MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 4

The magnitude of acceleration due to gravity decreases

0%
as the height from the surface of the earth increases
0%
as the depth from the surface of the earth increases
0%
as one moves from the pole of the earth to its equator
0%
All the above

Explanation

Function of g with respect to height from the ground is

$g_h = g(1 - \dfrac{2h}{R})$ for h <<R
So as 'h' increases, the acclelration due to gravity decreases.

Function of g with respect to depth from the surface is

$g_r = g (1 - \dfrac{x}{R} )$ x is the distance of the point from the surface of the earth.
As the depth increases, the value of g also decreases.

At the equator ], the net gravitational force is

$=$ Actual gravitational force - centrifugal force.
At the poles, the centrifugal force is zero and highest at the equator.

The escape velocity of a body on a planet depend on

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acceleration due to gravity on the planet
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the radius of the planet
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the mass of the planet
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All the above

Explanation

Escape velocity of an object on earth is given by

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

$v_e = \sqrt{gR}$

The escape velocity from a planet of mass (M), radius (R) and acceleration due to gravity (g) is given by:

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$2\sqrt {\dfrac {GM}{R}}$
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$\sqrt {2gR}$
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$\sqrt {\dfrac {2gM}{R}}$
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$\sqrt {\dfrac {2GM}{R}}$

Choose the correct statement(s). The acceleration due to gravity 'g' decreases if

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we go down from the surface of earth
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we go up from the surface of earth
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we go from the equator towards the poles on the surface of the earth
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the rotational velocity of the earth is increased

Explanation

The acceleration due to gravity 'g' decreases if we go down from the surface of the earth towards its center. As g is proportional to

$(d)^{-1}$ , we go up from the surface of earth

$g\alpha (R+h)^{-2}$ and the rotational velocity of the earth is increased

$g\alpha \omega.$

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 :

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Will go on increasing continuously
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Will go on decreasing continuously
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Will remain same
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Will first increases and then decreases

The ratio of escape velocity of a tennis ball to that of a basket ball on the surface of the earth is

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1:1
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1:2
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2:1
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4:1

Explanation

Equating KE and PE we get the expression for escape velocity as

$\sqrt{2GM/r}$ , where M = mass of earth , r = radius of earth and G = gravitational constant. ince escape velocity is independent of mass of ball, the ratio of escape velocity of a tennis ball to that of a basket ball on the surface of the earth is 1:1.

Gravity meters measures in units of mGal where 1 Gal (or 1 galileo) is the C.G.S unit of acceleration due to gravity. A gravity meter shows a reading of +300 mGal at a place where the normal 'g' is 9.78

$m s^{-2}$ . The correct 'g' at that place where the reading was taken is _____

$m s^{-2}$ .

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9.777
0%
9.783
0%
9.750
0%
9.810

Explanation

A body has weight due to the fall under the action of gravity. The unit of gravity is gal in C.G.S (named after Galileo). $1 Gal =\dfrac{1cm}{sec^{2}}$ .

$+300 mGal=0.3\times 10^{-2} ms^{-2}=0.003 m s^{-2}$

$\therefore g=9.78+0.003=9.783 m s^{-2}$

The distance from the surface of the earth at which above and below the surface acceleration due to gravity is same will be

0%
$\displaystyle h = \frac{\sqrt{5}- 1}{2} R$
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$\displaystyle h = \frac{\sqrt{3} - 1}{2}R$
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$\displaystyle h = \frac{\sqrt{3} - 1}{3} R$
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None of these

Explanation

Let the distance be

$d$ .
For the condition:

$g\left(1-\dfrac{ h }{ R }\right) =g\left(\dfrac { R }{ R+h } \right)^2$

$\left(1-\dfrac { h }{ R }\right) \left(1+\dfrac { h }{ R }\right)^{ 2 }=1$
Solving this we get:

$h=\dfrac { \sqrt { 5 } -1 }{ 2 } R$

At what altitude will the acceleration due to gravity be

$25\%$ of that at the earth's surface

$($ given radius of earth is

$R)$ ?

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$\dfrac{R}{4}$
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$R$
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$\dfrac{3R}{8}$
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$\dfrac{R}{2}$

Explanation

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,
$g = \dfrac{GMm}{R^2}$

$\textbf{Step2: Calculation of altitude}$
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
$R+h = 2R$
$h = R$

A satellite in earth orbit experiences a small drag force as it enters the earth's atmosphere. Two students were asked consequence of this
Student-A : The satellite would slow down as, it spirals towards earth due to work of frictional force.
Student-B : The satellite speed up due to earths gravitational pull as it spirals towards earth.

