case study class 9 physics gravitation

Case Study Questions Class 9 Science Gravitation

Case study questions class 9 science chapter 10 gravitation, cbse case study questions class 9 science – gravitation.

(3) Archimedes’ principle, stated as follows: When a body is immersed fully or partiallyin a fluid, it experiences an upward force thatis equal to the weight of the fluid displacedby it. The upward force is known as up thrust or buoyant force. In fact, all objects experience a force of buoyancy when they are immersed in a fluid. The magnitude of this buoyant force depends on the density of the fluid.Objects having density less than that of the liquid in which they are immersed float on the surface of the liquid. If the density of the object is more than the density of the liquid in which it is immersed then it sinks in the liquid. Hence body will float or sink depends upon difference between density of body and fluid.

(5) The force acting on an object perpendicular to the surface is called thrust. When you stand on loose sand, the force, that is, the weight of your body is acting on an area equal to area of your feet. When you lie down then the same force acts on an area equalto the contact area of your whole body, which is larger than the area of your feet. Thus, the effects of forces of the same magnitude on different areas are different. In the above cases, thrust is the same. But effects are different. Therefore the effect of thrust depends on the area on which it acts. The effect of thrust on sand is larger while standing than while lying. The thrust on unit area is called pressure.

(c) Both a and b

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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

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Case Study Questions of Chapter 10 Gravitation PDF Download

Case study Questions on Class 9 Science Chapter 10  are very important to solve for your exam. Class 9 Science Chapter 10 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

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

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 “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

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

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

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

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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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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

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

Chapter 10 Gravitation Class 9 Science NCERT Solutions

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Important Questions for CBSE Class 9 Science Chapter 11 - Work and Energy

• Class 9 Important Question
• Chapter 11: Work And Energy

CBSE Class 9 Science Chapter-11 Important Questions - Free PDF Download

Work and energy comprise one of the most fundamental units of physics. This chapter also sets the tone for the +2 syllabus. So, in this section, we will be dealing with the important questions of chapter Work and Energy Class 9 . We further aim to acknowledge children with all the binding terms, keeping in mind about the previous year questions.

Vedantu can help students to get hands-on the most essential topics from exam perspective. It also offers a widespread platform to know one’s ability through online interrogation from experts. You can also download free PDF for class 9 Science chapter 11 important questions. 

Register Online for Class 9 Science tuition on Vedantu.com to score more marks in your examination. Also download Class 9 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Download CBSE Class 9 Science Important Questions 2024-25 PDF

Also, check CBSE Class 9 Science Important Questions for other chapters:

Study Important Questions for Class 9 Science Chapter 11 – Work and Energy

1 Marks Questions

1. What is the power of a lamp which consumes 1000 J of electrical energy in 10 s?

Ans: Power = Work/Time = 1000Joules/10seconds = 100 Watts

2. Can an object displace without applying force?

3. What is the SI unit of power?

(a) $\mathrm{J} / \mathrm{s}$

(c) $\mathrm{s} / \mathrm{J}$

(d) $\mathrm{J} / \mathrm{s}^{2}$

Ans: (a) $\mathrm{J} / \mathrm{s}$

4. What will be the change in kinetic energy of a body if it starts from the rest?

Positive  

Negative  

It can be positive or negative, depending on the body mass.

Ans: (a) Positive.

5. Which of the following sets of quantities have similar units? 

(a) Power and Energy 

(b) Work and Power

(c) Energy and Work

(d) None of the above 

Ans: (c) Energy and work

6. A body is present at a height ‘h’. Which type of energy will it possess?

(a) Kinetic energy 

(b) Potential energy 

(c) Both kinetic and potential energy

(d) None 

Ans: (b) Potential energy

7. What will be the work done, if a body moves in the opposite direction of the direction in which force is applied?

(a) Positive 

(b) Negative 

(c) Zero 

(d) Cannot predicted

Ans: (b) Negative

8. What will be the work done, if the force is applied at an angle $\theta ?$

(a) $\quad W=F S \operatorname{Cos} \theta$ where $F=$ Force

(b) $\quad W=F S \theta$ where $S=$ Distance

(c) $W=F S \operatorname{Sin} \theta$ where $W=$ work

(d) None of the above

Ans: (a) $W=F S \cos \theta$ where $F=$ Force

9. A body having mass of 5 kg is lifted vertically upto the distance of 9 meters. What will be the work done?

(a) 450J 

(b) -450J 

(c) 45J 

(d) 540J 

Ans: (a) 450J

10. What is the relation between joules (J) and ergs (erg)?

(a) $\quad 1 J=10^{7} \mathrm{erg}$

(b) 1 erg $=10^{7} J$

(c) $1 J=10^{-7} e r g$

Ans: (a) $1 J=10^{7}$ erg

2 Marks Questions

1. When the work is said to be done?

Ans: When a force acts on an object and moves it in the same direction that of force then work is said to be done. 

2. What will be the expression for the work done when a force acts on an object in the direction of its motion.

Ans: Work done = Force × Displacement

If W is the work done, F is the force applied on object and d is the displacement, then the expression of work done will be

3. Explain 1 joule of work done.

Ans: When a force of 1 N (Newton) is applied on an object and that object displaces upto a distance of 1 m (meter) in the same direction of its displacement, then 1 joule (J) of work is done on the object.

4. How much work is done in ploughing a 15 m long field when a pair of bullocks applies a force of 140 N on the plough?

Ans: Since Work done (W) = Force (F) × Displacement (d)

Hence, Work done in ploughing (W) = 140 N × 15 m =  2100 J

5. The force acting on the object is 7 N, and the displacement of the object occurs in the direction of the force is 8 m. Suppose that force acts on the object through displacement, then how much work was done in this case?

