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Class 9 Science Case Study Questions PDF Download

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Class 9 Science Case Study Questions play a crucial role in the field of science education as they provide real-life scenarios for students to analyze, apply their knowledge, and develop problem-solving skills. This article aims to present a comprehensive collection of case study questions for Class 9 Science , covering various topics and concepts.

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CBSE Class 9 Science Exam will have a set of questions based on case studies in the form of MCQs. The CBSE Class 9 Science Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is where you should hang out.  CBSE Case Study Questions for Class 9  will provide you with detailed, latest, comprehensive & confidence-inspiring solutions to the maximum number of Case Study Questions covering all the topics from your  NCERT Text Books !

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CBSE Class 9th SCIENCE Chapterwise Case Study Question & Solution

Case study questions provide students with real-life scenarios that require critical thinking and application of scientific concepts. They help students understand the practical application of scientific principles and develop problem-solving skills in various scientific disciplines.

Chapterwise Case Study Questions for Class 9 Science

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked. For Science subjects, there would be 5 case-based sub-part questions, wherein a student has to attempt 4 sub-part questions.

• Case Study Questions for Chapter 1 Matter in Our Surroundings
• Case Study Questions for Chapter 2 Is Matter Around Us Pure?
• Case Study Questions for Chapter 3 Atoms and Molecules
• Case Study Questions for Chapter 4 Structure of Atom
• Case Study Questions for Chapter 5 The Fundamental Unit of Life
• Case Study Questions for Chapter 6 Tissues
• Case Study Questions for Chapter 7 Diversity in Living Organisms
• Case Study Questions for Chapter 8 Motion
• Case Study Questions for Chapter 9 Force and Laws of Motion
• Case Study Questions for Chapter 10 Gravitation
• Case Study Questions for Chapter 11 Work and Energy
• Case Study Questions for Chapter 12 Sound
• Case Study Questions for Chapter 13 Why do we Fall ill
• Case Study Questions for Chapter 14 Natural Resources
• Case Study Questions for Chapter 15 Improvement in Food Resources

The above  Case studies for Class 9 Science  will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Science Case Studies have been developed by experienced teachers of schools.studyrate.in for the benefit of Class 10 students.

Class 9 Maths Case Study Questions

Benefits of Case Studies in Science Education

Case studies offer several advantages over traditional teaching methods. Here are some key benefits:

• Real-World Application : Case studies present authentic scenarios, enabling students to understand how scientific concepts are applied in real-life situations.
• Critical Thinking : Analyzing case studies requires students to think critically, make connections, and apply scientific knowledge to solve problems.
• Interdisciplinary Approach : Case studies often involve multiple scientific disciplines, fostering an interdisciplinary understanding of complex issues.
• Engagement and Active Learning : Case studies actively engage students in the learning process, promoting active participation, discussion, and collaboration.
• Skill Development : Case studies develop essential skills such as analytical thinking, problem-solving, and effective communication of scientific concepts.

Importance of Practicing Case Study Questions

Practicing case study questions is crucial for Class 9 Science students to enhance their understanding and application of scientific concepts. Here’s why it is important:

• Application of Knowledge : Case studies allow students to apply their theoretical knowledge to practical situations, bridging the gap between theory and real-world scenarios.
• Developing Analytical Skills : Analyzing case studies improves students’ ability to identify relevant information, make connections, and draw logical conclusions.
• Problem-Solving Skills : Case studies present complex problems that require students to think critically and develop effective problem-solving strategies.
• Enhanced Exam Performance : Practicing case study questions familiarizes students with the format and types of questions they may encounter in exams, leading to improved performance.

Subjects Covered in the Case Study Questions for Class 9 Science

The case study questions for Class 9 Science cover the following subjects:

• Motion and Forces
• Light and Reflection
• Electricity
• Matter and Its Properties
• Atoms and Molecules
• Structure of the Atom
• Chemical Reactions
• Cell: The Fundamental Unit of Life
• Diversity in Living Organisms
• Natural Resources

Tips for Approaching Case Study Questions

To tackle case study questions effectively, consider the following tips:

• Read Carefully : Pay close attention to the details provided in the case study, as they hold crucial information for solving the problem.
• Analyze Methodically : Break down the problem into smaller components and analyze each part systematically.
• Apply Relevant Concepts : Identify the scientific principles relevant to the case study and apply them appropriately.
• Consider Multiple Perspectives : Explore different angles and viewpoints while proposing solutions, taking into account various scientific factors.
• Provide Justifications : Support your answers with scientific explanations and logical reasoning to strengthen your responses.

The Class 9 Science Case Study Questions provided in this article serve as a valuable resource for students seeking to enhance their scientific knowledge and problem-solving skills. By practicing these case studies, students can develop a deeper understanding of scientific concepts and their practical applications. Embrace this opportunity to engage with real-world scenarios and strengthen your scientific acumen.

Q1: Are the Class 9 Science Case Study Questions aligned with the official curriculum?

Yes, the Class 9 Science Case Study Questions presented in this article are aligned with the official curriculum. They cover relevant topics and concepts that students need to study for their exams.

Q2: Can practicing case study questions alone guarantee success in Class 9 Science exams?

Practicing case study questions is an important part of exam preparation, but it should be complemented with a thorough understanding of the subject matter. It is advisable to study the concepts in detail, refer to textbooks, and engage in other learning activities to achieve success in exams.

Q3: Where I Can get Class 9 Science Case Study Questions ?

You can practice Class 9 Science Case Study Questions on schools.studyrate.in for free.

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• Class 9 Science Case...

Class 9 Science Case Study Questions

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myCBSEguide App

Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you are wondering how to solve class 9 science case study questions, then myCBSEguide is the best platform to choose. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added Class 9 Science case study questions.

You can find a wide range of solved case studies on myCBSEguide, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

The rationale behind Science

Science is crucial for Class 9 students’ cognitive, emotional, and psychomotor development. It encourages curiosity, inventiveness, objectivity, and aesthetic sense.

In the upper primary stage, students should be given a variety of opportunities to engage with scientific processes such as observing, recording observations, drawing, tabulating, plotting graphs, and so on, whereas in the secondary stage, abstraction and quantitative reasoning should take a more prominent role in science teaching and learning. As a result, the concept of atoms and molecules as matter’s building units, as well as Newton’s law of gravitation, emerges.

Science is important because it allows Class 9 Science students to understand the world around us. It helps to find out how things work and to find solutions to problems at the Class 9 Science level. Science is also a source of enjoyment for many people. It can be a hobby, a career, or a source of intellectual stimulation.

Case study questions in Class 9 Science

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Examples of Class 9 science class case study questions

Class 9 science case study questions have been prepared by myCBSEguide’s qualified teachers. Class 9 case study questions are meant to evaluate students’ knowledge and comprehension of the material. They are not intended to be difficult, but they will require you to think critically about the material. We hope you find Class 9 science case study questions beneficial and that they assist you in your exam preparation.

The following are a few examples of Class 9 science case study questions.

