why is it difficult for a fireman to hold a hose, which ejects a large amount of water at a high velocity. Answer:  The water that is ejected out from the hose in the forward direction comes out with a large momentum and equal amount of momentum is developed in the hose in the opposite direction and hence the hose is pushed backward. It becomes difficult for a fireman to hold a hose which experiences this large momentum.

NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion (Hindi Medium)

Extra Questions from NCERT Textbook for Class 9 Science

Question 1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason. Answer:  When an object experiences a net zero external unbalanced force, in accordance with second law of motion its acceleration is zero. If the object was initially in a state of motion, then in accordance with the first law of motion, the object will continue to move in same direction with same speed. It means that the object may be travelling with a non-zero velocity but the magnitude as well as direction of velocity must remain unchanged or constant throughout.

Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain. Answer:  The carpet with dust is in state of rest. When it is beaten with a stick the carpet is set in motion, but the dust particles remain at rest. Due to inertia of rest the dust particles retain their position of rest and falls down due to gravity.

Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope? Answer:  In moving vehicle like bus, the motion is not uniform, the speed of vehicle varies and it may apply brake suddenly or takes sudden turn. The luggage will resist any change in its state of rest or motion, due to inertia and this luggage has the tendency to fall sideways, forward or backward. To avoid the fall of the luggage, it is tied with the rope.

Question 4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because (a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest. Answer:  (c) there is a force 6n the ball opposing the motion.

Question 9.What is the momentum of an object of mass m, moving with a velocity v? . (a) (mv) 2           (b) mv 2                     (c) 1/2 mv 2                   (d) mv Answer:  (d) mv

Question 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move. Answer:  The mass of truck is too large and hence its inertia is too high. The small force exerted on the truck cannot move it and the truck remains at rest. For the truck to attain motion, an external large amount of unbalanced force need to be exerted on it.

Question 17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions. Answer:  Rahul gave the correct reasoning and explanation that both the motorcar and the insect experienced the same force and a change in their momentum. As per the law of conservation of momentum. When 2 bodies collide: Initial momentum before collision = Final momentum after collision m 1  u 1 + m 2  u 2  = m 1  v 1 + m 2  v 2 The equal force is exerted on both the bodies but, because the mass of insect is very small it will suffer greater change in velocity.

Additional Exercises

Question 5. A large truck and a car, both moving with a velocity of magnitude v, have a head- on collision and both of them come to a halt after that. If the collision lasts for 1 s: (a) Which vehicle experiences the greater force of impact? (b) Which vehicle experiences the greater change in momentum? (c) Which vehicle experiences the greater acceleration? (d) Why is the car likely to suffer more damage than the truck? Answer: (a) During head on collision forces applied by truck and car are action-reaction forces. Hence both vehicles experience same (equal) force of impact. (b) Here initial velocity of both car and truck is same equal to v and final velocity of both is zero. But mass of truck is much more than that of car, hence change in momentum of truck is more than change in momentum of car. (c) For same force of impact, the acceleration of car will have greater magnitude because its mass is less. (d) Car suffers more damage than the truck, as acceleration of car is more, its velocity falls to zero in a shorter time and consequently, its momentum changes in a shorter time.

More Questions Solved

NCERT Solutions for Class 9 Science Chapter 9 Multiple Choice Questions

NCERT Solutions for Class 9 Science Chapter 9 Very Short Answer Type Questions

Question 1. Define force. Answer:  It is a push or pull on an object that produces acceleration in the body on which it acts. 4

Question 2. What is S.I. unit of force? Answer:  S.I. unit of force is Newton.

Question 3. Define one Newton. Answer:  A force of one Newton produces an acceleration of 1 m/s 2 on an object of mass 1 kg. . 1 N = 1 kg m/s 2

Question 4. What is balanced force? Answer:  When forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.

Question 5. What is frictional force? Answer:  The force that always opposes the motion of object is called force of friction.

Question 6. What is inertia? Answer:  The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia.

Question 7. State Newton’s first law of motion. Answer:  An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.

Question 8. State Newton’s second law of motion. Answer:  The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Question 9. What is momentum? Answer:  The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. The S. I. unit is kg m/s. (p = mv)

Question 10. State Newton’s III law of motion. Answer:  To every action, there is an equal and opposite reaction and they act on two different bodies.

Question 11. Which will have more inertia a body of mass 10 kg or a body of mass 20 kg? Answer:  A body of mass 20 kg will have more inertia.

Question 12. Name the factor on which the inertia of the body depends. Answer:  Inertia of a body depends upon the mass of the body.

Question  13. Name two factors which determine the momentum of a body. Answer:  Two factors on which momentum of a body depend is mass and velocity. Momentum is directly proportional to the mass and velocity of the body.

Question 14. What decides the rate of change of momentum of an object? Answer:  The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.

NCERT Solutions for Class 9 Science Chapter 9 Short Answer Type Questions

Question 2. What change will force bring in a body? Answer:  Force can bring following changes in the body:

• It can change the speed of a body.
• It can change the direction of motion of a body,
• It can change the shape of the body.

Question 3. When a motorcar makes a sharp turn at a high speed, we tend to get thrown to one side. Explain why? Answer:  It is due to law of inertia. When we are sitting in car moving in straight line, we tend to continue in our straight-line motion. But when an unbalanced force is applied by the engine to change the direction of motion of the motorcar. We slip to one side of the seat due to the inertia of our body.

Question 4. Explain why it is dangerous to jump out of a moving bus. Answer:  While moving in a bus our body is in motion. On jumping out of a moving bus our feet touches the ground and come to rest. While the upper part of our body stays in motion and moves forward due to inertia of motion and hence we can fall in forward direction. Hence, to avoid this we need to run forward in the direction of bus.

Question 5. Why do fielders pull their hand gradually with the moving ball while holding a catch? Answer:  While catching a. fast moving cricket ball, a fielder on the ground gradually pulls his hands backwards with the moving ball. This is done so that the fielder increases the time during which the high velocity of the moving ball decreases to zero. Thus, the acceleration of the ball is decreased and therefore the impact of catching the fast moving ball is reduced.

Question 6. In a high jump athletic event, why are athletes made to fall either on a cushioned bed or on a sand bed? Answer:  In a high jump athletic event, athletes are made to fall either on a cushioned bed or on a sand bed so as to increase the time of the athlete’s fall to stop after making the jump. This decreases the rate of change of momentum and hence the force.

Question 7. How does a karate player breaks a slab of ice with a single blow? Answer:  A karate player applied the blow with large velocity in a very short interval of time on the ice slab which therefore exerts large amount of force on it and suddenly breaks the ice slab.

Question 8. What is law of conservation of momentum? Answer:  Momentum of two bodies before collision is equal to the momentum after collision. In an isolated system, the total momentum remain conserved.

Question 9. Why are roads on mountains inclined inwards at turns? Answer:  A vehicle moving on mountains is in the inertia of motion. At a sudden turn there is a tendency of vehicle to fall off the road due to sudden change in the line of motion hence the roads are inclined inwards so that the vehicle does not fall down the mountain.

Question 10. For an athletic races why do athletes have a special posture with their right foot resting on a solid supporter? Answer:  Athletes have to run the heats and they rest their foot on a solid supports before start so that during the start of the race the athlete pushes the support with lot of force and this support gives him equal and opposite push to start the race and get a good start to compete for the race.

Question 11.Why do you think it is necessary to fasten your seat belts while travelling in your vehicle? Or How are safety belts helpful in preventing any accidents? Answer:  While we are travelling in a moving car, our body remains in the state of rest with respect to the seat. But when driver applies sudden breaks or stops the car our body tends to continue in the same state of motion because of its inertia. Therefore, this sudden break may cause injury to us by impact or collision. Hence, safety belt exerts a force on our body to make the forward motion slower.

Question 13. When you kick a football it flies away but when you kick a stone you get huh why? Answer:  This is because stone is heavier than football and heavier objects offer larger inertia. When we kick a football its mass is less and inertia is also less so force applied by our kick acts on it and hence it shows larger displacement but in case of stone, it has larger mass and offers larger inertia. When we kick (action) the stone it exerts an equal and opposite force (reaction) and hence it hurts the foot.

Question 14. If a person jumps from a height on a concrete surface he gets hurt. Explain. Answer:  When a person jumps from a height he is in state of inertia of motion. When he suddenly touches the ground he comes to rest in a very short time and hence the force exerted by the hard concrete surface on his body is very high, and the person gets hurt.

Question 15. What is the relation between Newton’s three laws of motion? Answer:  Newton’s first law explains about the unbalanced force required to bring change in the position of the body. Second law states/explains about the amount of force required to produce a given acceleration. And Newton’s third law explains how these forces acting on a body are interrelated.

