Gravitation Test

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Gravitation Test - 28

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Question 1
1 / -0

The Earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the Earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is:

A
$$\sqrt{2}$$

B
$$\dfrac{1} {\sqrt{2}}$$

C
$$\dfrac{1} {3}$$

D
$$\dfrac{1} {2}$$

Solution

According to question and by using COME
$$-\dfrac{GMm} {R + R} +\dfrac{1} {2}m\left (fv \right )^{2}$$ = $$0 + 0$$
$$\Rightarrow fv$$ = $$\sqrt{\dfrac{GM} {R}}$$ but $$v = \sqrt{\dfrac{2GM} {R}}$$
Therefore $$f\sqrt{\dfrac{2GM} {R}}$$ = $$\sqrt{\dfrac{GM} {R}} \Rightarrow f$$ = $$\sqrt{\dfrac{1} {2}}$$
Question 2
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

Let acceleration due to gravity at a place be $$g_d$$. If g is acceleration due to gravity at the surface then,
$$g_d=g(1-\frac {d}{R})$$
$$0.99g=g(1-\frac {d}{R})$$
$$\frac {d}{R}=(0.01)$$
$$d=0.01\times 6400=64 km$$ below the surface of earth
Question 3
1 / -0

Depth from the surface of the earth at which is acceleration due to gravity is 25% of acceleration due to gravity at the surface

A
1200 km

B
4000 km

C
3600 km

D
4800 km

Solution

g at the surface is given as $$g_s=\dfrac{4}{3}\pi GR\rho$$
and at a depth d it is given as $$g_d=\dfrac{4}{3}\pi G(R-d)\rho$$
Thus $$\dfrac{g_d}{g_s}=\dfrac{R-d}{R}$$
Thus for the depth where acceleration is 25% of the surface gravity we get $$g_d$$ as $$\dfrac{g}{4}$$
or
$$\dfrac{g}{4}=g\left ( 1-\dfrac{d}{R} \right )\Rightarrow d=\dfrac{3}{4}\times R = 4800 km$$ (R is 6371 km)
Question 4
1 / -0

The variation of g with height or depth (r) is shown correctly by the graph in the figure (where R = radius of the earth),

A

B

C

D

Solution

$$g_h$$ at height h is given by $$g_h = \dfrac{GM}{(R+h)^2} = GM(R+h)^{-2}$$ which denotes a parabola
$$g_x $$ at depth x is given by
$$g_x = g \left [ \dfrac{R-x}{R} \right ]$$
Which denotes is straight line. So, option A is correct.
Question 5
1 / -0

A man weighs $$60 kg$$ at earth's surface. At what height above the earth's surface his weight becomes $$30 kg$$? (radius of earth $$=6400 km$$).

A
$$1624 km$$

B
$$2424 km$$

C
$$2624 km$$

D
$$2826 km$$

Solution

Above the earth's surface, $$g_1 = \dfrac{g}{(1+\dfrac{h}{R})^2}$$
But changed weight = half of its weight on the earth.
$$\Rightarrow g_1 = \dfrac{g}{2}$$
$$\therefore h = 2624 km.$$
Question 6
1 / -0

At what altitude will the acceleration due to gravity be $$25\%$$ of that at the earth's surface $$($$given radius of earth is $$R)$$?

A
$$\dfrac{R}{4}$$

B
$$R$$

C
$$\dfrac{3R}{8}$$

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

Solution

Hint : Use the formula $$g'=\dfrac{GM}{(R+h)^{2}}$$
$$\textbf{Step1: Expression of acceleration due to gravity on earth's surface}$$
Force on the body placed on Earth's surface is
$$F = \dfrac{GMm}{R^2}$$
But, $$F = mg$$ hence,
$$g = \dfrac{GMm}{R^2}$$
where, variables have their usual meanings.
$$\textbf{Step2: Calculation of altitude}$$
Now, force on the body at geo-potential height say $$h$$ (altitude) where the acceleration due to gravity is $$25\%$$ of that at the earth's surface i.e.
$$\dfrac{25 g}{100} = \dfrac{g}{4}$$
use the formula $$g'=\dfrac{GM}{(R+h)^{2}}$$
Hence, we can write
$$\dfrac{g}{4} = \dfrac{GM}{(R+h)^2}$$

$$\dfrac{g}{4} = \dfrac{gR^2}{(R+h)^2}$$

$$(R+h)^2 = 4R^2$$
Taking roots for both sides we get
$$R+h = 2R$$
$$h = R$$
Question 7
1 / -0

