gravitation class 9 assignment pdf

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Gravitation Worksheet for Class 9 PDF with Answers

In Class 9 Science there is a chapter “Gravitation”, it is a crucial lesson for the students as they get to know all the basic topics of Gravitation. Since it is an important lesson, students should take Gravitation Worksheet Class 9 to better develop an understanding of the concepts explained.

It is very important for Class 9 students to practise questions of the chapter Gravitation because it will help them create their own exam strategy to score well in the upcoming final examination. 

In addition to that, Gravitation Worksheet Class 9 has quite interactive and creative tasks which boosts student’s creativity level.

Gravitation Worksheet With Solutions

The questions in the chapter Gravitation worksheet for Class 9 are provided with solutions. Through these solutions, students can easily solve all their doubts. By clearing the doubts of Gravitation, students can build a strong foundation. Accordingly students can also score good marks in questions related to the chapter Gravitation. The subject matter experts at Selfstudys has prepared Gravitation Worksheet With Solutions in a way that helps students answer all types of questions regardless of its difficulty.

Gravitation Worksheet Class 9 PDF

Class 9 students can easily download the Portable Document Format (PDF) of Gravitation worksheet with the help of Selfstudys website. This can help Class 9 students to understand all the topics and concepts of the chapter Gravitation which will help students increase their self-confidence level. A perfect level of self- confidence can easily decrease students' level of stress to prepare for the chapter Gravitation.

How to Download Gravitation Worksheet Class 9?

To look through the questions included in Gravitation worksheet Class 9, students can follow the given steps. These steps are the easiest one that one can follow to download Gravitation Worksheet Class 9. 

• Open the Selfstudys website. 
• Bring the arrow towards the CBSE which can be seen in the navigation bar. 
• A drop down menu will appear, select KVS NCERT CBSE Worksheet.

• A new page will appear, Class 9 from the list of classes.

• Click Science from the list of subjects.

• Again a new page will appear, now select Gravitation Worksheet from the list of chapters. 

Features of Gravitation Worksheet Class 9

Before solving questions from Gravitation worksheet Class 9, students should understand what makes Gravitation Worksheet Class 9 PDF special. Features of the worksheet are discussed below: 

• All Concepts are Covered: In Gravitation worksheet Class 9, all concepts and topics are covered in an elaborate manner in the questions format. Through this elaboration, students can understand all the topics of the chapter Earth in a better way.  
• Explained in an Easy Language: Answers of Gravitation Worksheet Class 9 is explained in an easy language which helps students easily understand the process of answering questions.  
• Varieties of Questions are Included: In the Class 9 Gravitation worksheet, varieties of questions are included. Through this students can solve all kinds of questions of the chapter Gravitation. 
• Eye Catching Format: Gravitation Worksheet of Class 9 is considered to be an eye catching one. This eye-catching format can attract many students to solve the questions of the chapter Gravitation.
• Solutions are Provided: For all the questions in the worksheet of Class 9 Gravitation, solutions are provided. Through the solutions, Class 9 students will be able to solve challenging questions which will help them develop a critical thinking capability.
• According to the Class 9 Syllabus: The questions in Class 9 Science Gravitation Worksheet are as per the Class 9 Syllabus and prescribed NCERT books. With the help of this, kids will be able to make their foundational understanding stronger.

How to Know If You're Ready for Gravitation Worksheet Class 9? 

First of all Class 9 students need to complete the chapter Gravitation from the main Science book. After covering all the topics, definitions and concepts from the chapter, students are totally ready to solve the questions from Gravitation Worksheet Class 9. Solving the questions from the Class 9 worksheet can help students to increase their conceptual understanding of Gravitation. 

What Is Gravitation Worksheet Class 9 and How to Use It?

Gravitation worksheet Class 9 is mainly given to students to practise a variety of questions. After practising questions from Gravitation worksheet, students can also look through the answers. Answers for all questions can help students to improve their practising skills. These skills can help Class 9 students to increase their level of understanding. 

Parents are advised to tell their Class 9rd kids to begin solving Gravitation Worksheet Class 9 the moment they finish their study of the chapter. Doing this, will help students brush up their all learning as well as be completed for upcoming annual exams or tests.

Advantages of Gravitation Worksheet Class 9

Solving questions from Gravitation worksheet Class 9, students can be benefited a lot. Those important advantages are: 

• Boosts Confidence Level: To solve questions from the Class 9 Gravitation worksheet can help students to boost their confidence. A perfect level of confidence can help students to improve the study process. 
• Assist the Preparation Process: Solving questions from the Class 9 Gravitation worksheet can help students in preparing for the chapter. 
• Helps in Self Evaluation: Through Gravitation worksheet, students can easily evaluate themselves. According to the self evaluation process, students can easily improve their preparation. 
• Builds a Strong Foundation: It is important for all students to solve questions from the chapter Gravitation. Regular solving questions from the worksheet can help students to build a strong foundation for the chapter Gravitation.  
• Enhances the Learning Process: Constant solving of questions from Gravitation Worksheet can enhance a student's learning process so that students can understand all topics easily.  
• Quick Revision: By solving questions from the Class 9 Gravitation worksheet, students can easily revise all the topics and concepts included in the chapter. 

Why Gravitation Worksheet Class 9 Is Right for You?

It is a must for students to exercise questions from Gravitation worksheet Class 9 as perfectly right for them. With the help of Gravitation worksheet Class 9 questions, students can increase their capability of solving questions in a different and creative manner.

Tips to Understand All Questions of Gravitation Worksheet Class 9 in a Better Way: 

Students should understand all questions of Gravitation worksheet Class 9 in a better way. Better understanding of questions can help students to score good marks in questions which are related to the chapter Gravitation. Those important tips are: 

• Finish Off The Chapter: First and foremost tip is to finish off the chapter Gravitation. Students need to complete each and every topic included in the lesson. 
• Practise Questions: After completing the chapter Gravitation, students need to practise questions from the Class 9 worksheet. Routine practice of questions can help students in identifying their strengths and weaknesses. 
• Note Down the Mistakes: While practising questions, it is very important that students note down their mistakes. Noting down the mistakes is very important as accordingly students can improve their preparation strategies. 
•  Correction of Mistakes: After noting down the mistakes, it is a must to correct all the mistakes made. Correction of mistakes can help students to solve worksheet questions in a better way. 
• Maintain a Positive Attitude: While solving questions from the Class 9 Gravitation worksheet, students need to maintain a positive attitude. A positive attitude can help students to remove stress and anxiety while preparing for the chapter Gravitation. 
• Remain Focused: To understand questions of Gravitation worksheet, students need to remain focused while preparing for it. 

Why Should Students Start Solving Gravitation Worksheet Class 9 From the PDF?

Gravitation Class 9 Worksheet is provided in the PDF so that students don’t need to search for them here and there. By solving the questions from Gravitation worksheet Class 9, students can understand the chapter in a fine way. Routine solving of questions from the Class 9 Science Worksheet can help students increase their comprehension skills. Comprehension skills will help in performing outstanding in the final examination. 