Class 9 science case study question 1

• due to its high compressibility
• large volumes of a gas can be compressed into a small cylinder
• transported easily
• all of these
• shape, volume
• volume, shape
• shape, size
• size, shape
• the presence of dissolved carbon dioxide in water
• the presence of dissolved oxygen in the water
• the presence of dissolved Nitrogen in the water
• liquid particles move freely
• liquid have greater space between each other
• both (a) and (b)
• none of these
• Only gases behave like fluids
• Gases and solids behave like fluids
• Gases and liquids behave like fluids
• Only liquids are fluids

Answer Key:

• (d) all of these
• (a) shape, volume
• (b) the presence of dissolved oxygen in the water
• (c) both (a) and (b)
• (c) Gases and liquids behave like fluids

Class 9 science case study question 2

• 12/32 times
• 18 g of O 2
• 18 g of CO 2
• 18 g of CH 4
• 1 g of CO 2
• 1 g of CH 4 CH 4
• 2 moles of H2O
• 20 moles of water
• 6.022  ×  1023 molecules of water
• 1.2044  ×  1025 molecules of water
• (I) and (IV)
• (II) and (III)
• (II) and (IV)
• Sulphate molecule
• Ozone molecule
• Phosphorus molecule
• Methane molecule
• (c) 8/3 times
• (d) 18g of CH ​​​​​4
• (c) 1g of H ​​​​​​2
• (d) (II) and (IV)
• (c) phosphorus molecule

Class 9 science case study question 3

• collenchyma
• chlorenchyma
• It performs photosynthesis
• It helps the aquatic plant to float
• It provides mechanical support
• Sclerenchyma
• Collenchyma
• Epithelial tissue
• Parenchyma tissues have intercellular spaces.
• Collenchymatous tissues are irregularly thickened at corners.
• Apical and intercalary meristems are permanent tissues.
• Meristematic tissues, in its early stage, lack vacuoles, muscles
• (I) and (II)
• (III) and (I)
• Transpiration
• Provides mechanical support
• Provides strength to the plant parts
• None of these
• (a) Collenchyma
• (b) help aquatic plant to float
• (b) Sclerenchyma
• (d) Only (III)
• (c) provide strength to plant parts

Cracking Class 9 Science Case Study Questions

There is no one definitive answer to Class 9 Science case study questions. Every case study is unique and will necessitate a unique strategy. There are, nevertheless, certain general guidelines to follow while answering case study questions.

• To begin, double-check that you understand the Class 9 science case study questions. Make sure you understand what is being asked by reading it carefully. If you’re unclear, seek clarification from your teacher or tutor.
• It’s critical to read the Class 9 Science case study material thoroughly once you’ve grasped the question. This will provide you with a thorough understanding of the problem as well as the various potential solutions.
• Brainstorming potential solutions with classmates or other students might also be beneficial. This might provide you with multiple viewpoints on the situation and assist you in determining the best solution.
• Finally, make sure your answer is presented simply and concisely. Make sure you clarify your rationale and back up your claim with evidence.

A look at the Class 9 Science Syllabus

The CBSE class 9 science syllabus provides a strong foundation for students who want to pursue a career in science. The topics are chosen in such a way that they build on the concepts learned in the previous classes and provide a strong foundation for further studies in science. The table below lists the topics covered in the Class 9 Science syllabus of the Central Board of Secondary Education (CBSE). As can be seen, the Class 9 science syllabus is divided into three sections: Physics, Chemistry and Biology. Each section contains a number of topics that Class 9 science students must study during the course.

CBSE Class 9 Science (Code No. 086)

Theme: Materials Unit I: Matter-Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics – shape, volume, density; change of state-melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter:  Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Particle nature and their basic units:  Atoms and molecules, Law of constant proportions, Atomic and molecular masses. Mole concept: Relationship of mole to mass of the particles and numbers. Structure of atoms:  Electrons, protons and neutrons, valency, the chemical formula of common compounds. Isotopes and Isobars.

Theme: The World of the Living Unit II: Organization in the Living World Cell – Basic Unit of life:  Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number. Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Theme: Moving Things, People and Ideas Unit III: Motion, Force and Work Motion:  Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, derivation of equations of motion by graphical method; elementary idea of uniform circular motion. Force and Newton’s laws:  Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Elementary idea of conservation of Momentum. Gravitation:  Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation:  Thrust and Pressure. Archimedes’ Principle; Buoyancy. Work, energy and power:  Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy. Sound:  Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme: Food Unit IV: Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

PRESCRIBED BOOKS:

• Science-Textbook for class IX-NCERT Publication
• Assessment of Practical Skills in Science-Class IX – CBSE Publication
• Laboratory Manual-Science-Class IX, NCERT Publication
• Exemplar Problems Class IX – NCERT Publication

myCBSEguide: A true helper

There are numerous advantages to using myCBSEguide to achieve the highest results in Class 9 Science.

• myCBSEguide offers high-quality study materials that cover all of the topics in the Class 9 Science curriculum.
• myCBSEguide provides practice questions and mock examinations to assist students in the best possible preparation for their exams.
• On our myCBSEguide app, you’ll find a variety of solved Class 9 Science case study questions covering a variety of topics and concepts. These case studies are intended to help you understand how certain principles are applied in real-world settings
• myCBSEguide is that the study material and practice problems are developed by a team of specialists who are always accessible to assist students with any questions they may have. As a result, students may be confident that they will receive the finest possible assistance and support when studying for their exams.

So, if you’re seeking the most effective strategy to study for your Class 9 Science examinations, myCBSEguide is the place to go!

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Case Study Questions of Class 9 Science PDF Download

Download PDF Case Study Questions of Class 9 Science to prepare for the upcoming CBSE Class 9 Exams Exam 2023-24. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added  Class 9 Science case study questions .

Case study questions are based on real or hypothetical scenarios that require students to analyze, evaluate, and apply scientific concepts to solve problems or make informed decisions. They often present a detailed context, providing students with the opportunity to demonstrate their understanding of the subject matter beyond basic recall.

Table of Contents

Class 9 Science: Case Study Questions

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Chapterwise Case Study Questions of Class 9 Science

• Case Study Questions for Chapter 1 Matter in Our Surroundings
• Case Study Questions for Chapter 2 Is Matter Around Us Pure?
• Case Study Questions for Chapter 3 Atoms and Molecules
• Case Study Questions for Chapter 4 Structure of Atom
• Case Study Questions for Chapter 5 The Fundamental Unit of Life
• Case Study Questions for Chapter 6 Tissues
• Case Study Questions for Chapter 7 Diversity in Living Organisms
• Case Study Questions for Chapter 8 Motion
• Case Study Questions for Chapter 9 Force and Laws of Motion
• Case Study Questions for Chapter 10 Gravitation
• Case Study Questions for Chapter 11 Work and Energy
• Case Study Questions for Chapter 12 Sound
• Case Study Questions for Chapter 13 Why do we Fall ill
• Case Study Questions for Chapter 14 Natural Resources
• Case Study Questions for Chapter 15 Improvement in Food Resources

You can find a wide range of solved case studies on cbseexperts, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

Class 9 Science Syllabus

Unit I: Matter-Nature and Behaviour

 Definition of matter; solid, liquid, and gas; characteristics – shape, volume, density; change of statementing (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation.

