Gravitation - chapter 10 NCERT Solutions for Class 9 Science by Md Farrukh Asif

 

NCERT Solutions for Class 9 Science

Chapter -10 Gravitation

NCERT Solutions for class 9 Science chapter-10 Gravitation

Chapter 10 of the NCERT Class 9 Science textbook, titled "Gravitation," covers the fundamental concepts of gravity and its effects on objects. The chapter delves into the following key topics:

1.       MATTER IN OUR SURROUNDINGS

2.       IS MATTER AROUND US PURE?

3.       ATOMS AND MOLECULES

4.       STRUCTURE OF THE ATOM

5.       THE FUNDAMENTAL UNIT OF LIFE

6.       TISSUES

7.       MOTION

8.       FORCE AND LAWS OF MOTION

9.       GRAVITATION

10.   WORK AND ENERGY

11.   SOUND

12.   IMPROVEMENT IN FOOD RESOURCES

1. **Introduction to Gravitation**: Understanding the concept of gravity and how it affects all objects with mass.

2. **Universal Law of Gravitation**: Newton's law states that every object in the universe attracts every other object with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

3. **Gravitational Force and Laws of Motion**: Exploring how gravitational force plays a role in the motion of objects, including the concepts of free fall and acceleration due to gravity.

4. **Mass and Weight**: Differentiating between mass (the amount of matter in an object) and weight (the force exerted by gravity on that mass).

5. **Free Fall**: Understanding objects falling under the influence of gravity alone, without any other force acting on them.

6. **Equations of Motion for Freely Falling Objects**: Applying the three equations of motion to objects in free fall.

7. **Gravitational Force of the Earth**: The concept of gravity specific to Earth, including how it keeps us grounded and affects the motion of celestial bodies.

8. **Gravitational Potential Energy**: The energy an object possesses due to its position in a gravitational field.

9. **Kepler's Laws of Planetary Motion**: Describing the motion of planets around the sun, including their elliptical orbits and the relationship between their orbital periods and distances from the sun.

Notes on Gravitation - Class 9

Introduction

• Gravitation: The force of attraction between any two bodies in the universe.
• Gravity: A particular case of gravitation where one body is Earth.
• The force of gravity: The force exerted by Earth on a body.

Universal Law of Gravitation

• Every object in the universe attracts every other object with a force which is:
• Directly proportional to the product of their masses.
• Inversely proportional to the square of the distance between them.
• Mathematically, F = G * (m1 * m2) / r^2
• F: Gravitational force
• G: Universal gravitational constant (6.67 x 10^-11 Nm^2/kg^2)
• m1, m2: Masses of the objects
• r: Distance between the centers of the objects

Acceleration Due to Gravity (g)

• The acceleration produced in a body falling freely under the action of gravity.
• Its value is approximately 9.8 m/s^2 on Earth.
• It is a vector quantity, directed towards the center of the Earth.
• It is independent of the mass of the falling body.

Mass and Weight

• Mass: The amount of matter contained in a body.
• Scalar quantity
• Measured in kilograms (kg)
• Remains constant everywhere in the universe.
• Weight: The force with which a body is attracted towards the center of the Earth.
• Vector quantity
• Measured in newtons (N)
• Varies from place to place.
• Weight = mass x acceleration due to gravity (W = mg)

Free Fall

• The motion of a body under the influence of gravity alone, without any other force acting on it.
• The acceleration of a freely falling body is equal to the acceleration due to gravity (g).
• Equations of motion can be applied to free fall by replacing 'a with 'g'.

Thrust and Pressure

• Thrust: The force acting on a body perpendicular to its surface.
• Pressure: The force acting on a unit area of a surface.
• Pressure = Thrust / Area
• The unit of pressure is Pascal (Pa)

Buoyancy

• The upward force exerted by a fluid on an object immersed in it.
• A buoyant force acts opposite to the weight of the object.
• Archimedes' principle: The buoyant force acting on an object equals the weight of the fluid displaced by the object.

 he mass per unit volume of a substance.

• Density = Mass / Volume
• The unit of density is kg/m^3

Relative Density

• The ratio of a substance's density to a reference substance's density.
• Usually, water is taken as the reference substance.
• Relative Density = Density of substance / Density of water
• Relative density has no units.

Important Points

• The value of g decreases with height above the Earth's surface and increases with depth below the Earth's surface.
• The value of g is different for different planets.
• Objects of different masses fall with the same acceleration due to gravity in a vacuum.

