**NCERT solutions of class 9 science chapter 10- Gravitation**

NCERT Solutions for Class 9 Science Chapter 10- Gravitation gives you an essential channel of knowledge on the ideas associated with one of the greatest natural laws of physics-gravitational force. NCERT Solutions for Class 9 Science Chapter 10 Gravitation provides clarifications of the concept of chapter 10-Gravitation and will help you in understanding the ideas systematically.

Gravity is an input for resolving problems of macrophysics that clarifies numerous things. From how our planet stays in elliptical orbits to why things tumble down. By studying NCERT Solutions for Class 9 Science Chapter 10 – Gravitation you will clear all the concepts regarding gravity. Content is made by profoundly qualified teachers and subject experts with many years of experience. Additionally, the NCERT solutions have been refreshed to incorporate the latest study material endorsed by the CBSE board.

Moreover, we guarantee that pertinent study material on NCERT Solutions Class 9 is routinely refreshed according to the standards and requirements that CBSE team of experts regularly search for the recruitment of board exams. This guarantees that the study material is custom-fitted to be class pertinent, yet without forfeiting the instructive remainder. Future study point additionally endeavors to give greatest enlightening worth without expanding the intricacy of points. This is accomplished by guaranteeing that the language is very simple and all technical languages are clarified as per the level of class.

**Download PDF -NCERT solutions of class 9 science chapter 10- Gravitaion**

**PDF of NCERT solutions of class 9 science chapter 10-Gravitation**

**NCERT solutions of class 9 science chapter 10- Gravitation**

The gravitational force was discovered by Newton when he was sitting under a tree and an apple fell on him, he observed that the earth attracts everything towards it, the question raised in his mind if the Earth attracts an apple. Can it not attract the moon?

He made experiments in his lab and observed the force between two objects in the whole of the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

**NCERT Solutions of class 9 science **

**NCERT solutions of class 9 science chapter 10- Gravitaion**

** ****Page-**** ****134**

**Q1.State the universal law of gravitation?**

Ans. According to the universal law of gravitation there exist a force of attraction between two objects in the universe, the magnitude of this force is proportional to the product of their masses and inversely proportional to the square of the distance between them. The magnitude of gravitational force is given as following.

** **

Where m & M are masses of both objects, d is the distance between the objects and G is the universal constant of gravitational force.

**Q2.Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the Earth.**

Ans. When the object exists on the surface of the Earth then d = r, the distance between the object and the earth will become = r, where ‘r’ is the radius of the earth. The gravitational force on the object will be as follows.

Substituting d = r** **

** ****Page-**** ****138**** **

**Q1.** **What are the differences between the mass of an object and its weight?**

**Mass. **Mass is the amount of substance contained in an object,It is constant everywhere in the Earth, moon or in the outer space. It is a scalar quantity or is denoted by the magnitude only. Mass is proportional to the amount of inertia of any object.Its SI unit is the kilogram.

**Weight- **Weight is the force of gravity by which the Earth (or any planet) attracts any object towards it. According to the Newton’s second law of motion

F = m × a

Where F is the force and a is the acceleration produced on an object.

In case of when the Earth attracts an object by a force F.

F = m × g, where g is the gravitational acceleration produced by the Earth to any object. Weight is denoted by W

W = m ×g ( g = 9.8 ms^{-1})

Weight is the vector quantity because it acts vertically downwards to the Earth and so it has magnitude as well as direction. Its SI unit is newton (N).

**Q2.** **Why is the weight of an object on the moon 1/6**** ****th of its weight on the earth?**

Ans. The mass of the moon is lesser than the earth due to which it exerts lesser force of attraction on objects.

Let the mass of an object be m. Let the mass of the moon M_{m}. Let the weight of the object at the moon is W_{m} and the radius of the moon is r_{m}.

According to the universal law of gravitational force, the weight of an object on the moon will be .

M_{m} = 7.36 × 10^{22} kg , r_{m} = 1.74 × 10^{6} m,

Let the mass of the earth is M_{e} , its radius is R_{e,} and the weight of the object W_{e} on the earth will be as follows.

