case study questions class 9 physics gravitation

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Case Study Questions Class 9 Science Gravitation

Case study questions class 9 science chapter 10 gravitation.

CBSE Class 9 Case Study Questions Science Gravitation. Important Case Study Questions for Class 9 Exam. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Gravitation.

At Case Study Questions there will given a Paragraph. In where some Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks or 4 marks.

CBSE Case Study Questions Class 9 Science – Gravitation

(1) Every object in the universe attracts every other object with a force which is proportional to the product of their masses (m1*m2) and inversely proportional to the square of the distance (d 2 ) between them. The force is along the line joining the centers of two objects.

(i)Gravitational force does not depend on

(a) Masses of objects

(b) Separation between objects

(c) Charges on objects

(d) None of these

(ii) Force of gravitation varies with masses of object as

(a) Product of masses

(b) Sum of masses

(c) Difference of masses

(iii) When mass of one body is doubled then force of gravitation will become

(a) Force will remain same

(b) Force will become double

(c) Force will become halved

(iv)What is universal gravitational constant? What is its SI unit?

(v) Two objects of masses 10kg and 20kg separated by distance 10m. What is gravitational force between them?

Answer key-1

(iv) The force of attraction between any two unit masses separated by a unit distance is called universal gravitational constant denoted by G measured in Nm 2 /kg 2 .

(v) Mathematically,

(2) We know that the earth attracts every object with a certain force and this force depends on the mass ( m ) of the object and the acceleration due to the gravity ( g ). The weight of an object is the force with which it is attracted towards the earth. Mathematically

Where, W= weight of object

m= mass of object

g= acceleration due to the gravitational force

As the weight of an object is the force with which it is attracted towards the earth, the SI unit of weight is the same as that of force, that is, Newton (N). The weight is a force acting vertically downwards; it has both magnitude and direction. We have learnt that the value of g is constant at a given place. Therefore at a given place, the weight of an object is directly proportional to the mass, say m, of the object, that is, W αm. It is due to this reason that at a given place, we can use the weight of an object as a measure of its mass. Answer the following questions.

(i) Unit of acceleration due to the gravity ( g ) is

(c) Newton(N)

(ii) Direction of weight of any object is 

(a) Always towards centre of earth

(b) Always away from centre of earth

(c) Weight don’t have direction

(iii) Which of the following has same unit

(a) Mass and weight

(b) Weight and force

(c) Velocity and acceleration

(iv) Whether weight is scalar quantity or vector quantity? Justify your answer.

(v) Differentiate between mass and weight.

Answer key-2

(iv) Weight is vector quantity as it has magnitude as well as direction which is always towards centre of a earth.

(v) Difference between mass and weight is given below

(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.

(i)The up thrust of the body is equal to the

(a) Mass of liquid

(b) Weight of liquid

(c) Weight of liquid displaced by body

(ii) If the density of the object is more than the density of the liquid in which it is immersed then

(a) It sinks in liquid

(b) It floats on liquid

(c) It comes out of liquid

(iii) When anybody immersed in liquid it experience a force called as

(a) Gravitational force

(b) Buoyancy force

(c) Nuclear force

(iv) State Archimedes’ principle.

(v) Why does cube of plastic released deep down under the water come up to surface of water?

Answer key-3

(iv) Archimedes’ principle, stated as follows: When a body is immersed fully or partially in a fluid, it experiences an upward force thatis equal to the weight of the fluid displaced by it.

(v) Cube of plastic released deep down under the water come up to surface of water because it has less density than that of water also it experiences upward buoyancy force on it due to liquid.

(4 ) As you know, the density of a substance is defined as mass of a unit volume. The unit ofdensity is kilogram per meter cube (kg/m -3 ). The density of a given substance, under specified conditions, remains the same. Therefore the density of a substance is one of its characteristic properties. It is different for different substances. For example, the density of gold is 19300 (kg/m -3 ) while that of water is 1000 (kg/m -3 ). The density of a given sample of a substance can help us to determine its purity. It is often convenient to express density of a substance in comparison with that of water. The relative density of a substance is the ratio of its density to that of water:

Density of a substance

Relative density =

Density of water

Since the relative density is a ratio of similar quantities, it has no unit. Answer the following.

(i) SI unit of density is

(b) kg /m 2

(c) kg /m 3

(ii) SI unit of relative density is

(b) No unit

(c) kg /s 3

(iii) Relative density of water is

(iv)Define relative density.

(v) Comment of relative density value of substance which

(1) Sink in water

(2) float on water

Answer key-4

(iv) The relative density of a substance is defined as the ratio of density of that substance to that of density of water.

(v) Relative density of substance which sinks in water is greater than one. That is density of that substance is more than density of water. While Relative density of substance which float on water is less than one. That is density of that substance is less than density of water.

(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.

Pressure=thrust/area. SI unit of pressure as N/m 2 or Pascal.

(i) SI unit of thrust is

(ii) We have two different areas A and B. where A>B. Then relation between pressure on A(P A ) and pressure on B (P B ) is

(a) P A> P B

(b) P B> P A

(c) P A= P A

(iii)SI unit of pressure is

(b) Pascal.

(c) Both a and b

(iv) The effects of forces of the same magnitude on different areas are different. True or false

(c) None of these

(v) Define pressure.

Answer key-5

(v) pressure is defined as thrust or perpendicular force acting per unit area

Case 3 III) question answer is buoyancy

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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

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

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

Gravitation Case Study Questions With Answers

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

Case Study/Passage-Based Questions

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

F ∝ m 1 m 2 and F ∝ 1/R 2

F =G m 1 m 2 /R 2

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

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

Answer: (d) everywhere

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

Answer: (b) attractive force

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

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

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

Answer: (b) 4 N

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

Answer: (a) Force of gravitation becomes 4 times

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

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

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

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

Answer: (c) zero

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

Answer: (c) constant

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

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

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

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

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

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

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

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Class 9 Science Case Study Questions

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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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Test: Gravitation- Case Based Type Questions - Class 9 MCQ

10 questions mcq test - test: gravitation- case based type questions, direction: read the following paragraph and choose the correct options to answer any four questions given below: 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. q. at what place on the earth’s surface is the weight of a body minimum.