What are Included in Gravitation Worksheet Class 9?

In Gravitation worksheet Class 9, questions from the chapter are included. After solving questions of the chapter Gravitation, students can also go through the answers included in the worksheet. Answers to these questions of Gravitation are explained in a detailed manner. As it can also help teachers to make students understand in a better and elaborate way. Through this students can easily identify their skills and flaws for the Science chapter Gravitation.

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NCERT Solutions for Class 9 Science Chapter 9 Gravitation

• 10th June 2023

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

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

CBSE Class 9 Science Gravitation Intext Questions (Solved)

PAGE NO. 102

Question 1: State the universal law of gravitation .

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

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

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

PAGE NO. 104

Question 1: What do you mean by free fall? 

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

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

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

PAGE NO. 106

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

Answer: 

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

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

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

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

PAGE NO. 109

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

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

Question 2: What do you mean by buoyancy? 

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

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

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

PAGE NO. 110

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

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

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

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

CBSE Class 9 Science Gravitation Exercise Questions and Answers

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 8: What is the acceleration of free fall?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence, the maximum height is 44.1 m.

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

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

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

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

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

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

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

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

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

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

Answer:  Density of the 500 g sealed packet

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

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• Gravitation Class 9 Notes CBSE Science Chapter 10 (Free PDF Download)
• Revision Notes

CBSE Class 9 Science Chapter 10 - Gravitation Revision Notes - Free PDF Download

Class 9 Science introduces you to the most fascinating chapters and concepts you have been studying for the past few years. In Class 9 Science, you will delve a little deeper and get to know more about various natural phenomena and learn how they happen. Class 9 Science Chapter 10 is on gravitation and the things happening around us related to it. To understand the chapter better, you can refer to the Gravitation Class 9 notes prepared by the highly experienced teachers at Vedantu.

You will be able to understand the concepts related to gravitation, law of gravitation from these revision notes and prepare the chapter well. Every section of these notes has been compiled in such a way that students can understand the new concepts, learn the formulas, and can solve various questions easily with the help of these notes. By referring to the Class 9 Science Notes Chapter 10 Gravitation, you will be able to prepare for the exam faster and learn the concepts better. Download the revision notes on your smart devices and use them at your convenience. Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. Maths Students who are looking for the better solutions, they can download Class 9 Maths NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Download CBSE Class 9 Science Revision Notes 2024-25 PDF

Also, check CBSE Class 9 Science revision notes for All chapters:

Access Class IX Science Chapter 10 - Gravitation Notes in 30 Minutes

Toss a stone from a great height. What are your observations? 

The stone, which was at first at rest, begins to move towards the ground and reaches its maximum speed right before it meets it. 

The stone is not travelling at a constant rate. Its speed fluctuates at all times, indicating that the stone is accelerating.

A force is necessary to cause an acceleration in a body, according to Newton's second law of motion. 

The stone was not pushed or pulled in any way. What was the source of the force? 

Sir Isaac Newton came up with the solution to this dilemma after seeing an apple fall from a tree. 

His thesis was that the apple is attracted to the Earth, and the Earth is attracted to the apple The Earth's force on the apple is enormous, and as a result, the apple arrives on Earth. 

The apple, on the other hand, is unable to draw the Earth since the force it exerts on it is insignificant. 

As a result, we can deduce that the acceleration caused by Earth's immense force of attraction is the cause of the stone's acceleration.

(image will be uploaded soon)

It is evident from the preceding example that this force of attraction ties our complicated universe together, keeps the moon revolving around the Earth, keeps all of the planets in their orbits around the Sun, and helps us walk correctly on the Earth's surface. 

The force of gravitation, or gravitation, is a form of attraction that exists between any two objects in the universe.

The force of gravity or gravity is the attraction or gravitational force between Earth (or any planet) and any other material objects in the cosmos.

The Universal Law of Gravitation or Newton's Law of Gravitation:

The universal law of gravitation is a mathematical relationship that Sir Isaac Newton proposed to measure the gravitational force.

According to this law “Every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, the direction of the force being along the line joining the masses”.

Consider two mass \[{{m}_{1}}\]and \[{{m}_{2}}\] objects separated by a distance \[d\] . The gravitational force \[F\] is proportional to the product of the masses, according to Newton's law.

\[\text{F }\propto \text{ }{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}\text{   }......\text{ }\left( \text{1} \right)\]

and inversely proportional to the square of the distance between the masses

\[\text{F }\propto \text{ }\frac{\text{1}}{{{\text{d}}^{\text{2}}}}\text{       }......\text{ }\left( \text{2} \right)\]

Two Objects of Masses \[{{m}_{1}}\] and \[{{m}_{2}}\] separated by a distance \[d\].

Inversely proportional is always represented as directly proportional to the reciprocal of that quantity.

Combining equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] , we get

$\text{F }\propto \text{ }\frac{{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}}{{{d}^{2}}}$ 

$\text{F }\propto \text{ }\frac{\text{G}{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}}{{{\text{d}}^{\text{2}}}}\text{    }.......\text{ }\left( \text{3} \right)$ 

Where $G$ is a proportionality constant known as the universal gravitational constant. 

$G$ is known as the universal constant because its value remains constant throughout the cosmos and is unaffected by object masses.

Universal Gravitational Constant:

The mathematical form of Newton's Law of Gravitation is

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

If \[{{\text{m}}_{\text{1}}}\text{ = }{{\text{m}}_{\text{2}}}=1\] , and \[\text{d = 1}\] , then

$\text{F = }\frac{\text{G  }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ 1}}{{{\text{1}}^{\text{2}}}}$

$\text{F = G}$ 

As a result, the universal gravitational constant can be defined as the gravitational force that exists between two unit masses separated by a unit distance.

\[SI\]unit of gravitational constant:

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

\[SI\] Unit of force \[F\] is \[N\], \[SI\] unit of distance is metres and that of mass is \[kg\].

\[SI\] Unit of \[\text{G = }\frac{\text{N}{{\text{m}}^{2}}}{\text{kg}\times \text{kg}}\]

\[SI\] Unit of \[\text{G = N}\frac{{{\text{m}}^{2}}}{\text{k}{{\text{g}}^{2}}}\] or \[\text{N}{{\text{m}}^{2}}\text{k}{{\text{g}}^{-2}}\]

The experimental value of \[G\] equal to \[\text{6}\text{.6734}\times \text{1}{{\text{0}}^{\text{11}}}\text{ }\frac{\text{N}{{\text{m}}^{\text{2}}}}{\text{k}{{\text{g}}^{\text{2}}}}\] was measured by Sir Henry Cavendish in \[1798\] .

Dependence of Gravitational Force on Mass:

The force of attraction is directly proportional to the mass of the body, according to Newton's law of gravitation.

