Gravitation Test

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Gravitation Test - 57

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Question 1
1 / -0

Correct form of gravitational law is:

A
$$\vec{F}=-\dfrac{Gm_{1}m_{2}}{r^{2}}$$

B
$$\vec{F}=-\dfrac{Gm_{2}m_{1}}{r^{2}}$$

C
$$\vec{F}=-\dfrac{Gm_{1}m_{2}}{r^{2}}\hat{r}$$

D
$$\vec{F}=-\dfrac{Gm_{1}m_{2}}{r^{3}}\vec{r}$$

Solution

Newton's law of gravitational states that energy point mass attracts every other point mass by a force acting along the line intersecting the two points given by $$F=G\dfrac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } $$ where $$F$$ is gravitational force acting between two objects $${ m }_{ 1 }$$, $${ m }_{ 2 }$$ are masses of two objects $$r$$ is distance between center of their masses and $$G$$ is gravitational constant.
Question 2
1 / -0

The mass and radius of moon are $$7.4 \times {10^{22}}kg$$ and $$17.4 \times {10^{6}}kg$$ . Find the acceleration due to gravity at the moon .$$(G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}})$$

A
$$ 1.8 m/s^2 $$

B
$$ 1.6 m/s^2 $$

C
$$ 1.620 m/s^ 2$$

D
$$ 1.720 m/s^2 $$

Solution

On the surface of the moon, the distance to the center of mass is the same as the radius: $$r = 1.74 \times {10}^{6} m = 17,40,000$$ m. The acceleration due to gravity on the surface of the moon can be found using the formula:
$$g=\dfrac{GM}{{r}^{2}}$$
$$g=\dfrac{\left(6.673\times{10}^{-11}\right)\left(7.4\times{10}^{22}\right)}{17400000}$$
The acceleration due to gravity on the surface of the moon is $$1.620$$ m$${s}^{2}$$.
Question 3
1 / -0

A mass of 3 kg is dropped from a tower of 125m high. After 3s its K.E. will be :-

A
1300 J

B
1050 J

C
750J

D
550J

Solution

Checking if the ball has reached the ground before or at $$t=3s$$.
So,
Applying $$2^{nd}$$ equation of motion
$$h=(u)(t)+\dfrac{g}{2}t^2$$
$$h=0(3)+4.905(3)^2=44.145$$m
As $$h < 125$$m, the mass is under free fall
So, till $$t=3s$$ work done by gravity
$$=mgh=3\times 10\times (44.145)$$
$$=1300$$J
As only gravity is acting on the mass under free fall
By work-Energy theorem,
W gravity $$=\Delta$$ KE
$$KE_f-KE_i=1300$$J
As, $$4=0$$
$$KE_i=0$$
So, $$KE(3s)=1300$$J
Option A is correct
Question 4
1 / -0

$$g_e$$ and $$g_p$$ denote the acceleration due to gravity on the surface of earth and another planet whose mass and radius are twice that of the earth, then

A
$$g_p = g_e$$

B
$$g_p = \dfrac{g_e}{2}$$

C
$$g_p = 2g_e$$

D
$$g_p = 3g_e$$

Solution
Question 5
1 / -0

A mass of 1kg on earth weighs ( 1/6) kg on moon. The radius of the moon is $$1.738\times 10^{6}$$m. The mass of the moon is ( g$$_{e}=9.8m/s^{2},G=6.67\times 10^{-11}Mn^{2}/kg^{2}$$)

A
$$7.4\times 10^{22}$$kg

B
$$4.35\times 10^{16}$$kg

C
$$4.4\times 10^{23}$$kg

D
$$7.4\times 10^{20}$$kg

Solution

$$\begin{array}{l} \frac { { GMe } }{ { { { { { Re } } }^{ 2 } } } } =\frac { { 6GMm } }{ { R{ m^{ 2 } } } } =9.8 \\ \Rightarrow Mm=\dfrac { { 9.8\times R{ m^{ 2 } } } }{ { 6G } } \\ =\dfrac { { 9.8\times 1.738\times 1.738\times { { 10 }^{ 12 } } } }{ { 6\times 6.67\times { { 10 }^{ -11 } } } } \\ =0.7396\times { 10^{ 23 } } \\ Hence, \\ option\, \, A\, \, is\, \, correct\, answer. \end{array}$$
Question 6
1 / -0

The tilt of Earth is _____ degrees.

