Gravitation Test

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Gravitation Test - 49

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Question 1
1 / -0

Two point masses having mass $$m_1$$ and $$m_2$$ respectively. If the distance between two masses is increased by a factor of $$2$$. Then gravitational force:

A
Decreases by a factor of $$4$$

B
Decreases by a factor of $$2$$

C
Increases by a factor of $$2$$

D
increases by a factor of $$4$$

E
Cannot be determined without knowing the masses

Solution

Gravitational force acting on bodies of masses $$m_1$$and $$m_2$$ separated by a distance $$r$$ is given by,
$$F=\dfrac{Gm_1m_2}{r^2}$$
When the distance between them doubles, $$r'=2r$$
$$ \implies F'=\dfrac{Gm_1m_2}{(2r)^2}=\dfrac{F}{4}$$
Therefore, the gravitational force decreases by a factor of $$4$$.
Question 2
1 / -0

A stone dropped from a height falls towards the earth because

A
The stone exerts a force of attraction on the earth

B
The earth exerts force of gravity on the stone

C
The stone attracts the earth towards itself

D
All the above statements are true

Solution

If we drop a piece of stone from some height, the stone falls down towards the earth. Since the stone starts moving downwards (it is in motion),therefore a force must be acting on it. This force is due to attraction between the earth and the stone and it is called the gravitational force of earth (or gravity of earth). Thus, a stone dropped from a height falls towards the earth because the earth exerts a force of attraction (called gravity) on the stone and pulls it down.
Option A and C are also correct but they do not explain why the stone falls to the ground.
Hence, only option B is correct.
Question 3
1 / -0

Two objects have the same mass and are located near each other at a distance $$r$$. If the mass of one of the objects is doubled and the mass of the other object is tripled, find out the change in gravitational attraction between them?

A
Decrease by a factor of $$1/6$$

B
Decrease by a factor of $$2/3$$

C
Increase by a factor of $$3/2$$

D
Increase by a factor of $$5$$

E
Increase by a factor of $$6$$

Solution

Gravitational force between two objects is given by

$$F = \dfrac{Gm_1m_2}{r^2}$$

Hence, In first case, the gravitational force of attraction is given by, $$F_1=\dfrac{Gmm}{r^2}$$

In second case, the gravitational force of attraction is given by, $$F_2=\dfrac{G(2m)(3m)}{r^2}=6F_1$$

Hence, the gravitational force of attraction increases by a factor of $$6$$.
Question 4
1 / -0

Two planet of mass $$m$$ and $$100m$$. If gravitational force exerted by planet of mass $$100m$$ on the planet of mass $$m$$ is $$F_1$$ and gravitational force exerted by planet of mass $$m$$ on the planet of mass $$100m$$ is $$F_2$$. Then which of the following is true?

A
$$F_1=100F_2$$

B
$$F_1=10F_2$$

C
$$F_1=F_2$$

D
$$F_2=10F_1$$

E
$$F_2=100F_1$$

Solution

The two forces form action-reaction pair. Thus they have same magnitude and equal to $$\dfrac{G(100m)m}{d^2}=F_1=F_2$$
Question 5
1 / -0

An astronaut visits another planet which has less gravitational acceleration than the earth. While on that planet, the astronaut will notice that his:

A
weight is less and his mass is greater

B
weight is the same and his mass is the same

C
weight is less and his mass is the same

D
weight is the same and his mass is less

E
weight is less and his mass is less

Solution

We know that mass of a body does not depend upon gravitational acceleration on a planet , therefore it will remain constant on other planet.
Now, weight of a body of mass $$m$$ on a planet is given by ,
$$W=mg_p$$ ,
where $$g_p$$ is gravitational acceleration on this planet.
The earth's gravitational acceleration is $$g=9.8 \ m/s^2$$
As the value of $$g_p$$ on other planet is lesser than that on earth ($$g_p<g$$), therefore weight of astronaut will decrease due to relation
$$W\propto g_p$$ .
Question 6
1 / -0

If the Earth were compressed in such a way that its mass remained the same, but the distance around the equator were just one-half what it is now, what would be the acceleration due to gravity at the surface of the Earth?

A
g/4

B
g/2

C
g

D
2g

E
4g

Solution

Acceleration due to gravity $$g = \dfrac{GM}{R^2}$$ where $$R$$ is the distance of equator from the centre
New distance of equator from the centre $$R' = \dfrac{R}{2}$$

Thus new acceleration due to gravity $$g' = \dfrac{GM}{(R/2)^2} =4g$$
Question 7
1 / -0

An astronaut stands on the top of a crater on the moon. He drops a rock from the top of the crater. Calculate the fraction of its speed to the final impact speed when it is midway between the top and bottom of the crater.

A
$$\dfrac{1}{4\sqrt{2}}$$

B
$$\dfrac{1}{4}$$

C
$$\dfrac{1}{2\sqrt{2}}$$

D
$$\dfrac{1}{\sqrt{2}}$$

Solution

Let the total drop distance be $$h$$.
Speed of rock when dropped through half the height:
$$v_1=\sqrt{2g \times \dfrac{h}{2}}$$
$$g$$ : acceleration due to gravity on the planet
Speed of rock when dropped through full height
$$v_2 = \sqrt{2gh}$$
Now,
$$\implies \dfrac{v_1}{v_2}=\dfrac{1}{\sqrt{2}}$$
Question 8
1 / -0

A purple ball rolls off a horizontal table with an initial speed of $$3.0 m/s$$. A red ball rolls off the same table with an initial speed of $$1.5 m/s$$. Both balls fly undhindered toward the horizontal floor that they hit at some horizontal distance from the edge of the table.
How does the horizontal distance for the purple ball compare to the horizontal distance for the red ball?