1. Two objects of mass \[m\]separated by a distance \[d\]:

If two objects of mass \[m\] are separated by a distance \[d\], the force between them is given by the relation are as follows;

${F_1}=\frac{\text{Gmm}}{{{\text{d}}^{\text{2}}}}$ 

${F_1}=\frac{\text{G}{{\text{m}}^{2}}}{{{\text{d}}^{\text{2}}}}$ 

2. Two objects of mass \[m\]and \[2m\] separated by a distance \[d\]:

When the mass of one of the two objects is doubled, force of attraction is given by the relation are as follows;

${{\text{F}}_{2}}\text{ = }\frac{\text{G}{{\text{m}}_{2}}\text{m}}{{{\text{d}}^{\text{2}}}}$

$\text{    = }\frac{\text{2G}{{\text{m}}^{2}}}{{{\text{d}}^{\text{2}}}}$

$\text{    = 2}{{\text{F}}_{\text{1}}}$ 

3. Two Objects of Mass \[2m\] Separated by a Distance \[d\]:

When the masses of both bodies are doubled, the force of attraction is given as;

${{\text{F}}_{3}}\text{ = }\frac{{{\text{G}}_{2}}{{\text{m}}_{2}}\text{m}}{{{\text{d}}^{\text{2}}}}$ 

${{\text{F}}_{3}}\text{ = }\frac{\text{4G}{{\text{m}}^{2}}}{{{\text{d}}^{\text{2}}}}$

${{\text{F}}_{3}}\text{ = } 4 {{\text{F}}_{\text{1}}}$

That is whenever the mass increases the force of attraction also increases.

Dependence of Gravitational Force on Distance:

The force of attraction between two bodies is inversely proportional to the square of their distance, according to the universal law of gravitation.

1. Force of Attraction Between Two Bodies of Mass \[m\] Separated by a Distance \[d\] :

${{\text{F}}_{1}}\text{ = }\frac{\text{G}{{\text{m}}_{1}}{{\text{m}}_{2}}}{{{\text{d}}^{\text{2}}}}$

$\text{    = }\frac{\text{G}{{\text{m}}^{2}}}{{{\text{d}}^{\text{2}}}}$

Here, two bodies of mass \[m\] are separated by a distance \[d\] and hence,

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

2. The Force of Attraction when the Distance is Doubled:

${{\text{F}}_{2}}\text{ = }\frac{\text{Gmm}}{{{\left( \text{2d} \right)}^{\text{2}}}}$

${F_2} = \frac{\text{G}{{\text{m}}^{2}}}{\text{4}{{\text{d}}^{\text{2}}}}$

Here, two bodies of mass \[m\] are separated by a distance \[2d\] and therefore,

${{\text{F}}_{2}}\text{ = }\frac{1}{4}\frac{\text{G}{{\text{m}}^{2}}}{{{\text{d}}^{\text{2}}}}$ 

${{\text{F}}_{2}}\text{ = }\frac{1}{4}{{F}_{1}}$

3. Force of Attraction When the Distance Between the Bodies is Increased Three Times:

${{\text{F}}_{3}}\text{ = }\frac{\text{Gmm}}{{{\left( \text{3d} \right)}^{\text{2}}}}$ 

$\text{    = }\frac{\text{G}{{\text{m}}^{2}}}{\text{9}{{\text{d}}^{\text{2}}}}$ 

Here, two bodies of mass \[m\] are separated by a distance \[3d\] and therefore,

${{\text{F}}_{3}}\text{ = }\frac{1}{9}\frac{\text{G}{{\text{m}}^{2}}}{{{\text{d}}^{\text{2}}}}$

${{\text{F}}_{3}}\text{ = }\frac{1}{9}{{F}_{1}}$

It Results That:

When the distance is doubled, the force is decreased to \[\frac{1}{4}\] th of its original value.

If the distance is extended three times, the force is reduced to \[\frac{1}{9}\] th of its original value. 

We can conclude from the preceding example that the force of attraction between the bodies varies inversely with the square of the distance between them.

Gravitational Force Between Two Light Objects:

Let us now calculate the force of gravitation existing between two unit masses separated by a unit distance.

${{\text{m}}_{\text{1}}}\text{ = 1 kg}$  

${{\text{m}}_{\text{2}}}\text{ = 1 kg}$ 

$\text{   d = 1 m}$

$\text{  G = 6}\text{.6734}\times \text{1}{{\text{0}}^{\text{-11}}}\frac{\text{N}{{\text{m}}^{\text{2}}}}{\text{k}{{\text{g}}_{\text{2}}}}$ 

Therefore, force can be calculated as;

$\text{F = }\frac{\text{G}{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}}{{{\text{d}}^{\text{2}}}}$

$\text{F = }\frac{\text{6}\text{.6734}\times \text{1}{{\text{0}}^{\text{-11}}}\text{ }\!\!\times\!\!\text{ 1 }\!\!\times\!\!\text{ 1}}{{{\text{1}}^{\text{2}}}}$

$\text{F = 6}\text{.6734}\times \text{1}{{\text{0}}^{\text{-11}}}\text{ N}$

This is a very weak force.

To find Force existing between two objects of masses \[\text{60 kg}\] and \[\text{100 kg}\] separated by a distance of \[\text{1 m}\] :

$\text{F = }\frac{\text{6}\text{.6734}\times \text{1}{{\text{0}}^{\text{-11}}}\text{ }\!\!\times\!\!\text{ 60 }\!\!\times\!\!\text{ 100}}{{{\text{1}}^{\text{2}}}}$

$\text{F = 6}\text{.6734}\times \text{1}{{\text{0}}^{\text{-8}}}\times \text{6 }$

$\text{F = 40}\text{.04}\times \text{1}{{\text{0}}^{\text{-8}}}\text{ N}$

This is also a very weak force.

It is clear from the preceding two examples why we do not feel the force produced by one object (on the Earth's surface) on the other.

Gravitational Force between Massive Objects:

Let's calculate the force of attraction between a \[\text{50 kg}\] object and the Earth. The Earth's mass is approximately \[\text{6}\times \text{1}{{\text{0}}^{\text{24}}}\text{ kg}\]. The Earth's distance from the object is roughly \[\text{64}\times \text{1}{{\text{0}}^{5}}\text{ m}\]. 

The force of attraction between the object and the Earth is,

$\text{   = }\frac{\text{6}\text{.6734}\times \text{1}{{\text{0}}^{\text{-11}}}\text{ }\!\!\times\!\!\text{ 6}\times \text{1}{{\text{0}}^{24}}\text{ }\!\!\times\!\!\text{ 50}}{{{\left( 64\times {{10}^{5}} \right)}^{\text{2}}}}$

$\text{   = }\frac{\text{6}\text{.6734 }\!\!\times\!\!\text{ 6 }\!\!\times\!\!\text{ 5}\times \text{1}{{\text{0}}^{-11}}\times \text{1}{{\text{0}}^{25}}}{{{\left( 64 \right)}^{2}}\times {{10}^{10}}}$

$\text{   = }\frac{\text{6}\text{.6734 }\!\!\times\!\!\text{ 6 }\!\!\times\!\!\text{ 5}\times \text{1}{{\text{0}}^{-11}}\times \text{1}{{\text{0}}^{25}}\times {{10}^{-10}}}{64\times 64}$

$\text{F = }\frac{\text{6}\text{.6734 }\!\!\times\!\!\text{ 6 }\!\!\times\!\!\text{ 5}\times \text{1}{{\text{0}}^{4}}}{64\times 64}$

$\text{ F  = }\frac{\text{6}\text{.6734 }\!\!\times\!\!\text{ 3}\times \text{1}{{\text{0}}^{5}}}{64\times 64}$

$\text{F = 488}\text{.7 N}$

This force is strong, we cannot ignore it.

