Kerala Syllabus Class 9 Physics Chapter 4 Gravitation Notes Solutions

The comprehensive approach in Kerala Syllabus 9th Standard Physics Textbook Solutions Chapter 4 Gravitation Notes Questions and Answers English Medium ensures conceptual clarity.

Std 9 Physics Chapter 4 Notes Solutions Gravitation

SCERT Class 9 Physics Chapter 4 Notes Solutions Kerala Syllabus Gravitation Questions and Answers

Class 9 Physics Chapter 4 Let Us Assess Answers Gravitation

Question 1.
If an object is lifted from the centre of the Earth to its surface, will the mass and weight of the object change? Justify the answer.
Answer:
Mass will not change.
If an object is lifted from the centre of the Earth to its surface, the weight of the object will change. At the centre of the Earth, weight of the object will be zero.(weight-mg, g=0 at the centre.). As it is lifted to the surface, weight increases.

Question 2.
The weight of an object of mass 5 kg is determined using a spring balance. If the object and the spring balance are dropped down together, what will be the weight of the object while falling down? What is the reason?
Answer:
Weight of the object while falling down will be zero. During falling the entire force of gravity is used to produce acceleration. So a freely falling body experience weightlessness.

Question 3.
Will there be a change in the mass and weight of an object brought to the Moon from the Earth? Justify the answer.
Answer:
Whether on Moon on Earth, mass of an object does not change, but weight changes. This is because weight of an object depends on g value, whereas mass do not. g value on moon is approximately 1/6^th of the value of g on earth.

Question 4.
An object is allowed to fall from the top of a tower of height 100 m. At the same time, another object was thrown vertically up with a velocity 25 m/s in order to collide with the object falling down (g_Earth = 10 m/s², g_Moon = 1.62 m/s²)
a) Calculate the time taken by them to collide.
b) Find out the height from the ground at which they collide.
c) Would the answers obtained in the above case change, if this activity was carried out on the Moon? Justify.
Answer:
u = 0 m/s
h = 100 m
t = time taken to collide
s = ut + \(\frac{1}{2}\) at²
s = ut + \(\frac{1}{2}\) gt² = 0 × t + \(\frac{1}{2}\) gt²
= \(\frac{1}{2}\) gt² = \(\frac{1}{2}\) × 10 × t²
s = 5t²
Distance covered by the vertically thrown object = 100 – s
u = 25m/s, g = -10m/s²
100 – s = ut + \(\frac{1}{2}\) gt²
100 – s = 25t +\(\frac{1}{2}\) × – 10 × t²
100 – s = 25t – 5t²
100 – 5t² = 25t – 5t²
100 = 25t – 5t² + 5t²
i.e, 25t = 100
t = \(\frac{100}{25}\) = 4s

b) s = ut + \(\frac{1}{2}\) gt²
s = 25 × 4 + \(\frac{1}{2}\) × – 10 × 4²
s = 100 – 80 = 20 m
They will coillide at a height 20 m

c) On the moon, g_moon = 1.62 m/s²
s = ut + \(\frac{1}{2}\) gt² = 0 × t + \(\frac{1}{2}\) gt²
s = \(\frac{1}{2}\) gt²
s = \(\frac{1}{2}\) × 1.62 × t²
s = 0.81t²
Distance covered by the stone = 100 – s
100 – s = 25t + \(\frac{1}{2}\) × -1.62 × t²
100 – s = 25t – 0.81t²
100 – 0.81t² = 25t – 0.81t²
100 = 25t – 0.81t² + 0.81t²
100 = 25t
t = \(\frac{100}{25}\) = 4s
s = ut + \(\frac{1}{2}\) gt²
=25 × 4 + \(\frac{1}{2}\) × – 1.62 × 4²
= 100 – 12.96 = 87.04m
They will collide at a height 87.04 m.

