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Representing a gravitational field - class-XII

Description: representing a gravitational field
Number of Questions: 44
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Tags: gravitation circular motion and gravitation gravitation: planets and satellites gravitational fields physics
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Electrical, as well as gravitational affects, can be thought to be caused by fields. Which of the following is true of an electrical or gravitational field?

  1. The field concept is often used to describe contact forces

  2. Gravitational or electric field does not exist in the space around an object

  3. Fields are useful for understanding forces acting through a distance

  4. There is no way to verify the existence of a force field since it is just a concept


Correct Option: C
Explanation:

In physics, concept of field is a model used to explain the influence that a massive body or charged particle extends into the space around itself, producing a force on another massive body or charged body placed in space. Thus, concept of field is used to explain gravitational and electrostatic phenomena.

A satellite is revolving in a circular orbit at a height $'h'$ from the earth's surface (radius of earth $R$;$h\ <\ <\ R$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to:(Neglect the effect of atmosphere.)

  1. $\sqrt{2gR}$

  2. $\sqrt{gR}$

  3. $\sqrt{gR/2}$

  4. $\sqrt{gR}\left(\sqrt{2}-1\right)$


Correct Option: D
Explanation:

height$=h$

No, the given formula,
Orbital velocity
$v=\cfrac{GM}{R+h}=\cfrac{GM}{R}$ as $h<<R$
Velocity required to escape
$\cfrac{1}{2}mv^{\prime 2}=\cfrac{GMm}{R+h}\v^{\prime}=\cfrac{2GM}{R+h}=\cfrac{2GM}{R}(h<<R)$
$\therefore$ Increase in velocity
$v^{\prime}=v=\sqrt{\cfrac{2GM}{R}}-\sqrt{\cfrac{GM}{R}}\ \quad=\sqrt{2gR}-\sqrt{gR}=\sqrt{gR}(\sqrt2-1)$

A force of 10 N of gravitational force in CGS units is  represented as

  1. $10 Dynes$

  2. $10^2 Dynes$

  3. $10^5 Dynes$

  4. $10^6 Dynes$


Correct Option: A
Explanation:

A force of 10 N of gravittional force in CGS units is represented by 10 Dynes.

Dyne is a CGS unit of force which accelerates a mass of one gram at the rate of one centimeter per second.

The presence of gravitational field is required for the heat transfer by

  1. $conduction$

  2. $striring\ of\ liquids$

  3. $natural\ convection$

  4. $radiation$


Correct Option: A
Explanation:

Presence of gravitational field is required for heat transfer by conduction.

The force experienced by a unit mass at a point in the gravitational field is called its

  1. gravitational intensity

  2. electric intensity

  3. magnetic intensity

  4. gravitational constant


Correct Option: A
Explanation:

The force experienced by a $unit $ mass at a point in the gravitational field $=F _g=mg=1\times g=g $ $ Newton $


We know that $g$ is called as gravitational intensity 

Hence correct answer is option $A $ 

The ratio of SI units to CGS of the gravitational intensity is

  1. $10^3:1$

  2. infinity

  3. zero

  4. $10^2:1$


Correct Option: D
Explanation:

Gravitational intensity is nothing but the acceleration due to gravity(g)

SI unit of g is $m/s^2=100cm/s^2$
But CGS unit of g is $cm/s^2$ 
So SI unit:CGS unit $=100:1$ 

Hence correct answer is option $D $ 

Which of the following represents the unit for gravitational intensity?

  1. N

  2. $kgm^{-2}$

  3. $Nkg^{-1}$

  4. $ms^{-3}$


Correct Option: C
Explanation:

Gravitational intensity is nothing but the acceleration due to gravity $=g=F _g/m$


Now we know that the SI unit of $F _g$ is $Newton(N) $ and that of $m$ is $kg$
So SI unit of $g=N/kg=Nkg^{-1}$ or $m/s^2$ 

Hence correct answer is option $C $ 

The space in which a body experiences a force by virtue of its mass, is called a/an

  1. magnetic field

  2. electric field

  3. gravitational field

  4. none of these


Correct Option: C
Explanation:

The region of space surrounding a body in which another body experiences a force of gravitational attraction is called a gravitational field.

The speed at which the gravitational field propagates is

  1. Equal to the speed of light in vacuum

  2. Less than the speed of light in vacuum

  3. More than the speed of light in vacuum

  4. Either less or more than the speed of light in

    vacuum


Correct Option: A
Explanation:

The speed of gravitational waves in the general theory of relativity is equal to the speed of light in vacuum.

