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Multiple Choice Questions
Question 1
Which of the following is true for Uniform Solid sphere of Mass M and radius b
(a) for r > b,Gravitational field decreases as r is increased
(b) for r< b, Gravitational field increased as r is increased
(c) Gravitational Field at r=0 is equal to 0
(d) Gravitational field at the r=b is $\frac {GM}{b^2}$
Solution
Gravitational Field at outside point is given by
$=\frac{GM}{r^2}$
Gravitational Field at internal point is given by
$=\frac{GM}{b^3}r$
So all the statement are true
Question 2
Let P denotes the Gravitational potential and F denotes the Gravitational Field
What all relation are true for a uniform spherical shell
(a) P≠0 inside the shell and it is constant inside the shell
(b) E≠0 inside the shell and it is constant inside the shell
(c) E=0 at all the point inside the shell
(d) none of these
Solution
Electric field inside the spherical shell is zero but Potential is not zero
Potential is same inside the shell
So a and c are correct
Paragraph Based Questions
(A) A thin rod of length L and mass M is placed along x axis.One end of rod is placed at origin and other end is at (L,0).
Let
i and
j are the unit vector across x and y axis
Question 3:
Find the Gravitational field at Point P( a,0) where a > L
(a) $\frac {GM}{a(L+a)}\mathbf{i}$
(b) $\frac {GM}{a(a-L)}\mathbf{i}$
(c) $\frac {GM}{L-a}\mathbf{i}$
(d ) None of these
Solution
Let us consider a small element of length dx at a distance x from the origin
Mass of the element=(M/L)dx
Field due to this mass element at point P and it will be directed towards x axis
$dE=\frac{GM}{L}\frac{dx}{(a-x)^2}$
Total field strength
$E=\int_{0}^{L}{\frac{GM}{L}\frac{dx}{(a-x)^2}}$
$=\frac{GM}{a(a-L)}$
Question 4
Find the gravitational potential at point P .We take gravitational potential to be zero at infinity
(a) $\frac {GM}{L} ln \frac {a+L}{a}$
(b) $\frac {GM}{L} ln \frac {a}{a -L}$
(c) $\frac {GM}{L} ln \frac {a-L}{a+L}$
(d) $\frac {GM}{L} ln \frac {a-L}{a}$
Solution
Let us consider a small element of length dx at a distance x from the origin
Mass of the element=(M/L)dx
Potential due to this mass element at point P
$dV=\frac{GM}{L}\frac{dx}{(a-x)}$
Total Potential
$V=\int_{0}^{L}{\frac{GM}{L}\frac{dx}{(a-x)}}$
$=\frac{GM}{L}ln{\frac{a-L}{a}}$
(B)Three particles A,B,C each of mass m are placed at the corner of the equilateral triangle of side a.
Question 5
Find the potential energy of the system
(a) $ - \frac {3Gm^2}{a}$
(a) $ - \frac {3Gm^2}{2a}$
(a) $ - \frac {Gm^2}{3a}$
(d) None of these
Solution
Potential energy of the system
U=Potential energy AB + Potential energy BC + Potential energy AC
$U=\frac{-3Gm^2}{a}$
Question 6
Find the Gravitational Potential at point A
(a)$-\frac {2Gm}{a}$
(b)$-\frac {Gm}{a}$
(c) $-\frac {3Gm}{a}$
(d) None of these
Solution
Potential at point A = Potential due to B + Potential due to C
$ =\frac{-Gm}{a}-\frac{Gm}{a}$
$ =\frac{-2Gm}{a}$
Question 7
Find the work done on this system if the side of the triangle is changed from a to 3a
(a) $\frac {4Gm^2}{a}$
(b) $\frac {3Gm^2}{a}$
(c) $\frac {Gm^2}{a}$
(d) $\frac {2Gm^2}{a}$
Solution
Potential energy of the system after the change
$ U=\frac{-3Gm^2}{3a}=\frac{-Gm^2}{a}$
So work done
$=\frac{-Gm^2}{a}-\frac{-3Gm^2}{a}$
$=\frac{2Gm^2}{a} $
Multiple Choice Questions
Question 8
Three concentric shells A,B ,C of uniform density have masses M
_{1} ,M
_{2} and M
_{3} respectively. There radii are R
_{1},R
_{2} and R
_{3 }respectively.