Question 5.
The gravitational force on the lunar surface is approximately 1/6 th that of the Earth.
a) What is the weight of an object of mass 10 kg on the Earth?
b) If this object is taken to the surface of the Moon, what will be its mass and weight?
Answer:
a) m = 10 kg
weight = mg 10 × 9.8 = 98 N
b) weight on Moon = 10 × 1.62 = 16.2 N
mass on Moon = 10 kg

Question 6.
Objects with a larger mass are attracted by the Earth more strongly than objects with a smaller mass. So, if an object with a larger mass and an object with a smaller mass are allowed to fall from the same height,
a) Which one will reach the ground first?
b) Justify the answer.
Answer:
a) Both reaches at the same time.
b) Acceleration due to gravity is g = \(\frac{G M}{R^2}\)
This equation is independent of the mass of the object. Whatever be the mass of the object, acceleration due to gravity will be the same at a place.

Question 7.
Explain the difference between mass and weight.
Answer:
Topic 3, Qn. 49, Page 144

Question 8.
The masses of a stone and a hydrogen-filled. balloon are equal. If both are placed on the same ground, will the force of attraction exerted by the Earth on them be the same? Justify the answer.
Answer:
A hydrogen-filled balloon will be much larger than a stone. So the distance to its centre from the centre of the Earth will be much greater. So the force of attraction exerted by the Earth on the balloon will be less.

Question 9.
A stone falling from the top of a tall building reaches the ground in 2 s (g = 9.8 m/s²).
a) Calculate the height of the building.
b) What will be the velocity of the stone just before touching the ground .
Answer:
a) u = 0, t = 2s, g = 9.8 m/s²
height = s = ut + \(\frac{1}{2}\) at²
s = ut + \(\frac{1}{2}\) gt²
= 0 × 2 + \(\frac{1}{2}\) × 9.8 × 2²
= 0 + 19.6
= 19.6 m

b) v = u + at
=u + gt
=0 + 9.8 × 2
= 19.6 m/s

Question 10.
Find examples of circular motion from the following and tabulate.

• Electrons revolving around the nucleus
• A child running a 100 m sprint
• Planets revolving around the Sun
• A train running along a railway track with no curves
• Moon orbiting around the Earth

Answer:

• Electrons revolving around the nucleus
• Planets revolving around the Sun
• Moon orbiting around the Earth

Question 11.
What will be the weight of an object of mass 10 kg on a planet having twice the mass and three times the radius of the Earth?
Answer:
m = 10kg, g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) = 9.8 m/s²
M_new = 2M
R_new = 3R
g = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\) = \(\frac{\mathrm{G} \times 2 \mathrm{M}}{(3 \mathrm{R})^2}\) = \(\frac{2}{9}\) × 9.8
= 2.2 m/s²
Weight = mg
=10 × 2.2 = 22N

Question 12.
The mass of a planet is half of the Earth and the radius is 1/4 times that of the Earth. The acceleration due to gravity of the planet is ¸ ……..times that of the Earth.
a) \(\frac{1}{4}\)
b) 4
c) \(\frac{1}{8}\)
d) 8
Answer:
M_new = \(\frac{M}{2}\)
R_new = \(\frac{R}{4}\)
g_E(g on earth) = \(\frac{\mathrm{GM}}{\mathrm{R}^2}\)
g_new = \(\frac{GM}{2}\) ÷ \(\left(\frac{\mathrm{R}}{4}\right)^2\)
= \(\frac{GM}{2}\) × \(\frac{16}{R^2}\)
= 8\(\frac{GM}{R^2}\)
So the acceleration due to gravity of the planet is 8 times that of the Earth.

