Gravitation Test

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Gravitation Test - 48

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Question 1
1 / -0

On a planet where $${ g }_{ planet }=0.2{ g }_{ earth }$$. What will be the difference in the height of column filled with mercury in a closed end manometer when the gas is filled withe pressure of $$2atm$$ on earth (Assuming:outside pressure to be $$1atm$$ on both planet; Volume of gas remain constant)

A
$$30.4cm$$

B
$$760cm$$

C
$$380cm$$

D
$$152cm$$

Solution

$$P=egh$$
$$2atm=760\times2$$ mm of hg
$$\Rightarrow\cfrac{P_1}{P_2}=\cfrac{e_1g_1h_1}{e_2g_2h_2}\\ \because P_1=P_2\& e_1=e_2\Rightarrow\cfrac{h_2}{h_1}=\cfrac{g_1}{g_2}\\ \quad\cfrac{h_2}{760\times2}=\cfrac{g}{0.2g}\\ \quad\Rightarrow h_2=\cfrac{760\times2}{0.2}=7600mm\\ \quad\quad=760cm$$
Question 2
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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
$$R/ 4$$

B
$$R$$

C
$$3R/ 8$$

D
$$R/ 2$$

Solution
Question 3
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A particle of mass M is situated at the centre of spherical shell of same mass and radius a. The gravitational potential at a point situated at a/2 distance from the centre will be

A
$$-\dfrac { 3\ GM }{ a } $$

B
$$-\dfrac { 2\ GM }{ a } $$

C
$$-\dfrac { GM }{ a } $$

D
$$-\dfrac { 4GM }{ a } $$

Solution

Let point $$A$$ is situated at $$a/2$$ distance from the center.
Gravitational Potential at point $$A$$ due to spherical shell, $$V_1=-\dfrac{GM}{a}$$,
Gravitational Potential at point $$A$$ due to mass $$M$$ at point $$O$$ , $$V_2=-\dfrac{GM}{(a/2)}=-\dfrac{2GM}{a}$$,
$$\therefore $$ Total Gravitational Potential $$=V_1+V_2= -\dfrac{GM}{a} -\dfrac{2GM}{a}=-\dfrac{3GM}{a}$$
Question 4
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Suppose(God forbid) due to some reason, the earth expands to make its volume eight-fold. What you expect your weight to be?

A
Two-fold

B
One-half

C
One-fourth

D
Unaffected

Solution

we know , gravity is
g= $$\dfrac{GM}{r^2}$$
the radius of new one will become r'=2r
then, g'=$$\dfrac{GM}{4r^2}$$
Question 5
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KEPLER'S LAWS
Which of the following orbits is a possible orbit for a planet?

A

B

C

D

Solution

Kepler’s first law of planetary motion is as follows:

All planets move about the sun in in an elliptical orbit having sun at one of its foci. Also a radius vector joining any planet to the sun sweeps out equal areas in equal interval of time.

An ellipse can be represented by the formula $$r=\dfrac{P}{1+\varepsilon \cos \theta }$$ where P is semi latus rectum, $$\varepsilon$$ is eccentricity and $$\theta $$ is the angle to the planet's current position from its closest approach.

In options (a) and (b) sun is not at a focus while in (c) the planet is not in orbit around the sun. Only (d) represents the possible orbit for a planet.
Question 6
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The height from earth's surface at which acceleration due to gravity becomes $$\dfrac {g}{4}$$ is (where $$g$$ is acceleration due to gravity on the surface of earth and $$R$$ is radius of earth).

A
$$\sqrt {2}R$$

B
$$R$$

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

D
$$2R$$

Solution

We know that $$g=\dfrac{GM_e}{R^2}$$

And at height h from the earth surface $$ g_1=\dfrac{GM_e}{(R+h)^2}$$

And here $$g_1=\dfrac{g}{4}$$

So, $$ \dfrac{g}{4}=\dfrac{GM_e}{(R+h)^2}$$

so, $$\dfrac{GM_e}{4R^2}=\dfrac{GM_e}{(R+h)^2}$$

Hence, $$\dfrac{1}{4R^2}=\dfrac{1}{(R+h)^2}$$

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

$$\implies\dfrac{R+h}{R}=2$$

$$\implies R+h=2R$$

$$\implies R=h$$

Hence, at height equal to R from earth surface gravity will become $$\dfrac{g}{4}$$.
Answer-(B)
Question 7
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'G' represents
I. Acceleration due to gravity.
II. Weight
III. Gravitational constant
Which combination is correct?

A
II and III only

B
I and III only

C
III only

D
I, II and III

Solution

$$G$$ denotes gravitational constant where as $$g$$ denotes acceleration due to gravity and weight is given by $$W=mg$$
Question 8
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A smooth and bottomless tunnel is dug through the centre of earth. A particle is released from the surface of earth into the tunnel. Time to reach centre of tunnel is (approximately) equal to (where $$R =$$ Radius of earth).

A
$$56.4\ minute$$

B
$$84.35\ minute$$

C
$$42.3\ minute$$

D
$$28.2\ minute$$

Solution

Force on particle at any $$x$$ is $$\dfrac{GM'm}{x^2}$$
But $$\dfrac{M'}{M}=\dfrac{x^3}{R^3}$$
$$\dfrac{GM x^3 m}{R^3 x^2}=(\dfrac{GMm}{R^3})x$$
Acceleration of particle, $$a=\dfrac{GM}{R^3}x=\dfrac{g}{R}x$$
It is an SHM motion, $$T=2 \pi \sqrt{\dfrac{R}{g}}$$
Time to reach center of earth $$=\pi \sqrt{\dfrac{R}{g}}$$
$$=3.14 \times \sqrt{\dfrac{6400 \times 10^3}{9.81}}=42.283$$ minutes
Question 9
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From a solid sphere of mass $$M$$ and radius $$R$$ a spherical portion of radius $$\dfrac {R}{2}$$ is removed, as shown in the figure. Taking gravitational potential $$V = 0$$ at $$r = \infty$$, the potential at the centre of the cavity thus formed is : $$(G =$$ gravitational constant).

A
$$\dfrac {-2GM}{3R}$$

B
$$\dfrac {-2GM}{R}$$

C
$$\dfrac {-GM}{2R}$$

D
$$\dfrac {-GM}{R}$$

Solution

By superposition principle, $$v_1=\dfrac{-GM}{2R^3}[3R^2-(\dfrac{R}{2})^2]$$
$$=-\dfrac{11GM}{8R^3}$$
Also, $$v_2=-\dfrac{3}{2} \dfrac{G (M/8)}{(R/2)}=\dfrac{-3GM}{8R}$$
The required potential is, $$v=v_1-v_2$$
$$=-\dfrac{11GM}{8R}-(-\dfrac{3GM}{8R})$$
$$V=-\dfrac{GM}{R}$$
Question 10
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A hole is drilled along the earth's diameter and a stone is dropped into it. When the stone is at the centre of the earth, it has.

A
Acceleration

B
Weight

C
Mass

D
Potential energy

Solution

At the center of the earth everything is weightless as acceleration and potential energy vanish. So it has only mass.