M_{e} = 5.98 × 10^{24}kg,

r_{e} = 6.37 × 10^{24}

Substituting these values in (i) and (ii), we get

Weight of the object on the moon

= (1/6) × the weight of the object on the earth

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** ****Page. 141**

**Q1.** **Why is it difficult to hold a school bag having a strap made of a thin and strong string?**

Ans. Schoolbag having a thin string occupies lesser area which results in larger pressure on our shoulders as we know that

Where the thrust is the force produced by the weight of bag to the strap and acts on the soldier, here in the case of handling a bag the thrust is acting in the perpendicular direction to the surface of the shouders which is in contact with the straps of the bag. Since pressure is inversely proportional to the area so thinner straps will apply more pressure.

**Q2.** **What do you mean by buoyancy ?**

Ans. The upward force exerted by the fluids on an immersing body is known as the force of buoyancy or upward thrust, it is because of this force every objects becomes lighter when immersed in a liquid. The force of buoyancy is proportional to the density of the liquid.

**Q3.** **Why does an object float or sink when placed on the surface of the water?**

Ans. When we placed an object into the water it will float if the density of the object is less than the water and it will sink if the density of the object is more than the water.

** Page 142**

**Q1.** **You find your mass to be 42 kg on a weighing machine .Is your mass more or less than 42 kg ?**

Ans. When we weigh ourself on the weighing machine a buoyant force acts on our body which is proportional to the volume of the body, so the reading of 42 kg is not quiet O.K our mass must be more than 42 kg. Let the force of buoyancy is f and actual waight is W, then reading will be W – f = 42, W = 42 +f,so actual waight will 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 ?**

Ans. Let the actual waight of cotton W_{C} and of iron bar W_{F} and force of buoyancy on them are B_{C} and B_{F} respectively. Volume of cotton is more so B_{C} > B_{F}. Readings of both are same on weighing machine i.e 100 kg. W_{C} – B_{C} = 100

W_{C} = 100 + B_{C}

W_{F} = 100 +B_{F}

B_{C} > B_{F}

So, W_{C} > W_{F, }cotton of 100 kg will be havier.

**EXERCISES**

**Q1.****How does the force of gravitation between two objects change when the distance between them is reduced to half ?**

**Ans. According to universal law of gravitation , the force of attraction between two objects is**

** **

**When distance reduced to half , d becomes d/2, substituting the distance then the gravitational force between them will become**

** ^{}**

** **

** **

**From equation (1) and equation (2) we have**

** **

F_{2} = 4F_{1}

Gravitational force will become 4 times of the previous one.

**Q2.** **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. The force by which the earth attracts every object is the weight (W) of the object.

W = m × g

According to second equation of motion .

** ^{}**

u = 0, Let object falls from a certain height H, so replacing s =H, a by the gravitational acceleration g, then

It is clear from this relation time taken by the object does not depend on the mass so heavy object and lighter object both will take same time if they fall simultaneously from certain height .

** Q3.** **What is the magnitude of the gravitational force between the earth and 1kg object on its surface?**

Ans. According to the universal law of the gravitational force

** **

Since the object is on the surface of the earth,so replacing d by r(radius of the earth)

M(mass of the earth) = 6 × 10^{24}kg, mass of the object ,m = 1kg, r = 6.4× 10^{6 }m

F = 9.8N

**Q4.****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. According to the universal law of the gravitational force, two objects have the masses and distance between them experience the same force or we can say the moon and the earth both attracts each other by the same force because the gravitational force is directly proportional to the product of their masses.

**Q5. I****f the moon attracts the earth. Why does the earth not move towards the moon?**

Ans. The gravitational force experienced by the moon and the earth is the same but the mass of the moon is lesser than the earth so as it is known to us that F = ma, where acceleration is inversely proportional to the mass so moon accelerates more compared to the earth and it moves around the earth.

