case study questions class 9 science gravitation

Class 9 Science Case Study Questions Chapter 10 Gravitation

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Case study Questions in Class 9 Science Chapter 10  are very important to solve for your exam. Class 9 Science Chapter 10 Class 9 Science Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 10 Gravitation

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In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Gravitation Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Science  Chapter 10 Gravitation

Case Study/Passage-Based Questions

Case Study 1: According to the universal law of gravitation, the force between two particles or bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between these particles or bodies. Consider two bodies A and B having masses m 1 and m 2 respectively. Let the distance between these bodies be R. The force of gravitation between these bodies is given by

F ∝ m 1 m 2 and F ∝ 1/R 2

F =G m 1 m 2 /R 2

Where G is constant and is known as the “universal gravitational constant”.

Newton’s law of gravitation is valid (a) in laboratory (b) only on the earth (c) only in our solar system (d) everywhere

Answer: (d) everywhere

Gravitational force is a (a) repulsive force (b) attractive force (c) neither (a) nor (b) (d) both (a) and (b)

Answer: (b) attractive force

Two particles of mass m 1 and m 2 , approach each other due to their mutual gravitational attraction only. Then (a) accelerations of both the particles are equal. (b) acceleration of the particle of mass m 1 is proportional to m 1 . (c) acceleration of the particle of mass m 1 is proportional to m 2 . (d) acceleration of the particle of mass m 1 is inversely proportional to m 1 .

Answer: (c) acceleration of the particle of mass m1 is proportional to m2.

The gravitational force between two bodies is 1 N. If the distance between them is made half, what will be the force? (a) 2 N (b) 4 N (c) 6 N (d) 7 N

Answer: (b) 4 N

How does the force of gravitation between two objects change when the distance between them is reduced to half? (a) Force of gravitation becomes 4 times (b) Force of gravitation becomes 9 times (c) Force of gravitation becomes 6 times (d) Force of gravitation becomes 12 times

Answer: (a) Force of gravitation becomes 4 times

Case Study 2: All freely falling bodies fall with a uniform acceleration due to gravity. As a result, all the equations of motion for the uniformly accelerated bodies moving in a straight line are applicable to the freely falling bodies.

The value of g is taken as positive when a body is (a) dropped from a certain height (b) moving in horizontal direction (c) both (a) and (b) (d) none of these

Answer: (c) both (a) and (b)

Velocity of an object at maximum height in case it has been thrown vertically upward is (a) maximum (b) minimum (c) zero (d) 9.8 m s –1

Answer: (c) zero

During free fall, the acceleration of the object is (a) zero (b) non-uniform (c) constant (d) none of these

Answer: (c) constant

Case Study 3: Gravitation is the force of attraction that exists between any two objects with mass. It is a fundamental force in nature that governs the motion of celestial bodies and plays a crucial role in our everyday lives. The force of gravity depends on the masses of the objects involved and the distance between them. According to the law of gravitation formulated by Sir Isaac Newton, the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This law explains the motion of planets, the falling of objects on Earth, and many other phenomena. The value of acceleration due to gravity is approximately 9.8 m/s^2 on the surface of the Earth. Understanding the concept of gravitation is essential for comprehending various phenomena in the universe and for solving problems related to motion and forces.

What is gravitation? a) The force of attraction between any two objects with mass b) The force of repulsion between any two objects with mass c) The force of attraction between a planet and its moon d) The force of attraction between the Earth and the Sun Answer: a) The force of attraction between any two objects with mass

What factors determine the force of gravity between two objects? a) Masses of the objects and the distance between them b) Speeds of the objects and the direction of their motion c) Shapes of the objects and their surface areas d) Colors of the objects and their temperature Answer: a) Masses of the objects and the distance between them

According to the law of gravitation, how is the force of attraction between two objects related to their masses and distance? a) Directly proportional to the masses and inversely proportional to the distance b) Directly proportional to the distance and inversely proportional to the masses c) Inversely proportional to both the masses and the distance d) Directly proportional to both the masses and the distance Answer: a) Directly proportional to the masses and inversely proportional to the distance

What does the law of gravitation formulated by Sir Isaac Newton explain? a) The motion of planets b) The falling of objects on Earth c) Both a) and b) d) None of the above Answer: c) Both a) and b)

What is the approximate value of acceleration due to gravity on the surface of the Earth? a) 9.8 cm/s^2 b) 9.8 m/s c) 9.8 km/h d) 9.8 m/s^2 Answer: d) 9.8 m/s^2

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 10 Gravitation with Answers Pdf free download has been useful to an extent. If you have any other queries about the CBSE Class 9 Science Gravitation Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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• Gravitation Class 9 Case Study Questions Science Chapter 9

Last Updated on August 26, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 9 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 science. In this article, you will find case study questions for cbse class 9 science chapter 9 Gravitation.

Table of Contents

Case Study Questions on Gravitation

Question 1:

Pallavi Mam was demonstrating an experiment in his class with the setup as shown in the figure below.

She took an eraser, sharpener, steel spoon, plastic ruler, pencil, compass and rubber band and asked the students to place them on the surface of the water. Students observed that a few objects float on the surface of water while a few sink in water.

Read the given passage carefully and give the answer of the following questions:

Q1. An object floats in a liquid if the buoyant force is: a. zero b. greater than its weight c. less than its weight d. equal to its weight

Difficulty Level: Easy

Q 2. An object sinks in a liquid if the buoyant force is: a. zero b. greater than its weight c. less than its weight d. equal to its weight

Q 3. The buoyant force on an object immersed in a liquid act: a. in the vertically upward direction b. in the vertically downward direction c. at an angle of 90 0  to the direction in which weight of the object acts. d. at an angle of 45 0  to the direction in which weight of the object acts.

Difficulty Level: Medium

Q 4. The magnitude of buoyant force depends on the: a. density of liquid b. volume of liquid c. weight of the object d. viscosity of liquid

Q 5. Magnitude of buoyant force is given by: a. Newton’s first law b. Archimedes’ principle c. Newton’s second law d. None of these

• (b) greater than its weight
• (c) less than its weight
• (a) in the vertically upward direction Explanation: The buoyant force on an object immersed in a liquid always acts vertically upwards, that is, in a direction opposite to the weight of the object.
• (a) density of liquid
• (b) Archimedes’ principle Explanation: Archimedes’ principle states that ‘when a body is immersed fully or partially in a fluid, it experiences an upward force (buoyant force) that is equal to the weight of the fluid displaced by it.’

Question 2:

What do aching feet, a falling apple and the orbit of the Moon have in common? Each is caused by the gravitational force. An apple falls from a tree because of the same force acting a few metres above Earth’s surface. And the Moon orbits Earth because gravity is able to supply the necessary centripetal force at a distance of hundreds of millions of metres. Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions.

The gravitational force is always attractive and it depends only on the masses involved and the distance between them (Newton’s universal law of gravitation).

Q 1. What is gravitational force?

Q 2. Briefly explain why Newton pondered over the existence of gravitation?

Q 3. Define the universal gravitational constant.

Q 4. State the value of G . Who obtained it for the first time?

Q 5. ‘Several phenomena of celestial bodies were believed to be unconnected but universal law of gravitation was successful to explain them.’ Mention any two phenomena.

• All objects (with mass) in the universe attract each other. This force of attraction between objects is called the gravitational force.
• It is said an apple fell on Newton’s head when he was sitting under a tree. He thought if Earth attracts an apple, can it also attract the Moon? Is the force same in both cases? This led to the study on gravitation.
• The universal gravitational constant is numerically equal to the force of attraction between two unit masses when they are separated by a unit distance as measured from their centres.
• The accepted value of G is 6.673 x 10 -11 N-m 2 kg -2 . The value of G was found out by Henry Cavendish by using a sensitive balance.
• Two phenomena explained by universal law of gravitation are: (i) the force that binds us to the Earth, and (ii) the motion of the Moon around the Earth.

Case study questions for other chapters of class 9 science is given below.

• Work and Energy Class 9 Case Study Questions Science Chapter 10
• Force and Laws of Motion Class 9 Case Study Questions Science Chapter 8
• Motion Class 9 Case Study Questions Science Chapter 7
• Tissues Class 9 Case Study Questions Science Chapter 6
• The Fundamental Unit of Life Class 9 Case Study Questions Science Chapter 5

Is Matter Around Us Pure Class 9 Case Study Questions Science Chapter 2

Matter in our surroundings class 9 case study questions science chapter 1.

We hope the given case study questions for Gravitation Class 9 helps you in your learning.

Helpful Links for CBSE Class 9 Science Preparation

• Chapter Tests for CBSE Class 9 Science
• Worksheets for CBSE Class 9 Science
• 100 Important Numerical Problems for CBSE Class 9 Physics
• 65 Important Case Study Questions for CBSE Class 9 Science

Topics from which case study questions may be asked

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

Gravitational Force is the force of attraction between any two objects in the universe.

Gravity is the force with which the Earth pulls the objects towards its centre.

Universal Law of Gravitation was given by Isaac Newton and states that, ‘the force of attraction between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of distance between them’.

The direction of the gravitational force is along the line joining the centres of two object.

For further practice on case study questions related to Gravitation Class 9 Science, we recommend exploring the link given below.

How to take quiz or test using the given link

It’s quite simple!

Step 1: Click on the given link. You will see the below screen.

Step 2: Fill in the necessary details. There is no need to register. Just fill your email and name and click on the button “Take Assessment”. The below screen will appear.

Step 3: Click on start assessment. Now you are ready to take test.

Frequently Asked Questions (FAQs) on Gravitation Case Study Questions

Q1: what are case study questions for cbse examinations.

A1: Case study questions in CBSE examinations typically involve scenarios or real-life examples, requiring students to apply their understanding of concepts to solve problems or analyze situations.

Q2: Why are case study questions important for understanding class 9 science chapters?

A2: Case study questions provide a practical context for students to apply theoretical knowledge to real-world situations, fostering deeper understanding and critical thinking skills.

Q3: How should students approach answering case study questions for CBSE?

A3: Students should carefully read the case study, identify the key issues or problems presented, analyze the information provided, apply relevant concepts and principles of chemical reactions and equations, and formulate well-supported solutions or responses.

Q4: Are there any resources available online for students to practice case study questions on class 9 science chapters for CBSE exams?

A4: Yes, several educational websites offer case study questions for CBSE students preparing for science examinations. We also offer a collection of case study questions for all classes and subject on our website. Visit our  website  to access these questions and enhance your learning experience. If you need more case study questions for your preparation, then you visit Physics Gurukul website.

Q5: How can students effectively prepare for case study questions on “Gravitation” for CBSE exams?

A5: Effective preparation strategies include regular revision of concepts, solving practice questions, analyzing case studies from previous exams, seeking clarification on doubts, and consulting with teachers or peers for guidance and support.

Q6: How can teachers incorporate case study questions on “Gravitation” class 9 science into classroom teaching?

A6: Teachers can integrate case studies into lesson plans, group discussions, or interactive activities to engage students in active learning, promote problem-solving skills, and facilitate a deeper understanding of “Force and Laws of Motion”.

Q7: When is the Importance of the Universal Law of Gravitation?

A7: The universal law of gravitation successfully explained several phenomena which are as follows: (i) the force that binds us to Earth, (ii) the motion of the Moon around the Earth, (iii) the motion of planets around the Sun, and (iv) the tides due to the Moon and the Sun.

Q8: What are the Applications of Archimedes’ Principle?

A8: It is used in: (i) designing ships and submarines. (ii) lactometers (used to determine the purity of a sample of milk). (iii) hydrometers (used for determining density of liquids).

Q9: If gravitational force acts between all objects, why don’t the two objects in a room move towards each other?

A9: The two objects in a room do not move towards each other because of their small masses, the gravitational force of attraction between them is very, very weak.

Q10: When a body is thrown upwards, its velocity becomes zero at the highest point. What will be its acceleration at this point?

A10: The acceleration at this point is equal to the value of g in the downward direction

Q11: How does the weight of a body vary from poles to equator?

A11: The weight of a body decreases from poles to equator because the force of gravitation is more at poles than at the equator.

Q12: What do you mean by buoyancy?

A12: The upward force exerted by a liquid on any object immersed in it is known as buoyant force or upthrust.

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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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Class 9 Science Case Study Questions

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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.
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Chapter 10 Class 9 - Gravitation

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Get NCERT Solutions, Notes, Solutions to Intext Questions, Examples of Chapter 10 Class 9 Gravitation free at Teachoo. 

