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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.

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CBSE Worksheets for Class 9 Science

CBSE Worksheets for Class 9 Science: One of the best teaching strategies employed in most classrooms today is Worksheets. CBSE Class 9 Science 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 Science Worksheets for the students of Class 9. All our CBSE NCERT Class 9 Science 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 Science will be useful to test your conceptual understanding.

Board: Central Board of Secondary Education(www.cbse.nic.in) Subject: Class 9 Science Number of Worksheets: 30

CBSE Class 9 Science Worksheets PDF

All the CBSE Worksheets for Class 9 Science 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 Science 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 Science Worksheet. CBSE Worksheets for Class 9 Science can also use like assignments for Class 9 Science students.

• CBSE Worksheets for Class 9 Science All Chapters Assignments
• CBSE Worksheets for Class 9 Science Diversity in Living Organisms Animals Assignment
• CBSE Worksheets for Class 9 Science Diversity in Living Organisms Plants Assignment
• CBSE Worksheets for Class 9 Science Experiments Assignment
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• CBSE Worksheets for Class 9 Science Fundamental Unit of Life Assignment
• CBSE Worksheets for Class 9 Science Gravitation Assignment
• CBSE Worksheets for Class 9 Science Improvement in Food Resources Assignment 1
• CBSE Worksheets for Class 9 Science Improvement in Food Resources Assignment 2
• CBSE Worksheets for Class 9 Science Law of Conservation of Mass Assignment
• CBSE Worksheets for Class 9 Science Matter around Us Assignment
• CBSE Worksheets for Class 9 Science Melting and Boiling Point Assignment
• CBSE Worksheets for Class 9 Science Physical and Chemical Changes Assignment
• CBSE Worksheets for Class 9 Science Preparation of Mixture and Compound Assignment
• CBSE Worksheets for Class 9 Science Seperation of Components of Mixture Assignment
• CBSE Worksheets for Class 9 Science Sound Assignment
• CBSE Worksheets for Class 9 Science Structure of Atom Assignment
• CBSE Worksheets for Class 9 Science Tissues Animals Assignment
• CBSE Worksheets for Class 9 Science Tissues Plants Assignment
• CBSE Worksheets for Class 9 Science Why do we fall Ill Assignment
• CBSE Worksheets for Class 9 Science Assignment 1
• CBSE Worksheets for Class 9 Science Assignment 2
• CBSE Worksheets for Class 9 Science Assignment 3
• CBSE Worksheets for Class 9 Science Assignment 4
• CBSE Worksheets for Class 9 Science Assignment 5
• CBSE Worksheets for Class 9 Science Assignment 6
• CBSE Worksheets for Class 9 Science Assignment 7
• CBSE Worksheets for Class 9 Science Assignment 8
• CBSE Worksheets for Class 9 Science Assignment 9
• CBSE Worksheets for Class 9 Science Assignment 10

Advantages of CBSE Class 9 Science Worksheets

• By practising NCERT CBSE Class 9 Science 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 Science Workbook Helps to improve subject-knowledge.
• CBSE Class 9 Science Worksheets encourages classroom activities.

Worksheets of CBSE Class 9 Science 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 Science Chapter wise question bank pdf and access it anytime, anywhere for free. Browse further to download free CBSE Class 9 Science Worksheets PDF .

Now that you are provided all the necessary information regarding CBSE Class 9 Science 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 Science revision worksheets . And more importantly, students should need to follow through the worksheets after completing their syllabus.  Working on CBSE Class 9 Science Worksheets will be a great help to secure good marks in the examination. So start working on Class 9 Science Worksheets to secure good score.

CBSE Worksheets For Class 9

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• NCERT Solutions for Class 9 Science Chapter 1 - Matters In Our Surroundings
• NCERT Solutions

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

The NCERT Answer for Class 9 Science Chapter 1 was written with the sole purpose of assisting students in obtaining more marks and improving their grades in mind. The answer for class 9 Science chapter 1 addresses students' academic expectations and needs. The themes in Chapter 1 on Matter in Our Surroundings are covered in accordance with the CBSE syllabus . The solution's strategy incorporates a step-by-step method that assists in improved knowledge of the issues.

NCERT Solution of class 9 Science chapter 1 has been created, keeping in mind the primary goal of helping students to secure more marks and improve grades. The academic requirements and needs of students have been addressed in the solution for class 9 Science chapter 1. The topics in chapter 1 related to Matter in Our Surrounding are discussed according to the CBSE syllabus. The approach in the solution involves a step-by-step process which aids in better comprehension of the topics.

CBSE class 9 Science chapter 1 solutions explain all the foundational concepts in detail for students to understand better. The solutions are made available to everybody over the Vedantu app. These free PDFs can be downloaded easily from Vedantu’s official portal. 