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A is correct, B is wrong
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B is correct, A is wrong
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both are correct
0%
both are wrong

Explanation

Let,

$U + K = c$ where c is some constant. As the body spirals down,

$U \downarrow, K \uparrow$ . Thus, speed of the satellite increases and the influence of air friction is negligable.

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

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angular velocity
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kinetic energy
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areal velocity
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potential energy

Satellites A and B are orbiting around the earth in orbits of ratio

$R$ and

$4R$ respectively. The ratio of their areal velocities is

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$1 : 2$
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$1 : 4$
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$1 : 8$
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$1 : 16$

Explanation

Using Kepler's second law we have:

$Areal \;velocity =\dfrac{Area swept}{time}=\dfrac{\dfrac{1}{2}r(rd\theta)}{dt}=\dfrac{1}{2}r^2\omega=\dfrac{1}{2}rv=constant$

Thus we get,

$\dfrac{1}{2}\sqrt{\dfrac{GM}{r}}r$ (from

$\dfrac{mv^2}{r}=\dfrac{GMm}{r^2}$ )
Thus we get
Areal velocity

$\propto \sqrt{r}$
Thus for the planets with radius in ratio of

$1:4$ we get radii in the ratio of

$1:2$

Two planets

$A$ and

$B$ have the same material density. If the radius of

$A$ is twice that of B, then the ratio of the escape velocity

$\displaystyle \dfrac{V_A}{V_B}$ is

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

Explanation

Let the density be d for both the planets. Given that

$R_A = 2R_B$
Now, mass of

$A,\ M_A = \dfrac{4d\pi {R_A}^3}{3} = \dfrac{32d\pi {R_B}^3}{3}$
similarly,

$M_B = \dfrac{4d\pi {R_B}^3}{3}$
Escape velocity for a planet is given by

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

So,

$V_A = \sqrt{\dfrac{2GM_A}{3R_A}} = \sqrt{\dfrac{64Gd\pi {R_B}^3}{6R_B}} = \sqrt{\dfrac{32Gd\pi {R_B}^2}{3}}$

Similarly,

$V_B = \sqrt{\dfrac{8Gd\pi {R_B}^2}{3}}$

Taking the ratio,

$\dfrac{V_A}{V_B} = \sqrt{\dfrac{32Gd\pi {R_B}^2}{3}} \times \sqrt{\dfrac{3}{8Gd\pi {R_B}^2}} =2$

A (nonrotating) star collapses onto itself from an initial radius R

$_i$ with its mass remaining unchanged. Which curve in figure best gives the gravitational acceleration a

$_g$ on the surface of the star as a function of the radius of the star during the collapse?

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a
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b
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c
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d

Explanation

$g\propto \dfrac{1}{R^2}$
Curve b represents this variation.

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

0%
$\displaystyle \dfrac{\omega^2 R^2}{g}$
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$\displaystyle \dfrac{\omega^2 R^2}{2g}$
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$\displaystyle \dfrac{2\omega^2 R^2}{g}$
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$\displaystyle \dfrac{\sqrt{Rg}}{g}$

Explanation

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

$g_e \left(1 - \dfrac{d}{R} \right) = g_e - \omega^2 R$

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

$\Rightarrow d = \dfrac{\omega^2 R}{g_e}$

Select the correct choice(s)

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The gravitational field inside a spherical cavity, within a spherical planet must be non zero and uniform.
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When a body is projected horizontally at an appreciable large height above the earth, with a velocity less than for a circular orbit, it will fall to the earth along a parabolic path.
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A body of zero total mechanical energy placed in a gravitational field will escape the field
0%
Earth's satellite must be in equatorial plane.

Explanation

Gravitational field is conservative field, total mechanical energy is conserved.
If a body is in gravitational field with

$U + K = 0 \rightarrow K = -U \rightarrow (1)$ i.e. Kinetic energy is non zero.So, it will move away from the field till its kinetic energy is also zero.
If

$K = 0 \rightarrow$ U = 0 from (1). But, U = 0 is at

$\infty$ and thus the body escapes the field.

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is

$V$ . Due to the rotation of planet about its axis the acceleration due to gravity

$g$ at equator is

$\dfrac{1}{2}$ of

$g$ at poles. The escape velocity of a particle on the pole of planet in terms of

$V$ .

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$V_e = 2V$
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$V_e = V$
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$V_e = V/2$
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$V_e = \sqrt{2}V$

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].