In this chapter, we will learn

What is Gravity ?

What is Universal Law of Gravitation

Important Natural Phenomena Occurring Due to Gravitation

What is Free Fall ?

What is Acceleration Due To Gravity

Deriving value of Acceleration due to Gravity

Different Equations of Motion for Free Falling Object

What is the Difference between Mass and Weight 

What is thrust ?

What is Pressure

What is buoyancy ?

Density And Relative Density Of An Object

Why do Objects float or sink in Water

Archimedes Principle

Different Factors Affecting Buoyancy

Why is gravity maximum at poles and minimum at equator ?

What is the SI unit of Thrust and Pressur e?

What is the SI Unit of g and G ?

What is the difference between gravity and gravitation ?

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NCERT Questions

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Gravitation

Class 9 - ncert science solutions, intext questions 1.

State the universal law of gravitation.

The universal law of gravitation states that every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.

Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

F = G Mm d 2 \dfrac{\text{Mm}}{\text{d}^2} d 2 Mm ​

F = force of attraction

G = constant of proportionality

M = mass of earth

m = mass of the object

d = distance between the earth's centre and object's centre

Intext Questions 2

What do you mean by free fall?

Earth attracts objects towards itself due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall.

What do you mean by acceleration due to gravity?

Whenever an object falls towards the earth, an acceleration is involved due to the earth's gravitational force. So, this acceleration is called the acceleration due to the gravitational force of the earth (or acceleration due to gravity). It is denoted by g and its value is 9.8 ms -2 .

Intext Questions 3

What are the differences between the mass of an object and its weight?

Differences between the mass of an object and its weight:

Why is the weight of an object on the moon 1 6 \dfrac{1}{6} 6 1 ​ th its weight on the earth?

The moon's gravitation force is determined by the mass and the size of the moon.

The mass of moon is 1 100 \dfrac{1}{100} 100 1 ​ times and its radius 1 4 \dfrac{1}{4} 4 1 ​ times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is 1 6 \dfrac{1}{6} 6 1 ​ th its weight on the earth.

Intext Questions 4

Why is it difficult to hold a school bag having a strap made of a thin and strong string?

Pressure is inversely proportional to the surface area on which the force acts. Incase of a strap made of a thin and strong string, the surface area of the strap in contact with the shoulder is less, hence, pressure on the shoulders is more which makes it difficult to hold the school bag.

What do you mean by buoyancy?

The property of liquid to exert an upward force on a body immersed in it, is called buoyancy.

Why does an object float or sink when placed on the surface of water?

An object floats or sinks in water because of the difference in densities of the substance and water. When the density of the substance is less than the density of water, then the upthrust of water is greater than the weight of the substance. Hence it floats.

When the density of substance is more than water, then the upthrust of water on the substance is less than the weight of the substance. Hence it sinks.

Intext Questions 5

You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

When we stand on a weighing machine, the weight acts downwards while the upthrust due to air acts upwards.

As, true weight = (apparent weight + up thrust)

So our apparent weight becomes less than the true weight. This apparent weight is measured by the weighing machine and hence, the mass indicated is less than the actual mass. So our actual mass will be more than 42 kg.

You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?

The cotton bag is heavier than an iron bar as it experiences a larger air thrust than the iron bar.

The cotton bag's density is less than that of the iron bar, so, the volume of the cotton bag is more compared to the iron bar. Hence, the cotton bag experience more upthrust due to the presence of air. So, in the presence of air, the cotton bag's true weight is more compared to the true weight of the iron bar.

How does the force of gravitation between two objects change when the distance between them is reduced to half?

F = G m 1 m 2 d 2 \dfrac{\text{m}_1\text{m}_2}{\text{d}^2} d 2 m 1 ​ m 2 ​ ​

m 1 = mass of first object

m 2 = mass of second object

d = distance between the two objects.

When distance is reduced to half then, d' = d 2 \dfrac{\text{d}}{2} 2 d ​

So, substituting we get,

F' = G m 1 m 2 ( d 2 ) 2 \dfrac{\text{m}_1\text{m}_2}{\Big(\dfrac{\text{d}}{2}\Big)^2} ( 2 d ​ ) 2 m 1 ​ m 2 ​ ​ = G m 1 m 2 d 2 4 \dfrac{\text{m}_1\text{m}_2}{\dfrac{\text{d}^2}{4}} 4 d 2 ​ m 1 ​ m 2 ​ ​ = 4G m 1 m 2 d 2 \dfrac{\text{m}_1\text{m}_2}{\text{d}^2} d 2 m 1 ​ m 2 ​ ​ = 4F

Hence, when the distance between the objects is reduced to half, the gravitational force increases four times.

Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). This acceleration remains constant and is independent of an object's mass. Hence, heavy objects do not fall faster than light objects.

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10 24 kg and radius of the earth is 6.4 × 10 6 m.)

Mass of the body (m) = 1 kg

Mass of the earth (M) = 6 × 10 24 kg

Radius of the Earth (R) = 6.4 × 10 6 m

G = 6.67 x 10 -11 Nm 2 kg -2

According to the formula for gravitational force we know,

F = G Mm R 2 \dfrac{\text{M}\text{m}}{\text{R}^2} R 2 M m ​

Substituting we get,

F = 6.67 × 1 0 − 11 × 6 × 1 0 24 × 1 ( 6.4 × 1 0 6 ) 2 = 6.67 × 1 0 13 6.4 × 6.4 × 1 0 12 = 9.77  N ≈ 9.8  N \text{F} = \dfrac{6.67 \times 10^{-11}\times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2} \\[1em] = \dfrac{6.67 \times 10^{13}}{6.4 \times 6.4 \times10^{12}} \\[1em] = 9.77 \text{ N} \approx 9.8 \text{ N} F = ( 6.4 × 1 0 6 ) 2 6.67 × 1 0 − 11 × 6 × 1 0 24 × 1 ​ = 6.4 × 6.4 × 1 0 12 6.67 × 1 0 13 ​ = 9.77  N ≈ 9.8  N

Hence, magnitude of the gravitational force = 9.8 N

The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

According to universal law of gravitation, two objects with masses attract each other with equal gravitational force, but in opposite directions.

This force is given by: F = G Mm R 2 \dfrac{\text{M}\text{m}}{\text{R}^2} R 2 M m ​

Hence, the earth attracts the moon with the same force with which the moon attracts the earth.

If the moon attracts the earth, why does the earth not move towards the moon?

According to the universal law of gravitation, two objects with masses attract each other with equal gravitational force, but in opposite directions. Hence, moon and earth both attract each other with the same force. Therefore, the earth does not move towards the moon.

What happens to the force between two objects, if

(i) the mass of one object is doubled?

(ii) the distance between the objects is doubled and tripled?

(iii) the masses of both objects are doubled?

(i) We know,

When mass of one object is doubled.

F' = G 2m 1 m 2 d 2 \dfrac{\text{2m}_1\text{m}_2}{\text{d}^2} d 2 2m 1 ​ m 2 ​ ​ = 2G m 1 m 2 d 2 \dfrac{\text{m}_1\text{m}_2}{\text{d}^2} d 2 m 1 ​ m 2 ​ ​ = 2F

Hence, when the mass of one object is doubled, the gravitational force increases two times.

(ii) When distance is doubled then, d' = 2r

F' = G m 1 m 2 ( 2 d ) 2 \dfrac{\text{m}_1\text{m}_2}{(2d)^2} ( 2 d ) 2 m 1 ​ m 2 ​ ​ = G m 1 m 2 4 d 2 \dfrac{\text{m}_1\text{m}_2}{4d^2} 4 d 2 m 1 ​ m 2 ​ ​ = F 4 \dfrac{\text{F}}{4} 4 F ​

Hence, when the distance between the objects is doubled, the gravitational force becomes one fourth of its original force.

Now, if it's tripled

F' = G m 1 m 2 ( 3 d ) 2 \dfrac{\text{m}_1\text{m}_2}{(3d)^2} ( 3 d ) 2 m 1 ​ m 2 ​ ​ =G m 1 m 2 9 d 2 \dfrac{\text{m}_1\text{m}_2}{9d^2} 9 d 2 m 1 ​ m 2 ​ ​ = F 9 \dfrac{\text{F}}{9} 9 F ​

Hence, when the distance between the objects is tripled, the gravitational force becomes one ninth of its original force.

(iii) If masses of both the objects are doubled, then

m 1 ' = 2m 1

m 2 ' = 2m 2

F' = G 2m 1 2m 2 d 2 \dfrac{\text{2m}_1\text{2m}_2}{\text{d}^2} d 2 2m 1 ​ 2m 2 ​ ​ = 4G m 1 m 2 d 2 \dfrac{\text{m}_1\text{m}_2}{\text{d}^2} d 2 m 1 ​ m 2 ​ ​ = 4F

Hence, when the masses of both the objects are doubled, the gravitational force becomes four times the original force.

What is the importance of universal law of gravitation?

The universal law of gravitation successfully explained several phenomena that were believed to be unconnected:

• The force that binds us to the earth
• The motion of the moon around the earth
• The motion of planets around the sun
• The tides due to the moon and the sun.

What is the acceleration of free fall?

The acceleration of free fall is denoted by g and its value on the surface of the earth is 9.8 ms -2 .

What do we call the gravitational force between the earth and an object?

Gravitational force between the earth and an object is known as the weight of the object.

Question 10

Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]

The weight of a body on the earth's surface (W) = mg (where m = mass of the body and g = acceleration due to gravity)

Given, value of g is larger at poles when compared to the equator. So, for the same mass of gold, weight will be less at the equator as compared to the poles. Therefore, Amit's friend will not agree with the weight of gold bought.

Question 11

Why will a sheet of paper fall slower than one that is crumpled into a ball?

A sheet of paper has a larger surface area in comparison to a crumpled paper ball. The larger surface area will suffer greater air resistance. Hence, a sheet of paper falls slower than the crumpled ball.

Question 12

Gravitational force on the surface of the moon is only 1 6 \dfrac{1}{6} 6 1 ​ as strong as gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the earth?

Acceleration due to earth's gravity g = 9.8 ms -2

Object's mass, m = 10 kg

Acceleration due to moon gravity = g m

Weight on the earth = W e

Weight on the moon = W m

Weight = mass x gravity

g m = 1 6 \dfrac{1}{6} 6 1 ​ g e (given)

So, W m = m g m = m x 1 6 \dfrac{1}{6} 6 1 ​ x g e

W m = 10 x 1 6 \dfrac{1}{6} 6 1 ​ x 9.8 = 16.34 N

W e = m x g e = 10 x 9.8

Hence, weight on moon and earth = 16.34 and 98 N respectively.

Question 13

A ball is thrown vertically upwards with a velocity of 49 m/s.

(i) The maximum height to which it rises,

(ii) The total time it takes to return to the surface of the earth.

Initial velocity (u) = 49 ms -1

Final velocity v at maximum height = 0

Acceleration due to earth gravity g = -9.8 ms -2 (negative as ball is thrown up).

Let H be the maximum height to which the ball rises.

By third equation of motion,

2gH = v 2 - u 2

2 × (-9.8) × H = 0 - (49) 2

-19.6 H = -2401

H = − 2401 − 19.6 \dfrac{-2401}{-19.6} − 19.6 − 2401 ​ = 122.5 m

Hence, the maximum height to which it rises = 122.5 m

(ii) Total time T = Time to ascend (T a ) + Time to descend (T d )

According to the first equation of motion,

0 = 49 + (-9.8) x T a

T a = 49 9.8 \dfrac{49}{9.8} 9.8 49 ​ = 5 s

Also, T d = 5 s

∴ T = T a + T d = 5 + 5 = 10 s

Hence, total time it takes to return to the surface of the earth = 10 s

Question 14

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Initial velocity (u) = 0

Tower height = total distance = 19.6 m

g = 9.8 ms -2

Final velocity (v) = ?

According to the third equation of motion,

v 2 = 2gs + u 2

v 2 = (2 x 9.8 x 19.6) + 0

v = 384.16 \sqrt{384.16} 384.16 ​

v = 19.6 ms -1

Hence, final velocity = 19.6 ms -1

Question 15

A stone is thrown vertically upward with an initial velocity of 40 ms -1 . Taking g = 10 ms -2 , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Initial velocity (u) = 40 ms -1

Final velocity (v) = 0

g = 10 ms -2

Max height = ?

v 2 = u 2 - 2gs

Note: [negative g as the object goes up]

0 = (40) 2 - 2 x 10 x s

⇒ 1600 = 20s

⇒ s = 1600 20 \dfrac{1600}{20} 20 1600 ​ = 80 m

Total Distance = s a + s d = 80 + 80 = 160 m

where, s a and s d are distance going up and coming down, respectively.