Nature of matter:  Elements, compounds, and mixtures. Heterogeneous and homogenous mixtures, colloids, and suspensions. Physical and chemical changes (excluding separating the components of a mixture).

Particle nature and their basic units:  Atoms and molecules, Law of Chemical Combination, Chemical formula of common compounds, Atomic and molecular masses.

Structure of atoms:  Electrons, protons and neutrons, Valency, Atomic Number and Mass Number, Isotopes and Isobars.

Unit II: Organization in the Living World

Cell – Basic Unit of life:  Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number.

Tissues, Organs, Organ System, Organism:  Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Unit III: Motio n, Force,  and Work

Motion:  Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, elementary idea of uniform circular motion.

Force and Newton’s laws:  Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration.

Gravitation:  Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation: Thrust and Pressure. Archimedes’ Principle; Buoyancy.

Work, Energy and Power:  Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy (excluding commercial unit of Energy).

Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Unit IV: Food Production

Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

Books for Class 9 Science Exams

Benefits of Case Study Questions

• Enhancing Analytical Skills : Case study questions challenge students to analyze complex scenarios, identify relevant information, and derive meaningful insights. By engaging with these questions, students develop critical analytical skills that are essential for scientific thinking and problem-solving.
• Promoting Critical Thinking : Case study questions encourage students to think critically and evaluate different perspectives. They require students to reason, make logical deductions, and justify their answers with supporting evidence. This process helps in honing their critical thinking abilities, enabling them to approach problems from multiple angles.
• Encouraging Practical Application of Concepts : By presenting real-world or hypothetical situations, case study questions promote the application of scientific concepts in practical scenarios. This application-based approach fosters a deeper understanding of the subject matter and helps students see the relevance of what they learn in the classroom to everyday life.

Case study questions of Class 9 Science provide students with an opportunity to apply their knowledge, enhance analytical skills, and think critically. By understanding the format, benefits, and effective strategies for answering case study questions, students can excel in this form of assessment. While challenges may arise, practicing time management, improving information extraction skills, and enhancing observation abilities will enable students to overcome these obstacles and perform well. Embracing case study questions as a valuable learning tool can contribute to a holistic understanding of scientific concepts and foster problem-solving abilities.

1. What is the purpose of case study questions in Class 9 Science?

Case study questions serve the purpose of evaluating a student’s understanding of scientific concepts, their ability to apply knowledge in real-life situations, and their analytical and critical thinking skills.

2. How can case study questions help improve analytical skills?

Case study questions require students to analyze complex scenarios, identify relevant information, and derive meaningful insights. Regular practice with such questions can significantly enhance analytical skills.

3. Are case study questions difficult to answer?

Case study questions can be challenging due to their comprehensive nature and the need for critical thinking. However, with practice and effective strategies, students can develop the skills necessary to answer them effectively.

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CBSE Class 9 Science Important Case Study Questions with Answers for Term 2 Exam 2022 (PDF)

Check important case study questions of cbse class 9 science to prepare for the cbse term 2 exam 2022. all these questions have been put together by subject experts..

CBSE Class 9 Term 2 Exam 2022:  Important case based questions for CBSE Class 9 Science are provided here students to prepare for the upcoming Term 2 Exam 2022. All the questions provided below are curated by the subject experts. These questions are really helpful to revise important concepts and prepare the case study questions for the exam. Answers to all questions have been provided for reference. So, students should practice the chapter-wise questions to clearly understand the right way to attempt the case based questions. Download the chapter-wise questions in PDF.

Check some of the important case study questions below:

Q. Read the following and answer the questions :

A student was asked by his teacher to verify the law of conservation of mass in the laboratory. He prepared 5% aqueous solutions of NaCl and Na 2 SO 4 . He mixed 10 mL of both these solutions in a conical flask. He weighed the flask on a balance. He then stirred the flask with a rod and weighed it after sometime. There was no change in mass.

• Was the student able to verify the law of conservation of mass?
• If not, what was the mistake committed by him?
• In your opinion, what he should have done?
• What is the molar mass of Na 2 SO 4 ?
• No, he could not verify the law of conservation of mass in-spite of the fact that there was no change in mass.
• No chemical reaction takes place between NaCl and Na 2 SO 4 . This means that no reaction actually took place in the flask.
• He should have performed the experiment by using aqueous solutions of BaCl 2 and Na 2 SO 4 . A chemical reaction takes place in this case and a white precipitate of BaSO 4 is formed.
• Will the weight of the precipitate be the same as that of the reactants before mixing?
• If not, what she should have done?
• Which law of chemical combination does this support?
• State the law of conservation of mass.
• No, it will not be the same.
• She should have weighed the total contents of the beaker after the reaction and not the precipitate alone.
• It supports the law of conservation of mass.
• Mass can neither be created nor destroyed during a chemical reaction.

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Class 9th Science - Motion Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Science Subject - Motion, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

QB365 - Question Bank Software

Motion case study questions with answer key.

9th Standard CBSE

Final Semester - June 2015

(ii) Find the Velocity of the boy.

(iii) A boy is sitting on a merry-go-round which is moving with a constant speed of 10m/s. This means that the boy is :

(iv) In which of the following cases of motion, the distance moved and the magnitude of displacement are equal ?

(ii) How far does it travel in 1 second ?

(iii) How far does it travel in 6 seconds ?

(iv) How long does it take to travel 240 m ?

(v) Which of the following statement is correct regarding velocity and speed of a moving body? (a) velocity of a moving body is always higher than its speed (b) speed of a moving body is always higher than its velocity (c) speed of a moving body is its velocity in a given direction (d) velocity of a moving body is its speed in a given direction

(ii) What type of motion is represented by BC ?

(iii) Find out the acceleration of the body.

(iv) Calculate the retardation of the body.

(v) Find out the distance travelled by the body from A to B.

*****************************************

Motion case study questions with answer key answer keys.

