Gravitation Test

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Gravitation Test - 43

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Question 1
1 / -0

A body weighs $$72\ N$$ on the surface of earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface?

A
$$72\ N$$

B
$$28\ N$$

C
$$16\ N$$

D
$$32\ N$$

Solution

Hint:- Use the formula of gravitational acceleration to check weight of the body on certain height.

Step-1 Note the given values
Let $$R_e$$ be the radius of earth
Given
height, $$h = \frac{R_e}2$$
Acceleration due to gravity at the surface of earth=g

Step -2 Calculate gravitational force on height h
Since h is comparable to $$R_e$$
So, $$g_{h}=\dfrac{GM}{(R_e+h) ^2}$$
Putting $$h=R_e/2$$,we get
$$\Rightarrow g_h=\dfrac{4}{9}\times \dfrac{GM}{R_e^2}=\dfrac{4g}{9}$$

Therefore weight at a height equal to half of the earth's radius $$ mg_h=\dfrac{4}{9}\times mg= \dfrac{4}9 \times {72 N}=32 N$$

Hence correct answer is option $$D$$
Question 2
1 / -0

At what depth from the surface of the earth (in terms of the radius of earth) the acceleration due to gravity will be: $$\dfrac {2g}{5}$$?

A
$$2/5R$$

B
$$3/5R$$

C
$$4/25R$$

D
$$9/25R$$

Solution

Gravity due to the earth is at depth $$g$$'=$$g(1-\dfrac{d}{R})$$,
Where $$d$$= depth from the surface of the earth, $$R$$= Radius of the earth,
Given that $$g$$'= $$\dfrac{2g}{5}$$
$$\Rightarrow \dfrac{2g}{5}=g(1-\dfrac{d}{R})$$,
Solving this $$\Rightarrow d=\dfrac {3}{5}R$$
Question 3
1 / -0

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

Acceleration on the earth's surface is $$g=\dfrac{GM}{R^2}$$ and the acceleration at height h will be $$g_h=\dfrac{GM}{(R+h)^2}$$
So, $$g_h=\dfrac{gR^2}{(R+h)^2}=\dfrac{g}{(1+h/R)^2}=\dfrac{980}{(1+64/6400)^2}=960.69 cm s^{-2}$$
Question 4
1 / -0

At a place, the value of 'g' is less by $$1$$% than its value on the surface of the Earth (Radius of Earth, $$R = 6400\ km$$). The place is:

A
$$64\ km$$ below the surface of the earth

B
$$64\ km$$ above the surface of the earth

C
$$30\ km$$ above the surface of the earth

D
$$32\ km$$ below the surface of the earth

Solution

$${ g }_{ d }=g\left( 1-\cfrac { d }{ R } \right) $$
Now, $${ g }_{ d }=g-\cfrac { g }{ 100 } =\cfrac { 99 }{ 100 } g\\ \therefore \cfrac { 99 }{ 100 } g=g\left( 1-\cfrac { d }{ 6400\times { 10 }^{ 3 } } \right) \\ or,\quad \cfrac { 99 }{ 100 } =1-\cfrac { d }{ 64\times { 10 }^{ 5 } } \\ \Rightarrow \cfrac { 1 }{ 100 } =\cfrac { d }{ 64\times { 10 }^{ 5 } } \\ \Rightarrow d=\cfrac { 64\times { 10 }^{ 5 } }{ 100 } =64\times { 10 }^{ 3 }m=64km$$.
Question 5
1 / -0

A spring balance is graduated when it is at sea level. If a body is weighed at consecutively increasing heights from earth's surface, the weight indicated by the balance:

A
will go on increasing continuously

B
will go on decreasing continuously

C
will remain same

D
will first increase and then decrease

Solution

The acceleration due to gravity will be decreasing with increasing of height from the earth's surface. Since the weight is the products of mass and acceleration due to gravity, so the weight will decrease with increasing height from earth's surface.
Question 6
1 / -0

At what height from the surface of the earth (in terms of the radius of earth) the acceleration due to gravity will be $$\dfrac {g}{100}?$$

A
$$10R$$

B
$$9R$$

C
$$100R$$

D
$$R/100$$

Solution

We have, $${ g }_{ h }=\dfrac { g }{ { \left( 1+\frac { h }{ R } \right) }^{ 2 } } $$
$${ g }_{ h }=\dfrac { g }{ 100 } $$

$$\therefore \quad \dfrac { g }{ 100 } =\dfrac { g }{ { \left( 1+\cfrac { h }{ R } \right) }^{ 2 } } \Rightarrow \dfrac { 1 }{ 100 } =\dfrac { 1 }{ { \left( 1+\cfrac { h }{ R } \right) }^{ 2 } }$$

$$100={ \left( 1+\cfrac { h }{ R } \right) }^{ 2 }\Rightarrow 10=1+\cfrac { h }{ R }$$

or, $$h=9R$$
Question 7
1 / -0

At a point very near to the earth's surface, the acceleration due to gravity is g. What will be the acceleration due to gravity at the same point if the earth suddenly shrinks to half its radius without any change in its mass?