You can also download NCERT Solutions for Class 9 Maths and NCERT Solution for Class 9 Science to help you to revise complete syllabus and score more marks in your examinations.

Access NCERT solutions for Class 9 Science Chapter – 1 Matter in our Surroundings

Intext exercise -1.

1. Which of the following are the matter?

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

Ans: Matter is anything that occupies space and has some mass. There are three states of matter called Solid, Liquid and Gas. On the basis of these three states, we can define that which of this is a matter:

Chair and almond are said to be in a solid state of matter as these have fixed shape., cold drink is in liquid state as it has the tendency to flow., air and the smell of perfume have gaseous particles which are free to move so this will also be considered as a gaseous state of matter..

While Love, hate, cold, smell and thought are not having any mass or neither do they occupy space these are just emotions or sensations felt by human beings so they are not considered as matter.

2. Give reasons for the following observation:

The smell of hot sizzling food reaches you several meters away, but to get the smell from cold food you have to go close.

Ans: The smell of hot sizzling food prepared by our mom reaches to us in our room from kitchen but if the food gets cold after some time, we did not feel any smell of that food this phenomenon can be defined on the basis of rate of diffusion which gets increases when the temperature get increases as high temperature increases the kinetic energy of food particles to get diffused in air. The temperature of hot food particles is high as compared to old one so its molecules get easily diffused in the air as compared to cold ones.

3: a diver is able to cut through water in a swimming pool. which property of matter does this observation show, ans: this can be explained on the properties of matter to attract the particles towards themselves and this will also decide their shape and rigidity. the force of attraction is highest in case of solid as compared to liquid and gas this defines that particles of solid are tightly bound to each other. while in case of liquid particles they have less forces of attraction which defines that there is space between the particles and due to this reason that we can cut them easily. that is why we can say that due to less forces of attraction between water molecules a diver is able to cut through water in a swimming pool., 4: what are the characteristics of particles of matter, ans: matter is anything that occupies space and has some mass. there are three states of matter called solid, liquid and gas., the main characteristics of matter can be described as follows:, particles of matter have space between them and the order of spacing is highest in gas after that liquid and solid have very less space between their particles., particles of matter are continuously moving in all the three states of matter., particles of matter attract each other with strong forces which help them to bind with each other. in solid particles attraction is very high whereas in liquid it is low and in gases it is quite low. , intext exercise -2.

1: The mass per unit volume of a substance is called density (density = mass/volume). Arrange the following in order of increasing density − air, exhaust from chimney, honey, water, chalk, cotton, and iron.

Ans: Density is depending on mass and volume hence higher the mass higher will be the density and out of these heavier particles have higher mass as compare to lighter one so the order of increasing density of given substances can be written as follows:

Air < Exhaust from chimney < Cotton < Water < Honey < Chalk < Iron.

Tabulate the differences in the characteristics of states of matter.

Matter is anything that occupies space and has some mass. There are three states of matter called Solid, Liquid and Gas.

Comment upon the following: rigidity, compressibility, fluidity, filling a gas container, shape, kinetic energy, and density.

Ans:  

Rigidity: it is defined as the tendency of matter to resist a change in shape..

Compressibility: The ability of matter to reduce in volume when any type of external force is applied on it. 

Fluidity: Tendency of particles to flow this property can be seen in case of liquid and gases which can also be known as fluids. 

Filling a Gas Container: Gases neither have definite shape nor have definite volume. Gases take the shape of that container in which it gets filled. Hence by filling a gas container, it means the attainment of shape of the container by the gas.

Shape: Shape corresponds to fixed volume and boundary. Only solids have a fixed shape.

Kinetic Energy: Particles which produce energy possessed due to its continuous motion.

Density: It is defined as the mass per unit volume.

3: Give Reasons:

A Gas Fills Completely the Vessel in Which it is Kept.

The gas particles have a tendency to move freely in all directions as they have very less force of attraction between their particles. Like water, gas can also take the shape of the container in which it is kept. Therefore, we can say that gas completely fills the vessel in which it is kept.

A Gas Exerts Pressure on the Walls of the Container.

The gas particles move freely due to its lesser forces of attraction between the particles. Therefore, these gaseous particles continuously collide with each other and with the walls of the container with a greater force. Pressure is known as the force produced by the gas particles per unit area. By this we can say that gas exerts pressure on the walls of the container.

A wooden table should be called a solid.

A wooden table is very rigid in nature which means that it has a definite shape and its shape cannot be changed easily and has definite volume too. The shape is fixed due to strong intermolecular forces hence it attains all the properties of solid therefore, it is considered as a solid.

We can easily move our hands in air, but to do the same through a solid block of wood, we need a karate expert.