Question 13.
A body falling freely from a certain height takes 50 s to reach the ground. How much time will the same object take to fall from the same height on another sphere having twice the radius and twice the mass of the Earth? (Answer: \(50 \sqrt{2} s\)).
Answer:
t = 50 S
s = ut + \(\frac{1}{2}\)at² = 0 × 50 + \(\frac{1}{2}\) × 10 × 50 × 50
= 0 + 12500 m
=12500 m
s =12500 m, M_new = 2M
g_E(g on earth) = \(\frac{GM}{R^2}\) = 10 m/s², R_new = 2R
g_E new = \(\frac{\mathrm{G} \times 2 \mathrm{M}}{(2 \mathrm{R})^2}\) = \(\frac{1}{2}\) × 10 = 5 m/s²
s = ut + \(\frac{1}{2}\) at²
12500 = 0 × t + \(\frac{1}{2}\) × 5 × t²
12500 = \(\frac{5 t^2}{2}\)
5t² = 12500 × 2
5t² = 25000
t² = \(\frac{25000}{5}\)
t² = 5000
t = \(\sqrt{5000}\) = \(\sqrt{2500 \times 2}\)
t = \(50 \sqrt{2} s\)

Question 14.
The mass of an object is 100 kg. Calculate its weight at the centre of the Earth, the polar region, the equatorial region, the Moon and the Jupiter (g on the Jupiter = 23.1 m/s²)
Answer:
m = 100 kg
At the centre, w = mg = 100 × 0 = 0 N
At the poles, w = mg = 100 × 9.83 = 983 N
At equator, w = mg 100 × 9.78 = 978 N
On Moon, w = mg = 100 × 1.62 = 162 N
On Jupiter, w=mg = 100 × 23.1 = 2310 N

Class 9 Physics Chapter 4 Extended Activities Answers Gravitation

Question 1.
Make a still model of the Solar System and exhibit it in the class. It should include the Moon and an artificial satellite orbiting the Earth.
Answer:
Hint
It can be fun and educational to make a still model of the Solar System that includes the Moon and a man-made satellite orbiting the Earth. Amodel of this kind can be built by following these steps.

Materials needed: Foam balls of various sizes (to represent the Sun, planets, and the Moon), small bead or tiny ball (to represent the artificial satellite), paints and brushes, thin metal rods or skewers, Styrofoam or cardboard base, glue, string or wire, markers or labels. Steps:

1. Prepare the base using Styrofoam or cardboard.
2. Use foam balls of varying sizes to represent the Sun, Moon, the planets and the artificial satellite. The largest ball will be the Sun, and the others will be the Moon and the planets and the smallest one will be the artificial satellite.
3. Paint the planets and the sun according to their appearance.
4. Attach each planet to the base using thin metal rods or skewers.
5. Use small pieces of paper or labels to label each planet, the Moon, and the artificial satellite and attach these labels to the base near the corresponding celestial bodies.

Arrange everything neatly and ensure all the rods and wires are securely attached.

Question 2.
The values of g in various planets are given. There is an object of mass 100 kg. Determine the weight of the object on these planets.

Planet	Acceleration due to gravity in m/s2(Approximate value)	Weight (N)
Earth	9.8
Mercury	3.7
Venus	8.9
Mars	3.7
Saturn	9.00
Uranus	8.7
Neptune	11.00

Answer:

Planet	Acceleration due to gravity in m/s2(Approximate value)	Weight (N)
Earth	9.8	9.8
Mercury	3.7	980
Venus	8.9	370
Mars	3.7	890
Saturn	9.00	370
Uranus	8.7	900
Neptune	11.00	870

Gravitation Class 9 Notes Questions and Answers Kerala Syllabus

Question 1.
Have you ever wondered why a stone thrown up and a bird’s feather fall to the ground?
Answer:
A stone thrown up and a bird’s feather fall to the ground due to the attraction of the Earth.

Question 2.
From where did the stone and the feather get the force they needed to fall?
Answer:
The stone and the feather get the force they needed to fall from the force of attraction of Earth.

Question 3.
Imagine dropping stones in wells at various places around the Earth. Aren’t stones attracted to the bottom of the well?