• A. At the equator
• B. At all places the weight will be equal
• C. At the north pole
• D. At the south pole

In case if a body is at a rest position the force acting on the body is equal to the weight of the body which is acting downward on the body and this is equal to the product of mass of the body and the acceleration due to gravity.

⇒ F = W = Mg

Where, W = weight of the body, M = mass of the body and g = acceleration due to gravity.

Now as we know that the acceleration due to gravity is not the same it is dependent on the latitude. I.e. it is increasing on increasing the latitude and vice-versa.

Now as we know that the latitude is minimum at the equator and maximum at the poles on the surface of the earth.

So the acceleration due to gravity is minimum at the equator and maximum at the poles on the surface of the earth.

Now the weight of the body is directly proportional to the acceleration due to the gravity on the surface of the earth.

So the weight of the body is maximum at the poles and minimum at the equator on the surface of the earth.

So this is the required answer.

Direction: Read the following paragraph and choose the correct options to answer any four questions given below: 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. Q. What is the SI unit of Mass?

• C. Kilogram

Direction: Read the following paragraph and choose the correct options to answer any four questions given below: 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. Q. Which of the following statement is true?

• A. The friend at equator will agree with the weight of gold bought at poles.
• B. The friend at equator will not agree with the weight of gold bought at poles.
• C. The friends at the poles will not agree with the weight of gold bought at poles.
• D. The friends at equator as well as at the poles will agree with the weight of gold bought at poles.

gold weight at poles is Wp = m X g p

gold weight at the equator is We = m X g e

Hence, we can say gold weight at poles is more than gold weight at the equator, Hence, a friend at the equator will not match with the weight of gold bought from a place situated on poles.

Direction: Read the following paragraph and choose the correct options to answer any four questions given below:

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.

Q. What is the relation between mass and weight?

• A. Weight (W) of the body is always less than the mass of the body.
• B. Weight (W) of the body is same as that of the mass of the body.
• C. Weight (W) of the body is inversely proportional to the mass of the body.
• D. Weight (W) of the body is directly proportional to the mass of the body.

While, mass m is the measure of how much matter is present in the object.

Weight is given by w = mg

where, g = acceleration due to gravity.

Q. The value of g is greater at the poles than at the equator, so the weight of gold at the equator will be

• A. more than the weight of gold at the poles.
• B. less than the weight of gold at the poles.
• C. same as the weight of gold at the poles.

Direction: Read the following paragraph and choose the correct options to answer the questions given below:

Seema was standing on a terrace of a building. She dropped a coin and a piece of paper simultaneously from the same height.

Q. Which of the two will touch the ground first?

• A. In air the coin will touch the ground first.
• B. The piece of paper will touch the ground first.
• C. Both will touch the ground simultaneously.
• D. None of them will touch the ground.

Q. Give reasons for your answer.

• A. The piece of paper will touch the ground first.
• B. In vacuum both will touch the ground together.
• C. The coin will touch the ground first.

If the coin and the piece of paper are dropped in vacuum, both of them will touch the ground at the same time.

Q. What will happen if they are dropped in vacuum?

Q. Which of the following statement is correct?

• A. In case of a free fall, there is no change in the direction of motion of the objects.
• B. In case of a free fall, there is always a change in the direction of motion of the objects.
• C. In case of a free fall, the direction of motion reverts.
• D. None of the above statement is correct.

Q. What is meant by free fall?

• A. This is the fall of body when many forces are acting on it from all the sides.
• B. This is the fall of the body from one medium to other.
• C. In this the body is prohibited from falling down.
• D. The falling of a body from a height towards earth under the gravitational force of earth (with no other force acting on it) is called free fall.

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Important Questions for Gravitation- Case Based Type Questions

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Gravitation

Class 9 - ncert science solutions, intext questions 1.

State the universal law of gravitation.

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

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

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

F = force of attraction

G = constant of proportionality

M = mass of earth

m = mass of the object

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

Intext Questions 2

What do you mean by free fall?

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

What do you mean by acceleration due to gravity?

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

Intext Questions 3

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

Differences between the mass of an object and its weight:

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

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

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

Intext Questions 4

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

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

What do you mean by buoyancy?

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

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

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

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

Intext Questions 5

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

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

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

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

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

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

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

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

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

m 1 = mass of first object

m 2 = mass of second object

d = distance between the two objects.

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

So, substituting we get,

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

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

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

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

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

Mass of the body (m) = 1 kg

Mass of the earth (M) = 6 × 10 24 kg

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

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

According to the formula for gravitational force we know,

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

Substituting we get,

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

Hence, magnitude of the gravitational force = 9.8 N

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

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

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

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

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

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

What happens to the force between two objects, if

(i) the mass of one object is doubled?

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

(iii) the masses of both objects are doubled?

(i) We know,

When mass of one object is doubled.

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

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

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

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

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

Now, if it's tripled

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

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

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

m 1 ' = 2m 1

m 2 ' = 2m 2

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

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

What is the importance of universal law of gravitation?

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

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

What is the acceleration of free fall?

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

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

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

Question 10

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

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

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

Question 11

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

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

Question 12

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

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

Object's mass, m = 10 kg

Acceleration due to moon gravity = g m

Weight on the earth = W e

Weight on the moon = W m

Weight = mass x gravity

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

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

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

W e = m x g e = 10 x 9.8

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

Question 13

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

(i) The maximum height to which it rises,

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

Initial velocity (u) = 49 ms -1

Final velocity v at maximum height = 0

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

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

By third equation of motion,

2gH = v 2 - u 2

2 × (-9.8) × H = 0 - (49) 2

-19.6 H = -2401

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

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

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

According to the first equation of motion,

0 = 49 + (-9.8) x T a

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

Also, T d = 5 s

∴ T = T a + T d = 5 + 5 = 10 s

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

Question 14

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

Initial velocity (u) = 0

Tower height = total distance = 19.6 m

g = 9.8 ms -2

Final velocity (v) = ?

