Gravitation Test

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Gravitation Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

The mass of earth is $$6\times 10^{24} \ kg$$ and that of the moon is $$7.4\times 10^{22} \ kg$$. If the distance between earth and moon is $$3.84\times 10^{6} \ km$$. Calculate the force exerted by the earth on the moon. $$(G = 6.7\times 10^{-11}N\ m^{2} kg^{-2})$$

A
$$6\times 10^{24} \ N$$

B
$$20\times 10^{17}\ N$$

C
$$36\times 10^{24} \ N$$

D
$$12\times 10^{17}\ N$$

Solution

Given, mass of earth $$m_e = 6 \times 10^{24} \ kg$$
mass of moon $$m_m = 7.4 \times 10^{22} \ kg$$
distance between earth and moon $$r = 3.84 \times 10^{6} \ km = 3.84 \times 10^{9} \ m$$
Gravitational constant $$G = 6.7 \times 10^{-11} \ Nm^2kg^{-2}$$

Using Newton's law of gravitation,

Force exerted by Earth on the Moon is given by,

$$F=\cfrac { G{ m }_{ e}{ m }_{ m } }{ { r }^{ 2 } } \\ \quad =\cfrac { 6.7\times { 10 }^{ -11 }\times 6\times { 10 }^{ 24 }\times 7.4\times { 10 }^{ 22 } }{ (3.84\times { 10 }^{ 9 })^2 } \\ =20 \times { 10 }^{ 17 } \ N$$
Question 2
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If you have to purchase $$100\ kg$$ weight of gold, which place would you prefer-the Earth or the Moon?
Similarly, if you have to sell gold, which place you prefer - The earth or the moon?

A
Earth, Moon

B
Moon, Moon

C
Earth, Earth

D
Moon, Earth

Solution

To buy $$100\ kg$$ of gold, a place with lower gravitational force will be preferred as a lot of gold will be needed to fill the $$100\ kg$$ quota. And to sell it, a place with higher gravitational force is preferred as that $$100\ kg$$ will seem more than $$100\ kg$$ in this place, thus giving the person excess of gold.
Thus gold must be purchased on the moon and sold on earth if a choice is given.
Question 3
1 / -0

Which of the following statements regarding gravitational force existing between two bodies is true?

A
The first body exerts attractive force on the second body while the second body exerts repulsive force on the first body

B
The gravitational force is zero when they are kept in vacuum

C
Force exerted by the first body on the second is not equal to the force exerted by the second body on the first

D
Force exerted by the first body on the second body is equal to the force exerted by the second body on the first

Solution

According to the universal law of gravitation, every object in the universe attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centers of two objects.
Therefore, the force exerted by the first body on the second body is equal to the force exerted by the second body on the first.
Question 4
1 / -0

If $$F$$ is the force between two bodies of masses $${ m }_{ 1 }$$ and $${ m }_{ 2 }$$ at certain separation, then what is the force between $$\sqrt { 5 } { m }_{ 1 }$$ and $$\sqrt { 3 } { m }_{ 2 }$$ at the same separation?

A
$$\sqrt { 5 } F$$

B
$${ F }/{ \sqrt { 15 } }$$

C
$$\sqrt { 15 } F$$

D
$$F$$

Solution

$$F=\frac { G{ m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } $$
Where r is the distance between $${ m }_{ 1 }$$ and $${ m }_{ 2 }$$
If $${ m }_{ 1 }\rightarrow \sqrt { 5 } { m }_{ 1 }$$
$${ m }_{ 2 }\rightarrow \sqrt { 3 } { m }_{ 2 }$$
$$\therefore \quad { F }^{ \prime }=\frac { G\sqrt { 5 } { m }_{ 1 }\sqrt { 3 } { m }_{ 2 } }{ { r }^{ 2 } } $$
$$=\sqrt { 15 } \frac { G{ m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } =\sqrt { 15 } F$$
Question 5
1 / -0

A particle is projected up with a velocity of 20 m $$s^{-1}$$ from a tower of height 25 m. Its velocity on reaching the ground is _________ m $$s^{-1}$$.

A
20

B
40

C
30

D
10

Solution

Taking upward direction to be positive.
Initial velocity   $$u = 20 \ m/s$$
Acceleration $$a = -g = -10 \ m/s^2$$
Displacement   $$S = -25 \ m$$
Using   $$v^2 - u^2 = 2aS$$
Or    $$v^2 - (20)^2 = 2(-10)(-25)$$
$$\therefore$$   $$v^2 -400 = 500$$
Or   $$v = \sqrt{900} = 30 \ m/s$$
Question 6
1 / -0

Find the gravitational force between the two atoms in a hydrogen molecule. Given that $$G = 6.67\times 10^{-11}N\ m^{2} kg^{-2}$$, mass of a hydrogen atom $$= 1.67\times 10^{-27}kg$$ and distance between the two atoms $$= 1^{\circ}A$$

A
$$2.86\times 10^{-34}N$$

B
$$3.86\times 10^{-34}N$$

C
$$1.86\times 10^{-34}N$$

D
$$1.86\times 10^{-14}N$$

Solution

Given,
Gravitational constant $$G=6\cdot 67\times { 10 }^{ -11 }N{ m }^{ 2 }{ kg }^{ -2 }$$