Ans:   As we know, Work done (W) = Force (F) × Displacement (d)

Thus, Work done in the given case (W) = 7 N × 8 m =  56 J

6. Define kinetic energy of an object.

Ans: The kinetic energy of an object is a kind of mechanical energy that exists in the object due to its state of motion (movement).

7. Write down the kinetic energy expression of an object.

Ans: If m is the mass of an moving object and v is its velocity, then the expression of its kinetic energy (KE) will be

$K.E=\frac{1}{2}mv^2$

8. Define power.

Ans: The rate by which work is done refers to power. It is expressed by P.

Power = Work done/Time

9. What is 1 watt of power?

Ans: When an object is doing work at the rate of 1 J/s, then the power of that body or object is 1 watt (where watt is the unit of power).

10. An object is thrown at an angle to the ground, moves along a curve and falls back to the ground. The start and end points of the object path are on the same horizontal line. How much work is done by the gravity on that object?  

Ans: There must be a displacement to calculate the work, but since the vertical displacement in this case is zero (because the start and end points are on the same horizontal line), the work done by gravity is zero.

11. How does the state of energy get changed when a battery lights up a bulb?

Ans: The chemical energy of the battery is converted into heat and light energy of the bulb in the given case.

12. Calculate the work done by the force that changes the velocity of a moving body from 5 ms -1 to 2 ms -1 . The body has a mass of 20 kg.

Ans: Since work done by force = Change in the kinetic energy of the moving body

Therefore, Work done by force = $\frac{1}{2}\times m(v_{1}^{2}-v_{2}^{2})$

=$\frac{1}{2}\times 20\times (5^2-2^2)$ =10×(25−4)=10×21 =210 J

13. An object having 10 kg weight is moved from point A to point B on the table. If the distance between A and B is horizontal, what work does gravity do to the object?  Give the reason for the answer. 

Ans: Since the work done by gravity on the object depends on the change in the vertical height of the object, the vertical height of the object will not change. Because the connection level of A and B is at the same height, the work done is zero.

14. The potential energy of an object decreases gradually in a free fall. How does this violate the law of conservation of energy? 

Ans: This does not violate the law of conservation of energy, because the potential energy of an object in free fall gradually decreases with gradual changes until the kinetic energy of the object maintains the state of free fall, that is, the total energy of the object remains conserved.

15. What energy conversion occurs when riding a bicycle?  

Ans: Our muscle energy is converted into mechanical energy while riding a bicycle.

16. Does energy transfer occur when you push a huge rock with all your strength without moving it? Where did the energy you applied go?  

Ans: As long as you push a big rock with all your strength and do not move it, energy transfer will not occur, because cell energy is only used for muscle contraction and relaxation, and also for releasing heat (sweating).

17. A household uses 250 units of energy in a month. How much energy is used  by that house in joules?

Ans: Energy consumption by a house = 250 kWh

Since, 1 kWh = 3.6× 10 6 J hence, 250kWh=250×3.6× 10 6 =9× 10 8 J

18. The output power of the electric heater is 1500 watts. How much energy does it consume in 10 hours?  

Ans: Power of electric heater (p) = 1500W = 1.5kW

Energy = Power × Time = 1.5kW × 10 hours = 15 kWh

19. An object of mass m moves at a constant speed v. How much work does the subject need to do to make it stable?  

Ans: For an object to be stationary, the work done must be equal to the kinetic energy of the moving object.

The kinetic energy of any object is equal to

$K.E=\frac{1}{2}mv^2$ , where m is the mass of the body and v is its velocity.

20. Sony said that even if different forces act on the object, the acceleration of the object can be zero. Do you agree with her, if yes, why?  

Ans: Yes, we agree with Soni, because the displacement of an object becomes zero when many balancing forces act on that object.

21. Calculate the energy (in kilowatt hours) consumed by four 500 W devices in 10 hours.

Ans: Since, Energy = Power × Time 

Hence, Energy consumed by four 500 W devices in 10 hours = 4 × 500 × 10 = 20000 Wh = 20 kWh

22. Free-falling objects will eventually stop when they hit the ground. What will happen to their kinetic energy? 

Ans: The object will eventually stop after it hits the ground in free fall, because its kinetic energy will be transferred to the ground when it hits the ground.

23. A large force acting on an object, and the displacement of that object is zero, what will be the work done?  

Ans: The work done on the body is defined as the force exerted on the body that causes a net displacement of the body.  

Work done = Force x Displacement  

If the force does not cause any displacement, the work done to the object is zero.

24. Write some differences between kinetic and potential energy.

Ans: Differences between kinetic and potential energy:

25. Describe the law of conservation of energy. 

Ans: The law of conservation of energy says that:

Energy cannot be produced or destroyed. It can only be transformed from one form to another.  

The energy of the universe is constant.

26. A person weighing 50 kg climbs the stairs with a height difference of 5 meters, within 4 seconds.  

What kind of work is done by that person?  

What is the average power of that person?

Ans: Mass of the man = 50 Kg 

Distance moved by that man = 5 meter 

Time taken to cover the given distance = 4s

Work Done = Force Acceleration

In this case, the increase in Potential energy = Work done =Mgh =50×10×5 =2500 J

Power =  work Done   Time Taken  = 2500 4 =625 Watts

27. Write differences between power and energy.

Ans: Differences between power and energy are given below:

28. Write down the expressions for

Potential energy of an object

Kinetic energy of an object

Ans: (a) The expression for Potential energy of an object = P.E = mgh 

Where, m = Mass of Body 

g = Acceleration due to gravity 

h = Height 

(b) The expression for Kinetic energy of an object = 1 2 m v 2

Where, m = Mass of body 

v = Velocity of body

29. If a force of 12.5 N is applied to complete a work of 100 J, what is the distance covered by the force?

Ans:   W = Work = 100 J 

F = Force = 12.5 N 

And S is the distance moved or displacement 

Since, Work done = Force Displacement

W=FS 100=12.5× S 100×10 12.5 =S 1000 125 =S 8 m=S (Displacement)

30. A car weighing 1800 kg is moving at a speed of 30 m/s when braking. If the average braking force is 6000 N, it is determined that the vehicle has traveled to a standstill distance. What is the distance at which it becomes stable?