According to the third equation of motion,

v 2 = 2gs + u 2

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

v = 384.16 \sqrt{384.16} 384.16 ​

v = 19.6 ms -1

Hence, final velocity = 19.6 ms -1

Question 15

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

Initial velocity (u) = 40 ms -1

Final velocity (v) = 0

g = 10 ms -2

Max height = ?

v 2 = u 2 - 2gs

Note: [negative g as the object goes up]

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

⇒ 1600 = 20s

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

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

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

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

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

Question 16

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

Mass of the sun (m) = 2 × 10 30 kg

Average distance R = 1.5 × 10 11 m

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

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

Question 17

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

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

Initial velocity u = 0

Distance travelled = x

Time taken = t

According to the second equation of motion,

S = ut + 1 2 \dfrac{1}{2} 2 1 ​ gt 2

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

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

(b) When another stone is projected vertically upwards,

Initial velocity u = 25 ms -1

Distance travelled = (100 - x)

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

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

From equations (1) and (2)

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

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

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

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

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

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

Question 18

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

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

(b) the maximum height it reaches, and

(c) its position after 4s.

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

(a) Final velocity (v) = 0

From first equation of motion

v = u – gt up

⇒ u = v + gt up

⇒ u = 0 + (9.8 x 3)

⇒ u = 29.4 ms -1

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

(b) According to the second equation of motion,

S = ut - 1 2 \dfrac{1}{2} 2 1 ​ gt 2

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

Hence, maximum height stone reaches is 44.1 m

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

Distance travelled in another 1 sec = s'

g is +ve as ball is going down

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

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

Question 19

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

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

Question 20

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

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

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

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

Question 21

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

Volume of substance = 20 cm 3

Mass of substance = 50 g

Density of water = 1 gcm -3

Density of the substance = ?

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

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

Question 22

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

Volume of sealed packet = 350 cm 3

Mass of sealed packet = 500 g

Density of the sealed packet = ?

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

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

According to Archimedes Principle,

Displaced water volume = Volume of packet

Volume of water displaced = 350 cm 3

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

Hence, mass of displaced water = 350 g

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• Important Questions for CBSE Class 9 Science Chapter 10 - Gravitation

CBSE Class 9 Science Chapter-10 Important Questions - Free PDF Download

Important Questions for CBSE Class 9 Science Chapter 10 - Gravitation is provided in this article in the form of free-to-download pdf notes. 

Gravity, or gravitation, is a physical phenomenon that is observed for all objects which have mass. It is the phenomenon that governs the movement of heavenly bodies such as stars and planets and keeps one stuck on the ground and makes objects fall from a height. Gravity or Gravitation is one of the four fundamental forces of nature. Solving the important questions of gravitation of Class 9 will help the students boost their confidence and get ready for their final exam. The questions on gravitation for Class 9  Science will provide the students, the essential insights covered in the chapter.

So, download the free pdf and start solving the important questions of CBSE Class 9 Science Chapter 10 Gravitation.

Download CBSE Class 9 Science Important Questions 2024-25 PDF

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

CBSE Class 9 Science Chapter 10 - Gravitation - Topics Covered 

Following are the topics that are covered in the CBSE Class 9 Science Chapter 10:

Gravitation

Universal Law of Gravitation

Acceleration Due to Gravity (g)

Difference between G and  g

Mass and Weight

Weight of an Object on the Surface of the Moon

Important Questions for CBSE Class 9 Science Chapter 10 - Gravitation - Benefits of Gravitation Important Questions Class 9

The Class 9 Science Chapter 10 important questions will help the students learn and thoroughly understand the chapter. Here is the list of benefits of solving the gravitation Class 9 questions. 

The gravitation Class 9 important questions will help students understand the questions patterns and learn how to study for the exam.

It will also help the students to improve the speed and accuracy in solving the papers.

The Gravitation Class 9 questions are informational. They present all the important exercises on the gravitation chapters in a detailed manner and save a lot of time during the exam preparation.

Detailed answers and explanations provided by Vedantu will help the students in understanding the concepts clearly.

The Class 9 Science Chapter 10 important questions are completely free to use and download.

Since the important questions are taken from old question papers, they are highly accurate.

Important Questions for CBSE Class 9 Science Chapter 10 - Gravitation - Key Takeaways 

Gravitation chapter is an important topic in the Physics syllabus. Physics being the fundamental study of physical objects governs each and every aspect of our life. In the broader understanding of Physics, gravitation plays an important piece of the jigsaw puzzle. A sound understanding of gravitation will help in developing a comprehensive knowledge base of physics.

In the gravitation chapter, the students will learn about the universal law of gravitation, Acceleration, Free Fall, Mass, Weight, Thrust and Pressure, Archimedes’ Principle, and Relative Density. The gravitation Class 9 important questions will also benefit students during the revision period to keep a record of the covered topics as well as the uncovered ones. Solving the important questions of gravitation Class 9 will help the students get a basic idea of the chapter which will help them score well in the exam. 

Tips to Study CBSE Science Class 9  

Following are some of the tips that will help Class 9 Science students:

Students are required to understand the concepts well enough in order to understand the true concepts of the chapters. 

The students are expected to practise the questions and answers from the back of the chapter.

They must solve all the problems sums that are related to the practical problems of the chapters. 

The formula must be learned well by them.

Questions and answers must also be studied from the NCERT Solutions and Sample question papers. 

Revision is a must as well. 

Well, I hope the content was helpful to the students, they can solve the problems from the Important Questions for CBSE Class 9 Chapter 10 . Students are required to download the free pdf and practise the questions and answers asked in this study material. Students must also take note of the tips and suggestions suggested by us, it will definitely help them in the exam. 

Important Related Links for CBSE Class 9 

How are vedantu's important questions helpful.

Vedantu’s Important Questions for Class 9 Science Chapter 10 - Gravitation are the finest material to understand and practice the topics in the best way.

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These important questions help you understand how questions are asked, improve your speed, and give you detailed exercises to practice. Vedantu's answers make learning easy, and the best part is, they're all free to access and download and super reliable since they're from old exam papers. In addition to these important questions, you can further enrich your study materials by accessing a wide array of NCERT Class 12 Question Papers for various subjects. For a comprehensive revision approach, don't forget to explore the last-minute revision notes provided on Vedantu's website. These combined resources will be instrumental in helping you excel in your academic endeavors and grasp the subject matter with confidence. 