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$\displaystyle\left [ \dfrac{2Gm}{r} \right ]^{1/2}\leq c$
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$\displaystyle\left [ \dfrac{2gm}{r} \right ]^{1/2}= c$
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$\displaystyle\left [ \dfrac{2Gm}{r} \right ]^{1/2}\geq c$
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$\displaystyle\left [ \dfrac{gm}{r} \right ]^{1/2}\geq c$

Explanation

For a uniform spherical mass to be a black hole, total energy of the "particle + spherical mass" system

$\leq 0$
Let the mass of particle be

$m_1$ moving with its maximum speed

$c$ .

$\therefore$

$\dfrac{1}{2}m_1 c^2 + (-\dfrac{Gmm_1}{r}) \leq 0$

$\implies \ c\leq\bigg[\dfrac{2Gm}{r}\bigg]$

A particle of mass

$10 g$ is kept on the surface of a uniform sphere of mass

$100 kg$ and radius

$10 cm$ . Find the work to be done against the gravitational force between them to take the particle far away from the sphere

0%
$13.34\times 10^{-10}J$
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$13.33\times 10^{-10}J$
0%
$6.67\times 10^{-9}J$
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$6.67\times 10^{-10}J$

Explanation

$W=\Delta PE=\left ( \dfrac{-GMm}{\infty } \right )-\left ( \dfrac{-GMm}{R} \right )=\dfrac{6.67\times10^-11 \times 100\times 10\times 10^{-3}}{1}= 6.67\times10^{-10} J$

Find the height above the surface of the earth where weight of a body becomes half.

0%
$\displaystyle \frac{R}{2}$
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$(\sqrt{2} - 1) R$
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$\displaystyle \frac{R}{(\sqrt{2} + 1)}$
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$\displaystyle \frac{R}{\sqrt{2}}$

Explanation

Given, weight of the body becomes half.

Let that height be

$h$ .

$\dfrac { mg }{ 2 } =\dfrac { GMm }{ { \left( R+h \right) }^{ 2 } }$

And,

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

Hence,

$h=\left( \sqrt { 2 } -1 \right) R$

Multiplying numerator and denominator by

$\sqrt 2+1$

$h = \dfrac{R}{\sqrt 2 +1}$

A body is projected up with a velocity

$2$ times the escape velocity

$v_{e}$ from the surface of the earth. The velocity of the body at a point far away the earth's attraction is

0%
$v_{es}$
0%
$\sqrt{7} v_{es}$
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$\sqrt{2} v_{es}$
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$\sqrt{3} v_{es}$

Explanation

$v_e = (\dfrac{2GM}{R})^{0.5}$
now,

$V = 2v_e = 2(\dfrac{2GM}{R})^{0.5} = (\dfrac{8GM}{R})^{0.5}$

By conservation of energy,

$\dfrac{-GMm}{R} +\dfrac{8GMm}{2R} = 0 + \dfrac{mv^2}{2}$

solving this, we get

$v = (\dfrac{3 \times 2GM}{R})^{0.5} = 3^{0.5}v_e$

A planet of radius

$R = 1/10 \times$ (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth

$R/5$ on it and lower a wire of the same length and of linear mass density

$10^{-3} kgm^{-3}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth

$= 6 \times 10^6 m$ and the acceleration due to gravity on Earth is

$10 ms^{-2}$ )

0%
$96 N$
0%
$108 N$
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$120 N$
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$150 N$

Explanation

$R_e$ is the radius of the earth

$g_e=\dfrac{GM_e}{R_e^2}=G\dfrac{4}{3}\pi R_e\rho$
Now g at the mentioned planet is given as

$g_p=\dfrac{GM}{r^2}$ , here

$m=\dfrac{4}{3}\pi r^3\rho$
Thus

$g_p=\dfrac{4}{3}r\rho$
Thus we get

$F=\displaystyle\int^R_{4R/5}\lambda g_p dr = 108 N$

A planet revolves about the sun in elliptical orbit. The arial velocity

$\displaystyle \left ( \frac{dA}{dt} \right )$ of the planet is

$4.0 \times 10^{16} m^2/s$ . The least distance between planet and the sun is

$2 \times 10^{12}$ m. Then the maximum speed of the planet in km/s is

0%
10
0%
20
0%
40
0%
none of these

Explanation

We have:

$\dfrac{dA}{dt}=\dfrac{1}{2}r^2\dfrac{d\theta}{dt}$

$4\times 10^{16}=\dfrac{1}{2}(2\times 10^{12})^2\dfrac{d\theta}{dt}$

$\omega=\dfrac{d\theta}{dt}=2\times 10^{-8}$

Using,

$v=r\omega$

We get:

$v=2\times 10^{12}\times 2\times 10^{-8}=40\ km/s$

The change in the value of

$g$ at a height

$h$ above the surface of the earth is the same as at a depth d below the surface of the earth.When both

$h$ and d are much smaller than the radius of earth,then which one of the following is true?