Question 16. Give any three examples in daily life which are based on Newton’s third law of motion. Answer:  Three examples based on Newton’s third law are :

• Swimming: We push the water backward to move forward. action – water is pushed behind reaction – water pushes the swimmer ahead
• Firing gun: A bullet fired from a gun and the gun recoils. action – gun exerts force on the bullet reaction – bullet exerts an equal and opposite force on the gun
• Launching of rocket action – hot gases from the rocket are released reaction – the gases exert upward push to the rocket

NCERT Solutions for Class 9 Science Chapter 9 Long Answer Type Questions

Question 2. State all 3 Newton’s law of motion. Explain inertia and momentum. Answer: Newton’s I law of motion: An object remains in a state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force. Newton’s II law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the-force. Newton’s III law of motion: To every action, there is an equal and opposite reaction and they act on two different bodies. Inertia: The natural tendency of an object to resist a change in their state of rest or of uniform motion is called inertia. Momentum: The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. Its S.I. unit is kgm/s. p = m x v

Question 3. Define force. Give its unit and define it. What are different types forces? Answer:   Force: It is a push or pull on an object that produces acceleration in the body on which it acts. A force can do 3 things on a body (a) It can change the speed of a body. (b) It can change the direction of motion of a body. (c) It can change the shape of the body. The S.I. unit of force is Newton. Newton: A force of one Newton produces an acceleration of 1 m/s 2 on an object of mass 1 kg. 1N = 1kg m/s 2 Types of forces:

• Balanced force: When the forces acting on a body from the opposite direction do not change the state of rest or of motion of an object, such forces are called balanced forces.
• Unbalanced force: When two opposite forces acting on a body move a body in the direction of the greater force or change the state of rest, such forces are called as unbalanced force.
• Frictional force: The force that always opposes the motion of object is called force of friction.

Question 4. What is inertia? Explain different types of inertia. Give 3 examples in daily life which shows inertia. Answer: Inertia: The natural tendency of an object to resist change in their state of rest or of motion is called inertia. The mass of an object is a measure of its inertia. Its S.I. unit is kg. Types of inertia: Inertia of rest: The object at rest will continue to remain at rest unless acted upon by an external unbalanced force. Inertia of motion: The object in the state of uniform motion will continue to remain in motion with same speed and direction unless it is acted upon by an external unbalanced force. . Three examples of inertia in daily life are:

• When we are travelling in a vehicle and sudden brakes are .applied we tend to fall forward.
• When we shake the branch of a tree vigorously, leaves fall down.
• If we want to remove the dust from carpet we beat the carpet so that dust fall down.

NCERT Solutions for Class 9 Science Chapter 9 Activity-Based Questions

Question 1.

• Make a pile of similar carom coins on a table, as shown in the figure.

Question 2.

• Set a five-rupees coin on a stiff card covering an empty glass tumbler standing on a table as shown in the figure.
• Give the card a sharp horizontal flick with a finger. If we do it fast then the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia.

Answer:  The force applied on the card due to flicking changes the inertia of the card but the coin resist a change and stay at the rest i.e. inertia of rest and due to gravity falls down in the tumbler.

Question 3.

• Place a water-filled tumbler on a tray.
• Hold the tray and turn around as fast as you can.
• We observe that the water spills. Why?

Answer:  The water-filled in tumbler and tray are at rest. On moving/turning around the tray at faster speed the water spills because the tray and the tumbler comes into motion while the water in the tumbler remain at inertia of rest.

Question 4.

• Request two children to stand on two separate carts as shown on the next page.
• Give them a bag full of sand or some other heavy object. Ask them to play a game of catch with the bag.
• Does each of them receive an instantaneous reaction as a result of throwing the sand bag (action)?
• You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other.

Question 5.

• Take a big rubber balloon and inflate it fully. Tie its neck using a thread. Also using adhesive tape, fix a straw on the surface of this balloon.
• Pass a thread through the straw and hold one end of the thread in your hand or fix it on the wall.
• Ask your friend to hold the other end of the thread or fix it on a wall at some distance. The arrangement is shown in the figure below.
• Now remove the thread tied on the neck of balloon. Let the air escape from the mouth of the balloon.

Question 6.

• Take a test tube of good quality glass material and put a small amount of water in it. Place a stop cork at the mouth of it.
• Now suspend the test tube horizontally by two strings or wires as shown in the figure on next page.
• Heat the test tube with a burner until water vaporises and the cork blows out.

NCERT Solutions for Class 9 Science Chapter 9 Value-Based Questions

Question 1. Class V students were playing cricket with the cork hall in the school campus. Charu a senior student told them about the accidents that can occur due to cork ball in the campus and also advised them to bring soft cosco ball to play the game. (a) Why it was safe to play with soft ball and not with hard cork ball? (b) A player pulls his hands backwards after holding the ball shot at high speed. Why? (c) What value of Charu is seen in this act? Answer: (a) The soft ball will have less inertia as compared to the heavy ball and it would not hurt the players. (b) By pulling the hand backwards it reduces the force exerted by the ball on hands. (c) Charu showed the value of being responsible and helpful by nature.

Question 2. Saksham saw his karate expert friend breaking a slate. He tried to break the slate but Saksham’s friend stopped him from doing so and told him that it would hurt, one needs lot of practice in doing so. (a) How can a karate expert break the slate without any injury to his hand? (b) What is Newton’s third law of motion? (c) What value of Saksham’s friend, is seen in the above case? Answer: (a) A karate player applies the blow with large velocity in a very short interval of time on the slate, therefore large force is exerted on the slate and it breaks. (b) To every action there is an equal and opposite reaction, both act on different bodies. Saksham’s friend showed the value of being responsible and caring friend.

Free Resources

Quick Resources

Class 9 Physics Notes, MCQ’s, Exercise Q, Numerical

Knowing what is important and what’s not, saves a lot of time. So, have a look at class 9 physics notes containing numerical, short questions, long questions and multiple choice question. We provide you with everything that is related to Physics. From quality notes to books and from exercises to past papers. We recommend you preparing your exams from our notes because we have quality notes and also important questions to share with you. Past papers really give us an understanding of how board exams thing works. So, have a look just scroll through it once, you will be amazed at what a good quality content we have to share. Currently, you can see physics 9th class notes.

Generally, 9th class students face issues when it involves finding the class 9 physics notes. For an equivalent purpose, this site was created wherever the Pakistani students are supplied with the things that they have to urge higher marks in exams. As a result, you are free to download the notes containing numerical, short questions, long questions and multiple choice question below and get the foremost of it. Not only keep this notes restricted to you but also, you must share along with your friends and loving ones. You might be thinking if I share with others, they’d get high marks. This life does not work like this. If you share along with your friends, Allah will reward you.

In addition to, if your friends study in 9th class, you should share this blog or helping site wherever much helpful content about obtaining nice marks in exams are given. If you face any drawback or any problem, send an email to [email protected] .

This Post Has 168 Comments

Send me all the numerical of class 9th physics

Snd mcqs of CHP 4 5 6

Please send me all numerical for CHP 1 3 4 6

Can you make a pdf of the important numerical of all the chapter of 9th physics

Send me numericals of chapter 1

Appreciated❤

G bas theek hi han Sara hi likha howa ha important to koi nhi ha is ma ….Kal pepper ha or aj ye Mili ha

This happens kiu k paper me sirf SLO question ate h ya phir conceptual — wo sidde sidde questions nai dete

exam me ase question hi ate han.but ab kese aye gay,i dont know.may allah help all of us

Please send urdu medium subjects 9th class

I need physics and math notes for 9th cladd

You can download it from here.

I need physics notes for class 9th send me pdf please

Please provide me physics all chapters solution pdf

Best learning app thanks

How can I download it please help

Download option is not available please help

very good notes

Your notes are very effective in learning.

It is really helpful

Your notes very helpful and interesting

Really excellent notes thanks

Plzz send numerical physics öth and 10th

AOA please ineed numerical chap 2 physics

Hello Sir please give me 9th class chemistry notes

search class 9th chemistry notes and download them

really this website is good for me i like it very much

From where we have to download ?

Just choose the chapter and download it from there it will take some time

Hi assalamualikum contact me on Instagram @javaid_zahoor7

Assalam o Alaikum! Your notes are really helpful, I appreciate your efforts. Have you prepared Urdu Medium notes for class 9th/10th or not yet?

Really very good notes.

These are helpfull but these are not opening in my pdf file

Right. no doubt.👌👌

Really helpful…keep up the good work👍👍

Some mcqs are wrongly answered plz correct them.zaheer awan lecturer Phyiscs I.e chptr 3 viii mcq chptr 4 i mcq And many more

Asslam u Alaikum i need Math physics of 9th nd 1oth notes ..please

Waliakum-o-Asalam, you can download it.

Kindly share

I need 9th class notes. But there is no option of downloading

I need notes that have more headings for biology and chemistry

Best notes I like this❤️😘

How many chapters are included in physics and maths according to Sindh board

It’s really good.

Sir please upload physics short questions

yes sir plz upload

sarah tumharay pass physics ka send up ka paper hai 9th class ka

PLEASE HELP ME

I need physics, chemistry and biology notes in English medium

Sir haven’t you any notes for this badge of 9th class 2020, l’m talking about the sort of notes in which only smart syllabus is included ???

