NCERT Solutions for Class 9 Science Chapter 9 - Gravitation
Have you ever wondered why, when you throw objects in the air, they end up falling back on the ground or why astronauts do not fall in space? This would be answered by the science of gravitation, which is an interesting subject covered in Class 9 Science Chapter 9 Gravitation. This chapter dwells on the hidden power that not only attracts everything to the earth, but also controls the movement of the planets and satellites in space. Some of the main concepts that students learn include gravitational force, free fall, buoyancy, thrust and pressure, the Principle of Archimedes, mass and weight, which are fundamental to understanding the action of forces in air and fluids.
This Story also Contains
- Class 9 Science Chapter 9 - Gravitation Question Answers: Download Solution PDF
- NCERT Solutions for Class 9 Science Chapter 9: Solved Intext Questions
- NCERT Solutions for Class 9 Science Chapter 9 - Gravitation: Exercise Questions
- Gravitation NCERT Science Class 9 Chapter 9 Topics
- NCERT Solutions for Class 9 Chapter 9 Physics: Important Formulae
- Approach to Solve Questions of Class 9 NCERT Chapter 9: Gravitation
- Benefits of NCERT Class 9 Science Chapter 9 - Gravitation question answers
- How Can NCERT Solutions for Class 9 Science Chapter 9 Help in Exam Preparation?
- NCERT Solutions for Class 9 Science Chapter-wise
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The NCERT Solutions for Class 9 Science Chapter 9 - Gravitation are designed by the subject experts to give step-by-step solutions to all the questions in the text, and exercises to make it clear and accurate. These NCERT solutions include significant formulas, fundamental concepts, and additional practice problems to promote problem-solving skills. The solutions align with the current CBSE Class 9 syllabus and, besides being a good preparation for school examinations, they provide a solid foundation to high-stakes examinations such as NSEJS and Olympiads. Through the use of these perfectly designed NCERT Solutions for Class 9 Science Chapter 9 - Gravitation, students will be able to develop a better conceptual grasp, work out corrections faster and score better grades without fear.
Also Read
Class 9 Science Chapter 9 - Gravitation Question Answers: Download Solution PDF
Class 9 Science Chapter 9 - Gravitation question answers give the solutions to all NCERT textbook questions in a clear and step-by-step manner, and this helps the students to get a better grasp on such important concepts as gravitational force, free fall, thrust, buoyancy, mass and weight and Archimedes' Principle. These class 9 science chapter 9 Gravitation question answers are structured according to the latest CBSE syllabus to enable students to enhance their conceptual understanding, practice, as well as excel in exams. Get the free PDF of Gravitation NCERT Solutions by clicking below.
NCERT Solutions for Class 9 Science Chapter 9: Solved Intext Questions
Chapter 9 Gravitation Solved in-text questions offer explanation and step-wise answers to in-text questions to reinforce understanding of concepts. Such solutions are beneficial in assisting students in relating theoretical understanding with practical life so that studying and ace exams can be easy and successful.
Gravitation class 9 NCERT solutions - Topic 9.1 Gravitation
Q1. State the universal law of gravitation.
Answer:
The universal law of gravitation states that every body in the universe attracts every other body by virtue of its mass. This force is directly proportional to the product of the masses of the two bodies and inversely proportional to the square of the distance between them.
Let there be two bodies of masses
$
F_g=G \frac{m_1 m_2}{r^2}
$
where G is the universal gravitational constant and is equal to $6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$
Answer:
Let ME
$
F=G \frac{M_E m}{d^2}
$
Mass of Earth, $M_E=5.972 \times 10^{24} \mathrm{~kg}$
d would be approximately equal to the radius of the Earth.
Radius of Earth $=6.378 \times 10^6 \mathrm{~m}$
$
G=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}
$
$F=6.67 \times 10^{-11} \times \frac{5.972 \times 10^{24} \times m}{\left(6.378 \times 10^6\right)^2}$
$F=9.8 m N$
Gravitation class 9 NCERT solutions - Topic 9.2 Free fall
Q1. What do you mean by free fall?
Answer:
We say an object is freely falling when it is dropped from some height and is attracted by the gravitational force of Earth only, and is under the influence of no other considerable force.
Q2. What do you mean by the acceleration due to gravity?
Answer:
Each object on the Earth is under the influence of the gravitational force of the Earth. The acceleration due to the Earth's gravitational force is known as acceleration due to gravity.
Gravitation class 9 NCERT solutions - Topic 9.4 Weight
Q1. What are the differences between the mass of an object and its weight?
