# NCERT Solutions Class 9 Science Chapter 10

## NCERT Solutions for Class 9 Science Chapter 10: Gravitation

Science introduces a lot of unique concepts to students, and it also opens numerous opportunities in the field of Engineering and Sciences. Students who share an interest in this subject find it challenging to understand the concepts. Class 9 is a stage where students have an opportunity to increase their knowledge and score better on the exam. It is also essential as it prepares them for the higher classes.

Class 9 Science NCERT Solutions Chapter 10: Gravitation provides students with necessary insights into the concepts involved in the chapter. It has detailed answers and explanations provided by our subject matter experts. Gravity is a fascinating chapter and students will learn how our planet stays in orbit and why things fall. Further, students will learn about velocity, acceleration, force, work, and energy. This chapter prepares students with the fundamental concepts of studying advanced physics.

Our Extramarks NCERT Solutions for Class 9 Science Chapter 10 are available for the students. The Extramarks NCERT solutions have been updated to include the latest content prescribed by the CBSE board. The topics are explained in a simple language without using the  technical jargons. . Using the NCERT solutions meant for class 9 Science chapter 10, students can comprehend each concept with illustrative examples. Extramarks is one of India's leading online learning platforms, as lakhs of students refer to our study material. Students pursuing primary and secondary can sign up at Extramarks’  website and explore a repository of study materials. Students just need to pick the right resource to step up their learning experience. Our subject matter experts always follow the latest CBSE format to frame the solutions so that students can get a better hold on the subject. Extramarks doesn’t believe in rote learning, in fact it follows the latest CBSE guidelines to provide experiential learning to students.

Students can visit our website for updates on the latest exam-related news. Further, students can also refer to NCERT solutions class 10, NCERT solutions class 11, and NCERT solutions class 12.

## Key Topics Covered in NCERT Solutions for Class 9 Science Chapter 10:

NCERT Solutions for Class 9 Science Chapter 10 elaborates new terms and definitions related to gravitation. This will discuss why the objects fall on the ground and do not float away from the surface. Students will learn what gravitational force means and how it can be defined using specific terms. If students discover any difficulty understanding these new concepts, they can refer to our Extramarks NCERT Solutions for Class 9 Science Chapter 10.

Some of the key topics featured in NCERT Solutions for Class 9 Science Chapter 10 are:

### Gravitation

This chapter will discuss gravitation  and the universal laws of gravitation. It is also essential to examine the effects of gravity on objects. Students will also learn how weight changes from one place to another and what conditions are required for objects to allow things to float on the water.

Gravitation, also known as gravity, states that  every object in the universe attracts every other object with a force called the gravitational force. It is the force of attraction between two bodies. Although objects attract specific energy, most are too far apart to observe. Gravity's range is infinite, but it becomes less effective as things move further away. Some of the features of gravitation elaborated in NCERT Solutions for Class 9 Science Chapter 10:

• Gravitation is the force that causes the ball's descent.
• For example, the sun keeps the planets orbiting around the sun because of gravity.
• Gravity is the force that causes the rock to fall downwards. .

### The universal law of gravitation

Newton's Law of Universal Gravitation explains that all particles attract each other with a force proportional to the product of masses but is inversely proportional to the distance between them.

Where F is the gravitational force between bodies, m 1 and 2 respectively are the masses of bodies, or the distance between two bodies' centres, and G is the universal gravitational constant.

The universal gravitation constant is the constant proportionality (G), as shown in the equation. Henry Cavendish determined the exact value of G experimentally. It is G = 6.673x10 -11N m /kg .

The Universal Gravitational Law can explain nearly everything, ranging from how an apple falls from a tree and also  to why the moon revolves around the earth. The law of universal gravitation is beautiful.

Students can refer to NCERT Solutions for Class 9 Science Chapter 10 and understand the law of gravitation in detail.

### Free fall

Free fall is a scenario when an object is moving only under the effect of the gravitational pull of Earth. Once any external force is exerted on the object, the motion of the object will be constantly accelerated. This is called freefall acceleration.

Newton's law explains that mass is a significant quantity. Although we think mass and weight are the same, they are not. The gravitational force applied to an object with a particular mass is called weight. You can calculate the object's weight by multiplying its mass (m) with the acceleration due to gravity (g). The Earth's surface measured gravitational acceleration is approximately 980 cm/second/second.