I need class 9th kpk physics notes

yes plz need them cause finding included syllabus from these is very difficult

Plz if you upload these notes together not separate. …

yes it will be easy for us to study

Sir I need physics notes and numerical of 9th class please sr

Your MCQs are very helpful but please make MCQs of Numerical as this session 2020-21 my Aga Khan Board Exams are taken as MCQs so please make MCQs of numericals in Physics. Thank you

it’s really good and very helpful for students there is no complex answers it’s really useful for everyone

good notes and very useful thank u keep it up

Great Work,… I Downloaded all notes for my brother. Keep it up. Thanks.

best thing is that u can download

these notes helped me alot…… thanks

I need english grammar chapter translation notes

Thanx For Notes Really Helpfull Notes💞

It really very useful for me.

Where are numerical problems

💞 Wonderful notes of both Math and Physics. Whenever I have a problem I consult them. Thanks a lot. Your effort is appreciated. Best Regards, Grade 9 student.

Very help full to me

I want pak studies new notes of class 9th

sindh board hai ye????

Being a physics teacher .

I recommend many of students to have a copy of mcqs attached above.

Gud luck further.

Plz make notes of full book in one pdf file…..

I appreciate your all efforts you made. One of the best website ever. Please keet it up. 😍

Sir !can you please upload examples of chapters

It is very easy and very helpful notes 🙂

Sindh or Punjab board is this

Assalam u alikum , sir plzz mail me islamiat notes and chemistry notes of class 9 and 10

Sir please class 12 1st chapter numericals please

So much better 💯

anyone plzz tell the answer of question 4.5

Awesome! very helpful

thanks very helpful

Physica numericals

thank you very much it must be help full for new board sceem

Where is solved numericals

hello sir salam i am sadaf from class 9

Search in the site

You will find in💖

Best page for gaining good marks💖

Plz send me numerical of ch 1 and 3

there is a mistake in mcqs no 5 of ch 7 the correct ans is (weight of displaced liquid) according to its definition

ok, we’ll correct it

Sir, how can I download this there is no download option here kindly help me

I need physics all chapter notes

Always find this helpful

There is mistake in unit 5 mcqs number 4 it’s answer is 1600N not 160N

There is no mistake found in it. Answer is correct.

it’s answer is 160N not 1600N. we have to multiply 1.6 with 100 that gives us 160N You can calculate it also to check .

In numericals there are some extra zeros

Assalamu Alaikum Respected Sir , please I want to download these notes chapter wise but no pdf option is given here on your site. Hope you will understand that each student cannot avail internet service every time. Please make it possible to be downloaded. Jazakallahu khaira

you can download it.

Sir, Please make guides which contain these all chapters so we can download it easily.

Please upload the numericals

This helps me a lot…But I have a question…Do your notes are verified? I mean that everything you had written in it is correct and on the 9th board…Will it is accepted, need a quick response plz, so that I will start learning from here. JAZAK ALLAH

Yes, it is correct. You can prepare from it.

Great website 💯😁

Good work sir 👍

Yesss bohat hi achay notes ha

Assalamoalaikum sir. sir I need math notes and physics plz send link for downloading

I wonder if you should open a school!

Kindly send me a numerical notes of class. 9th physics

Share your contact details

2 chapter of physics problem number 11 Balochistan board

download kaise karien

Can someone share the link to download these notes of 9 and 10 Physics please?

I need pak studies class 9 notes in english

Quite helpful, Thanks

Please send me chapter 5 numericals

Plz upload national book Federal textbook of physics class 9

Aoa, I need short question notes of 9th class general science Urdu medium

I need physics all chapter note…??

i really like the notes

sir kia isi mesa board ka exam aya ga ya extra bhi kuch karna hoga? kindly reply soon

I need Pakistan studies new notes please

Thanks for this app it is v.v usefull for us ❤️

This is a very helpful notes I hope all students get 100% ‼️ mark in it ❤️

nice is so useful for us

Nice notes 💯

Great 👍 job sir thank you soooooo much for your hard work.

You are provideing such a best note’s , Great job!

Need practice questions for physics

Need full short and long and mcqs solve book

9th class physics numerical ist chapter

Is this same are sindh text book jamshoro or diffrent slybuss??

Need latest question paper according to board pattern

Plz open the download options

I enjoy from your website

Best yarr this is the masterpiece we want I lost my book and I got full marks thanks free ilm love you 😍💝💞I don’t know how to thankyou

I need full chapter notes

I appreciate you .. very satisfying..

Where do you understand by the zero error of a measuring instrument? answer bhejo

Wow And Amazing This is Really Amazing Thankyou Thankyou so much

Yaahh guys the author is right you can download it but his mistake is that he didn’t tell you the method of download it So you can download the PDF of it by clicking on the top right corner of the notes page😊 It works guys go & do that😎😊

ASSLAM O ALAIKUM.THERE IS A POBLEM IN CH.4 4.1 NUMERICAL.KINDLY CHECK IT.

This is really amazing 👏

Best hota hain online test

Best hota hain online test Yar

Leave a Reply Cancel reply

NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions for Class 9 Science (physics) Chapter 8 Motion are given below. In these solutions, we have answered all the intext and exercise questions provided in NCERT class 9 science textbook. Class 9 NCERT Solutions Science Chapter 8 provided in this article are strictly based on the CBSE syllabus and curriculum. Students can easily download these solutions in PDF format for free from our app.

Class 9 Science Chapter 8 Textbook Questions and Answers

INTEXT QUESTIONS

PAGE NO. 100

Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer: Yes, zero displacement is possible if an object has moved through a distance. Suppose a body is moving in a circular path and starts moving from point A and it returns back at same point A after completing one revolution, then the distance will be equal to its circumference while displacement will be zero.

Question 2: A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Given, side of the square field = 10 m Therefore, perimeter = 10 m × 4 = 40 m Farmer moves along the boundary in 40 second Time = 2 minutes 20 second = 2 × 60 + 20 = 140 s Since, in 40 second farmer moves 40 m

Therefore, in 1 second distance covered by farmer = 40 ÷ 40 = 1m.

Therefore, in 140 second distance covered by farmer = 1 × 140 m = 140 m

Thus, after 2 minute 20 second the displacement of farmer will be equal to 14.1 m north east from initial position.

Question 3: Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.

Answer: (a) Not true

Displacement can become zero when the initial and final position of the object is the same.

(b) Not true

Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.

PAGE NO. 102

Question 1: Distinguish between speed and velocity

Answer: Speed has only magnitude while velocity has both magnitude and direction. So, speed is a scalar quantity but velocity is a vector quantity.

Question 2: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer: The magnitude of average velocity of an object will be equal to its average speed in the condition of uniform velocity in a straight line motion.

Question 3: What does the odometer of an automobile measure?

Answer: In automobiles, odometer is used to measure the distance.

Question 4: What does the path of an object look like when it is in uniform motion?

Answer: In the case of uniform motion, the path of an object will look like a straight line.

Question 5: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3×10 8 ms -1 .

Answer: Here we have, speed = 3 × 10 8 m/s

Time = 5 minute = 5 × 60 s = 300 s

Using, Distance = Speed × Time ⇒ Distance = 3 × 10 8 × 300 m = 900 × 10 8 m = 9.0 × 10 10 m

PAGE NO 103

Question 1: When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Answer: (i) A body is said in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitude in equal interval of time.

(ii) A body is said in non-uniform acceleration when its motion is along a straight line and its velocity changes by unequal magnitude in equal interval of time.

Question 2: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Question 3: A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

PAGE NO. 107

Question 1: What is the nature of the distance – time graphs for uniform and non-uniform motion of an object?

Answer: (a) The slope of the distance-time graph for an object in uniform motion is a straight line. (b) The slope of the distance-time graph for an object in non-uniform motion is not a straight line.

Question 2: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer: When the slope of distance-time graph is a straight line parallel to time axis, the object is stationary.

Question 3: What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Answer: When the graph of a speed time graph is a straight line parallel to the time axis, the object is moving with constant speed.

Question 4: What is the quantity which is measured by the area occupied below the velocity- time graph?

Answer: The quantity of distance is measured by the area occupied below the velocity time graph.

PAGE NO 109-110

Question 1: A bus starting from rest moves with a uniform acceleration of 0.1ms –2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer:  Here we have, Initial velocity (u) = 0 m/s Acceleration (a) = 0.1ms –2   Time (t) = 2 minute = 120 seconds 

(a) The speed acquired: We know that, v = u + at ⇒ v = 0 + 0.1 × 120 m/s ⇒ v = 12 m/s Thus, the bus will acquire a speed of 12 m/s after 2 minute with the given acceleration.

(b) The distance travelled:

Thus, bus will travel a distance of 720 m in the given time of 2 minute.  

Question 2: A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s 2 . Find how far the train will go before it is brought to rest.

Question 3: A trolley, while going down an inclined plane, has an acceleration of 2 cm/s 2 . What will be its velocity 3 s after the start?

Answer: Here we have, Initial velocity, u = 0 m/s Acceleration (a) = 2 cm/s 2 = 0.02 m/s 2 Time (t) = 3 s Final velocity, v = ?

We know that, v = u + at            Therefore, v = 0 + 0.02 × 3 m/s           ⇒ v = 0.06 m/s Therefore, the final velocity will be 0.06 m/s after start.