A
$$90$$

B
$$23.5$$

C
$$0$$

D
$$35.2$$

Solution
Question 7
1 / -0

The density of cooking oil is $$0.6g/c{m^3}$$ What is the mass of $$18c{m^3}$$ of the cooking oil?

A
$$0.5 g$$

B
$$10.8 g$$

C
$$18.9 g$$

D
$$20.0 g$$

Solution

Given,
$$V=18cm^3$$
$$\rho=0.6g/cm^3$$
Density, $$\rho=\dfrac{m}{V}$$
$$m=\rho V$$
$$m=0.6\times 18$$
$$m=10.8g$$
Question 8
1 / -0

Newton's law of gravitation:

A
is not applicable out side the solar system

B
is used to govern the motion of satellite and planets

C
control the rotational motion of satellites and planets

D
control the rotational motion of electrons in atoms.
Question 9
1 / -0

The mass of body is measured to be $$12$$ $$ kg$$ on earth. If it is taken to the moon, then its mass will be

A
$$2 kg$$

B
$$72 kg$$

C
$$32 kg$$

D
$$12 kg$$

Solution

The mass of the body does not change by taking the body from the earth to the moon. It is a property of the object and truly invariant. Since the mass of the body is $$12\ kg$$ on earth, its mass will be $$12\ kg$$ on the moon.
Question 10
1 / -0

The value of universal gravitational constant on earth for a particle of mass 5 kgs is

A
$$6.67\times { 10 }^{ -11 }$$

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

C
$$5\times 6.67\times { 10 }^{ -11 }$$

D
$$6.67\times { 10 }^{ -23 }$$

Solution

A universal gravitational constant is a constant number that does not depends on the masses of any planets or objects. And it is equal to $$G=6.67\times 10^{-11}$$ $$Nm^2kg^{-2}$$
Option A

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Gravitation Test - 57

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Gravitation Test - 57

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Question 1

$$\vec{F}=-\dfrac{Gm_{1}m_{2}}{r^{2}}$$

$$\vec{F}=-\dfrac{Gm_{2}m_{1}}{r^{2}}$$

$$\vec{F}=-\dfrac{Gm_{1}m_{2}}{r^{2}}\hat{r}$$

$$\vec{F}=-\dfrac{Gm_{1}m_{2}}{r^{3}}\vec{r}$$

Solution

Question 2

$$ 1.8 m/s^2 $$

$$ 1.6 m/s^2 $$

$$ 1.620 m/s^ 2$$

$$ 1.720 m/s^2 $$

Solution

Question 3

1300 J

1050 J

750J

550J

Solution

Question 4

$$g_p = g_e$$

$$g_p = \dfrac{g_e}{2}$$

$$g_p = 2g_e$$

$$g_p = 3g_e$$

Solution

Question 5

$$7.4\times 10^{22}$$kg

$$4.35\times 10^{16}$$kg

$$4.4\times 10^{23}$$kg

$$7.4\times 10^{20}$$kg

Solution

Question 6

$$90$$

$$23.5$$

$$0$$

$$35.2$$

Solution

Question 7

$$0.5 g$$

$$10.8 g$$

$$18.9 g$$

$$20.0 g$$

Solution

Question 8

is not applicable out side the solar system

is used to govern the motion of satellite and planets

control the rotational motion of satellites and planets

control the rotational motion of electrons in atoms.

Question 9

$$2 kg$$

$$72 kg$$

$$32 kg$$

$$12 kg$$

Solution

Question 10

$$6.67\times { 10 }^{ -11 }$$

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

$$5\times 6.67\times { 10 }^{ -11 }$$

$$6.67\times { 10 }^{ -23 }$$

Solution

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