The unit of (V) Gravitational Potential is

  1. $\dfrac{\text{Joule}}{kg}$

  2. $\text{Joule}$

  3. $\dfrac{\text{Joule}}{s}$

  4. $\dfrac{kg}{\text{Joule}}$


Correct Option: A
Explanation:

Gravitational potential is defined as the Gravitational energy per unit mass of the object on the planet. Unit of energy is always joule in SI units. Hence that of potential will be Joule/kg.

Value of gravitational constant, $'G'$ is

  1. $6.674 08 \times 10^{-11} m^3 kg^{-1} s^{-2}$

  2. $4.674 08 \times 10^{-11} m^3 kg^{-1} s^{-2}$

  3. $6.674 08 \times 10^{11} m^3 kg^{-1} s^{-2}$

  4. $6.674 08 \times 10^{-11} m^3 kg^{-2} s^{2}$


Correct Option: A
Explanation:

The value of G in SI units is $6.67408\times {10^{-11}}$

And the unit is $ Newton.metre^2/kg^2 $
Newton is $kg.m.s^{-2} $
Hence it becomes $ m^3.kg^{-1}.s^{-2} $

Gravitational field is 

  1. directly proportional to square of the distance between two masses.

  2. inversely proportional to square of the distance between two masses.

  3. directly proportional to the distance between two masses.

  4. inversely proportional to the distance between two masses.


Correct Option: B
Explanation:

The gravitational field is the region in which the gravitational force can be experienced or its presence can be felt. The intensity of gravitational field is the force acting on a unit mass of a body. 

Gravitational force is given by $Force = \dfrac{GMm} {R^2}$, 
where $G$ is gravitational constant, $M$ is the mass of the body which is creating the gravitational force, $m$ is the mass of the body which is undergoing gravitational force and $R$ is the distance between them. You can see that this force is inversely proportional to the square of the distance between the two masses. Gravitational field is given by gravitational force / mass of the 2nd body i.e. $m$. 
$Field = \dfrac{GM}{R^2}$. Hence it is inversely proportional to square of the distance between two masses.

The unit of gravitational field is 

  1. $\dfrac{N}{kg}$

  2. $\dfrac{N}{s}$

  3. $\dfrac{kg}{s^2}$

  4. $\dfrac{N}{kg m^2}$


Correct Option: A
Explanation:

The gravitational field is the region in which the gravitational force can be experienced or its presence can be felt. The intensity of gravitational force is the force acting on a unit mass of a body. The gravitational field is mathematically given by, gravitational force divided by the mass of the body. And the unit of the gravitational field is the ratio of the unit of force to that of mass, that is, $N/kg$.

What is gravitational field?

  1. A gravitational field is a region where any other body that has mass will experience a force of attraction.

  2. A gravitational field is a region where any other body that has mass will experience a force of repulsion.

  3. A gravitational field is a region where any other body that has mass will experience no force of attraction.

  4. A gravitational field is a region where any other body that has mass will experience no force of repulsion.


Correct Option: A
Explanation:

Point A is the definition of the gravitational field. 

Gravitational force is always attractive in nature and hence there is no question of repulsion.

Which of the following option is/are correct?

  1. If acting at a single point, the gravitational force on an extended object can be treated as its centre of gravity

  2. If the gravitational field is nonuniform across the object then it can be treated as its centre of mass.

  3. If acting at multiple point, the gravitational force on an extended object can be treated as its centre of gravity

  4. If the gravitational field is uniform across the object then it can be treated as its centre of mass.


Correct Option: A,D
Explanation:

The gravitational force on an extended object can be treated as its center of gravity when acted on a single point. Also, it is uniform across its center of mass.

Determine the gravitational force of two particle of mass $3kg$ and $7 kg$ separated by a distance $2m$.

  1. $35 \times 10^{-11} N$

  2. $25 \times 10^{-11} N$

  3. $5 \times 10^{-11} N$

  4. $3.5 \times 10^{-11} N$


Correct Option: A
Explanation:
Gravitational force of two particle of mass $3kg$ and $7kg$ separated by a distance $2m$ is given by: 
$F=G\dfrac{Mm}{r^2} = 6.674× 10^{-11} \times \dfrac{ 3 \times 7}{2^2}=35 \times 10^{-11} N$

How far from the centre of the Moon is the Earth-Moon neutral point, where the Earth and the Moon's gravitational field strengths are equal in magnitude but opposite in direction?