Ans: M = Mass of the car = 1800 Kg 

V = Velocity of the car = 30 m/s 

F = Force applied while braking = 6000 N 

KE= 1 2 m v 2

KE = 1 2 1800×900

KE=810000 J

KE of car = Work done by the car = Force Displacement 

810000=6000× Displacement

810000 6000 = Displacement

135 m= Displacement

3 Marks Questions

1. The kinetic energy of an object with mass m moving at a speed of 5 m per second is 25 J. If its speed doubles, what is its kinetic energy? What is its kinetic energy when its speed triples?

Ans: K.E. of the object = $\frac{1}{2}\times m\times (5)^2$

$25=\frac{1}{2}\times m\times 25$

$m=(25\times y^2)/25=2kg$

If velocity is doubled,

$K.E=\frac{1}{2}\times 2\times 10^2=200/2=100J$ i.e. K.E. will become four times

If velocity is increased three times

$K.E=\frac{1}{2}\times 2\times 15^2=225J$ i.e. K.E. will become nine times.

2. What do you understand about average power?

Ans: The agent may not always be able to complete the same amount of work in a given time period. In other words, the power of this work will change over time. Therefore, in this case, we can take the average power of the work done by the body per unit time (that is, the total energy consumed divided by the total time).

3. Take a look at the steps below. Based on your understanding of the word "work", prove whether the work will proceed.  

Suma swims in the pond.  

The donkey carries a heavy load.  

The windmill draws water from the well.  

Green plants perform photosynthesis.  

The trains are pulled by engines. 

Drying food grains in the sun.  

Sailing boats are powered by wind.

Ans: The work is said to be done when a force acts on an object and moves in the direction of the force. According to this explanation, the following activities were taken in which work will be proceeded:

4. An object weighing 40 kg rises to a height of 5 m above the ground. What is its potential energy? If you let an object fall, find the kinetic energy when it is in the middle.

Ans: Potential energy of the object $=P . E=m g h=40 \times 10 \times 5=2000$ Joules 

Height at which objeot is present when it is in the middle $=2.5 \mathrm{~m}$

As the object is thrown from the rest, hence, its initial velocity $=0$

Since $v^{2}=u^{2}+2 g h$

$ \begin{array}{l} v^{2}=0+2 \times 10 \times 2.5 \\ v^{2}=50 \\ \text { Kinetic energy }=\dfrac{1}{2} \times m \times v^{2} \\ K E=\dfrac{1}{2} \times 40 \times 50=1000 \mathrm{~J} \end{array} $

5. A satellite is moving around the earth. What will be the work done by the force of gravity on that satellite? Give justification.

Ans: The displacement made by the object is perpendicular to the force direction as it is moving on a round path.

$ \begin{array}{l} \theta=90^{\circ} \\ W=F \times s \operatorname{Cos} \theta \\ W=F \times s \operatorname{Cos} 90^{\circ} \end{array} $

$W=F \times 0=0$

Therefore, work done is zero.

6. A person will feel tired if he puts a bundle of hay on his head for 30 minutes. What will be the work done by the person? Prove your answer.

Ans: When a person lifts a bundle of hay above their head for 30 minutes and feels tired, they exert an upward force, and the bundle of hay moves forward perpendicular to the direction of the applied force, so the displacement is zero.

$W=F \times s \operatorname{Cos} \theta$

$ \begin{aligned} W &=F \times s \operatorname{Cos} 90^{\circ} \\ W &=F \times 0=0 \end{aligned} $

Hence, no work done.

7. The law of conservation of energy is explained by discussing the energy changes that occur when we move the pendulum laterally and swing it. Why does the pendulum eventually stop? What happens to the energy and does it violate energy conservation law?

Ans: Bob will eventually stop due to the friction created by the air and the rigid support that holds the thread in place. This does not violate the law of conservation of energy, because mechanical energy can be converted into another unusable form of energy for some useful work. This energy loss is called energy dissipation.

8. How much work is done to stop a car of weight 1500 kg moving with a velocity 60 km/h?

Ans: Given that, initial velocity of a car 

= $60 \mathrm{~km} / \mathrm{h}=(60 \times 1000) / 60 \times 60=50 / 3 \mathrm{~ms}^{-1}$

The object is stopped, thus, its final velocity $=0$

Initial kinetic energy $=\dfrac{1}{2} \times m \times v^{2}$

$K E=\dfrac{1}{2} \times 1500 \times(50 / 3)^{2}=208333.30 J$

Final kinetic energy $=\dfrac{1}{2} \times 1500 \times 0=0$

Therefore, work done $=$ change in kinetic energy $=208333.30-0=208333.30 J$

9. In each of the following cases, the force F acts on an object of mass m. The direction of the object's movement is from west to east and is indicated by the longest arrow. Check the given diagram carefully to see if the work done by the force is negative, positive, or zero.

Ans: (i) Since the displacement in the first figure is perpendicular to the direction of the force, the work done is zero.  

(ii) Since the displacement in the second figure is in the direction of the force, the work done is positive.  

(iii) Since the displacement in the third diagram is opposite to the applied force, the work done is negative.

