NCERT Solutions Class 9 Science Chapter 10 Gravitation have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 9 Science have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 9 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 9 Science are an important part of exams for Class 9 Science and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 9 Science and also download more latest study material for all subjects. Chapter 10 Gravitation is an important topic in Class 9, please refer to answers provided below to help you score better in exams

## Chapter 10 Gravitation Class 9 Science NCERT Solutions

Class 9 Science students should refer to the following NCERT questions with answers for Chapter 10 Gravitation in Class 9. These NCERT Solutions with answers for Class 9 Science will come in exams and help you to score good marks

### Chapter 10 Gravitation NCERT Solutions Class 9 Science

**Chapter 10 – Gravitation**

**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 m1 and m2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:

*F* = G*m*_{1}*m*2/*r*_{2}

Where, G is the universal gravitation constant given by:

G = 6.67x10^{-11} Nm^{-2}kg^{-2}

**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 ME 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:

*F* = G*m*_{1}*m*2/*r*_{2}

**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}.

**Chapter 10 – Gravitation**

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

**Answer:**

S.No | Mass | Weight | |

I. | Mass is the quantity of matter contained in the body. | Weight is the force of gravity acting on the body. | |

II. | It is the measure of inertia of the body. | It is the measure of gravity. | |

III. | Mass is a constant quantity. | Weight is not a constant quantity. It is different at different places. | |

IV. | It only has magnitude. | It has magnitude as well as direction. | |

V. | Its SI unit is kilogram (kg). | Its SI unit is the same as the SI unit of force, i.e., Newton (N). |

**Question 2: Why is the weight of an object on the moon 1/6th 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,

*W*_{E} = GM_{E}*m* /R_{E}^{2}

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:

*W*_{M} = GM_{M}^{m} /*R*_{M}

*W*_{M}/*W _{E}* = M

_{M}R

_{E}2 / M

_{E}R

_{M}

^{2}

Where , ME = 5.98 x 10^{24}kg. M_{M}= 7.36 x10^{22}kg

RE = 6.4 x 10^{6} m, R_{M} = 1.74 x 10^{6} m

∴ WM /WE = 7.36 x 10^{22} x(6.37 x 10^{6})^{2} / 5.98 x 10^{24} x(1.74 x 10^{6})^{2} = 0.165 ≈ 1/6

Therefore, weight of an object on the moon is 1/6 of its weight on the Earth.

**Chapter 10 – Gravitation**

**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.

**Chapter 10 – Gravitation**

**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 iron bar is heavier than the bag of cotton. 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 iron bar is more that that of the cotton bag.

**Chapter 10 – Gravitation**

**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 (r) between them, i.e., F ∝ 1/r2

If distance r becomes r/2, then the gravitational force will be proportional to 1/(r/2)^{2} = 4/r^{2}

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

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

*F* = G*MM/r ^{2}*

Where,

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}

Since the object is on the surface of the Earth, r = radius of the Earth (R)

r = R = 6.4 × 10^{6} m

Gravitational force, *F* = G*MM/*R^{2}

= 6.7 x 10^{-11} x 6 x 10^{24} x1 / (6.4 x 10^{6})^{2} = 9.8N

**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:**

(i) Doubled (ii) One-fourth and one-ninth (iii) four times

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

F = Gm_{1}m_{2}/r^{2}

(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 m s−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 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 x Weight of an object on the Earth Also,

Weight = Mass × Acceleration

Acceleration due to gravity, g = 9.8 m/s2

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 x98 = 16.3N

**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) 122.5 m (ii) 10 s

According to the equation of motion under gravity:

v^{2} − u^{2} = 2 gs

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

u = 49 m/s

During upward motion, g = − 9.8 m s^{−2}

Let h be the maximum height attained by the ball.

Hence,

0-(49)^{2} = 2x(-9.8) x h

h = 49 x 49/2 x 9.8 = 122.5m

Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:

v = u + gt

We get,

0 = 49 + t x (-9.8)

9.8t = 49

t = 49/9.8 = 5s

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} = 2 gs

Where,

u = Initial velocity of the stone = 0

v = Final velocity of the stone

s = Height of the stone = 19.6 m

g = Acceleration due to gravity = 9.8 m s−2

∴ v^{2} − 02 = 2 × 9.8 × 19.6

v^{2} = 2 × 9.8 × 19.6 = (19.6)^{2}

v = 19.6 m s^{− 1}

Hence, the velocity of the stone just before touching the ground is 19.6 m s^{−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}= 2 gs

Where,

u = Initial velocity of the stone = 40 m/s

v = Final velocity of the stone = 0

s = Height of the stone

g = Acceleration due to gravity = −10 m s

^{−2}

Let h be the maximum height attained by the stone.

