Gravitation Test

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Gravitation Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

The mass of earth is $$80$$ times that of moon and its diameter is double that of moon. If the value of acceleration due to gravity on earth is $$9.8 m/s^{2}$$, then the value of acceleration due to gravity on moon will be

A
$$0.98\ ms^{-2}$$

B
$$0.49\ ms^{-2}$$

C
$$9.8\ ms^{-2}$$

D
$$4.9\ ms^{-2}$$

Solution

$$g = \dfrac {GM}{R^{2}}$$
$$9.8 = \dfrac {GM}{R^{2}}$$
$$g' = \dfrac {GM \times 4}{80\times R^{2}}$$
$$= \dfrac {GM}{R^{2}} \times \dfrac {1}{20}$$
$$= \dfrac {9.8}{20}$$
$$g' = 0.49\ ms^{-2}$$.
Question 2
1 / -0

The gravitational force between two bodies is $$6.67\times 10^{-7}N$$ when the distance between their centres is 10 m. If the mass of first body is 800 kg, then the mass of second body is :

A
1000 kg

B
1250 kg

C
1500 kg

D
2000kg

Solution

$$F=\displaystyle \frac{Gm_{1}m_{2}}{R^{2}}$$
In the above formula,
$$G=6.67\times10^{-11},\; m_1=800kg\;and\;R=10m$$
$$\Rightarrow 6.67\times { 10 }^{ -7 }=\dfrac { 6.67\times { 10 }^{ -11 }\times 800\times { m }_{ 2 } }{ 100 }$$
$$\Rightarrow m_2=1250\;kg$$
Question 3
1 / -0

A body of mass $$5\ kg$$ is taken into space. Its mass:

A
Remains $$5\ kg$$

B
Becomes $$10\ kg$$

C
Becomes $$2\ kg$$

D
Becomes zero

Solution

Mass is a scalar quantity that is independent of its position and configuration in the particular inertial frame. So, it remains constant everywhere in space.
Question 4
1 / -0

In the Newton's gravitational law, $$F = \dfrac{G M m}{d^{2}} $$, the quantity G:

A
depends on the value of g at the place of observation

B
is used only when the earth is one of the two masses

C
is greatest at the surface of the earth

D
is universal constant in nature

Solution

From Newton's law of gravitation, the force of attraction between bodies $$ F= \dfrac{GMm}{r^2} $$.
Here G is called the universal gravitational constant as the value of G is same whether the measurements are made on earth or any other planet.
The value of gravitational constant is $$ G = 6.67 \times 10^{-11} N/m^{2}$$.
Question 5
1 / -0

Two spherical balls each of mass $$1 kg$$ are placed $$1 cm$$ apart. The gravitational force of attraction between them is

A
$$6.67\times 10^{-7}$$ N

B
$$6.67\times 10^{-4}$$ N

C
$$6.67\times 10^{-2}$$ N

D
$$6.67\times 10^{-9}$$ N

Solution

The gravitational force of attraction is given by:
$$ F = \dfrac{Gm_1m_2}{r^2} $$

Here $$m_1=m_2 = 1\ kg$$ and $$r = 1$$ cm $$ = 0.01\ m$$

$$ F = \dfrac{6.67 \times 10^{-11}\times 1^2}{(0.01)^2} $$

$$\implies F = 6.67 \times 10^{-7} N$$
Question 6
1 / -0

The mass of a ball is four times the mass of another ball. When these balls are separated by a distance of $$10cm$$, the gravitational force between them is $$6.67\times 10^{-7}N$$ . The masses of the two balls are

A
10 kg, 20 kg

B
5 kg, 20 kg

C
20 kg, 30 kg

D
20 kg, 40 kg

Solution

$$ F = 4G M^{2} / a^{2} $$
$$ M = \sqrt{F \times a^{2}/4G} $$
$$a= 0.1\ m $$, $$F= 10000G$$
$$ M = \sqrt{10000\times0.01/4} = \sqrt{25} = 5kg $$
Masses are $$5kg$$ and $$20kg$$.
Question 7
1 / -0

