Gravitation Test

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Gravitation Test - 51

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Question 1
1 / -0

A man weighs $$80$$kg on the surface of earth of radius R. At what height above the surface of earth his weight will be $$40$$kg?

A
$$2637.6km$$

B
$$3.675km$$

C
$$8765km$$

D
$$(\sqrt{2}+1)R$$

Solution

On surface of earth $$g$$ is $$9.81m/{ s }^{ 2 }$$. The weight is given by $$80kg$$.
The gravitational force $$\left( g \right) =\dfrac { G\times { m }_{ 1 }\times { m }_{ 2 } }{ { d }^{ 2 } } $$ where $$G$$ is gravitational constant distance between two objects that weight $${ M }_{ 1 }$$ and $${ M }_{ 2 }$$. Considering weight on earth as $${ M }_{ 1 }$$.
$$\left( g \right) =\dfrac { h\times { m }_{ 1 } }{ { d }^{ 2 } } $$
Radius of earth is $$6371km$$ which will be halved $$2637.6km$$.
Therefore for a person going on earth to weight $$40kg$$. He has to move $$2637.6km$$ above earth's surface.
Question 2
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A planet is revolving in an elliptical orbit around the Sun. Its closest distance from the Sun is $$r_{min}$$ and the farthest distance is $$r_{max}$$. If the velocity of the planet at the distance of the closest approach is $$v_1$$ and that at the farthest distance from the Sun is $$v_2$$, then $$v_1/v_2$$.

A
$$\dfrac{r_{max}}{r_{min}}$$

B
$$\dfrac{r_{min}}{r_{max}}$$

C
$$\dfrac{r_{min}+r_{max}}{r_{max}-r_{min}}$$

D
None

Solution

Since angular momentum is conserved therefore
$${ I }_{ 1 }{ w }_{ 1 }={ I }_{ 2 }{ w }_{ 2 }$$
or $${ MR }^{ 2 }{ w }_{ 1 }={ MR }^{ 2 }{ w }_{ 2 }$$
But $${ w }_{ 1 }=\dfrac { { V }_{ 1 } }{ { r }_{ min } } $$
$${ w }_{ 2 }=\dfrac { { V }_{ 2 } }{ { r }_{ max } } $$
$$\therefore$$ $${ r }_{ min }^{ 2 }\times \dfrac { { V }_{ 1 } }{ { r }_{ min } } ={ r }_{ max }^{ 2 }\times \dfrac { { V }_{ 2 } }{ { r }_{ max } } $$
$${ V }_{ 1 }{ r }_{ min }={ V }_{ 2 }{ r }_{ max }$$
$${ V }_{ 2 }=\dfrac { { V }_{ 1 }{ r }_{ min } }{ { r }_{ max } } $$
$$\dfrac { { V }_{ 2 } }{ { V }_{ 1 } } =\dfrac { { r }_{ min } }{ { r }_{ max } } $$
and $$\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =\dfrac { { r }_{ max } }{ { r }_{ min } } $$
$$\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =\dfrac { { r }_{ max } }{ { r }_{ min } } $$
Question 3
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Directions For Questions

A planet is revolving around the sun in an elliptical orbit. Its $$KE$$ different for different points and the total energy is negative. Its linear momentum is not conserved. The eccentricity decides the shape of the orbit.

...view full instructions

Net torque on the planet is

A
Constant at all points

B
Zero at all point

C
Maximum at $$A$$

D
Minimum at $$D$$

Solution

Only force acting as gravitational force which is passing through the centres of both.
So, Net torque $$=0$$.
Question 4
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Two identical particles of mass $$m$$ are placed at a distance $$r$$ from each other. If their separation is doubled, then the effect on gravitational constant will be

A
Gravitational constant remains same

B
Gravitational constant becomes quadrupled

C
Gravitational constant becomes 1/4th of the actual one

D
Gravitational constant becomes doubled

Solution

According to the Newton's law of gravitation, gravitational force between two objects is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

$$ F \propto \dfrac{m_1m_2}{r^2}$$

Hence, $$ F = \dfrac{Gm_1m_2}{r^2}$$, where $$G$$ is the proportionality constant , known as the Universal Gravitational constant.

Gravitational constant does not depend on masses or the distance between them

Changing the distance between the masses, changes the force between them.

Hence, The correct option is (A)
Question 5
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The value of universal gravitational constant on earth for a particle of mass 5 kg 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

Gravitational constant depends only on the mass of the earth and not on the mass of the particle on the earth. Hence its values remains constant = $$6.67 \times 10^{-11}$$

The correct option is (a)
Question 6
1 / -0

Directions For Questions

A planet is revolving around the sun in an elliptical orbit. Its $$KE$$ different for different points and the total energy is negative. Its linear momentum is not conserved. The eccentricity decides the shape of the orbit.