Air particles have very less forces of attraction between their particles so they have large space in between them. But wood has very little space between the particles due to its high force of attraction therefore, wood is considered to be of rigid nature. Due to this reason, we can easily move our hands in air but the same will not happen through a solid block of wood. For this we need a karate expert.

4: Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why.

Ans: as we know that density is defined as the mass per unit volume. this corresponds with the increase of the volume of any substance density will decrease as they are inversely proportional to each other. .

Ice is solid in nature therefore it contains strong intermolecular forces which tightly bound them and they contain lesser volume but on the other hand liquid has tendency to move freely due to weak intermolecular forces and contain large volume.

From this we can say that water has larger volume and lesser density so it has a tendency to float on water.

Intext Exercise -3

1: convert the following temperature to celsius scale:.

Celsius and Kelvin are two main scales to measure the temperature. By subtracting the 273K from the given value we can get the value in degrees Celsius. The formula corresponds to degree Celsius can be shown as:

\[X{}^\circ C=(X-273)\]

\[300K=(300-273){}^\circ C=27{}^\circ C\]

 573 K

$573K=(573-273){}^\circ C=300{}^\circ C$

2: What is the physical state of water at:

$\mathbf{\text{25}{{\text{0}}^{\text{o}}}\text{C}}$

Ans: Physical state corresponds to the state of matter whether it exists in solid, liquid or gas.

$\text{25}{{\text{0}}^{\text{o}}}\text{C}$– As we know that water starts boiling at 100°C and above this temperature water exists in gaseous state.         So, this defines that water at 250°C exists in a gaseous state.

$\mathbf{{{100}^{o}}C}$

${{100}^{o}}C$ – It is the starting point where water starts boiling so at this temperature water exists in both liquid and gaseous states.

3: For any substance, why does the temperature remain constant during the change of state?

Ans: this can be explained as the whole heat which we are providing to the substance to increase its temperature is used to break the intermolecular forces of attraction between them. this heat will also correspond to the latent heat i.e., the heat which gets absorbed or released during change of state. hence all the energy gets used so temperature remains constant. , 4: suggest a method to liquefy atmospheric gases., ans: atmospheric gases can be defined as the gases present in the atmosphere. it can be liquified i.e., converted into liquid by applying suitable conditions of applying pressure and by reducing their temperature., intext exercise -4, 1: why does a desert cooler cool better on a hot dry day, ans: this can be explained by the process called evaporation, this is the process in which the liquid particles absorb energy from the surroundings and cause cooling. the rate of evaporation generally depends on the amount of water vapour present in the air. if the amount of water vapor present in air is more than the rate of evaporation is more or vice-versa. on a hot dry day, the amount of water vapor present in air is quite low so this will evaporate easily and make its surroundings cooler. thus, from this we can say that a desert cooler cools better on a hot dry day as compared to rainy one., 2: how does water kept in an earthen pot (matka) become cool during summers, ans: an earthen pot or matka is generally made up of sand particles in which many tiny pores exist and this helps the water inside the pot to evaporate and surroundings makes the water cool. this is the reason why people kept the water in an earthen pot during summers., 3: why does our palm feel cold when we put some acetone or petrol or perfume on it, ans: acetone, petrol or perfume are considered as organic compounds which are volatile in nature whereas volatile substances are those which easily get vaporized and go through the process of evaporation. we know that during the process of evaporation particles of these organic liquids absorb energy from the surroundings or the surface of the palm and make the surroundings or surface of the palm somewhat cool. this is the reason why our palm feels cold when we put some acetone, petrol or perfume on it. , 4: why are we able to sip hot tea or milk faster from a saucer than a cup, ans: this can also be explained on the basis of rate of evaporation as we know that evaporation produces a cooling effect and evaporation depends on the surface area, larger the surface area higher the evaporation. as in saucer the area is larger as compared to cup so evaporation will be high in case of greater surface area. thus, we can say that liquid cools faster in a saucer than in a cup and due to this reason, we are able to sip hot tea or milk faster from a saucer than a cup., 5: what type of clothes should we wear in the summer, ans: in summer we usually sweat so we have to wear cotton or light-colored clothes because cotton or light-colored clothes can absorb more sweat from our body and transfers the sweat which is in the form of liquid to the atmosphere and makes the evaporation process faster. evaporation process causes a cooling effect which makes our body cool in cotton clothes as compared to synthetic or woolen ones. , ncert exercise, 2: convert the following temperature to kelvin scale:.

$\mathbf{25{}^\circ C}$

Celsius and Kelvin are two main scales to measure the temperature. By adding the 273K from the given value we can get the value in degrees Celsius. The formula corresponds to degree Celsius can be shown as:

\[0{}^\circ C=273K\]

\[{{27}^{o}}C=(27+273)K=300K\]

$\mathbf{373{}^\circ C}$

${{373}^{o}}C=(373+273)K=646K$

3: Give reason for the following observations.