Displacement = 0 [as the stone comes to the same point from where it was thrown ].

Hence, total distance covered = 160 m and displacement = 0, as the first point is the same as the last point

Question 16

Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10 24 kg and of the Sun = 2 × 10 30 kg. The average distance between the two is 1.5 × 10 11 m.

Mass of the sun (m) = 2 × 10 30 kg

Average distance R = 1.5 × 10 11 m

F = 6.67 × 1 0 − 11 × 6 × 1 0 24 × 2 × 1 0 30 ( 1.5 × 1 0 11 ) 2 \dfrac{6.67 \times 10^{-11}\times 6 \times 10^{24} \times 2 \times 10^{30} }{(1.5 \times 10^{11})^2} ( 1.5 × 1 0 11 ) 2 6.67 × 1 0 − 11 × 6 × 1 0 24 × 2 × 1 0 30 ​ = 3.56 x 10 22 N

Hence, magnitude of the gravitational force = 3.56 x 10 22 N

Question 17

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 ms -1 . Calculate when and where the two stones will meet.

(a) In the case, when the stone falls from the top of the tower,

Initial velocity u = 0

Distance travelled = x

Time taken = t

According to the second equation of motion,

S = ut + 1 2 \dfrac{1}{2} 2 1 ​ gt 2

⇒ S = 0 x t + 1 2 \dfrac{1}{2} 2 1 ​ gt 2

⇒ S = 1 2 \dfrac{1}{2} 2 1 ​ gt 2    [Eq 1]

(b) When another stone is projected vertically upwards,

Initial velocity u = 25 ms -1

Distance travelled = (100 - x)

Using S = ut + 1 2 \dfrac{1}{2} 2 1 ​ gt 2

S' = 25t - 1 2 \dfrac{1}{2} 2 1 ​ gt 2    [Eq 2]

From equations (1) and (2)

S + S' = 1 2 \dfrac{1}{2} 2 1 ​ gt 2 + 25t - 1 2 \dfrac{1}{2} 2 1 ​ gt 2

100 = 1 2 \dfrac{1}{2} 2 1 ​ gt 2 + 25t - 1 2 \dfrac{1}{2} 2 1 ​ gt 2

t = 100 25 \dfrac{100}{25} 25 100 ​ = 4 sec.

Hence, after 4 secs, the two stones will meet.

S = 1 2 \dfrac{1}{2} 2 1 ​ x 10 x 4 2 = 5 x 16 = 80m.

Hence, after 4 sec, 2 stones meet a distance of 80 m from the top.

Question 18

A ball thrown up vertically returns to the thrower after 6 s. Find

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

(c) its position after 4s.

Ball returns to thrower in 6s. So, time up + time down = 6 s hence, time up = 3 s.

(a) Final velocity (v) = 0

From first equation of motion

v = u – gt up

⇒ u = v + gt up

⇒ u = 0 + (9.8 x 3)

⇒ u = 29.4 ms -1

Hence, the velocity with which it was thrown up = 29.4 ms -1 .

(b) According to the second equation of motion,

S = ut - 1 2 \dfrac{1}{2} 2 1 ​ gt 2

S = (29.4 x 3) - ( 1 2 \dfrac{1}{2} 2 1 ​ x 9.8 x 3 x 3) = 88.2 - 44.1 = 44.1 m

Hence, maximum height stone reaches is 44.1 m

(c) In 3 sec, it reaches the maximum height.

Distance travelled in another 1 sec = s'

g is +ve as ball is going down

S = 0 + ( 1 2 \dfrac{1}{2} 2 1 ​ x 9.8 x 1 x 1) = 4.9 m

Hence, in 4 sec the ball will be 4.9 m from the top or 39.2 m (i.e., 44.1 - 4.9) from the bottom.

Question 19

In what direction does the buoyant force on an object immersed in a liquid act?

The buoyant force acts vertically upwards on the object that is immersed in a liquid.

Question 20

Why a block of plastic when released under water come up to the surface of water?

When plastic is immersed in water, two forces act on it:

• Gravitational force or its weight in vertically downward direction
• Buoyant force in vertically upward direction.

As the density of plastic is less than that of water, the buoyant force is greater than its weight. Hence, the plastic comes up to the surface.

Question 21

The volume of 50 g of a substance is 20 cm 3 . If the density of water is 1 gcm -3 , will the substance float or sink?

Volume of substance = 20 cm 3

Mass of substance = 50 g

Density of water = 1 gcm -3

Density of the substance = ?

Density = Mass Volume \dfrac{\text{Mass}}{\text{Volume}} Volume Mass ​ = 50 20 \dfrac{50}{20} 20 50 ​ = 2.5 gcm -3

As density of the substance is greater than density of water. Hence, the substance will sink.

Question 22

The volume of a 500 g sealed packet is 350 cm 3 . Will the packet float or sink in water if the density of water is 1 g cm -3 ? What will be the mass of the water displaced by this packet?

Volume of sealed packet = 350 cm 3

Mass of sealed packet = 500 g

Density of the sealed packet = ?

Density = Mass Volume \dfrac{\text{Mass}}{\text{Volume}} Volume Mass ​ = 500 350 \dfrac{500}{350} 350 500 ​ = 1.42 gcm -3

As density of the packet is greater than density of water. Hence, the sealed packet will sink.

According to Archimedes Principle,

Displaced water volume = Volume of packet

Volume of water displaced = 350 cm 3

Mass of water displaced = ρ x V = 1 × 350 = 350 g

Hence, mass of displaced water = 350 g

Extra Questions for Class 9 Science Chapter 10 Gravitation

Extra questions for Class 9 Science Chapter 10 Gravitation with answers is given below. Our subject expert prepared these solutions as per the latest NCERT textbook. These questions will be helpful to revise the all topics and concepts. CBSE Class 9 extra questions are the most simple and conceptual questions that are prepared by subject experts for the students to study well for the final exams. By solving these extra questions, students can be very efficient in their exam preparations.

Gravitation Class 9 Science Extra Questions and Answers

Very short answer questions.

1: What is the S.I. unit of thrust? Answer:  Newton.

2: What is the S.I. unit of pressure? Answer:  The S.I. unit of pressure = N/m 2  = Pascal.

3: Define thrust. Answer:  The net force exerted by a body in a particular direction is called thrust.

4: Define pressure. Answer:  The force exerted per unit area is called pressure.

5: Why is it easier to swim in sea water than in river water?

Answer:  The density of sea water is more due to dissolved salts in it as compared to the density of river water. Hence the buoyant force exerted on the swimmer by the sea water is more which helps in floating and makes swimming easier.

6: Why a truck or a motorbike has much wider tyres? Answer:  The pressure exerted by it can be distributed to more area, and avoid the wear and tear of tyres.

7: Why are knives sharp? Answer:  To increase the pressure, area is reduced, As pressure ∝ 1/Area hence the pressure or force exerted on a body increases.

8: Why is the wall of dam reservoir thicker at the bottom? Answer:  The pressure of water in dams at the bottom is more, to withstand this pressure the dams have wider walls.

9: Why do nails have pointed tips? Answer:  The force exerted when acts on a smaller area, it exerts larger pressure. So the nails have pointed tips.

10: While swimming why do we feel light? Answer:  The swimmer is exerted by an upward force by water, this phenomenon is called buoyancy and it makes the swimmer feel light.

11: Define density and give its unit. Answer:  The density of a substance is defined as mass per unit volume. Its unit is kg/m 3 .

12: What is relative density? Answer:  The relative density of a substance is the ratio-of its density to that of water. Relative density = density of a substance/density of water

Short Answer Type Questions

1: A ship made of iron does not sink but the iron rod sinks in water, why?

Answer:  The iron rod sinks due to high density and less buoyant force exerted by the water on it, but in case of ship the surface area is increased, the upthrust experienced by the body is more. So it floats on water.

2: Camels can walk easily on desert sand but we are not comfortable walking on the sand. State reason.

Answer:  Camels feet are broad and the larger area of the feet reduces the force/ pressure exerted by the body on the sand. But when we have to walk on the same sand, we sink because the pressure exerted by our body is not distributed but is directional.

3: What is lactometer and hydrometer? Answer:  Lactometer is a device used to find the purity of a given sample of milk. Hydrometer is a device used to find the density of liquids.

4: The relative density of silver is 10.8. What does this mean? Answer:  It means that the density of silver is 10.8 times more than that of water.

5: The relative density of gold is 19.3. The density of water is 10 3  kg/m 3 ? What is the density of gold in S.I. unit?

Answer:  Relative density of gold = 19.3 Relative density of gold = Density of gold/Density of water

∴ Density of gold = Relative density of gold × Density of water = 19.3 × 10 3  Kg/m 3 =19300 Kg/m 3

6: State Archimedes’ principle.

Answer:  Archimedes’ principle—When a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it. It is used in designing of ships and submarines.

7: Two cork pieces of same size and mass are dipped in two beakers containing water and oil. One cork floats on water but another sink in oil. Why?

Answer:  The cork floats on water because the density of cork is less than the density of water, and another cork sinks in the oil because the density of cork is more than the oil.

8: What are fluids? Why is Archimedes’ principle applicable only for fluids? Give the application of Archimedes’ principle.

Answer:  Fluids are the substances which can flow e.g., gases and liquids are fluids. Archimedes’ principle is based on the upward force exerted by fluids on any object immersed in the fluid. Hence it is applicable only for fluids. Applications of Archimedes’ principle: 1. It is used in designing of ship and submarine. 2. It is used in designing lactometer, used to determine the purity of milk. 3. To make hydrometers, used to determine the density of liquids.

Long Answer Type Questions

1: With the help of an activity prove that the force acting on a smaller area exerts a larger pressure?

Answer:  Consider a block of wood kept on a table top. The mass of the wooden block is 5 kg. Its dimension is 40 cm x 20 cm x 10 cm. Now, we have to find the pressure exerted by the wooden block on the table top by keeping it vertically and horizontally. The mass of the wooden block = 5 kg Weight of the wooden block applies a thrust on the table top

∴ Thrust = F = m × g = 5 × 9.8 m/s 2   = 49 N

(case a) — when the wooden box is kept vertically with sides 20 cm × 10 cm. Area of a side = length × breadth = 20 cm × 10 cm = 200 cm 2   = 0.02 m 2

(case b) — When the block is kept horizontally with side 40 cm × 20 cm Area = length × breadth = 40 cm × 20 cm = 800 cm 2   = 0.08 m 2

∴ The pressure exerted by the box in case (a) is more as compared to the pressure exerted in case

(b). The area is reduced and the pressure exerted is more. This shows that pressure ∝ 1/area. Pressure will be larger if the area is reduced.

Application: • Nails have pointed tips. • Knives have sharp edges. • Needles have pointed tips.

Value Based Questions

1: A milkman sold his milk in the city and always carried lactometer with him. The customers trusted him and his business flourished. (a) What is lactometer? (b) What is the principle of working of lactometer? (c) What value of milkman is seen in this case?

Answer:  (a) Lactometer is a device that measures the purity of milk. (b) The principle of lactometer is ‘Archimedes’ principle’. It states that when a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it. (c) Milkman is very honest and trustworthy.

2: Reeta was wearing a high heel shoe for a beach party, her friend told her to wear flat shoes as she will be tired soon with high heels and will not feel comfortable, (a) Why would one feel tired with high heel shoes on beach? (b) Give the unit of pressure. (c) What value of Reeta’s friend is seen in the above act?

Answer:  (a) The high heel shoes would exert lot of pressure on the loose sand of beach and will sink more in the soil as compared to flat shoes. Hence large amount of force will be required to walk with heels. (b) Unit of pressure is Pascal. (c) Reeta’s friend showed the value of being helpful, concerned and intelligent.

3: In the school fair, there was a game in which one need to find the heaviest ball without holding them in hand. Three balls were given and few disposable glasses were kept. Tarun saw his friend struggling to win the game but he was unable to find the heaviest ball. Tarun helped him by dipping the three balls one by one in the glass’es full of water upto the brim and finally they won the game.