Now let us calculate the force of attraction between Earth and the Sun. The mass of the Earth \[\text{= 6}\times \text{1}{{\text{0}}^{\text{24}}}\text{ kg}\]

The mass of the Sun \[\text{= 1}\text{.99}\times \text{1}{{\text{0}}^{30}}\text{ kg}\]

The distance between the Earth and the Sun \[\text{= 15}\times \text{1}{{\text{0}}^{10}}\text{ m}\]

$\text{   = }\frac{\text{6}\text{.6734}\times \text{1}{{\text{0}}^{\text{-11}}}\text{ }\!\!\times\!\!\text{ 6}\times \text{1}{{\text{0}}^{24}}\text{ }\!\!\times\!\!\text{ 1}\text{.99}\times \text{1}{{\text{0}}^{30}}}{{{\left( 15\times {{10}^{10}} \right)}^{\text{2}}}}$

$\text{   = }\frac{\text{6}\text{.6734 }\!\!\times\!\!\text{ 6 }\!\!\times\!\!\text{ 1}\text{.99}\times \text{1}{{\text{0}}^{-11}}\times \text{1}{{\text{0}}^{24}}\times \text{1}{{\text{0}}^{30}}\times \text{1}{{\text{0}}^{-20}}}{225}$

$\text{F = }\frac{\text{6}\text{.6734 }\!\!\times\!\!\text{ 6 }\!\!\times\!\!\text{ 1}\text{.99}\times \text{1}{{\text{0}}^{23}}}{225}$

$\text{F = 3}\text{.541 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{22}}}\text{ N}$

This force is very large and it is this force that keeps the planets in their respective orbits.

Conclusion: 

When two objects of ordinary size are considered, the gravitational force is quite tiny, but when at least one of the items is huge, the force is very large.

Centre of Gravity:

Every particle is drawn to the centre of the earth, as we all know. A body is made up of number of particles. Because the body is small in comparison to the earth, the gravitational attraction acting on these particles can be considered parallel to one another, as illustrated in the diagram.

A single force acting vertically in the downward direction can be replaced by a single force acting through a fixed location called the body's centre of gravity. 

The resulting force is equivalent to the body's weight.

As a result, the centre of gravity is the point through which the body's weight acts regardless of its position. 

For bodies which are of regular shape and which have uniform density, the centre of gravity lies at the geometrical centre of the body.

The geometrical centre of gravity of bodies of regular shape and uniform density is located at the geometrical centre of the body.

Application of Newton's Law of Gravitation:

One of the important applications of Newton's law is to estimate masses of binary stars. A binary star is a system of two stars orbiting around their common centre of mass.

Any irregularity in the motion of a star indicates that it might be another star or a planet going around the stars. This regularity in the motion of a star is called a wobble.

Mass and Weight:

Mass and weight are sometimes confused, yet they are two distinct numbers. Let's try to figure out what makes them different.

The mass of a body is defined as the amount of matter contained in it. 

The kilogramme is the \[SI\] unit of mass \[\left( kg \right)\]. 

Mass is a scalar quantity. 

The amount of matter contained in a body does not vary with time or location, i.e., the mass of a body remains constant throughout the cosmos. However, the masses of the two bodies might differ significantly. 

A pan balance is used to determine a person's mass.

Weight is defined as the force with which an object is pulled towards the centre of the Earth. 

\[\text{Weight of a body = force exerted by the Earth = mg}\] (according to Newton's second law of motion)

\[\text{W = mg}\]

\[SI\] unit of weight is Newton.

For example, the weight of a body having a mass of \[1kg\] is 

$\text{W = mg}$

$\text{W = 1}\times \text{9}\text{.8}$

$\text{W = 9}\text{.8 N}$

We know that \[\text{kg wt}\] is commonly used as the unit of weight.

\[\text{1kg weight}\]is the force with which an object of mass \[1kg\] is pulled towards the Earth.

       \[\text{W = mg}\]

$\text{1 kg wt = 1}\times \text{9}\text{.8}$

$\text{1 kg wt = 9}\text{.8 N}$

Spring balance is used to determine weight. 

Weight fluctuates from location to place because it is affected by gravity's acceleration. 

At the poles, a body weighs more than in the equator, and at the centre of the Earth, a body's weight becomes zero because gravity's acceleration is zero.

Difference Between Mass and Weight:

To show that weight of a body on moon is \[\frac{1}{6}\] th its weight on Earth:

Let \[m\] be the mass of a body on Earth. Its weight on Earth is given by the equation is as follows;

 ${{\text{W}}_{e}}\text{ = m}{{\text{g}}_{e}}$

${{\text{W}}_{\text{e}}}\text{ = }\frac{\text{mG}{{\text{M}}_{\text{e}}}}{{{\text{R}}_{\text{e}}}^{\text{2}}}\text{     ----}\left( \text{1} \right)$

\[\left[ {{\text{g}}_{\text{e}}}\text{ = }\frac{\text{G}{{\text{M}}_{\text{e}}}}{{{\text{R}}_{\text{e}}}^{\text{2}}} \right]\]

The weight of the same body on moon \[\left( Wm \right)\] is given by,

${{\text{W}}_{m}}\text{ = m}{{\text{g}}_{m}}$

${{\text{W}}_{m}}\text{ = }\frac{\text{mG}{{\text{M}}_{m}}}{{{\text{R}}_{m}}^{\text{2}}}\text{     ----}\left( 2 \right)$

\[\left[ {{\text{g}}_{m}}\text{ = }\frac{\text{G}{{\text{M}}_{m}}}{{{\text{R}}_{m}}^{\text{2}}} \right]\]

By dividing equation \[\left( 2 \right)\] by equation \[\left( 1 \right)\] we get,

$\frac{{{\text{W}}_{m}}}{{{\text{W}}_{e}}}\text{ = }\frac{\text{mG}{{\text{M}}_{m}}}{{{\text{R}}_{m}}^{\text{2}}}\times \frac{{{\text{R}}_{e}}^{\text{2}}}{\text{mG}{{\text{M}}_{e}}}$

 $\frac{{{\text{W}}_{m}}}{{{\text{W}}_{e}}}\text{ =}\frac{{{\text{M}}_{m}}{{\text{R}}_{e}}^{\text{2}}}{{{\text{R}}_{m}}^{\text{2}}{{\text{M}}_{e}}}$