0%
$a=h/2$
0%
$d=3h/2$
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$d=2h$
0%
$h=d$

Explanation

$\displaystyle{{g}'=g \left ( 1-\frac{2h}{R} \right )=g \left ( 1-\frac{d}{R}\right )\therefore d=2h}$

$75 kg$ astronaut is repairing Hubble telescope at a height of

$600 km$ above the surface of the earth. Find his weight there.

0%
$740 N$
0%
$700 N$
0%
$650 N$
0%
$610 N$

Explanation

$\displaystyle mg' = mg \left ( 1 - \dfrac{2h}{R} \right )$

$= 75 \times 10 (1 - 0.185) = 610 N$

If the orbital speed of moon is increased by

$41.4\%$ then moon will

0%
leave its orbit and will escape out.
0%
fall on earth.
0%
attract all bodies on earth towards it.
0%
have time period equal to 27 days.

Explanation

Velocity of the planet is $v=\sqrt{\dfrac{GM}{a}}$

Kinetic Energy is $K=\dfrac{1}{2}m{v}^{2}=\dfrac{GMm}{2a}$

Potential Energy is $U=-{\dfrac{GMm}{a}}$

$Total\ energy= K+U= -{\dfrac{GMm}{2a}}$

${v}= \sqrt{\dfrac{GM}{a}}$

So if $,\quad v=1+0.414 v=1.414v= \sqrt{2} v= \sqrt{\dfrac{2GM}{a}}$

Escape Velocity is $v_e=\sqrt{\dfrac{2GM}{a}}$

So the moon will escape.

At what height over the earth's pole does the freefall acceleration decreases by 1% ?

0%
$64 km$
0%
$16 km$
0%
$8 km$
0%
$32 km$

Explanation

$mg'=\dfrac{GMm}{R^2+h^2}$

$mg=\dfrac{GMm}{R^2}$

$\dfrac{g'}{g}=\dfrac{R^2}{R^2+h^2}$

$\dfrac{g'}{g}=(R^2)(R^2+h^2)^{-2}$

$\dfrac{g'}{g}=(1+\dfrac{h^2}{r^2})^{-2}$

Expanding using binomial expansion and neglecting higher terms,

$\dfrac{g'} {g}= \displaystyle \left ( 1 - \dfrac{2h}{R} \right ) = \dfrac{99}{100} or h = 32 km$

Find the work done to take a particle of mass m from surface of the earth to a height equal to

$2R$ .

0%
$2$ $mg R$
0%
$\displaystyle \frac{mg R}{2}$
0%
$3 mg R$
0%
$\displaystyle \frac{2mgR}{3}$

Explanation

$Work\quad Done\quad =\quad \Delta \quad PE\\ PE=\quad \dfrac { GMm }{ r } \\ \Delta PE=GMm(\dfrac { 1 }{ R } -\dfrac { 1 }{ 3R } )=\dfrac { 2GMm }{ 3R } =\dfrac { 2 }{ 3 } mgR$

Let

$V$ and

$E$ denote the gravitational potential and gravitational field at a point. It is possible to have

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$V = 0$ and $E= 0$
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$V = 0$ and $E\neq 0$
0%
$V \neq 0$ and $E = 0$
0%
$V \neq 0$ and $E\neq 0$

Explanation

A) At

$\infty$ both

$V$ and

$E$ are zero.
B) Let,

$V_ \infty = \dfrac{GM}{R}$ for an unit mass. So,

$V_R = 0$ i.e. at the radius R of solid sphere(mass M) and

$E_R = \dfrac{GM}{R^2}$
C) Inside a spherical shell

$V =\dfrac{-GM}{R}$ and

$E =0$
D) Consider a point at a distance of

$r$ from mass

$m$ . Then,

$V =\dfrac{-GM}{r}$ and

$E =\dfrac{GM}{r^2}$ .

The escape velocity for a planet is

$v_e$ . A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be

0%
$v_a$
0%
$1.5 v_e$
0%
$\sqrt{1.5} v_e$
0%
$2 v_e$

Explanation

Within the earth's body the gravitational potential at distance

$r$ from the center is given as

$V=2\pi G\rho(r^2-3R^2)/3$ .

where

$\rho$ is the density of earth,

$R$ is the earth's radius.