10. In the given force-displacement plot, calculate the work done in the time interval of

0 < x < 2 m 

2 < x < 6 m 

0 < x < 9 m 

Ans: The area of force-displacement plot gives the work done, therefore,

a) For time interval of $0<x<2 \mathrm{~m}$

Work Done $=$ Area of triangle OAE

$ W=\dfrac{1}{2} \times \text { Base } \times \text { Height } $

$W=\dfrac{1}{2} \times O E \times A E$

$W=\dfrac{1}{2} \times 2 \times 8$

$W=8 \mathrm{~J}$

b) For time interval of $2<x<6 \mathrm{~m}$

Work Done = Area of rectangle ABED

$W=$ Length $\times$ Breadth

$W=A B \times B D$

$W=4 \times 8$

c) Fortime interval of $0<x<9 \mathrm{~m}$

Work Done $=$ Area of triangle $\mathrm{DBC}$ 

$\mathrm{W}=\dfrac{1}{2} \times \mathrm{B}$ $\times$ Height 

$W=\dfrac{1}{2} \times D C B D$ 

$W=\dfrac{1}{2} \times 3 \times 8$ 

11. Derive the expression of the kinetic energy of an object. Calculate the kinetic energy of a 5 kg object moving at a speed of 2.5 ms -1 .

Ans: The kinetic energy of the body is defined as energy with the dignity of body movement. 

An object with mass m is at rest. The force F N acting on it will cause acceleration ms -2 , assuming the velocity is v ms -1 and covering the distance s m. 

Now from the third equation of motion

$ \begin{array}{l} v^{2}=u^{2}+2 a s \\ v^{2}=0+2 a s \\ 0=\dfrac{v^{2}}{2 s} \\ (u=0 \because \text { body starts from rest }) \end{array} $

From Newton's second law,

$F=\dfrac{m \times v^{2}}{2 s}=\dfrac{m v^{2}}{2 s}$ 

Work Done on the moving Body = Kinetic energy 

$W=$ Force $\times$ Distance 

$W=\dfrac{m v^{2}}{2 s} \times s=\dfrac{1}{2} m v^{2}$ 

Mass of the body $=5 \mathrm{Kg}$ 

Velocity of the body $=2.5 \mathrm{~m} /\mathrm{s}$ 

$K E=\dfrac{1}{2} m v^{2}$ 

$K E=\dfrac{1}{2} \times 5 \times(2.5)^{2}$ 

$K E=\dfrac{3125}{2 \times 100}$ 

$K E=\dfrac{1562.5}{100}$ 

12. A stone is thrown with a velocity of 40 m/s in upward direction.

The potential and kinetic energy of that stone will be equal at what height?

If the stone’s mass is 10 kg, what will be its potential energy?

Ans: Given that, the initial velocity of stone $=u=0$

And its final velocity $=v=40 \mathrm{~m} / \mathrm{s}$

Let the mass of the body $=M$

(a) Kinetic energy of the body $=\dfrac{1}{2} m v^{2}$

And its potentialenergy $=\mathrm{Mgh}$

Now, $K E=P E$

$ \begin{array}{l} \dfrac{1}{2} m v^{2}=m g h \\ \dfrac{1}{2} m \times(40)^{2} m \times g \times h \\ \dfrac{1600}{2}=g h \\ 800=g h \end{array} $ $ \begin{array}{l} \dfrac{800}{10}=h \\ 80 m=h \end{array} $

(b) $\quad P E=m g h$

P E=10 \times 10 \times 80=8000 \mathrm{~J}

13. A body having mass 5 kg and constant velocity 12 m/s is lifted upwards. Calculate:

Force applied in lifting the body

Work done in lifting the body

What will happen to the work done?

Ans: Given that, mass of the body $=m=5 K g$

And height upto which it lifted upwards $=h=12 \mathrm{~m}$.

$\mathrm{g}=$ Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^{2}$

a) $\mathrm{PE}=\mathrm{mgh}$

$P E=5 \times 12 \times 10$ $P E=600 \mathrm{~J}$

b) Force $=$ ?

Work done = Potential energy of the Body

Force $\times$ Distance Moved $=600$

$F \times 12=60$ $F=50 \mathrm{~N}$

c) The work done is stored as the potential energy while lifting the body.

14. Get the expression of the potential energy of an object. Calculate PE for a body of 10 kg which is resting at a height of 10 m.

Ans: The potential energy of an object with mass $=\mathrm{m} \mathrm{kg}$, at height above the ground $=\mathrm{h}$ $\mathrm{m}$

Gravitational force of attraction on that body $=\mathrm{mg} \mathrm{N}$

To lift that body to $B$ height at $h$ m above the ground.

Force applied to lift this body with a constant velocity $=\mathrm{mg} \mathrm{N}$

Distance moved by the body after applying force = $\mathrm{h} \mathrm{m}$

Work done in lifting the body from a to $B$ distance $=$ Force $\times$ Distance

\mathrm{W}=m g \times h=m g h

Energy cannot be destroyed, hence, this energy is stored as potential energy in the stone.

$ \begin{array}{l} m=10 \mathrm{Kg} \\ g=10 \mathrm{~m} / \mathrm{s}^{2} \\ h=10 \mathrm{~m} \\ \mathrm{P} E=\mathrm{mgh} \\ \mathrm{PE}=m g h=10 \times 10 \times 10=1000 \mathrm{Joules} \end{array} $

15. Prove that the total energy of a ball, having mass m, remains conserved when it is thrown downwards from a height of h.

Ans: According to the law of energy conservation, energy can neither be created nor destroyed, it can only be transformed from one form to another.  

Consider a ball with a mass of m stationary at point A at an elevation h from the ground.

Total energy of ball at position A will be:

Potential energy of the ball = mgh

The body is at rest, hence, its KE = 0

Total energy of ball at position $A=K E+P E=0+m g h=m g h$

Total energy of the ball at ground (position B) will be:

When the body strikes to the ground, its elevation is equal to zero, hence, its potential energy wrt ground $=0$

Velocity of ball when it strikes to the ground (position $\mathrm{B})=$ ?