$\frac{{{\text{W}}_{m}}}{{{\text{W}}_{e}}}\text{ =}\left( \frac{{{\text{M}}_{m}}}{{{\text{M}}_{e}}} \right){{\left( \frac{{{\text{R}}_{e}}}{{{\text{R}}_{m}}} \right)}^{2}}\text{      ----}\left( \text{3} \right)$

But we know that , \[{{\text{M}}_{\text{e}}}\text{ = 100}{{\text{M}}_{m}}\] and \[{{\text{R}}_{\text{e}}}\text{ = 4}{{\text{R}}_{m}}\] , therefore

$\frac{{{\text{M}}_{\text{m}}}}{{{\text{M}}_{\text{e}}}}\text{ = }\frac{\text{1}}{\text{100}}$

$\frac{{{\text{R}}_{\text{e}}}}{\text{ }{{\text{R}}_{\text{m}}}}\text{ = 4}$ 

By substituting above values in equation \[\left( 3 \right)\], we get

$\frac{{{\text{W}}_{m}}}{{{\text{W}}_{e}}}\text{ =}\left( \frac{1}{100} \right){{\left( 4 \right)}^{2}}$

$\frac{{{\text{W}}_{m}}}{{{\text{W}}_{e}}}\text{ =}\frac{16}{100}$

$\frac{{{\text{W}}_{m}}}{{{\text{W}}_{e}}}\text{ =}\frac{1}{6.25}$

$\frac{{{\text{W}}_{m}}}{{{\text{W}}_{e}}}\text{ =}\frac{1}{6}$ 

\[{{\text{W}}_{m}}\text{ = }\frac{1}{6}{{\text{W}}_{e}}\]

That is the weight of a body on moon is \[\frac{1}{6}\] th its weight on Earth.

Weightlessness:

We frequently hear that astronauts in space experience weightlessness. What exactly does this imply? 

Let us show weightlessness with a simple experiment. Suspend a stone from a spring balance, and the weight of the stone is displayed on the spring balance's pointer. 

Allow the stone, as well as the spring balance, to fall freely. 

The spring balance registers $0$ weight, showing that the stone is devoid of weight. 

Does this imply that the stone has no weight? 

The stone, on the other hand, is in a state of weightlessness because it is falling freely. 

When the weight of one object is balanced against the weight of another, the body becomes aware of its own weight.

Let us now attempt to explain why an astronaut in a spaceship feels weightless. 

When an astronaut is orbiting the Earth in a spaceship, both the person and the spaceship are in free fall towards the Earth. 

During a free fall, both go downhill with the same acceleration, which is equal to gravity's acceleration. 

As a result, the astronaut exerts no force on the spaceship's sides or floor, and the spaceship's sides and floor do not push the astronaut up. 

As a result, the astronaut feels weightless while orbiting the Earth in a spaceship.

Cotton takes up more room than iron, hence $\text{0}\text{.5 kg}$ of cotton takes up more space than $\text{0}\text{.5 kg}$ of iron.

Iron particles are tightly packed, whereas cotton particles are loosely packed. There is more iron packed into a given volume. 

This explains why iron is heavier than cotton of the same volume.

A substance's density is defined as the mass per unit volume of the substance.

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

$\text{D = }\frac{\text{M}}{\text{v}}$

Where, $\text{D}$ represents the density, $\text{M}$ mass and $\text{v}$ volume.

$\text{SI}$ unit of density is $\frac{\text{kg}}{{{\text{m}}^{\text{3}}}}$

When specific criteria are met, a substance's density remains constant. 

As a result, one of a material's distinguishing characteristics is its density, which may be used to determine the purity of any substance.

Relative Density of a Substance:

We utilise the relation $\text{D = }\frac{\text{M}}{\text{v}}$ to determine the density of a substance or an item by determining the mass and volume of the substance. 

Only if the thing has a regular shape is this possible.

Measuring the size of an object with an irregular shape is difficult. 

In such instances, we express the object's density in terms of water density.

The relative density of a substance is the ratio of its density to the density of water at $\text{4}$ degrees Celsius. The relative density of water is assumed to be one.

What does it mean when someone says gold's relative density is $\text{19}\text{.3}$ ?

It means that gold has $\text{19}\text{.3}$ times the density of water of the same volume.

Objects with a relative density less than one float in water, while those with a density larger than one sink.

$\text{Relative density of a substance = }\frac{\text{Density of the substance}}{\text{Density of water at }{{\text{4}}^{\text{0}}}\text{C}}$

$\text{                                                  = }\frac{\left( \frac{\text{Mass of the substance}}{\text{Volume of the substance}} \right)}{\left( \frac{\text{Mass of water}}{\text{Volume of water}} \right)}$

$\text{                                                  =}\frac{\text{Mass of the substance}}{\text{Volume of the substance}}\text{ }\!\!\times\!\!\text{ }\frac{\text{Volume of water}}{\text{Mass of water}}$

Now, if we take equal amounts of the material and water, we get the following:

\[\text{Relative density of a substance = }\frac{\text{Mass of the substance}}{\text{Volume of an equal volume of water}}\]

The relative density has no unit because it is a ratio of two identical quantities.

Thrust and Pressure:

We defined force as an external agent that modifies the direction of motion, speed, or shape of the body at the start of this chapter. 

We were simply talking about the forces acting at a spot on a body the whole time. 

Consider the forces at work in a given location.

If you want to hang a poster on your classroom bulletin board, you must apply force to the head of the drawing pin, which is perpendicular to the surface of the bulletin board. 

This force, which is acting perpendicular to the surface, is referred to as thrust.

The force operating on a body perpendicular to its surface is known as thrust.

The Newton is the $\text{SI}$ unit of thrust $\left( \text{N} \right)$ .

Let's see if there's a relationship between the applied force (thrust) and the region on which it acts.

Hold a pin erect in the middle of a stack of papers. Another pin should be placed next to it, upside down, with its flat head resting on the pile.

Place a flat object, such as a duster, on both of these pins to press them down. We notice that the erect pin pierces through the stack of papers.

This is because the force operating on the erect pin is applied over a limited area, but the force acting on the second pin is applied over a broad area. 

A thin yet durable string strap is used to hold your bag.

Now, using a wide cloth band as a strap, raise the same bag. A school bag with a wide textile band is more comfortable to carry than one with a tiny strip. 

This is because the weight of the books is dispersed over a larger area of the shoulder in the second example, exerting less force.

As can be seen from the examples above, the efficiency of the applied force is dependent on the region on which it acts.

There is now a requirement to define a new physical quantity known as pressure.

The force operating on a unit area is known as pressure.

$\text{Pressure = }\frac{Force}{Area}$ 

$\text{Pressure = }\frac{Thrust}{Area}$

The $\text{SI}$ unit of pressure is $\frac{N}{{{m}^{2}}}$ .

$\frac{N}{{{m}^{2}}}$ is known as Pascal $\left( Pa \right)$ in honor of the French Scientist Blaise Pascal.

\[\text{1 }\frac{\text{N}}{{{\text{m}}^{\text{2}}}}\text{ = 1 Pascal}\]

Because a kilopascal is a very small unit, we frequently utilise it. 