Naphthalene balls disappear with time without leaving any solid.

This phenomenon can be explained on the basis of sublimation which defines that solid is directly converted into gaseous form without turning it into liquid. naphthalene is one that substances which undergo through the process of sublimation easily at room temperature. that is why we can say that naphthalene balls disappear after some time without leaving any solid., we can get the smell of perfume sitting several meters away., gaseous particles have very less internuclear forces due to which its molecules are very free to move and it possesses high kinetic energy. due to this reason particles of perfumes diffuse into the atmosphere and its molecules will mix in the environment which enables us to smell the perfume from several meters away., 4: arrange the following substances in increasing order of forces of attraction between particles − water, sugar, oxygen., ans: forces of attraction is the attracting power of molecules which keep them together and intermolecular forces are very strong in case of solid as compared to liquid or gas. , here sugar is said to be solid and contains higher forces of attraction., water is liquid comparatively containing lesser forces of attraction but higher as compared to gases., oxygen is a gas which contains very less attraction between forces., thus, the increasing order of forces of attraction between the particles of water, sugar and oxygen is.

Oxygen < Water < Sugar

5: What is the physical state of water at:

$\mathbf{{{25}^{o}}C}$

Ans: Physical state corresponds to the state of matter whether it exists in solid, liquid or gas. 

$25{}^\circ C$– As we know that water melts at 0°C and above this temperature water exists in liquid state.         So, this defines that water at 25°C exists in a liquid state

$\mathbf{{{0}^{o}}C}$

Ans: $0{}^\circ C$– It is the temperature at which water starts melting i.e., water gets converted into liquid from ice so at this temperature water exists as both solid and liquid state. 

Ans: $100{}^\circ C$– It is the starting point where water starts boiling so at this temperature water exists in both liquid and gaseous states.

6: Give two reasons to justify:

water at room temperature is a liquid.

We find water is in liquid state at room temperature this can be justified as follows:

Water does not have any fixed shape; it can take the shape of the container in which it is kept and water has definite volume.

It has a tendency to flow.

Have weak intermolecular forces between their particles.

These all describe the property of liquid so we can say that water is liquid at room temperature.

An Iron Almirah is a Solid at Room Temperature.

Iron almirah kept in our room is said to be solid due to following reasons:

It has a fixed shape and definite volume.

It contains strong intermolecular forces between their particles.

Rigid in nature, difficult to compress.

These all describe the property of solid so we can say that almirah is solid at room temperature.

7: Why is ice at 273 K more effective in cooling than water at the same temperature?

Ans: here condition given that both ice and water are at same temperature i.e. 273 k. but ice at 273 k has less energy as compared to water this can be explained on the basis of latent heat of fusion which is possessed by water as an additional energy but ice does not have such type of energy. therefore, we can say that at 273 k ice is more effective in cooling as compared to water., 8: what produces more severe burns, boiling water or steam, ans: steam and water both are said to be at the same temperature i.e., 373 k. but steam contains more energy as compared to boiling water. this can be explained on the basis of latent heat of fusion which is possessed by water as an additional energy. therefore, steam produces more severe burns than boiling water., 9: name a, b, c, d, e and f in the following diagram showing change in its state..

A is the process of converting solid into liquid is called Melting.

B is the process which converts liquid into gaseous state, this is called vaporisation.

C in which gases get converted into liquid this is called condensation.

D is the process which converts liquid into solid. It is called solidification.

E and F are the processes which convert solid into gas or vice versa is known as sublimation. 

You can opt for Chapter 1 - Matter in Our Surroundings NCERT Solutions for Class 9 Science PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Science

Chapter 1 - Matter in Our Surroundings

Chapter 2 - Is Matter Around us Pure

Chapter 3 - Atoms and Molecules

Chapter 4 - Structure of Atom

Chapter 5 - The Fundamental Unit of Life

Chapter 6 - Tissues

Chapter 7 - Diversity in Living Organisms

Chapter 8 - Motion

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

Important Concepts Covered in Chapter 1 - Matter in Our Surroundings of Class 9 Science NCERT Solutions 

The first chapter delves into the meaning of the matter, its definition, and the various physical states of matter. The matter is defined as everything that takes up space and has mass. For example, water and sugar, sand and sugar, hydrogen and oxygen, and so forth. The matter is made up of minuscule, microscopic particles. Because matter particles have space between them, they are attracted to one another.

Matter in Our Surrounding is the first chapter of Class 9 Science and it covers the following concepts.

Class 9 Science Chapter 1 Matter in Our Surroundings

Even though the topic of Matter in Our Surroundings can be very wide in scope, NCERT solution of Class 9 Science Chapter 1 strictly adheres to the CBSE syllabus. Here are the topics covered –

Matter in Our Surroundings

The introductory chapter elucidates the concept of matter, its definition and the and the physical states in which it exists in nature.