Answer:
The following are the differences between the mass of a body and its weight
| Mass | Weight |
| (i) Mass is the amount of matter contained in the body. | (i) Weight is the gravitational force experienced by the body. |
| (ii) Mass of a body is always constant. | (ii) Weight of a body depends on the place where it is. |
| (iii) Mass is the measure of the inertia of the body. | (iii) Weight is the measure of the gravitational force acting on the body. |
| (iv) Mass only has magnitude. | (iv) Since weight is a force, it naturally has a direction associated with it as well |
| (v) Its SI unit is the kilogram (kg). | (v) Its SI unit is Newton (kg ms-2) |
Q2. Why is the weight of an object on the moon (1/6)th its weight on the Earth?
Answer:
The weight of an object on the moon would be given by
$
W_M=G \frac{M_M m}{r_M^2}
$
where $M_M$ is the mass of the moon, $m$ is the mass of the body, $r_M$ is the radius of the moon, and $G$ is the gravitational constant.
The weight of an object on the Earth would be given by
$
W_E=G \frac{M_E m}{r_E^2}
$
where $M_E \ and\ r_E$ are the mass and radius of the Earth, respectively.
$
\frac{W_M}{W_E}=\frac{M_M r_E^2}{M_E r_M^2}
$
The above ratio is approximately equal to 1/6, and this is why the weight of an object on the moon is (1/6) th of its weight on the Earth.
Gravitation class 9 NCERT solutions - Topic 9.5 Thrust and Pressure
Q1. 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 strap made of a thin and strong string because the surface area of the string in contact with the shoulders is very less, and due to this, its weight applies a large pressure on the shoulders.
Q2. What do you mean by buoyancy?
Answer:
When an object is placed in a fluid, it displaces a volume of liquid equal to its own volume. Due to this, the liquid exerts an upward force on the body called the Buoyant force. This tendency of a liquid to exert the upward buoyant force is called buoyancy.
Q3. Why does an object float or sink when placed on the surface of the water?
Answer:
When an object is placed on the surface of the water, it displaces a certain volume of water. If the density of the object is less than that of water, the buoyant force due to this displacement of water is equal to the weight of the object, and it floats on the surface of the water. If the density of the object is more than the density of water, the volume of water displaced would be equal to the volume of the object itself and the buoyant force acting upwards due to this displacement of water would be less than the weight of the object, and the object would sink
Gravitation class 9 NCERT solutions - Topic 9.6 Archimedes' Principle
Q1. You find your mass to be 42kg on a weighing machine. Is your mass more or less than 42kg?
Answer:
The weighing scale shows the reading according to the weight applied to it. The weight on the weighing scale would be slightly less than our weight because there is a small upward force acting on us due to the buoyancy of the atmosphere, and the reading would be slightly lower than our actual weight; therefore, our mass must be more than 42 kg.
Q2. 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 the other. Can you say which one is heavier and why?
Answer:
The volume of the bag of cotton would be much more than the iron bar and the upward buoyant force on the bag of cotton would be more than that acting on the iron bar and therefore the value is shown in case of the bag of cotton must be lesser than the actual value by a larger amount and therefore the bag of cotton is heavier than the iron bar.
NCERT Solutions for Class 9 Science Chapter 9 - Gravitation: Exercise Questions
Chapter 9 Gravitation exercise solutions have important concepts covered, such as gravitational force, free fall, mass, weight, buoyancy, and more, covered in detailed explanations. The solutions are aimed at making complex topics easy and answering accurately to enable students to practice effectively in exams.
Answer:
Let there be two bodies of masses m1 and m2, and let the distance between them be r. The gravitational force F between them would be given by
$F=G \frac{m_1 m_2}{r^2}$
Let the distance between them be halved by two. The gravitational force between them would now be given by
$\begin{aligned} & F^{\prime}=G \frac{m_1 m_2}{\left(\frac{r}{2}\right)^2} \\ & F^{\prime}=4 G \frac{m_1 m_2}{r^2} \\ & F^{\prime}=4 F\end{aligned}$
The force of gravitation between two objects would increase by 4 times if the distance between them is halved.
Answer:
The gravitational force acting is definitely more in the case of a heavier object than a light object, but the acceleration depends on the ratio of the force acting on the body to the mass of the body and since the gravitational force acting on a body is proportional to its mass, the acceleration due to gravity is the same for all bodies irrespective of their masses.