**Q6. 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 the universal law of the gravitational force we have

^{}

If the mass of one object is doubled then the force F_{2}

will exist between the objects

From equation (i) and equation (ii)

**F _{2} = 2F_{1}**

**(ii)** The gravitational force between two objects is

The distance between two objects is doubled,so the distance between them will become =2d

From equation (i) and equation (ii)

4F_{2} = F_{1}

F_{2} = F_{1}/4, the force between them will become 1/4 of the previous one.

If the distance between them become = d/3

From equation (i) and equation (iii)

9F_{2} = F_{1}

F_{2 }= F_{1}/9

F_{2} =F_{1}/9, force between them will become 1/9 th.

(iii) If the masses of both objects are doubled, then their masses becomes 2m and 2M

The gravitational force between both objects will become

From equation (i) and equation (iv)

F_{2} =4F_{1}

The gravitational force between both objects will become 4 times of the previous one

**Q7.What is the importance of universal law of gravitation?**

Ans. All the planets are bounded by the force of gravitation of the Sun ,due to this force they are moving around the Sun.The moon is moving around the earth due to the gravitational force beween the earth and the moon.The tides arises in the ocean due to the force attracted by the moon to the earth.

**Q8.What is the acceleration of free fall ?**

Ans. Free fall means when the object is moving downwards under the control of gravity only. It is accelerated by a constant value of 9.8 m/s².

**Q9. What do we call the gravitation force between the earth and an object ?**

Ans. The gravitation force between the earth and an object is called the weight of the object.

**Q10.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 equater. Will the friend agree with the weight of gold bought ?If not .why?**

Ans. We have g = GM/r², the earth is flattened at the poles so the radius r will be lesser at the poles so the value of g will be more in poles and so the weight of gold will be lesser at the equator compared to poles. Therefore Amit’s friend will not agree to buy the gold from him.

**Q11.Why will a sheet of paper fall slower than one that is crumpled into a ball ?**

Ans. A sheet of paper has a larger area so it will face larger air resistance which makes it to move slower than when it is crumpled into a ball.

**Q12. 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 10kg object on the moon and on the earth?**

Ans. Weight of the object on the earth

= mg =10 ×9.8 =98N

In the moon the gravitational acceleration = g/6

The weight on the moon

=10 ×9.8/6

=98/6

=16.33N

Hence, the weight of a 10kg object on the moon is 16.33 N and on the earth is 98N

**Q13.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.(i) Let the maximum height to which it rises= h

Applying the second equation of the motion

v² = u² + 2gh

0 = 49² ─ 2× 9.8 x h

2 × 9.8h = 49 ×49

h = 49 49/2 × 9.8

h =122.5m

(ii) time of ascent = time of descent =t

v = u ─ gt

0 = 49 ─ 9.8 t

t =49/9.8

t =5 s

So time taken by the ball in returning to the surface of the earth

2 ×5=10 s

**Q14. A stone is released from the top of a tower of height 19.6m. Calculate its final velocity just before touching the ground .**

Ans.The initial velocity of the stone- u =0, h = 19.6 m

Let its final velocity before touching the ground = v

Applying third equation of the motion

v² = u² +2gh

v² = 0 + 2 × 9.8 ×19.6

v² = 19.6 × 19.6

v = 19.6m/s

**Q15.A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10m/s². Find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?**

Ans. Initial velocity of stone,u = 40 m/s, g = 10 m/s², Final velocity of stone,v= 0

Let the maximum height achieved by the stone = h

Applying the third equation of motion

v² = u² – 2gh

0 = 40² – 2 × 10h

20h = 1600

h = 80 m

Height of ascent = height of descent

So net displacement in total journey = 80 + (-80) = 0

And total distance covered by the stone = 2 × 80 = 160 m

**Q16.Calculate the force of gravitation beween 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.**

Ans. The gravitational force between two objects of the masses m and M which are distant apart by the distance d is given as following.