Physical State of Matter 

The three physical states of matter are discussed – solid, liquid and gas. The point is elaborated with instances, one of which exemplifies the property of matter when a diver cuts through water.  

Characteristics of Particles of Matter  

In this chapter, students get to know various characteristics of particles of matter. It discusses that particles of matter – (1) have intermolecular spaces, (2) are in continuous motion and (3) experiences force between one another. 

States of Matter 

Chapter 4 deals with the features of states of matter – compressibility, rigidity, fluidity, kinetic energy, shape and density. 

Can Matter Change its State?

The occurrence of change in state of matter is discussed through the example of ice converting into water and eventually into gaseous form. 

Evaporation 

The concept of evaporation is exemplified through the question as to why a desert cooler functions better on a hot, dry day as opposed to other weather conditions.

Why is it Beneficial for Students to Study from NCERT solutions class 9 Science Matter in Our Surroundings?

Teachers and mentors have always stressed on the importance of self-study in order for a student to succeed. The material that is referred to in the course of self-study also holds a critical value. Here are some of the benefits of studying NCERT solution of Class 9 Science Chapter 1 that would provide students with that much-needed competitive edge – 

The solutions are framed by domain experts possessing much experience. The language of the material has been purposely kept lucid; however, the content remains comprehensive. It enables students to gain in-depth knowledge on the topic 

If there is any concern about the relevance and accuracy of the Class 9 Science Ch 1 NCERT solutions, it can be put to rest. The solution strictly adheres to the CBSE curriculum, without any scope for deviation

It is ensured that the basic concepts are suitably explained so that students remain clear on the topic

Ample problem exercises are provided so that students get sufficient practice 

The NCERT Solutions for Matter in Our Surroundings (Chapter 8) for Class 9 are given in this article.

Conclusion  

This solution contains all of the chapter's questions so that students may prepare for their examinations properly. We have also provided NCERT activities to help pupils grasp the kind or pattern of questions in the CBSE Class 9 test.

While your devotion and constant practice will put you on the right track, NCERT answers will only pave the road. Therefore, get the answer right now and get started on your preparation!

FAQs on NCERT Solutions for Class 9 Science Chapter 1 - Matters In Our Surroundings

1. What are the different characteristics of state of matter explained in NCERT solutions class 9 Science chapter 1?

The different characteristics as elaborated in CBSE Class 9 Science Chapter 1 solutions are based on six parameters - (1) Shape, (2) Volume, (3) Rigidity or Fluidity, (4) Intermolecular force, (5) Intermolecular space, and (6) Compressibility. 

A few of the properties listed in Matter in Our Surroundings Class 9 NCERT answers suggest that solids have a set form, volume, and are rigid, whereas liquids have no fixed shape or volume and are less rigid than solids. Gas, on the other hand, has no definite shape or volume and is not hard.

2. Which are the chapters included in Class 9 Science as per the NCERT Science Book?

There are altogether fifteen chapters included in CBSE class 9 Science – (1) Matter in Our Surroundings, (2) Is Matter Around Us Pure, (3) Atoms and Molecules, (4) Structure of Atom, (5) The Fundamental Unit of Life, (6) Tissues, (7) Diversity in Living Organisms, (8) Motion, (9) Force and Laws of Motion, (10) Gravitation, (11) Work and Energy, (12) Sound, (13) Why do We Fall Ill, (14) Natural Resources, (15) Improvement in Food Resources.

3. Can NCERT Solutions be an effective way to prepare Science Class 9 Chapter 1?

Yes, NCERT Solutions will help the studnets to prepare for chapter 1 of Class 9 pretty well. While there can be a multitude of ways to prepare Ch 1 Class 9 Science, it is only when such preparation is done in a systematic manner, can an optimal output be gained. It is advised that instead of starting with a topic in Class 9 Chapter 1 solution of one’s preference, proceed according to the flow of the chapter.

4. How many questions are present in each exercise of NCERT Solutions for Class 9 Science Chapter 1?

There are five exercises in the NCERT Solutions for Class 9 Science Chapter 1. In Exercise 1, there are four questions. In Exercise 2, there are four questions. Exercise 3 also consists of four questions. In Exercise 4, there are four questions. Lastly, the NCERT exercise has nine questions. All of these questions are provided along with their solutions in the NCERT Solutions for Class 9 Science Chapter 1. 

5. Are the NCERT Solutions for Class 9 Science Chapter 1 sufficient for the exam preparation?

Yes. To do well in the test, students must study and prepare through self-study in addition to what is given in school. Students must use the best study resources provided on Vedantu for self-study. Students should refer to NCERT Answers for Class 9 Science Chapter 1 and others when studying for the Class 9 Science test. They are developed by qualified teachers and are simple to grasp.