FAQs on Important Questions for CBSE Class 9 Science Chapter 10 - Gravitation

1. State the difference between the mass of an object and its weight?

The differences between the mass of an object and its weight are that the mass is the quantity of matter in the body, whereas the weight is the force of gravity acting on the body. The mass is the measure of inertia of the body, and the weight is the measure of gravity. The mass has magnitude, whereas weight has magnitude as well as direction. Mass is a constant quantity, and weight is not a constant quantity. It is different in different places. The SI unit of mass is the kilogram (kg) and the SI unit of weight is Newton (N).

2. What are the benefits of Gravitation Class 9 important questions?

The important questions on the gravitation chapter give an in-depth idea about the chapters and the question patterns that will come in the science exam. These questions are selected from previous years’ question papers and the answers are drafted by subject matter experts at Vedantu. Solving the questions papers will benefit the students to prepare adequately for their exams, understand the subject, and score better marks in the exam.

3. How do you solve the questions of Class 9 Chapter Gravitation of Science?

It is easy to solve the problems in Class 9 of the chapter Gravitation after you understand the theory and derivation of the different formulae. Once you are knowledgeable about the ideas, it will be a gentle cruise to practise the numerical problems. There are a number of formulas in the chapter that you might find difficult to understand the applications. Hence, to resolve your doubts, you can refer to NCERT Solutions by Vedantu ( vedantu.com ) and Review Notes.

4. What is gravitation?

Gravity, also known as gravitation, is the force that pulls two bodies (that have mass) together.

The chapter is important from the perspective of the exams; hence, it is recommended you solve all the numerical questions. The important questions provided by Vedantu can be a great aid to you for preparing and revving the chapter. Since these questions cover all the fundamental topics, you will have near to no burden of missing out on any vital concept.

5. What is the acceleration due to gravity according to the Chapter Gravitation from Class 9?

Acceleration due to gravity is the acceleration experienced by a body during free fall. It is denoted by ‘g’ and measured in m/s². There are numerous numericals based on the formulas under this topic. The numerical problems are often worded in a twisted manner and can confuse the students, so it will be beneficial to solve them all to learn the application of the formulas correctly in each question.

6. Explain buoyancy with an example.

Buoyancy is the overall upward force exerted on an object that is immersed in a fluid. This happens because of the pressure of the fluid. For example, a wooden plank floats in water. There are many other examples of buoyancy that we experience in our day-to-day life. You can learn more about them in the chapter itself. It is vital to learn the definition word by word so you retain the keywords in it.

7. Is Chapter 10 of Class 9th Science difficult?

Gravitation is a chapter that contains ideas related to our daily lives. When you link the content to things you already do, it is much easier to learn and retain the concepts in the chapter. Though the numerical portion of the chapter may be difficult to grasp, the chapter as a whole is simple to understand. You can get a fine grip on numerical problems by practising them on a regular basis.

CBSE Class 9 Science Important Questions

Cbse study materials.

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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Class 9 Science Gravitation

• 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).

• 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.

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

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

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

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Gravitation

Ncert solutions for chapter 10 gravitation class 9 science.

According to Universal Law of gravitation, the gravitational force of attraction between any two objects of mass  M  and  m  is proportional to the product of their masses and inversely proportional to the square of distance r between them So, force  F  is given by 

So as the mass of any one of the object is doubled the force is also doubled.

(ii) The force F is inversely proportional to the distance between the objects. So if the distance between two objects is doubled then the gravitational force of attraction between them is reduced to one fourth(1/4) of its original value. Similarly, if the distance between two objects is tripled, then the gravitational force of attraction becomes one ninth(1/9) of its original value.

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

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.

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.

According to question,   M S  = Mass of the Sun = 2×10 30  kg M E  = Mass of the Earth = 6×10 24  kg R  = Average distance between the Earth and the Sun = 1.5×10 11  m From Universal law of gravitation,

Let  t  be the point at which two stones meet and let  h  be their height from the ground. Height of the tower is H = 100 m (Given) It is clear from the question that we need to calculate time when the two stones met. After calculating time, we will also be able to calculate the distance. 

NCERT Solutions for Chapter 4 The Age of Industrialisation Class 10 History

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• Is Matter Around Us Pure
• Atoms and Molecules
• Structure of the Atom
• The Fundamental Unit of Life

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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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• Textbook Solutions
• Lakhmir Singh and Manjit Kaur Solutions
• Lakhmir Singh Class 9
• Class 9 Physics
• Chapter 3 Gravitation

Lakhmir Singh Solutions Class 9 Physics Chapter 3 Gravitation

Lakhmir Singh Solutions for Class 9 Physics Chapter 3 Gravitation is provided in this article. We know that the earth attracts every object towards it with the help of the force of attraction, which is known as gravitation or gravitational pull.

According to the universal law of gravitation, every object in the universe attracts every other object by the force of attraction, which is directly proportional to the product of the masses of the objects and is inversely proportional to the square of the distance between them. This is mathematically given as:

• F is the force of the gravitational pull
• G is the constant known as the gravitational constant
• M is the mass of object 1
• m is the mass of object 2
• d is the distance between object 1 and object 2

The universal law of gravitation is responsible for the motion of all the planets around the sun and for the motion of the moon around the earth. It can be said that the law of gravitation binds us to the earth.

When an object falls from a certain height which is under the influence of gravitational force, it is known as a free fall. The speed with which the object falls towards the ground is dependent on the acceleration due to gravity which is denoted by g and is equal to 9.8 m/s 2 . The unit of g is m/s 2 . There are two factors that affect the value of g, and they are as follows:

• The radius of the earth. As the radius of the earth increases from the poles towards the equator, the value of g becomes greater at the poles than at the equator.
• As the height increases, the value of g decreases.

It is important to understand that mass and weight also play an important role. Mass is defined as the quantity of matter that is contained in a body, while weight is defined as the force with which an object gets attracted towards the centre of the earth.

Lakhmir Singh Solutions for Class 9 Physics Chapter 3 Gravitation

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Very Short Answer Type Questions

Q1. What is the value of gravitational constant G

a) on the earth

b) on the moon

The value of gravitational constant G on the earth as well as on the moon = 6.67× 10 -11 Nm 2 /kg 2 .