Answer:
Yes, stones are attracted to the bottom of the well.

Question 4.
Do the people standing on the opposite hemisphere of the Earth fall down? Isn’t this due to the attraction of the Earth?

Answer:
No, the people standing on the opposite hemisphere of the Earth do not fall down. This is due to the attraction of the Earth.

Activity
Fasten a spring balance to the grill of a window. Pull its hook with your hand.

Question 5.
Why did the spring stretch?
Answer:
The spring is stretched due to the force applied by the hand.

Question 6.
What is the reading on the spring balance?
Answer:
(Measurement shown by the spring balance is to be taken)

Question 7.
Isn’t this the force that we applied?
Answer:
Yes, this is the force that we applied.

Question 8.
What is the unit of force?
Answer:
The unit of force is newton (N).

Activity

Answer:
Suspend a mass of 100 g from a spring balance.

Question 9.
Why did the spring stretch?
Answer:
Because a force is experienced on the spring.

Question 10.
Which is the force that pulled down the 100 g mass?
Answer:
The force exerted by the Earth.

Question 11.
What is the reading on the spring balance?
Answer:
1 N

Question 12.
Isn’t this the force that attracted the object to the Earth?
Answer:
Yes, this is the force that attracted the object to the Earth.

Question 13.
Why does the spring balance experience more stretching force?
Answer:
Since the force exerted by the Earth is greater, the spring balance experiences more stretching force.

Question 14.
What happens to the force of attraction as mass increases? (increases/decreases)
Answer:
increases

Question 15.
If so, write down a factor that influences the force of attraction.
Answer:
Mass
The force of attraction on an object at different positions on the Earth is given in the table.

Mass of the object (kg)	Height from the surface of the Earth (m)	Attractive Force (N)
100	On the surface (0)	980
100	1,00,000 m	950
100	10,00,000 m	730

Analyse the table and answer the following questions.

Question 16.
Where did the object of mass 100 kg experience a greater force of attraction? (on the surface /at a height of 1,00,000 m at a height of 10,00,000 m)
Answer:
The object of mass 100 kg experience a greater force of attraction on the surface.

Question 17.
As the distance from the Earth to the object increases, the force of attraction exerted by the Earth (increases/decreases)
Answer:
decreases
Tides are caused by the influence of gravitational force on the Earth by the Moon and the Sun.

Question 18.
If the Sun and the Moon exert a force on the Earth, wouldn’t the other celestial bodies of the universe also exert a mutual force of attraction between them?
Answer:
Yes

Question 19.
Complete the table given below based on Newton’s Universal Law of Gravitation and answer the questions given below.

Answer:

Question 20.
Two mutually attracting objects are placed at a fixed distance between them. If the mass of one of them is doubled, how many times will the force of attraction between them be?
Answer:
The force of attraction between them will be doubled.
F = \(\mathrm{G} \frac{m_1 \times m_2}{d^2}\)
F(new) = \(G \times \frac{2 m_1 \times m_2}{d^2}\)
= 2 × G × \(\frac{m_1 \times m_2}{d^2}\)
= 2F

Question 21.
What if the mass of one object is doubled and the mass of the other tripled?
Answer:
The force of attraction between them will be increased by 6 times.
F = \(\mathrm{G} \frac{m_1 \times m_2}{d^2}\)
F(new) = G × \(\frac{2 m_1 \times 3 m_2}{d^2}\)
= 6 × G × \(\frac{m_1 \times m_2}{d^2}\) = 6F

Question 22.
What if the distance between the objects is doubled?
Answer:
The force of attraction between them will be de creased by (1/4) times.
F = G\(\frac{m_1 \times m_2}{d^2}\)
F(new) = G × \(\frac{m_1 \times m_2}{(2 d)^2}\)
= \(\frac{1}{4}\) × G × \(\frac{m_1 \times m_2}{d^2}\)
= \(\frac{1}{4}\)F

Question 23.
What if the distance between the objects is halved?
Answer:
The force of attraction between them will be increased by 4 times.