(i) (b) 2 m/s Total distance travelled is 100 m + 100 m = 200 m and the total time taken is 50 s + 50 s = 100 s. \(\text { Speed of boy }=\frac{\text { Distance travelled }}{\text { Time taken }}=\frac{200 \mathrm{~m}}{100 \mathrm{~s}}=2 \mathrm{~m} / \mathrm{s}\) (ii) (d) 0 m/s The boy runs 100 m towards East and then 100 m towards West and reaches at the starting point, his home. So, the displacement will be 100 m – 100 m = 0 m. The total time taken is 50 s + 50 s = 100 s. \(\text { Velocity of boy }=\frac{\text { Displacement }}{\text { Time taken }}=\frac{0 \mathrm{~m}}{100 \mathrm{~s}}=0 \mathrm{~m} / \mathrm{s}\) (iii) (c) in accelerated motion (iv) (a) if the car is moving on straight road (v) (c) 2r

(i) (b) 30 m/s Average speed = total distance travelled/total time taken = 150/5 = 30 m/s (ii) (b) 30 m Time = 1 s Distance = (average speed)(time) = 30 m/s x 1s = 30 m (iii) (c) 180 m  Time = 6 s Distance = (average speed)(time) = 30 m/s x 6s = 180m (iv) (d) 8s Distance = 240m Time = Distance/average speed = 240/30 = 8s (v) (d) velocity of a moving body is its speed in a given direction

(i) (b) uniform acceleration OA is a straight line graph between speed and time, and it is sloping upwards from O to A. Therefore, the graph line OA represents uniform acceleration. (ii) (c) negative acceleration BC is a straight line graph between speed and time which is sloping downwards from B to C. Therefore, BC represents uniform retardation (or negative acceleration). (iii) (a) 1.5 m/s 2 The slope of speed-time graph OA will give us the acceleration of the body. Thus, Acceleration = Slope of line OA = AD/OD Now, in the given graph, we find that AD = 6 m/s and OD = 4 seconds. So, putting these values in the above relation, we get : Acceleration = 6 m/s / 4 s = 1.5 m/s 2 (iv) (d) 1 m/s 2 The slope of speed-time graph BC will be equal to the retardation of the body. So, Retardation = Slope of line BC = BE/EC Now, in the graph given to us, we find that BE = 6 m/s and EC = 16 – 10 = 6 seconds. So, putting these values in the above relation, we get : Retardation = 6m/s / 6 s = 1 m/s 2 (v) (c) 36 m Distance travelled from A to B = Area under the line AB and the time axis = Area of rectangle DABE = DA × DE Now, from the given graph, we find that DA = 6 m/s and DE = 10 – 4 = 6 s. Therefore, Distance travelled from A to B = 6 × 6 = 36 m

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Case study questions class 9 science chapter 8 motion, cbse case study questions class 9 science – motion.

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Class 9 Physics Chapter 1 – Motion

Class 9 Physics Chapter 2 – Force And Laws Of Motion

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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 .

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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 .

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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.

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

NCERT Solutions for Class 9 Science (physics) Chapter 9 Force and Laws of 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 9 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 9 Textbook Questions and Answers

INTEXT QUESTIONS

PAGE NO-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:  Inertia is the measure of the mass of the body. The greater is the mass of the body; the greater is its inertia and vice-versa.

(a) Mass of a stone is more than the mass of a rubber ball for the same size. Hence, inertia of the stone is greater than that of a rubber ball.

(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is greater than that of the bicycle.

(c) Mass of a five rupees coin is more than that of a one-rupee coin. Hence, inertia of the five rupees coin is greater than that of the one-rupee coin.

Question 2: In the following example, try to identify the number of times the velocity of the ball changes:

“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.

Also identify the agent supplying the force in each case.

Answer: The velocity of football changes four times. First, when a football player kicks to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth when the goalkeeper kicks the football towards a player of his own team. Agent supplying the force –First case – First player Second case – Second player Third case – Goalkeeper Fourth case – Goalkeeper

Question 3: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer: When the branch of the tree is shaken, the branch moves in a to-and-fro motion. However, the inertia of the leaves in attached to the branch resists the motion of the branch. Therefore, the leaves that are weakly attached to the branch fall off due to inertia whereas the leaves that are firmly attached to the branch remain attached.

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: In a moving bus, passengers are in motion along with bus. When brakes are applied to stop a moving bus, bus comes in the position of rest. But because of tendency to be in the motion a person falls in forward direction. Similarly, when a bus is accelerated from rest, the tendency to be in rest, a person in the bus falls backwards.

PAGE NO 126-127

Question 1: If action is always equal to the reaction, explain how a horse can pull a cart.

Answer: A horse pushes the ground in the backward direction. According to Newton’s third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.

Question 2: Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer: When large amount of water is ejected from a hose at a high velocity, according to Newton’s Third Law of Motion, water pushes the hose in backward direction with the same force. Therefore, it is difficult for a fireman to hold a hose in which ejects large amount of water at a high velocity.

Question 3: From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 . Calculate the initial recoil velocity of the rifle.

Answer: Mass of the rifle, m 1 = 4 kg   Mass of the bullet, m 2 = 50g = 0.05 kg  Recoil velocity of the rifle = v 1   Bullet is fired with an initial velocity, v 2 = 35 m/s Initially, the rifle is at rest. Thus, its initial velocity, v = 0

Total initial momentum of the rifle and bullet system = (m 1 + m 2 )v = 0   

Total momentum of the rifle and bullet system after firing: = m 1 v 1 + m 2 v 2 = 4(v 1 ) + 0.05×35 = 4v 1 + 1.75

According to the law of conservation of momentum:

Total momentum after the firing = Total momentum before the firing

⇒ 4v 1 + 1.75 = 0 ⇒ v 1 = -1.75/4 ⇒ v 1 = – 0.4375 m/s 

The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.

Question 4: Two objects of masses 100g and 200g are moving along the same line and direction with velocities of 2 and 1 , respectively. They collide and after the collision, the first object moves at a velocity of 1.67 . Determine the velocity of the second object.

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: Yes, it is possible. An object moving in some direction with constant velocity will continue in its state of motion as long as there are no external unbalanced forces acting on it. In order to change the motion of the object, some external unbalanced force must act upon it.

Question 2: When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer: Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to resist their state of rest. According to Newton’s first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet.

Question 3: Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer: Luggage kept on the roof of a bus has the tendency to maintain its state of rest when bus is in rest and to maintain the state of motion when bus is in motion according to Newton’s First Law of Motion. When bus will come in motion from its state of rest, in order to maintain the position of rest, luggage kept over its roof may fall down. Similarly, when a moving bus will come in the state of rest or there is any sudden change in velocity because of applying of brake, luggage may fall down because of its tendency to remain in the state of motion. This is the cause that it is advised to tie any luggage kept on the roof a bus with a rope so that luggage can be prevented from falling down.

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 on the ball opposing the motion. 

Explanation: When ball moves on the ground, the force of friction opposes its movement and after some time ball comes to the state of rest.

Question 5: A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.)

Question 6:   A stone of 1 kg is thrown with a velocity of 20 m s -1  across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Hence, the force of friction between the stone and the ice is – 4 N. Here negative sign shows that force is being applied in the opposite direction of the movement of the stone.

Question 7: A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force; (b) the acceleration of the train

Answer: Mass of the automobile vehicle, m= 1500 kg Final velocity, v = 0 m/s Acceleration of the automobile, a = –1.7 ms –2

From Newton’s second law of motion:

Force = Mass × Acceleration = 1500 × (–1.7) = – 2550 N

Hence, the force between the automobile and the road is –2550 N, in the direction opposite to the motion of the automobile.

Question 8: An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms -2 ?

Question 10: Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Answer: A force of 200 N is applied in the forward direction. Thus, from Newton’s third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the fictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet.

Question 11: Two objects, each of mass 1.5 kg are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms -1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

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: Because of the huge mass of the truck, the force of static friction is very high. The force applied by the student is unable to overcome the static friction and hence he is unable to move the truck. In this case, the net unbalanced force in either direction is zero which is the reason of no motion happening here. The force applied by the student and the force because of static friction are cancelling out each other. Hence, the rationale given by the student is correct.