Question 4:  A racing car has a uniform acceleration of 4 m/s 2 . What distance will it cover in 10 s after start?

Thus, racing car will cover a distance of 200 m after start in 10 s with given acceleration.

Question 5: A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s 2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Thus, stone will attain a height of 1.25 m and time taken to attain the height is 0.5 s.

Question 1: An athlete completes one round of circular track of diameter 200 m in 40 sec.  What will be the distance covered and the displacement at the end of 2 minutes 20 sec? 

Answer: Time taken = 2 min 20 sec = 140 sec. Radius, r = 100 m. In 40 sec the athlete complete one round.    

So, in 140 sec the athlete will complete = 140 ÷ 40 = 3.5 round.

At the end of his motion, the athlete will be in the diametrically opposite position.

Displacement = diameter = 200 m.

Question 2: Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 sec and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging.

(a) from A to B (b) from A to C?

Answer: For motion from A to B: Distance covered = 300 m = Displacement = 300 m.

Time taken = 150 sec.

(a) We know that,   Average speed  = Total distance covered ÷ Total time taken  = 300 m ÷ 150 sec = 2 ms -1  

Average velocity = Net displacement ÷ time taken = 300 m ÷ 150 sec = 2 ms -1

(b) For motion from A to C:

Distance covered = 300 + 100 = 400 m.

Displacement = AB – CB = 300 – 100 = 200 m.

Time taken = 2.5 min + 1 min = 3.5 min = 3.5 × 60 min = 210 sec.

Therefore,        Average speed = Total distance covered ÷ Total time taken  = 400 ÷ 210 = 1.90 ms -1 .

Average velocity = Net displacement ÷ time taken  = 200 m ÷ 210 sec = 0.952ms -1 . 

Question 3: Abdul, while driving to school, computes the average speed for his trip to be 20 kmh -1 . On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed of Abdul’s trip? 

Question 4: A motor boat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms -2 for 8.0 s. How far does the boat travel during this time?

Answer:  Here,      u = 0 m/s                  a = 3 ms -2                t = 8 s

Using,  s = ut + ½ at 2   ⇒ s = 0 × 8 + ½ × 3×8 2 ⇒ s = 96 m.

Question 5: A driver of a car travelling at 52 kmh -1 applies the brakes and accelerates uniformly in the opposite direction. The car stops after 5 s. Another driver going at 34 kmh -1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for two cars. Which of the two cars travelled farther after the brakes were applied?  

Answer: In in the following graph, AB and CD are the time graphs for the two cars whose initial speeds are 52 km/h(14.4 m/s) and 34 km/h(8.9 m/s), respectively.

Distance covered by the first car before coming to rest = Area of triangle AOB = ½ × AO × BO = ½ × 52 kmh -1 × 5 s = ½  (52 × 1000 × 1/3600) ms -1 × 5 s = 36.1 m 

Distance covered by the second car before coming to rest = Area of triangle COD = ½ × CO × DO = ½ × 34 km h -1 × 10 s = ½ × (34 × 1000 × 1/3600) ms -1 ×10 s = 47.2 m

Thus, the second car travels farther than the first car after they applied the brakes.

Question 6: Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

Answer: (a) B is travelling fastest as he is taking less time to cover more distance. (b) All three are never at the same point on the road. (c) Approximately 6 kms.   [as 8 – 2 = 6] (d) Approximately 7 kms. [as 7 – 0 = 7] 

Question 7: A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms -2 , with what velocity will it strike the ground? After what time will it strike the ground?

Answer: Given, initial velocity of the ball (u) = 0 (since it began at the rest position) Distance travelled by the ball (s) = 20m Acceleration (a) = 10 ms -2

As per the third motion equation, v 2 = u 2 +2as ⇒ v 2 = 2 × (10ms ‒2 ) × (20m) + 0 ⇒ v 2  = 400m 2 s ‒2 ⇒ v = 20ms -1

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation, t = (v-u)/a = (20-0)ms ‒1  / 10ms ‒2 = 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

Question 8: The speed – time graph for a car is shown in Figure:

(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

Answer: Shaded area representing the distance travelled is as follows:

(a) The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

Distance = 1/2 × 4 × 6 = 12 m

Therefore, the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6 th  to the 10 th  second.

Question 9: State which of the following situations are possible and give an example of each of the following:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving with an acceleration but with uniform speed.

(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer: (a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is impossible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

Question 10: An artificial is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hrs to revolve around the earth.   

Answer: Here, 

Given, radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2 × π × 42250 km = 265571.42 km

Time taken for the orbit = 24 hours

Therefore, speed of the satellite = 265571.42 ÷ 24 = 11065.4 km.h -1

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.

Class 9 Science NCERT Solutions Chapter 8 Motion

CBSE Class 9 Science NCERT Solutions Chapter 8 helps students to clear their doubts and to score good marks in the board exam. All the questions are solved by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 9 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

NCERT Solutions for Class 9 Science Chapter 8 PDF

Below we have listed the topics discussed in NCERT Solutions for Class 9 Science Chapter 8. The list gives you a quick look at the different topics and subtopics of this chapter.

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Save my name, email, and website in this browser for the next time I comment.

NCERT Solutions for Class 9 Science Chapter 1 Matter in Our Surroundings

Ncert solutions for class 9 science chapter 1 matter in our surroundings.

Chapter 1 Matter in Our Surroundings NCERT Solutions for Class 9 Science

Contact Form

Talk to our experts

1800-120-456-456

Matter in Our Surroundings Class 9 Notes CBSE Science Chapter 1 (Free PDF Download)

• Revision Notes
• Chapter 1 Matter In Our Surroundings

CBSE Class 9 Science Chapter 1 - Matter in Our Surroundings Revision Notes - Free PDF Download

Matter can be defined as something that has mass and occupies space. For instance, air, water, oxygen, fruits, etc. All these are considered as matter in our surroundings and are classified as solids, liquids and gases. They are made up of microscopic particles called molecules which are tightly, loosely and very loosely packed, respectively. Being the elementary lesson of Class 9 Science , you must be thorough on the same. In this regard, Matter in Our Surroundings Class 9 Notes by Vedantu can be of immense help. Along with textbooks, make sure to refer to this material for revision purposes and also achieve desired scores in exams. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for the better solutions, they can download Class 9 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

The revision notes of Chapter 1, Class 9 Science subject are present here. The students can simply study these notes before the exam to get a comprehensive overview of teh whole chapter.

We also have provided the free pdf to download which will help the students to revise the chapter anytime, with or without the internet.

Download CBSE Class 9 Science Revision Notes 2024-25 PDF

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

Access Class 9 Science Chapter 1 - Matter in Our Surrounding Notes

Introduction

Everything around us is formed of the matter: a pencil, a pen, a table, the food we consume, the clothes we wear, the walls of our homes. But what is the matter?

Anything that occupies space has mass, and can be sensed by our senses is considered the matter. In other words, the term "matter" refers to all of the substances and materials that make up the cosmos.

Composition of Matter

According to ancient Indian philosophers, the matter is made up of five constituents or tattvas, according to studies found in our sacred books and scriptures.

Illustration 1. How many different ways did ancient Indian philosophers classify matter?

a. $2$ 

b. $6$ 

c. $7$ 

d. $5$ 

Ans: $\left( D \right)$

Matter is made up of Particles  

Now that we have defined matter let us ask ourselves the question – What is a matter made up of? 

All matter comprises very small particles. 

All matter can be broken up in a similar manner to get very small particles. 

Hence we now conclude that all matter is made up of small particles.

(Image will be uploaded soon)

Illustration 2. Which of the following are matters? 

Chair, air, love, smell, hate, almonds, thought, cold, cold-drink, the smell of perfume.

Ans: chair, air, almond, cool drink

Properties of Matter 

Small particles of matter make up all matter. Some features are shared by all of these particles. A theory called Kinetic Theory of Matter explains forth these features.

Simply said, The Kinetic Theory of Matter States is a theory that describes how matter changes throughout time.

a. All matter is made up of tiny particles.

b. There is space between these particles.

c. The particles are in constant motion.

d. The particles are attracted to one another.

Particles of Matter have space between them 

Small particles make up matter, and these particles have small spaces between them.

These areas are not visible to the naked eye, yet particles of other matter can pass through them without changing their volume.

Particles of Matter are continuously moving 

Particles in the matter are constantly moving. Three types of motion were seen in the matter particles.

a. Translatory Motion - It occurs when particles move in straight lines and change direction without losing energy after interacting with another particle or the container's wall. When compared to liquids, translational motion is greatest in gases and least in solids.

b. Rotational motion: When particles travel about their own axis, this is known as rotational motion . This motion is comparable to the earth's rotation around its axis. In gases and liquids, the rotational motion will be quite high.

c. Vibrational Motion - When particles move back and forth around a central point. Solids have the greatest amount of motion because the particles are held in a hard framework.