Its acceleration $=0-\mathrm{g} \mathrm{m} / \mathrm{s}^{2}=-\mathrm{g} \mathrm{m} / \mathrm{s} 2$

Total energy of the body on ground $=K E+P E=m g h+0=m g h$

Total energy of ball at point Cwill be:

Say that, the ball falls through $x$ and be at $C$ during its fall.

Elevation of the body at $\mathrm{C}=\mathrm{h}-\mathrm{x}$

Potential energy at $\mathrm{C}=\mathrm{mg}(\mathrm{h}-\mathrm{x})$

Let the velocity at position $C$ will be $v$

$ \begin{array}{l} v^{2}=u^{2}+2 a s \\ v^{2}=2(-g)(-x)=2 g x \\ K E=\dfrac{1}{2} m v^{2}=\dfrac{1}{2} \times m \times 2 g x=m g x \end{array} $

Total energy at $C=m g(h-x)+m g x=m g h$

It means that the total energy at all points of the fall is always the same.

16. Define power. Prove that the power = force x speed. Can you calculate the power of a 10 kg object accelerating at a speed of 10 m/s 2 and reaching a velocity of 5 m/s?

Ans: Power is the rate of work done.

\text { Power }=\dfrac{\text { Work Done }}{\text { Time taken }}

Unit of Power is watt (w)

$ \begin{array}{l} P=\text { Power }=\dfrac{\text { Work Done }(v)}{\text { ime taken }(t)} \\ P=\dfrac{\text { Force }(F) \times \text { Displacement }(S)}{\text { Time taken }(t)} \\ P=\dfrac{F \times S}{t} \\ \dfrac{S}{t}=v=\text { Velocity } \end{array} $

Mass of the object $(\mathrm{M})=10 \mathrm{Kg}$

Acceleration of the object $(a)=10 \mathrm{~m} / \mathrm{s}^{2}$

And its velocity $(\mathrm{v})=5 \mathrm{~m} / \mathrm{s}$

$ \begin{array}{l} P=F \times v \\ P=F v \\ P=M \times a \times v \\ P=10 \times 10 \times 5 \\ P=500 W \end{array} $

17. What does the unit of electrical energy mean? When the meter displays 400 energy units, how much energy is consumed in joules?

Ans: The unit of electrical energy is defined as the energy consumed (or consumed) by an electrical device with an output power of $1 \mathrm{~kW}$ in one hour.

400 units $=144 \times 10^{7} J$ or 1 unit $=1 k w h$

Now, $1 k w h=1000 w \times 3600 s=3.6 \times 10^{6} w s$

$ \begin{array}{l} 1 w=\dfrac{1 \text { Joule }(J)}{1 \operatorname{second}(S)} \\ 1 \text { unit } \operatorname{mm} K w h=\dfrac{3.6 \times 10^{6} J \times S}{S}=3.6 \times 10^{6} J \end{array} $

So, if 400 units of electrical energy is consumed then,

$ \begin{array}{l} 1 \text { Unit }=3.6 \times 10^{6} J \\ 400 \text { Units }=3.6 \times 10^{6} \times 400 J \\ 400 \text { units }=144 \times 10^{7} J \end{array} $

CBSE Important Questions Class 9 Science Chapter 11

Definition of work.

In simple terms, when any displacement happens, work is said to be done. Work is a form of energy. When a considerable amount of force is applied to someone or something, which carries displacement, upon the subject where force has applied, work is said to be done.

Therefore, two conditions must satisfy for a work to be done:

Applying force is must

Displacement must take place

There can be multiple causes of displacement as per the direction:

(Image to be added soon)

What is Energy?

Energy is typically defined as the capacity to do a certain amount of work. These various kinds typically fall under class 9 work and energy important questions. Let’s discuss in more detail about energy and its distinctive forms.

Which are the Forms of Energy

There are several forms of energy, namely, kinetic, potential, electrical, thermal, nuclear, chemical etc. Let's discuss these in detail.

Kinetic Energy: Our body possesses a certain amount of energy, when in motion. This energy is called kinetic energy. It is always directly proportional to speed. The more is the velocity, the more the work is is required to do to manage that velocity.

Consider an object of mass m moves with uniform velocity u. Then the displacement s will take place because of the constant force F on an object.

We already know the formula for work is, W= F X s

Potential Energy: A body presents a certain amount of energy due to its position or shape. This is called potential energy. It can better be explained through a diagram below

Gravitational Potential: When an object is raised to a specific height h, then the work is said to be done on it against gravity. So such an object is said to acquire GPE.

Therefore, GPE= work done in raising a body from the ground to a point against gravity.

Consider a body of mass m raised through height h, from the ground

The force required to raise the object will be the same as its weight mg.

Derivation:

Object gains energy= work done on it  

Therefore, the work done will be W. This work is done against the gravity.                                                                                      

Attaining same height= same amount of work done, different paths don’t matter.                                                                                                                                                                                                                                                            

Mechanical Energy: kinetic and potential energy together combines to form mechanical energy.

Law of Conservation of Energy

This topic will primarily cover most of the important questions of chapter work and energy class 9. This proves that energy cannot be created or destroyed. It only passes on from one object to another with the same or different form. This can better be explained from a picture.

Also, potential and kinetic energy combine to become constant.

Potential energy happens to be maximum when an object of mass m is bought to height h, ultimately nullifying the kinetic energy. Likewise, the potential energy decreases when the height is decreasing and kinetic energy increase with the increase in v. After that, h will be negligible when the object reaches the ground and velocity will be maximum. So kinetic energy will be equal to or greater than the potential energy. 

Both potential and kinetic energy are inversely proportional. This chapter is exceptionally integral to know about, from an examination perspective. One can also expect to get considerable work and energy class 9 important questions as per previous year analysis. This topic will also be there in the upper classes, so leaving it today can be a hurdle for tomorrow.

