Gravitation Test

Result

Gravitation Test - 21

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The maximum value of $$g$$ is

A
At the poles

B
At the top of the Mounts Everest

C
At the equator

D
Below the sea level

Solution

The value of $$'g'$$ depends upon the distance from the centre of the earth; lesser thea distance higher the value. At the poles this distance is least, hence $$g$$ is maximum there.
Question 2
1 / -0

A body weighs 60 kg on the earth's surface. What would be its weight at the centre of the earth?

A
60 kg-wt

B
6 kg-wt

C
60$$\times$$9.8 kg-wt

D
zero

Solution

The weight at the centre of the earth would be zero. As the body moves closer to the centre of the earth the mass of the earth between the centre of the earth and the body keeps decreasing. This causes the force acting from the centre of the earth on the body to decrease.
Question 3
1 / -0

A raised hammer possesses

A
K.E. only

B
Gravitational P.E,

C
Electrical energy

D
sound energy

Solution

The energy possesed by a body due to virtue of its position is called potential energy.
A raised hammer will not have any velocity so it won't have any kinetic energy however it will posses potential energy due to the work done on the hammer by the gravity.

Hence correct answer is option $$B $$
Question 4
1 / -0

The variation of g with height or depth (r) is shown correctly by which of the following graphs. (where R is the radius of the earth).

A

B

C

D

Solution

Variation of g with height (r) $$=\dfrac{GM}{(R+r) ^2}$$
$$\Rightarrow g $$ $$\alpha $$ $$r^{-2}$$

Variation of g with depth (r) $$=\dfrac{GM}{R ^2} (1- \dfrac{r}{R}) $$
$$\Rightarrow g $$ $$\alpha $$ $$r$$

Hence correct answer is option $$A $$
Question 5
1 / -0

The evidence to show that there must be force acting on Earth and directed towards the Sun is.

A
phenomenon of day and night

B
apparent motion of the Sun around the Earth

C
revolution of Earth around the Sun

D
deviation of the falling bodies towards east

Solution

The earth revolves around the sun due to gravitation pull of the sun. Due to this gravitational attraction between this celestial body, centripetal force is generated which binds the solar system together. Hence revolution of earth round the sun is the evidence to show that there ust be force acting on earth nd directed towards the sun.
Question 6
1 / -0

Which of the following units can be used to express G?

A
N Kg$$^{-2}$$

B
Nm$$^{-2}$$ Kg$$^{2}$$

C
Nm$$^{2}$$ Kg$$^{-2}$$

D
Nm$$^{-2}$$ Kg$$^{-3}$$

Solution

Gravitational force between two masses $$m$$ and $$M$$, separated by a distance $$r$$ is given by $$F = \dfrac{GMm}{r^2}$$
Thus, $$G = \dfrac{Fr^{2}}{Mm}$$
We know that the unit of force ($$F$$) is newton (N), the unit of mass ($$m\ or\ M$$) is kg and the unit of distance ($$r$$) is meter(m).
$$\text{The unit of G} = \dfrac{Nm^2}{kg^2} = Nm^2kg^{-2}$$
Question 7
1 / -0

At a height equal to earth's radius, above the earth's surface, the acceleration due to gravity is

A
g

B
$$\displaystyle \frac{g}{2}$$

C
$$\displaystyle \frac{g}{4}$$

D
$$\displaystyle \frac{g}{8}$$

Solution

Acceleration due to gravity = $$g = \dfrac{GM_{earth}}{R^2} $$, $$R$$ = distance from the centre of the earth.
At a height = R above earth's surface $$g = \dfrac{GM_{earth}}{(R+R)^2} = \dfrac{GM_{earth}}{4R^2} = \dfrac{g}{4}$$ Option C.
Question 8
1 / -0

The universal law of gravitation must be applicable to

A
The earth and the moon.

B
The planets around the Sun.

C
Any pair of bodies.

D
The earth and the apple.

Solution

The Law of Gravitation is universal which means that it applies to all the bodies in the Universe which have some mass for example earth and moon, planets around sun, apple and earth and almost everything. Therefore, option C is correct.
Question 9
1 / -0

Calculate the weight of the block at a distance of four times of Earth radii from the center of the Earth. It is given that g is the acceleration due to gravity at the surface of Earth and mass of the block is m.

A
16mg

B
4mg

C
2mg

D
$$\dfrac{1}{4}$$mg

E
$$\dfrac{1}{16}$$mg

Solution

Acceleration due to gravity at earth's surface $$g = \dfrac{GM_{earth}}{R^2} $$,
$$R$$ = distance from the centre of the earth.
At a height $$h= 4R$$ from earth's center $$g' = \dfrac{GM_{earth}}{(4R)^2} = \dfrac{GM_{earth}}{16R^2} = \dfrac{g}{16}$$
So, weight $$W' = mg' = \dfrac{mg}{16}$$
Question 10
1 / -0

Fill in the blanks:
Value of gravitational constant (G) on moon is __________ as on earth.

A
Greater

B
Smaller

C
Same

D
None

Solution

G is an universal constant and it remains same everywhere. It does not depend on mass or radius of the body.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Gravitation Test - 21

Result

Gravitation Test - 21

Score

-

Rank

-

SHARING IS CARING

Question 1

At the poles

At the top of the Mounts Everest

At the equator

Below the sea level

Solution

Question 2

60 kg-wt

6 kg-wt

60$$\times$$9.8 kg-wt

zero

Solution

Question 3

K.E. only

Gravitational P.E,

Electrical energy

sound energy

Solution

Question 4

Solution

Question 5

phenomenon of day and night

apparent motion of the Sun around the Earth

revolution of Earth around the Sun

deviation of the falling bodies towards east

Solution

Question 6

N Kg$$^{-2}$$

Nm$$^{-2}$$ Kg$$^{2}$$

Nm$$^{2}$$ Kg$$^{-2}$$

Nm$$^{-2}$$ Kg$$^{-3}$$

Solution

Question 7

g

$$\displaystyle \frac{g}{2}$$

$$\displaystyle \frac{g}{4}$$

$$\displaystyle \frac{g}{8}$$

Solution

Question 8

The earth and the moon.

The planets around the Sun.

Any pair of bodies.

The earth and the apple.

Solution

Question 9

16mg

4mg

2mg

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

$$\dfrac{1}{16}$$mg

Solution

Question 10

Greater

Smaller

Same

None

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.