At the center of earth,

$r=0$ .

So the potential is

$V_c=-2\pi G\rho\times3R^2/3=-3GM/2R$ .

where

$M=\rho\times4\pi R^3/3$

At very large distance, potential

$V_i=0$

So gain in kinetic energy = change in potential energy.

$\Rightarrow mv^2/2=0-(-3GM/2R)=3GM/2R=1.5GM/R\Rightarrow v=\sqrt{1.5}v_e$

where

$v_e=\sqrt{GM/R}$ is the escape velocity.

Let

$V_G$ and

$E_G$ denote gravitational potential and field respectively, the it is possible to have

0%
$V_G = 0, E_G = 0$
0%
$V_G \neq 0, E_g = 0$
0%
$V_G = 0, E_g \neq 0$
0%
$V_G \neq 0, E_G \neq 0$

Explanation

A) At $\infty$ both $V$ and $E$ are zero.

B) Let, $V_ \infty = \dfrac{GM}{R} \rightarrow V_R = 0$ and $E_R = \dfrac{GM}{R^2}$

C) Inside a spherical shell $V =\dfrac{-GM}{R}$ and $E =0$

D) Consider a point at a distance of $r$ from mass $m$ . Then, $V =\dfrac{-GM}{r}$ and $E =\dfrac{GM}{r^2}$ .

Use the assumptions of the previous question. An object weighed by a spring balance at the equator gives the same reading as a reading taken at a depth d below the earth's surface at a pole

$(d << R)$ . The value of

$d$ is

0%
$\displaystyle \dfrac{\omega^2 R^2}{ge}$
0%
$\displaystyle \dfrac{\omega^2 R^2}{2g}$
0%
$\displaystyle \dfrac{2\omega^2 R^2}{g}$
0%
$\displaystyle \dfrac{\sqrt{Rg}}{\omega}$

Explanation

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 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,

$g_e \left(1 - \dfrac{d}{R} \right) = g_e - \omega^2 R$

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

$\Rightarrow d = \dfrac{\omega^2 R}{g_e}$

Find the height at which the weight will be same as at the same depth from the surface of the earth.

0%
$\displaystyle \frac{R}{2}$
0%
$\sqrt{5} R - R$
0%
$\displaystyle \frac{\sqrt{5} R - R}{2}$
0%
$\displaystyle \frac{\sqrt{3}R - R}{2}$

Explanation

Let the height and depth be 'd'

$\Rightarrow \quad \dfrac { g }{ { \left( 1+\dfrac { d }{ R } \right) }^{ 2 } } =g\left( 1-\dfrac { d }{ R } \right)$

$\Rightarrow { \left( 1+\dfrac { d }{ R } \right) }^{ 2 }\left( 1-\dfrac { d }{ R } \right) =1$

$\Rightarrow { \left( 1+\dfrac { { d }^{ 2 } }{ { R }^{ 2 } } +\dfrac { 2d }{ R } \right) }\left( 1-\dfrac { d }{ R } \right) =1\\ \Rightarrow { \left( \dfrac { d }{ R } \right) }^{ 3 }+{ \left( \dfrac { d }{ R } \right) }^{ 2 }-{ \left( \dfrac { d }{ R } \right) }=0\\ \Rightarrow \quad d=\dfrac { \sqrt { 5 } R-R }{ 2 }$

A small mass

$m$ is moved slowly from the surface of the earth to a height

$h$ above the surface. The work done (by an external agent) in doing this is

0%
$mgh$ , for all values of $h$
0%
$mgh$ , for $h < < R$
0%

$\dfrac{1}{2}$ mgR, for h $=$ R
0%
$- \dfrac{1}{2}$ mgR, for h $=$ R

Explanation

For, $h<<R$ force is constant and is equal to mg

Therefore, work done is mgh in moving by distance $h$

For, $h=R$

gravitational potential at x is $-\dfrac { GM }{ x }$

Work done in moving from $x=R$ to $x=2R$ is $-\dfrac { GMm }{ 2R } +\dfrac { GMm }{ R } =\dfrac { GMm }{ 2R }$

which is equal to $\dfrac { 1 }{ 2 } mgR$

Consider a planet moving in an elliptical orbit round the sun. The work done on the planet by the gravitational force of the sun

0%
is zero in any small part of the orbit.
0%
is zero in some parts of the orbit.
0%
is zero in one complete revolution.
0%
is zero in no part of the motion.