R
_{3} > R
_{2} > R
_{1}.
A particle of mass m is placed at placed at a distance d
Match the column
Column A ( it tells us about the location)
(P) R
_{1} < d < R
_{2}
(Q) d < R
_{1}
(R) R
_{1} < R
_{2} < d < R
_{3}
(S) R
_{1} < R
_{2} < R
_{3} < d
Column B ( it tells us about the gravitational force experienced)
(D) $\frac {G(M_1 + M_2 + M_3)m}{d^2}$
(E) $\frac {G(M_1 + M_2)m}{d^2}$
(F) $\frac {GM_1m}{d^2}$
(G) 0
Solution
For $R_1 < d < R_2$
Shell A will exert force on mass m
$d < R_1$
mass is inside all the shell .So no force
$R_1 < R_2 < d < R_3$
Shell A and B will exert force on mass m
$R_1 < R_2 < R_3 < d$
Shell A , B and C will exert force on mass m
Question 9
A satellite of mass m is orbiting the earth in a circular orbit of radius R.
Statement I: Angular Momentum of the satellite varies as $\frac {1}{\sqrt {R}}$
Statement II: Linear momentum of the satellite varies as 1/R
Statement III: Kinetic energy varies as 1/R
Statement IV: Frequency of revolution varies as $\frac {1}{R^{3/2}}$
Which of the following is correct
(a) All the statements are correct
(b) Statement I,II,III are correct only
(c) Statement II,IV are correct only
(d) Statement I,II,IV are correct only
Solution
Kinetic energy=$\frac{1}{2}mv^2$
We know that
$v=\sqrt{\frac{GM}{R}}$
Where M is the mass of the earth
So KE varies with 1/R
Angular momentum=$mvR=m\sqrt{GMR}$
So it varies with $\sqrt R$
Linear momentum=mv
So it varies as $\frac{1}{\sqrt R}$
Frequency of revolutions
$=\frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{GM}{R^3}}$
So frequency varies as $\frac{1}{R^{3/2}}$
So c is correct
Question 10
The force of gravitation is
(a) Repulsive
(b) Electrostatic
(c) Conservative
(d) Non-conservative
Solution
(c)
Question 11
Mass M is divided into two parts bM and (1-b)M. For a given separation, the value of b for which the gravitational attraction between the two pieces becomes maximum is
(a) 1/2
(b) 3/5
(c) 1
(d) 2
Solution
Answer is (a)
$F \alpha bm \times (1-b)m= bm^2(1-b)$
For maximum force $\frac {dF}{dx}=0$
$\frac {dF}{dx}=m^2-2bm^2=0$
or b=1/2
Question 12
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of earth
(a) The acceleration of S is always directed towards the centre of the earth
(b) The angular momentum of S about the centre of the earth changes in direction but its magnitude remains constant
(c) The total mechanical energy of S varies periodically with time
(d) The linear momentum of S remains constant in magnitude
Solution
(a) is correct
Question 13
A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above the earth?s surface ($R_{Earth}=6400km$) will approximately be
(a) 1/2 h
(b) 1 h
(c) 2 h
(d) 4 h
Solution
Correct Answer: C
$\frac {T_2}{T_1}=(\frac {r_2}{r_1})^{3/2}$
$T_2=24(\frac {6400}{36000})^{3/2} =2 \ hour$
Question 14
Which of the following options are correct?
(a) Acceleration due to gravity decreases with increasing altitude.
(b) Acceleration due to gravity increases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity increases with increasing latitude.
(d) Acceleration due to gravity is independent of the mass of the earth.