Note: This is a basic outline of the Gravitation chapter. For a deeper understanding, refer to your textbook and practice numerical problems.

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Question  and Answer begins here….

Question  1. State the universal law of gravitation.
Answer:

According to Newton’s universal law of gravitation :
Every mass in this universe attracts every other mass with a force that is directly proportional to the product of two masses and inversely proportional to the square of the distance between them.

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:

The formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth is given below:

F = Mm/d²

F = magnitude of gravitational force

G = Universal gravitation constant

M = mass of earth

m = mass of object

d = distance of the object from the center of the earth

Question  3. What do you mean by free fall?
Answer:

It is the object falling towards earth under the influence of the attraction force of earth or gravity.

Question  4. What do you mean by acceleration due to gravity?
Answer:

During free fall any object that has mass experiences force toward the center of the earth and hence acceleration works as well. “acceleration experienced by an object in its free fall is called acceleration due to gravity.” It is denoted by g.

Question  5. What are the differences between the mass of an object and its weight?
Answer:

 

S. No.	Mass	Weight
1	Mass is the quantity of matter contained in the body.	Weight is the force of gravity acting on the body.
2	It is the measure of inertia of the body.	It is the measure of gravity.
3	Mass is a constant quantity.	Weight is not a constant quantity. It is different at different places.
4	It only has magnitude.	It has magnitude as well as direction.
5	Its SI unit is kilogram (kg).	Its SI unit is the same as the SI unit of force, i.e., Newton (N).

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

Answer:

 The mass of an object remains the same whether on Earth or moon but the value of acceleration on the moon is 1/6^th of the value of acceleration on Earth. Because of this weight of an object on the moon is 1/6^th its weight on the Earth.

Question  7. 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 a bag of that kind will make its weight fall over a small area of the shoulder and produce greater pressure which makes holding the bag difficult and painful.

As the pressure increases, it is directly inversely proportional to its mass.

Question  8. What do you mean by buoyancy?
Answer:

It is the upward force experienced by an object when it is immersed in a fluid.

Question  9. Why does an object float or sink when placed on the surface of water?
Answer:

As an object comes in contact with the surface of a fluid it experiences two types of forces: gravitational force or gravity that pulls the object in a downward direction and the second force is the force of buoyancy that pushes the object in an upward direction.
It is these two forces that are responsible for an object to float or sink
i.e. if gravity >buoyancy then the object sinks
if gravity <buoyancy then object floats

In other words:

·         An object sinks in water if its density is greater than that of water.

·         An object floats in water if its density is less than that of water.

Question  10. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:

The force of gravitation between two objects is inversely proportional to the square of the distance between them therefore the gravity will become four times if the distance between them is reduced to half.

Question  11. 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:

In the free fall of objects the acceleration in velocity due to gravity is independent of the mass of those objects hence a heavy object does not fall faster than a light object.

Question  12. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (The mass of the earth is 6 x 1024 kg and the radius of the earth is 6.4 x 106 m.)
Answer:

F = G.M.m/d²

= (6.6 x 10^-11 x 6 x 10²⁴ x 1)  /(6.4 x 10⁶)²

= 9.81 N

Question  13. 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  14. If the moon attracts the Earth, why does the Earth not move towards the moon?
Answer:

Earth does not move towards the moon because the mass of the moon is very small compared to that of Earth.

Question  15. 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:

From the Universal law of gravitation, the force exerted on an object of mass m by the earth is given by

F= GMm/r2

(i) The force between two objects will be doubled.

Now m is 2m

F’= GM2m/r2

 (ii) the force between two objects will become 1/4th and 1/9th of the present force.

(iii) the force between two objects will become four times the present force. Force F is directly proportional to the product of both the masses. So, if both the masses are doubled then the gravitational force of attraction becomes four times the original value.

Question  16. What is the importance of the universal law of gravitation?
Answer:

The universal law of gravitation is important due to the following:

i) this law explains well the force that binds us to earth.

ii) this law describes the motion of planets around the sun.

iii) this law justifies the tide formation on Earth due to the moon and sun.

iv) this law gives reason for the movement of the moon around the earth.

Question  17. What is the acceleration of free fall?
Answer:
The acceleration of free fall is  g = 9.8 m/s2(on earth)

Question  18. What do we call the gravitational force between the earth and an object?
Answer: Weight.

