Gravitation Test

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Gravitation Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

SI unit of G is $$Nm^{2}kg^{-2}$$.Which of the following can also be used as the SI unit of G?

A
$$m^3 kg^{-1} s^{-2}$$

B
$$m^2 kg^{-2} s^{-1}$$

C
$$m kg^{-1} s^{-1}$$

D
$$m^3kg^{-3}s^{-2}$$

Solution

Here $$N$$ represents Newton.
so we can write unit of force in place of newton, also $$F=ma$$
so we can write $$G=M^1L^1t^{-2}L^2M^{-2}=M^{-1}L^3T^{-2}$$
SI unit for mass is kilogram(kg), for length is meter(m), for time is sec(s)
so we can write unit for G in SI as $$kg^{-1}m^3s^{-2}$$
so best possible answer is option A.
Question 2
1 / -0

The force of gravity on a body of mass W on the surface of the earth is

A
W

B
$$W \times \ g$$

C
$$\dfrac {W}{g}$$

D
$$W\times g^2$$

Solution

The force acting on an object of mass m on the earth's surface due to gravitational interaction with the earth has magnitude mg where g = 9.8 ms$$^{2}$$.
So the force of gravity on a body of mass $$W$$ on the surface of the earth is $$W \times g.$$
Question 3
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Choose the correct statement.

A
Gravity and gravitation are same

B
Gravity is a particular case of gravitation

C
Acceleration due to gravity is a scalar quantity

D
Different heavenly bodies have different values of G

Solution

Gravitation is the attractive force existing between any two objects that have mass. The force of gravitation pulls objects together. Gravity is the gravitational force that occurs between the earth and other bodies.
so gravity is a subset of gravitation.
acceleration is a vector quantity so option C is wrong.
G is a universal constant, always constant so option D is wrong.
Hence, option B is the only correct option.
Question 4
1 / -0

The ratio of the value of G in SI units to CGS units is

A
$$10^3:1$$

B
$$10^2:1$$

C
$$10^{-2}:1$$

D
$$10^{-3}:1$$

Solution

$$\text{Quantity}$$ $$\text{SI System}$$ $$\text{CGS System}$$
Force Newton (N) dyne
Distance metre (m) centi-meter (cm)
Mass kilogram (kg) gram (g)
The unit of G can be calculated by putting the unit of all the quantities in the expression of the force due to gravity.
$$\dfrac{Gm_1m_2}{r^2}=F$$ $$\Rightarrow G=\dfrac{F\times r^2}{m_1\times m_2}$$

So we get, $$\text{Unit of G}=\dfrac{\text{Unit of force}\times \text{(Unit of distance)}^2}{\text{(Unit of mass)}^2}$$

$$\text{SI unit of G}=\dfrac{N.m^2}{kg^2}$$ and $$\text{CGS unit of G}=\dfrac{dyne.cm^2}{g^2}$$

We know that $$1\ N = 10^5\ dyne$$, $$1\ kg = 10^3\ g$$ and $$1\ m = 10^2\ cm$$

$$\underline{\text{Required ratio:}}$$
$$\dfrac{SI}{CGS}=\dfrac{N.m^2.g^{-2}}{dyne.cm^2.kg^{-2}}=\dfrac{10^{5}\ dyne\times 10^4\ cm^2\times g^{-2}}{dyne\times cm^2\times 10^{-6}\ g^{-2}}=10^3 : 1$$

So the answer is option A.
Question 5
1 / -0

The gravitational force between two stones of mass 1 kg each, separated by a distance of 1 m in vacuum is.

A
zero

B
$$6.675 \times 10^{-6}N$$

C
$$8.326 \times 10^{-8}N$$

D
$$6.675 \times 10^{-11}N$$

Solution

force due to gravity is given by $$F=\dfrac{Gm_1m_2}{r^2}$$
where G is gravitational constant $$G=6.675\times10^{-11}$$
also $$m_1=m_2=1kg$$, $$r=1m$$
so $$F=\dfrac{G1.1}{1^2}=$$$$G=6.675\times10^{-11}N.$$
so correct option is option D.
Question 6
1 / -0

If the symbols had their actual meaning, $$\dfrac{GM}{R^2}$$ will be equal to

A
$$g^2$$

B
$$\dfrac {1}{g^2}$$

C
$$g$$

D
$$\dfrac {1}{g}$$

Solution

The gravitational force acting on two objects is given by $$F=\dfrac{GMm}{R^2}$$
Also, we know that $$F=mg$$, where $$g$$ is the acceleration due to gravity.