Mass of hydrogen atom $${ m }_{ 1 }={ m }_{ 2 }=1\cdot 67\times { 10 }^{ -27 }kg$$

Distance between the two atoms $$r=1\ { A^o } ={ 10 }^{ -10 }\ m$$

$$\therefore \quad F=\dfrac { { Gm }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } =\dfrac { { 6\cdot 67\times 10 }^{ -11 }\times { 1\cdot 67\times 1\cdot 67\times 10 }^{ -27 }{ \times 10 }^{ -27 } }{ { 10 }^{ -10 }\times { 10 }^{ -10 } } $$

$$F=1\cdot 86\times { 10 }^{ -34 }\ N$$
Question 7
1 / -0

The acceleration due to gravity on the surface of a planet, whose mass and diameter are double that of the earth, is _____ times the acceleration due to gravity on the surface of the earth.

A
$$1$$

B
$$0.5$$

C
$$2$$

D
$$0.25$$

Solution

Acceleration due to the gravity at the surface of the earth $$=g_e=\dfrac{GM_e}{R_e^2} $$
It is given that $$M_p=2M_e,R_p=2R_e$$
Acceleration due to the gravity at the surface of the planet $$g_p=\dfrac{GM_p}{R _p^2}=\dfrac{G(2M_e)}{(2R_e)^2} =\dfrac{GM_e}{2 R_e^2}=\dfrac{g_e}{2}$$

Hence correct answer is option $$B $$
Question 8
1 / -0

The gravitational force with which the earth attracts the moon :

A
is less than the force with which the moon attracts the earth

B
is equal to the force with which the moon attracts the earth

C
is greater than the force with which the moon attracts the earth

D
varies with the phases of the moon

Solution

Hint:- Gravitational force follows Newton’s third law.
Explanation:-
$$\bullet$$ Gravitational force between two bodies follows Newton’s third law of motion,i.e, every action has equal and opposite reaction .
$$\bullet$$ It means the force with which the earth attracts the moon is equal to the force with which the moon attracts the earth.
$$\textbf{Hence option B corect}$$
Question 9
1 / -0

Two persons having mass $$50$$ kg each, are standing such that their centre of gravities are $$1\ m$$ apart. Calculate the force of gravitation between them.
(Take $$G = 6.67\times 10^{-11}N\ m^{2} kg^{-2}$$, mass of earth $$M = 6\times 10^{24} kg$$, Radius of earth $$R = 6.4\times 10^{6}m$$)

A
$$2.67\times 10^{-7}N$$

B
$$1.67\times 10^{-7}N$$

C
$$1.32\times 10^{-7}N$$

D
$$1.67\times 10^{-9}N$$

Solution

When both masses are equal then,
As we know that Gravitational force is given by:
$$F=\cfrac { { GM }^{ 2 } }{ { R }^{ 2 } } =\cfrac { 6.67\times { 10 }^{ -11 }\times 50\times 50 }{ 1\times 1 } =1.67\times { 10 }^{ -7 }N$$
Question 10
1 / -0

The ratio of SI unit of G to its CGS unit is _______.

A
$$100 : 1$$

B
$$1000 : 1$$

C
$$10 : 1$$

D
$$10000 : 1$$

Solution

We know that SI unit of G $$=m^3/s^2kg=(100cm) ^3/s^2(1000g)=10^6cm^3/10^3gs^2=10^3cm^3/gs^2$$

Also the CGS unit of G $$=cm^3/gs^2$$
$$\Rightarrow \dfrac{SI}{CGS}=1000$$

Hence correct answer is option $$B$$

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Gravitation Test - 51

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Gravitation Test - 51

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SHARING IS CARING

Question 1

$$6\times 10^{24} \ N$$

$$20\times 10^{17}\ N$$

$$36\times 10^{24} \ N$$

$$12\times 10^{17}\ N$$

Solution

Question 2

Earth, Moon

Moon, Moon

Earth, Earth

Moon, Earth

Solution

Question 3

The first body exerts attractive force on the second body while the second body exerts repulsive force on the first body

The gravitational force is zero when they are kept in vacuum

Force exerted by the first body on the second is not equal to the force exerted by the second body on the first

Force exerted by the first body on the second body is equal to the force exerted by the second body on the first

Solution

Question 4

$$\sqrt { 5 } F$$

$${ F }/{ \sqrt { 15 } }$$

$$\sqrt { 15 } F$$

$$F$$

Solution

Question 5

20

40

30

10

Solution

Question 6

$$2.86\times 10^{-34}N$$

$$3.86\times 10^{-34}N$$

$$1.86\times 10^{-34}N$$

$$1.86\times 10^{-14}N$$

Solution

Question 7

$$1$$

$$0.5$$

$$2$$

$$0.25$$

Solution

Question 8

is less than the force with which the moon attracts the earth

is equal to the force with which the moon attracts the earth

is greater than the force with which the moon attracts the earth

varies with the phases of the moon

Solution

Question 9

$$2.67\times 10^{-7}N$$

$$1.67\times 10^{-7}N$$

$$1.32\times 10^{-7}N$$

$$1.67\times 10^{-9}N$$

Solution

Question 10

$$100 : 1$$

$$1000 : 1$$

$$10 : 1$$

$$10000 : 1$$

Solution

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