Answer:
Mass of Earth $M_E$ is $6 \times 10^{24} \mathrm{~kg}$
Mass of the body m is 1 kg
The radius of the Earth $R_E$ is $6.4 \times 10^6 m$.
Universal gravitational constant, $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$
The magnitude of the gravitational force between the Earth and the body would be
$
\begin{aligned}
& F=G \frac{M_E m}{r_E^2} \\
& F=6.67 \times 10^{-11} \times \frac{6 \times 10^{24} \times 1}{\left(6.4 \times 10^6\right)^2} \\
& F=9.77 N
\end{aligned}
$
Answer:
The Earth attracts the moon with the same force as the moon attracts the Earth.
We know from the third law of motion that each force has an equal and opposite force, and the universal gravitational law also states the same, i.e. the gravitational force of attraction between two bodies is the same.
Q5. If the moon attracts the Earth, why does the Earth not move towards the moon?
Answer:
The moon and the Earth attract each other with the same gravitational force. It is because of the much larger mass of the Earth than the mass of the Moon that the Earth does not move towards the Moon.
Q6. (i) What happens to the force between two objects if the mass of one object is doubled?
Answer:
As the force between the objects is directly proportional to the product of the masses of the objects, if the mass of one object is doubled, the force between them will also double.
Answer:
The force between two objects is inversely proportional to the square of the distance between them. Therefore, if the distance between the objects is doubled and tripled, the force between them would become one-fourth and one-ninth of the initial value, respectively.
Q.6.(iii) What happens to the force between two objects if the masses of both objects are doubled?
Answer:
As the force between the objects is directly proportional to the product of the masses of the objects, if the masses of both objects are doubled, the force will become four times the initial value.
Q7. What is the importance of the universal law of gravitation?
Answer:
The importance of the universal law of gravitation lies in the fact that it proves that every object in the universe is attracted by every other object in the universe by virtue of their masses.
Q8. What is the acceleration of free fall?
Answer:
The acceleration of free fall on Earth is $9.8 \mathrm{~ms}^{-2}$ and is the same for all objects, i.e. its value is independent of their masses.
Q9. What do we call the gravitational force between the Earth and an object?
Answer:
The gravitational force between the Earth and an object is called the weight of the object.
Answer:
As the value of g is more at the poles than at the equator, the weight of the same amount of gold would be more at the poles than at the equator, and therefore the friend will not agree with the weight of gold bought.
Q11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Answer:
A sheet of paper has much more area than the same paper crumbled, and due to this, the sheet experiences more air resistance and thus falls at a speed slower than when it is crumbled.
Answer:
The weight of an object on Earth is given by W = mg, where m is the mass of the object and g is the gravitational acceleration.
g = 9.8 ms-2
Weight of a 10 kg object on earth = 10 X 9.8 = 98 N
As the gravitational force on the surface of the moon is only 1/6th of that on Earth, the gravitational acceleration on the moon would be equal to g/6.
The weight of an object of mass 10 kg on the moon is therefore given as follows
$
\begin{aligned}
& W=10 \times \frac{g}{6} \\
& W=10 \times \frac{9.8}{6}
\end{aligned}
$
$
W=16.33 N
$
Answer:
The initial velocity of the ball is u = 49 ms-1
Final velocity at the highest point would be v = 0
The magnitude of the acceleration is equal to the acceleration due to gravity, g
Acceleration, a = -g = -9.8 ms-2
Let the maximum height to which it rises be s
Using the third equation of motion, we have
$\begin{aligned} & v^2-u^2=2 a s \\ & s=\frac{v^2-u^2}{2 a} \\ & s=\frac{0^2-49^2}{2 \times-9.8} \\ & s=122.5 \mathrm{~m}\end{aligned}$
The ball would rise to a maximum value of 112.5 m.
Answer:
Let the time taken by the ball to reach the highest point be t
$\begin{aligned} & v=u+a t \\ & 0=49+(-9.8) \times t \\ & t=5 s\end{aligned}$
At the same time, t would be taken to come back to the ground from the highest point.
Therefore, the total time it takes to return to the surface of the earth = 2t = 10 s.