(a) Why did Tarun told his friend to dip the balls one by one in completely filled glass of water? (b) Name the principle used here. (c) What value of Tarun is reflected in this case?

Answer: (d) Tarun wanted to measure the amount of water displaced by each ball when dipped in water. (b) The principle used is ‘Archimedes’ principle’. (c) Tarun showed the value of being helpful, kind and intelligent.

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

Chapter 10 Gravitation Class 9 Science NCERT Solutions

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NCERT Class 9 Science Chapter 10 Gravitation Extra Questions

Ncert class 9 science chapter 10 gravitation extra questions and answers.

Class 9 Science Chapter 10 Extra Inside Questions and Answers – Gravitation. Here in this Page Class IX Students can Learn Extra Questions & Answer 10th Chapter Science fully Inside.

Class 9 Science Chapter 10 Extra Question with Answer – Gravitation

Very short answer type questions:.

1.) Is gravitational force depend on separation between objects?

2.) Why does the ball moves towards earth when it dropped from top of tower?

4.) State the centripetal force which binds earth in its orbit.

7.) On what factors does the weight depend?

10.) What is the meaning of ‘Eureka’?

Answer: Eureka means ‘I have got it’.

Answer: Yes, relative density of platinumis greater than water.

14.) State the quantity which depend on force and area.

Short Answer Type Questions:

Answer: As we know that, gravitational force is inversely proportional to square of distance between two objects.

If d is the separation between two objects then the magnitude of force is,

Acceleration due to gravity = g =9.8 m/s².

Force = 980 N.

M – Kilogram.

5.) If a force of 20 N is exerted on an object having mass 4 kg then calculate acceleration produced in it.

Acceleration = 20 / 4.

According to 1st equation of motion

According to 3rd law of motion, Force exerted by table on wooden block is equal to force exerted by wooden block on table.

Force = mass × acceleration

Weight = 49000 N.

14.) A man has mass 70 kg then calculate its weight on moon.

Force = (0.2 × 0 –0.2 ×60)/0.2

Long Answer Type Questions

Area = (22/7 ) × 1.4²

Answer: Given, Density of brick = 15000 kg/m³.

Mass = 11250 kg.

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NCERT Solutions for Class 9 Science Chapter 9 Gravitation

• 10th June 2023

NCERT Solutions for Class 9 Science Chapter 9 Gravitation provides detailed answers for all in-text and exercise Questions. These solutions contain an in-depth explanation of each topic involved in the chapter. Students studying in class 9 can access these solutions for free in PDF format.

All these solutions are prepared by expert teachers and updated for the current academic session. NCERT Solutions for Class 9 Science Chapter 9 Gravitation help students to understand the fundamental concepts given in class 9 Science textbook. We have prepared the answers to all the questions in an easy and well-structured manner. It helps students to grasp the chapter easily.

CBSE Class 9 Science Gravitation Intext Questions (Solved)

PAGE NO. 102

Question 1: State the universal law of gravitation .

Answer: The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. 

For two objects of masses m 1 and m 2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:

Question 2: Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth. 

PAGE NO. 104

Question 1: What do you mean by free fall? 

Answer: Gravity of the Earth attracts every object towards its centre. When an object is released from a height, it falls towards the surface of the Earth under the influence of gravitational force. The motion of the object is said to have free fall. 

Question 2: What do you mean by acceleration due to gravity? 

Answer: When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s 2 . 

PAGE NO. 106

Question 1: What are the differences between the mass of an object and its weight? 

Answer: 

Question 2: Why is the weight of an object on the moon (1/6)th its weight on the earth? 

Answer: Let M E be the mass of the Earth and m be an object on the surface of the Earth. Let R E be the radius of the Earth. According to the universal law of gravitation, weight W E of the object on the surface of the Earth is given by,

 Let M M and R M be the mass and radius of the moon. Then, according to the universal law of gravitation, weight W M of the object on the surface of the moon is given by:

Therefore, weight of an object on the moon is of its weight on the Earth. 

PAGE NO. 109

Question 1: Why is it difficult to hold a school bag having a strap made of a thin and strong string? 

Answer: It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large. 

Question 2: What do you mean by buoyancy? 

Answer:  The upward force exerted by a liquid on an object immersed in it is known as buoyancy. When you try to immerse an object in water, then you can feel an upward force exerted on the object, which increases as you push the object deeper into water. 

Question 3: Why does an object float or sink when placed on the surface of water? 

Answer: If the density of an object is more than the density of the liquid, then it sinks in the liquid. This is because the buoyant force acting on the object is less than the force of gravity. On the other hand, if the density of the object is less than the density of the liquid, then it floats on the surface of the liquid. This is because the buoyant force acting on the object is greater than the force of gravity. 

PAGE NO. 110

Question 1: You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg? 

Answer: When you weigh your body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value. 

Question 2: You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why? 

Answer: The bag of cotton is heavier than an iron bar. This is because the surface area of the cotton bag is larger than the iron bar. Hence, a more buoyant force acts on the bag than that on an iron bar. This makes the cotton bag lighter than its actual value. For this reason, the iron bar and the bag of cotton show the same mass on the weighing machine, but actually the mass of the cotton bag is more than that of the iron bar. 

CBSE Class 9 Science Gravitation Exercise Questions and Answers

Question 1: How does the force of gravitation between two objects change when the distance between them is reduced to half? 

Answer: According to the universal law of gravitation, gravitational force (F) acting between two objects is inversely proportional to the square of the distance between them, i.e., 

Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

Question 2: Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object? 

Answer: All objects fall on the ground with constant acceleration, called acceleration due to gravity (in the absence of air resistance). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

Question 3: What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10 24 kg and radius of the earth is 6.4 × 10 6 m). 

Answer: According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by:

Question 4: The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why? 

Answer: According to the universal law of gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the moon with an equal force with which the moon attracts the earth. 

Question 5: If the moon attracts the earth, why does the earth not move towards the moon? 

Answer: The Earth and the moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the moon. Hence, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. For this reason, the Earth does not move towards the moon. 

Question 6: What happens to the force between two objects, if (i) the mass of one object is doubled? (ii) the distance between the objects is doubled and tripled? (iii) the masses of both objects are doubled?

Answer: According to the universal law of gravitation, the force of gravitation between two objects is given by:

(i) F is directly proportional to the masses of the objects. If the mass of one object is doubled, then the gravitational force will also get doubled. 

(ii) F is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value. Similarly, if the distance is tripled, then the gravitational force becomes one-ninth of its original value. 

(iii) F is directly proportional to the product of masses of the objects. If the masses of both the objects are doubled, then the gravitational force becomes four times the original value. 

Question 7: What is the importance of universal law of gravitation? 

Answer: The universal law of gravitation proves that every object in the universe attracts every other object. 

Question 8: What is the acceleration of free fall?

Answer: When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 ms −2 , which is constant for all objects (irrespective of their masses). 

Question 9: What do we call the gravitational force between the Earth and an object? 

Answer: Gravitational force between the earth and an object is known as the weight of the object. 

Question 10:  Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator]. 

Answer:  Weight of a body on the Earth is given by W = mg Where, m = Mass of the body  g = Acceleration due to gravity 

The value of g is greater at poles than at the equator. Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought. 

Question 11: Why will a sheet of paper fall slower than one that is crumpled into a ball?

Answer: When a sheet of paper is crumbled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper. 

Question 12: Gravitational force on the surface of the moon is only (1/6)  as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth? 

Answer: Weight of an object on the moon =  (1/6)× Weight of an object on the Earth   Also,   Weight = Mass × Acceleration  Acceleration due to gravity, g = 9.8 m/s 2   Therefore, weight of a 10 kg object on the Earth = 10 × 9.8 = 98 N  And, weight of the same object on the moon = (1/6) × 98 = 16.3 N

Question 13: A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate:  (i)  the maximum height to which it rises.  (ii) the total time it takes to return to the surface of the earth. 

Answer: (i) According to the equation of motion under gravity v 2 − u 2 = 2gs  Where, u = Initial velocity of the ball  v = Final velocity of the ball  s = Height achieved by the ball  g = Acceleration due to gravity  At maximum height, final velocity of the ball is zero, i.e., v = 0 m/s and u = 49 m/s  During upward motion, g = − 9.8 m s −2   Let h be the maximum height attained by the ball.   Hence, using 𝑣 2 − 𝑢 2 = 2𝑔𝑠, we have,  

(ii) Let t be the time taken by the ball to reach the height 122.5 then according to the equation of motion v = u + at, we get

But, Time of ascent = Time of descent  Therefore, total time taken by the ball to return = 5 + 5 = 10 s 

Question 14: A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. 

Answer:  According to the equation of motion under gravity v 2 − u 2 = 2gs  Where,  u = Initial velocity of the stone = 0 m/s  v = Final velocity of the stone  s = Height of the stone = 19.6 m  g = Acceleration due to gravity = 9.8 ms −2  

Now using v 2 − u 2 = 2gs, we get v 2 − 0 2 = 2 × 9.8 × 19.6  ⇒ v 2 = 2 × 9.8 × 19.6 = 19.6) 2   ⇒ v = 19.6 ms −1   Hence, the velocity of the stone just before touching the ground is 19.6 ms −1 .  

Question 15: A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s 2 , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? 

Answer: According to the equation of motion under gravity v 2 − u 2 = 2gs  Where,  u = Initial velocity of the stone = 40 m/s  v = Final velocity of the stone = 0 m/s  s = Height of the stone   g = Acceleration due to gravity = −10 ms −2   Let h be the maximum height attained by the stone.  

Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m  Net displacement during its upward and downward journey = 80 + (−80) = 0. 

Question 16: Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10 24 kg and of the Sun = 2 × 10 30 kg. The average distance between the two is 1.5 × 10 11 m. 

Answer: According to the universal law of gravitation, the force of attraction between the Earth and the Sun is given by

Hence, the force of gravitation between the Earth and the Sun is 3.57 × 10 22 𝑁

Question 17: A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet. 

Answer: Let the two stones meet after a time t. 

When the stone dropped from the tower  Initial velocity, u = 0 m/s  Let the displacement of the stone in time t from the top of the tower be s.   Acceleration due to gravity, g = 9.8 ms −2   From the equation of motion, 

Initial velocity, u = 25 ms −1   Let the displacement of the stone from the ground in time t be 𝑠′.   Acceleration due to gravity, g = −9.8 ms −2   Equation of motion, 

The combined displacement of both the stones at the meeting point is equal to the height of the tower 100 m. 

In 4 s, the falling stone has covered a distance given by (1) as 𝑠 = 4.9 × 4 2 = 78.4 𝑚 Therefore, the stones will meet after 4 s at a height (100 – 78.4) = 20.6 m from the ground.

Question 18: A ball thrown up vertically returns to the thrower after 6 s. Find  (a) the velocity with which it was thrown up,  (b) the maximum height it reaches, and  (c) its position after 4 s. 

Answer: (a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey.  Hence, it has taken 3 s to attain the maximum height. 

Final velocity of the ball at the maximum height, v = 0 m/s  Acceleration due to gravity, g = −9.8 ms −2 Using equation of motion, v = u + at, we have  0 = u + (−9.8 × 3)  ⇒ u = 9.8 × 3 ⇒ u = 29.4 m/s  Hence, the ball was thrown upwards with a velocity of 29.4 m/s. 

(b) Let the maximum height attained by the ball be h.  Initial velocity during the upward journey, u = 29.4 m/s  Final velocity, v = 0 m/s Acceleration due to gravity, g = −9.8 ms −2   Using the equation of motion, 

Hence, the maximum height is 44.1 m.

(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.   In this case,  Initial velocity, u = 0 m/s Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3 s = 1 s. 

Now, total height = 44.1 m  This means the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds. 

Question 19: In what direction does the buoyant force on an object immersed in a liquid act? 

Answer: An object immersed in a liquid experiences buoyant force in the upward direction. 

Question 20: Why does a block of plastic released under water come up to the surface of water? 

Answer: Two forces act on an object immersed in water. One is the gravitational force, which pulls the object downwards, and the other is the buoyant force, which pushes the object upwards. If the upward buoyant force is greater than the downward gravitational force, then the object comes up to the surface of the water as soon as it is released within the water. Due to this reason, a block of plastic released under water comes up to the surface of the water. 

Question 21: The volume of 50 g of a substance is 20 cm 3 . If the density of water is 1 g cm −3 , will the substance float or sink? 

Answer: If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid. 