Question 13: A hockey ball of mass 200 g travelling at 10 m s -1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s -1 . Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Question 14: A bullet of mass 10 g travelling horizontally with a velocity of 150 ms -1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer: Now, it is given that the bullet is travelling with a velocity of 150 m/s.

Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s

Final velocity, v = 0 (since the bullet finally comes to rest)

Time taken to come to rest, t = 0.03 s Acceleration of the bullet, a = ?

Hence, the distance of penetration of the bullet into the block is 2.25 m.

Force, F = Mass × Acceleration

Mass of the bullet, m = 10 g = 0.01 kg

Acceleration of the bullet, a = 5000 m/s 2

So, the force exerted by the wooden block on the bullet: F = ma = 0.01×5000 = 50 N

Therefore, Penetration of bullet in wooden block = 2.25 m Force exerted by wooden block on bullet = – 50 N. Here negative sign shows that force is exerted in the opposite direction of bullet.

Question 15: An object of mass 1 kg travelling in a straight line with a velocity of 10 ms -1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Question 16: An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms -1 to 8 ms -1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Question 17: Akhtar, Kiran and Rahul were riding in a motorocar 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 the 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: According to the law of conservation of momentum:

Momentum of the car and insect system before collision = Momentum of the car and insect system after collision

Hence, the change in momentum of the car and insect system is zero.

The insect gets stuck on the windscreen. This means that the direction of the insect is reversed. As a result, the velocity of the insect changes to a great amount. On the other hand, the car continues moving with a constant velocity. Hence, Kiran’s suggestion that the insect suffers a greater change in momentum as compared to the car is correct. The momentum of the insect after collision becomes very high because the car is moving at a high speed. Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to the mass of the insect.

Rahul gave a correct explanation as both the car and the insect experienced equal forces caused by the Newton’s action-reaction law. But, he made an incorrect statement as the system suffers a change in momentum because the momentum before the collision is equal to the momentum after the collision.

Question 18: How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s -2 .

ADDITIONAL EXERCISES

Question A1: The following is the distance-time table of an object in motion:          

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?

(b)What do you infer about the forces acting on the object?

Answer: (a) The distance covered by the object at any time interval is greater than any of the distances covered in previous time intervals. Therefore, the acceleration of the object is increasing.

(b) As per the second law of motion, force = mass × acceleration. Since the mass of the object remains constant, the increasing acceleration implies that the force acting on the object is increasing as well

Question A2: Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms -2 . With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort)

Answer: Mass of the motor car = 1200 kg

Only two persons manage to push the car. Hence, the acceleration acquired by the car is given by the third person alone.

Acceleration produced by the car, when it is pushed by the third person, a = 0.2 m/s 2

Let the force applied by the third person be F .

Force = Mass × Acceleration ⇒ F = 1200 × 0.2 ⇒ F = 240 N

Thus, the third person applies a force of magnitude 240 N.

Hence, each person applies a force of 240 N to push the motor car.

Question A3:   A hammer of mass 500 g, moving at 50 ms -1 , strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

The hammer strikes the nail with a force of -2500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.

Question A4: A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Question A5: A large truck and a car, both moving with a velocity of magnitude , have a headon collision and both of them come to a halt after that. If the collision lasts for 1s:

(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?

(c) From the first equation of motion, acceleration produced in a system is independent  of the mass of the system. The initial velocity, the final velocity, and the time of impact remain the same in both cases. Hence, both the car and the truck experience the same amount of acceleration.

(d) According to Newton’s third law of motion, for every action there is an equal and  opposite reaction that acts on different bodies. Since the truck experiences a greater force of impact (action), this larger

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

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NCERT Solutions for Class 9 Science Chapter 9: Force and Laws Of Motion

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* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 8.

NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion, are prepared to help students clear their doubts and understand concepts thoroughly. Class 9 Solutions of Science is a beneficial reference material that helps students to clear doubts instantly in an effective way. NCERT Solutions for Class 9 Science are designed in a student-friendly way and are loaded with questions, activities, and exercises that are CBSE and competitive exam-oriented. 

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• 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 8 Motion
• Chapter 10 Gravitation
• Chapter 11 Work and Energy
• Chapter 12 Sound
• Chapter 13 Why Do We Fall ill
• Chapter 14 Natural Resources
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Access Answers of Science NCERT class 9 Chapter 9: Force and Laws Of Motion(All intext and exercise questions solved)

Intext Questions – 1   Page: 118

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-rupee coin and a one-rupee coin?

Since inertia is dependent on the mass of the object, the object with the greater mass will hold greater inertia. The following objects hold greater inertia because of their mass.

• Five-Rupee coin

2. In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case.

The velocity of football changes four times.

First, when a football player kicks a football to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth, when the goalkeeper kicks the football towards his team player.

Agent supplying the force:

a) The First case is the First player

b) The Second case is the Second player

c) The Third case is Goalkeeper

d) The Fourth case is Goalkeeper

3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

When the branch of the tree is shaken, the branch moves in a to-and-fro motion. However, the inertia of the leaves in attached to the branch resists the motion of the branch. Therefore, the leaves that are weakly attached to the branch fall off due to inertia whereas the leaves that are firmly attached to the branch remain attached.

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?

Initially, when the bus accelerates in a forward direction from a state of rest, the passengers experience a force exerted on them in the backward direction due to their inertia opposing the forward motion.

Once the bus starts moving, the passengers are in a state of motion in the forward direction. When the brakes are applied, the bus moves towards a position of rest. Now, a force in the forward direction is applied on the passengers because their inertia resists the change in the motion of the bus. This causes the passengers to fall forwards when the brakes are applied.

Intext Questions – 2 Page: 126,127

1. If action is always equal to the reaction, explain how a horse can pull a cart.

When the horse walks forward (with the cart attached to it), it exerts a force in the backward direction on the Earth. An equal force in the opposite direction (forward direction) is applied on the horse by the Earth. This force moves the horse and the cart forward. As a result, the cart moves forward.

2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton’s third law of motion. As a result of the backward force, the stability of the fireman decreases. Hence, it is difficult for him to remain stable while holding the hose.

3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s –1 . Calculate the initial recoil velocity of the rifle.

Given, the Bullet’s mass (m 1 ) = 50 g

The rifle’s mass (m 2 ) = 4kg = 4000g

Initial velocity of the fired bullet (v 1 ) = 35 m/s

Let the recoil velocity be v 2 .

Since the rifle was initially at rest, the initial momentum of the rifle = 0

The total momentum of the rifle and bullet after firing = m 1 v 1 + m 2 v 2

As per the law of conservation of momentum, the total momentum of the rifle and the bullet after firing = 0 (same as initial momentum)

Therefore, m 1 v 1 + m 2 v 2 = 0

The negative sign indicates that recoil velocity is opposite to the bullet’s motion.

4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms –1 and 1 ms –1 , respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms –1 . Determine the velocity of the second object.