Mass is the measurement of how much material is contained in an object. Weight measures the gravitational force exerted upon the material in a gravity field. Thus, weight and mass are proportional to each acceleration due to gravity being the proportionality constant.

### Mass

It is one of the fundamental quantities of Physics and the most fundamental property of matter. Further, mass is the measurement of the volume of weight within a body. Kilogram (kg) is the SI unit for mass. A body's mass does not change.

In extreme cases, a large amount of energy is taken away from the  body. A nuclear reaction is an example of this. It converts a small amount of matter into vast amounts of energy. This reduces the substance's mass.

Calculating mass is complicated. Therefore, students can refer to our NCERT Solutions for Class 9 Science Chapter 10. We have provided examples with step-by-step explanations.

### Weight

It measures the force of gravity acting upon a body.

Here is the formula to calculate your weight:

w = mg

Because weight is a force, its SI unit is the same as force. Hence, SI Unit of weight (N) The expression of weight depends on both mass and acceleration due to gravity. While the mass may not change, the acceleration due to gravity can vary depending on where it is placed. Let's look at this example to understand the concept.

### Thrust and  Pressure

Thrust is the force acting on an object perpendicularly to its surface. The area of contact determines the effect of thrust. Further, the area of thrust is the pressure per unit area. SI unit is the pascal. A smaller force exerts more pressure on a smaller area than a force acting on the same area.

Archimedes  Principle

Archimedes' principle states, "If a body is submerged in a fluid, it will experience a force equal to the fluid's weight." This principle is used in ship design, determination of purity, and lactometers to determine milk quality. Further, students can refer to our NCERT Solutions for Class 9 Science Chapter 10 to understand the principles of Archimedes.

### Relative  Density

Density is defined as the amount of mass per unit volume of matter. Every substance has a unique density. We need a tall glass, honey, water, and coconut oil to understand density.

The difference in density and specific gravity is 1 gram per 1 cm at room temperatures and pressure. This is the density of water. It is used as a standard. The relative density or specific gravity is how the density of other materials (usual liquids is also calculated close to this.

Specific gravity refers to the ratio between the mass of a substance and that of a reference substance. Let's take, for example, the honey density, would be approx. 1.42 gram / cm3. This would mean that its specific gravity is 1.42/1 = 1.42. Specific gravity is a ratio. Therefore, specific gravity doesn't have a unit.

It will tell us if a substance will sink or float. For example, a sense with a specific gravity below 1 gram / cm3 will float, while one higher than 1 gram / cm3 will sink. For more elaborative examples, students can refer to our NCERT Solutions for Class 9 Science Chapter 10.

## NCERT Solutions for Class 9 Science Chapter 10: Exercise &  Solutions

NCERT Solutions for Class 9 Science Chapter 10 are available on  Extramarks’ website. The NCERT solutions are helpful resources that help the students cover the entire syllabus and provide an in-depth analysis of the topics. This allows students to assess their learning immediately and strengthen their foundation at an early stage.. The solutions are reliable, accurate, detailed, and prepared by our subject matter experts. These solutions are designed in such a way that the answers are self-explanatory, and will clarify your doubts making it easier to understand the concepts quickly and thoroughly. Further, students can develop the ability to respond to high-end questions effectively.

Class 9 Science chapter 10 Gravitation helps build a great foundation of concepts. In addition, the chapter will teach students the prime characteristics of gravitational force and its importance. Further, students will get to study the changes in the velocity of a free  falling body. It will also discuss the difference between gravitational constant and gravitational acceleration. Eventually, students will understand how the equations of motion can be related to this chapter.

Students can click on the links below for specific questions and solutions:

• Chapter 10: Exercise 10.1 Solutions 2 Questions
• Chapter 10: Exercise 10.2 Solutions 2 Questions
• Chapter 10: Exercise 10.3 Solutions 2 Questions
• Chapter 10: Exercise 10.4 Solutions 3 Questions
• Chapter 10: Exercise 10.5 Solutions 22 Questions

Along with this, students can also refer to other solutions for primary and secondary classes:

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT Solutions Class 10
• NCERT Solutions Class 11
• NCERT Solutions Class 12

## NCERT Exemplar for Class 9 Science:

Science is an exciting subject with a lot of unique concepts. However, students may find it challenging to remember the key concepts. Therefore, solving different types of questions with a higher difficulty level will be beneficial for the students. It will help students to increase their knowledge, and they will be able to strategize their exam preparation. In addition, it also helps them to remember the key concepts and definitions.