Particles of Matter attract each other 

1. The force with which they attract one another differs depending on the matter.

2. The force is modest in some types of materials (waste paper, matchsticks) (as we can tear or break them easily).

3. The force is large in other types of material (iron nail) (as we cannot break the nail easily).

Illustration 3. When sugar dissolves in water, what happens to it? What happens to the sugar? What does the dissolution of sugar in water tell you about the nature of matter?

a. When sugar dissolves in water, the solid sugar crystals are broken up into microscopic particles.

b. The sugar particles interact with the water particles in the gaps between them (to form a sugar solution).

c. Sugar dissolving in water indicates that the stuff (in this case, sugar and water) is made up of minute particles. There are voids between the particles of stuff (in this case, water).

Diffusion 

“The mixing and spreading out of a substance with another substance due to the movement or motion of its particles is called diffusion.” 

The process of one substance diffusing into another continues until a homogenous mixture is achieved. Let's have a look at an example.

Put a crystal of potassium permanganate (purple colour) in one of the beakers that is full of water. Gradually, you'll notice that the purple-colored crystal begins to diffuse or dissolve into water, and after a while, it turns purple.

Diffusion in Gases 

Gases have a very fast diffusion rate. Because gas particles move very swiftly in all directions, this is the case.

Examples \[1:\]

Even from a long distance, the smell of food being prepared in the kitchen reaches us.

The smell of hot, sizzling food reaches us even when we are a long way away, but we must approach close to get the smell of cold food.

This is because the rate of diffusion of hot gases is substantially faster than the rate of diffusion of cold gases released by cold food.

Example \[2:\] 

When someone opens a bottle of perfume in one corner of a room, the scent soon travels throughout the space.

When a perfume bottle is opened, the liquid perfume soon turns into vapour (or gas).

The scent vapours flow quickly in all directions in the air, mixing with the air particles and spreading across the room.

Example 

The diffusion of a strong-smelling chemical (ethyl mercaptan) found in the cooking gas into the air detects the leaking of cooking gas (LPG) in our houses.

Diffusion in Liquids 

Liquid diffusion is slower than gas diffusion. This is due to the fact that particles in liquids move slower than particles in gases.

Solid in Liquid 

When a crystal of potassium permanganate is placed in the bottom of a beaker of water, the purple colour of the potassium permanganate progressively spreads throughout the water.

The liquid in Liquid: 

When a drop of ink is dropped into a beaker of water, the colour of the ink spreads across the entire water in the beaker; this is due to the diffusion of ink particles into water.

Gases like carbon dioxide and oxygen are necessary for aquatic plants and animals to survive. The carbon dioxide and oxygen gases in the air (or atmosphere) diffuse into and dissolve in water (ponds, lakes, and rivers). Aquatic plants use dissolved carbon dioxide to prepare food through photosynthesis, whereas aquatic animals breathe using dissolved oxygen in the water.

Diffusion in Solids 

Solid-state diffusion is an extremely slow process.

Example : 

If we write something on a blackboard and then leave it filthy for a long time (say, 10 to 15 days), cleaning the blackboard becomes quite tough. This is owing to the fact that certain chalk particles have dispersed into the backboard's surface.

When two metal blocks are closely linked together and left undisturbed for several years, the particles of one metal permeate into the other metal. Gases dissipate quickly. A gas's rate of diffusion is proportional to the square root of its density.

Force of Attraction (or Cohesion) 

Between the particles of matter, there is an attractive force that binds them together. The force of attraction is the attraction between particles of the same substance (or cohesion).

In general, the force of attraction is greatest in solid matter particles and least in gaseous matter particles.

Illustration 4. Analyse the effects of diffusion in different states of matter, such as solid, gas, and liquid.

Ans: $\text{Solid}<\text{Liquid}<\text{Gases}$ 

         Slow        Fast            Very Fast

States of Matter

Solids have a definite volume and shape. They are more difficult to break than liquids and gases.

Liquids have a specific volume but not a specific shape. They take on the shape of the container they're housed in.

Gases do not have a defined shape or volume. They take up all of the available space and take on the shape of the container in which they are kept.

Plasma - At extremely high temperatures, the plasma state is a fused and ionic condition of matter (like the core of the sun, stars). Because it is made up of positive ions and a pool of electrons, the fused ionic mass is neutral. Around 99 per cent of the universe is made up of fused ionic matter.

Illustration 5. 

a. Give two reasons why wood is a solid material.

b. ‘A material has a known volume but no known shape.' Indicate if the substance is solid, liquid, or gaseous.

c. Describe the physical state of matter that can be squeezed readily.

d. ‘A substance has both a definite shape and a defined volume.' Which physical state does this statement represent? A substance has neither a fixed shape nor a fixed volume. State whether it is a solid, a liquid or a gas. 

e. Give two reasons to justify that: 

i. Water is a liquid at room temp. 

ii. An iron almirah is solid.

a. Wood has

i. fixed shape, and 

ii. fixed volume 

b. Liquid 

c. Gas 

d. Solid 

e. Gas 

i. Fixed volume but no fixed shape 

ii. Fixed shape and fixed volume.

Rigid and Fluid 

Rigid is a word that denotes "unbending" or "inflexible." Because it is unbending or inflexible, a stone is stiff. Fluid is defined as "a material that flows easily" and requires the use of a vessel (container) to keep it contained.

A solid is a kind of stuff that is unyielding. Solids have a tendency to keep their shape when subjected to external force due to their rigidity. As a result, rigidity is the primary distinguishing feature of solids. As a result, rigidity is the primary distinguishing feature of solids. Solids don't need to be kept in a container. Two common solids are a brick and a log of wood.

A liquid is a fluid type of stuff that fills the container's lower half. Liquids must be kept in a container because they are fluids. Because liquids have a well-defined surface, they can be stored in an open container. The liquid will not spontaneously escape from the open container. Water and milk are two prevalent liquids found in our environment.

Gas is a form of stuff that fills the entire container in which it is contained. Gases, like liquids, require a container to keep them contained. Because gas has no open surface, it must be stored in a closed container. If gas is kept in an open container, it will escape. Gases are frequently stored in airtight gas cylinders because of this. Cooking gas (LPG), for example, is stored in airtight metal cylinders. We can conclude from this discussion that fluids include both liquids and gases. Fluidity is a property of liquids and gases that allows them to flow smoothly. When exposed to external stress, liquids and gases change shape quickly due to their fluidity.

Illustration 6. Which of the following is a rigid form of matter

Ans: Ether and alcohol

Interconversion of the state of matter

Changing the temperature, pressure, or both can cause matter to change its physical condition.

a. Melting is the transformation of a solid into a liquid.

b. Solidification is the process of turning a liquid into a solid.

c. The process of converting a liquid to a gas is known as vaporisation.

d. Condensation is the process of turning a gas into a liquid.

e. Sublimation is the process of converting a solid to a gas.

Note: While increasing pressure in gas will not change the physical condition of the gas, it will bring the particles closer together, causing the gas to liquefy.

Vaporization is promoted by lowering pressure over a liquid's surface.

Illustration 7. When solid carbon dioxide is exposed to air, which of the following factors is responsible for the change in state?

a. Increase in pressure

b. Decrease in pressure

c. Increase in temperature

d. Decrease in temperature

Ans: $(a)$ Decrease in pressure; Increase in temperature

Effect of change of Temperature and Pressure 

We can change the physical condition of matter in two ways: 

a. by changing the temperature; and 

b. by changing the pressure

A solid can be changed to a liquid state by raising the temperature, and a liquid may be converted to a gaseous state by lowering the temperature.

Melting (Fusion) 

Melting is the transformation of a solid substance into a liquid when it is heated (or fusion).

Melting of the substance refers to the temperature at which a solid melts and transforms into a liquid at atmospheric pressure.

The heat energy in a solid substance causes its particles to vibrate more vigorously. At the melting point, a solid's particles have enough kinetic energy to overcome the strong forces of attraction that keep them in fixed places, and they break apart into small groups. And the solid transforms into a liquid.

The greater the force of attraction between the particles of a solid substance, the higher its melting point. The melting point of iron metal, for example, is extremely high (1535 degrees celsius), indicating that the force of attraction between the particles of iron is extremely strong.

Boiling (Vaporisation) 

Boiling is the transformation of a liquid substance into a gas when heated rapidly.

The boiling point of a liquid is the temperature at which it boils and transforms rapidly into a gas at atmospheric pressure.

Condensation 

When a gas (or vapour) is cooled sufficiently, the process of turning it into a liquid is termed condensation.

Condensation of steam occurs when steam (or water vapour) cools and converts to water (or condensation of water vapour).

It's the polar opposite of vaporisation. (Boiling)

Freezing 

Freezing is the process of turning a liquid (solidification) into a solid by chilling it, the reverse of melting. 

When a liquid cools, its particles lose energy, slowing its movement.

If the liquid is sufficiently chilled (to the point of freezing), each particle ceases to move and vibrates in a fixed location. The liquid freezes and solidifies at this point. 

As a result of the preceding discussion, we can conclude that changing the temperature can change the state of matter.

Effect of the change in Pressure on the state of matter 

Short particles separated by small distances make up matter.

Interparticle distances are exceedingly short in the solid-state.

The inter-particle distances in liquids are slightly greater than in solids.

When compared to liquids or solids, interparticle distances are greatest in the gaseous state.