The force exerted and the area over which the force acts are the two factors that determine pressure.

Buoyancy and Archimedes' Principle:

It is general knowledge that when bodies are submerged in water or any other liquid, they appear lighter. 

While bathing, we notice that as soon as the mug of water rises over the water's surface, it becomes noticeably heavier. 

When a fish is taken out of the water, it looks to be heavier in the air than it was in the water. 

Let's have a look at why that is.

Because the liquid or water exerts an upward force on the items immersed in it, they appear to be lighter in water or any other liquid. 

Let's see if there is an apparent loss of weight while immersed in water by conducting an experiment.

Attach a stone to one of the spring balance's ends. Suspend the spring balance in the manner depicted in the diagram.

Experimental Set Up to Prove Archimedes' Principle:

Take note of the spring balance reading. Let's call it ${W_1}$. 

Now carefully immerse the stone into a jar of water and record the reading on the spring balance. 

The spring balance's reading continues to drop until it is entirely submerged in water. 

The weight of the stone is determined by the reading on the spring balance.

We may deduce that the weight of the object is reducing as it is dropped in water because the reading continues to decrease. 

The apparent weight loss indicates that a force is working on the object in an upward direction, causing it to lose weight.

The buoyant force is the upward force exerted on an object immersed in a liquid that causes the object to appear to lose weight.

Buoyancy is defined as a liquid's tendency to exert an upward force on an object placed in it, causing it to float or rise.

Factors Affecting the Buoyant Force:

We know that an iron nail sinks when placed on the surface of water, whereas an iron ship floats. This is due to the ship's larger size or volume.

When an iron nail and a cork of equal mass are placed in water, the iron nail sinks because the density of the iron nail is greater than that of the water, whilst the density of the cork is lower.

When the density of the liquid exceeds the density of the body's material, the body floats due to the buoyant force it exerts, and vice versa.

The buoyant force experienced by a body while submerged in a liquid is dependent on the volume of the body and the density of the liquid, as shown in the examples above.

Archimedes' Principle:

Archimedes investigated the up thrust acting on a body when it is partially or entirely submerged in a fluid, conducting various tests, and finally stating the Archimedes' Principle.

When a body is partially or completely immersed in a fluid, it feels an up thrust (buoyant force) equal to the weight of the liquid displaced, according to this principle.

Experiment to Verify Archimedes' Principle:

Using a physical balance, determine the mass $\left( m \right)$ of a clean and dry beaker. 

Suspend a stone from a spring balance to determine its weight. Fill a Eureka can (a beaker with a spout towards the top) with water until it reaches the spout. Place the mass $m$ beaker under the spout.

Gently lower the solid into the Eureka can, suspended from spring balance, until the stone is entirely immersed in water. 

When submerged in water, the stone displaces a particular amount of water. 

The spring balance registers a lower value, indicating that the solid is up thrust. The water that has been displaced is collected in the beaker.

The mass of the water and beaker is calculated using the physical balance. Let's call it ${{m}_{1}}$. 

Therefore,  

\[\text{Amount of water displaced = }{{\text{m}}_{\text{1}}}\text{- m }\]

When the apparent loss of weight of the solid in water is compared to the amount of water displaced, they are found to be equal. As a result, this experiment proves Archimedes' Principle.

Application of Archimedes' Principle:

It's employed in the construction of ships and submarines. 

This principle underpins lactometers and hydrometers, which are used to determine the density of liquids and measure the purity of a sample of milk.

Gravitation: Summary of Class 9 Chapter 10 Revision Notes

In the previous classes, you have studied gravitation multiple times. You know why the planets are revolving around the sun and can also explain why an apple falls on the ground rather than going upwards. Hence, you have a simpler idea of how gravitation works. Let us take a quick look into the chapter to understand what Class 9 Science Chapter 10 notes are going to offer you.

The notes will describe properly the Universal Law of Gravitation. This hypothesis has been stated as law as every object in the entire universe follows it to attract or get attracted by the other celestial bodies. According to the law, the gravitational force between two bodies can be calculated. The terms included in the law are properly explained in Chapter 10 Science Class 9 notes so that you can relate the expressions of the law with the concept. Concentrate on every term explained so that you can use it as a formula to solve the problems later.

Calculating acceleration due to gravity will become a  lot easier. After understanding the law of gravitation, the chapter will proceed to this section. Here, you will find out how to calculate the speed of a freely falling body. If you refer to the Science Ch 10 Class 9 notes, you will find out how the earth attracts all bodies towards its center. This acceleration due to gravity can also be used in Newton’s Laws of Motions. Learn how acceleration due to gravity is calculated and defined.

After understanding the law of gravitation and acceleration in the Class 9 Ch 10 Science notes, you will learn the concept of the moon’s revolution around the earth without falling into the earth. As per this section, all bodies travel in a straight line in space when not disturbed or a force is applied. Proceeding further, you will discover how a centripetal force acts that make a body rotate around an axis of another body even if there is a gravitational force in between them. With the help of Class 9 Science Ch 10 notes, you can easily grab hold of this concept and learn how centripetal force acts.

This chapter will also describe mass, weight, and their differences. Learn what motion and free fall are. Use the terms you have learned here to define these terms. Understand how these terms are characterized by the signs. In fact, all the equations relate to the laws of motion can be used to calculate free fall and gravitational motion. With the help of Gravitation Chapter Class 9 notes, you will define thrust, pressure, and will be able to explain Archimedes Principle. Here, you will understand what relative density is and how it can be used. Proceed to Kepler’s Law and understand how a planet moves in its orbit around the sun.

Why are Revision Notes for Class 9 Chapter 10 - Gravitation Important?

Vedantu’s Revision Notes for Class 9 Chapter 10 - “Gravitation” are the finest material to understand and practice the topics in the best way.

These notes focus on key concepts and formulas, ensuring you save time and focus on what truly matters.

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For a rich understanding of Gravitation in Class 9, Vedantu's well-crafted Gravitation Class 9 notes, designed by experienced educators, are your valuable study companions. These notes simplify the challenges of gravitation, breaking them down into digestible sections for an easier grasp of concepts, formulas, and problem-solving. Diving into Chapter 10 of Class 9 Science Notes on Gravitation not only streamlines exam readiness but also gives a deeper subject understanding.

FAQs on Gravitation Class 9 Notes CBSE Science Chapter 10 (Free PDF Download)

1. What is the S.I. unit of gravitation in NCERT Class 9 Gravitation chapter?

The SI unit of gravity is the newton per kilogram (N/kg). Gravity is the force of attraction between two objects with mass. You can learn more about the S.I. unit of gravitation in revision notes provided by Vedantu.

2. What is the formula for gravity in NCERT class 9 gravitation?

$F = \dfrac{Gm_1m_2}{r^2}$. This is the formula of gravity in class 9 gravitation chapter, learn more about this from the revision notes that is provided by Vedantu.

The strength of the gravitational force between two objects depends on their masses and how far apart they are. The more massive an object is, the stronger its gravitational force. The closer two objects are, the stronger their gravitational force. The gravitational constant is a fundamental constant of nature that determines the strength of the gravitational force between any two objects.