6. What is the matter in our surroundings?

Matter in our surroundings is the first topic covered in Chapter 1 of Class 9 Science. In this chapter, students learn what matter is and which entities can not be considered as matter. The five basic elements that constitute matters in our surroundings are fire, earth, sky, water and air. The building blocks of matter are very small particles that can only be seen under a microscope. 

7. Write in short answer format what is matter according to the syllabus of Class 9.

Matter is defined as everything that fills space and has mass. Matter may be perceived by humans through their senses. Water, earth, fire, sky, and air are components that make up matter. It is made up of microscopic particles that can only be seen under a microscope. These are the fundamental building components of matter. Light, heat, electricity, and magnetism are not considered matter since they do not occupy space or have mass.

8. How to download the NCERT Solutions Class 9 Science Chapter 1?

Students can use NCERT Solutions for Class 9 Science Chapter 1 to download the NCERT Solutions Class 9 Science Chapter 1 from Vedantu (vedantu.com). These solutions contain the answers to all the questions from Chapter 1 of the Class 9 Science NCERT textbook in a concise and structured manner. These are available free of cost on Vedantu (vedantu.com). Students can download these using the Vedantu app as well. By going through the solutions, students can learn how to write their answers in the exam. Apart from the solutions, students can also find other study material on Vedantu, including important questions, revision notes and previous year question papers.

NCERT Solutions for Class 9

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Students can refer to Assignments for Class 9 Science available for download in Pdf. We have given below links to subject-wise free printable Assignments for Science Class 9 which you can download easily. All assignments have a collection of questions and answers designed for all topics given in your latest NCERT Books for Class 9 Science for the current academic session. All Assignments for Science Grade 9 have been designed by expert faculty members and have been designed based on the type of questions asked in standard 9 class tests and exams. All Free printable Assignments for NCERT CBSE Class 9, practice worksheets, and question banks have been designed to help you understand all concepts properly. Practicing questions given in CBSE NCERT printable assignments for Class 9 with solutions and answers will help you to further improve your understanding. Our faculty have used the latest syllabus for Class 9. You can click on the links below to download all Pdf assignments for class 9 for free. You can get the best collection of Kendriya Vidyalaya Class 9 Science assignments and questions workbooks below.

Class 9 Science Assignments Pdf Download

CBSE NCERT KVS Assignments for Science Class 9 have been provided below covering all chapters given in your CBSE NCERT books. We have provided below a good collection of assignments in Pdf for Science standard 9th covering Class 9 questions and answers for Science. These practice test papers and workbooks with question banks for Class 9 Science Pdf Download and free CBSE Assignments for Class 9 are really beneficial for you and will support in preparing for class tests and exams. Standard 9th students can download in Pdf by clicking on the links below.

Subjectwise Assignments for Class 9 Science

Benefits of Solving Class 9 Science Assignments

• The best collection of Grade 9 assignments for Science have been provided below which will help you in getting better marks in class tests and exams.
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• We have provided step by step solutions for all questions in the Class 9 assignments so that you can understand the solutions in detail.
• We have provided single click download links to all chapterwise worksheets and assignments in Pdf.
• Class 9 practice question banks will support to enhance subject knowledge and therefore help to get better marks in exam

FAQs by Science Students in Class 9

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CBSE Worksheets for Class 9 Science

CBSE Worksheets for Class 9 Science: One of the best teaching strategies employed in most classrooms today is Worksheets. CBSE Class 9 Science 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 CoolGyan have come up with Kendriya Vidyalaya Class 9 Science Worksheets for the students of Class 9. All our CBSE NCERT Class 9 Science 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 Science will be useful to test your conceptual understanding. Board: Central Board of Secondary Education(www.cbse.nic.in) Subject: Class 9 Science Number of Worksheets: 30

CBSE Class 9 Science Worksheets PDF

All the CBSE Worksheets for Class 9 Science 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 Science 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 Science Worksheet. CBSE Worksheets for Class 9 Science can also use like assignments for Class 9 Science students.

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• CBSE Worksheets for Class 9 Science Assignment 10

Advantages of CBSE Class 9 Science Worksheets

• By practising NCERT CBSE Class 9 Science Worksheet , students can improve their problem solving skills.
• Helps to develop the subject knowledge in a simple, fun and interactive way.
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Worksheets of CBSE Class 9 Science are devised by experts of CoolGyan 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 Science Chapter wise question bank pdf and access it anytime, anywhere for free. Browse further to download free CBSE Class 9 Science Worksheets PDF . Now that you are provided all the necessary information regarding CBSE Class 9 Science 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 Science revision worksheets . And more importantly, students should need to follow through the worksheets after completing their syllabus. Working on CBSE Class 9 Science Worksheets will be a great help to secure good marks in the examination. So start working on Class 9 Science Worksheets to secure good score.