Solution
(a) and (c) are correct
Question 15
Two bodies of masses $m_1$ and $m_2$ are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is
(a)$\sqrt {2G \frac {m_1 - m_2}{r}}$
(b)$\sqrt {2G \frac {m_1 + m_2}{r}}$
(c)$\sqrt {2G \frac {m_1 m_2}{r}}$
(d)$\sqrt {2G \frac {r}{m_1m_2}}$
Solution
Let velocities of these masses at r distance from each other be $v_1$ and $v_2$ respectively.
Law of conservation of Momentum
$m_1v_1-m_2v_2=0$ or $m_1v_1=m_2v_2$
By conservation of energy
change in P.E.=change in K.E.
$\frac {Gm_1m_2}{r}=\frac {1}{2}m_1v_1^2+\frac {1}{2}m_1v_2^2 $
On solving equation ,we get
$v_1=\sqrt {\frac {2Gm_2^2}{r(m_1+m+2)}}$
and $v_2=\sqrt {\frac {2Gm_1^2}{r(m_1+m+2)}}$
Now relative velocity of approach at a separation distance r = $|v_1| + |v_2| = \sqrt {2G \frac {m_1 + m_2}{r}}$
Question 16
If the mass of sun were ten times smaller and gravitational constant G were ten times larger in magnitudes-
(a) walking on ground would became more difficult.
(b) the acceleration due to gravity on earth will not change.
(c) raindrops will fall much faster.
(d) airplanes will have to travel much faster
Solution
(a), (c), (d)
Subjective Questions
Question 17
A cosmic body A of mass m moves to the sun with velocity v
_{0} ( when far from sun) and aiming parameter d the arm of the velocity v
_{0} relative to the center of the sun.
(a) Find the angular momentum of the body about the sun at the starting point
(b) Find the minimum distance which this body will get to the sun
(c) Find the velocity at minimum separation
(d) Find the total energy of the system at the minimum distance
Mass of the sun =M
_{s}
Solution
At the minimum separation with the sun,the cosmic body’s velocity is perpendicular to its position vector relative to the sun
Let v be the velocity of the cosmic at minimum separation
And r be minimum separation distance
Now Angular momentum about Sun at starting
$=mv_0d$
Angular momentum at minimum separation
$= mvr$
Since no torque acts through out the motion.,angular momentum is conserved
$mv_0d= mvr$
or
$v=\frac{v_0d}{r}$
Now applying law of conservation of energy
$\frac{1}{2}mv_0^2=\frac{1}{2}mv^2-\frac{GM_sm}{d}$
Substituting the value of v in the above equation
$v_0^2r^2+2GM_sr-v_0^2d^2=0$
Or
$r=\frac{GM_s}{v_0^2}\left(\sqrt{1+\frac{d^2v_0^2}{G^2M_s^2}}-1\right)$
So velocity at minimum separation
$v=\frac{v_0^3d}{GM\left(\sqrt{1+\frac{d^2v_0^2}{G^2M_s^2}}-1\right)}$
Question 18
A planet of mass M moves along a ellipse around the sun so that its maximum and minimum distance from the sun are equal to a and b respectively
(a) Find the velocities at the minimum and maximum distance
(b) Find the Total energy of the system
(c) Find the angular momentum of the planet relative to the center of the sun
(d) At a point P when at a distance r
_{0} from the sun,the angle between velocity vector and radius vector is θ. Find the velocity at the point P
(e) find the ecentricty and semi major axis of the ellipse
Mass of the sun =m
Solution
At minimum and maximum distance positions of the planet,the velocity vector and position vector makes right angles with each other
Let $v_1$ and $v_2$ be the velocities at minimum and maximum position
Since the force is central,Angular momentum is conserved