** **

** **** **

F = 35.73 × 10^{21}

F = 3.57 × 10^{22}N

So the net force between the Sun and the earth is =3.57 × 10^{22}N

**Q17. A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25m/s. Calculate when and where the two stones will meet.**

Ans. The initial velocity of the stone is,u = 0

Let the height of the stone achieved by the stone thrown upward when they meet to each other = h at time t the distance traveled by the stone falling downwards to the ground = 100 ─ h at the same time.

Let us start from the falling stone , u = 0

Applying the second equation of the motion

Replacing a by g, u = 0

The stone moving upward , u = 25 m/s

From (1) and (2)

25 t = 100

t =4

Both of the stone will meat after 4 second

Substituting t=4 in equation (i)

**h = 100 ─ 4.9 × 16**

**h = 100 ─ 78.4**

**h =21.6m**

**Both stone will meet after 4 s and at the height from surface of the earth 21.6 m**

**Q18. A ball is 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.

(a) Let the velocity of ball with which it was thrown = u

As we know , time of ascent = time of descent

t= 6/2 = 3 s

So, the time to achieve the maximum height = h

v = u ─ gt

0 = u ─ 9.8 ×3

u = 29.4 m/s

Hence,velocity of the ball with which it was thrown 29.4 m/s

(b)Let the maximum height the ball reaches= h

Applying the second equation of the motion

h = 88.2 ─ 4.9 x9

h = 88.2 ─ 44.1

h =44.1

The maximum height the stone will reach = 44.1 m

(c) Let the height of the stone after 4 s = h’

h’ =117.6 ─ 4.9 × 16

h’ = 117.6 ─ 78.4

h’ = 39.2

h’ = 39.2

The maximum height the stone will reach after 4 s= 39.2 m

**Q19. In what direction does the buyant force on an object immersed in a liquid act?**

** **Ans.The direction of the buoyant force on an immersed object in a liquid acts in an upward direction.

**Q20. Why does a block of plastic released under water come up to the surface of the water?**

** **Ans. The density of the plastic box is less than the density of water so the force of buoyancy is more than the weight of a block of plastic therefore it will float on the surface of the water.

**Q21. The volume of 50 gm of a substance is 20cm ^{3}. If the density of water is 1 g.cm^{-3}.will the substance float or sink?**

**Ans. Density of the substance = mass/volume=M/V = 50/20 = 2.5 g/cm**

^{3}

The density of the water is =1 g/cm^{3}

The density of the substance is more than the density of water so it will sink.

**Q22.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?**

Ans. The density of the sealed packet is

=M/V= 500/350

=1.42 g/cm^{-3}.

density of the water = 1 g cm^{-3}

It is clear that density of the object > density of the water

So, the sealed packet will sink into the water.

**Study notes of Maths and Science NCERT and CBSE from class 9 to 12**

**Summary of the chapter Gravity**

Gravity is one of the most important chapter of class 9 CBSE ,in this chapter you will study about universal law of gravitational force discovered by Newton.

According to Newton’s observation,the force exist between two bodies .This force is proportional to the product of their masses and inversely proportional to the square of the distance between them.

Let the mass of both objects are m and M ,distance between them is d

From (i) and (ii), we have

Where G is the universal constant of gravitational force

The value of gravitational force is 6.67 ×10^{-11 }m³kg^{-1}s^{-2}

All the heavenly bodies in the universe are under the impact of this balance force F, the Sun is moving along it solar system around the galactic centre of the milky way,all the planets in solar system are moving around the sun and all the setellites are moving around their corresponding planets.

**The gravitational force of an object on the earth**

Let the mass of an object is m and the mass of the earth is M,then gravitaional force F between object and the earth is

Here, the distance between the object and the earth is, d = r,where r is the radius of the earth

According to Newton’s second law of motion,the waight of the object is expressed as following

F = mg (where g is gravitational acceleration )

Waight of the object = Gravitational force on the object applied by the earth

Therefore after evaluating g ,we can compute the gravitational force on the object applied by the earth.

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**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

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**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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