Chapter Summary 

CBSE Class 9 Science Chapter 11 - Work and Energy dives into the fascinating world of motion and power. It's all about understanding how things work and why they move. The chapter introduces the concept of energy, which is like the fuel that makes everything happen. We explore the connection between force, work, and energy, unraveling the secrets behind everyday activities. From the basics of what 'work' means in physics to the different forms of energy, it's a journey into the forces that make our world go round.

Benefits of using Vedantu for Class 9 Chapter 11 - Work and Energy

Mastering Class 9 Chapter 11 - Work and Energy becomes seamless with Vedantu's tailored resources. These curated tools encompass NCERT solutions that delve into the core principles of work and energy. Through concise explanations and practice materials, Vedantu elevates understanding, fostering confidence and competence in tackling scientific concepts. Here are the Benefits of using Vedantu for Class 9 Chapter 11 - Work and Energy: 

Focus on key topics for efficient studying.

Prepares students for exams and reduces anxiety.

Reinforces understanding of fundamental concepts.

Teaches effective time management.

Enables self-assessment and progress tracking.

Strategic approach for higher scores.

Covers a wide range of topics for comprehensive understanding.

Supports exam preparation and boosts confidence.

Reviewing all the crucial questions for Class 9 Chapter 11 Science - Work and Energy provides students with a solid grasp of the chapter's topics. The extra and important questions for Class 9 Chapter 11 Science - Work and Energy engage in a concept-focused discussion, encompassing all chapter themes. This question-and-answer method proves time-saving during exam prep, offering an efficient way to revise the chapter and enhance understanding. Practising these important questions streamlines preparation and boosts confidence for the upcoming exams.

Important Related Links for CBSE Class 9 

FAQs on Important Questions for CBSE Class 9 Science Chapter 11 - Work and Energy

1. Which are the most important questions of work and energy class 9?

Ans: It is always suggested to go through the whole chapter diligently. A number of questions are likely to come from the chapter, law of conservation of energy. Other than the important applications, students are also advised to polish their numerical abilities. As, maximum numerical questions often come from this chapter.

2. Which are the best references for understanding physics class 9?

Ans: One can always rely on NCERT for the best references and maximum questions. This book covers a wide range of all the crucial topics from exam perspective. Other than that, Vedantu also offers a great online platform for better understanding of a subject. Student can either delve into the notes and PDF provided on a particular topic or attend a live session with our experts.

3. Are the important questions of chapter work and energy class 9 helpful from future perspective?

Ans: The chapter work and energy is definitely crucial for higher studies. More importantly, one has to make up your mind from here about the stream one is going to select. And if you are planning to go with science for higher studies, you can’t escape the topic. It will cover the wide portion of the higher secondary. So, not just for today, this chapter is essentially important for the future studies.

4. How are work and energy-related in Chapter 11 from Class 9 Science?

Ans: All sorts of work are forms of energy. Work performed on a body is stored as energy. Greater energy is required to accomplish more work. For example, in any kind of work, you do need force, and that force comes from the energy stored in your body.

Chapter 11 of Class 9 Science gives the students various information along with easy to comprehend examples that makes the chapter relatable and fun to learn.  

5. What is work according to Chapter 11 from Class 9 Science?

Ans: When force is applied to displace the object it is known as work. It is calculated with the product of force and displacement and measured in joule. Students should focus on learning the key definitions and formulas from this chapter as they are very scoring. The numerical problems are easy once you practice them on a regular basis. You can effectively retain the chapter if you learn it using real-life examples given in the chapter for references.

6. What is energy according to Chapter 11 of Class 9 Physics?

Ans: Energy is said to be the ability to perform a task. Like work, it is also measured in Joules and the formula for Potential Energy is mgh that is the product of mass, acceleration due to gravity, and height. There are different forms in which energy exists. The students gain more in-depth insight on the various types of energy like kinetic, electromagnetic, and so on in the chapter along with examples for references.

7. Can one create energy?

Ans: It is stated by the Law of conservation of energy that ‘Energy can neither be created nor destroyed’. We can only convert energy from one form to another using different methods. For example, the potential energy of a substance is converted into kinetic energy when we introduce the substance to speed. Solar power is used to produce electricity by saving energy in the solar panels. There are numerous other examples that prove that energy cannot be created but only converted.

8. What are the various types of energy?

Ans: Energy exists in several forms, like kinetic energy, mechanical energy, chemical energy, electrical energy, elastic energy, thermal energy, heat energy, gravitational energy, magnetic energy, nuclear energy, and so on. Some of these energies can be converted in other forms through different man-made processes. For example, kinetic energy is converted into electric energy through hydropower plants. You can get a deeper understanding of these categories of energy in Chapter 11 of Class 9th Science on Vedantu.

CBSE Class 9 Science Important Questions

Cbse study materials.

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

NCERT Solutions for Class 9 Science (physics) Chapter 10 Gravitation are given below. In these solutions, we have answered all the intext and exercise questions provided in NCERT class 9 science textbook. Class 9 NCERT Solutions Science Chapter 10 provided in this article are strictly based on the CBSE syllabus and curriculum. Students can easily download these solutions in PDF format for free from our app.

Class 9 Science Chapter 10 Textbook Questions and Answers

INTEXT QUESTION

PAGE NO 134

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. 

Answer:  Let M E be the mass of the Earth and ‘m’ be the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation:

PAGE NO 136

Question 1: What do you mean by free fall? 

Answer: 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.

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 138

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

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

PAGE NO 141

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: An object float or sink 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.

PAGE NO 142

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 iron bar. This is because the surface area of the cotton bag is larger than the iron bar. Hence, 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. 