Question 24.
What if the distance between the objects is quartered?
Answer:
The force of attraction between them will be increased by 16 times.

Question 25.
Calculate the gravitational force of attrac- tion between two children of masses 40 kg and 50 kg when they are 2 m apart.

Answer:
m₁ = 40 kg
m₂ = 50 kg
d = 2m
F = \(\mathrm{G} \times \frac{m_1 \times m_2}{d^2}\)
F = \(\frac{6.67 \times 10^{-11} \times 40 \times 50}{2^2}\)
F = 500 × 6.67 × 10^-11
= 3335 × 10^-11 N
F = 3.335 × 10³ × 10^-11N
F = 3.335 × 10^-8N
F = 0.0000003335 N

Question 26.
This force of attraction is not experiences in everyday life. why?
Answer:
The force of attraction between two persons is not felt because it is very feeble. This force is very small that it cannot even be compared to other forces like frictional force, magnetic force etc. Therefore, this force is not experienced in everyday life.

Question 27.
Do both of the children experience the same force of attraction?
Answer:
Here, the children of mass 40 kg and 50 kg attract each other with a force of 3.335 × 10⁸ N. This means that both of them are experiencing the same force of attraction.

Question 28.
If the force of attraction of the Earth on the Moon is F, what will be the force of attraction of the Moon on the Earth?
Answer:
F itself.

Question 29.
A heavy body and a light body are dropped down together from a certain height. Which one reached first?
Answer:
Both reached simultaneously.

Question 30.
Which one will experience greater acceleration?
Answer:
Both will experience the same acceleration.
According to Newton’s second law of motion, F = ma
IfF is the force of attraction of the Earth and m is the mass of the object, then a is the acceleration due to the force of attraction of the Earth.

Question 31.
What is the unit of g?
Answer:
m/s²

Question 32.
Is the mass of the object included in this equation?
Answer:
No, mass of the object is not included in this equation.

Question 33.
If mangoes and leaves fall down from a mango tree at the same time, will they reach the ground together? What will be the reason?
Answer:
No, if mangoes and leaves fall down from a mango tree at the same time, they will not reach the ground together. It is due to the influence of air that leaves fall down slowly.

Question 34.
A piece of paper and a coin are dropped down together from the same height. What do you observe?
Answer:
The coin reaches the ground first and the piece of paper later.

Question 35.
Repeat the above activity after crumpling the paper. What difference do you observe now?
Answer:
The coin and the crumpled paper reach the ground simultaneouly.

Question 36.
Is the acceleration due to due to gravity the same everywhere on the Earth?
Answer:
No

Question 37.
Observe the figure. Does the Earth have a perfect spherical shape?

Answer:
No, Earth does not have a perfect spherical shape.

Question 38.
Which region of the Earth is the farthest from its centre?
(polar region/equatorial region)
Answer:
Equatorial region

Question 39.
Which region lies closer to the centre of the Earth? (polar region/equatorial region)
Answer:
Polar region

Question 40.
How does the value of g vary as the distance from the Earth’s centre to the surface changes? Verify your answer using the equation.
Answer:
As the distance from the Earth’s centre to the surface increases, the value of g decreases.

Question 41.
Where is the value of g maximum?
(at the polar region /at the equatorial region)
Answer:
At the polar region

Question 42.
The forces of attraction on an object at the centre of the Earth, from all sides of the Earth are equal. If so, what will be the value of g at the centre?
Answer:
Zero

Question 43.
Gravitational force is a force of mutual attraction. When the engine ceases working, the aeroplane falls down to the Earth. But the Earth does not move towards the aeroplane even though the aeroplane attracts the Earth. Why?