3. Who discovered gravity? Class 9 NCERT chapter 10 Gravitation?

Sir Isaac Newton is credited with discovering gravity. He was inspired to study gravity after seeing an apple fall from a tree. Newton realized that the same force that pulls apples to the ground also keeps the moon in orbit around the Earth and the planets in orbit around the sun. Go through the revision notes provided by Vedantu to understand more about this chapter.

STUDY MATERIALS FOR CLASS 9

Assignment - Gravitation, Class 9, Science PDF Download

MATCH THE COLUMN SINGLE COLUMN (MATRIX)

FILL IN THE BLANKS :

1. .................... is the force of attraction between any two bodies in the universe.

2. .................... is the force of attraction between a body and a planet. 3. Acceleration due to gravity .................... with height from the surface of the earth. 4. Acceleration due to gravity .................... with depth from the surface of the earth. 5. Acceleration due to gravity is a maximum at the .................... 6. .................... of a body changes from place to place but its .................... remains constant. 7. The upward force experienced by a body immersed partially or fully in a fluid is called .................... 8. Density of a substance is defined as the ratio of the mass of a body to its .................... 9. Relative density of a substance is defined as the ratio of the density of the substance to the density of water at .................... 10. In cgs the relative density of a substance is .................... equal to its density in gcm–3. 11. The combined weight of the sinker and cork is .................... than the weight of the sinker alone. 12. For a body to float the density of the floating object should be .................... than or equal to the density of the liquid in which it is to float.

13. The centre of ..........is a point, where the total upthrust, due to fluid displaced by part ...........of body acts. 14. A fish .................... by squeezing out air, from its .................... 15. When a body is partly or wholly immersed in a .................... , it experience an .................... 16. An iceberg floats with .................... of its volume below .................... 17. The unit of upthrust in SI is .................... 18. Lactometer is used to measure .................... of .................... 19. A hydrometer sinks .................... in water than in pure milk. 20. The density of hot air is .................... that of cold air.

UNDERSTANDING BASE QUESTIONS :-

1. In what sense does the moon fall towards the earth ? Why does it not actually fall on the surface of earth ? 2. If the earth attracts an apple. Does the apple also attract the earth? If so, why does the earth not move towards the apple ? 3. If the force of gravity somehow vanished away, why would we sent flying in space ? 4. There are two kinds of balances, that is, a beam balance and a spring balance. If both the balances give same measure of a given body on the surface of the earth, will they give same measure on the surface of moon ? Explain your answer. 5. A bag of sugar weighs W at some place on the equator. If this bag is taken to Antarctica, will it weigh same, more or less ? Give a reason for your answer. 6. Which is greater : the attraction of Earth for 1 kg of lead or attraction of 1 kg lead for earth ? 7. The weight of a man on the earth is 100 kg. Does this weight on the moon increase or decrease ? 8. When dropped from the same height a body reaches the ground quicker at poles than at the equator. Why ? 9. What would be the effect on the weight of a body if the earth stopped rotating ? 10. A man at the top of a tower throws and object horizontally whereas he simply drops another. Will these two objects reach the earth at the same time ? 11. If gravitation is the power by which all bodies tend to towards each other, then why do all bodies tend towards the centre of the earth ? 12. What is weightlessness ? 13. Why is the bottom part of the foundation of a building made wider ? 14. Why cutting instruments are sharpened ? 15. Explain why it is difficult to push a tin can into water keeping its mouth upwards than when its mouth is kept downwards towards the water. 16. A piece of ice is gently placed on the surface of water filled in glass tumbler, so the water rises to the brim. What will happen to the level of water when the ice melts ? Will the water overflow ? If not, explain with reason. 17. You are provided with a hollow iron ball of volume 20 cm3, a mass of 15 g and a solid iron ball of mass 20 g. Both are placed on the surface of water contained in a large tub. Which will float ? Give reason for your answer. 18. Explain the following : (i) Icebergs floating in the sea are dangerous for the ship. (ii) An egg sinks in fresh water but floats in a strong solution of salt. (iii) An iron nail sinks in water but a ship made of iron floats. (iv) A ship sinks to a great depth in river water than in sea water. (v) It is easier for a man to swim in sea water than in river water. (vi) A dead body floats, with its head immersed in water. 19. A solid brass cylinder tied to a thread is hanging from the hook of a spring balance. The cylinder is gradually dipped into the water contained in a jar. What change do you expect in the reading of the spring balance ? Give reason for your answer. 20. If a solid of the same density as that of a liquid is placed in it, what will happen to the solid ?

HIGH ORDER THINKING SKILL QUESTIONS

1. Newton's law of gravitation is called universal law. Why ? 2. According to Newton's law of gravitation, every, particle of matter attracts every other particle. But bodies on the surface of earth never move towards each other on account of this force of attraction, why ? 3. What is the nature of motion of an object falling freely under the action of gravity ? 4. Will a body weight more on equator or on pole ? 5. Earth is continuously pulling the moon towards its centre, still it does not fall to the earth, why ? 6. How does acceleration due to gravity depend on the mass of planet ? 7. If the diameter of the earth becomes two times its present value and its mass remains unchanged, then how would the weight of an object on the surface of the earth be affected ? 8. Choose the correct alternative : (a) Acceleration due to gravity increases/decreases with increasing altitude. (b) Acceleration due to gravity increases/decreases with increasing depth. (c) Acceleration due to gravity is independent of the mass of the earth/mass of the body. 9. A person can jump higher on the surface of the moon than on the earth. Why ? 10. A glass jar contains a liquid of density 'd' upto a height of 'h' at a place where acceleration due to gravity is 'g', the atmospheric pressure is PA : (i) What is the pressure at the free surface of the liquid ? (ii) Write an expansion for the total pressure at the base of the jar. 11. Explain why a gas bubble released at the bottom of a lake grows in size as it rises to the surface of the lake. 12. A body dipped into a liquid experiences an upthrust. State the factors on which the upthrust on the body depends. 13. Explain briefly why a balloon filled with helium gas rises in air. Why does the balloon rise to a particular height above the ground and does not rise further ? 14. Will a body weigh more in air or in vacuum when weighed with a spring balance ? Give a reason for your answer.

NUMERICAL PROBLEMS

1. What is the gravitational acceleration of a spaceship at a distance equal to two Earth's radius from the centre of the Earth? 2. A boy on a cliff 49 m high drops a stone. One second later, he throws a second stone after the ftrst. They both hit the ground at the same time. With what speed did he throw the second stone ? 3. A stone drops from the edge of the roof. It passes a window 2 m high in 0·1 s. How far is the roof above the top of the window? 4. A particle is dropped from a tower 180 m high. How long does it take to reach the ground? What is the velocity when it tourhes the ground ? Take g = 10 m/s 2 . 5. To estimate the height of a bridge over a river, a stone is dropped freely on the river from the bridge. The stone takes 2 s to touch the water surface in the river. Calculate the height of the bridge from the water level. Take g = 9·8 m/s 2 . 6. How much would a 70 kg man weigh on moon? What will be his mass on Earth and Moon? Given g on Moon = 1·7 m/s 2 .