The density of the substance is more than the density of water (1 g cm −3 ).  Hence, the substance will sink in water. 

Question 22: The volume of a 500 g sealed packet is 350 cm 3 . Will the packet float or sink in water if the density of water is 1 g cm −3 ? What will be the mass of the water displaced by this packet? 

Answer:  Density of the 500 g sealed packet

The density of the substance is more than the density of water (1 𝑔/𝑐𝑚 3 ). Hence, it will sink in water.  The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g. 

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NCERT Solutions for Class 9 Science Chapter 9- Gravitation

• NCERT Solutions
• Chapter 10 Gravitation

NCERT Solutions for Gravitation Class 9 Questions and Answers FREE PDF Download

Class 9 Science Ch 9 explores the concept of gravitational force, understanding its effects on celestial bodies like planets and moons, and Archimedes’ Principle. Vedantu’s Class 9 Gravitation NCERT Solutions solves all the questions in the chapter and helps students navigate through complex concepts with clarity and precision.  Access Vedantu's Gravitation Class 9 solutions for step-by-step explanations and problem-solving strategies and enhance your learning experience.

Download Vedantu's Science Class 9 Gravitation NCERT Solutions, revised to align with the Class 9 Science syllabus . Start your academic journey with Vedantu and pave your way towards academic excellence.

Quick Insights for NCERT Solutions for Class 9 Chapter 9 Science Gravitation

Class 9 Science Ch 9 comprehends the concept of gravitation, the force of attraction that exists between any two objects with mass, and elucidates Newton's Law of Universal Gravitation, a cornerstone of classical physics. 

Ch 9 Science Class 9 explores the effects of gravitation in maintaining the stability of celestial bodies such as planets, stars, and galaxies.

Class 9 Gravitation Question Answer delves into the concept of acceleration due to gravity, the distinction between weight and mass, and Archimedes’ Principle.

Gravitation Class 9 Questions And Answers develop proficiency in solving numerical problems related to the motion of objects under the influence of the earth's gravitational force, Pressure, and Thrust.

Vedantu offers additional resources such as class notes, important concepts, formulas, and exemplar solutions to reinforce learning and ensure a strong grasp of foundational scientific principles.

Access NCERT Solutions For Class 9 Science Chapter 9 – Gravitation

Intext exercise 1.

1. State the universal law of gravitation.

Ans: Every object in the universe attracts every other object with some force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centres of two objects.

Let the two objects \[A\] and \[B\]of masses \[M\]and \[m\] lie at a distance \[d\] from each other. Let the force of attraction between two objects be \[F\].

\[F=\frac{GMm}{{{r}^{2}}}\]

Where, 

\[G\]is the universal gravitation constant which is given by:

\[G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}\]

2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Ans: Let the mass of the Earth be \[M\] and the mass of an object on its surface be \[m\]. If \[R\]is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (\[F\]) that acts between the Earth and the object can be given by the relation:

\[F=\frac{GMm}{{{R}^{2}}}\].

Intext Exercise 2

1. What do you mean by free fall?

Ans: Each object is drawn towards the centre of the Earth by its gravity. When any object is released from a certain height, under the impact of gravitational force, it falls to the Earth's surface. The movement of the object is said to be in free fall.

2. What do you mean by acceleration due to gravity?

Ans: When any object falls freely from a certain height towards the earth's surface, its velocity changes with respect to time. This change in velocity causes acceleration. This acceleration is known as the acceleration due to gravity (\[g\]). The value of acceleration due to gravity is \[9.8m{{s}^{-2}}\].

Intext Exercise 3

1. What are the differences between the mass of an object and its weight?

Ans: The difference between the mass of an object and its weight is given in the table below:

2. Why is the weight of an object on the moon \[\frac{1}{6}th\] its weight on the earth?

Ans: Let the mass of the Earth be \[{{M}_{E}}\] and the mass of an object on the surface of earth \[=m\] and the radius of earth \[{{R}_{E}}\].

According to the Universal law of gravitation, weight \[{{W}_{E}}\] of the object on the surface of the earth is given by,

\[{{W}_{E}}=\frac{G{{M}_{E}}m}{{{R}_{E}}^{2}}\]

Let \[{{M}_{M}}\] and \[{{R}_{M}}\] be the mass and radius of the moon. Then, according to the universal law of gravitation, weight \[{{W}_{M}}\] of the object on the surface of the moon is given by:

\[{{W}_{M}}=\frac{G{{M}_{M}}m}{{{R}_{M}}^{2}}\]

So, ratio of weight of object on moon to weight on earth is

\[\frac{{{W}_{M}}}{{{W}_{E}}}=\frac{{{M}_{M}}{{R}_{E}}^{2}}{{{M}_{E}}{{R}_{M}}^{2}}\]

Where, \[{{M}_{E}}=5.98\times {{10}^{24}}kg\] 

\[{{M}_{M}}=7.36\times {{10}^{22}}kg\]

\[{{R}_{E}}=6.4\times {{10}^{6}}m\]

\[{{R}_{M}}=1.74\times {{10}^{6}}m\] 

Substituting the values in the ratio,

\[\Rightarrow \frac{{{W}_{M}}}{{{W}_{E}}}=\frac{7.36\times {{10}^{22}}\times {{\left( 6.37\times {{10}^{6}} \right)}^{2}}}{5.98\times {{10}^{24}}\times {{\left( 1.74\times {{10}^{6}} \right)}^{2}}}\]

\[\Rightarrow \frac{{{W}_{M}}}{{{W}_{E}}}=0.165\approx \frac{1}{6}\]

Hence, the weight of an object on the moon is \[\frac{1}{6}th\] of its weight on the Earth.

Intext Exercise 4

1. Why is it difficult to hold a school bag with a strap made of a thin and strong string?

Ans: Pressure can be given by the formula,

\[P=\frac{F}{A}\]

Pressure is inversely proportional to the surface area on which the force is acting. The smaller is the surface area, the larger will be the pressure on the surface on which the force is being acted upon. In the case of a thin strap of the school bag, the contact surface area is very less. Hence, the pressure exerted on the shoulder is very high. Therefore, it becomes difficult to hold a school bag with a thin strap.

2. What do you mean by buoyancy?

Ans: The liquid exerts an upward force on any object when it is immersed in a liquid or fluid. The tendency of the liquid to exert such an upward force on the object is called buoyancy, and the upward force which is exerted on the object by the liquid is called the buoyant force.

3. Why does an object float or sink when placed on the surface of the water?

Ans: If the density of an object is greater than the density of the liquid, it will sink into the liquid. This is due to the buoyant force which is acted by the object is less than the force of gravity. 

On the contrary, if the density of the object is less than the density of the liquid, it floats on the liquid's surface. This is because the force that is acting on the object is greater than the force of gravity.

Intext Exercise 5

1. You find your mass to be \[42\] kg on a weighing machine. Is your mass more or less than \[42\]kg?

Ans: An upward force acts on our body when we weigh our body while standing on a weighing machine. The buoyant force is which is a upward force that is acting. Consequently, the body is pushed up slightly, resulting in the weighing machine showing less reading than the real value.

2. You have a bag of cotton and an iron bar, each indicating a mass of \[100kg\] when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?

Ans: Weight measured \[=\] Actual weight \[-\] buoyant force

Therefore, Actual weight \[=\] Weight measured \[+\]buoyant force

As the surface area of the cotton, the bag is greater than the iron bar, more buoyant force acts on the bag than that on the iron bar. Hence, the mass of the cotton bag is more than that of the iron bar.

NCERT Exercise

1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Ans: According to the universal law of gravitation, the gravitational force (\[F\]) acting between two objects of mass \[{{m}_{1}}\]and \[{{m}_{2}}\], separated by a distance ‘\[r\]’ is given by

\[F=\frac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\]

Where \[{{m}_{1}}\]and \[{{m}_{2}}\]are the masses of two bodies and \[r\]is the distance between them, \[G\] is the universal gravitational constant.

When the distance is reduced to half, i.e., \[{r}'=\frac{r}{2}\]

\[\Rightarrow F=\frac{G{{m}_{1}}{{m}_{2}}}{{{\left( \frac{r}{2} \right)}^{2}}}\]

\[\Rightarrow F=\frac{G{{m}_{1}}{{m}_{2}}}{\frac{{{r}^{2}}}{4}}\]

\[\Rightarrow F=\frac{4G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}\]

Hence, if the distance is reduced to half, then the gravitational force becomes four times that of the previous value.

2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

Ans: All the objects fall towards the ground with constant acceleration, called acceleration due to gravity (if there is no air resistance present). It is constant and independent of the mass of the object. Hence, heavy objects do not fall faster than light objects.

3. What is the magnitude of the gravitational force between the earth and a \[1kg\]object on its surface? (Mass of the earth is \[6\times {{10}^{24}}kg\] and radius of the earth is \[6.4\times {{10}^{6}}m\]).

Ans: According to the Universal law of gravitation, the gravitational force exerted on an object of mass \[m\]is given by:

Mass of Earth, \[M=6\times {{10}^{24}}kg\]

Mass of object, \[m=1kg\]

Universal gravitational constant, \[G=6.7\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}\]

Since the object is on the surface of the Earth, \[r=\]radius of the Earth (\[R\])

\[r=R=6.4\times {{10}^{6}}m\]

Gravitational force,

\[\Rightarrow F=\frac{6.7\times {{10}^{-11}}\times 6\times {{10}^{24}}\times 1}{{{\left( 6.4\times {{10}^{6}} \right)}^{2}}}=9.8N\].

The magnitude of the gravitational force between the earth and a \[1kg\]object on its surface is \[9.8N\].

4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

Ans: According to the Universal law of gravitation, two objects attract each other and according to Newton's third law of motion, the force of attraction between two objects is the same but acts in the opposite direction. Thus, the earth attracts the moon with the same force as the moon exerts on earth but the force acts in the opposite direction.

5. If the moon attracts the earth, why does the earth not move towards the moon?

Ans: The Earth and the moon experience equal gravitational forces acting towards each other.

By Newton's Second Law, \[F=ma\]

\[\Rightarrow a=\frac{F}{m}\]

For a certain force, acceleration is inversely proportional to the mass of an object.

\[a\propto \frac{F}{m}\]

Mass of the Earth \[>>\] Mass of the moon.

Hence, the acceleration experienced by earth due to the gravitational pull of the moon is very small when compared to that experienced by the moon due to earth. That is why the Earth does not move towards the moon.

6. What happens to the force between two objects, if

a) The mass of one object is doubled?

Ans: According to the universal law of gravitation, the force of gravitation between two objects is given by: \[F=\frac{GMm}{{{r}^{2}}}\]

\[F\]is directly proportional to the product of masses of the two objects.

\[F\propto Mm\]

If the mass of one object is doubled, then the gravitational force will also change to double the original.

b) The distance between the objects is doubled and tripled?

Ans: \[F\]is inversely proportional to the square of the distance between the objects.

If the distance between the objects is doubled, then the gravitational force becomes one-fourth of its original value. Also, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.

c) The masses of both objects are doubled?

Ans: \[F\]is directly proportional to the product of masses of the objects.

If the masses of both the objects are doubled, then the gravitational force becomes four times the original value.

7. What is the importance of the universal law of gravitation?

Ans: The universal law of gravitation states that every object in the universe attracts every other object.

The force of gravitation binds us to the earth.

It is the cause for the motion of the moon around the earth and planets around the sun.

It results in the formation of tides due to the moon and the Sun. High tide occurs at the side where the moon pulls towards itself.

8. What is the acceleration of free fall?

Ans: A free-falling object is an object that is falling due to gravity without any air resistance. When it falls, there is a variation in velocity with respect to time that is associated with it.

Acceleration of free fall is denoted by \[g\]and its value on the surface of the earth is \[9.8m{{s}^{-2}}\], which is constant for all objects (irrespective of their masses).

9. What do we call the gravitational force between the Earth and an object?

Ans: The gravitational force between the earth and an object is called the weight of that object. It is equal to the product of acceleration due to the gravity and mass of the object.

10. Amit buys a few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? (Hint: The value of g is greater at the poles than at the equator).

Ans: Weight of a body on the Earth is given by:

\[m=\]Mass of the body

\[g=\]Acceleration due to gravity

The shape of Earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of \[g\]becomes greater at the poles than at the equator. Since the value of \[g\] is greater at the poles than the equator.

Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.

11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

Ans: When a sheet of paper is crumbled into a ball, then its surface area becomes much lesser than the surface area of a plain non-crumpled sheet of paper.