Answer:
Initial velocity u = 0
Acceleration, a = g = 9.8 ms-2
Distance travelled, s = 19.6 m
Let the final velocity be v
According to the third equation of motion
$\begin{aligned} & v^2-u^2=2 a s \\ & v^2-0^2=2 \times 9.8 \times 19.6 \\ & v=\sqrt{2 \times 9.8 \times 19.6} \\ & v=19.6 m s^{-1}\end{aligned}$
Its final velocity just before touching the ground will be 19.6 ms-1
Answer:
Initial velocity $u=40 \mathrm{~ms}^{-1}$
Acceleration $\mathrm{a}=-\mathrm{g}=-10 \mathrm{~ms}^{-2}$
Final velocity at the highest point would be $\mathrm{v}=0$
Let the maximum height reached be s
As per the third equation of motion
$
\begin{gathered}
v^2-u^2=2 a s \\
s=\frac{v^2-u^2}{2 a} \\
s=\frac{0^2-40^2}{2 \times-10} \\
s=80 \mathrm{~m}
\end{gathered}
$
The net displacement would be zero as the stone would return to the point from where it was thrown.
The total distance covered by the stone = 2s = 160m
Answer:
Mass of the Earth, $M_E$ $=6 \times 10^{24} \mathrm{~kg}$
Mass of the Sun, $M_S$ $=2 \times 10^{30} \mathrm{~kg}$
Distance between the Earth and the Sun, $d=1.5 \times 10^{11} m$
Universal gravitational constant, $G=6.67 \times 10^{-11} \mathrm{Nkg}^{-2} \mathrm{~m}^2$
The force of gravitation between the Earth and the Sun would be given as
$
\begin{aligned}
& F=G \frac{M_E M_S}{d^2} \\
& F=6.67 \times 10^{-11} \times \frac{6 \times 10^{24} \times 2 \times 10^{30}}{\left(1.5 \times 10^{11}\right)^2} \\
& F=3.56 \times 10^{22} N
\end{aligned}
$
Answer:
Let the distance travelled by the stone which is dropped from the top up to the instant when the two stones meet be x
Initial velocity u = 0
Acceleration $\mathrm{a}=\mathrm{g}=9.8 \mathrm{~m} \mathrm{~s}^{-2}$
Using the second equation of motion
$
\begin{aligned}
& s=u t+\frac{1}{2} a t^2 \\
& x=0 \times t+\frac{1}{2} \times 9.8 \times t^2 \\
& x=4.9 t^2--(i)
\end{aligned}
$
The distance travelled by the stone which is projected vertically upwards from the ground up to the instant when the two stones meet would be equal to 100 - x
$
\text { Initial velocity }=25 \mathrm{~m} \mathrm{~s}^{-1}
$
Acceleration $\mathrm{a}=-\mathrm{g}=-9.8 \mathrm{~m} \mathrm{~s}^{-2}$
Using the second equation of motion
$
\begin{aligned}
& s=u t+\frac{1}{2} a t^2 \\
& 100-x=25 \times t+\frac{1}{2} \times(-9.8) \times t^2 \\
& 100-x=25 t-4.9 t^2 \\
& x=4.9 t^2-25 t+100-(i i)
\end{aligned}
$
Equating $x$ from (i) and (ii), we get
$
\begin{gathered}
4.9 t^2=4.9 t^2-25 t+100 \\
25 t=100 \\
t=4 s \\
x=4.9 t^2 \\
x=4.9 \times 4^2 \\
x=78.4 \mathrm{~m} \\
100-x=100-78.4=21.6 \mathrm{~m}
\end{gathered}
$
The stones meet after a time of 4 seconds at a height of 21.6 meters from the ground.
Q18. 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
Answer:
(a) Let the ball be thrown with initial velocity u
Time taken to get back to the thrower = 6 s
Time taken to reach the highest point is t = 6/2 = 3 s
Final velocity at the highest point is v = 0
Acceleration $\mathrm{a}=-\mathrm{g}=-9.8 \mathrm{~m} \mathrm{~s}^{-2}$
Using the first equation of motion
$
\begin{aligned}
& v=u+a t \\
& 0=u-9.8 \times 3 \\
& u=29.4 m s^{-1}
\end{aligned}
$
(b) Let the maximum height it reaches be s
Using the second equation of motion
$\begin{aligned} & s=u t+\frac{1}{2} a t^2 \\ & s=29.4 \times 3-4.9 \times 3^2 \\ & s=44.1 \mathrm{~m}\end{aligned}$
(c) Out of the 4 seconds, 3 have been spent in reaching the highest point
The distance travelled by the ball in the next 1 second is s' given by
$\begin{aligned} & s^{\prime}=u t+\frac{1}{2} g t^2 \\ & s^{\prime}=0 \times+4.9 \times 1^2 \\ & s^{\prime}=4.9 m\end{aligned}$
Distance from the ground after 4 seconds = s - s' = 44.1 - 4.9 = 39.2 m
The position of the ball after 4 seconds is 39.2 m from the ground.