Answer:
To understand the reason for this, we calculate the gravitational force between an aeroplane and the Earth when the aeroplane of mass 10000 kg is at a height 10 km above the surface of the Earth.
m = 10000 kg = 10⁴kg
Height from the surface of the earth, d=10 km
=10000 m=104 m
F = \(G \frac{M m}{R^2}\)

The acceleration of the aeroplane produced by the above force
F= mg
g = \(\frac{F}{M}\) = 97400/10000 = 9.74 m/s²
The aeroplane exerting the same force on the Earth as that exerted by the Earth on the aeroplane.
Acceleration of the Earth due to the force exerted by the aeroplane
g = \(\frac{F}{M}\) = \(\frac{97400}{6 \times 10^{24}}\) = 1.6 × 10^-20 m/s²
g=0.000000000000000000016 m/s².

The acceleration of the Earth is almost zero. Although the forces of attraction between the aeroplane and the Earth are equal, the Earth will not experience any considerable acceleration.

Question 44.
An artificial satellite of mass 10000 kg stops working and falls down to the Earth. We know that attraction between objects is mutual. The satellite attracts the Earth with the same force with which the Earth attracts the satellite.
Height from the Earth = 5000 m,
Radius of the Earth, R = 6.4 × 10⁶ m
a) What is the acceleration of the satellite?
b) What is the acceleration of the Earth?
(Mass of the Earth = 6 × 10²⁴ kg)
Answer:
Mass of the artificial satellite,
m = 10000 kg =10+ kg
Mass of the Earth; M = 6 × 10²⁴ kg
Distance, d=Radius of the Earth + Height from the Earth
= 6.4 × 10⁶ + 5000 = 6400000 + 5000 = 6405000 m

Question 45.
An object of 10 kg is allowed to fall to the ground from a height of 20 m.
a) How long will it take to reach the ground?
b) Calculate the time required to reach the ground, if it is on the Moon.
g_(Earth) = 10 m/s², g_(Moon) = 1.62 m/s²
Answer:
m = 10 kg
u = 0
S = 20 m
g= 10 m/s²
a) s = ut + \(\left(\frac{1}{2}\right)\) at² = u t + \(\left(\frac{1}{2}\right)\) gt²
20 = 0 × t + \(\left(\frac{1}{2}\right)\) 10 × t²
20 = \(\left(\frac{1}{2}\right)\) 10 × t²
20 = 5 × t²
t² = 20/5 = 4
t = 2s

b) g_(Moon) = 1.62 m/s²
s = ut + \(\left(\frac{1}{2}\right)\) at² = u t + \(\left(\frac{1}{2}\right)\) gt²
20 = 0 × t + \(\left(\frac{1}{2}\right)\) 1.62 t²
20 = 0.81 × t²
t² = 20 / 0.81 = 24.69
t = 5 s (approximate value)

Question 46.
A stone is thrown vertically upwards from the lunar surface. If the stone returns in 6 s.
a) What is the initial velocity of the stone?
b) What is the distance that can be covered by the stone?
c) What will be the position of the stone after 4.s?
Answer:
Time taken to reach the maximum height,
t = 6/2 = 3 s
v = 0
u =?
a = g = 1.62 m/s²
a) v=u+at
0 = u+ -1.62 × 3
0 = u-1.62 × 3
0 = u – 4.86
u = 4.86 m/s

b) v² = u2 + 2 gs
0 = u² +2 × 1.62 × s
0 = (4.86)² + 3.24 × s
-3.24 s = (4.86)²
s = (4.86)² / -3.24 = – 7.29 m
s = 7.29 m
Total distance 7.29 m + 7.29 m = 14.58 m

c) t = 4 s
s = ut + \(\left(\frac{1}{2}\right)\) at² = ut + \(\left(\frac{1}{2}\right)\) gt²
= 4.86 × 4 + \(\left(\frac{1}{2}\right)\) × -1.62 × 4²
= 19.44 – 12.96
= 6.48 m
The stone will be at a height 6.48 m.
The stone will be at a height 6.48 m.