The value of G always remains constant irrespective of the location.

Q2. When an object is dropped from a height, it accelerates and falls down. Name the force which accelerates the object.

The force that accelerates the object is the gravitational force of the earth.

Q3. Give the formula for the gravitational force F between two bodies of masses M and m kept at a distance d from each other.

The gravitational force F between two bodies of masses M and m kept at a distance d from each other is given as:

F = G × (m × M)/d 2

Where G = 6.67× 10 -11 Nm 2 /kg 2

Q4. What force is responsible for the earth revolving round the sun?

Gravitational force is responsible for the earth revolving round the sun.

Q5. What name has been given to the force with which two objects lying apart attract each other?

Gravitational force attracts the two objects lying apart from each other.

Q6. What type of force is involved in the formation of tides in the sea?

The formation of tides in the sea is because of the gravitational force.

Q7. Which force is responsible for holding the solar system together?

Gravitational force is responsible for holding the solar system together.

Q8. What is the weight of a 1kg mass of the earth? When g = 9.8 m/s 2 .

Weight, w = mg = (1) (9.8) = 9.8N

Q9. On what factor/factors does the weight of a body depend?

The weight of a body depends on the mass and on the acceleration due to gravity which varies from place to place.

Q10. As the altitude of a body increases, do the weight and mass both vary?

The weight of the body varies with altitude, whereas the mass of an object remains constant.

Q11. If the same body is taken to places having different gravitational field strengths, then what will vary: its weight or mass?

The weight of the body varies as the mass remains constant.

Q12. Which force is responsible for the moon revolving around the earth?

The revolution of the moon around the earth is because of gravitational force.

Q13. If the mass of an object is 10kg, what is its weight?

Weight, W = mg = (10)(9.8) = 98N

Q14. The weight of a body is 50N. What is its mass?

Weight, W = mg

Mass, m = W/g = 50/9.8 = 5.102kg

Q15. A body has a weight of 10kg on the surface of the earth. What will be its weight when taken to the centre of the earth?

At the centre of the earth, the value of g is zero. Therefore, the weight of the body at the centre of the earth will be zero.

Q16. Write down the weight of a 50kg mass on the earth.

= (50)(9.8)

Q17. If the weight of a body on the earth is 6N, what will it be on the moon?

The weight of the body on the moon will be one-sixth that of the earth. Therefore, the weight of the body on the surface of the moon will be 1N.

Q18. State whether the following statements are true or false.

a) A falling stone also attracts the earth.

b) The force of gravitation between two objects depends on the nature of the medium between them.

c) The value of G on the moon is about one-sixth of the value of G on the earth.

d) The acceleration due to gravity acting on a freely falling body is directly proportional to the mass of the body.

e) The weight of an object on the earth is about one-sixth of the weight on the moon.

Q19. Does the acceleration produced in a freely falling body depend on the mass of the body?

No, the acceleration produced in a freely falling body does not depend on the mass of the body.

Q20. Name the scientist who gave the three laws of planetary motion.

Johannes Kepler gave the three laws of planetary motion.

Q21. Name the scientist who explained the motion of planets on the basis of the gravitational force between the sun and the planets.

The motion of the planets on the basis of the gravitational force between the sun and the planets was explained by Newton.

Q22. State Kepler’s law which is represented by the relation r 3 ∝ T 2 .

The above relation is used in Kepler’s law of periods which states that the cube of the mean distance of a planet from the sun is directly proportional to the square of times it takes to move around the sun.

Q23. Which of Kepler’s laws of planetary motion led Newton to establish the inverse-square rule for gravitational force between two bodies?

Kepler’s third law of planetary motion led to the establishment of Newton’s gravitational force between the two bodies, which is the inverse-square rule.

Q24. Name the property of earth which is responsible for extremely small acceleration being produced in it as a result of attraction by other small objects.

The extremely large mass of the earth is responsible for extremely small acceleration that is produced in small objects.

Q25. What is the acceleration produced in a freely falling body of mass 10kg?

The acceleration produced in a freely falling body of mass 10kg is 9.8 m/s 2 .

Q26. Fill in the blanks with suitable words:

a) The acceleration due to gravity on the moon is about ………. of that on the earth.

b) In order that the force of gravitation between two bodies may become noticeable and cause motion, one of the bodies must have an extremely large ………….

c) The weight of an object on the earth is about ………… of its weight on the moon.

d) The weight of an object on the moon is about ……….. of its weight on the earth.

e) The value of g on the earth is about ……….. of that on the moon.

f) If the weight of a body is 6N on the moon, it will be about ………… on the earth.

a) One-sixth

c) Six times

d) One-sixth

e) Six times

Short Answer Type Questions

Q27. Explain what is meant by the equation:

g = G (M/R 2 )

where the symbols have their usual meanings.

The equation is the acceleration produced by the earth which is also known as acceleration due to gravity.

G is the gravitational constant

M is the mass of the earth

R is the radius of the earth

Q28. a) What do you mean by the term free fall?

b) During a free fall, will heavier objects accelerate more than the lighter ones?

a) The term free fall refers of the falling of an object towards the earth from a certain height towards the earth under the gravitation force of the earth.

b) Acceleration of an object is independent of the mass of the body during free fall.

Q29. Can we apply Newton’s third law to the gravitational force? Explain your answer.

Yes, we can apply Newton’s third law to the gravitational force as the earth exerts a force of attraction on an object such that the force exerted by other objects on the earth is equal and in the opposite direction.

Q30. Give reason for the following:

The force of gravitation between two cricket balls is extremely small, but that between a cricket ball and the earth is extremely large.

We know that the force of gravitation between two bodies is directly proportional to the product of their masses. We also know that the mass of the cricket ball is very small when compared to earth; the force of gravitation between the two cricket balls and between the ball and the earth will be extremely large.

Q31. Describe how the gravitational force between two objects depends on the distance between them.

The gravitational force F between the two bodies of masses M and m is kept at a distance d from each other such that:;

F = G (m.M/d 2 )

While the force between the two bodies is inversely proportional to the square of the distance between them, which is given as:

F ∝ 1/d 2

Therefore, as the distance between the two bodies is doubled, the gravitational force becomes one-fourth and when the distance is halved, the gravitational between these two bodies becomes four times.