When a satellite moves around the earth in a certain orbit, the quantity which remains constant is:

A
angular velocity

B
kinetic energy

C
areal velocity

D
potential energy

Solution

Hint: Use the Keplers law
Explanation:

According to Kepler's rules of planetary motion, when a satellite/planet moves around the Earth/Sun in a certain orbit, aerial velocity remains constant.
Correct option is (C) areal velocity
Question 8
1 / -0

Let $$\omega$$ be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth a below earth's surface at a pole $$(d << R)$$ The value of $$d$$ is

A
$$\displaystyle \dfrac{\omega^2 R^2}{g}$$

B
$$\displaystyle \dfrac{\omega^2 R^2}{2g}$$

C
$$\displaystyle \dfrac{2\omega^2 R^2}{g}$$

D
$$\displaystyle \dfrac{\sqrt{Rg}}{g}$$

Solution

Let $$g_e$$ be the acceleration due to gravity on surface of earth, $$g'$$ be the acceleration due to gravity at depth $$d$$ below the surface of earth and is given by
$$g^{'} = g_e \left(1 - \dfrac{d}{R} \right)$$ ............................(1)
The acceleration due to gravity of an object on latitude(below the surface of earth) is
$$g^{'} = g_e - \omega^2 R \cos \phi$$
where, $$\phi$$ is the solid angle at the center of earth sustained by the object.
At the equator, $$\phi = 0^\circ$$
$$\therefore \cos \phi = 1$$
Hence,
$$g' = g_e - \omega^2 R$$ ...............(2)
From (1) and (2), we can write
$$ g_e \left(1 - \dfrac{d}{R} \right) = g_e - \omega^2 R$$

$$ \Rightarrow\dfrac{dg_e}{R} = \omega^2 R$$

$$ \Rightarrow d = \dfrac{\omega^2 R}{g_e}$$
Question 9
1 / -0

The condition for a uniform spherical mass $$M$$ of radius $$r$$ to be a black hole is $$(G=$$ gravitational constant, $$g=$$ acceleration due to gravity $$)$$.

A
$$\sqrt{\dfrac {2GM}{r}} \leq c$$

B
$$\sqrt{\dfrac {gM}{r}} \leq c$$

C
$$\sqrt{\dfrac {2GM}{r}} \geq c$$

D
$$\sqrt{\dfrac {gM}{r}} \geq c$$

Solution

For black hole, light even can not escape from the surface of the star.
Since $$v_{esc} = \sqrt{ \dfrac{2GM}{r}}$$, putting $$v_{esc} =c$$ where c is the velocity of light $$c = \sqrt{ \dfrac{2GM}{r}}$$,
Hence, $$\sqrt{ \dfrac{2GM}{r}} \ge c$$
Question 10
1 / -0

A spring balance is graduated on sea level. If a body is weighted 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 increases and then decreases

Solution

A spring balance is graduated on sea level. If a body is weighed with at consecutively increasing heights from earth's surface, the weight indicated by the balance will go on decreasing continuously. This is because value of g decreases with height.

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Re-Attempt Weekly Quiz Competition

Gravitation Test - 28

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Gravitation Test - 28

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SHARING IS CARING

Question 1

$$\sqrt{2}$$

$$\dfrac{1} {\sqrt{2}}$$

$$\dfrac{1} {3}$$

$$\dfrac{1} {2}$$

Solution

Question 2

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 3

1200 km

4000 km

3600 km

4800 km

Solution

Question 4

Solution

Question 5

$$1624 km$$

$$2424 km$$

$$2624 km$$

$$2826 km$$

Solution

Question 6

$$\dfrac{R}{4}$$

$$R$$

$$\dfrac{3R}{8}$$

$$\dfrac{R}{2}$$

Solution

Question 7

angular velocity

kinetic energy

areal velocity

potential energy

Solution

Question 8

$$\displaystyle \dfrac{\omega^2 R^2}{g}$$

$$\displaystyle \dfrac{\omega^2 R^2}{2g}$$

$$\displaystyle \dfrac{2\omega^2 R^2}{g}$$

$$\displaystyle \dfrac{\sqrt{Rg}}{g}$$

Solution

Question 9

$$\sqrt{\dfrac {2GM}{r}} \leq c$$

$$\sqrt{\dfrac {gM}{r}} \leq c$$

$$\sqrt{\dfrac {2GM}{r}} \geq c$$

$$\sqrt{\dfrac {gM}{r}} \geq c$$

Solution

Question 10

Will go on increasing continuously

Will go on decreasing continuously

Will remain same

Will first increases and then decreases

Solution

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