AP 9th Class Physics 9th Lesson Questions and Answers Gravitation

AP State Board new syllabus AP Board Solutions Class 9 Physics 9th Lesson Gravitation Questions and Answers.

AP 9th Class Physical Science 9th Lesson Questions and Answers Gravitation

9th Class Physics 9th Lesson Gravitation Questions and Answers (Exercise)

Question 1.
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
The square of the distance between two particles is inversely proportional to the gravitational force.
∴ The gravitational force increases 4 times (Quadra Pulled).

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:
Irrespective of the mass acceleration due to gravity is same on all objects, so all objects (heavier, lighter) comes down with same acceleration.

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:
Gravitational force between 1 kg mass object and earth :
Mass of earth M=6 ×10²⁴; Mass of object m =1kg
Radius of earth r =6.4 × 10⁶ m
Gravitational constant G =6.67 × 10^-11 Nm² kg^-2
By universal law of Gravitation F= \(\mathrm{G} \frac{\mathrm{Mm}}{\mathrm{r}^2}\)

F=0.2084 × 0.046875 × 10³ ⇒ F=9.76875 ms^-2  i.e F=9.8 ms^-2

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:

• Gravitational force with which a body Aattracts another body B is equal in magnitude and opposite in direction to the gravitational force with which a body B attracts the body A.
• Thus, the magnitude of force with which the earth attracts the moon is equal to the magnitude of the force with which the moon attracts the earth.
• Thus, both the earth and the moon attract each other with equal forces.

Question 5.
If the moon attracts the earth, why does the earth not move towards the moon?
Answer:
The mass of the moon is smaller than the mass of the earth, hence the force exerted by the moon is not sufficient to move the earth towards it.

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) We know, Force of Gravitation F= \(\frac{\mathrm{GMm}}{\mathrm{r}^2}\)
When the mass of one object is doubled.
Let the mass of the object one be doubled.
New mass of object one = 2 M ⇒ F = \(\frac{\mathrm{G} \times 2 \mathrm{M} \times \mathrm{m}}{\mathrm{r}^2}=\frac{2 \mathrm{GMm}}{\mathrm{r}}\) = 2F
∴ If mass of one object is doubled, then force is also doubled.

ii) The distance between the object is doubled and tripled.

Force decreases 4 times.
Similarly, if r=3 r, then force decreases 9 times.

iii) Both masses are doubled.
M=2m ; m=2 M

Question 7.
What is the importance of Universal Law of Gravitation?
Answer:
The gravitational force plays an important role to explain various phenomena in nature.
i) All the planets revolve around the sun due to the gravitational force between the sun and the planets.

The force required by a planet to move around the sun in elliptical path (known as centripetal force) is provided by the gravitational force of attraction between the planet and the sun. Thus, gravitational force is responsible for the existence of the solar system. If the gravitational force stops to act, then the planets will move along the tangent to the point of the elliptical path.

ii) Tides in oceans are formed due to the gravitational force between the moon and the water in oceans. Let us see how ocean tides take place. The positions of the earth and the moon are explained below.

The moon exerts large gravitational force on the earth. The distance between the moon and the water in ocean at B is less than the distance between the moon and the earth (A). So, the moon pulls ocean water (B) with large force than the moon pulls the earth (A). Therefore, the water in ocean (B) is raised towards the side of the moon. As a result of this, high ocean tide is caused.

iii) Gravitational force between a planet and its satellite (i.e., moon) decides whether a planet has a moon or not. Since the gravitational force of the planets like mercury and venus is very small, therefore, these planets do not have any satelite or moon.

iv) Artificial and natural satellites revolve around the earth due to the gravitational force between the earth and the satellite. The gravitational force between the earth and the satellite provides a necessary centripetal force to the satellite to move in a circular path around the earth.

v) The atmosphere (envelope of gases) of the earth is possible due to gravitational force of the earth. The various gases forming atmosphere are bound to:the earth due to gravitational pull of earth.

vi) Rainfall and snowfall is possible only due to gravitational force of the earth. Rain drops and snow flakes fall on to the earth due to gravitational pull of earth.

vii) We stay on the earth due to the gravitational force between the earth and us.

Question 8.
What is the acceleration of free fall?
Answer:
The acceleration of free fall a =9.8 ms^-2

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:
No, the gravitational force at the poles is greater than at the equator. So, Amith friend should not agree with this he will get less gold at the equator.

Question 11.
Why will a sheet of paper fall slower than one that is crumbled into a ball ?
Answer:
The surface area of sheet of a paper is greater than that of a crumbled paper.
So, sheet of paper experiences more air resistive force due to this it comes down slower than the crumbled paper ball.