Q32. What happens to the gravitational force between two objects when the distance between them is:

a) Doubled?

a) When the distance between the two objects is doubled, the gravitational force becomes one-fourth.

b) When the distance between the two objects is halved, the gravitational force becomes four times.

Q33. State two applications of the universal law of gravitation.

Following are the two applications of the universal law of gravitation:

a) It is used for determining the masses of the sun, the earth, and the moon.

b) It is also used for discovering new planets and stars.

Q34. Explain why if a stone held in our hand is released, it falls towards the earth.

When the stone held in our hand is released, it falls towards the earth because the earth exerts a force of attraction on the stone pulling it towards itself.

Q35. Calculate the force of gravitation between two objects of masses 50kg and 120kg respectively kept at a distance of 10m from one another.

F = G. (m.M/d 2 )

G = 6.7 × 10 -11 Nm 2 /kg 2

Substituting the values,

F = 4.02 × 10 -9 N

Q36. What is the force of gravity on a body of mass 150 kg lying on the surface of the earth? Assume mass of the earth = 6 × 10 24 kg, radius of the earth = 6.4 × 10 6 m, and G = 6.7 × 10 -11 Nm 2 /kg 2 .

F = G.(m.M/d 2 )

Force due to gravity,

Q37. The mass of the sun is 2 × 10 30 kg, and the mass of the earth is 6 × 10 24 kg. If the average distance between the sun and the earth be 1.5 × 10 8 km, calculate the force of gravitation between them.

Distance, d = 1.5×10 8 km = 1.5×10 11 m

Mass of the sun, m = 2×10 30 kg

Mass of the earth, M = 6×10 24 kg

Force of gravitation, F = G(m.M/d 2 )

F = 3.57×10 22 N

Q38. A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone is 9.8 m/s 2 directed towards the ground, calculate the initial velocity of the same with which it is thrown upwards.

Initial velocity of the stone, u = ?

Final velocity of stone, v = 0

Acceleration due to gravity, g = -9.8 m/s 2

Time, t = 3s

Using the relation, v = u + gt

We get, u = 29.4 m/s

Q39. A stone falls from a building and reaches the ground in 2.5 seconds later. How high is the building?

Initial velocity, u = 0 m/s

Acceleration due to gravity, g = 9.8 m/s 2

Time taken to reach the ground, t = 2.5 s

Height, h = ?

s = ut + ½ gt 2

s = 30.65 m

Q40. A stone is dropped from a height of 20m.

a) How long will it take to reach the ground?

b) What will be its speed when it hits the ground?

a) Using the relation:

t 2 = 20/5 = 4

b) For a freely falling body:

v 2 = u 2 + 2gh

Q41. A stone is thrown vertically upwards at a speed of 20m/s. How high will it go before it beings to fall?

Initial velocity, u = 20 m/s

Final velocity, v = 0

Using the relation v 2 = u 2 + 2gh

Q42. When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 meters.

a) What was the initial speed of the ball?

b) How much time is taken by the ball to reach the highest point?

Initial velocity, u = ?

Acceleration due to gravity, g = -10 m/s 2

Height, h = 5m

a) For a freely falling body:

b) Using relation, v = u + gt

Q43. Write the difference between the mass and weight of an object.

Following is the difference between the mass and weight of an object:

a) The mass of an object is defined as the quantity of matter that is contained in it, while the weight of an object is the force with which the object gets attracted towards the centre of the earth.

b) The SI unit of mass is kg, while the SI unit of weight is N.

c) The mass of an object remains unchanged while the weight of an object varies from place to place as it is dependent on acceleration due to gravity.

Q44. Can a body have mass but no weight? Give reasons for your answer.

Yes, a body can have a mass but no weight. This is because the weight of the body varies from place to place, with the value of acceleration due to gravity.

Q45. A force of 20N acts upon a body whose weight is 9.8N. What is the mass of the body, and how much is its acceleration?

Weight = 9.8N

Force, F = ma

a = 20 m/s 2

Q46. A stone resting on the ground has a gravitational force of 20N acting on it. What is the weight of the stone? What is its mass?

Weight of the stone = gravitational force acting on it = 20N

Therefore, m = 2kg

Q47. An object has a mass of 20 kg on the earth. What will be its

b) weight on the moon.

a) The mass of an object on the moon will be 20 kg.

b) The weight of the object on the moon will be (20)(1.6) = 32N.

Q48. Which is more fundamental, the mass of a body or its weight? Why?

The mass of the body is more fundamental than the weight because the mass of the body is constant and does not vary from place to place.

Q49. How much is the weight of an object on the moon as compared to its weight on the earth? Give reason for your answer.

The weight of an object on the moon, as compared to its weight on the earth is one-sixth of its weight on the earth. This is because the acceleration due to gravity on the moon is one-sixth of that on the earth.

Long Answer Type Questions

Q50. a) Define the mass of a body. What is the SI unit of mass?

b) Define the weight of a body. What is the SI unit of weight?

c) What is the relation between the mass and weight of a body?

a) The mass of the body is defined as the quantity of matter that is contained in it. The SI unit of mass is kg.

b) The weight of the body is defined as the force with which the body gets attracted towards the centre of the earth. The SI unit is N.

c) The relation between the mass and weight of a body is W = mg.

Q51. a) State the universal law of gravitation. Name the scientist who gave this law.

b) Define gravitational constant. What are the units of gravitational constant?

a) The universal law of gravitation states that every body in the universe attracts every other body with a force F which is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.

This law was given by Sir Isaac Newton.

b) The gravitational constant G is equal to the force of gravitation which exists between the two bodies of unit masses that are kept at a unit distance from each other.

The unit of gravitational constant is Nm 2 /kg 2 .

G = F × d 2 /(m × M)

Q52. a) What do you understand by the term ‘acceleration due to gravity of earth’?

b) What is the usual value of acceleration due to the gravity of earth?

c) State the SI unit of acceleration due to gravity.

a) Acceleration due to gravity is the uniform acceleration produced by the body that is falling freely due to the gravitational force of the earth.

b) Usual value of acceleration due to the gravity of earth is 9.8 m/s 2 .

c) SI unit of acceleration due to gravity is m/s 2 .