Therefore,

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

h = 40 x 40 /20 = 80m

Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m

Net displacement of the stone 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:

F = GM

_{sun}M

_{Earth}/ R

^{2}

Where,

M_{Sun} = Mass of the Sun = 2 × 10^{30} kg

M_{Earth} = Mass of the Earth = 6 × 10^{24} kg

R = Average distance between the Earth and the Sun = 1.5 × 10^{11} m

G = Universal gravitational constant = 6.7 × 10^{−11} Nm^{2} kg^{−2}

F = 6.7 x 10^{-11} x 2 x 10^{30} x 6 x 10^{24} / (1.5 x 10^{11})^{2} = 3.5 x 10^{22}N

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

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

(i) 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 be s.

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

From the equation of motion,

s = ut+1/2gt^{2}

= 0xt+1/2x9.8xt^{2}

∴ s = 4.9t^{2}

(ii) For the stone thrown upwards:

Initial velocity, u = 25 m s^{−1}

Let the displacement of the stone from the ground in time t be s'.

Acceleration due to gravity, g = −9.8 m s^{−2}

Equation of motion,

s = ut + 1/2gt^{2}

= 25t - 1/2 x9.8 x t^{2}

∴ s = 25t = 4.9t^{2 }

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

∴ s + s'=100

1/2gt^{2 }+ 25t - 1/2gt^{2}= 100

∴ t = 100/25 = 4s

In 4 s, the falling stone has covered a distance given by equation (1) as

s = 1/2 x10x4^{2} = 80m

Therefore, the stones will meet after 4 s at a height (100 − 80) = 20 m from them 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) 29.4 m/s (b) 44.1 m (c) 39.2 m above the ground

(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

Acceleration due to gravity, g = −9.8 m s^{−2}

Equation of motion, v = u + gt will give,

0 = u + (−9.8 × 3)

u = 9.8 × 3 = 29.4 ms^{− 1}

Hence, the ball was thrown upwards with a velocity of 29.4 m s^{−1}.

(b) Let the maximum height attained by the ball be h.

Initial velocity during the upward journey, u = 29.4 m s^{−1}

Final velocity, v = 0

Acceleration due to gravity, g = −9.8 m s^{−2}

s = ut + 1/2at^{2}

From the equation of motion,

h = 29.4 x 3+1/2 x-9.8 x(3)^{2} = 44.1m

(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

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.

Equation of motion, s = ut + 1/2gt^{2} will give,

s = 0 x t + 1/2 x 9.8x 1^{2} = 4.9m

Total height = 44.1 m

This means that 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 cm3. 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.

Here, density of the substance = Mass of substance / Volume of substance = 50/20 = 2.5g cm^{-1}

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 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?**

**Answer: **Density of the 500 g sealed packet = Mass of packet/ Volume of packet= 500 / 350 = 1.428g cm^{-1 }

The density of the substance is more than the density of water (1 g cm^{-1} ). 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.

**Question. Ether (density 0.71 g/cc), water (density 1.00 g/cc) and mercury (density 13.6 g/cc) are 3 liquids which do not mix with each other. They are filled in a container as shown here.A piece of diamond (density 3.5 g/cc) is dropped into the liquid. Where will it come to rest?**

(a) In the ether layer

(b) In the water layer.

(c) Between the water and the mercury layers.

(d) In the mercury layer (it will sink to the bottom

**Answer : C**

**Question. A paper and a stone are dropped from the top of a building. Which one will reach the ground first and why?**

(a) The stone, because it is heavier (air resistance plays no part.)

(b) The stone, only because it faces much less air resistance

(c) The paper, because it is lighter (air resistance plays no part.)

(d) The paper, only because it faces much less air resistance

**Answer : B**

**Question. A body has a mass of 2 kg. When will the mass of the body change?**

(a) When the body is taken to the moon

(b) When the body is dropped from a height

(c) When the body is being pulled along a smooth surface.