Assuming that the first object is object A and the second one is object B, it is given that:

Mass of A (m 1 ) = 100g

Mass of B (m 2 ) = 200g

Initial velocity of A (u 1 ) = 2 m/s

Initial velocity of B (u 2 ) = 1 m/s

Final velocity of A (v 1 ) = 1.67 m/s

Final velocity of B (v 2 ) =?

Total initial momentum = Initial momentum of A + initial momentum of B

= m 1 u 1 + m 2 u 2

= (100g) × (2m/s) + (200g) × (1m/s) = 400 g.m.sec -1

As per the law of conservation of momentum, the total momentum before collision must be equal to the total momentum post collision.

v 2 = 1.165 m/s

Therefore, the velocity of object B after the collision is 1.165 meters per second.

Exercises Page: 128,129

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.

Yes, it is possible. An object moving in some direction with constant velocity will continue in its state of motion as long as there are no external unbalanced forces acting on it. In order to change the motion of the object, some external unbalanced force must act upon it.

2. When a carpet is beaten with a stick, dust comes out of it. Explain.

When the carpet is beaten with a stick, the stick exerts a force on the carpet which sets it in motion. The inertia of the dust particles residing on the carpet resists the change in the motion of the carpet. Therefore, the forward motion of the carpet exerts a backward force on the dust particles, setting them in motion in the opposite direction. This is why the dust comes out of the carpet when beaten.

3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

When some luggage is placed on the roof of a bus which is initially at rest, the acceleration of the bus in the forward direction will exert a force (in the backward direction) on the luggage. In a similar manner, when a bus which is initially in a state of motion suddenly comes to rest due to the application of brakes, a force (in the forward direction) is exerted on the luggage.

Depending on the mass of the luggage and the magnitude of the force, the luggage may fall off the bus due to inertia. Tying up the luggage will secure its position and prevent it from falling off the bus.

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.

When the ball rolls on the flat surface of the ground, its motion is opposed by the force of friction (the friction arises between the ground and the ball). This frictional force eventually stops the ball. Therefore, the correct answer is (c).

If the surface of the level ground is lubricated (with oil or some other lubricant), the friction that arises between the ball and the ground will reduce, which will enable the ball to roll for a longer distance.

5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if it’s mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

Given, distance covered by the truck (s) = 400 meters

Time taken to cover the distance (t) = 20 seconds

The initial velocity of the truck (u) = 0 (since it starts from a state of rest)

6. A stone of 1 kg is thrown with a velocity of 20 ms -1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Given, Mass of the stone (m) = 1kg

Initial velocity (u) = 20m/s

Terminal velocity (v) = 0 m/s (the stone reaches a position of rest)

Distance travelled by the stone (s) = 50 m

As per the third equation of motion

v² = u² + 2as

Substituting the values in the above equation we get,

0² = (20)² + 2(a)(50)

-400 = 100a

a = -400/100  =  -4m/s² (retardation)

We know that

Substituting above obtained value of a = -4 in F = m x a

F = 1 × (-4) = -4N

Here the negative sign indicates the opposing force which is Friction

7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train

(a) Given, the force exerted by the train (F) = 40,000 N

Force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction)

Therefore, the net accelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N

(b) Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg

The total mass of the train is 18000 kg.

As per the second law of motion, F = ma (or: a = F/m)

Therefore, acceleration of the train = (net accelerating force) / (total mass of the train)

= 35,000/18,000 = 1.94 ms -2

The acceleration of the train is 1.94 m.s -2 .

8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms -2 ?

Given, mass of the vehicle (m) = 1500 kg

Acceleration (a) = -1.7 ms -2

As per the second law of motion, F = ma

F = 1500kg × (-1.7 ms -2 ) = -2550 N

Hence, the force between the automobile and the road is -2550 N, in the opposite direction of the automobile’s motion.

9. What is the momentum of an object of mass m, moving with a velocity v?

(a) (mv) 2 (b) mv 2 (c) ½ mv 2 (d) mv

The momentum of an object is defined as the product of its mass m and velocity v

Momentum = mass x velocity

Hence, the correct answer is mv i.e option (d)

10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Since the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. This implies that the magnitude of opposing frictional force is equal to the force exerted on the cabinet, which is 200 N. Therefore, the total friction force is -200 N.

11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms -1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

Mass of first object, m 1 = 1.5 kg

Mass of second object, m 2 = 1.5 kg

Velocity of first object before collision, v 1 = 2.5 m/s

The velocity of the second object which is moving in the opposite direction, v 2 = -2.5 m/s

We know that,

Total momentum before collision = Total momentum after collision

m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v

1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5)v

3.75 – 3.75 = 3v

Therefore, the velocity of the combined object after the collision is 0 m/s

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.

Since the truck has a very high mass, the static friction between the road and the truck is high. When pushing the truck with a small force, the frictional force cancels out the applied force and the truck does not move. This implies that the two forces are equal in magnitude but opposite in direction (since the person pushing the truck is not displaced when the truck doesn’t move). Therefore, the student’s logic is correct.

13. A hockey ball of mass 200 g travelling at 10 ms –1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms –1 . Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Given, mass of the ball (m) = 200g

Initial velocity of the ball (u) = 10 m/s

Final velocity of the ball (v) = – 5m/s

Initial momentum of the ball = mu = 200g × 10 ms -1 = 2000 g.m.s -1

Final momentum of the ball = mv = 200g  × –5 ms -1 = –1000 g.m.s -1

Therefore, the change in momentum (mv – mu) = –1000 g.m.s -1 – 2000 g.m.s -1 = –3000 g.m.s -1

This implies that the momentum of the ball reduces by 1000 g.m.s -1 after being struck by the hockey stick.

14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s –1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Given, mass of the bullet (m) = 10g (or 0.01 kg)

Initial velocity of the bullet (u) = 150 m/s

Terminal velocity of the bullet (v) = 0 m/s

Time period (t) = 0.03 s

To find the distance of penetration, the acceleration of the bullet must be calculated

Let the distance of penetration be s

As per the first law of motion

0 = 150 + a (0.03)

a = -5000 ms -2

v 2 = u 2 + 2as

0 = 150 2 + 2 x (-5000)s

F = 0.01kg × (-5000 ms -2 )

15. An object of mass 1 kg travelling in a straight line with a velocity of 10 ms –1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Given, mass of the object (m 1 ) = 1kg

Mass of the block (m 2 ) = 5kg

Initial velocity of the object (u 1 ) = 10 m/s

Initial velocity of the block (u 2 ) = 0

Mass of the resulting object = m 1 + m 2 = 6kg

Velocity of the resulting object (v) =?

Total momentum before the collision = m 1 u 1 + m 2 u 2 = (1kg) × (10m/s) + 0 = 10 kg.m.s -1

As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum post the collision. Therefore, the total momentum post the collision is also 10 kg.m.s -1

Now, (m 1 + m 2 ) × v = 10kg.m.s -1

The resulting object moves with a velocity of 1.66 meters per second.

