Gravitation Test

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Gravitation Test - 61

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Question 1
1 / -0

Calculate the ratio of weights of a body when it is (radius of the Earth is 6400km.)
1. 200 km above the surface of the Earth and
2. 200 km below the surface of the Earth.

A
0.97

B
1.2

C
2

D
0

Solution

The acceleration Due to Gravity variation on :-
at an height 'h' $$ g_{1}^{1} = g(\dfrac{R}{R+h})^{2} $$ at adept 'd'$$ g_{2}^{1} = g(1-\dfrac{d}{R}) $$
where g : at surface and R : Radius of Earth
an weight of body, w = mg', mg',$$g' $$ accln. due to gravity at that point.
Ratio of weights :-
$$ \dfrac{w_{1}}{w_{2}} = (\dfrac{R}{R+h})^{2}\times (\dfrac{R}{R-d}) $$
From question, R = 6400km, d = h = 200km so,
$$ \dfrac{w_{1}}{w_{2}} = \dfrac{(64000)^{3}}{(6400+200)^{2}(6400-200)} = \dfrac{(6400)^{3}}{(6600)^{2}(6200)} \simeq 0.97 $$ (Ans)
Question 2
1 / -0

If the Earth were to shrink so that the density becomes eight times the present density without any change in its mass, then 'the weight of a person on the surface of Earth

A
becomes 8 times

B
becomes 4 times

C
becomes 2 times

D
remains same as before

Solution
Question 3
1 / -0

The height at which the acceleration due to gravity becomes $$g/9$$ (where $$g=the$$ acceleration due to gravity on the surface of the earth) in terms of $$R$$, the radius of the earth, is

A
$$R/2$$

B
$$\sqrt { 2 } R$$

C
$$2R$$

D
$$\dfrac { R }{ \sqrt { 2 } }$$

Solution
Question 4
1 / -0

If the change in the value of g at a height h above the surface of the earth is same as at depth below the surface of the earth, then (h >>R)

A
$$x = h$$

B
$$x = 2h$$

C
$$x = h/2$$

D
$$x = h^2$$

Solution
Question 5
1 / -0

A hole is drilled from the surface of earth to its centre. A particle is dropped from rest at the surface of earth. The speed of the particle when it reaches the centre of the earth in terms of its escape velocity on the surface of earth $$v_{e}$$ is :

A
$$\dfrac {v_{e}}{2}$$

B
$$v_{e}$$

C
$$\sqrt {2}v_{e}$$

D
$$\dfrac {v_{e}}{\sqrt {2}}$$

Solution

Lets $$m=$$ mass of the particle

$$M=$$ mass of the earth

$$R=$$ radius of the earth

$$v=$$ speed of particle when it reaches the center of the earth

applying conservation of total energy,

$$(K.E)_1+(P.E)_1=(K.E)_2+(P.E)_2$$

$$0+(-\dfrac{GMm}{R})=\dfrac{1}{2}mv^2+(-\dfrac{3GMm}{2R})$$

$$-\dfrac{GM}{R}+\dfrac{3GM}{2R}=\dfrac{1}{2}v^2$$

$$v^2=\dfrac{GM}{R}$$

$$v=\sqrt{\dfrac{GM}{R}}$$. . . . (1)

We know that the escape velocity of the particle from the earth is given by,

$$v_e=\sqrt{\dfrac{2GM}{R}}$$. . . . .(2)

$$v_e=\sqrt{2}\sqrt{\dfrac{GM}{R}}$$

$$v_e=\sqrt{2} v$$ (from equation 1)

$$v=\dfrac{v_e}{\sqrt{2}}$$

The correct option is D.
Question 6
1 / -0

A tunnel is dug along a diameter of the planet. A particle is dropped from the surface of the planet and reaches the center of the planet with speed $$V$$. The escape velocity from surface of planet is

A
$$\sqrt {2}V$$

B
$$2\ V$$

C
$$\sqrt {5}V$$

D
$$\dfrac {V}{2}$$

Solution
Question 7
1 / -0

The weight of the body at earth's surface is W. At a depth halfway to the center of the earth, it will be? (assuming uniform density in the earth)

A
$$W $$

B
$$\dfrac{W}{2} $$

C
$$\dfrac {W}{ 3 } $$

D
$$\dfrac { W} {8 } $$

Solution

$${ G }^{ 1 }=g\left( 1-\dfrac { d }{ R } \right) \quad \longrightarrow \left( 1 \right) $$
Now $$d$$ that is depth is $$\dfrac { R }{ 2 } $$ substituting $$d=\dfrac { R }{ 2 } $$ in eq $$(1)$$
$${ G }^{ 1 }=g\left( 1-\dfrac { R }{ 2R } \right) =g\left( 1-\dfrac { 1 }{ 2 } \right) =\dfrac { g }{ 2 } $$
So, $${ g }^{ 1 }=\dfrac { g }{ 2 } $$
Multiplying both sides by $$M$$
$${ mg }^{ 1 }=\dfrac { mg }{ 2 } $$
Weight is nothing but mass $$\times$$ gravity
Weight at half way depth $$=\dfrac { W }{ 2 } $$
Question 8
1 / -0

The magnitude of gravitational potential energy of a body at a distance 'r' from the center of the earth is V. Its weight at a distance '2r' from the center of the earth is

A
$$\frac { V }{ r } $$

B
$$\frac { V }{ 4r } $$

C
$$\frac { V }{ 2r } $$

D
$$\frac { 4V }{ r } $$

Solution
Question 9
1 / -0

If the acceleration due to gravity at a height $$h$$ from the surface of the earth is $$96\%$$ less than its value on the surface, then $$h$$ is (where $$R$$ is radius of the earth).

A
$$.5*R$$

B
$$2R$$

C
$$3R$$

D
$$4R$$

Solution

the gravitational acceleration at a height $$h$$ above earth's surface is calculated as:

$$g_h = \dfrac{GM}{(R+h)^2}$$
where,
$$M$$ is the mass of the earth
$$R$$ is the radius of the earth
$$G$$ is the universal Gravitational constant

at the surface, $$g = \dfrac{GM}{R^2}$$

Rearranging $$g_h$$ using the above equations as follows

$$\therefore g_h = \dfrac{GM}{R^2}\times\dfrac{R^2}{(R+h)^2} = g \left( \dfrac{R}{R+h} \right)^2$$

In the given question, $$g_h$$ is $$96\%$$ less than its value on the surface $$\implies 100-96 = 4\%$$ of its value on surface.

$$\therefore g_h = 0.04g $$

$$\therefore g\left( \dfrac{R}{R+h} \right)^2 = 0.04g$$

$$\therefore \dfrac{R}{R+h} = \sqrt{0.04} = 0.2 = \dfrac{1}{5}$$

$$\therefore 5R = R+h$$

$$\therefore h = 5R-R = 4R$$

The gravitational acceleration is $$96\%$$ less than its value on the surface of the earth at a height of $$4R$$, where $$R$$ is the radius of the earth.
Question 10
1 / -0

Ratio of time period of satellites orbiting around earth at a distance of about 9r and 4r from earth surface is : ( where r is the radius of earth)

A
4:9

B
9:4

C
27:8

D
16:27

Solution