Q53. a) Is the acceleration due to the gravity of earth ‘g’ a constant? Discuss.

b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 10 22 kg and a radius of 1.74 × 10 6 m. Which satellite do you think it could be?

a) No, the value of acceleration due to gravity is not constant at all the places on the surface of the earth. This is because the radius of the earth is minimum at the poles while maximum at the equator. The value of g is maximum at the poles and minimum at the equator. The value of g increases as we go from the centre of the earth towards the poles, and the value of g decreases as we go inside the earth.

b) Acceleration due to gravity,

g = G M/R 2

Mass, M = 7.4 × 10 22 kg

Radius, R = 1.74 × 10 6 m

Gravitational constant, G = 6.7 × 10 -11 Nm 2 /kg 2

Therefore, substituting the values, we get g = 1.637 m/s 2 .

Q54. State and explain Kepler’s laws of planetary motion. Draw diagrams to illustrate these laws.

Kepler’s first law states that the planets move in an elliptical orbit around the sun, with the sun at one of the foci of the elliptical orbit.

Kepler’s second law states that each planet revolves around the sun in such a way that the line joining the planet to the sun sweeps over equal areas in equal intervals of time.

Kepler’s third law states that the cube of the mean distance of a planet from the sun is directly proportional to the square of the time it takes to move around the sun.

r 3 ∝ T 2

Q55. The mass of a planet is 6 × 10 24 kg, and its diameter is 12.8 × 10 3 km. If the value of the gravitational constant is 6.7 × 10 -11 Nm 2 /kg 2 , calculate the value of acceleration due to gravity on the surface of the planet. What planet could this be?

Acceleration due to gravity,

Mass, M = 6 × 10 24 kg

Diameter = 12.8 × 10 3 km = 12.8 × 10 6 m

Radius, R = 6.4 × 10 6 m

Substituting in the formula, g = 9.8 m/s 2 .

Multiple Choice Questions

Q56. An object is thrown vertically upwards with a velocity u, the greatest height h to which it will rise before falling back is given by:

The correct answer is b) u 2 /2g

Q57. The mass of the moon is about 0.012 times that of the earth, and its diameter is about 0.25 times that of the earth. The value of G on the moon will be:

a) less than that on the earth

b) more than that on the earth

c) same as that on the earth

d) about one-sixth of that on the earth

The correct answer is c) same as that on the earth

Q58. The value of g on the surface of the moon:

a) is the same as on the earth

b) is less than that on the earth

c) is more than that on the earth

d) keeps changing day by day

The correct answer is b) is less than that on the earth

Q59. The atmosphere consisting of a large number of gases, is held to the earth by:

c) earth’s magnetic field

The correct answer is d) gravity

Q60. The force of attraction between two unit point masses separated by a unit distance is called:

a) gravitational potential

b) acceleration due to gravity

c) gravitational field strength

d) universal gravitational constant

The correct answer is d) universal gravitational constant

Q61. The weight of an object at the centre of the earth of radius R is:

b) R times the weight at the surface of the earth

c) infinite

d) 1/R 2 times the weight at the surface of the earth

The correct answer is a) zero

Q62. Two objects of different masses falling freely near the surface of the moon would:

a) have the same velocities at any instant

b) have different accelerations

c) experience forces of the same magnitude

d) undergo a change in their inertia

The correct answer is a) have the same velocities at any instant

Q63. The value of acceleration due to the gravity of earth:

a) is the same on the equator and poles

b) is the least on poles

c) is the least on the equator

d) increases from the pole to the equator

The correct answer is c) is the least on the equator

Q64. The law of gravitation gives the gravitational force between:

a) the earth and a point mass only

b) the earth and the sun only

c) any two bodies having some mass

d) any two charged bodies only

The correct answer is c) any two bodies having some mass

Q65. The value of quantity G in the formula for gravitational force:

a) depends on mass of the earth only

b) depends on the radius of the earth only

c) depends on both the mass and radius of the earth

d) depends neither on mass nor on the radius of the earth

The correct answer is d) depends neither on mass nor on the radius of the earth

Q66. Two particles are placed at some distance from each other. If keeping the distance between them unchanged, the mass of each of the two particles is doubled, the value of gravitational force between them will become:

a) ¼ times

b) ½ times

The correct answer is c) 4 times

Q67. In the relation F = (G)(M)(m/d2) the quantity G:

a) depends on the value of g at the place of observation

b) is used only when the earth is one of the two masses

c) is the greatest on the surface of the earth

d) is of the same value irrespective of the place of observation

The correct answer is d) is of the same value irrespective of the place of observation

Q68. The gravitational force of attraction between two objects is x. Keeping the masses of the objects unchanged, if the distance between the objects is halved, then the magnitude of the gravitational force between them will become:

The correct answer is d) 4x

Q69. An apple of mass 100g falls from a tress because of gravitational attraction between the earth and the apple. If the magnitude of the force exerted by the earth on the apple is F1 and the magnitude of the force exerted by the apple on the earth is F2, then:

a) F1 is very much greater than F2

b) F2 is very much greater than F1

c) F1 is only a little greater than F2

d) F1 and F2 are exactly equal

The correct answer is d) F1 and F2 are exactly equal

Q70. According to one of Kepler’s laws of planetary motion:

a) r 2 ∝ T 3

b) r ∝ T 2

c) r 3 ∝ T 2

d) r 3 ∝ 1/T 2

The correct answer is c) r 3 ∝ T 2

Also, visit Lakhmir Singh Solutions for Class 9 Physics to get complete solutions for all chapters.

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HBSE Class 11 Latest Physics Syllabus 2024-25: Download The PDF For Free!

Hbse class 11 physics syllabus 2024-2025: the physics curriculum of the 11th class is available on the official website of the haryana board of school education. you can also download the free pdf along with the question paper design to start preparing for the exam. .

General Instructions For The Exam 

• There will be an Annual Examination based on the entire syllabus.
• The annual examination (Theory) will be of 70 Marks whereas the Practical examination will be 30 marks (Internal). Therefore, the Total annual evaluation (70+30) will be of 100 marks. For Practical examination, the criteria are as follows:

HBSE Class 11 Physics Criteria For Practical Examination 

Unit wise chapters and their weightage, hbse class 11 physics total units and chapters , unit i: physical world and measurement.