Question 12.
Gravitational force on the surface of the moon is only \(\frac{1}{6}\) 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:
Object :
Mass m =10kg ; Acceleration due to gravity g =9.8 ms^-1
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:
Object :
Intial Velocity u = 49 ms^-1
finial Velocity v = ms^-1
g = -9.8ms^-2

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:
Object (stone):

Initial velocity u = 0 ms^-1
Acceleration due to gravity g = 9.8 ms^-2 Displacement s = 19.6 m
From kinematics, v² – u² = 2gs
v² – (0)² = 2 x 9.8 x 19.6 v² = (19.6)² v = 19.6 ms^-1
Velocity of stone before it touching the ground is 19.6 m^-1

Question 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 ?
Answer:
Stone: Initial velocity u = 40 ms^-1
Acceleration due to gravity g = – 10 ms^-2; Final velocity v = 0

i) From kinematics, v²– u² = 2gh
(0)² – (40)² = 2 (-10) (h)
– (40 x 40) = – 20h

Maximum height h = 80m

ii) The stone after reaching the maximum height falls down and reach the ground.
∴ Displacement of the stone =0
Total distance =80+80=160m

Question 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.
Answer:
Earth: Mass M_e = 6 x 10²⁴ kg
Sun : Mass M_s = 2 x 10³⁰ kg
Gravitational Constant G = 6.67 x 10^-11 Nm² kg^-2
Distance between Sun & Earth (r) = 1.5 x 10¹¹ m
= 6.67 × 2.666 × 2 × 10^-11 × 10² ×10³⁰
= 35.56444 × 10²¹
F =3.556444 × 10²² N
[Force of Attraction between Sun & Earth]

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:
i) Stone (S-I):
Initial velocity u=0
Displacement s₁=100 m
Final velocity = v

Stone (S – II) :
Initial velocity u = 25 ms^-1
Final velocity v=0
Acceleration due to gravity g =9.8 ms^-2
Let they both meet at distance =x m
From kinematics, then s₁= 100-x

ii) Distance transversed by two stones:
s₂=ut + \(\frac{1}{2}\) gt²

x =100-80=20m
s₁=100 – x = 100-20=80 m
The two stones will meet at 80 m from top. (or) at 20 m from ground.

Question 18.
A ball thrown up vertically returns to the thrower after 6s. Find
a) the velocity with which it was thrown up,
b) the maximum height it reaches and
c) its position after 4s
Answer:
Ascent (Ball thrown upwards) :
Initial velocity = u;
Final velocity v = 0;
Acceleration due to gravity g =-9.8ms^-2
Time of Ascent = Time of descent
[t₁+t₂=6 sec] [t₁=t₂=3sec]

a) t = 3 sec
From Kinematics, v=u+gt
0=u+(-9.8)(3)=u-29.4 ⇒ u =29.4ms^-1
The ball is thrown with an initial velocity of 29.4 ms^-1

After 4 sec it is at distance of 4.9 m from the top.

Question 19.
In what direction does the buoyant force on an object immersed in a liquid act ?
Answer:
The buoyant force acts in the vertically upward direction.

Question 20.
Why does a block of plastic released under water come surface of water ?
Answer:
The plastic box experience a greater upthrust (Buoyant force) than its weight. This makes it to come up to the surface of the water.

Question 21.
The volume of 50g of a substance is 20 cm³. If the density of water is 1gcm^-3, will the substance float or sink ?
Answer:
Substance : Mass m = 50 g; Volume v = 20 cm³
Density of water ρ₁ = 1g cm^-3

From (1) & (2) ρ₂>ρ₁
∴ The density of substance is greater than density of water.
So, the substance sinks.

Question 22.
The volume of  500g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1g cm³ ? What will be the mass of the water displaced by this packet?
Answer:
Sealed packet : Mass m=500g; Volume V = 350 cm³ ………………… (1)
Density of water ρ₁=1g cm^-3
Density of packet
From (1) & (2) ρ₂>ρ₁
The density of the substance is greater than density of water.
So, the packet, will sink.

b) Mass = Density × Volume ⇒ m=1 × 350=350 ×g
The packet will displaces 350 g of water.

9th Class PS 9th Lesson Questions and Answers (InText)

Question 1.
State the Universal Law of Gravitation.
Answer:
Universal law of Gravitation : The gravitational force of attraction between any two particles is directly proportional to the product of masses of the particles and is 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 gravitational force between two uniform objects is directed along the line joining their centres.

• Let A & B two particles of masses m₁ & m₂ at a distance r
• Let the force of attraction between A & B is F.
• According to universal law of Gravitation,

Note : The value of G is G = 6.67 x 10^-11 Nm² kg^-2

Question 3.
What do you mean by free fall ?
Answer:
A body is said to be falling freely or in a state of free fall, when it is under the influence of gravity alone and no other force acts on it.