$ M _E= 6.0 \times 10^{24} kg \ \ \  M _M = 7.4 \times 10^{22} kg$
The radius of Moon's orbit (assumed to be circular) is: $3.8\times 10^{8} m$.

  1. $3.8 \times 10^{2} m$

  2. $38 \times 10^{6} m$

  3. $38 \times 10^{4} m$

  4. $28 \times 10^{6} m$


Correct Option: B
Explanation:

Strength of gravity $\cfrac { Gm }{ { r }^{ 2 } } $

Let ${ g } _{ e }$ be the value of $Gm$ of earth $=4\times { 10 }^{ 14 }\cfrac { { m }^{ 3 } }{ { s }^{ 2 } } $
      ${ g } _{ m }$ be the value of $Gm$ of moon $=5\times { 10 }^{ 12 }\cfrac { { m }^{ 3 } }{ { s }^{ 2 } } $
Let $d\rightarrow$distance between earth and moon $=380\times { 10 }^{ 6 }m$
Let equilibirum point from earth be at a distanexe :- $x$
       $\cfrac { { g } _{ e } }{ { x }^{ 2 } } =\cfrac { { g } _{ m } }{ { \left( d-x \right)  }^{ 2 } } $
      ${ g } _{ e }\left( { x }^{ 2 }-2xd+{ d }^{ 2 } \right) ={ g } _{ m }{ x }^{ 2 }\ \left( { g } _{ e }-{ g } _{ m } \right) { x }^{ 2 }-{ 2dg } _{ e }x+{ d }^{ 2 }{ g } _{ e }=0$
Solving the quadratic formula
$x=\cfrac { { 2dg } _{ e }\pm \sqrt { 4{ d }^{ 2 }{ g } _{ e }^{ 2 }-{ 4d }^{ 2 }{ g } _{ e }\left( { g } _{ e }-{ g } _{ m } \right)  }  }{ 2\left( { g } _{ e }-{ g } _{ m } \right)  } =\cfrac { { dg } _{ e }\pm d\sqrt { { g } _{ e }{ g } _{ m } }  }{ { g } _{ e }-{ g } _{ m } } \approx 342\times { 10 }^{ 6 }m$
$\therefore$ Distance from moon :-$d-x=38\times { 10 }^{ 6 }m$

Gravitational field is directed

  1. towards the earth

  2. away from earth

  3. has no direction

  4. in a specific direction making angle with earth


Correct Option: A
Explanation:

Force if gravity on an object due to Earth always act towards the Earth as it is attractive in nature. Therefore, when an object us thrown up, the force of gravity acts towards the Earth. 

Due to a mass distribution, the gravitational field is $\dfrac{k}{x^3} $ along x-axis where $k$ is a constant. If the gravitational potential is taken to be at infinity, then the gravitational potential at $x$ is 

  1. $\dfrac{k}{x}$

  2. $\dfrac{k}{2x^2}$

  3. $\dfrac{k}{x^4}$

  4. $\dfrac{k}{x^6}$


Correct Option: B
Explanation:

$V=-\int { Edx } =-\int { \cfrac { K }{ { x }^{ 3 } } dx=K\left( \cfrac { 1 }{ 2{ x }^{ 2 } }  \right)  } =\cfrac { K }{ { 2x }^{ 2 } } $

The gravitational intensity is denoted by :

  1. $g$

  2. $G$

  3. $E$

  4. none of these


Correct Option: A
Explanation:

Gravitational intensity is a vector quantity related to the condition at any point under gravitational influence the measure of which is the gravitational force exerted upon a unit mass placed at the point in question. It is denoted by $g.$ 

$g = GM/$$R^2$, where $M$ is the mass of earth and $R$ is the radius of earth.

A gravitational field is

  1. a field of gravitons

  2. a field of massive particles

  3. the force field that exists in the space around every mass or group of masses.

  4. Force exerted on an unit charge


Correct Option: C
Explanation:

Gravitational field is the force field that exists in the space around every mass or group of masses. Mathematically, it is F/m

A gravitational field is a model used to explain the influence that a massive body extends into the space around itself, producing a force on another massive body. 