NCERT exemplar introduces different types of questions with a higher level of difficulty. Questions that appear in the exemplar are multiple-choice questions, fill in the blanks, match the pair, and true or false. Thus, students can explore a repository of educational materials at Extramarks and challenge themselves to advanced questions to have an enhanced learning experience. All questions asked in CBSE annual exams are based on the NCERT books. Thus, students need to have a good knowledge of the NCERT exemplar. It will be easier for  students to crack the toughest competitive exams.

A few questions are from NCERT Solutions for Class 9 Science Chapter 10. It helps students get a clear idea of how to find the acceleration due to gravity, the mass of the body, and calculate gravitational force.

## Key Features of NCERT Solutions for Class 9 Science Chapter 10:

NCERT Solutions for Class 9 Science Chapter 10 are an excellent way for students to grasp the main topics. These solutions help students understand concepts and to help them plan their strategies to succeed in exams. Referring to Extramarks NCERT solutions is quite beneficial for students.  These solutions are designed in such a way that the answers are self-explanatory, and will clarify your doubts making it easier to understand the concepts quickly and thoroughly.

The key benefits and features of NCERT Solutions for Class 9 Science Chapter 10 are listed here:

• Students can understand the concepts and increase their knowledge.
• The solutions are revised and updated regularly by the subject matter experts who have meticulously explained each chapter and are  explained in the light of current topics.
• Students can easily access these NCERTsolutions on the Internet.
• It can be referred to by students while preparing for exams.
• The NCERT Solutions for Class 9 Science Chapter 10 are prepared by faculty experts who know how to explain concepts clearly and logically to students. So by studying from our NCERT solutions students are able to understand the topic easily and can also revise the syllabus in detail before the exam.
• There are comprehensive notes explained with easy-to-understand language, which helps students to comprehend the more complex topics and concepts.
• The solutions are provided by Extramarks which is known for providing the best online learning resources for Class 1 to Class 12 students.
• The solutions give you a good overview of the chapter, what the key topics in this chapter are, and how you can successfully prepare for these topics in the exam.
• This NCERT solution for class 9 Science chapter 10 will help students revise concepts with examples provided, so they can easily understand and remember the concepts explained in these solutions after studying them.

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

Ans-

$\begin{array}{l}\text{Gravitational force}\left(\text{F}\right)\text{acting between two objects is inversely proportional to the square of the distance}\\ \left(\text{r}\right)\text{between them}\text{.}\\ \text{Mathematically,}\\ \text{F}\propto \frac{1}{{\text{r}}^{\text{2}}}\\ \text{When distance them becomes}\frac{r}{2}\text{then,}\\ \text{F}\propto \text{}\frac{1}{{\left(\frac{\text{r}}{2}\right)}^{2}}\text{}\propto \text{}\frac{4}{{r}^{2}}\\ \text{Thus, if the distance is reduced to half, the gravitational force weill become four times larger}\text{.}\end{array}$

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

Ans-

All the objects fall towards the ground under the acceleration due to gravity and this acceleration is constant and does not depend upon the mass of the falling object. Thus, a heavy object does not fall faster than a light object.

Q.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 × 1024 kg and radius of the earth is 6.4 × 106 m.)

Ans-

$\begin{array}{l}\text{Given, Mass of the object, m = 1 kg}\\ {\text{Mass of the Earth, M = 6 × 10}}^{\text{24}}\text{kg}\\ \text{According to universal law of gravitation,}\\ \text{Force, F =}\frac{\text{GMm}}{{\text{r}}^{\text{2}}}\text{}...\left(\text{i}\right)\\ \text{Where, G = universal gravitational constant}\\ {\text{= 6.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ \text{Since, the object is on the earth surface hence,}\\ {\text{r = radius of earth = 6.4×10}}^{\text{6}}\text{m}\\ \text{On putting these values in}\left(\text{i}\right)\text{, we have}\\ {\text{= 6.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\text{×}\frac{\left({\text{6×10}}^{\text{24}}\text{kg}\right)\left(\text{1 kg}\right)}{{\left({\text{6.4 ×10}}^{\text{6}}\text{m}\right)}^{\text{2}}}\\ \text{= 9.8 N}\end{array}$

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

Ans-

The universal law of gravitation states that two objects attract each other with equal forces, but in opposite directions. Hence, the Earth attracts the moon with an equal force with which the moon attracts the earth.

