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NEET • PHYSICS • GRAVITATION

Gravitation practice · Medium

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Showing 31-60 of 135 questions
31 Medium
31 of 135
Analyzing +4 | -1
The correct option is
View answer and explanation
Correct option: B
The time taken in travelling DAB  DAB\; is greater than that for BCDBCD
Option B (The time taken in travelling DABDAB is greater than that for BCDBCD) is correct because the gravitational potential is lower closer to the planet, resulting in higher speeds and shorter travel times for paths closer to the planet.
32 Medium
32 of 135
Application +4 | -1
A spherical planet has uniform density π2×104kgm3\frac{\pi }{2} \times {10^4}\frac{{kg}}{{{m^3}}} . Find out the minimum period for a satellite in a circular orbit around it in seconds. [use  G=203×10−11N−m2kg2]\left[ {use\,\,G = \frac{{20}}{3} \times {{10}^{ - 11}}\frac{{N - {m^2}}}{{k{g^2}}}} \right]
View answer and explanation
Correct option: B
3000
Option B (3000) is correct because the minimum period is related to the density and gravitational constant.
33 Medium
33 of 135
Analyzing +4 | -1
Statement I: Earth continuously attracts the moon towards its centre and yet the moon does not move towards the centre of earth.

Statement II: The gravitational force of atteraction between the moon and earth provides the necessary centripetal force.
View answer and explanation
Correct option: D
If Statement - I is false but Statement - II is true.
Option D (Statement - I is false but Statement - II is true) is correct because the moon *does* move towards the center of the earth (it's constantly falling), and the gravitational force provides the centripetal force.
34 Medium
34 of 135
Application +4 | -1
A geostationary satellite is orbiting the earth at a height of 5R5R above the surface of the earth, RR being the radius of the earth. The time period of another satellite in hours at a height of 2R2R from the surface of the earth is
View answer and explanation
Correct option: C
626\sqrt 2
Option C (626\sqrt 2) is correct because it is derived from Kepler's Third Law, relating the time period to the orbital radius.
35 Medium
35 of 135
Remembering +4 | -1
LANDSAT series of satellites move in near polar orbits at an altitude of
View answer and explanation
Correct option: C
918 km
Option C (918 km) is correct because LANDSAT satellites are designed to orbit at this specific altitude for optimal Earth observation.
36 Medium
36 of 135
Understanding +4 | -1
The acceleration due to gravity is gg at a point distant rr from the centre of earth of radius RR. If r<Rr < R, then
View answer and explanation
Correct option: A
g∝rg \propto r
Option A (g∝rg \propto r) is correct because inside the Earth, the acceleration due to gravity is directly proportional to the distance from the center.
37 Medium
37 of 135
Understanding +4 | -1
The escape velocity for a body projected vertically upwards from the surface of earth is 11  kms−111\;\mathrm{km}{\mathrm{s}^{ - 1}}. If the body is projected at an angle of 450{45^0} with the vertical, the escape velocity will be
View answer and explanation
Correct option: C
11  kms−111\;\mathrm{km}{\mathrm{s}^{ - 1}}
Option C (11  kms−111\;\mathrm{km}{\mathrm{s}^{ - 1}}) is correct because escape velocity is independent of the angle of projection.
38 Medium
38 of 135
Application +4 | -1
In a satellite, if the time period of revolution is TT, then kinetic energy is proportional to
View answer and explanation
Correct option: D
T−2/3{T^{ - 2/3}}
Option D (T−2/3{T^{ - 2/3}}) is correct because the kinetic energy of a satellite is inversely proportional to the period raised to the power of 2/3.
39 Medium
39 of 135
Analyzing +4 | -1
Spot the wrong statement:
The acceleration due to gravity 'g' decreases if
View answer and explanation
Correct option: C
We go from the equator towards the poles on the surface of the earth
Option C (We go from the equator towards the poles on the surface of the earth) is the wrong statement because 'g' *increases* as we move from the equator towards the poles due to the Earth's shape.
40 Medium
40 of 135
Understanding +4 | -1
If ge,  gh and gd{g_e},\;{g_h}\,and\,{g_d} be the accelerations due to gravity at earth’s surface, a height hh and at depth dd respectively. Then
View answer and explanation
Correct option: B
ge>gh<gd{g_e} > {g_h} < {g_d}
Option B (ge>gh<gd{g_e} > {g_h} < {g_d}) is correct because gravity decreases with height and increases with depth initially.
41 Medium
41 of 135
Analyzing +4 | -1
The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity
View answer and explanation
Correct option: D
cannot be zero at any point.
Option D (cannot be zero at any point) is correct because even with varying density, gravity will always exert a force.
42 Medium
42 of 135
Analyzing +4 | -1
A body of mass mm is moving in a circular orbit of radius RR about a planet of mass MM. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius R2\frac{R}{2}. and the other mass, in a circular orbit of radius 3R2\frac{3 R}{2}. The difference between the final and initial total energies is :
View answer and explanation
Correct option: C
−GMm6R-\frac{G M m}{6 R}
Option C (−GMm6R-\frac{G M m}{6 R}) is correct because it represents the difference between the final and initial total energies after the mass splits and moves into different orbits.
43 Medium
43 of 135
Understanding +4 | -1
An astronaut orbiting the earth in a circular orbit 120 km above the surface of earth, gently drops a spoon out of space-ship. The spoon will
View answer and explanation
Correct option: C
Will move along with space-ship
Option C (Will move along with space-ship) is correct because the spoon has the same initial velocity as the spaceship and will continue in the same orbit.
44 Medium
44 of 135
Application +4 | -1
A cyclist turns around a curve at 15 miles/hour. If the turns at double the speed, the tendency to overturn is
View answer and explanation
Correct option: B
Quadrupled
Option B (Quadrupled) is correct because the tendency to overturn is proportional to the square of the velocity.
45 Medium
45 of 135
Application +4 | -1
If gravitational force on a body of mass 1.5 kg at point is 45N, then the intensity of the gravitational field at that point is
View answer and explanation
Correct option: C
30  N  kg−130\mathrm{\;N\;k}{\mathrm{g}^{ - 1}}
Option C (30  N  kg−130\mathrm{\;N\;kg}^{ - 1}) is correct because the intensity of the gravitational field is the gravitational force per unit mass.
46 Medium
46 of 135
Remembering +4 | -1
Statement(A): For Geostationary
satellites, height above the earth is
about 35,800 km.