16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms –1 to 8 ms –1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Given, mass of the object (m) = 100kg

Initial velocity (u) = 5 m/s

Terminal velocity (v) = 8 m/s

Time period (t) = 6s

Now, initial momentum (m × u) = 100kg × 5m/s = 500 kg.m.s -1

Final momentum (m × v) = 100kg × 8m/s = 800 kg.m.s -1

Therefore, the object accelerates at 0.5 ms -2 . This implies that the force acting on the object (F = ma) is equal to:

F = (100kg) × (0.5 ms -2 ) = 50 N

Therefore, a force of 50 N is applied on the 100kg object, which accelerates it by 0.5 ms -2 .

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 the 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.

As per the law of conservation of momentum, the total momentum before the collision between the insect and the car is equal to the total momentum after the collision. Therefore, the change in the momentum of the insect is much greater than the change in momentum of the car (since force is proportional to mass).

Akhtar’s assumption is partially right. Since the mass of the car is very high, the force exerted on the insect during the collision is also very high.

Kiran’s statement is false. The change in momentum of the insect and the motorcar is equal by conservation of momentum. The velocity of insect changes accordingly due to its mass as it is very small compared to the motorcar. Similarly, the velocity of motorcar is very insignificant because its mass is very large compared to the insect.

Rahul’s statement is completely right. As per the third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. However, Rahul’s suggestion that the change in the momentum is the same contradicts the law of conservation of momentum.

18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms –2 .

Given, mass of the dumb-bell (m) = 10kg

Distance covered (s) = 80cm = 0.8m

Initial velocity (u) = 0 (it is dropped from a position of rest)

Acceleration (a) = 10ms -2

Terminal velocity (v) =?

Momentum of the dumb-bell when it hits the ground = mv

As per the third law of motion

v 2 – u 2 = 2as

Therefore, v 2 – 0 = 2 (10 ms -2 ) (0.8m) = 16 m 2 s -2

The momentum transferred by the dumb-bell to the floor = (10kg) × (4 m/s) = 40 kg.m.s -1

Additional Exercises Page: 130

1. The following is the distance-time table of an object in motion:

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero? (b) What do you infer about the forces acting on the object?

(a) The distance covered by the object at any time interval is greater than any of the distances covered in previous time intervals. Therefore, the acceleration of the object is increasing.

(b) As per the second law of motion, force = mass × acceleration. Since the mass of the object remains constant, the increasing acceleration implies that the force acting on the object is increasing as well

2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms -2 . With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort)

Given, mass of the car (m) = 1200kg

When the third person starts pushing the car, the acceleration (a) is 0.2 ms -2 . Therefore, the force applied by the third person (F = ma) is given by:

F = 1200kg × 0.2 ms -2 = 240N

The force applied by the third person on the car is 240 N. Since all 3 people push with the same muscular effort, the force applied by each person on the car is 240 N.

3. A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Given, mass of the hammer (m) = 500g = 0.5kg

Initial velocity of the hammer (u) = 50 m/s

Terminal velocity of the hammer (v) = 0 (the hammer is stopped and reaches a position of rest).

Time period (t) = 0.01s

a = -5000ms -2

Therefore, the force exerted by the hammer on the nail (F = ma) can be calculated as:

F = (0.5kg) * (-5000 ms -2 ) = -2500 N

As per the third law of motion, the nail exerts an equal and opposite force on the hammer. Since the force exerted on the nail by the hammer is -2500 N, the force exerted on the hammer by the nail will be +2500 N.

4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Initial velocity (u) = 90 km/hour = 25 meters/sec

Terminal velocity (v) = 18 km/hour = 5 meters/sec

Time period (t) = 4 seconds

Therefore, the acceleration of the car is -5 ms -2 .

Initial momentum of the car = m × u = (1200kg) × (25m/s) = 30,000 kg.m.s -1

Final momentum of the car = m × v = (1200kg) × (5m/s) = 6,000 kg.m.s -1

Therefore, change in momentum (final momentum – initial momentum) = (6,000 – 30,000) kg.m.s -1

= -24,000 kg.m.s -1

External force applied = mass of car × acceleration = (1200kg) × (-5 ms -2 ) = -6000N

Therefore, the magnitude of force required to slow down the vehicle to 18 km/hour is 6000 N

NCERT Solution Class 9 Science Chapter 9 explains what is a force and its application with various examples. It also explains the three laws of motion, along with examples and mathematical formulations. Conservation of Momentum is explained in a simple way. Various activities are incorporated into the chapter to explain the concepts in an easy way. NCERT Class 9 Science Chapter 9 Force and Laws of Motion also describes the natural tendency of objects to resist a change in their state of rest or of uniform motion, known as inertia, in a detailed way. NCERT Class 9 Science Chapter 9 Force and laws of motion are covered under unit 3 Motion, Force and Work. Refer to previous years’ question papers and sample papers to know the type of questions that appear, along with the allotment of marks and patterns. The topics that are covered under this chapter include:

• Inertia and Mass (4 questions)
• Conservation of Momentum (4 questions)
• Post-Chapter Exercises (18 Questions)

NCERT Solutions For Class 9 Science Chapter 9: Force and Laws Of Motion

• NCERT Solutions for Class 9 explains motion and the causes of motion in a detailed way.
• 1st law of motion, 2nd law of motion and 3rd law of motion are explained with illustrations and mathematical formulations.
• Exercise covers all the topics of the chapter and helps students in gaining confidence to write the CBSE exam.
• Concepts like conservation of momentum, inertia and mass are described in brief.

Key Features of NCERT Solutions for Class 9 Science Chapter 9: Force and Laws Of Motion

• A simple and easily understandable method is followed in NCERT Solutions to make students grasp concepts.
• NCERT Solutions offer comprehensive answers to all the questions to help students in their preparations.
• Provides completely solved solutions to all the questions present in the chapter.
• These solutions will be useful for CBSE exams, Science Olympiads, and other competitive exams.

Disclaimer:

Dropped Topics-  9.6 Conservation of Momentum, Activity 9.5, 9.6, Example 9.6, 9.7, 9.8 and Box item ‘Conservation Laws’.

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Case Study and Passage Based Questions for Class 9 Science Chapter 8 Motion

• Last modified on: 2 years ago
• Reading Time: 3 Minutes

Case Study/Passage Based Questions:

Question 1:

Read the following paragraph and any four questions from (i) to (v).

Distance is the length of the actual path covered by an object, irrespective of its direction of motion. Displacement is the shortest distance between the initial and final positions of an object in a given direction. Distance is a scalar quantity. Displacement is a vector quantity. Distance covered can never be negative. It is always positive or zero. Displacement may be positive, negative or zero.

(i) ______________ is the actual path covered by an object. (a) Speed (b) Motion (c) Velocity (d) Distance

(ii) _______________ is the shortest distance between the initial and final positions of an object. (a) Displacement (b) Acceleration (c) Distance (d) Motion

(iii) Which of the following is a scalar quantity? (a) Displacement (b) Distance (c) Velocity (d) Acceleration

(iv) Distance covered ____________ (a) can never be negative. (b) can never be positive. (c) can never be zero. (d) can be positive or negative.