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

Ans-

The Earth and the moon exert equal gravitational forces to each other. Due to heavier mass of Earth, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. Therefore, the Earth does not move towards the moon.

Q.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?

Ans-

(i) According to universal law of gravitation, force is directly proportional to the product of the masses. If the mass of one object is doubled, then the gravitational force will also get doubled.

(ii) According to universal law of gravitation, force is inversely proportional to the square of the distance between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value. Also, if the distance gets tripled, then the gravitational force becomes one-ninth of its original value.

(iii) Force is directly proportional to the product of the masses. When masses of both the objects are doubled, then the gravitational force becomes four times.

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

Ans-

Importance of the universal law of gravitation

(i) It binds us to the earth.

(ii) It explains the motion of moon around the earth.

(iii) It explains the motion of planets around the sun.

Q.8 What is the acceleration of free fall?

Ans-

An object falling towards the earth under the effect of gravitation is said to be in free fall. Acceleration of free fall also called acceleration due to gravity is 9.8 ms−2 near the surface of earth. It is constant for all the objects.

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

Ans-

Gravitational force acting between the earth and an object is called the weight of the object.

Q.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.]

Ans-

As the gold purchased at the equator weighs less than at the poles so the value of acceleration due to gravity (g) for an object is greater at poles than at the equator. Therefore, Amit’s friend will not agree with the weight of the gold bought.

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

Ans-

When the paper is crumbled in the shape of a ball, its surface area decreases. Thus, air does not exert too much resistance to its motion and it falls faster than the sheet of paper.

Q.12 Gravitational force on the surface of the moon is only MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0le9yqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaqGXaaabaGaaeOnaaaaaaa@376B@

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?

Ans-

$\begin{array}{l}\text{Let,}\text{\hspace{0.17em}}{\text{Weight of an object on the moon = W}}_{\text{m}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Weight}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}{\text{an object on the Earth = W}}_{\text{E}}\\ \text{Hence,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{W}}_{\text{m}}\text{=}\frac{\text{1}}{\text{6}}{\text{× W}}_{\text{E}}\\ \text{Also, Weight = Mass × Acceleration}\\ \text{Acceleration due to gravity, g = 9}{\text{.8 ms}}^{\text{-2}}\\ \text{Hence, weight of a 10 kg object on the Earth}\\ \text{= 10 kg × 9}{\text{.8 ms}}^{\text{-2}}\text{= 98}\text{\hspace{0.17em}}\text{N}\\ \text{And,}\text{\hspace{0.17em}}\text{weight}\text{\hspace{0.17em}}\text{of the same object on the moon}\\ \text{=}\frac{\text{1}}{\text{6}}\text{× 98}\text{\hspace{0.17em}}\text{N =16}\text{.3}\text{\hspace{0.17em}}\text{N}\end{array}$

Q.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.

Ans-

$Given, u = 49 ms -1 , Here, v = final velocity of the ball = 0 Let, s = maximum height achieved by the ball On using third equation of motion, V 2 =u 2 + 2 as Where, a = g = acceleration due to gravity It would be negative, as the ball has thrown upward On putting these values, we have, 0 = ( 49 ms -1 )×( 49 ms -2 ) 2×9 .8 ms -2 = 122.5 m Let, t = time taken by the ball to reach at height of 122.5 m. On using the first equation of motion under gravity , v = u + at On putting the values, 0 = 49 ms -1 + t × ( -9.8 ) t = 49 ms -1 9 .8 ms -2 = 5 s Here, Time of ascent = Time of descent Hence, total time taken by the ball of return = 5 s + 5 s = 10 s$

Q.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.

Ans-

$Given, inital velocity, u = 0 height of the stone, h = 19.6 m acceleration due to gravity, g = 9.8 ms -2 Let, final velocity of the stone, v = ? On using the third equation of motion under gravity, v 2 = u 2 + 2gh On putting the values, v 2 = 0 + 2 (9.8 ms -2 )(19.6 m) v 2 = ( 19 .6ms -1 ) 2 v = 19.6 ms -1$

Q.15 A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Ans-

$Given, inital velocity, u = 40 ms -1 final velocity, v = 0 ms -1 gravitational acceleration for upward motion, g = -10 ms -2 Let h be the height attained by the stone Using third equation of motion under gravity, v 2 = u 2 +2gh On putting the values, 0 = ( 40 m s -1 ) 2 +2( -10 m s -2 )h ∴ h = 1600 m 2 s -2 20 m s -2 =80 m Now, total distance covered by the stone after reaching on the earth = 80 m + 80 m = 160 m. The net displacement of the stone when it reaches to the ground = 80 m – 80 m = 0.$

Q.16 Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m.