As a result, it can be shown that when pressure is applied to matter, the effect on solids is insignificant because the particles are so close together.

In liquids, the effect of pressure will be minimal.

Because the interparticle distances are vast, the effect of pressure on gases will be the greatest.

As a result, when pressure is applied to gases, the particles begin to move closer together. The attractive forces between the particles increase as the particles get closer together.

This rise in attracting forces aids the gas's transition of state. When enough pressure is applied, the attraction forces build to the point where the physical state transforms from gaseous to liquid.

The reverse can be expected to happen if the pressure on a gas is decreased.

Illustration 8. Define melting process

Ans: Melting is the transformation of a solid substance into a liquid when it is heated.

Latent Heat 

The heat that a substance needs to change its condition without increasing its temperature. It's called latent heat (hidden heat) because it's buried in the substance undergoing a state transition and doesn't show up as a rise in temperature.

 “During a transition of state, the latent heat is used up in overcoming the force of attraction between the particles of the substance. It has no effect on the kinetic energy of the substance's particles. And since the substance's temperature does not rise.”

Illustration 9. What is the latent heat of fusion of ice?

Ans: $3.34\times {{10}^{5}}j/kg$ 

Latent heat of Vaporization and Fusion 

There are two types of latent heat: 

i. Latent heat of fusion 

ii. Latent heat of vaporization 

Latent heat of Vaporization

The latent heat of vaporisation is the amount of heat in Joules necessary to turn a unit quantity of 1 kg liquid into vapours without a temperature change.

Experiments have shown that it takes \[22.5\text{ }\times \text{ }{{10}^{5}}\] joules of heat to convert 1 kilogramme of water (at its boiling point, \[\text{100 }\!\!{}^\circ\!\!\text{ C}\]) to steam at the same temperature. As a result, water's latent heat of vaporisation is \[22.5\text{ }\times \text{ }{{10}^{5}}\] joules per kilogramme (or \[22.5\text{ }\times \text{ }{{10}^{5}}\]J/kg).

“If the liquid freezes to create a solid and steam condenses to form water, the substance will emit an equal amount of latent heat of fusion and vaporisation.”

The Latent Heat of Vaporization varies depending on the substance.

Latent heat of Fusion (Solid to Liquid) 

It is the amount of heat in Joules required to transform one kilogramme of solid into liquid form without causing a temperature increase.

Experiments have shown that to turn 1 kilogramme of ice into water at the same temperature (\[\text{0 }\!\!{}^\circ\!\!\text{ C}\]), \[3.34\times {{10}^{5}}\] J of heat is required.

So, latent heat of fusion of ice is \[3.34\times {{10}^{5}}\] J/ Kg. 

Different substances have different Latent Heat of Fusion.

Illustration 10. Why the temperature of melting ice does not rise even though heat is being supplied continuously.

Ans: Because ice is solid, its particles are attracted to one another by strong forces. These attraction forces keep the particles tightly packed in solid ice. The heat we give ice during melting is completely consumed by overcoming the forces of attraction between ice particles, causing them to loosen up and become liquid water. As a result of this heat not increasing the kinetic energy of particles, no temperature rise occurs during the melting of ice. However, once all of the ice has melted to create water, additional heating increases the kinetic energy of water particles, causing the temperature of the water to rapidly rise.

Sublimation

Sublimation is defined as the transformation of a solid into vapours on heating and back to a solid on cooling.

$\text{Solid}\rightleftarrows \text{Vapour (or Gas)}$ 

Ammonium chloride, iodine, camphor, naphthalene, and anthracene are some of the common substances that sublimate.

Solid carbon dioxide is yet another example of sublimation (which is commonly known as dry ice).

Carbon dioxide gas is formed when solid carbon dioxide (or dry ice) sublimates.

Illustration 11. 

a. When heated, which of the following substances sublimates:

iv. Camphor

v. Sodium Chloride

b. What happens to the heat energy that has been delivered once a solid has melted? 

c. A substance's melting point is lower than room temperature. Predict the state of its physical state. 

d. Is it permissible to refer to ammonia in its gaseous state as vapours?

e. What is the name of the chemical reaction that converts a solid into a gas?

f. During a substance's change of state, which of the following energy is absorbed?

i. Specific Heat

ii. Latent heat

iii. Heat of solution

g. Identify one common chemical that can change state when heated or cooled.

a. Camphor and iodine 

b. It is converted into latent heat of fusion 

c. It is a liquid. 

d. No, it is not 

e. It is called sublimation 

e. Latent heat 

Evaporation

The phenomenon of evaporation occurs when a liquid transforms to a gaseous state below its boiling point.

Water molecules are attracted to other water molecules in all directions, but the water molecules near the surface of the water are only dragged inward, which is below the water's surface.

Note: Evaporation is a phenomenon that occurs in all liquids in theory. But, in general, when we talk about evaporation, we're talking about water evaporation.

Vapour is a substance that can remain in a gaseous state at a temperature where it would ordinarily be a solid or liquid.

Examples of solids that can exist as a vapour: camphor, naphthalene 

Factors Affecting Evaporation 

Evaporation depends on temperature, surface area and weather conditions 

a. As the quantity of water molecules at the surface grows, evaporation increases if the surface area of the water is big. As a result, more water molecules are likely to break out once they have enough kinetic energy.

b. As the temperature approaches the boiling point of water, evaporation increases. The kinetic energy of the molecules increases as the temperature rises. The extra kinetic energy required by surface molecules to break loose or evaporate is reduced as a result. As a result, evaporation increases.

c. Evaporation decreases in excessively humid weather because the air is saturated with water molecules.

d. As water evaporates, the air just above the water surface becomes saturated with water molecules.

Illustration 12. What effect does temperature and surface area have on evaporation?

Ans: Evaporation increases as the temperature and surface area increase.

Cooling Effect 

How Does Evaporation Cause Cooling? 

When a liquid evaporates, the energy is extracted from the liquid. As a result, it continues to cool. The liquid absorbs the energy lost by the surroundings, causing them to cool. During the summer, for example, air coolers are used to provide forced cooling.

Illustration 13. Make a note of the cooling mechanism.

Ans: As a gas particle's energy drops due to cooling, the particle's moment slows down. The particles also become significantly closer to one another, resulting in the intermolecular attraction force. The gas contracts as a result of this.

Matter in our Surroundings Notes Free PDF Download

Class 9 Science Notes Chapter 1 PDF material is entirely free and can be easily availed from Vedantu’s website or application. So download the same and make it a revision guide for both unit test and final examinations. By downloading the Chapter 1 Science Class 9 Notes, you no longer need to prepare a list of essential topics and explanations by yourself. Besides the usage of simple and lucid language, this material is also drafted as per the latest CBSE curriculum, which makes it a perfect revision guide.

Class 9 Chapter 1 Science Notes – Summary

The notes of Ch 1 Science Class 9 begins with the introduction about the matter in our surroundings, its composition, what are they made up of, etc. Although you have learnt this in your previous classes, you must study it again to have a proper grasp over this concept. The next part of Class 9 Science Chapter 1 Notes explain the properties of matter. As you all know that matter is made up of tiny particles, and have unique characteristics. Additionally, these characteristics of matter are provided by a theory termed as Kinetic Theory of Matter which states the following:

Particles of matter have space between them.

Particles of matter are continuously moving – rotational and vibrational motion.

Particles of matter attract each other.

In the next section of Science Class 9 Chapter 1 Notes, students will get to know about diffusion; especially the provided diagram makes it easy to comprehend. You will also learn that diffusion occurs at a fast rate in gases. The reason for this is elaborated with suitable examples in the PDF material of Matter in our Surroundings Class 9 Notes. Besides, diffusion in liquids and solids are also discussed precisely that will help you to understand this concept clearly.

Information about the three states of matter – solid, liquid and gas are correctly laid out in this Class 9 Matter in our Surroundings Notes PDF alongside their various properties. Plus a tabular representation of differences between solid, liquid and gases is also given.

Inter Conversion of the States of Matter

With a change in pressure and temperature, matter can change from one form to another form. This section of Class 9 Science Ch 1 Notes deal regarding the same. You will get to know about all the various changes of states like melting, solidification, vaporisation, condensation, sublimation, etc in the Matter in our Surroundings Class 9 Notes.

Latent Heat and Latent Heat of Vaporisation and Fusion

This section of Class 9 th Science Chapter 1 Notes discuss latent heat and different types of the same – latent heat of fusion and latent heat of vaporisation. The different values of latent heat are also given which you can learn in the Matter in our Surroundings Class 9 Notes to write accurate answers.

Sublimation, Evaporation and Cooling Effect

In the last section of Science Chapter 1 Class 9 Notes, you will get an insight about sublimation, evaporation and cooling effect. All these state change processes are supported with appropriate examples.

Make sure to download notes of Science Class 9 Chapter 1 PDF file today from Vedantu’s official website or the app. Along with topic wise explanation, few questions and solutions are also provided in Class 9 Ch 1 Science Notes that will assist you during quick recapitulation before the exams.