The universal law of gravitation gives the gravitational force between

A
the Earth and a point mass only

B
the Earth and the Sun only

C
any two bodies having some mass

D
two charged bodies only

Solution

The gravitational force between any two bodies having some mass can be determined from Newton's law of gravitation.
The force of attraction between bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
$$ F= G\dfrac{Mm}{r^2} $$.
Here, G is called the universal gravitational constant.
Question 8
1 / -0

Moon is revolving in a circular orbit with a uniform velocity $$V_{0}$$. If the gravitational force suddenly disappears, the moon will

A
continue to move in the same orbit

B
move with a velocity $$V_{0}$$ tangentially to the orbit

C
fall down freely

D
ultimately comes to rest

Solution

If the gravitational pull on the moon suddenly disappears, the centripetal force acting on the moon is no longer present and hence, the moon escapes the orbit on a tangential path at velocity equal to $$V_0$$ as no force is available to make it revolve in a circle.
Question 9
1 / -0

If the gravitational force of earth suddenly disappears, then which of the following is correct?

A
weight of the body is zero

B
mass of the body is zero

C
both mass and weight become zero

D
neither the weight nor the mass is zero

Solution

Mass = $$m$$
Acceleration due to gravity = $$g$$
Weight of a body is given by, $$(W) = m\times g$$
When gravitational force disappears $$g$$ becomes zero, but the mass remains the same.
So, $$W = m\times g=m\times 0$$
Hence, $$W = 0$$
Correct option will be $$(A)$$
Question 10
1 / -0

If suddenly the gravitational force of attraction between earth and satellite revolving around it becomes zero, then the satellite will

A
Continue to move in its orbit with same velocity

B
Move tangential to the original orbit with the same velocity

C
Becomes sationary in its orbit

D
Move towards the earth

Solution

For a body revolving around any object, it is bound by the centripetal force by the object on the body.
This force is along the line joining the body and the object. But the velocity of the body will be perpendicular to this force, that is tangential.
In this case, when the gravitational force becomes zero, there is no binding on the satellite to revolve around the Earth. Since it was moving with a velocity '$$v$$' in the tangential direction, it continues to move in the same direction with the same velocity '$$v$$'.
Example: Keep a stone on a rotating wheel. Stone will fly off the wheel tangential to the wheel's radius.

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Gravitation Test - 16

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Gravitation Test - 16

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Question 1

$$0.98\ ms^{-2}$$

$$0.49\ ms^{-2}$$

$$9.8\ ms^{-2}$$

$$4.9\ ms^{-2}$$

Solution

Question 2

1000 kg

1250 kg

1500 kg

2000kg

Solution

Question 3

Remains $$5\ kg$$

Becomes $$10\ kg$$

Becomes $$2\ kg$$

Becomes zero

Solution

Question 4

depends on the value of g at the place of observation

is used only when the earth is one of the two masses

is greatest at the surface of the earth

is universal constant in nature

Solution

Question 5

$$6.67\times 10^{-7}$$ N

$$6.67\times 10^{-4}$$ N

$$6.67\times 10^{-2}$$ N

$$6.67\times 10^{-9}$$ N

Solution

Question 6

10 kg, 20 kg

5 kg, 20 kg

20 kg, 30 kg

20 kg, 40 kg

Solution

Question 7

the Earth and a point mass only

the Earth and the Sun only

any two bodies having some mass

two charged bodies only

Solution

Question 8

continue to move in the same orbit

move with a velocity $$V_{0}$$ tangentially to the orbit

fall down freely

ultimately comes to rest

Solution

Question 9

weight of the body is zero

mass of the body is zero

both mass and weight become zero

neither the weight nor the mass is zero

Solution

Question 10

Continue to move in its orbit with same velocity

Move tangential to the original orbit with the same velocity

Becomes sationary in its orbit

Move towards the earth

Solution

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