Explanation

The work done is zero at points A and B, since the force acts along the AB axis and the motion is perpendicular to it(

$W=|F||S| \cos \theta$ )

A spaceship is released in a circular orbit near the Earth's surface. How much additional velocity will have to be given to the spaceship in order to escape out of this orbit?

0%
$3.28 \quad m/s$
0%
$3.28 \times 10^3 m/s$
0%
$3.28 \times 10^7 m/s$
0%
$3.28 \times 10^{-3} m/s$

Explanation

Here, orbital velocity and escape velocity are

$v_0 = \sqrt{gR}$

$v_e = \sqrt{2gR}$
Hence, required additional velocity will be,

$v = v_e - v_o$

$v = \sqrt{2gR} - \sqrt{gR}$

$v = (\sqrt 2 - 1) \sqrt{gR}$

$v = (1.414 - 1)\sqrt{(10)(6400 \times 10^3)}$ ..........

$($ since,

$g = 10 ms^{-2}$ and

$R = 6400 \times 10^3 m)$

$v = (0.414)(8) \times 10^3$

$v = 3.3 \times 10^3 ms^{-1}$

The gravitational potential difference between the surface of a planet and a point

$20 m$ above the surface is

$2 Joule/Kg$ . If the gravitational field is uniform then the work done in carrying a

$5 Kg$ body to a height of

$4 m$ above the surface is

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$2 Joule$
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$20 Joule$
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$40 Joule$
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$10 Joule$

Explanation

since it is given that gravitational field is uniform
gradient of potential

$=\dfrac { Potential\ diff. }{ Dist.\ between\ points }=\dfrac { 2 }{ 20 } =0.1\ J/Kg s$
Total work done

$=mass\times distance\times pot.\ gradient=5\times 4\times 0.1=2\ Joule$

The magnitude of the potential energy per unit mass of an object at the surface of the earth is given

$U$ . Then, the escape velocity for the object is given by

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$v_e = 2U$
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$v_e = \sqrt{U}$
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$v_e = \sqrt{2U}$
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$v_e = (2U)^2$

Explanation

Potential energy of unit mass of body at earth surface U =

$\dfrac{-GM(1)}{R} = \dfrac{-GM}{R}$
escape velocity

$v_e = (\dfrac{2GM}{R})^{0.5} = (2U)^{0.5}$

The value of G for two bodies in vacuum is

$6.67 \times 10^{-11} N-m^2/Kg^2$ . Its value in a dense medium of density

$10^{10} gm/cm^3$ will be:

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$6.67 \times 10^{-11} N-m^2 /Kg$
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$6.67 \times 10^{-31} N-m^2 /Kg$
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$6.67 \times 10^{-21} N-m^2 /Kg$
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$6.67 \times 10^{-10} N-m^2 /Kg$

Explanation

G is a Universal Gravitational constant and is same every where in the universe. Thus,
G

$= 6.67 \times 10^{-11} N.m^{2}.kg^{-2}$

The potential energy of a rocket of mass

$100\ kg$ at height

$10^7\ m$ from earth surface is

$4 \times 10^9\ Joule$ . The weight of the rocket at height

$10^9$ will be

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$4 \times 10^{-2} N$
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$4 \times 10^{-3} N$
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$8 \times 10^{-2} N$
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$8 \times 10^{-3} N$

Explanation

Distance

$R = r + h = 0.64 \times 10^{7} + 10^{7} = 1.64 \times 10^7 m$
Now

$,\ U = \dfrac{GMm}{R} = \dfrac{GM \times 100}{1.64 \times 10^7} = GM \times .609 \times 10^{-5}$

But, given

$U = 4 \times 10^9 \Rightarrow GM \times 0.609 \times 10^{-5} = 4 \times 10^9$

$GM = 6.57 \times 10^{14}$
now at height

$10^9 m$ ,

$R = r+h = 100.64 \times 10^7$
weight

$mg = \dfrac{GMm}{R^2} = \dfrac{6.57 \times 10^{14} }{(100.64 \times 10^7)^{2}} = 4 \times 10^{-2}$ N.

A tunnel is dug along a diameter of earth. The force on a particle of mass

$m$ and distance

$x$ from the centre in this tunnel will be :

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$\displaystyle \frac{GM_e m}{R^3 x}$
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$\displaystyle \frac{GM_e mR^3}{x}$
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$\displaystyle \frac{GM_e mx}{R^2}$
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$\displaystyle \frac{GM_e mx}{R^3}$

Explanation

Field at a distance

$x$ from the center will be equal to field at a distance

$R-x$ from the earth's surface.