Question 4.
What do you mean by acceleration due to gravity ?
Answer:
The acceleration caused due to the force of gravity is known as acceleration due to gravity.
Note : The acceleration due to gravity on earth is 9.8 ms^-2

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

Mass	Weight
1. It is the amount matter in a body.	1. It is the gravitational force of the  earth on the body.
2. It is a scalar quantity.	2. It is a vector quantity.
3. It is constant in the Universe.	3. It varies from place to place.
4. It’s S.I. unit is kgs.	4. It’s S.I. unit is Newton (N).
5. m = \(\frac{F}{a}\)	5. W = mg
6. It is measured in common balance.	6. It is measured in spring balance.

Question 6.
Why is the weight of an object on the moon \(\frac{1}{16}\) th its weight on the earth ?
Answer:
Reasons why the weight of an object on the moon is 1/6 th its weight on the earth :

• The moon has less mass than the Earth, so it exerts less gravitational force on objects on its surface.
• The gravity on the moon is only about 1/6th that of the Earth’s gravity.
• The weight of an object is directly proportional to the force of gravity acting on it, so a decrease in the force of gravity on the moon results in a decrease in an object’s weight.
• Example : An object that weighs 60 kilograms on Earth would weigh only 10 kilograms on the moon because 60 kg multiplied by 1/6th is equal to 10 kg.

Question 7.
Why is it difficult to hold a school hag having a strap made of a thin and strong string ?
Answer:
A thin strap exert a greater pressure on the hand than a boarder strap on the school bag.

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

• Buoyancy refers to the upward force that a fluid exerts on an object that is partially or completely submerged in it.
• This force is equal to the weight of the fluid that the object displaces. In other words, buoyancy is the ability of an object to float in a fluid.

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

• If the density of the object is greater than density of water, then it will sink in water.
• If the density of the object is less than density of water, then the object will float.

Question 10.
You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42kg ?
Answer:

• When you are on a weighing machine the air surrounding you create a buoyant force, so our actual mass will be more than 42 kg.
• However, if we ignore buyoyant force the mass is a constant quantity therefore it cannot be more (or) less than 42 kg

Question 11.
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 mass of iron & cotton both are same however the volume of cotton is greater than iron, so it experiences a greater upthrust than iron. So, cotton is heavier than iron.

Example:

Question 1.
The mass of the earth is 6 x 10²⁴ kg and that of the moon is 7.4 x 10²² If the distance between the earth and the moon is 3.84 x 10⁵ km, calculate the force exerted by the earth on the moon. (Take G = 6.7 x 10^-11 Nm² kg^-2)
Answer:
The mass of the earth, M = 6 x 10²⁴ kg
The mass of the moon, m = 7.4 x 10²² kg
The distance between the earth and the moon,
d = 3.84 x 10⁵ km
= 3.84 x 10⁵ x 1000 m
= 3.84 x 10⁸ m
G = 6.7 x 10^-11 Nm²  kg^-2
From Eq. (4), the force exerted by the earth on the moon is

Thus, the force exerted by the earth on the moon is 2.02 × 10²⁰ N

Question 2.
A car falls off a ledge and drops to the ground in 0.5s. Let g =10ms^-2 (for simplifying the calculations).
i) What is its speed on striking the ground?
ii) What is its average speed during the 0.5 s ?
iii) How high is the ledge from the ground?
Answer:
Time, t = 1/2 second; Initial velocity, u =0 ms^-1
Acceleration due to gravity, g=10 ms^-2
Acceleration of the car, a=+10 ms^-2. (downward)

i) speed
v = at
v=10 ms^-2 × 0.5 s=5 ms^-1
Thus, i) its speed on striking the ground = 5 ms^-1
ii) its average speed during the 0.5 s = 2.5 ms^-1
iii) height of the ledge from the ground = 1.25 m

Question 3.
An object is thrown vertically upwards and rises to a height of 10 m. Calculate i) the velocity with which the object was thrown upwards and ii) the time taken by the object to reach the highest point.
Answer:
Distance travelled, s = 10 m;
Final velocity, v = 0 ms^-1
Acceleration due to gravity, g = 9.8 ms^-2
Acceleration of the object, a = – 9.8 ms^-2(upward motion)
i) v² = u² + 2as
0 = u² + 2 x (-9.8 ms^-2) x 10 m
– u² = – 2 x 9.8 x 10 m²s^-2
u = \(\sqrt{196}\) ms⁴ = 14 ms⁴

ii) v = u + at
0 = 14 ms^-1 – 9.8 ms^-2 x t
t = 1.43 s

Thus, i) Initial velocity, u = 14 ms^-1 and
ii) Time taken, t = 1.43 s.