On dividing both the equations, we get
$$\dfrac{GM}{R^2}=g$$

Hence. option C is correct.
Question 7
1 / -0

The gravitational force between two objects placed at a distance r is proportional to

A
$$r$$

B
$$r^2$$

C
$$\dfrac {1} {r^2}$$

D
$$\dfrac {1}{r}$$

Solution

The force of gravitation between the two bodies $$=\dfrac{GMm}{r^2} $$
Where G is gravitational constant
r is separation between the two bodies
M is mass of one body
m is mass of the other body

From the above expression we can easily see that the force is proportional to the inverse of the square of the distance between the two bodies i.e. $$\dfrac{1 }{r^2}$$

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

How much would a $$W \ kg-wt$$ man weigh on the moon in terms of gravitational units?

A
$$\dfrac {W}{6} \ kg-wt$$

B
$$6W \ kg-wt$$

C
$$W \ kg-wt$$

D
$$zero$$

Solution

We know that weight of a body is given by $$W=mg$$

So $$W_{earth} = mg$$

The gravity on the moon is $$\dfrac{1}{6}th$$ that of the earth.

Hence, $$W_{moon} = m\dfrac{g}{6}$$

Hence, a man weighing $$W \ kg-wt$$ on the earth , would weigh $$\dfrac{W}{6} \ kg-wt$$ on the moon
Question 9
1 / -0

How much would a $$60\ kg$$ boy weigh on the moon?
Given: $$g_{moon}=\dfrac {g_{earth}}{6}$$

A
$$10\ kg-f$$

B
$$6\ kg-f$$

C
$$\dfrac {1}{6}\ kg-f$$

D
Zero

Solution

Weight of the boy on moon $$=m\times g_{moon}=60\times\dfrac{g_{earth}}{6}=10kgf$$

Hence correct answer is option $$A $$
Question 10
1 / -0

As a body moves from the equator to the poles, the value of g experienced by the body

A
decreases

B
increases

C
remains unchanged

D
first decreases then increases

Solution

The shape of the earth is not perfectly spherical. It is flattened at the poles and bulged out at the equator. So earth has more radius at the equator in compare to at poles.
As we know $$g=\dfrac{GM}{R^2}$$, where G: Universal gravity constant, R: Radius of earth.
Because radius at equator is more, acceleration due to gravity will be more at pole.
Hence when a body moves from the equator to the poles, the value of g experienced by the body increases.

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$$\text{Quantity}$$	$$\text{SI System}$$	$$\text{CGS System}$$
Force	Newton (N)	dyne
Distance	metre (m)	centi-meter (cm)
Mass	kilogram (kg)	gram (g)

Gravitation Test - 25

Result

Gravitation Test - 25

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Question 1

$$m^3 kg^{-1} s^{-2}$$

$$m^2 kg^{-2} s^{-1}$$

$$m kg^{-1} s^{-1}$$

$$m^3kg^{-3}s^{-2}$$

Solution

Question 2

W

$$W \times \ g$$

$$\dfrac {W}{g}$$

$$W\times g^2$$

Solution

Question 3

Gravity and gravitation are same

Gravity is a particular case of gravitation

Acceleration due to gravity is a scalar quantity

Different heavenly bodies have different values of G

Solution

Question 4

$$10^3:1$$

$$10^2:1$$

$$10^{-2}:1$$

$$10^{-3}:1$$

Solution

Question 5

zero

$$6.675 \times 10^{-6}N$$

$$8.326 \times 10^{-8}N$$

$$6.675 \times 10^{-11}N$$

Solution

Question 6

$$g^2$$

$$\dfrac {1}{g^2}$$

$$g$$

$$\dfrac {1}{g}$$

Solution

Question 7

$$r$$

$$r^2$$

$$\dfrac {1} {r^2}$$

$$\dfrac {1}{r}$$

Solution

Question 8

$$\dfrac {W}{6} \ kg-wt$$

$$6W \ kg-wt$$

$$W \ kg-wt$$

$$zero$$

Solution

Question 9

$$10\ kg-f$$

$$6\ kg-f$$

$$\dfrac {1}{6}\ kg-f$$

Zero

Solution

Question 10

decreases

increases

remains unchanged

first decreases then increases

Solution

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