Question  19. Amit buys a 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:

The 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 the 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  20. 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  21. The gravitational force on the surface of the moon is only 1/6 as strong as the gravitational force on the earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?
Answer:

value of gravity on earth g = 9.8 m/s2

value of gravity on moon = 1/6th of earth = 9.8/6 = 1.63 m/s2

weight of object on moon = m x 1.63 = 10 x 1.63 = 16.3 N

weight of object on earth = m x 9.8 = 10 x 9.8= 98 N

Question  22. 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:

v² - u² = 2gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, the 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 - 492 = 2×9.8×h
⇒ h = 492/(2×9.8)
⇒ h = 2401/19.6 = 122.5

Let t be the time taken by the ball to reach the height of 122.5 m, then according to the equation of motion:
(i) v = u + gt
We get,
0 = 49 + t x (- 9.8)
9.8t = 49
t = 49 / 9.8 = 5s
Also,
(ii) Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s
Question  23. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity.
Answer:

According to the equation of motion under gravity:
v² − u² = 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
∴ v² − 0² = 2 × 9.8 × 19.6
v² = 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  24. 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?
Answer:
According to the equation of motion under gravity:
v² − u² = 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×h×(-10)
h = (40×40)/20 = 80 m

Therefore, the total distance covered by the stone during its upward and downward journey

= 80 + 80 = 160 m

Question  25. Calculate the force of gravitation between the Earth and the Sun, given that the mass of the Earth = 6 x 1024 kg and of the Sun= 2x 1030 kg. The average distance between the two is 1.5 x 1011 m.
Answer:

According to Question, 
MS = Mass of the Sun = 2×1030 kg
ME = Mass of the Earth = 6×1024 kg
R = Average distance between the Earth and the Sun = 1.5×1011 m
From the Universal law of gravitation, F= GMm/r2

Putting the above values,

F = 6.67x10^-11 (6x10²⁴) x (2 x 10³⁰)/(1.5 x 10¹¹)²

= 6.67x 10^-11 12x10⁵⁴/2.25x10²²

= 6.67 x 12 x 10²¹/2.25

= 3.56x1022 N

Question  26. 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:
Suppose both the stones will meet after t seconds.

Let t be the point at which two stones meet and let h be their height from the ground.
Height of the tower is H = 100 m (Given)

It is clear from the Question  that we need to calculate the time when the two stones met. After calculating time, we will also be able to calculate the distance. 

Now, first,, consider the stone which falls from the top of the tower.
Initial velocity (u) = 0
So, the distance covered by this stone at time t can be calculated using the equation of motion

Question  27. A ball thrown vertically returns to the thrower after 6 seconds. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
Answer:
(a) time taken by ball to reach maximum height(t) = 6/2 = 3 s

v = u + gt

0 = u + (-9.8) x 3

u = 29.4 m/s (the velocity with which it was thrown up)

 (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
From the equation of motion, s= ut + 1/2 at2
h= 29.4 × 3 + 1/2 × -9.8 × (3)2 = 44.1 m
(c) its position after 4 s will be:

The ball attains the maximum height after 3 seconds. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0
The position of the ball after 4 s of the throw is given by the distance traveled by it during its downward journey in 4 s − 3 s = 1 s.
Equation of motion, s= ut + 1/2 gt2 will give,
s = 0 × t + 1/2 × 9.8 × 12 = 4.9 m
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  28. In what direction does the buoyant force on an object immersed in a liquid act?
Answer:

In the upward direction only.

Question  29. Why does a block of plastic released under water come up to the surface of water?
Answer:

Since of plastic has a density much as compared to water i.e. weight of plastic is less than the buoyant force experienced by it therefore a block of plastic released under water comes up to the surface of the water/floats.

Question  30. 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:
Density of that substance (d) = mass/volume = 50/20 = 2.5 g / cm³

since the density of the substance (2.5) is greater than the density of water (1) therefore it will sink.

Question  31. The volume of a 500 g sealed packet is 350 cm 3. Will the packet float or sink in water if the density of water is 1 g cm–3? What will be the mass of the water displaced by this packet?
Answer:
Density of the packet = mass/volume = 500/350 = 1.428 g / cm3

Since the density of the packet is more than the density of water it will sink. And packet will displace water equal to its volume :

volume of water displaced by packet =350 cm³(volume of packet)

mass of water displaced = volume of water displaced x density of water

= 350 x 1 = 350 g

 *** See You Again ***

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