(d) The mass of the body will not change unless it is cut or broken

**Answer : D**

**Question. When a solid cube made of wax (density 0.9 g/cc) is placed in a beaker of alcohol (density 0.8 g/cc), it sinks (see figure 1). The same cube when placed in water (density 1g/cc), it floats (see figure 2). What will happen if the cube of wax is placed in vegetable oil (density 0.9 g/cc, almost the same as the wax itself)?**

(a) It will sink to the bottom (same as figure 1

(b) It will float (same as figure 2).

(c) It is not possible that a solid and liquid have the same density.

(d) The solid will stay in the liquid at any point it is placed without sinking or floating.

**Answer : D**

**Question. The drawing shows an apple falling to the ground. In which of the three positions does gravity act on the apple?**

(a) 2 only

(b) 1 and 2 only

(c) 1 and 3 only

(d) 1, 2, and 3

**Answer : D**

**Question. The density of cork is 0.2 g/cc. A 10 cm tall cubical piece of cork floats on water (density: 1 g/cc) as shown. Which of the diagrams correctly represents the same piece of cork when placed in methanol whose density is 0.8g/cc?**

**Answer : D**

**Question. The 'mobile' shown is completely balanced. The sticks and strings are weightless. If the mass of P is 30 grams, what is the mass of Q?**

(a) 20g

(b) 30g

(c) 40g

(d) 60g

**Answer : A**

**Question. A block of wood is cut as shown in the figure.What will happen to its mass, volume and density?**

(a) Mass and density will remain the same but volume will decrease.

(b) Mass and volume will remain the same but density will decrease.

(c) Mass remains the same but density and volume will decrease.

(d) Density will remain the same but mass and volume will decrease.

**Answer : D**

**Question. Generally the density of a solid is higher than its liquid form. This is NOT true for:**

(a) Wax

(b) Water

(c) Iron

(d) Sodium

**Answer : B**

**Question. A beam is mounted on the fulcrum as shown. A 5 kg weight on pan P is balanced by placing a 2 kg weight on pan Q. Now a 2 kg weight is added to pan P. What must be added to Q to maintain the balance?**

(a) 0.8 kg

(b) 1.2kg

(c) 2 k

(d) 5 kg

**Answer : A**

**Question. An apple is falling from a tree. At which of the points shown is its speed the highest?**

(a) A

(b) B

(c) C

(d) D

**Answer : C**

NCERT Solutions Class 9 Science Chapter 1 Matter in Our Surroundings |

NCERT Solutions Class 9 Science Chapter 2 Is Matter Around Us Pure |

NCERT Solutions Class 9 Science Chapter 3 Atoms and Molecules |

NCERT Solutions Class 9 Science Chapter 4 Structure of the Atom |

NCERT Solutions Class 9 Science Chapter 5 The Fundamental Unit of Life |

NCERT Solutions Class 9 Science Chapter 6 Tissues |

NCERT Solutions Class 9 Science Chapter 7 Diversity in Living Organisms |

NCERT Solutions Class 9 Science Chapter 8 Motion |

NCERT Solutions Class 9 Science Chapter 9 Force and Laws of Motion |

NCERT Solutions Class 9 Science Chapter 10 Gravitation |

NCERT Solutions Class 9 Science Chapter 11 Work and Energy |

NCERT Solutions Class 9 Science Chapter 12 Sound |

NCERT Solutions Class 9 Science Chapter 13 Why Do We Fall Ill |

NCERT Solutions Class 9 Science Chapter 14 Natural Resources |

NCERT Solutions Class 9 Science Chapter 15 Improvement in Food Resources |

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

NCERT Solutions Class 9 Science Chapter 10 Gravitation is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 9 Science textbook online or you can easily download them in pdf.

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

The Class 9 Science NCERT Solutions Chapter 10 Gravitation are designed in a way that will help to improve the overall understanding of students. The answers to each question in Chapter 10 Gravitation of Science Class 9 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 10 Gravitation Class 9 chapter of Science so that it can be easier for students to understand all answers.

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

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#### Chapter 10 Gravitation Class 9 NCERT Solution Science

These solutions of Chapter 10 Gravitation NCERT Questions given in your textbook for Class 9 Science have been designed to help students understand the difficult topics of Science in an easy manner. These will also help to build a strong foundation in the Science. There is a combination of theoretical and practical questions relating to all chapters in Science to check the overall learning of the students of Class 9.

#### Class 9 NCERT Solution Science Chapter 10 Gravitation

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