Question 47.
What is the weight of an object of mass 50 kg?
Answer:
Weight = mg = 50 kg × 9.8 m/s² = 490 N
This is also known as 50 kgwt.

Question 48.
The mass of an object is 10 kg. Calculate its weight on the Earth. What would be its weight if it were on the moon? (g_Moon 1.62 m/s²)
Answer:
Weight on the Earth = mg = 10 × 9.8 = 98 kgm/s² = 98 N
g_(Moon) = 1.62 m/s²
Weight on the Moon = mg_(Moon) = 10 × 1.62 = 16.2 N

Question 49.
The mass of an object remains the same everywhere In the universe. Then what about its weight? Compare mass and weight and complete the table.
Answer:

Mass	Weight
Measured using common balance Unit kg Mass of an object is the quantity of matter contained in it. Scalar quantity Mass of an object is same everywhere	Measured using spring balance Unit newton (N) or kgwt The weight of the object on the Earth is the gravitational force exerted by the Earth on that object. Vector quantity Weight of an object will vary from place to place

Question 50.
When goods shipped from Kochi to England were weighed in England using the same spring balance used in Kochi, the weight was found to be 20 N more. What could be the reason?
Answer:
England is situated near polar region and Kochi near equatorial region. Value of g is greater at polar region than at equatorial region. So weight (w = mg) will be more when weighed at England.

Question 51.
Is an object heavier at the poles or at the equator? Justify your answer.
Answer:
An object is heavier at the poles. Value of g is greater at polar region than at equatorial region. So weight (w=mg) will be more at the poles.

Question 52.
What will be the weight of an object at the centre of the Earth?
Answer:
At the centre of the Earth, objects are attracted equally in all directions. The resultant force will be zero there. The value of g is zero, so weight of an object at the centre of the Earth will be zero.

Question 53.
Hook a 20 g mass on a spring balance and hold it. Bring it down quickly What will be the change in the reading at this time?
(increases/decreases)

Answer:
decreases
If these are allowed to fall freely, the reading can be seen as zero.

Weightlessness
We know that a freely falling object has acceleration. The force required for acceleration is provided by the force of gravity. If the entire force of gravity is used to provide acceleration, the freely falling object will be weightless.

Question 54.
What are the instances in which weightlessness is experienced?
Answer:

• For a person who orbits the Earth in space stations.
• For a coconut falling from a coconut tree.

Question 55.
Why does a freely falling body experience weightlessness?
Answer:
During falling the entire force of gravity is used to produce acceleration. So a freely falling body experience weightlessness.

Question 56.
Isn’t the motion of a freely falling stone in a straight line?
Answer:
Yes

Question 57.
What type of motion do artificial satellites have? Write down examples for such types of motion.
Answer:

• Whirling a stone tied to a string.
• Rotation of fan blades.

Question 58.
Does this object have uniform velocity though it has uniform speed? Why?
Answer:
No, this object does not have uniform velocity. Though it has uniform speed, its direction is changing continuously. So it has non uniform velocity.

Question 59.
Does this object experience any force?
Answer:
Yes

Question 60.
If the string is released from the hand, in which direction will the stone move? Won’t it be along the tangent?

Answer:
Yes, it will be along the tangent.

Question 61.
Who gave this force? Isn’t it our hand?
Answer:
It is our hand who gave this force.

Question 62.
If there is no centripetal force, can there be circular motion?
Answer:
No, if there is no centripetal force, there can not be circular motion.
It is due to the lack of centripetal force that vehicles negotiating curves on a road tend to skid or roll off the curve. Mass and speed of the vehicle and curvature of the road are the factors that influence the tendency of the vehicle to roll over. The path followed by the objects moving under centripetal force will be circular or curved.

Question 63.
If so, from where do the artificial satellites orbiting around the Earth get their centripetal force?
Answer:
It is the Earth’s gravitational force that is acting