Hence, the upward force exerted by air on the sheet is greater as compared to the one exerted on the ball. Hence the sheet falls slower as compared to a paper ball.

12.Gravitational force on the surface of the moon is only \[\frac{1}{6}\] as strong as the gravitational force on the Earth. What is the weight in newtons of a \[10kg\]object on the moon and on the Earth?

Ans: It is provided that, \[Weight\text{ }of\text{ }an\text{ }object\text{ }on\text{ }the\text{ }moon=\frac{1}{6}\times Weight\text{ }of\text{ }an\text{ }object\text{ }on\text{ }the\text{ }Earth\]

\[Weight=\,Mass\times Acceleration\]

Acceleration due to gravity, \[g=9.8m{{s}^{-2}}\]

Therefore, the weight of a 10 kg object on the Earth \[=10\times 9.8N=98N\]

Weight of the same object on the moon \[=\frac{1}{6}\times 9.8N=16.3N\]

13. A ball is thrown vertically upwards with a velocity of \[49m{{s}^{-1}}\]. Calculate

a) The maximum height to which it rises.

Ans: According to the equation of motion under gravity:

\[{{v}^{2}}-{{u}^{2}}=2gh\]

\[u=\]Initial velocity of the ball

\[v=\]Final velocity of the ball

\[h=\]Height achieved by the ball

At maximum height, final velocity of the ball is zero, i.e., \[v=0\]

\[u=49m{{s}^{-1}}\]

During upward motion, \[g=9.8m{{s}^{-2}}\]

Let \[h\] be the maximum height attained by the ball.

\[\Rightarrow {{\left( 0 \right)}^{2}}-{{\left( 49 \right)}^{2}}=2\times \left( -9.8 \right)\times h\]

\[\Rightarrow h=\frac{49\times 49}{2\times 9.8}=122.5\]

b) The total time it takes to return to the surface of the earth.

Ans: Let \[t\] be the time taken by the ball to reach the height \[122.5m\], then according to the equation of motion:

Substituting the values and solving,

\[\Rightarrow 0=49+t\times \left( -9.8 \right)\]

\[\Rightarrow 9.8t=49\]

\[\Rightarrow t=\frac{49}{9.8}=5s\]

Time of ascent = Time of descent

Therefore, the total time taken by the ball to return is \[5+5=10s\].

14. A stone is released from the top of a tower of height \[19.6m\]. Calculate its final velocity just before touching the ground.

\[u=\]Initial velocity of the stone \[=0\]

\[v=\]Final velocity of the stone

\[s=\]Height of the stone \[=9.6m\]

g = Acceleration due to gravity \[=9.8m{{s}^{-2}}\]

\[\Rightarrow {{v}^{2}}-{{0}^{2}}=2\times 9.8\times 19.6\]2

\[\Rightarrow {{v}^{2}}=2\times 9.8\times 19.6={{\left( 19.6 \right)}^{2}}\]

\[\Rightarrow v=19.6m{{s}^{-1}}\]

Hence, the velocity of the stone just before touching the ground is \[19.6m{{s}^{-1}}\].

15. A stone is thrown vertically upward with an initial velocity of \[40m{{s}^{-1}}\]. Taking \[g=10m{{s}^{-2}}\], find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

\[u=\]Initial velocity of the stone \[=40m{{s}^{-1}}\]

\[v=\]Final velocity of the stone\[=0\]

\[s=\]Height of the stone 

g = Acceleration due to gravity \[=-10m{{s}^{-2}}\]

Let \[h\] be the maximum height attained by the stone.

\[\Rightarrow 0-{{\left( 40 \right)}^{2}}=2\times h\times \left( -10 \right)\]

\[\Rightarrow h=\frac{40\times 40}{20}=80m\]

Therefore, the total distance covered by the stone during its upward and downward journey is \[80+80=160m\].

The net displacement of the stone during its upward and downward

journey is \[80+\left( -80 \right)=0m\].

16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth \[=6\times {{10}^{24}}kg\] and of the Sun \[=2\times {{10}^{30}}kg\]. The average distance between the two is \[1.5\times {{10}^{11}}m\].

Ans: According to the Universal l law of gravitation, the force of attraction between the Earth and the Sun is given by:

\[{{M}_{Sun}}=\]Mass of the Sun \[=2\times {{10}^{30}}kg\]

\[{{M}_{Earth}}=\]Mass of the Earth \[=6\times {{10}^{24}}kg\]

\[R=\] Average distance between the Earth and the Sun \[=1.5\times {{10}^{11}}m\]

\[G=\]Universal gravitational constant \[=6.7\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}\].

\[F=\frac{G{{M}_{Sun}}{{M}_{Earth}}}{{{R}^{2}}}\]

\[\Rightarrow F=\frac{6.7\times {{10}^{-11}}\times 2\times {{10}^{30}}\times 6\times {{10}^{24}}}{{{\left( 1.5\times {{10}^{11}} \right)}^{2}}}\]

\[\Rightarrow F=3.57\times {{10}^{22}}N\]

Hence, the force of gravitation between the earth and the sun is \[3.57\times {{10}^{22}}N\].

17. A stone is allowed to fall from the top of a tower \[100m\] high and at the same time another stone is projected vertically upwards from the ground with a velocity of \[25m{{s}^{-1}}\]. Calculate when and where the two stones will meet.

Ans: Let the two stones meet after time \[t\]from the start.

a) For the stone dropped from the tower:

Initial velocity, \[u=0\].

Let the displacement of the stone in time t from the top of the tower be s.

From the equation of motion,

\[s=ut+\frac{1}{2}g{{t}^{2}}\]

\[\Rightarrow s=0\times t+\frac{1}{2}\times 9.8\times {{t}^{2}}\]

\[\Rightarrow s=4.9{{t}^{2}}\]……. (1)

b) For the stone thrown upwards:

Initial velocity, \[u=25m{{s}^{-1}}\]

Let the displacement of the stone from the ground in time \[t\]be \[{s}'\].

Equation of motion,

\[{s}'=ut+\frac{1}{2}g{{t}^{2}}\]

\[\Rightarrow {s}'=25t-\frac{1}{2}\times 9.8\times {{t}^{2}}\]

\[\Rightarrow {s}'=25t-4.9{{t}^{2}}\]…… (2)

The combined displacement of both the stones at the meeting point is equal to the height of the tower \[100\]m.

\[s+{s}'=100\]…… (3)

Substituting equation (1) and (2) in (3),

\[4.9{{t}^{2}}+25t-4.9{{t}^{2}}=100\]

\[\Rightarrow 25t=100\]

\[\Rightarrow t=\frac{100}{25}=4s\]

In \[4s\], the falling stone has covered a distance given by equation (1) as

\[s=\frac{1}{2}\times 9.8\times {{4}^{2}}=78.4m\]

Therefore, the stones will meet after \[4s\] at a height \[\left( 100-78.4 \right)=21.6m\] from the ground.

18. A ball thrown up vertically returns to the thrower after \[6s\]. Find

a) The velocity with which it was thrown up,

Ans: Time of ascent is equal to the time of descent. The ball takes a total of \[6s\]for its upward and downward journey.

Hence, time taken for upward journey, \[t=\frac{6}{2}=3s\]

Final velocity of the ball at the maximum height, \[v=0\]

Equation of motion, \[v=u+gt\]will give,

\[\Rightarrow 0=u+\left( -9.8\times 3 \right)\]

\[\Rightarrow u=9.8\times 3=29.4m{{s}^{-1}}\]

Hence, the ball was thrown upwards with a velocity of \[29.4m{{s}^{-1}}\].

b) The maximum height it reaches 

Ans: Let the maximum height attained by the ball be \[h\].

Initial velocity during the upward journey, \[u=29.4m{{s}^{-1}}\] 

Final velocity, \[v=0\]

Acceleration due to gravity, \[g=-9.8m{{s}^{-2}}\]

\[s=ut+\frac{1}{2}a{{t}^{2}}\]

\[h=29.4\times 3+\frac{1}{2}\times \left( -9.8 \right)\times {{\left( 3 \right)}^{2}}=44.1m\] 

c) Its position after \[4s\].

Ans: Ball attains the maximum height after \[3s\]. After attaining this height, it will start falling downwards.

In this case, Initial velocity, \[u=0\]

Position of the ball after \[4s\] of the throw is given by the distance

travelled by it during its downward journey in \[4s-3s=1s\]

\[s=0\times t+\frac{1}{2}\times 9.8\times {{1}^{2}}=4.9m\]

Total height \[=44.1m\]

This means that the ball is \[44.1m-4.9m=39.2m\] above the ground

after \[4\]seconds.

19. In what direction does the buoyant force on an object immersed in a liquid act?

Ans: An object immersed in a liquid is acted upon by the buoyant force in the vertically upward direction.

20. Why does a block of plastic released under water come up to the surface of water?

Ans: The number of forces acting on a certain item in water are two. The first one is the gravitational force pulling down the object, and the other is the buoyant force pushing up the object. If the buoyant force acting in the upward direction is higher than the gravitational force that is acting downward, then the object goes up to the water's surface as quickly as it is released into water. That is why a block of plastic released under the water comes up to the surface of the water.

21. The volume of \[50g\]of a substance is \[20c{{m}^{3}}\] . If the density of water is \[1gc{{m}^{-3}}\], will the substance float or sink?

Ans: If the density of an object is more than the density of a liquid, then it sinks in the liquid. If the density of an object is less than the density of a liquid, then it floats

\[Density\text{ }of\text{ }the\text{ }substance=\frac{Mass\text{ }of\text{ }the\text{ }substance}{Volume\text{ }of\text{ }the\text{ }substance}\]

\[\Rightarrow Density\text{ }of\text{ }the\text{ }substance=\frac{50}{20}\]

\[\Rightarrow Density\text{ }of\text{ }the\text{ }substance=2.5gc{{m}^{-3}}\].

The density of the substance \[>\] The density of water \[\left( 1gc{{m}^{-3}} \right)\].

Hence, the substance will sink in water.     

22. The volume of a \[500g\] sealed packet is \[350c{{m}^{3}}\]. Will the packet float or sink in water if the density of water is \[1gc{{m}^{-3}}\]? What will be the volume of the water displaced by this packet?

Ans: If the density of an object is greater than the density of a liquid, then the object will sink in the liquid. If the density of an object is less than the density of a liquid, then it will float on the surface of the liquid.

\[Density\text{ }of\text{ }the\text{ }500\text{ }g\text{ }sealed\text{ }packet=\frac{Mass\text{ }of\text{ }the\text{ }packet}{Volume\text{ }of\text{ }the\text{ }packet}\]

\[\Rightarrow Density\text{ }of\text{ }the\text{ }500\text{ }g\text{ }sealed\text{ }packet=\frac{500}{350}\]

\[\Rightarrow Density\text{ }of\text{ }the\text{ }500\text{ }g\text{ }sealed\text{ }packet=1.428gc{{m}^{-3}}\]

The density of the substance is more than the density of water \[\left( 1gc{{m}^{-3}} \right)\].

Hence, the object will sink in water.

Clearly, the mass of water displaced by the packet can be considered equal to the volume of the packet\[=0.350g\].

Class 9th Science Gravitation Class 9 - Quick Overview of Detailed Structure of Topics

Class 9 Science Ch 9 Gravitation - Important Formula and Concepts

The universal law of gravitation: The force of attraction between any two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them. 

Archimedes’ Principle: When a body is immersed fully or partially in a fluid, it experiences an upward force that is equal to the weight of the fluid displaced by it. 

Where \[G\]is the universal gravitation constant, which is given by:

Weight of the object on the moon = (1/6) × its weight on the earth.

Pressure= Thrust/Area

Benefits of Vedantu’s NCERT Class 9 Science Gravitation Question Answer

Vedantu’s solutions explain all key concepts covered in Ch 9 Science Class 9, including the Universal Law of Gravitation, acceleration due to gravity, the distinction between mass and weight, free fall, and gravitational fields.

Class 9 Science Gravitation Question Answers are presented in a step-by-step format, making it easier for students to follow and understand the problem related to the motion of objects under the influence of the gravitational force of the earth, pressure, and Thrust. 

Vedantu’s NCERT Gravitation Class 9 Questions And Answers help students prepare effectively for their exams. 

The solutions include various types of questions, from multiple-choice to descriptive, ensuring comprehensive exam readiness.