  1. True

  2. False


Correct Option: A
Explanation:
  • The gravitational field is the actually the force experimented by a unit mass due to a body of mass M.
  • So when a unit mass is kept at a distance 'r' from mass 'M' then it experiences a force which is equal to $\dfrac{GM}{r^2}$[as m=1],which is nothing but the gravitational field of mass M.
  • Its units are $\dfrac{N}{kg}$
  • Hence the above statement is true.

If the distance between two particles is reduced to half, the gravitational attraction between them will be

  1. Halved

  2. Quadrupled

  3. Doubled

  4. Reduced to a quarter


Correct Option: B
Explanation:

Gravitational force $=\dfrac { G{ m } _{ 1 }{ m } _{ 2 } }{ { r }^{ 2 } } ={ F } _{ 1 }$

If distance made $0$ half of original value
${ F } _{ 2 }=\dfrac { G{ m } _{ 1 }{ m } _{ 2 } }{ { \left( \dfrac { r }{ 2 }  \right)  }^{ 2 } } $
${ F } _{ 2 }=\dfrac { 4G{ m } _{ 1 }{ m } _{ 2 } }{ { r }^{ 2 } } $
${ F } _{ 2 }=4{ F } _{ 1 }$

Under the force of gravity, a heavy body falls quicker than a light body (neglect air resistance).

  1. True

  2. False


Correct Option: B
Explanation:

According to newton's laws the time taken to fall is not related to mass

$S=ut+\dfrac { 1 }{ 2 } { at }^{ 2 }$
${ V }^{ 2 }-{ u }^{ 2 }=2as$
It have not to do anything with mass so both lighter and heavier body will taken same time to fall from certain point.

Three particles of masses $2m, m$ and $2m$ are at the vertices $A, B$ and $C$ of an equilateral triangle $ABC$ of side length $'l'$. Then the intensity if gravitational field at the mid point of side $BC$ is:-

  1. $\dfrac{\sqrt{208}}{3} \dfrac{Gm}{l^2}$

  2. $\dfrac{\sqrt{59}}{3} \dfrac{Gm}{l^2}$

  3. $\dfrac{\sqrt{142}}{3} \dfrac{Gm}{l^2}$

  4. $\dfrac{\sqrt{308}}{3} \dfrac{Gm}{l^2}$


Correct Option: A

PRESSURE AND KINETIC INTERPRETATION OF TEMPERATURE
At what temperature the mean kinetic energy of hydrogen molecules increases to such that they will escape out of the gravitational field of earth for over?
take $({ v } _{ c }=11.2km/sec)$

  1. 12075 K

  2. 10000 K

  3. 20000 K

  4. 10075 K


Correct Option: D

The gravitational field in a region is given by $\vec {g} = 2\hat {i} + 3\hat {j} m/s^{2}$. The work done in moving a particle of mass $1\ kg$ from $(1, 1)$ to $\left (2, \dfrac {1}{3}\right )$ along the time $3y + 2x = 5$ is

  1. Zero

  2. $20\ J$

  3. $-15\ J$

  4. $18\ J$


Correct Option: C

At some planet gravitational acceleration is $1.96m/sec^{ -2 }$. If is safe to jump from a height of 2 m on earth, then what should be the corresponding safe height for jumping on the planet:

  1. 5 m

  2. 2 m

  3. 10 m

  4. 20 m


Correct Option: C
Explanation:

G on earth=9.8

ratio between earth and planet$=9.8:1.96$
= 5
so the safe height on the planet is 5 time greater than earth because
gravitational pull of that planet is 5 time less than earth
so required height = height on earth $\times$ 5
$2 \times 5$m
=10 m is the safe height on that planet.
Hence,
option $C$ is correct answer.

A body is acted upon by a constant force directed towards a fixed point. The magnitude of the force varies inversely as the square of the distance from the fixed point then path can be described by an equation similar to:

  1. $y=mx+c$

  2. ${ x }^{ 2 }+{ y }^{ 2 }={ r }^{ 2 }$

  3. $y=c{ x }^{ 2 }$

  4. none of these


Correct Option: D
Explanation:

The force field is similar to gravitational field.
So, the path is parabolic or elliptical or hyperbolic.
The exact trajectory will depend on the masses and  energy of the particle.
Option(A) is equation of a straight line, it is ruled out.
Option(B) is equation for a circle and is also eqn of an ellipse with length of major axis equal to minor axis.
Option(C) is eqn of a parabola.
Hence, (D) is the best option.