Question 4.
Mass of an object is 10 kg. What is its weight on the earth ?
Answer:
Mass, m = 10 kg; Acceleration due to gravity, g = 9.8 ms^-2
W = m x g = 10 kg x 9.8 ms^-2 = 98 N
Thus, the weight of the object is 98 N.

Question 5.
An object weighs 10 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon ?
Answer:
We know, Weight of object on the moon = (1/6) x its weight on the earth.
That is, \(W_m=\frac{W_e}{6}=\frac{10}{6} N\) = 1.67 N
Thus, the weight of object on the surface of the moon would be 1.67 N.

Question 6.
A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm x 20 cm x 10 cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions
(a) 20 cm x 10 cm and
(b) 40 cm x 20 cm.

Answer:
The mass of the wooden block = 5 kg The dimensions = 40 cm x 20 cm x 10 cm
Here, the weight of the wooden block applies a thrust on the table top That is,
That = F = m x g = 5kg x 9.8 ms⁰= 49 N
Area 6f a side = length x breadth
= 20 cm x 10 cm = 200 cm² = 0.02 m² 49 N
From Eq. Pressure = \(\frac{\text { thrust }}{\text { area }}\), Pressure = \(\frac{49 \mathrm{~N}}{0.02 \mathrm{~m}^2}\)=2450 Nm^-2
When the block lies on its side of dimensions 40 cm × 20 cm, it exerts the same thrust.
Area = length × breadth
= 40cm ×20 cm =800 cm² = 0.08m²
From Eq. Pressure = \(\frac{\text { thrust }}{\text { area }}\), Pressure =\(\frac{49 \mathrm{~N}}{0.08 \mathrm{~m}^2}\)= 612.5 Nm^-2
The pressure exerted by the side 20 cm × 10 cm is 2450 Nm^-2 and by the side 40 cm × 20 cm is 612.5 Nm^-2.

9th Class Physical Science Chapter 9 Questions and Answers (Lab Activities)

(Page No. 28) Activity -9.1

1. Procedure:

• Take a piece of thread.
• Tie a small stone at one end. Hold the other end of the thread and whirl it round, as shown in the adjacent figure.
• Note the motion of the stone.
• Release the thread.
• Again, note the direction of motion of the stone.

Observations:

• The motion of the stone is a circular path with a velocity of constant magnitude.
• The direction of motion of the stone is tangential.

(Page No. 28) Activity -9.2

2. Procedure:

• Take a stone.
• Throw it upwards.
• It reaches a certain height and then it starts falling down.

Self – Interpretation.

(Page No. 34) Activity -9.3

3. Procedure:

1. Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building. Observe whether both of them reach the ground simultaneously.

2. We see that paper reaches the ground little later than the stone. This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the paper is more than the resistance offered to the stone. If we do the experiment in a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate.
Self – Interpretation.

(Page No. 34) Activity -9.4

4. Procedure:

• Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper.
• Put it in a bucket filled with water. You see that the bottle floats.
• Push the bottle into the water. You feel an upward push. Try to push it further
  down. You will find it difficult to push deeper and deeper. This indicates that water exerts a force on the bottle in the upward direction. The upward force exerted by the water goes on increasing as the bottle is pushed deeper till it is completely immersed.
• Now, release the bottle. It bounces back to the surface.

Observations:

1. Does the force due to the gravitational attraction of the earth act on this bottle ?
Answer: Yes, gravitational force of earth acts on the bottle.

2. If so, why doesn’t the bottle stay immersed in water after it is released ?
Answer: The bottle doesn’t gets immersed due to buyoyant force (upward force) exerted by water.

3. How can you immerse the bottle in water ?
Answer: By making it heavier.

(Page No. 44) Activity 9.5

5. Procedure:

• Take a beaker filled with water.
• Take an iron nail and place it on the surface of the water.

Observation:
Observe what happens ?
Answer: The nail sinks, as the downward force is greater than buyoyant force on the nail.

(Page No. 46) Activity 9.6

6. Procedure :

• Take a beaker filled with water
• Take a piece of cork and an iron nail of equal mass.
• Place them on the surface of water

Observation : Cork floats and iron nail sinks this is due to difference in their density.

(Page No. 46) Activity 9.7

7. Procedure :

• Take a piece of stone and tie it to one end of a rubber string or a spring balance.
• Suspend the stone by holding the balance or the string as shown in the figure (a).
• Note the elongation of the string or the reading on the spring balance due to the weight of the stone.
• Now, slowly dip the stone in the water in a container as shown in the figure (b).

Observation :
Observe what happens to elongation of the string or the reading on the balance ?
Answer: As we immerse the stone further the weight of decreases until it immersed completely.