Gravitation Test

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Gravitation Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The value of g on the earth's surface is $$980 cm s^{-2}$$. Its value at a height of 64km from the earth's surface is:

A
$$960.40 cm s^{-2}$$

B
$$984.90 cm s^{-2}$$

C
$$982.45 cm s^{-2}$$

D
$$977.55 cm s^{-2}$$

Solution

$$g=980 cm s^{-2}$$
$$g_h=g(1-\dfrac {2h}{R})=980(1-\dfrac {2(64)}{6400})$$
$$=960.40 cm s^{-2}$$
Question 2
1 / -0

In the relation F= $$\dfrac{G M m}{r^{2}}$$, 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

$$G$$ is the universal gravitational constant which remains constant at all places in the universe. $$G$$ is equivalent to the force of attraction between two bodies of unit mass and unit distance apart.
Question 3
1 / -0

Acceleration due to gravity ---------- with depth from the surface of the earth.

A
Decreases

B
Increases

C
Remains constant

D
Data insufficient

Solution

$$g=(\dfrac{4\pi G\rho_o}{3})(1-\dfrac{d}{R})$$ where $$\rho_o$$ is the density of earth,$$R$$ is the radius of Earth, $$d$$ is depth
So as d increases acceleration due to gravity decreases
Question 4
1 / -0

The force of attraction between two unit point masses separated by a unit distance is called

A
Gravitational potential

B
Acceleration due to gravity.

C
Gravitational field

D
Universal gravitational constant.

Solution

Universal gravitational constant is the force of attraction between two bodies of unit mass and at a unit distance from each other.
Question 5
1 / -0

A body has a weight of $$10 kg$$ on the surface of the Earth. What will be its mass and weight when taken to the centre of the Earth?

A
$$10$$ kg, zero

B
zero, zero

C
$$10$$ kg, $$10$$g

D
zero, $$10$$g

Solution

Mass is constant and hence will be $$10 kg$$.
W=mg and $$g=\dfrac{4 \pi G \rho_0}{3}(1- \dfrac{d}{R})$$ where R is the radius of earth, $$\rho_0$$ is the density of earth, d is the depth.
At the center of earth $$d=R$$, and hence $$g=0$$ and $$W=0$$
Question 6
1 / -0

The weight of an object at the centre of the earth of radius $$R$$ is

A
Zero

B
Infinite

C
$$R$$ times the weight at the surface of the earth.

D
$${1/R^2}$$ times the weight at surface of the earth.

Solution

Gravitational pull for any point below the earth's surface is given by $$g= \dfrac{4\pi G(R_e-d) \rho}{3}$$
At center of earth depth $$d= R_e$$
Therefore, $$g= \dfrac{4\pi G(R_e-R_e) \rho}{3}$$ = 0
Question 7
1 / -0

Where will a body weigh minimum?

A
At a height of $$100 m$$ above the earth's surface

B
At the earth's surface

C
At a depth of $$100 m$$ below the earth's surface

D
At the centre of the earth

Solution

Weight at center of earth is 0. $$g=\dfrac{4\pi}{3} {G\rho (R-d)}$$
Thus, for $$d=R$$ i.e. at centre of earth $$g=0$$ and hence $$W=0$$
Question 8
1 / -0

The three laws of planetary motion were given by:

A
Einstein

B
Kepler

C
Copernicus

D
Tycho Brahe
Solution
Answer is B.

In the early 1600s, Johannes Kepler proposed three laws of planetary motion.
Kepler's three laws of planetary motion can be described as follows:
- The path of the planets about the sun is elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses)
- An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas)
- The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)
Question 9
1 / -0

The value of G does not depend on

A
nature of the interacting bodies

B
size of the interacting bodies

C
mass of the interacting bodies

D
all the above

Solution

G or the Universal gravitational constant remains constant throughout the universe and its value is $$6.67 \times 10^{-11} m^3kg^{-1} s^{-2}$$. It is not affected by any factor.
$$F = \dfrac{GmM}{r^2} \ \Rightarrow F \propto \dfrac{Mm}{r^2}$$
G is a proportionality constant and is independent of all variables.
Question 10
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.

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Gravitation Test - 19

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Gravitation Test - 19

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Question 1

$$960.40 cm s^{-2}$$

$$984.90 cm s^{-2}$$

$$982.45 cm s^{-2}$$

$$977.55 cm s^{-2}$$

Solution

Question 2

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 3

Decreases

Increases

Remains constant

Data insufficient

Solution

Question 4

Gravitational potential

Acceleration due to gravity.

Gravitational field

Universal gravitational constant.

Solution

Question 5

$$10$$ kg, zero

zero, zero

$$10$$ kg, $$10$$g

zero, $$10$$g

Solution

Question 6

Zero

Infinite

$$R$$ times the weight at the surface of the earth.

$${1/R^2}$$ times the weight at surface of the earth.

Solution

Question 7

At a height of $$100 m$$ above the earth's surface

At the earth's surface

At a depth of $$100 m$$ below the earth's surface

At the centre of the earth

Solution

Question 8

Einstein

Kepler

Copernicus

Tycho Brahe

Solution

Question 9

nature of the interacting bodies

size of the interacting bodies

mass of the interacting bodies

all the above

Solution

Question 10

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

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

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

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

Solution

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