Ans-

$\begin{array}{l}{\text{Given, Mass of the Sun, m}}_{\text{1}}{\text{= 2 × 10}}^{\text{30}}\text{kg}\\ {\text{Mass of the Earth, m}}_{\text{2}}{\text{= 6 × 10}}^{\text{24}}\text{kg}\\ \text{Average distance between the Earth and the Sun, r = 1}{\text{.5 × 10}}^{\text{11}}\text{m}\\ \text{Universal gravitational constant, G = 6}\text{.67}×{\text{10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ \text{As per universal law of gravitation,}\\ \text{Force, F = G}\frac{{\text{m}}_{\text{1}}{\text{m}}_{\text{2}}}{{\text{r}}_{\text{2}}}\\ \text{On putting the given values,}\\ \text{=}\left(\text{6}\text{.67}×{\text{10}}^{\text{11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\right)\frac{\left(2×{10}^{30}\text{kg}\right)×\left(6×{10}^{24}\text{kg}\right)}{{\left(1.5×{10}^{11}\text{m}\right)}^{2}}\\ \text{= 3}{\text{.57 × 10}}^{\text{22}}\text{N}\end{array}$

Q.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.

Ans-

$\begin{array}{l}\text{}\end{array}$ $Let, two stones meet at point O in time t. For the stone dropped from the tower. Initial velocity, u = 0 Let the displacement of the stone in time t from the top of the tower = s g = 10 ms -2 On using the second equation of motion under gravity, s = ut + 1 2 gt 2 or, = 0 + 1 2 ( 10 ms -2 ) t 2 or, s = 5t 2 …(i) For the stone thrown upward Initial velocity, u = 25 ms -1 Let the displacement of the stone form the ground in time t = s’ Acceleration due to gravity, g = -10 ms -2 On using the second equation of motion under gravity, s = ut + 1 2 gt 2 or, = 25t + 1 2 ( -10 ms -2 ) t 2 or, s’ = ( 25 ms -1 ) t – 5t 2 ….( ii ) Now, from the figure, s+s’ = 100 m ..( iii ) ∴ 5t 2 +25t – 5t 2 =100 m ⇒ t = 100 m 25 ms -1 = 4 s Hence, Both the stones will meet after 4 s. Now, from ( 1 ), distance covered by first stone in 4 s s = 5t 2 = 5 ( 4 s ) 2 = 80 m From ( iii ), distance covered by second stone in 4 s, s’ = 100 m – 80 m = 20 m$