Conclusion 

In conclusion, the Matter in Our Surroundings Class 9 Notes CBSE Science Chapter 1 provides a comprehensive understanding of the fundamental concepts related to matter. This chapter introduces students to the basic properties of matter, including its physical nature, states, and characteristics. The notes offer a concise and organised overview of the chapter, aiding students in grasping the key concepts effectively.

By studying these notes, students can gain knowledge about the different states of matter, such as solids, liquids, and gases, and the changes they undergo under various conditions. They also learn about the concept of the particle nature of matter and the behaviour of particles at the microscopic level.

FAQs on Matter in Our Surroundings Class 9 Notes CBSE Science Chapter 1 (Free PDF Download)

1. Is matter in our surroundings pure Class 9 notes?

Not all matter surrounding us exists in its pure form. The chapter discusses the difference between a substance and a mixture. A substance is present in its pure form, whereas, a mixture is a combination of two or more pure substances that can be separated into other types of matter through physical processes. For instance, seawater is a mixture of water and salt. Salt can be separated from it by the process of evaporation.

2. What are the basics of Class 9 Science, Chapter 1 - ‘Matter in Our Surroundings?’

Class 9 Science, Chapter 1 - "Matter in Our Surroundings," covers the fundamentals of Physical Chemistry. It starts with an introduction to matter and the three states of matter: solids, liquids, and gases. The chapter then dives into the process through which matter may change states. Some key ideas taught include the boiling and melting points. As the chapter progresses, the distinction between the latent heat of fusion and the latent heat of vapourization becomes obvious. The chapter concludes with a discussion of evaporation and the elements that influence it.

3. Will Revision Notes help ace Class 9 Science, Chapter 1 - ‘Matter in Our Surroundings?’

Yes, Revision Notes of Class 9 Chapter 1 are the ultimate and end saviour for before the exams. The students are required to go through other study material as well.

The first step is to properly read NCERT Class 9 Science, Chapter 1 - "Matter in Our Surroundings." Study the chapter attentively and attempt to grasp the fundamental principles. Clarify your doubts and reason through everything you encounter in the chapter to ensure a thorough comprehension of all the principles. Finally, using Vedantu's Revision notes, answer the back exercise questions. All issues should be reviewed on a regular basis. Answer a large number of previous year problems to learn this chapter and do well on the Science exam.

4. What are the best Revision Notes for Class 9 Science, Chapter 1 - ‘Matter in Our Surroundings?’

The best Revision Notes for Class 9 Science, Chapter 1 - ‘Matter in Our Surroundings,' can be accessed by visiting the page CBSE Class 9 Science Revision Notes Chapter 1 on the Vedantu website. Vedantu's Revision Notes are the most credible and productive study material from an examination viewpoint. If you study from these notes, you don't have to look for other study materials. These notes have zero errors and have been prepared by the best faculty of Chemistry teachers in India. All the important topics of this chapter are covered in simple language. These PDFs are available at free of cost on the Vedantu website and the Vedantu app.

5. Can I use these revision notes even in offline mode? 

Yes, by pre-downloading the pdf of revision notes the students can easily study the notes late on even when they are offline. 

STUDY MATERIALS FOR CLASS 9

• NCERT Solutions
• NCERT Class 9
• NCERT 9 Science
• Chapter 8: Motion

NCERT Solutions for Class 9 Science Chapter 8: Motion

Ncert solutions class 9 science chapter 8 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.

NCERT Solutions for Class 9 Science Chapter 8 Motion is designed with the intention of clarifying doubts and concepts easily. Class 9 NCERT Solutions in Science is a beneficial reference and guide that helps students clear doubts instantly in an effective way. 

Download Exclusively Curated Chapter Notes for Class 9 Science Chapter – 8 Motion

Download most important questions for class 9 science chapter – 8 motion.

NCERT Solutions for Class 9 Science approaches students in a student-friendly way and is loaded with questions, activities, and exercises that are CBSE exam and competitive exam-oriented. NCERT Solutions for Class 9 Science is the contribution of our faculty, having vast teaching experience. It is developed keeping in mind the concept-based approach along with the precise answering method for CBSE examinations. Refer to NCERT Solutions for Class 9 for best scores in CBSE and competitive exams. It is a detailed and well-structured solution for a solid grip on the concept-based learning experience. NCERT for Class 9 Science Solutions is made available in both web and PDF format for ease of access.

• Chapter 1 Matter in Our Surroundings
• Chapter 2 Is Matter Around Us Pure?
• Chapter 3 Atoms and Molecules
• Chapter 4 Structure of the Atom
• Chapter 5 The Fundamental Unit of Life
• Chapter 6 Tissues
• Chapter 7 Diversity in Living Organisms
• Chapter 9 Force And Laws Of Motion
• Chapter 10 Gravitation
• Chapter 11 Work and Energy
• Chapter 12 Sound
• Chapter 13 Why Do We Fall Ill?
• Chapter 14 Natural Resources
• Chapter 15 Improvement in Food Resources

carouselExampleControls111

Previous Next

Access Answers of Science NCERT class 9 Chapter 8: Motion  (All intext and exercise questions solved)

Intext Questions – 1   Page: 100

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Yes, an object which has moved through a distance can have zero displacement if it comes back to its initial position.

Example: If a person jogs in a circular park which is circular and completes one round. His initial and final position is the same.

Hence, his displacement is zero.

2. A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Side of the given square field = 10m

Hence, the perimeter of a square = 40 m

Time taken by the farmer  to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of 1 m

Distance covered by the farmer in 2 min 20 sec = 1 x 140 = 140 m

The total number of rotations taken by the farmer to cover a distance of 140 m = total distance/perimeter

At this point, let us say the farmer is at point B from the origin O

Therefore,  from Pythagoras theorem, the displacement s = √(10 2 +10 2 )

s =  10 √ 2

s = 14.14 m

3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.

Neither of the statements is true.

(a) Given statement is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero.

(b) Given statement is false because the displacement of an object can be equal to, but never greater than the distance travelled.

Intext Questions – 2   Page: 102

1. Distinguish between speed and velocity.

2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Since average speed is the total distance travelled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance travelled is equal to the displacement.

3. What does the odometer of an automobile measure?

An odometer, or odograph, is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.

4. What does the path of an object look like when it is in uniform motion?

The path of an object in uniform motion is a straight line.

5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10 8 m/s.

Given that the signal travels in a straight line, the distance between the spaceship and the ground station is equal to the total distance travelled by the signal.

5 minutes = 5*60 seconds = 300 seconds.

Speed of the signal = 3 × 10 8 m/s.

Therefore, total distance = (3 × 10 8 m/s) * 300s

= 9*10 10 meters.

Intext Questions – 3   Page: 103

1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Uniform Acceleration:  When an object is travelling in a straight line with an increase in velocity at equal intervals of time, then the object is said to be in uniform acceleration.

The free-falling of an object is an example of uniform acceleration.

Non-Uniform Acceleration:  When an object is travelling with an increase in velocity but not at equal intervals of time is known as non-uniform acceleration.

Bus moving or leaving from the bus stop is an example of non-uniform acceleration.

2. A bus decreases its speed from 80 km h –1 to 60 km h –1 in 5 s. Find the acceleration of the bus.

Given, the initial velocity (u) = 80km/hour = 80000m/3600s= 22.22 m.s -1

The final velocity (v) = 60km/hour = 60000m/3600s= 16.66 m.s -1

Time frame, t = 5 seconds.

Therefore, acceleration (a) =(v-u)/t = (16.66 m.s -1 – 22.22 m.s -1 )/5s

= -1.112 m.s -2

Therefore, the total acceleration of the bus is -1.112m.s -2 . It can be noted that the negative sign indicates that the velocity of the bus is decreasing.

3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h –1 in 10 minutes. Find its acceleration.

Given parameters

Initial velocity (u) = 0

Final velocity (v) = 40 km/h

v = 40 × (5/18)

v = 11.1111 m/s

Time (t) = 10 minute

t = 60 x 10

Acceleration (a) =?

Consider the formula

11.11 = 0 + a × 600

11,11 = 600 a

a = 11.11/600

a = 0.0185 ms -2

Intext Questions – 4   Page: 107

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

For uniform motion, the distance-time graph is a straight line. On the other hand, the distance-time graph of an object in non-uniform motion is a curve.

The first graph describes the uniform motion and the second one describes the non-uniform motion.

2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

The distance-time graph can be plotted as follows.

When the slope of the distance-time graph is a straight line parallel to the time axis, the object is at the same position as time passes. That means the object is at rest.

3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

The speed-time graph can be plotted as follows.

Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion.

4. What is the quantity which is measured by the area occupied below the velocity-time graph?

Considering an object in uniform motion, its velocity-time graph can be represented as follows.

Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by OA*OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:

Area under the velocity-time graph = velocity*time.

Substituting the value of velocity as displacement/time in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.