CBSE Worksheets 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

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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

= 2/[(3 + 2)/60]

= 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.

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Matter in Our Surroundings Chapter 1 Class 9 Science Assignments

Please refer to Matter in Our Surroundings Chapter 1 Class 9 Science Assignments below. We have provided important questions and answers for Matter in Our Surroundings which is an important chapter in Class 9 Science. Students should go through the notes and also learn the solved assignment with exam solved questions provided below. All examination and class tests questions are as per the latest syllabus and books issued by CBSE, NCERT, and KVS. We have also provided Class 9 Science Assignments for all chapters on our website.

Chapter 1 Matter in Our Surroundings Class 9 Science Assignments

Question. Predict the physical state of melting point of a substance is below the room temperature.

Question. We can get the smell of perfume sitting several metres away, why?

This is because perfumes diffuse very fast and can reach to people sitting several metres away.

Question. The boiling point of alcohol is 78°C. What is this temperature on Kelvin scale?

K = °C + 273 = 78 + 273 = 351 K

Question. We can easily move our hand in the air but to do the same through a solid block of wood. We need a karate expert. Why?

In air, the inter-particle attractive forces are negligible and hence, it is easy to separate the particles in air and we can easily move our hand in air. The interparticle forces are very strong in solids. So, it is not easy to separate the particles and it is not easy to move our hand through a solid block of wood.

Question. Give two examples of diffusion.

Milk drops dissolved in water and perfume sprayed in a room.

Question. Express the boiling point of water in Celsius as well as Kelvin scale.

100°C and 373 K.

Question. Name the state of water at 100 degree Celsius, zero degree Celsius and 4 degree Celsius.

The state of water at 100 degree Celsius is gas, at 0 degree Celsius it is solid and at 4 degree Celsius it is liquid.

Question. Name two processes from which it may be concluded that the particles of a gas move continuously.

Compressibility and Brownian movement.

Question. Is it possible to turn a liquid into vapour without heating?

Yes, by the process of evaporation as evaporization of water occur below the boiling point under atmospheric pressure.

Question. What is diffusion?

The intermingling of molecules of one substance with that of the other is called diffusion.

Question. What happens to the rate of diffusion if the temperature is increased?

With increased temperature, the rate of diffusion also increases as the particles gain energy and vibrate more.

Question. What is dry ice?

Solid carbon dioxide obtained by cooling and applying pressure on carbon dioxide gas. It does not melt so it is called dry ice.

Question. What is humidity?

The air holds water vapour, this air with water is called humid air and the amount of water vapour present in the air is called humidity.

Question. What is normal atmospheric pressure?

The atmospheric pressure at sea level is 1 atmosphere and taken as the normal atmospheric pressure.

Question. Write the SI unit of temperature?

Question. Give the temperature at which water exists in two different phases/states.

(i) At 0°C water can be in solid or in liquid state. (ii) At 100°C water can be in liquid or in gaseous state.

Question. What are fluids?

The states of matter that can flow due to less intermolecular force of attraction are liquids and gases and are called fluids.

Question. Define matter.

Anything that occupies space and has mass and is felt by senses is called matter.

Question. Why are light and sound not considered as matter?

Light and sound are not considered as matter because they have no mass and do not occupy space.

Question. What happens if you put copper sulphate crystals in water?

Copper sulphate crystals mixed between the spaces of molecules of water and disappear.

Question. Give state of a matter if this substance has neither a fixed shape nor a fixed volume.

Question. What do you mean by vapour?

A substance that is found in gaseous state only at room temperature is called vapour.

Question. Is it true to say that fluorescent tube contains plasma? Explain Answer.  It is right to say that fluorescent tube contains plasma. As fluorescent tube has helium or some other rare gas.The particles of the gas get ionized in the presence of high voltage applied. These charged particles are called plasma which glows.

Question. A karate expert can easily move his hand through a solid block of wood but we cannot. Why? Answer.  In a solid block of wood, the inter-particle forces are very strong and hence, it is not easy to separate the particles. Therefore, it is not easy to move our hand through a solid block of wood, only a karate expert can do it as he has expertise in this.

Question. What is latent heat of fusion? Answer.  The heat required to change 1 kg of a solid substance into liquid state at the melting point of the substance. For example : Amount of heat required to melt ice at 0°C into water, at 0°C will be known as the latent heat of fusion of ice.

Question. What is compressibility? How it is negligible in solids? Answer.  Compressibility is the ability of a substance to be reduced to its volume under pressure. Solids are incompressible as their particles are held together. So, we can tell that compressibility is negligible in solids.

Question. Two cubes of ice are pressed hard between two palms and after releasing the pressure, the cubes join together. Why? Answer.  Pressure is directly proportional to temperature when we apply pressure, temperature increases then the ice in contact melts and it turns into water. When pressure is removed, the temperature decreases again and melted ice again freezes. Hence, cubes join together.