Chapter 1: Units and Measurement

UNIT II: Kinematics

Chapter 2: Motion in a Straight Line

Instantaneous Velocity and Speed, Acceleration, Kinematic Equations for Uniformly Accelerated Motion, x-t, v-t and a-t graphs

Chapter 3: Motion in a Plane

UNIT III: Laws of Motion

Chapter 4: Laws of Motion

UNIT IV: Work, Energy and Power

Chapter 5: Work, Energy and Power

UNIT V: Motion of System of Particles and Rigid Body

Chapter 6: Systems of Particles and Rotational Motion

UNIT VI: Gravitation

Chapter 7: Gravitation

UNIT VII: Properties of Bulk Matter

Chapter 8: Mechanical Properties of Solids

Stress and Strain, Hooke’s Law, Stress-Strain Curve, Elastic Moduli: Young’s Modulus, Shear Modulus, Bulk Modulus, Poisson’s Ratio, Elastic Potential Energy in a Stretched Wire, Applications of Elastic Behaviour of Materials.

Chapter 9: Mechanical Properties of Fluids

Pressure: Pascal’s Law, Variation of Pressure with Depth, Atmospheric Pressure and Gauge Pressure, Hydraulic Machines, Streamline Flow, Bernoulli’s Principle: Speed of Efflux: Torricelli’s Law, Dynamic Lift, Viscosity: Stokes’ Law, Surface Tension: Surface Energy, Surface Energy And Surface Tension, Angle of Contact, Drops and Bubbles, Capillary Rise.

Chapter 10: Thermal Properties of Matter

UNIT VIII: Thermodynamics

Chapter 11: Thermodynamics

UNIT IX: Behaviour of Perfect Gases and Kinetic Theory of Gases

Chapter 12: Kinetic Theory

UNIT X: Oscillations and Waves

Chapter 13: Oscillations

Periodic and Oscillatory Motions: Period and Frequency, Displacement, Simple Harmonic Motion, Simple Harmonic Motion and Uniform Circular Motion, Velocity and Acceleration in Simple Harmonic Motion, Force Law for Simple Harmonic Motion, Energy in Simple Harmonic Motion, The Simple Pendulum.

Chapter 14: Waves

PRACTICALS:

• Record at least 08 Experiments [with 4 from each section], to be performed by the students.
• Record at least 6 Activities [with 3 each from section A and section B], to be performed by the students.
• The Report of the project carried out by the students.
• To measure the diameter of a small spherical/cylindrical body.
• To measure dimensions of a given regular body of known mass and hence find its density.
• To measure the internal diameter & depth of a given beaker and find its volume
• To measure the diameter of a given wire.
• To measure the thickness of a given sheet.
• To measure the volume of an irregular lamina.

3. To determine the radius of curvature of a given spherical surface by a spherometer.

4. To measure the mass of two different objects using a beam balance.

5. To find the weight of a given body using the parallelogram law of vector.

6. Using a simple pendulum, plot L-T and L-T2 graphs. Find the effective length of a second’s pendulum using the appropriate Graph.

7. To study the relationship between free of limiting friction and normal reaction and to find the coefficient of friction between a block and a horizontal surface.

1. To determine Young’s Modules of elasticity of the material of a given wire.

2. To find the force constant and effective mass of a helical spring by plottingT2- m graph using the Method of oscillations.

3. To study the variation in volume with pressure for a sample of air at constant temperature by plotting graphs between P&V and P&1/V.

4. To determine the surface tension of water by capillary rise method.

5. To determine the coefficient of viscosity of a given viscous liquid by measuring the terminal velocity of a given spherical body.

6. To study the relationship between the temperature of a hot body and time by plotting a cooling curve.

7. To study the relation between the length of a given wire and tension for constant frequency using a sonometer.

8. To find the speed of sound in air at room temperature using a resonance tube by two resonance positions.

HBSE Class 11 Physics Question Typology

Scholars can easily take a look at the syllabus designed for the Class 12 Science stream. The syllabus is made as per the latest curriculum. 

HBSE Class 11 Physics Syllabus 2024-2025 Free PDF Download 

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Case Study and Passage Based Questions for Class 9 Science Chapter 9 Force and Laws of Motion

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

Case Study/Passage Based Questions:

Question 1:

Read the following and answer any four questions from (i) to (v) given below :

In the figure below the card is flicked with a push. It was observed that the card moves ahead while coin falls in glass.

(i) Give reason for the above observation. (a) The coin possesses inertia of rest, it resists the change and hence falls in the glass. (b) The coin possesses inertia of motion; it resists the change and hence falls in the glass. (c) The coin possesses inertia of rest, it accepts the change and hence falls in the glass. (d) The coin possesses inertia of rest, it accepts the change and hence falls in the glass.

(ii) Name the law involved in this case. (a) Newton’s second law of motion. (b) Newton’s first law of motion. (c) Newton’s third law of motion. (d) Law of conservation of energy

(iii) If the above coin is replaced by a heavy five-rupee coin, what will be your observation. Give reason. (a) Heavy coin will possess more inertia so it will not fall in tumbler. (b) Heavy coin will possess less inertia so it will fall in tumbler. (c) Heavy coin will possess more inertia so it will fall in tumbler. (d) Heavy coin will possess less inertia so it will not fall in tumbler.

(iv) Name the law which provides the definition of force. (a) Law of conservation of mass (b) Newton’s third law. (c) Newton’s first law (d) Newton’s second law.

(v) State Newton’s first law of motion. (a) Energy can neither be created nor be destroyed, it can be converted from one form to another, total amount of energy always remains constant. (b) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force. (c) For every action in nature there is an equal and opposite reaction. (d) The acceleration in an object is directly related to the net force and inversely related to its mass.

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Case Study and Passage Based Questions for Class 9 Science Chapter 10 Gravitation

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Case Study and Passage Based Questions for Class 9 Science Chapter 11 Work and Energy

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Case Study and Passage Based Questions for Class 9 Science Chapter 13 Why Do We Fall Ill

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