Potantial (V) at a point in space is given by $v = x^2 + y^2 + z^2$. Gravitational field at a point (x, y, z) is 

  1. $-2 x \hat{i} - 2 y \hat{j} - 2 z \hat{k}$

  2. $2 x \hat{i} + 2 y \hat{j} + 2 z \hat{k}$

  3. $x \hat{i} + y \hat{j} - z \hat{k}$

  4. $-x \hat{i} - y \hat{j} - z \hat{k}$


Correct Option: A
Explanation:

$v=x^2+y^2+z^2$

$E = \dfrac{{ - dv}}{{dx}}$

$E =  - \left[ {2\hat x + 2y\hat j + 2z\hat k} \right]$

$E =  - 2\hat x - 2y\hat j - 2z\hat k$
So, option $A$ is correct.

A large object is placed at exactly $65$% of the distance to the moon from the earth. Find out correct statement about the object ?

  1. Fall to the sun

  2. Fall to the moon

  3. Fall to the earth

  4. Remain in the same place

  5. Drift out of the solar system


Correct Option: C
Explanation:

The mass of moon is $1.2$% of that of earth. Hence $M _{moon}=0.012M _{earth}$

The acceleration due to gravity of the earth=$\dfrac{GM _{earth}}{(0.65R)^2}$
The acceleration due to gravity of the moon=$\dfrac{GM _{moon}}{(0.35R)^2}$
Hence $g _{earth}>g _{moon}$, and thus object falls towards the earth.

What is the time period satellite near the earth surface (neglect the height of orbit of satellite from the surface of ground)?

  1. $30.53 \ minutes$

  2. $50.38 \ minutes$

  3. $52.68 \ minutes$

  4. $84.75 \ minutes$


Correct Option: D
Explanation:

$T =$ Time period

$R=$ radius of earth $= 6400000m$
Gravitational constant $= G=$ $ 6.67\times {10}^{-11} Nm^2/kg^2$
$M=$ mass of earth $=$ $6\times{10}^{24}kg$

$T=$ $2\pi\sqrt{\dfrac {R^3}{GM}}$
Substitute and calculate
Whatever answer you get, divide it by $60$ so that it gets converted into minutes.

Two block of mass $10 kg$ and $20kg$ is separated by a distance $100 km$. What is the gravitational field (F) if $G=6.674 08 \times  10^{-11} m^3 kg^{-1} s^{-2}$?

  1. $13.34 \times 10^{-16} N$

  2. $1.334 \times 10^{-11} N$

  3. $ 1334 \times 10^{-16} N$

  4. $1.334 \times 10^{-18} N$


Correct Option: D
Explanation:
Given : $m _1 = 10kg$  $m _2 = 20 kg$  $r =10^5 m$
Gravitational force between the blocks $F = \dfrac{G m _1 m _2}{r^2}$
$\therefore$ $F = \dfrac{6.67408 \times 10^{-11} \times 10\times 20}{(10^5)^2}$
$\implies$ $1.334 \times 10^{-18} N$

If the radius of the earth were to shrink and its mass were to remain the same, the acceleration due to gravity on the surface of the earth with?

  1. Increase

  2. Decrease

  3. Remains same

  4. Zero


Correct Option: A
Explanation:

Gravitational force between two masses = $\quad K\dfrac { m1m2 }{ { r }^{ 2 } } \quad \quad $

Where

K is universal gravitational constant $m _1$ and $m _2$ are the masses $r$ is the distance between the masses.

Now , the acceleration due to $ m1 $ being earth will be = $ K\dfrac { m1 }{ { r }^{ 2 } } \quad \quad $

If the radius decreases and the mass remains same the acceleration will increase.

What is the gravitational field strength at the surface of Jupiter (mass $1.9\times 10^{27} kg$, radius $7.1\times 10^7 m$?

  1. $25 N kg^{-1}$

  2. $25 N kg^{-2}$

  3. $35 N kg^{-1}$

  4. $55 N kg^{-2}$


Correct Option: A
Explanation:

Gravitational field = $ K\dfrac { m1 }{ { r }^{ 2 } } \quad $

where $K$ is universal gravitational constant $m _1$ is the mass $r$ is the distance. 