Q19. In what direction does the buoyant force on an object immersed in a liquid act?
Answer:
The buoyant force acts on an object in the vertically upward direction opposite to that of the gravitational force.
Q20. Why does a block of plastic released under water come up to the surface of the water?
Answer:
The density of plastic is less than that of water and due to which the upwards acting buoyant force is more than the downwards acting gravitational force. Due to this, a block of plastic released under water comes up to the surface of the water.
Answer:
Mass of the given amount of substance = 50 g
Volume of the given amount of substance = 20cm3
Density of the given substance is $\rho$
$
\begin{aligned}
& \rho=\frac{50}{20} \\
& \rho=2.5 \mathrm{gcm}^{-3}
\end{aligned}
$
As the given substance has a higher density than that of water, it will sink in water.
Answer:
Mass of the packet = 500 g
Volume of the packet = 350cm3
The density of the packet is given by
$\begin{aligned} & \rho=\frac{500}{350} \\ & \rho=1.428 \mathrm{gcm}^{-3}\end{aligned}$
As the density of the packet is less than that of water, it will sink in water.
Volume of the water displaced by the packet = volume of the packet = 350 cm3
Mass of the water displaced by the packet = Volume of the water displaced by the packet x Density of water
= 350 X 1
= 350 g
The mass of water displaced is less than the mass of the packet, so the packet will sink.
Gravitation NCERT Science Class 9 Chapter 9 Topics
Chapter 9 Gravitation talks about the universal law of gravitation, objects in motion under gravity and the notion of free fall, mass, and weight. It also teaches the concept of thrust, pressure, buoyancy, Archimedes' principle and relative density, enabling the student to correlate real-life events with concepts of science.
10.1 Gravitation
10.1.1 Universal law of gravitation
10.1.2 Importance of the universal law of gravitation
10.2 Free fall
10.2.1 To calculate the value of g
10.2.2 Motion of objects under the influence of the gravitational force of the Earth
10.3 Mass
10.4 Weight
10.4.1 Weight of an object on the Moon
10.5 Thrust and pressure
10.5.1 Pressure in fluids
10.5.2 Buoyancy
10.5.3 Why do objects float or sink when placed on the surface of water?
10.6 Archimedes’ principle
NCERT Solutions for Class 9 Chapter 9 Physics: Important Formulae
Class 9 Science Chapter 9 Gravitation: Important Formulae is a fast and easy way to find the necessary information on how to revise the most important formulae of gravitation, equations of motion under gravity, pressure in fluids, thrust, buoyancy, and Archimedes' Principle. These formulae are clarified and have examples, and students find it easy to apply the formulae in both numerical problems and conceptual questions. These are important in helping one revise in a hurry, as well as being a strong basis to help in exams.
Gravitational Force:
$
F=\frac{G \cdot m_1 \cdot m_2}{r^2}
$
Acceleration due to gravity:
$
g=\frac{G \cdot m}{r^2},\left(\text { on earth }, g \approx 9.8 \mathrm{~m} / \mathrm{s}^2\right)
$
Weight:
$
W=m \cdot g
$
Pressure:
$
P=\frac{F}{A}
$
Liquid pressure:
$
P=h \cdot \rho \cdot g
$
Buoyant force:
$
F_b=\rho \cdot V \cdot g
$
Approach to Solve Questions of Class 9 NCERT Chapter 9: Gravitation
How to solve questions of Class 9 Science Chapter 9 - Gravitation entails having an insight into the fundamental concepts such as the gravitational force, free fall, acceleration due to gravity, buoyancy, thrust, and Archimedes' Principle. Students are expected to pay attention to the use of clear definitions, proper application of formulas, and connection of the concepts and real-life examples to answer questions efficiently. The systematic practice of both theoretical and numerical problems will assist in mastering this chapter and recording a high score in the exams.
1. Read the question carefully and identify what is given (mass, height, area, force, etc.).
2. Note the given values with proper units.
3. Choose the correct formula, like:
- $F=\frac{G \cdot m_1 \cdot m_2}{r^2}$ (gravitational force)
- $W=m \cdot g$ (weight)
- $P=\frac{F}{A}$ (pressure)
- $p=h \cdot \rho \cdot g$ (liquid pressure)
- $F_b=\rho \cdot V \cdot g$
4. Substitute values into the formula step by step.
5. Use correct units (e.g., $N, m^2, k g / m^3$ ).
6. Draw a small diagram(FBD) if needed, especially for buoyancy or floating objects.
7. Solve neatly, and check your final answer.
Benefits of NCERT Class 9 Science Chapter 9 - Gravitation question answers
The Gravitation Class 9 question answers give well-organised and step-by-step solutions that simplify the complicated concepts. The class 9 science chapter 9 Gravitation question answers assist the students in reinforcing fundamentals, practising significant numericals, and preparing efficiently for school exams as well as competitive tests.