(v) Which of the following is vector quantity? (a) Displacement (b) Velocity (c) Acceleration (d) All of these

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• Poverty as a Challenge Class 9 Case Study Social Science Economics Chapter 3

Last Updated on September 10, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 9 social science. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 9 social science. In this article, you will find case study for CBSE Class 9 Social Science Economics Chapter 3 Poverty as a Challenge. It is a part of Case Study Questions for CBSE Class 9 Social Science Series.

Table of Contents

Case Study Questions on Poverty as a Challenge Class 9

Read the following passage and answer the questions:

Poverty in India also has another aspect or dimension. The proportion of poor people is not the same in every state. Although state level poverty has witnessed a secular decline from the levels of early seventies, the success rate of reducing poverty varies from state to state. Recent estimates show while the all India Head Count Ratio (HCR) was 21.9 per cent in 2011-12 states like Madhya Pradesh, Assam, Uttar Pradesh, Bihar and Odisha had above all India poverty level. Bihar and Odisha continue to be the two poorest states with poverty ratios of 33.7 and 32.6 per cent respectively. Along with rural poverty, urban poverty is also high in Odisha, Madhya Pradesh, Bihar and Uttar Pradesh.

In comparison, there has been a significant decline in poverty in Kerala, Maharashtra, Andhra Pradesh, Tamil Nadu, Gujarat and West Bengal. States like Punjab and Haryana have traditionally succeeded in reducing poverty with the help of high agricultural growth rates. Kerala has focused more on human resource development. In West Bengal, land reform measures have helped in reducing poverty. In Andhra Pradesh and Tamil Nadu, public distribution of foodgrains could have been responsible for the improvement.

Q. 1. Which are the poorest states in India having low poverty ratios?

Ans. The two poorest states in India having low poverty ratios are Bihar and Odisha with the ratios of 33.7 and 32.6 per cent respectively.

Q. 2. In which states, rural as well as urban poverty are high?

Ans. Rural as well as urban poverty are high in the states of Odisha, Madhya Pradesh, Bihar and Uttar Pradesh.

Q. 3. Which states have shown a significant decline in poverty?

Ans. The states like Kerala, Maharashtra, Andhra Pradesh,Tamil Nadu, Gujarat and West Bengal have shown a significant decline in poverty.

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The french revolution class 9 case study social science history chapter 1, topics from which case study questions may be asked.

Here is a list of topics from which case study questions may be asked.

• Two Typical Cases of Poverty—Urban and Rural
• Poverty as Seen by the Social Scientists
• Poverty Line
• Poverty Estimates
• Vulnerable Groups
• Inter-state Disparities
• Global Poverty Scenario

Poverty is a situation in which a person fails to satisfy his basic needs. His income is insufficient to buy minimum subsistence.

Human poverty refers to means of denial of all types of opportunities, e.g., economics, social and physical opportunities to live a tolerable life.

NREGA was passed in September 2005. The Act provides 100 days assured employment every year to every rural household in 200 districts.

Frequently Asked Questions (FAQs) on Poverty as a Challenge Class 9 Case Study

Q1: what are case study questions.

A1: Case study questions are a type of question that presents a detailed scenario or a real-life situation related to a specific topic. Students are required to analyze the situation, apply their knowledge, and provide answers or solutions based on the information given in the case study. These questions help students develop critical thinking and problem-solving skills.

Q2: How should I approach case study questions in exams?

A2: To approach case study questions effectively, follow these steps: Read the case study carefully: Understand the scenario and identify the key points. Analyze the information: Look for clues and relevant details that will help you answer the questions. Apply your knowledge: Use what you have learned in your course to interpret the case study and answer the questions. Structure your answers: Write clear and concise responses, making sure to address all parts of the question.

Q3: What are the benefits of practicing case study questions from your website?

A3: Practicing case study questions from our website offers several benefits: Enhanced understanding: Our case studies are designed to deepen your understanding of historical events and concepts. Exam preparation: Regular practice helps you become familiar with the format and types of questions you might encounter in exams. Critical thinking: Analyzing case studies improves your ability to think critically and make connections between different historical events and ideas. Confidence: Practicing with our materials can boost your confidence and improve your performance in exams.

Q4: What are the important keywords in this chapter “Poverty as a Challenge”?

A4: Important keywords for CBSE Class 9 Poverty as a Challenge are given below: Poverty: It is a situation under which a man is unable to afford minimum necessities of life, health and efficiency. Poverty Line: It is an estimated income that is required to purchase the minimum subsistence needs of a person. Mass Poverty: It is a situation in which large section of people in an economy is deprived of basic necessities. Human Poverty: It refers to shortfall or lack of income. It is the denial of economic, political, social and physical opportunities to lead long, healthy and creative life. Poverty Ratio: Extent of poverty in a country is measured by the number of people who live below the poverty line and their ratio to the total population. Relative Poverty: It refers to the income or asset position of one class of people compared to the others or individual or vis-à-vis others.

Q5: Describe poverty trends in India since 1973.

A5: As per the data, there is a substantial decline in poverty ratio in India from 55% in 1973 to 36% in 1993. There was further decline from 36% in 1993 to 26% in 2000. Although the number of poor people remained stable (about 320 million) in the earlier two decades (1973 to 1993), there was a significant reduction in the number of the poor to about 260 million till 2000. It may also be noted that poverty ratio always remained higher in rural areas compared to urban areas. If the present trend continues, the people below poverty line may come down to less than 20% in the next few years

Q6: What is Poverty?

A6: Poverty is inability to afford minimum basic necessities of life such as, food, clothing and shelter.

Q7: Why is calorie requirement higher in rural areas than in urban areas?

A7: The calorie requirement is higher in rural areas since people are engaged in more physical labour than in urban areas.

Q8: Why does the poverty line vary from time and place?

A8: The poverty line is estimated on the basis of income incurred on consumption of minimum calorie intake. Since that varies from region to region and the consumption expenditure is subject to prices of food items, the poverty line also varies.

Q9: Why do different countries use different poverty lines?

A9: Different countries use different poverty lines because the basic needs of different countries are different. For example, a person may be considered poor in America if he is not having a car, whereas in India owning of a car is still considered a luxury.

Q10: How can we reduce regional poverty?

A10: Measures to reduce regional poverty: Several states of India like Odisha, Bihar, Madhya Pradesh, Uttar Pradesh, Nagaland, etc., are economically very poor. To remove this regional poverty, the following measures may be suggested: (a) More Central assistance and grants should be given to backward states. (b) Special concessions may be given for investments in backward areas. (c) Public sector enterprises should be set up in backward states.

Q11: Name any two poorest states in India.

A11: Poorest states in India are Bihar and Orissa (Odisha).

Q12: Are there any online resources or tools available for practicing “ Poverty as a Challenge” case study questions?

A12: We provide case study questions for CBSE Class 9 Social Science on our website. Students can visit the website and practice sufficient case study questions and prepare for their exams.

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