Class 9 Gravitation Question Answers are prepared by Vedantu Master Teachers with a deep understanding of the curriculum and examination patterns. This ensures that the content is accurate, reliable, and aligned with the latest syllabus.

Using Class 9 Gravitation NCERT Solutions, students can save time by quickly finding answers and explanations for their doubts and questions. This allows them to allocate more time to practice and revision.

Vedantu’s Class 9 Chapter 9 Science solutions are available online, making them accessible anytime and anywhere. This flexibility supports continuous learning and allows students to study independently.

Related Study Materials for Class 9 Science Ch 9 Gravitation

Vedantu’s NCERT Class 9 Gravitation NCERT Solutions is an important study material. They provide clear explanations and step-by-step solutions, helping students grasp important concepts like the laws of gravitation, mass, weight, and gravitational force. Focusing on these areas is essential, as they form the chapter's foundation. In previous years, around 7-8 questions on gravitation have been asked in exams, highlighting its importance. By using Gravitation Class 9 Questions And Answers, students can effectively prepare, clear their doubts, and perform well in their exams, ensuring a strong understanding of gravitation.

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FAQs on NCERT Solutions for Class 9 Science Chapter 9- Gravitation

1. How can students understand the features of Gravitational Force Properly?

Chapter 9 Science Class 9 is an important part of the Science syllabus. Focus on the classroom sessions and concentrate on what the teachers are explaining. Study the chapter unit-wise and clear your doubts by using the Science Class 9 NCERT Solutions provided by Vedantu. You will surely understand these newfound concepts well.

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You must practise regularly using the NCERT Solutions Class 9 Science Chapter 9 as a reference and become more efficient. Your speed will automatically increase as you can remember the formulas properly.

3. Why do students prefer using NCERT Solutions for Class 9 Science Chapter 9?

By using the NCERT Solutions for Class 9 Science Gravitation, a student can save time in finding the right answers. They can focus better in preparing the chapter and score higher in the exams by following the ideal answering format recommended by the experts.

4. Why does the Earth not move towards objects due to Gravitation according to Chapter 9 Gravitation of Class 9 Science? 

Newton's third law states, “Every action has its equal and opposite reaction”. It means the force applied by an object on the Earth is equal to the force applied by the earth on the object, but we know that acceleration is inversely proportional to mass. This means when the acceleration is increased, the mass is decreased, or when the mass is increased, the acceleration is decreased. As the mass of the earth is large, the acceleration due to an object is small or negligible. Therefore, it's not noticeable. And the Earth doesn’t seem to be moving.

5. What are the different applications of Archimedes' principle?

The different applications of Archimedes' principle include the following:

It is used in designing ships and submarines.

Lactometers used to determine the purity of a milk sample and hydrometers used to determine the density of a liquid are based on this principle. 

6. Why does the Earth doesn’t move towards objects due to Gravitation according to Chapter 9 Gravitation of Class 9 Science? 

As per Newton's third law, “Every action has its equal and opposite reaction”. It means the force applied by an object on the Earth is equal to the force applied by the earth on the object, but we know that acceleration is inversely proportional to mass. This means when the acceleration is increased, the mass is decreased or when the mass is increased, the acceleration is decreased. As the mass of the earth is large, the acceleration due to an object is small or negligible. Therefore, it's not noticeable. And the Earth doesn’t seem to be moving.

7. What are topics covered in Vedantu’s Class 9 Gravitation NCERT Solutions?

Vedantu’s NCERT Class 9 Science Gravitation Question Answer covers all key concepts such as Newton’s Law of Gravitation, the universal law of gravitation, the relationship between gravitational force, mass, and distance, free fall, mass, weight, and the concept of acceleration due to gravity.

8. Why should I refer to Vedantu’s NCERT Solutions for Gravitation Class 9?

Vedantu’s NCERT Solutions for Gravitation Class 9 offers detailed explanations and step-by-step answers to textbook problems, making it easier to understand complex concepts. They also provide diagrams and illustrative examples that aid in better comprehension.

NCERT Solutions for Class 9 Science

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NCERT Solutions for Class 9 Science Chapter 10 - Gravitation

Ncert solutions class 9 science chapter 10 – cbse free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.

NCERT Solutions for Class 9 Science Chapter 10 Gravitation provides you with the necessary insights into the concepts involved in the chapter. Detailed answers and explanations provided by us in NCERT Solutions will help you in understanding the concepts clearly.

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Gravity is a fascinating topic that explains many things, from how our planet stays in orbit to why things fall down. Explore Science Chapter 10 – Gravitation of NCERT Solutions for Class 9  to learn everything you need to know about gravity. Content is crafted by highly qualified teachers and industry professionals with decades of relevant knowledge. Moreover, the solutions have been updated to include the latest content prescribed by the CBSE board.

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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 9 Force And Laws Of Motion
• 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

NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

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Access Answers to NCERT Class 9 Science Chapter 10 – Gravitation ( All In text and Exercise Questions Solved)

Exercise-10.1 page: 134.

1. State the universal law of gravitation.

The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.

Consider F as the force of attraction between an object on the surface of earth and the earth

Also, consider ‘m’ as the mass of the object on the surface of earth and ‘M’ as the mass of earth

The distance between the earth’s centre and object = Radius of the earth = R

Therefore, the formula for the magnitude of the gravitational force between the earth and an object on the surface is given as

F = G Mm/R 2

Exercise-10.2 Page: 136

1. What do you mean by free fall?

Earth’s gravity attracts each object to its center. When an object is dropped from a certain height, under the influence of gravitational force it begins to fall to the surface of Earth. Such an object movement is called free fall.

2. What do you mean by acceleration due to gravity?

When an object falls freely from a certain height towards the earth’s surface, its velocity keeps changing. This velocity change produces acceleration in the object known as acceleration due to gravity and denoted by ‘g’.

The value of the acceleration due to gravity on Earth is,

Exercise-10.3 Page: 138

1. What are the differences between the mass of an object and its weight?

The differences between the mass of an object and its weight are tabulated below.

2. Why is the weight of an object on the moon 1/6th its weight on the earth?

The mass of the moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitational attraction on the moon is about one-sixth when compared to earth. The moon’s gravitation force is determined by the mass and the size of the moon. Hence, the weight of an object on the moon is 1/6th its weight on the earth. The moon is far less massive than the Earth and has a different radius(R) as well.

Exercise-10.4 Page: 141

1. Why is it difficult to hold a school bag having a strap made of a thin and strong string?

It is tough to carry a school bag having a skinny strap because of the pressure that is being applied on the shoulders. The pressure is reciprocally proportional to the expanse on which the force acts. So, the smaller the surface area, the larger is going to be the pressure on the surface. In the case of a skinny strap, the contact expanse is quite small. Hence, the pressure exerted on the shoulder is extremely huge.

2. What do you mean by buoyancy?

The upward force possessed by a liquid on an object that’s immersed in it is referred to as buoyancy.

3. Why does an object float or sink when placed on the surface of water?

An object floats or sinks when placed on the surface of water because of two reasons.

(i) If its density is greater than that of water, an object sinks in water.

(ii) If its density is less than that of water, an object floats in water.

Exercise-10.5 Page: 142

1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?

A weighing machine measures the body weight and is calibrated to indicate the mass. If we stand on a weighing machine, the weight acts downwards while the upthrust due to air acts upwards. So our apparent weight becomes less than the true weight. This apparent weight is measured by the weighing machine and therefore the mass indicated is less than the actual mass. So our actual mass will be more than 42 kg.

2. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier and why?

The correct answer is the cotton bag is heavier than an iron bar.  The bag of cotton is heavier than the bar of iron. The cotton bag experiences a larger air thrust than the iron bar. Therefore, the weighing machine indicates less weight than its actual weight for the cotton bag. The reason is

True weight = (apparent weight + up thrust)

The cotton bag’s density is less than that of the iron bar, so the volume of the cotton bag is more compared to the iron bar. So the cotton bag experience more upthrust due to the presence of air.

Therefore, in the presence of air, the cotton bag’s true weight is more compared to the true weight of the iron bar.

Exercises-10.6 Page: 143

1. How does the force of gravitation between two objects change when the distance between them is reduced to half?

Consider the Universal law of gravitation,

According to that law, the force of attraction between two bodies is

m 1 and m 2 are the masses of the two bodies.

G is the gravitational constant.

r is the distance between the two bodies.

Given that the distance is reduced to half then,

Therefore once the space between the objects is reduced to half, then the force of gravitation will increase by fourfold the first force.

2. Gravitational force acts on all objects in proportion to their masses. Why then does a heavy object not fall faster than a light object?

All objects fall from the top with a constant acceleration called acceleration due to gravity (g). This is constant on earth and therefore the value of ‘g’ doesn’t depend on the mass of an object. Hence, heavier objects don’t fall quicker than light-weight objects provided there’s no air resistance.

3. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10 24 kg and radius of the earth is 6.4 × 10 6 m.)

From Newton’s law of gravitation, we know that the force of attraction between the bodies is given by

4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?

The earth attracts the moon with a force same as the force with which the moon attracts the earth. However, these forces are in opposite directions. By universal law of gravitation, the force between moon and also the sun can be

d = distance between the earth and moon.

m 1 and m 2 = masses of earth and moon respectively.

5. If the moon attracts the earth, why does the earth not move towards the moon?

According to the universal law of gravitation and Newton’s third law, we all know that the force of attraction between two objects is the same, however in the opposite directions. So the earth attracts the moon with a force same as the moon attracts the earth but in opposite directions. Since earth is larger in mass compared to that of the moon, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. Therefore, for this reason the earth does not move towards the moon.

6. What happens to the force between two objects, if

(i) The mass of one object is doubled?

(ii) The distance between the objects is doubled and tripled?

(iii) The masses of both objects are doubled?

According to universal law of gravitation, the force between 2 objects (m 1 and m 2 ) is proportional to their plenty and reciprocally proportional to the sq. of the distance(R) between them.

If the mass is doubled for one object.

F = 2F, so the force is also doubled.

If the distance between the objects is doubled and tripled

If it’s doubled

F = (Gm 1 m 2 )/(2R) 2

F = 1/4 (Gm 1 m 2 )/R 2

Force thus becomes one-fourth of its initial force.

Now, if it’s tripled

F = (Gm 1 m 2 )/(3R) 2

F = 1/9 (Gm 1 m 2 )/R 2

Force thus becomes one-ninth of its initial force.

If masses of both the objects are doubled, then

F = 4F, Force will therefore be four times greater than its actual value.

7. What is the importance of universal law of gravitation?

The universal law of gravitation explains many phenomena that were believed to be unconnected:

(i) The motion of the moon round the earth

(ii) The responsibility of gravity on the weight of the body which keeps us on the ground

(iii) The tides because of the moon and therefore the Sun

(iv) The motion of planets round the Sun

8. What is the acceleration of free fall?

Acceleration due to gravity is the acceleration gained by an object due to gravitational force. On Earth, all bodies experience a downward force of gravity which Earth’s mass exerts on them. The Earth’s gravity is measured by the acceleration of the freely falling objects. At Earth’s surface, the acceleration of gravity is 9.8 ms -2 and it is denoted by ‘g’. Thus, for every second an object is in free fall, its speed increases by about 9.8 metres per second.

9. What do we call the gravitational force between the earth and an object?

The gravitation force between the earth and an object is called weight. Weight is equal to the product of acceleration due to the gravity and mass of the object.

10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]

The weight of a body on the earth’s surface;

W = mg (where m = mass of the body and g = acceleration due to gravity)

The value of g is larger at poles when compared to the equator. So gold can weigh less at the equator as compared to the poles.

Therefore, Amit’s friend won’t believe the load of the gold bought.

11. Why will a sheet of paper fall slower than one that is crumpled into a ball?

A sheet of paper has a larger surface area when compared to a crumpled paper ball. A sheet of paper will face a lot of air resistance. Thus, a sheet of paper falls slower than the crumpled ball.

12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newton’s of a 10 kg object on the moon and on the earth?

Given data:

Acceleration due to earth’s gravity = g e or g = 9.8 m/s 2

Object’s mass, m = 10 kg

Acceleration due to moon gravity = g m

Weight on the earth= W e

Weight on the moon = W m

Weight = mass x gravity

g m = (1/6) g e (given)

So W m = m g m = m x (1/6) g e

W m = 10 x (1/6) x 9.8 = 16.34 N

W e = m x g e = 10 x 9.8

13. A ball is thrown vertically upwards with a velocity of 49 m/s.

(i) The maximum height to which it rises,

(ii) The total time it takes to return to the surface of the earth.