...view full instructions

Velocity of the planet is minimum at

A
$$C$$

B
$$D$$

C
$$A$$

D
$$B$$

Solution

As, here gravitational force is the only acting force which always passes through centre i.e $$C$$ about centre of sun $$=0$$.
So, angular momentum about centre of sun $$=$$ conserved
i.e $$MVr=$$ constant
$$\Rightarrow$$ Velocity will be minimum when $$'r'$$ will be maximum
$$'r'$$ is maximum at $$D$$.
$$\Rightarrow$$ $$'V'$$ will be minimum at $$D$$.
Question 7
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If the gravitational constant is expressed in terms of $$dynes\ m^{-2} kg^{2}$$, how will the value of G change:

A
$$6.67 \times 10^{-11} dynes \ kg^2 m^{-2}$$

B
$$6.67 \times 10^{-8} dynes \ kg^2 m^{-2}$$

C
$$6.67 \times 10^{-6} dynes \ kg^2 m^{-2}$$

D
$$6.67 \times 10^{-3} dynes \ kg^2 m^{-2}$$

Solution

'G' is expressed as
$$G=6.67\times { 10 }^{ -11 }N.{ m }^{ -2 }.{ kg }^{ +2 }\\ \therefore 1N={ 10 }^{ 5 }dyne\\ 1m={ 10 }^{ 2 }cm\\ 1kg={ 10 }^{ 3 }g.$$
$$G=6.67\times { 10 }^{ -11 }\times { 10 }^{ 5 }dynes.kg.{ m }^{ -2 }$$ [As,$$1N={ 10 }^{ 5 }$$dyne]
Question 8
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A spring balance is graduated on sea level. If a body of mass 5 kg is weighed with this balance and the balance is taken to a height of 360kms and the object is weighed again, then the weight of the object

A
Will be more than 5 kg

B
Will be less than 5 kg

C
Will remain same

D
Will first increase and then decrease

Solution

The apparent acceleration due to gravity is given by $$g'=g(1-2h/R)$$. As h increases, g' decreases. Thus weigh will go on decreasing continously

The option (b) is correct
Question 9
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At what height over the earths pole, the free fall acceleration decreases by one percent (assume the radius of earth to be 6400 km)

A
32 km

B
80 km

C
1.293 km

D
64 km

Solution

Acceleration due to gravity at poles = $$9.8 m/s^2$$

The apparent acceleration due to gravity at a height h is given by $$g'=g(1-2h/R)$$.

Rearranging the terms, we get, $$(g'-g)/g=2h/R \implies h= (R/2) (\delta g/g) = (6400/2) \times (1/100) = 32 km$$.

The correct option is (a)
Question 10
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Which of the following laws are conserved, if the areal acceleration is zero

A
Law of conservation of angular velocity

B
Law of conservation of angular momentum

C
Law of conservation of angular acceleration

D
Law of conservation of angular displacement

Solution

If angular momentum is conserved, areal velocity is constant or areal acceleration is zero

Thus angular momentum is conserved is the right option

The correct option is (b)

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Gravitation Test - 51

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Gravitation Test - 51

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Question 1

$$2637.6km$$

$$3.675km$$

$$8765km$$

$$(\sqrt{2}+1)R$$

Solution

Question 2

$$\dfrac{r_{max}}{r_{min}}$$

$$\dfrac{r_{min}}{r_{max}}$$

$$\dfrac{r_{min}+r_{max}}{r_{max}-r_{min}}$$

None

Solution

Question 3

Constant at all points

Zero at all point

Maximum at $$A$$

Minimum at $$D$$

Solution

Question 4

Gravitational constant remains same

Gravitational constant becomes quadrupled

Gravitational constant becomes 1/4th of the actual one

Gravitational constant becomes doubled

Solution

Question 5

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

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

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

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

Solution

Question 6

$$C$$

$$D$$

$$A$$

$$B$$

Solution

Question 7

$$6.67 \times 10^{-11} dynes \ kg^2 m^{-2}$$

$$6.67 \times 10^{-8} dynes \ kg^2 m^{-2}$$

$$6.67 \times 10^{-6} dynes \ kg^2 m^{-2}$$

$$6.67 \times 10^{-3} dynes \ kg^2 m^{-2}$$

Solution

Question 8

Will be more than 5 kg

Will be less than 5 kg

Will remain same

Will first increase and then decrease

Solution

Question 9

32 km

80 km

1.293 km

64 km

Solution

Question 10

Law of conservation of angular velocity

Law of conservation of angular momentum

Law of conservation of angular acceleration

Law of conservation of angular displacement

Solution

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