Q.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

Ans-

$\begin{array}{l}\text{(a)\hspace{0.17em}Time of ascent of the ball = Time of descent of the ball}\\ \text{As total time of journey = 6 s}\\ \therefore \text{\hspace{0.17em}Time of ascent = 3 s}\\ {\text{For upward motion, v = 0, g = -9.8 ms}}^{\text{-2}}\\ \text{Now, by using the relation, v = u + gt}\\ {\text{or, 0 = u + (-9.8 ms}}^{\text{-2}}\text{) × 3\hspace{0.17em}s}\\ {\text{or,\hspace{0.17em}u = 29.4 ms}}^{\text{-1}}\\ {\text{Therefore,\hspace{0.17em}the ball was thrown upwards with a velocity of 29.4 ms}}^{\text{-1}}.\\ \left(\text{b}\right)\\ {\text{Here,\hspace{0.17em}Initial velocity, u = 29.4 ms}}^{\text{-1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Final velocity, v = 0}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Acceleration due to gravity, g = -9.8 ms}}^{\text{-2}}\\ \text{Let, the maximum height = h}\\ \text{on\hspace{0.17em}using the relation}\\ \text{h = ut +}\frac{\text{1}}{\text{2}}{\text{\hspace{0.17em}gt}}^{\text{2}}\\ \text{On putting the given values,}\\ {\text{h = 29.4 ms}}^{\text{-1}}\text{× 3 s +}\frac{\text{1}}{\text{2}}\left({\text{-9.8\hspace{0.17em}m\hspace{0.17em}s}}^{\text{-2}}\right){\left(\text{3\hspace{0.17em}s}\right)}^{\text{2}}\text{= 44.1\hspace{0.17em}m}\\ \text{(c)}\\ \text{After 3 s,\hspace{0.17em}the ball attains the maximum height}\\ {\text{At maximum height, initial velocity, u = 0 m s}}^{\text{-1}}\\ \text{Position of the ball after 4 s of the throw is given by the}\\ \text{distance travelled by it during its downward}\\ \text{journey in (4s – 3s) = 1 s}\\ \text{On\hspace{0.17em}using the relation for downward motion,}\\ \text{s\hspace{0.17em}=\hspace{0.17em}ut +}\frac{\text{1}}{\text{2}}{\text{at}}^{\text{2}}\\ \text{s\hspace{0.17em}=\hspace{0.17em}0\hspace{0.17em}×\hspace{0.17em}t\hspace{0.17em}+}\frac{\text{1}}{\text{2}}\text{×\hspace{0.17em}}\left({\text{9.8\hspace{0.17em}ms}}^{\text{-2}}\right){\left(\text{1s}\right)}^{\text{2}}\\ \text{=\hspace{0.17em}4.9\hspace{0.17em}m}\\ \text{As total height = 44.1 m}\\ \text{Therefore, the\hspace{0.17em}}\mathrm{h}\text{eight\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}ball\hspace{0.17em}from\hspace{0.17em}the\hspace{0.17em}ground\hspace{0.17em}after ​4\hspace{0.17em}s would\hspace{0.17em}be\hspace{0.17em}}\\ \text{=\hspace{0.17em}}\left(\text{44.1\hspace{0.17em}m – 4.9​\hspace{0.17em}m}\right)\text{= 39.2\hspace{0.17em}m.}\end{array}$

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

Ans-

The buoyant force acts in the upward direction for an object immersed in a liquid.

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

Ans-

A plastic block under the water experiences two forces. One is the gravitational force, which pulls the block downwards and the other is the buoyant force which pushes the block upwards. If the upward buoyant force is greater than the downward gravitational force, then the block comes up to the surface of the water as soon as it is released within water.

Q.21 The volume of 50 g of a substance is 20 cm3. If the density of water is 1 g cm−3, will the substance float or sink?

Ans-

$\begin{array}{l}\text{Density}\text{of}\text{the}\text{substance}=\frac{\text{Mass}\text{of}\text{the}\text{substance}}{\text{Volume}\text{of}\text{the}\text{substance}}\\ =\frac{50 g}{20{\text{cm}}^{3}}=\text{2}.\text{5}{\text{gcm}}^{-3}\end{array}$

if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid. Conversely, the density of an object is more than the density of a liquid, then it sinks in the liquid.

Here, the density of the substance is more than the density of water i.e., 1 g cm−3. Therefore, the substance will sink into the water.

Q.22 The volume of a 50 g sealed packet is 350 cm3. 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?

Ans-

$\begin{array}{l}\text{Density of packet =}\frac{\text{Mass}}{\text{Volume}}\text{=}\frac{\text{500 g}}{{\text{350 cm}}^{\text{3}}}\text{=1}{\text{.428 gcm}}^{\text{-3}}\\ \text{Here, the density of the packet is more than the density of water}\text{.}\\ \text{Therefore, it will sink into the water}\text{.}\\ \text{The mass of water displaced}\\ \text{= volume of packet × density of water}\\ {\text{= 350 cm}}^{\text{3}}{\text{× 1 gcm}}^{\text{-3}}\text{=350g}\end{array}$

1. How to score full marks in NCERT solutions class 9 science chapter 10?

Understanding Science concepts would be easier if you have followed NCERT solutions to clarify your concepts to solve long and short answer questions, MCQs and intext and end text questions. Also, don’t forget to take notes and practise with sample papers.  Chapter 10 Gravitation Science chapter will give you the highest grades if your understanding of all topics is good.  You will be pleasantly surprised by the excellent grades in the board exams

2. How to use NCERT Solutions for Class 9 Science Chapter 10?

These Solutions give students the information they need to comprehend the chapters clearly and concisely.

1. The NCERT solutions are written and edited by subject matter experts who meticulously researched and prepared concise, accurate and to the point answers of each and every chapter to help the students get excellent results in their exams.. Students swear by Extramarks solutions and it’s an integral part of their NCERT syllabus. If you wish to step up your preparation, just the NCERT books won’t suffice.