$g_x=g(1-\displaystyle\frac{(R-x)}{R})=\frac{gx}{R}$

$F=\displaystyle\frac{mgx}{R}=\frac{mx}{R}\frac{GM_e}{R^2}=\frac{GM_e}{R^3}mx$

At a height

$h$ above the surface of earth, the change in value of

$g$ is same as that at a depth

$x$ below the surface on the earth. Both

$x$ and

$h$ are very small in comparison to the radius of the earth. Then,

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$x = h$
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$x = 2h$
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$2x = h$
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$x= h^2$

Explanation

Let,

$g_h$ be the acceleration due to gravity at height

$h.$

$g_x$ be the acceleration due to gravity at depth

$x.$
The acceleration due to gravity on the surface of Earth is

$g = \dfrac{GM}{R^2}$
where

$, R$ is the radius o Earth

$, M$ is mass of Earth and

$G$ is gravitational constant.

$\therefore g \propto \dfrac{1}{R^2}$

$\Rightarrow g_h \propto \dfrac{1}{(R+h)^2}$
Therefore

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

$\dfrac{g_h}{g} = \dfrac{1}{(1+\frac{h}{R})^2}$

$\dfrac{g_h}{g} = \left(1+\dfrac{h}{R}\right)^{-2}$

$\dfrac{g_h}{g} = \left(1-\dfrac{2h}{R}\right)$

$g_h = g - \dfrac{2gh}{R}$

$g - g_h = \dfrac{2gh}{R}$
Also, the acceleration due to gravity at depth

$x,$

$g_x = \dfrac{4}{3}G\rho (R-x)$

$\Rightarrow g_x \propto (R-x)$
Therefore

$\dfrac{g_x}{g} = \dfrac{R-x}{R}$

$\dfrac{g_x}{g} = 1 - \dfrac{x}{R}$

$g_x = g - \dfrac{gx}{R}$

$g - g_x = \dfrac{gx}{R}$
If the change in the value of

$g$ at height

$h$ above earth surface is the same as that at depth

$x$ i.e.

$g - g_h = g - g_x$

$\dfrac{2gh}{R} = \dfrac{gx}{R}$

$\therefore x = 2h$

The value of acceleration due to gravity at height

$h$ from earth surface will become half its value on the surface if (R

$=$ radius of earth)

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$h = R$
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$h = 2R$
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$h = (\sqrt{2} - 1)R$
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$h = (\sqrt{2} + 1)R$

Explanation

The gravitational force acting on mass m on the surface of Earth is

$F = \dfrac{GMm}{R^2}$

$mg = \dfrac{GMm}{R^2}$

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

$gR^2 = GM$ .............

$(1)$
where,

$g$ is acceleration due to gravity,

$G$ is gravitational constant,

$M$ is mass of Earth and R is radius of Earth.
Now, value of g at height h is

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

As per the problem,

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

$g(R+h)^2 = 2 gR^2$ ................from

$(1)$

$(R+h)^2 = 2 R^2$
Taking roots from both sides, we get

$R+h = \sqrt 2 R$

$\therefore h = \sqrt 2 R - R$

$\therefore h = (\sqrt 2 - 1)R$

If the change in the value of

$g$ at height

$h$ above earth surface is the same as that at depth

$x (x \quad or\quad h < R_e),$ then

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$x = h^2$
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$x = h$
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$x = \dfrac{h}{2}$
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$x = 2h$

Explanation

Let

$g_h$ be the acceleration due to gravity at height

$h$ .

$g_x$ be the acceleration due to gravity at depth

$x$
The acceleration due to gravity on the surface of Earth is

$g = \dfrac{GM}{R^2}$
where,

$R$ is the radius o Earth

$, M$ is mass of Earth and

$G$ is gravitational constant.

$\therefore g \propto \dfrac{1}{R^2}$

$\Rightarrow g_h \propto \dfrac{1}{(R+h)^2}$
Therefore

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

$\dfrac{g_h}{g} = \dfrac{1}{\left(1+\dfrac{h}{R}\right)^2}$

$\dfrac{g_h}{g} = \left(1+\dfrac{h}{R}\right)^{-2}$

$\dfrac{g_h}{g} = \left(1-\dfrac{2h}{R}\right)$

$g_h = g - \dfrac{2gh}{R}$

$g - g_h = \dfrac{2gh}{R}$
Also, the acceleration due to gravity at depth

$x,$

$g_x = \dfrac{4}{3}G\rho (R-x)$

$\Rightarrow g_x \propto (R-x)$
Therefore

$\dfrac{g_x}{g} = \dfrac{R-x}{R}$

$\dfrac{g_x}{g} = 1 - \dfrac{x}{R}$

$g_x = g - \dfrac{gx}{R}$

$g - g_x = \dfrac{gx}{R}$
If the change in the value of

$g$ at height

$h$ above earth surface is the same as that at depth

$x$ i.e.