Intext Questions – 5 Page: 109,110

1. A bus starting from rest moves with a uniform acceleration of 0.1 m s -2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

(a) Given, the bus starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) = 0.1 m.s -2

Time = 2 minutes = 120 s

Acceleration is given by the equation a=(v-u)/t

Therefore, terminal velocity (v) = (at)+u

= (0.1 m.s -2 * 120 s) + 0 m.s -1

= 12 m.s -1 + 0 m.s -1

Therefore, terminal velocity (v) = 12 m/s

(b) As per the third motion equation, 2as = v 2 – u 2

Since a = 0.1 m.s -2 , v = 12 m.s -1 , u = 0 m.s -1 , and t = 120 s, the following value for s (distance) can be obtained.

Distance, s =(v 2 – u 2 )/2a

=(12 2 – 0 2 )/2(0.1)

Therefore, s = 720 m.

The speed acquired is 12 m.s -1 and the total distance travelled is 720 m.

2. A train is travelling at a speed of 90 km h –1 . Brakes are applied so as to produce a uniform acceleration of –0.5 m s -2 . Find how far the train will go before it is brought to rest.

Given, initial velocity (u) = 90 km/hour = 25 m.s -1

Terminal velocity (v) = 0 m.s -1

Acceleration (a) = -0.5 m.s -2

As per the third motion equation, v 2 -u 2 =2as

Therefore, distance traveled by the train (s) =(v 2 -u 2 )/2a

s = (0 2 -25 2 )/2(-0.5) meters = 625 meters

The train must travel 625 meters at an acceleration of -0.5 ms -2 before it reaches the rest position.

3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s -2 . What will be its velocity 3 s after the start?

Given, initial velocity (u) = 0 (the trolley begins from the rest position)

Acceleration (a) = 0.02 ms -2

Time (t) = 3s

As per the first motion equation, v=u+at

Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms -2 )(3s)= 0.06 ms -1

Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s -1

4. A racing car has a uniform acceleration of 4 m s -2 . What distance will it cover in 10 s after start?

Given, the car is initially at rest; initial velocity (u) = 0 ms -1

Acceleration (a) = 4 ms -2

Time period (t) = 10 s

As per the second motion equation, s = ut+1/2 at 2

Therefore, the total distance covered by the car (s) = 0 * 10m + 1/2 (4ms -2 )(10s) 2

= 200 meters

Therefore, the car will cover a distance of 200 meters after 10 seconds.

5. A stone is thrown in a vertically upward direction with a velocity of 5 m s -1 . If the acceleration of the stone during its motion is 10 m s –2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Given, initial velocity (u) = 5 m/s

Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)

Acceleration = 10 ms -2 in the direction opposite to the trajectory of the stone = -10 ms -2

As per the third motion equation, v 2 – u 2 = 2as

Therefore, the distance travelled by the stone (s) = (0 2 – 5 2 )/ 2(10)

Distance (s) = 1.25 meters

As per the first motion equation, v = u + at

Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a

=(0-5)/-10 s

Time taken = 0.5 seconds

Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.

Exercises Page: 112,113

1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Given, diameter of the track (d) = 200m

Therefore, the circumference of the track (π*d) = 200π meters.

Distance covered in 40 seconds = 200π meters

Distance covered in 1 second = 200π/40

Distance covered in 2minutes and 20 seconds (140 seconds) = 140 * 200π/40 meters

= (140*200*22)/(40* 7) meters = 2200 meters

Number of rounds completed by the athlete in 140 seconds = 140/40 = 3.5

Therefore, the final position of the athlete (with respect to the initial position) is at the opposite end of the circular track. Therefore, the net displacement will be equal to the diameter of the track, which is 200m.

Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Given, distance covered from point A to point B = 300 meters

Distance covered from point A to point C = 300m + 100m = 400 meters

Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds

Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds

Displacement from A to B = 300 meters

Displacement from A to C = 300m – 100m = 200 meters

Average speed = total distance travelled/ total time taken

Average velocity = total displacement/ total time taken

Therefore, the average speed while traveling from A to B = 300/150 ms -1 = 2 m/s

Average speed while traveling from A to C = 400/210 ms -1 = 1.9 m/s

Average velocity while traveling from A to B =300/150 ms -1 = 2 m/s

Average velocity while traveling from A to C =200/210 ms -1 = 0.95 m/s

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km.h –1 . On his return trip along the same route, there is less traffic and the average speed is 30 km.h –1 . What is the average speed for Abdul’s trip?

Distance travelled to reach the school = distance travelled to reach home = d (say)

Time taken to reach school = t 1

Time taken to reach home = t 2

therefore, average speed while going to school = total distance travelled/ total time taken = d/t 1 = 20 kmph

Average speed while going home = total distance travelled/ total time taken = d/t 2 = 30 kmph

Therefore, t 1 = d/20 and t 2 = d/30

Now, the average speed for the entire trip is given by total distance travelled/ total time taken

= (d+d)/(t 1 +t 2 )kmph = 2d/(d/20+d/30)kmph

= 120/5 kmh -1 = 24 kmh -1

Therefore, Abdul’s average speed for the entire trip is 24 kilometers per hour.

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s –2 for 8.0 s. How far does the boat travel during this time?

Given, initial velocity of the boat = 0 m/s

Acceleration of the boat = 3 ms -2

Time period = 8s

As per the second motion equation, s = ut + 1/2 at 2

Therefore, the total distance travelled by boat in 8 seconds = 0 + 1/2 (3)(8) 2

= 96 meters

Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.

5. A driver of a car travelling at 52 km h –1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h –1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

The speed v/s time graphs for the two cars can be plotted as follows.

The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.

Therefore, displacement of the first car = area of triangle AOB

= (1/2)*(OB)*(OA)

But OB = 5 seconds and OA = 52 km.h -1 = 14.44 m/s

Therefore, the area of the triangle AOB is given by: (1/2)*(5s)*(14.44ms -1 ) = 36 meters

Now, the displacement of the second car is given by the area of the triangle COD

= (1/2)*(OD)*(OC)

But OC = 10 seconds and OC = 3km.h -1 = 0.83 m/s

Therefore, area of triangle COD = (1/2)*(10s)*(0.83ms -1 ) = 4.15 meters

Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters. Therefore, the first car (which was traveling at 52 kmph) travelled farther post the application of brakes.

6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

(a) Since the slope of line B is the greatest, B is travelling at the fastest speed.

(b) Since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.

(c) Since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.

Since the initial point of an object C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 km

When B passes A, the distance between the origin and C is 8km

Therefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km

(d) The distance that object B has covered at the point where it passes C is equal to 9 graph units.

Therefore, total distance travelled by B when it crosses C = 9*(4/7) = 5.14 km

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s -2 , with what velocity will it strike the ground? After what time will it strike the ground?

Given, initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance travelled by the ball (s) = 20m

Acceleration (a) = 10 ms -2

As per the third motion equation,

v 2 – u 2 = 2as

= 2*(10ms -2 )*(20m) + 0

v 2 = 400m 2 s -2

Therefore, v= 20ms -1

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation,

Therefore, t = (v-u)/a

= (20-0)ms -1 / 10ms -2

= 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

8. The speed-time graph for a car is shown in Fig. 8.12

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?

The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

(1/2)*4*6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6 th to the 10 th second.

9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is possible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

Circular motion is an example of an object moving with acceleration but with uniform speed.

An object moving in a circular path with uniform speed is still under acceleration because the velocity changes due to continuous changes in the direction of motion.

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Given, the radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2*π*42250km = 265571.42 km

Time is taken for the orbit = 24 hours

Therefore, speed of the satellite = 11065.4 km.h -1

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.

NCERT Class 9 Science Chapter 8 explains the concept of motion, and types of motion with relevant examples for a clear understanding of the concept. It explains the causes of phenomena like sunrise, sunset, and changing of the seasons. It helps students understand uniform and non-uniform motion. Distance-time graphs and velocity-time graphs, which are considered important concepts for examination, are explained in an easy way in NCERT Solutions . It describes how the acceleration of an object is the change in velocity per unit time. NCERT Class 9 Science Chapter 8 is covered under Unit III: Motion, Force and Work and can get you maximum marks.

• NCERT Solutions for Class 9 explains motion in terms of distance moved or displacement.
• Uniform and non-uniform motions of objects are explained through the graph and examples.
• Uniform circular motion concept is made understandable in a simple way.
• Problems on acceleration, velocity, and average velocity are also solved.

Key Features of NCERT Solutions for Class 9 Science Chapter 8: Motion

• A simple and easily understandable approach is followed in NCERT Solutions to make students aware of topics.
• Provides complete solutions to all the questions present in the respective NCERT textbooks.
• NCERT Solutions offers detailed answers to all the questions to help students in their preparations.
• These solutions will be useful for CBSE exams, Science Olympiads, and other competitive exams.

Disclaimer:

Dropped Topics –  8.5 Equations of motion by graphical method, 8.5.1 Equation for Velocity–Time Relation, 8.5.2 Equation for Position–Time relation and 8.5.3 Equation for Position– Velocity.

Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 8

What will i learn from chapter 8 of ncert solutions for class 9 science, how will the ncert solutions for class 9 science chapter 8 help us to score well in the cbse exam, how can i solve the problems from the ncert solutions for class 9 science chapter 8 effortlessly.

this is very useful to the children who scared from physics . I am so HELPFUL from this site for my exams as my exam are coming.

Very useful

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.