Initial velocity u = 49 m/s

Final speed v at maximum height = 0

Acceleration due to earth gravity g = -9.8 m/s 2 (thus negative as ball is thrown up).

By third equation of motion,

2gH = v 2  – u 2

2 × (- 9.8) × H = 0 – (49) 2

– 19.6 H = – 2401

H = 122.5 m

Total time T = Time to ascend (T a ) + Time to descend (T d )

0 = 49 + (-9.8) x T a

Ta = (49/9.8) = 5 s

Also, T d = 5 s

Therefore T = T a + T d

14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Initial velocity

Tower height = total distance = 19.6m

g = 9.8 m/s 2

Consider third equation of motion

v 2 = u 2 + 2gs

v 2 = 0 + 2 × 9.8 × 19.6

v 2 = 384.16

v = √(384.16)

v = 19.6m/s

15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s 2 , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Initial velocity u = 40m/s

g = 10 m/s 2

Max height final velocity = 0

0 = (40) 2 – 2 x 10 x s

s = (40 x 40) / 20

Maximum height s = 80m

Total Distance = s + s = 80 + 80

Total Distance = 160m

Total displacement = 0 (The first point is the same as the last point)

16. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10 24 kg and of the Sun = 2 × 10 30 kg. The average distance between the two is 1.5 × 10 11 m.

Mass of the sun m s = 2 × 10 30 kg

Mass of the earth m e = 6 × 10 24 kg

Gravitation constant G = 6.67 x 10 -11 N m 2 / kg 2

Average distance r = 1.5 × 10 11 m

Consider Universal law of Gravitation

17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.

(i) When the stone from the top of the tower is thrown,

Initial velocity u’ = 0

Distance travelled = x

Time taken = t

(ii) When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 – x)

From equations (a) and (b)

5t 2 = 100 -25t + 5t 2

t = (100/25) = 4sec.

After 4sec, two stones will meet

x = 5t 2 = 5 x 4 x 4 = 80m.

Putting the value of x in (100-x)

= (100-80) = 20m.

This means that after 4sec, 2 stones meet a distance of 20 m from the ground.

18. A ball thrown up vertically returns to the thrower after 6 s. Find

(a) The velocity with which it was thrown up,

(b) The maximum height it reaches, and

(c) Its position after 4s.

g = 10m/s 2

Total time T = 6sec

T a = T d = 3sec

(a) Final velocity at maximum height v = 0

From first equation of motion:-

v = u – gt a

u = v + gt a

= 0 + 10 x 3

The velocity with which stone was thrown up is 30m/s.

(b) From second equation of motion

The maximum height stone reaches is 45m.

(c) In 3sec, it reaches the maximum height.

Distance travelled in another 1sec = s’

The distance travelled in another 1sec = 5m.

Therefore in 4sec, the position of point p (45 – 5)

= 40m from the ground.

19. In what direction does the buoyant force on an object immersed in a liquid act?

The buoyant force on an object that is immersed in a liquid will be in a vertically upward direction.

20. Why a block of plastic when released under water come up to the surface of water?

The density of plastic is lesser than that of water. Therefore, the force of buoyancy on plastic block will be greater than the weight of plastic block. Hence, the acceleration of plastic block is going to be in the upward direction. So, the plastic block comes up to the surface of water.

21. The volume of 50 g of a substance is 20 cm 3 . If the density of water is 1 g cm –3 , will the substance float or sink?

To find the Density of the substance the formula is

Density = (Mass/Volume)

Density = (50/20) = 2.5g/cm 3

Density of water = 1g/cm 3

Density of the substance is greater than density of water. So the substance will sink.

22. The volume of a 500 g sealed packet is 350 cm 3 . Will the packet float or sink in water if the density of water is 1 g cm –3 ? What will be the mass of the water displaced by this packet?

Density of sealed packet = 500/350 = 1.42 g/cm 3

Density of sealed packet is greater than density of water

Therefore the packet will sink.

Considering Archimedes Principle,

Displaced water volume = Force exerted on the sealed packet.

Volume of water displaced = 350cm 3

Therefore displaced water mass = ρ x V

Mass of displaced water = 350g.

NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

Chapter 10 – Gravitation is a part of Unit 3 – Motion, Force and Work, which carries a total of 27 out of 100. Usually, 2 or 3 questions do appear from this chapter every year, as previous trends have shown.

The topics usually covered under this chapter are:

• Universal Law of Gravitation and its Importance
• Characteristics of Gravitational Forces
• Concept of Free Fall
• Difference between Gravitation Constant and Gravitational Acceleration

Often times, the term gravity and gravitation are used interchangeably and this is wrong. However, these two terms are related to each other but their implications are quite different. Academically, Chapter 10 – Gravitation is an important concept as it carries a considerable weightage in the CBSE exam. Therefore, ensure that all relevant concepts, formulas and diagrams are studied thoroughly.

Explore how gravitation works at the molecular level, discover its applications and learn other related important concepts by exploring our NCERT Solutions.

Key Features of NCERT Solutions for Class 9 Science Chapter 10 – Gravitation

• Solutions provided in an easy-to-understand language
• Qualified teachers and their vast experience helps formulate the solutions
• Questions updated to the latest prescribed syllabus
• A detailed breakdown of the most important exam questions
• Access to additional learning resources like sample papers and previous year question papers

Disclaimer:

Dropped Topics –  Following Box Items: a. Brief Description of Isaac Newton, b. How did Newton guess the inverse– square rule? 10.7 Relative Density and Example 10.7.

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RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

July 4, 2018 by Prasanna

These Solutions are part of NCERT Solutions for Class 9 Science . Here we have given NCERT Solutions for Class 9 Science Chapter 10 Gravitation. LearnInsta.com provides you the Free PDF download of NCERT Solutions for Class 9 Science (Physics) Chapter 10 – Gravitation solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 10 – Gravitation Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

More Resources

• NCERT Solutions for Class 9 Science
• HOTS Questions for Class 9 Science
• Value Based Questions in Science for Class 9
• NCERT Exemplar Solutions for Class 9 Science
• Previous Year Question Papers for CBSE Class 9 Science

NCERT TEXT BOOK QUESTIONS

IN TEXT QUESTIONS

Question 1. State the universal law of gravitation. (CBSE Sample Paper, CBSE 2010, 2011, 2012, 2013, 2015) Answer: The force of attraction between two particles or objects is

• directly proportional to the product of the masses of the objects and
• inversely proportional to the square of the distance between them.

Question 3. What do you mean by free fall ?

What is meant by free fall ? (CBSE 2010, 2011, 2012, 2013) Answer: When an object falls towards the earth under the influence of gravitational force alone, then the object is in free fall.

Question 4. What do you mean by acceleration due to gravity ? (CBSE 2011, 2012, 2013) Answer: The acceleration with which an object falls freely towards the earth is known as acceleration due to gravity. It is denoted by ‘g’.

Question 5. What is the difference between the mass of an object and its weight ? (CBSE 2010, 2011, 2012, 2013) Answer:

NCERT CHAPTER END EXERCISE

Question 2. Gravitational force acts on all objects in proportion to their masses. Why, then, a heavy object does not fall faster than a light object ? (CBSE 2015) Answer: The acceleration with which a body falls towards the earth is constant (= 9.8 m s -2 ) and independent of the mass of the body. Thus, all bodies fall with the same acceleration irrespective of their masses. That is why, a heavy body does not fall faster than the light body.

Question 4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth ? Why ? Answer: Gravitational force with which a body A attracts another body B is equal in magnitude and opposite in direction to the gravitational force with which a body B attracts the body A. Thus, the magnitude of force with which the earth attracts the moon is equal to the magnitude of the force with which the moon attracts the earth. Thus, both the earth and the moon attract each other with equal forces.

Question 5. If the moon attracts the earth, why does the earth not move towards the moon ? (CBSE 2011, 2013, 2015) Answer: . The acceleration produced in the earth due to the force exerted on it by the moon is very small as the mass of the earth is very large. Hence, the movement of the earth towards the moon is not noticed.

Question 6. What happens to the force between two objects, if

• the mass of one object is doubled ?
• the distance between the objects is doubled and tripled ?
• the masses of both the objects are doubled ? (CBSE 2011, 2013, 2014)

• F is doubled if m 1 or m 2 is doubled.
• F becomes 1/4 times the original value if distance (r) is doubled and F becomes 1/9 times the original value if r is tripled.
• F becomes four times the original value if both m 1 and m 2 are doubled.

Question 7. What is the importance of universal law of gravitation ? Answer: The gravitational force plays an important role in nature

• All the planets revolve around the sun due to the gravitational force between the sun and the planets. The force required by a planet to move around the sun in elliptical path (known as centripetal force) is provided by the gravitational force of attraction between the planet and the sun. Thus, gravitational force is responsible for the existence of the solar system.
• Tides in oceans are formed due to the gravitational force between the moon and the water in oceans.
• Gravitational force between a planet and its satellite (i.e., moon) decides whether a planet has a moon or not. Since the gravitational force of the planets like mercury and venus is very small, therefore, these planets do not have any satellite or moon.
• Artificial and natural satellites revolve around the earth due to the gravitational force between the earth and the satellite. The gravitational force between the earth and the satellite provides a necessary centripetal force to the satellite to move in a circular path around the earth.
• The atmosphere (envelope of gases) of the earth is possible due to gravitational force of the earth.
• Rainfall and snowfall is possible only due to gravitational force of the earth.
• We stay on the earth due to the gravitational force between the earth and us.

Question 8. What is the acceleration of free fall ? Answer: Acceleration of free fall = 9.8 m s -2 ≈ 10 m s -2 .

Question 9. What do we call the gravitational force between the earth and an object ? Answer: Force of gravity.

Question 10. . Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought ? If not, why ? (Hint: The value of g is greater at the poles than at the equator) (CBSE 2012, 2013) Answer: Weight = mg Since value of g is greater at the poles than at the equator, so the weight of gold at the poles will be greater than the weight of gold at the equator. Hence, his friend will say that the weight of the gold is less than as told by Amit.

Question 11. Why will a sheet of paper fall slower than one that is crumpled into a ball ? Answer: Since the area of a sheet of paper is more than the area of the paper crumpled into a small, therefore, a sheet of paper will experience a large opposing force due to air than the ball, while falling down. Hence, a sheet of paper falls slower than one that is crumpled into a ball.

Question 12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth.

Question 17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet ? Answer: Let t = time after which both stones meet . S = distance of the stone dropped from the top of tower (100 – S) = distance travelled by the projected stone.

Get 100 percent accurate NCERT Solutions for Class 9 Science Chapter 10 (Gravitation) explained by expert Science teachers. We provide solutions for the questions given in Class 9 Science textbook as per CBSE Board guidelines from the latest NCERT book for Class 9 Science. The topics and sub-topics in  Chapter 10 Gravitation

• 10.1 Gravitation
• 10.1.1 UNIVERSAL LAW OF GRAVITATION
• 10.1.2 IMPORTANCE OF THE UNIVERSAL LAW OF GRAVITATION
• 10.2 Free Fall
• 10.2.1 TO CALCULATE THE VALUE OF g
• 10.2.2 MOTION OF OBJECTS UNDER THE INFLUENCE OF GRAVITATIONAL FORCE OF THE EARTH
• 10.4 Weight
• 10.4.1 WEIGHT OF AN OBJECT ON THE MOON
• 10.5 Thrust and Pressure
• 10.5.1 PRESSURE IN FLUIDS
• 10.5.2 BUOYANCY
• 10.5.3 WHY OBJECTS FLOAT OR SINK WHEN PLACED ON THE SURFACE OF WATER?
• 10.6 Archimedes’ Principle
• 10.7 Relative Density.

We cover all exercises in the chapter given below:-

• EXERCISE 10.1 – 2 Questions with Solutions
• EXERCISE 10.2 – 2 Questions with Solutions
• EXERCISE 10.3 – 2 Questions with Solutions
• EXERCISE 10.4 – 3 Questions with Solutions
• EXERCISE 10.5 – 2 Questions with Solutions
• EXERCISE 10.6 – 22 Questions with Solutions.

Download the free PDF of Chapter 10 Gravitation or save the solution images and take the print out to keep it handy for your exam preparation.

Hope given NCERT Solutions for Class 9 Science Chapter 10 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.