A
The horizontal distance for the purple ball is twice as much as that for the red ball

B
The horizontal distance for the purple ball is the same as the horizontal distance for the red ball

C
The horizontal distance for the purple ball is four times as much as that for the red ball

D
The horizontal distance for the purple ball is half as much as that for the red ball

E
We need to know the relative masses of the balls to answer this question

Solution

As both the balls are in free fall therefore they will take equal time to reach to the floor if air resistance is neglected , let the time taken is $$t $$ ,
now horizontal distance for purple ball $$x_{p}=3.0\times t$$ ,
horizontal distance for red ball $$x_{r}=1.5\times t$$ ,
therefore , $$x_{p}/x_{r}=3.0/1.5=2$$ ,
or $$x_{p}=2x_{r}$$
the horizontal distance for purple ball is twice of the red ball .
Question 9
1 / -0

If $$F$$ is the gravitational force between two objects of masses $${m}_{1}$$ and $${m}_{2}$$ separated by a distance $$R$$. Find out the new magnitude of the gravitational force between them if we double the distance between them ?

A
$${F}/{4}$$

B
$${F}/{2}$$

C
$$F$$

D
$$2F$$

E
$$4F$$

Solution

Gravitational force between two masses $$m_{1}$$ and $$m_{2}$$ separated by a distance $$R$$ is given by
$$F=Gm_{1}m_{2}/R^{2}$$
if the distance is doubled then new gravitational force will be
$$F'=Gm_{1}m_{2}/\left(2R\right)^{2}$$
or $$F'=Gm_{1}m_{2}/4R^{2}$$
or $$F'=F/4$$
Question 10
1 / -0

Rank the arrangements of masses given in the table below according the force between masses, greatest first.
The first column in the table tells the mass of one of the objects in each arrangements, the second column gives the mass of the second object , and the third column gives the distance between the centers of the objects.
$$m_1$$ $$m_2$$ r
Arrangement 1 M M d
Arrangement 2 2M M 2d
Arrangement 3 3M M 2d

A
1,2,3

B
1, 2 and 3 tie

C
1, 3, 2

D
3, 2, 1

E
2 and 3 tie, 1

Solution

The gravitational force $$F$$ between two masses $$m_{1} , m_{2}$$ separated by a distance $$r$$ is given by ,
$$F=K\dfrac{m_{1}m_{2}}{r^{2}}$$ ,

For first arrangement,
$$F_{1}=K\dfrac{M\times M}{d^{2}}=K\dfrac{M^{2}}{d^{2}}$$ ,

For second arrangement ,
$$F_{2}=K\dfrac{2M\times M}{(2d)^{2}}=2K\dfrac{M^{2}}{4d^{2}}$$ ,

or $$F_{2}=K\dfrac{M^{2}}{2d^{2}}$$ ,

For third arrangement ,
$$F_{3}=K\dfrac{3M\times M}{(2d)}^{2}=3K\dfrac{M^{2}}{4d^{2}}$$ ,
it is clear that $$F_{1}>F_{3}>F_{2}$$
therefore the order from greatest first is 1,3,2.

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	$$m_1$$	$$m_2$$	r
Arrangement 1	M	M	d
Arrangement 2	2M	M	2d
Arrangement 3	3M	M	2d

Gravitation Test - 49

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Gravitation Test - 49

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Question 1

Decreases by a factor of $$4$$

Decreases by a factor of $$2$$

Increases by a factor of $$2$$

increases by a factor of $$4$$

Cannot be determined without knowing the masses

Solution

Question 2

The stone exerts a force of attraction on the earth

The earth exerts force of gravity on the stone

The stone attracts the earth towards itself

All the above statements are true

Solution

Question 3

Decrease by a factor of $$1/6$$

Decrease by a factor of $$2/3$$

Increase by a factor of $$3/2$$

Increase by a factor of $$5$$

Increase by a factor of $$6$$

Solution

Question 4

$$F_1=100F_2$$

$$F_1=10F_2$$

$$F_1=F_2$$

$$F_2=10F_1$$

$$F_2=100F_1$$

Solution

Question 5

weight is less and his mass is greater

weight is the same and his mass is the same

weight is less and his mass is the same

weight is the same and his mass is less

weight is less and his mass is less

Solution

Question 6

g/4

g/2

g

2g

4g

Solution

Question 7

$$\dfrac{1}{4\sqrt{2}}$$

$$\dfrac{1}{4}$$

$$\dfrac{1}{2\sqrt{2}}$$

$$\dfrac{1}{\sqrt{2}}$$

Solution

Question 8

The horizontal distance for the purple ball is twice as much as that for the red ball

The horizontal distance for the purple ball is the same as the horizontal distance for the red ball

The horizontal distance for the purple ball is four times as much as that for the red ball

The horizontal distance for the purple ball is half as much as that for the red ball

We need to know the relative masses of the balls to answer this question

Solution

Question 9

$${F}/{4}$$

$${F}/{2}$$

$$F$$

$$2F$$

$$4F$$

Solution

Question 10

1,2,3

1, 2 and 3 tie

1, 3, 2

3, 2, 1

2 and 3 tie, 1

Solution

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