G$ 6.67\times { 10 }^{ -11 }\dfrac { 1.9({ 10 }^{ 27 }) }{ { (7.1\times { 10 }^{ 7 }) }^{ 2 } } \quad \quad $ 

   =  $\quad 25N{ kg }^{ -1 }\quad $

Both earth and moon are subject to the gravitational force to the sun. As observed from the sun, the orbit of the moon 

  1. will be elliptical.

  2. will not be strictly elliptical because the total gravitational force on it is not central.

  3. is not elliptical but will necessarily be closed curve.

  4. deviates considerably from being elliptical due to influence of plants other than earth.


Correct Option: B
Explanation:

If the moon were to experience only the gravitation of the earth only then the path would have been elliptical but in the present case moon experience the gravitational force of sun as well which distorts the path.

The fours basic forces in nature are.
I. Gravitational force
II. Electromagnetic force
III. Strong nuclear force
IV. Weak nuclear force
The relative magnitudes of these forces are in the order of.

  1. III $>$ II $>$ I $>$ IV

  2. III $>$ II $>$ IV $>$ I

  3. I $>$ II $>$ III $>$ IV

  4. III $>$ I $>$ II $>$ IV


Correct Option: B
Explanation:
Actually the order of the four basic forces in nature in order of their relative magnitudes are strong nuclear force $>$ Electromagnetic force $>$ Weak nuclear force $>$ Gravitational force.

Read the following statements.
I. A magnetic field is unable to penetrate into a superconductor.
II. An electric field can be cut off by a screen of conducting material
III. Gravitational field can be freely transmitted through all bodies
Which statement is false?

  1. III

  2. II

  3. I

  4. None of these


Correct Option: D
Explanation:

A. stronout feels weightless inside an artificial satellite because nothing is stopping their fall due to force of gravity.

Three particles each of mass m are kept at verticles of an equilateral triangle of side L. The gravitational field at centre due to these particle is:

  1. Zero

  2. $\frac{3GM}{L^2}$

  3. $\frac{9GM}{L^2}$

  4. $\frac{12}{\sqrt{3}}\frac{GM}{L^2}$


Correct Option: A
Explanation:

We know that for an equilateral triangle the line joining the center of gravity to each vertex of the triangle are each at angle $120^0$ and we also know that the field by each mass will be along these lines joining the center and vertex.

Also as the triangle is equilateral so all the vertex will be at same distance from the center, so the filed produced by each mass will be same.
Now as the field produced are eqaul in magnitude and at angle $120^0$ so 
net field will be $zero.$

The gravitational field lines are

  1. Directed inwards towards a particle

  2. Directed outwards from a particle

  3. Directed along the particle's motion

  4. Directed perpendicular to the particle's motion


Correct Option: A
Explanation:

The gravitational field lines are directed inwards towards a particle because at any point on the earth's field, a body will feel a force directed towards the center of the earth. The field lines becomes more spread out as the distance form the earth increases, which indicates the diminishing strength of the field

Two masses m and  100m are kept at points A and B. The gravitational field lines

  1. Will be crowded at A than B

  2. Will be crowded at B than A

  3. Will be crowded equally at A and B

  4. Diverge from both the masses


Correct Option: B
Explanation:

The strength of a gravitational field is given by the number of lines crowding at a point. This field strength is given by $GM/R^2$. Thus, it is proportional to the mass of the object

Larger the mass, more the field intensity and hence the number of lines of forces

Thus, the correct option is (b)

The field strength for a planet A of mass M and radius R is F. In another planet, the density is found to be 27 times the density of the planet A and the radius of the new planet is one third of A. Then,

  1. Number of lines of force in both A and B are same

  2. Number of lines of force in A is more than B

  3. Number of lines of force in B is more than A

  4. Number of lines of force cannot be determined with this information


Correct Option: A
Explanation:

Number of lines of force is given by the flux $\phi=\int (g.dA)=(GM/R^2)4 \pi R^2 = 4 \pi GM$

Thus, number of lines of forces is proportional to M. 

So, $M _A= \rho (4 \pi R^3)/3$ AND $M _B=27 \rho (4 \pi [R/3]^3)/3=\rho (4 \pi R^3)/3$

The mass of both the planets are same AND hence the flux is also same for both the planets

The correct option is (a)

There are _____ gravitational lines of force inside a spherically symmetric shell

  1. Infinitely many

  2. Zero

  3. Varying number depending upon surface area

  4. Varying number depending upon volume


Correct Option: B
Explanation:

As there is no gravitational field in the shell, there are zero gravitational lines of force inside a spherically symmetric shell.

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