Applying law of conservation of angular momentum at minimum and maximum distance
$Mv_1a =Mv_2b$
Or $v_1a=v_2b$ -----(1)
Applying law of conservation of energy at minimum and maximum distance
$\frac{-GMm}{a}+\frac{1}{2}Mv_1^2=\frac{-GMm}{b}+\frac{1}{2}Mv_2^2$ ----(2)
Solving equation 1 and 2,we get
$v_1=\sqrt{\frac{2Gmb}{a(a+b)}}$
$v_2=\sqrt{\frac{2Gma}{b(a+b)}}$
Total energy of the system
$=\frac{-GMm}{a}+\frac{1}{2}Mv_1^2$
Substituting the value $v_1$
We get
Total energy
$=\frac{-GMm}{(a+b)}$
Angular momentum of the planet about Sun
$L=Mav_1$
Substituting the value of v1
$L=M\sqrt{\frac{2Gmab}{a+b}}$
Applying law of conservation of angular momentum at Point P and minimum distance point
$Mr_0vsin{\theta}=Mav_1$
Substituting the value of v1,
$v=\frac{a}{r_0sin{\theta}}\sqrt{\frac{2Gmb}{a(a+b)}}$
$v=\sqrt{\frac{2Gmab}{r_0^2{sin}^2{\theta}(a+b)}}$
Ecentricty is given by
$e=\frac{b-a}{b+a}$
Semi major axis=$\frac{a+b}{2}$
Question 19
A hole were drilled completely through the earth along a diameter.A particle of mass m is released into the hole from the surface of the earth
(a) find the force acting on the mass m at distance r from the center
(b) Show that the motion will be simple harmonic and Find out the time period of it
Let M be the mass of the Earth
R be the radii of the Earth
We assume that density is Uniform
Solution
Suppose the earth is divided in spherical shell.
For the spherical shell whose radii is less than r,Force on the particle m would be
$dF=\frac{-GmdM}{r^2}$
The force will be towards the center so the minus sign
For the spherical shell whose radii are greater then r,the force would be zero as the point will be internal to them
So total Force
$F=\frac{-Gm}{r^2}\int d M$
Where $\int d M$ is the mass of the sphere up to the radii r
Now mass per unit volume of Earth
$=\frac{M}{\frac{4}{3}\pi R^3}$
So
$\int{dM=\frac{M}{\frac{4}{3}\pi R^3}\frac{4}{3}\pi r^3}$
So force would be
$F=-\frac{GMmr}{R^3}$
Now the above equation can be written as
$ma=\frac{-GMm}{R^3}r$
Or
$a=-kr$
Where $k=\frac{GM}{R^3}$
Since acceleration is directed toward center and proportional to r, The particle will follow a Simple harmonic motion
Time period
$T=2\pi\sqrt{\frac{m}{k}}$
Question 20
Two satellite M and N are revolving around the earth in coplanar concentric orbits.
The Time period of the satellite are 2 h and 16 H respectively. The radius of the orbit of the satellite N is 3.2X 10
^{4} Km.The satellite are moving in opposite direction.
At t=0, the position of the satellite is shown in below figure
Find out the following things
(a) The time period of the satellite M
(b) Orbital velocity of both the satellite
(c) Find the angular velocity of satellite N as observed by M at t=8 hour
Solution
From Keplers law we know that
$T^2\alpha R^3$
So
$\frac{T_1}{T_2}=\left(\frac{R_1}{R_2}\right)^{3/2}$
Or
$R_1=\left(\frac{T_1}{T_2}\right)^{2/3}R_2$
$=\left(\frac{2}{16}\right)^{2/3}3.2*10^4$
$=.8*10^4$
Orbital Velocity of M
$v_1=\frac{2\pi R_1}{T_1}=\frac{2\pi X.8*10^4}{2}=.8\pi X10^4 \ Km/hr$
Orbital Velocity of N
$v_2=\frac{2\pi R_{21}}{T_2}=\frac{2\pi X3.2*10^4}{16}=.4\pi X10^4 \ Km/hr$
At t=8 Hour,the satellite are closest to each other.
So Angular velocity
$\omega=\frac{v_1+v_2}{R_2-R_1}=.5\pi \ rad/sec$
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