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NCERT Book Class 9 Science Chapter 9 Gravitation (PDF) (New 2023-24)

NCERT Book Class 9 Science Chapter 9 Gravitation is here. This is the downloadable PDF format of Chapter 9 – Gravitation from the NCERT Book for Class 9 Science (new 2023-24 version) from book whose name is Science . You can also use the NCERT Solutions available at oneedu24.com to solve questions of Gravitation.

NCERT Book Class 9 Science Chapter 9 Gravitation

The direct links to download pdf of complete Chapter 9 – Gravitation from the NCERT Book for Class 9 Science (book name – Science) are given below.

NCERT Book Class 9 Science Chapter 9 Gravitation PDF Download Link – Click Here to Download Complete Chapter

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NCERT Book Class 9 Science Chapter 9 Gravitation PDF

The the complete chapter is as follows.

NCERT Book for Class 9 Science

After you have read Chapter 9 Gravitation, you can read all other lessons of class 9 Science at oneedu24, from following links.

• Chapter 1 – Matter in our surroundings
• Chapter 2 – Is matter around us pure?
• Chapter 3 – Atoms and molecules
• Chapter 4 – Structure of the atom
• Chapter 5 – The fundamental
• Chapter 6 – Tissues
• Chapter 7 – Motion
• Chapter 8 – Force and laws of motion
• Chapter 9 – Gravitation
• Chapter 10 – Work and energy
• Chapter 11 – Sound
• Chapter 12 – Improvement in food resources
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NCERT Book Class 9 Science Chapter 9 Gravitation – Key Highlights

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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

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NCERT Solutions for Class 9 Science Chapter 10 PDF

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• CBSE Notes For Class 9
• Class 9 Science Notes
• Chapter 10: Gravitation

Gravitation Class 9 CBSE Notes - Chapter 10

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

CBSE Class 9 Science Gravitation Notes

Introduction to gravitation.

This chapter discusses gravitation and the Universal Law of Gravitation. The motion of objects under the influence of gravitational force on Earth is also examined closely. Students will also understand how weight varies from place to place and the conditions required for objects to float on water.

What Is Gravitation?

Gravitation or just gravity is the force of attraction between any two bodies. All the objects in the universe attract each other with a certain amount of force, but in most cases, the force is too weak to be observed due to the very large distance of separation. Besides, gravity’s range is infinite but the effect becomes weaker as objects move away. Some examples of gravity are:

• The force that causes the ball to come down is known as gravity
• Gravity keeps the planets in orbit around the sun.
• Gravity is the force that causes a rock to roll downhill.

Chapter Summary Video

Type of Forces

There are four fundamental forces in the universe and they are:

• Gravitational force
• Electromagnetic force
• Strong nuclear force
• Weak nuclear force

Gravitational Force

Gravitational force is the weakest force out of the four forces. When gravitational force is considered for massive objects, such as the sun, or giant planets, the gravitational force is considered to be strong as the masses of these objects are also large. On an atomic level, this force is considered weak.

Electromagnetic Force

The electromagnetic force is a type of physical interaction that occurs between electrically charged particles. It acts between charged particles and is the combination of magnetic and electrical forces. The electromagnetic force can be attractive or repulsive.

Strong Nuclear Force

The strong force holds together quarks, the fundamental particles that make up the protons and neutrons of the atomic nucleus, and further holds together protons and neutrons to form atomic nuclei.

Weak Nuclear Force

Weak force is the force existing between the elementary particles which are responsible for certain processes to take place at a low probability.

The Universal Law of Gravitation

Newton’s Law of gravitation states that every object in the universe attracts every other object by a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

where G is the universal gravitation constant.

Value of G = 6.673*10 -11 Nm 2 Kg -2

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Acceleration due to Gravity

Plug the values of G (6.673*10 -11 Nm 2 Kg -2 )

M(mass of Earth) = 6 * 10 24  kg  and R= 6 * 10 6 m , to get the value of gas ≈ 9.8ms -2

This is the acceleration due to gravity and the acceleration felt by any freely falling body towards the Earth.

The value of g keeps changing due to the variation of Earth’s radius.

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The Moon’s Falling – Moon’s Revolution around Earth

The moon revolves around the Earth due to centripetal force, which is the force of gravity of the Earth.  If the force of attraction between the Earth and the moon ceases, then the moon will continue to travel in a straight-line path tangential to its orbit around the Earth.

For more information on Why the Moon Does Not Fall, watch the below video

Centripetal Force

When a body undergoes circular motion, it experiences a force that acts towards the centre of the circle. This centre-seeking force is called a centripetal force. Centripetal force is given by the following equation:

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Free Fall and Motion

When an object is under free fall, acceleration due to gravity is constant at g = 9.8ms -2 .

Value of g does not depend on mass i.e any object big or small experiences the same acceleration due to gravity under free fall. All three equations of motion are valid for freely falling objects as it is under uniform motion.

The sign of convention → towards earth g is +ve / away from earth g is -ve.

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Weight and Mass

The mass of an object is the measure of its inertia and is constant throughout the universe. The weight of an object keeps changing as the value of g changes. Weight is nothing but a force of attraction of the Earth on an object and is given by the following equation:

The weight of an object on the Moon is 1/6 times the weight on Earth.

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Thrust and Pressure

Force acting on an object perpendicular to the surface is called thrust. The effect of thrust depends on the area of contact. The pressure is thrust per unit area. SI unit is Pascal (Pa). Force acting on a smaller area applies more pressure than the same force acting on a larger area.

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Pressure in Fluids

The pressure exerted by a fluid in a container is transmitted undiminished in all directions on the walls of the container.

Archimedes’ Principle – Why Objects Float or Sink

The upward force exerted by a fluid on an object is known as upthrust or buoyant force.

The magnitude of buoyancy depends on the density of the fluid.  If the density of an object is less than the fluid, it will float. If the density of the object is greater than the fluid, it will sink.

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

Relative Density

The story of gravity – introduction to gravitation: kepler’s laws.

In astronomy, Kepler’s laws of planetary motion are three scientific laws describing the motion of planets around the sun.

• Kepler’s first law – The law of orbits
• Kepler’s second law – The law of equal areas
• Kepler’s third law – The law of periods

The orbit of a planet is an ellipse with the sun as its foci.  The line joining the planets and the sun sweeps equal areas in equal intervals of time.

Cube of a mean distance of a planet from the sun ∝ Square of orbital time period T.

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Physics Formula for Class 9 Science

Frequently Asked Questions on CBSE Class 9 Science Notes Chapter 10 Gravitation

What is gravity.

Gravity is a natural phenomenon by which all masses and energy are attracted to one another and to Earth.

What is relative density?

The ratio of the density of a substance to the density of a standard substance under specified conditions.

What is meant by free fall?

When an object falls under the sole influence of gravity, the object is said to be under ‘free fall’.

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