Question. What is the reason that “Ice has lower density than water”? Answer. The mass per unit volume of a substance is called density (density = mass/volume). The density of substance decreases as the volume of a substance increases. Space between particles increases when water changes into ice. These spaces are larger as compared to the spaces present between the particles of water. Thus, the volume of ice become greater as compared to the water. Hence, the density of ice become lower than that of water. And, a substance with lower density than water can float on water. Thus, ice floats on water.

Question. Why does the temperature remain constant during the change of state, for any substance? Answer. On increasing the temperature of solids, the kinetic energy of the particles increases which is used up in changing the state as it overcome the forces of attraction between the particles, therefore, the temperature remains constant during the change of state.

Question. Why cannot you smell its perfume at a short distance when incense stick is not lighted? Answer.  The particles of the perfume (matter) do not have sufficient energy to drift through the air. Thus, we cannot smell it at a few steps away from incense stick.

Question. How can you show that evaporation causes cooling? Answer.  When we put some acetone on our hand, after some time we will feel coolness on our hand because the acetone absorbs kinetic energy from our hand and evaporates and evaporation causes cooling.

Question. What do you mean by latent heat of vaporization? Answer.  The latent heat of vaporization of a liquid is the quantity of heat in joules required to convert 1 kilogram of the liquid to vapour or gas at its boiling point, without any change in temperature.

Question. Explain, why solids have fixed shape but liquids and gases do not have fixed shape? Answer.  Solids have fixed shape due to strong intermolecular force of attraction between them. The liquids and gases have molecules with less intermolecular force of attraction, and hence they can flow and take shape of the container.

Question. Liquids and gases can be compressed but it is difficult to compress solids. Why? Answer.  Liquids and gases have intermolecular space; on applying pressure externally on them the molecules can come closer thereby minimizing the space between them. But in case of solids, there is no intermolecular space to do so.

Question. Is it not proper to regard the gaseous state of ammonia as vapours? Explain. Answer.  The gaseous state of a substance can be regarded as vapours only in case it is a liquid at room temperature. Since ammonia is a gas at room temperature, its gaseous state cannot be regarded as vapours. Naphthalene is volatile solid and has a tendency to sublime. So, it changes into vapours completely, thus disappear into the air and no solid is left.

Question. Cotton is solid but it floats on water. Why? Answer.  Cotton has large number of pores where air is trapped.This process reduces cotton’s density and increase the volume. Therefore, cotton floats on water. But, when these pores get filled with water, it starts sinking.

Question. Why do we see water droplets on the outer surface of a glass containing ice-cold water? Answer.  If we take some ice-cold water in a glass, after some time we will see small droplets of water deposited on the outer walls of the glass. Because water vapour present in air come into the contact of cold wall of glass, lose energy and converted into liquid state which can be seen in the form of small droplets.

Question. Pressure and temperature determine the state of a substance. Explain this in detail. Answer. Any matter, i.e. solid, liquid or gas when experiences an increase in temperature then they change their state. Example :

When we take ice cubes in a beaker or heat them slowly, the temperature increases and ice melts to form liquid. We heat this liquid further it will become steam. On lowering down the temperature of any matter,show change in their state. Example :

We take the steam that is coming out of boiling water and allow it to cool down, it condenses to form water and on further cooling of this water we get ice. On applying pressure and reducing temperature we can liquefy gases or change them into solid. Example : We take carbon dioxide gas, reduce its temperature and apply lot of pressure on it so that it changes into solid carbon dioxide, called dry ice, which is used as refrigerant for cooling. If pressure on it is decreased it directly changes into gas. In LPG cylinders, the petroleum gas is cooled and with lot of pressure changes it into liquid state. While using this LPG, we release the pressure exerted on it and hence, it comes out in the form of gas.

Question. Give difference between Evaporation and Boiling. Answer.     Evaporation                                            Boiling            1. It takes place at any       It takes place at definite temperature called boiling point of liquid. place. 2. Temperature of liquid      Temperature of liquid does not change during boiling. decreases during evaporation. 3. Evaporation is a              Boiling is the bulk phenomenon; it takes place in the whole mass of the liquid. surface phenomenon; it takes place only at the surface of the liquid.  4. Evaporation is a slow       Boiling is a rapid and violent process. and silent process. Question. Explain the inter-conversion of three states in terms of force of attraction and kinetic energy of the molecules. Answer. The force working between the particles of a matter is called intermolecular force. Intermolecular forces are strong in solids and the particles are close to each other and thus make the substance rigid. In liquids,intermolecular force is less than solids and more than gases. So, they cannot have rigid shape and kinetic energy of the molecules is not enough to hold gas in open container.

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