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

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 ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). 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:  Given, Mass of Earth, M = 6 × 10 24 kg  Mass of object, m = 1 kg  Universal gravitational constant, G = 6.7 × 10 −11 Nm 2 kg −2   Radius of the Earth, R = 6.4 × 10 6 m 

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:

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

Answer: 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 force that binds North American nation to the world (iii) The tides because of the moon and therefore the Sun (iv) The motion of planets round the Sun

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  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: Surface area of a Sheet which is crumpled into a ball, is much smaller than the surface area of a plain or flat sheet. Therefore, despite both experience same force of gravity, the plain or flat sheet of paper will have to face more air resistance than the crumpled ball, so it will fall slower than the sheet crumpled into a ball.

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   Max. Height attained by the ball (s) = ?

∴ Max. Height attained by the ball (s) = 122.5 m

(ii) Let t be the time taken by the ball to reach the height 122.5 m, then according to the first equation of motion

Time for upward journey of the ball will be the same as time for downward journey i.e., t = 5 s.

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, v 2 = u 2 + 2as ⇒ v 2 − u 2 = 2as ⇒ v 2 − 0 2 = 2 × 9.8 × 19.6  ⇒ v 2 = 2 × 9.8 × 19.6 ⇒ v 2 = 19.6 × 19.6 ⇒ v 2 = (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:  Given 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   the maximum height attained by the stone (s) = ? 

According to the equation of motion under gravity, v 2 − u 2 = 2gs 

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. 

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  

When the stone thrown upwards  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  

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 4s 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 4s.

Answer: (a) The velocity with which ball was thrown up : Acceleration due to gravity, g = – 9.8 ms –2   As the total time taken in upward and return journey by the ball is 6 s. Therefore, The upward journey, t = 6/2 s = 3 s Final velocity, v = 0 ms –1 Initial velocity, u = ?

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 s.  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  

Hence, the maximum height is 44.1 m.

(c) Ball attains the maximum height after 3s. 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 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 (1𝑔/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., 350g. 

Class 9 Science NCERT Solutions Chapter 10 Gravitation

CBSE Class 9 Science NCERT Solutions Chapter 10 helps students to clear their doubts and to score good marks in the board exam. All the questions are solved by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 9 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

NCERT Solutions for Class 9 Science Chapter 10 PDF

Below we have listed the topics discussed in NCERT Solutions for Class 9 Science Chapter 10. The list gives you a quick look at the different topics and subtopics of this chapter.

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NCERT Solutions for Class 9 Science Chapter 10 Gravitation

NCERT Solutions Class 9 Science Chapter 10 Gravitation – Here are all the NCERT solutions for Class 9 Science Chapter 10. This solution contains questions, answers, images, step by step explanations of the complete Chapter 10 titled Gravitation of Science taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Science, then you must come across Chapter 10 Gravitation. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Science Chapter 10 Gravitation in one place. For a better understanding of this chapter, you should also see Chapter 10 Gravitation Class 9 notes , Science.

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

Topics and Sub Topics in Class 9 Science Chapter 10 Gravitation:

• Gravitation
• Thrust and Pressure
• Archimedes’ Principle
• Relative Density

These solutions are part of NCERT Solutions for Class 9 Science . Here we have given NCERT Solutions for Class 9 Science Chapter 10 Gravitation.

In – Text Questions Solved

Questin 2. What do you mean by buoyancy? Answer:  The upward force exerted by any fluid (liquid, gas) on an object is known as upthrust or buoyancy.

More Resources for CBSE Class 9

NCERT Solutions

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

Questin 3. Why does an object float or sink when placed on the surface of water? Answer:  The density of the objects and water decides the floating or sinking of the object in water. The density of water is 1 gm/cm3.

• If the density of an object is less than the density of water then the object will float.
• If the density of an object is more than the density of water then the object will sink.

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

Class 9 Science NCERT Textbook – Page 142 Questin 1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg? Answer:  The weighing machine actually measures the weight of the body as the acceleration due to gravity ‘g’ is acting on the body. Hence the mass reading of 42 kg given by a weighing machine is same as the actual mass of the body. As mass is the quantity of inertia, it remains the same.

Questions From NCERT Textbook for Class 9 Science

Question 19. In what direction does the buoyant force on an object immersed in a liquid act? Answer:  The buoyant force on an object immersed in a liquid acts upwards, i.e. opposite to the direction of the force exerted by the object.

Question 20. Why does a block of plastic released under water come up to the surface of water? Answer. The floating or sinking of a body in the water is decided by the density of both the body and water’s buoyant force acting on the body by the liquid. The density of plastic is less than the water and the buoyant force exerted by water on the plastiq block is greater than the force exerted by plastic on the water.

NCERT Solutions for Class 9 Science Chapter 10 Gravitation and Floatation (Hindi Medium)

More Questions Solved 

NCERT Solutions for Class 9 Science Chapter 10 Multiple Choice Questions

Choose the correct option:

• The device used to measure the purity of milk is (a) hydrometer (b) lactometer (d) hygrometer (d) maltometer
• The cork floats while the nail sinks in the water, this is due to (a) density of cork is more than nail (b) density of nail is more than cork. (c) density of cork is less than the density of water. (d) density of iron is less than the density of water.
• The relative density of silver is 10.8 and the density of water is 1o 3 kg/m 2 . The density of silver is (a) 1.8 x 1o 4  N/m 3  (b) 10.8 x 1o 3 N/m 3 (c) 1.8 x 1o 4 kg/m 3 (d) 10.8 x 1o 4 kg/m 3
• Buoyant force exerted by different fluids on a given body is (a) same (b) different (c) zero  (d) negligible
• Liquid A is denser than liquid B, a body of wood is dipped in both the liquids? The buoyant force experienced by the body in (a) liquid A is more (b) liquid B is more (c) liquid A is less    (d) none of the above Answer. 1 -(b), 2—(c), 3—(b), 4-(b), 5—(a).