A
$$2\ g$$

B
$$4\ g$$

C
$$g$$

D
$$3\ g$$

Solution

Acceleration due to gravity on the earth's surface is $$g=\dfrac{GM}{R^2}$$
If earth suddenly shrinks to half its radius, the acceleration due to gravity becomes, $$g'=\dfrac{GM}{(R/2)^2}=4GM/R^2=4g$$
Question 8
1 / -0

As the altitude increases, the acceleration due to gravity:

A
Remains constant

B
Becomes zero

C
Decreases

D
Increases

Solution

$$g=\frac{Gm}{R^{2}}$$
Let acceleration due to gravity at a height $$h$$ =$$g_{1}$$
$$g_{1}=\frac{Gm}{(R+h)^{2}}$$
$$\frac{g_{1}}{g}=(1+\frac{h}{R})^{-2}$$
Expanding by binomial theorem and neglecting higher powers
$$\frac{g_{1}}{g}=(1-2\frac{h}{R})$$
$$g_{1}=g(1-2\frac{h}{R})$$
So as altitude increases acceleration decreases
Question 9
1 / -0

If earth's radius were to hypothetically shrink by $$1\%$$, the value of $$G$$ would:

A
Shrink by $$1\%$$

B
Expand by $$1\%$$

C
Remain the same

D
Shrink by $$0.01\%$$

Solution

According to the universal law of gravitation, every object in the universe attracts every other objects with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centers of two objects.

$$F=G\dfrac{m_1m_2}{d^2}$$

where $$G$$ is the constant of proportionality and is called the universal gravitation constant. It does not depend on the radius.
So, if the earth's radius shrinks by $$1\%$$, the value of $$G$$ would remain the same.
Question 10
1 / -0

The value of universal gravitational constant $$G$$ is-

A
$$6.67\times 10^{-11} \dfrac{Nm^2}{kg}$$

B
$$6.67\times 10^{-11} \dfrac{Nm^2}{kg^2}$$

C
$$66.7\times 10^{-11} \dfrac{Nm^2}{kg^2}$$

D
$$66.7\times 10^{-11} \dfrac{Nm^2}{kg}$$

Solution

Gravitational force between two objects of mass $$m_1$$ and $$m_2$$ separated by a distance $$r$$, $$F = \dfrac{Gm_1 m_2}{r^2}$$
$$\implies$$ $$G = \dfrac{F r^2 }{m_1 m_2}$$
Value of universal gravitational constant $$G$$ is $$6.67\times 10^{-11}$$ $$\dfrac{Nm^2}{kg^2}$$.

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Gravitation Test - 43

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Gravitation Test - 43

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Question 1

$$72\ N$$

$$28\ N$$

$$16\ N$$

$$32\ N$$

Solution

Question 2

$$2/5R$$

$$3/5R$$

$$4/25R$$

$$9/25R$$

Solution

Question 3

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

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

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

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

Solution

Question 4

$$64\ km$$ below the surface of the earth

$$64\ km$$ above the surface of the earth

$$30\ km$$ above the surface of the earth

$$32\ km$$ below the surface of the earth

Solution

Question 5

will go on increasing continuously

will go on decreasing continuously

will remain same

will first increase and then decrease

Solution

Question 6

$$10R$$

$$9R$$

$$100R$$

$$R/100$$

Solution

Question 7

$$2\ g$$

$$4\ g$$

$$g$$

$$3\ g$$

Solution

Question 8

Remains constant

Becomes zero

Decreases

Increases

Solution

Question 9

Shrink by $$1\%$$

Expand by $$1\%$$

Remain the same

Shrink by $$0.01\%$$

Solution

Question 10

$$6.67\times 10^{-11} \dfrac{Nm^2}{kg}$$

$$6.67\times 10^{-11} \dfrac{Nm^2}{kg^2}$$

$$66.7\times 10^{-11} \dfrac{Nm^2}{kg^2}$$

$$66.7\times 10^{-11} \dfrac{Nm^2}{kg}$$

Solution

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