$g - g_h = g - g_x$

$\dfrac{2gh}{R} = \dfrac{gx}{R}$

$\therefore x = 2h$

At some planet,

$g = 1.96\ ms^{-2}$ . If it is safe to jump from a height of

$2\ m$ on earth, then the corresponding safe height on that planet is

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$2\ m$
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$5\ m$
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$10\ m$
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$20\ m$

Explanation

Gravitation potential on the planet

$, P.E._P = mgh_P$

$\Rightarrow P.E._P = m\times 1.96h_P$
Gravitation potential on the earth

$, P.E._E = mgh_E$

$\Rightarrow P.E._E = m\times 9.8\times 2$

$\Rightarrow P.E._E = P.E._E$

$\Rightarrow m\times 1.96h_P=m\times 9.8\times 2$

$\Rightarrow h_P = \dfrac{m\times 9.8\times 2}{m\times 1.96} = 10\ m$

${ W }_{ 1 }$ ,

${ W }_{ 2 }$ and

${ W }_{ 3 }$ represent the work done in moving a particle from

$A$ to

$B$ along three different paths

$1$ ,

$2$ and

$3$ , respectively, (as shown in the figure) in the gravitational field of a point mass

$m$ , find the correct relation between

${ W }_{ 1 }$ ,

${ W }_{ 2 }$ and

${ W }_{ 3 }$

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${ W }_{ 1 }>{ W }_{ 2 }>{ W }_{ 3 }$
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${ W }_{ 1 }={ W }_{ 2 }={ W }_{ 3 }$
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${ W }_{ 1 }<{ W }_{ 2 }<{ W }_{ 3 }$
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${ W }_{ 2 }>{ W }_{ 1 }>{ W }_{ 3 }$

Explanation

In a conservative field work done does not depend on the path, and gravitational field is a conservative field.

$\therefore W_1=W_2=W_3$

$g$ is same at a height

$h$ and at a depth

$d$ , then

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${ R }={ 2d }$
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${ d }={ 2h }$
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${ h }={ d }$
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none

Explanation

We know,

$g_h=g(1-\dfrac{2h}{R})$ and

$g_d=(1-\dfrac{d}{R})$
When

$g_d=g_h$ then,

$(1-\dfrac{2h}{R})=1-\dfrac{d}{R}$

$\therefore (\dfrac{-2h}{R})=-\dfrac{d}{R}$

$d=2h$

A missile is launched with a velocity less than the escape velocity from earth. The sum of its kinetic energy and potential energy is

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positive
0%
negative
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zero
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may be positive or negative depending on the direction of projection

If the escape velocity from the surface of a spherical planet is given by

$2 \displaystyle \sqrt{\frac{GM}{K}}$ , where

$M$ is the mass of the planet and

$K$ is a constant, then the radius of the planet is

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$\dfrac{K}{2}$
0%
$K$
0%
$2K$
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$4K$

Explanation

Escape velocity of planet with radius

$R,\quad v_e = \sqrt{\dfrac{2GM}{R}}$
given,

$v_e = 2\sqrt{\dfrac{GM}{K}}$
So,

$v_e = 2\sqrt{\dfrac{GM}{K}} =\sqrt{\dfrac{2GM}{R}}$ and solving it we get

$R = \dfrac{K}{2}$

The distances from the centre of the earth, where the weight of a body is zero and one-fourth that of the weight of the body on the surface of the earth are (assume

$R$ is the radius of the earth)

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$0$ , $\displaystyle \frac { R }{ 4 }$
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$0$ , $\displaystyle \frac { 3R }{ 4 }$
0%
$\displaystyle \frac { R }{ 4 }$ , $0$
0%
$\displaystyle \frac { 3R }{ 4 }$ , $0$

Explanation

The weight of the body at the centre of the earth is equal to zero because

$g_{centre}=g\left(1-\dfrac{d}{R}\right)=g\left(1-\dfrac{R}{R}\right)=0$

$\dfrac{g_{1}}{g}=\left(1-\dfrac{d}{R}\right)=\dfrac{1}{4}\Rightarrow d=\dfrac{3R}{4}$
So from the centre,

$d=\dfrac{R}{4}$

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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