- Concept Clarity- Guides students to have perfect ideas about concepts such as gravity, free fall, mass, weight, thrust, pressure, Archimedes' principle, and buoyancy.
- Step-By-Step Solutions- Gives explanations to numerical problems in a step-by-step manner that makes problem-solving an easy task.
- Exam Preparation- Discusses every in-text and exercise question according to the NCERT, thus providing complete preparation for the exams.
- Increases Confidence- Enhances conceptual ability and problem-solving skills, so the students will know how to write correct answers in exams.
- Time-saving- Provides prepared, systematic solutions, resulting in a fast and efficient revision.
- Supports Higher Studies- Establishes a robust base of Physics concepts that can be used in higher grades and competitive exams such as NEET, JEE and Olympiads.
How Can NCERT Solutions for Class 9 Science Chapter 9 Help in Exam Preparation?
Gravitation Class 9 question answers, Chapter 9 Science is crucial in the preparation of exams by giving clear and explanatory answers to the basic concepts of the forces of gravitation, gravitational force, free fall and the distinction between mass and weight mentioned in earlier chapters. These Gravitation class 9 question answers assist students in step-by-step solutions of numerical problems to facilitate them in mastering calculation with the gravitational constant and acceleration due to gravity. The class 9 science Gravitation question answers allow the students to learn how the concept of gravitation applies to objects on the earth and the space by simplifying complex theories into simple words and answers that come in the form of examples. This comprehensive conceptual clarity and practice questions that are exam-oriented, make students prepared to handle the board exams and competition at a reasonable level with the confidence that they are highly prepared to answer the questions of the latest CBSE syllabus both theoretically and numerically. The class 9 science Gravitation question answers also address similar concepts, such as buoyancy and the principle of Archimedes, which would further reinforce the understanding of physics concepts among the students to prepare them to pass examinations in science comprehensively.
NCERT Solutions for Class 9 Science Chapter-wise
The chapter-wise NCERT Solutions for Class 9 Science provide detailed and accurate answers to all in-text and exercise questions. These solutions are designed as per the latest NCERT guidelines, helping students understand concepts clearly, practice effectively, and score well in exams.
NCERT Solutions for Class 9 - Subject Wise
Also Read,
- NCERT notes for Class 9 Science Chapter 9 Gravitation
- NCERT Exemplar for Class 9 Science Solutions Chapter 9
Also Check NCERT Books and NCERT Syllabus here:
Frequently Asked Questions (FAQs)
Q: How is it hard to hold a school bag with a narrow strap, according to the NCERT solution?
A:
This is explained in the solution by means of the pressure concept. Force per unit area (P = F/A) is the definition of pressure. The surface area of a thin strap is quite small. For the same amount of force (the weight of the bag), this tiny area puts a great deal of pressure on the shoulder. The same force is distributed over a greater area by a broader strap, which reduces pressure and improves carrying comfort.
Q: What is gravitation for class 9 NCERT solutions?
A:
The force of attraction existing between two objects in the universe is referred to as gravitation. The Universal Law of Gravitation, according to which Newton developed his arguments, is based on the fact that this force is proportional to the masses of the objects and the distance between them, thus making objects fall to Earth and planets orbit around the Earth.
Q: What is free fall?
A:
When an object plunges solely under the effect of gravity on Earth without air opposition, the object is considered to be in a free fall.
Q: What is the Archimedes principle?
A:
According to it, an immersed body exerts an upward force (buoyancy force) that acts upon the body and is equal to the weight of the fluid displaced by the body.
Q: Why do bodies weigh less in water?
A:
Since water produces an upward force of buoyancy upon the body and, therefore, acts against its weight, it tends to diminish its weight.
Q: What is the variation of the value of g on earth?
A:
Depending on the rotation and shape of the Earth the acceleration due to gravity is most at the poles and least at the equator.
Q: Why all objects fall with the same rate in vacuum?
A:
There is no air resistance found in a vacuum, and this means that the acceleration of all objects in the vacuum at a given gravitational force (g=9.8m/s2) is unaffected by their mass.
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