NCERT Solutions for Class 9 Science Chapter 10 Very Short Answer Type Questions

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

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

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

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

Question 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.

Question 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.

Question 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.

Question 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.

Question 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.

Question 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.

Question 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 .

NCERT Solutions for Class 9 Science Chapter 10 Short Answer Type Questions

Question 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

Question 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.

Question 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.

Question 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. T

Question 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.

Question 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.

Question 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:

• It is used in designing of ship and submarine.
• It is used in designing lactometer, used to determine the purity of milk,
• To make hydrometers, used to determine the density of liquids.

NCERT Solutions for Class 9 Science Chapter 10 Long Answer Type Questions

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

NCERT Solutions for Class 9 Science Chapter 10 Activity -Based Questions

Question 1.

• Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with water. You see that the bottle floats.
• Push the bottle into the water. You feel an upward push. Try to push it further down. You will find it difficult to push deeper and deeper. This indicates that water exerts a force on the bottle in the upward direction. The upward force exerted by the water goes on increasing as the bottle is pushed deeper till it is completely immersed.
• Now, release the bottle. It bounces back to the surface.
• Does the force due to the gravitational attraction of the earth act on this bottle? If so, why doesn’t the bottle stay immersed in water after it is released? How can you immerse the bottle in water?

Question 2.

• Take a beaker filled with water.
• Take an iron nail and place it on the surface of the water.
• Observe what happens.

Question 3.

• Take a piece of cork and an iron nail of equal mass.
• Place them on the surface of water.

Answer:  The iron nail sinks as.the density of nail is more and the downward force exerted on nail is more than the buoyant force. The cost floats as the density of cost is less and the buoyant force exerted on it is more than the downward force.

Question 4.

• Take a piece of stone and tie it to one end of a rubber string or a spring balance.

• Note the elongation of the string or the reading on the spring balance due to the weight of the stone.
• Now, slowly dip the stone in the water in a container as shown in Fig. (b).
• Observe what happens to the elongation of the string or the reading on the balance. Observations :
• In Fig. (a) the elongation of the string is 6 cm.
• In Fig. (b) when the stone is dipped in water the length of string reduced to 5 cm.
• The length of the string in case (b) decreases due to the upward force exerted by water on the stone called as buoyant force.

NCERT Solutions for Class 9 Science Chapter 10 Value-Based Questions

Question 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.

Question 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.

Question 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.

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Chapter 10 Class 9 - Gravitation

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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 Solutions
• NCERT Class 9
• NCERT 9 Science
• Chapter 10: Gravitation

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.

Download Exclusively Curated Chapter Notes for Class 9 Science Chapter – 10 Gravitation

Download most important questions for class 9 science chapter – 10 gravitation.

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.

Furthermore, we ensure that relevant content on  NCERT Solutions Class 9 is regularly updated as per the norms and prerequisites that examiners often look for in the CBSE exam. This ensures that the content is tailored to be class relevant without sacrificing the informational quotient. BYJU’S also strives to impart maximum informational value without increasing the complexity of topics. This is achieved by ensuring that the language is simple and that all technical jargon is explained at the required school level.

• 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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CBSE Class 9 Physics -Gravitation -Study Notes

CBSE Class 9  Science CBSE Class 9 Social Science CBSE Class 9 Maths

Notes and Study Materials

• Concepts Of Gravitation
• Concepts of Buoyancy
• Gravitation Master File
• Ncert Solution
• Numerical Example
• Lakhmir Singh Solution

Examples and Exercise

• Gravitation Assignment
• Unsolved Examples
• Test Paper 1
• Test Paper 2
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Problems Based on Class 9 Science Chapter 10 Gravitation

• Last modified on: 2 years ago
• Reading Time: 6 Minutes

Q.1. An object weighs 10N when measured on the surface of the earth. What would be its weight when measured on the surface of the Moon?

Q.2. Mass of an object is 10kg. What is its weight on Earth?

Q.3. What is the mass of an object whose weight is 49N?

Q.4. A body has a weight of 10 kg on the surface of earth. What will be its mass and weight when taken to the centre of earth?

Q.5. An object is thrown vertically upwards and rises to a height of 10m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.

Q.6. A stone drops from the edge of a roof. It passes a window 2m high in 0.1s. How far is the roof above the top of the window?

Q.7. Calculate the force of gravity acting on your friend of mass 60kg. Given mass of earth = 6 x 10 24 kg and radius of Earth = 6.4 x 10 6 m.

Q.8. A force of 20N acts upon a body weight is 9.8N. What is the mass of the body and how much is its acceleration?

Q.9. If a planet existed whose mass was twice that of Earth and whose radius 3 times greater, how much will a 1kg mass weigh on the planet?

Q.10. A boy on cliff 49m high drops a stone. One second later, he throws a second stone after the first. They both hit the ground at the same time. With what speed did he throw the second stone?

Q.11. A stone is dropped from the edge of a roof. (a) How long does it take to fall 4.9m ? (b) How fast does it move at the end of that fall? (c) How fast does it move at the end of 7.9m? (d) What is its acceleration after 1s and after 2s?

Q.12. How much would a 70 kg man weigh on moon? What will be his mass on earth and moon? Given g on moon = 1.7 m/s 2 .

Q.13. The Earth’s gravitational force causes an acceleration of 5 m/s 2 in a 1 kg mass somewhere in space. How much will the acceleration of a 3 kg mass be at the same place?

Q.14. A force of 2 kg wt. acts on a body of mass 4.9kg. Calculate its acceleration.

Q.15. A particle is thrown up vertically with a velocity of 50m/s. What will be its velocity at the highest point of the journey? How high would the particle rise? What time would it take to reach the highest point? Take g = 10 m/s 2 .

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