Statement(B): For Polar satellites,
height above the earth is about 500 to
800 km.

Statement(C): For Geostationary
satellites, T = 24 hours.

Statement(D): For Polar satellites, time
period is around 100 minutes.
View answer and explanation
Correct option: D
All are true
Option D (All are true) is correct because all the statements about Geostationary and Polar satellites are factually accurate.
47 Medium
47 of 135
Remembering +4 | -1
What is the distance of a geostationary satellite from the Earth's centre?
View answer and explanation
Correct option: A
4.22×104 km4.22 \times 10^4 \mathrm{~km}
Option A (4.22×104 km4.22 \times 10^4 \mathrm{~km}) is correct because this is the approximate distance from the Earth's center at which a geostationary satellite orbits.
48 Medium
48 of 135
Remembering +4 | -1
Earth binds the atmosphere because of
View answer and explanation
Correct option: A
Gravity
Option A (Gravity) is correct because Earth's gravitational pull is the primary force holding the atmosphere.
49 Medium
49 of 135
Application +4 | -1
If orbital velocity of planet is given by v=GaMbRcv = {G^a}{M^b}{R^c}, then
View answer and explanation
Correct option: B
a=1/2,  b=1/2,  c=−1/2a = 1/2,\;b = 1/2,\;c = - 1/2
Option B (a=1/2,  b=1/2,  c=−1/2a = 1/2,\;b = 1/2,\;c = - 1/2) is correct because the orbital velocity formula is v=GMR=G1/2M1/2R−1/2v = \sqrt{\frac{{GM}}{R}} = G^{1/2}M^{1/2}R^{-1/2}.
50 Medium
50 of 135
Analyzing +4 | -1
A satellite is moving around the earth in a circular orbit. The following statements are given

(i)It is moving with a constant velocity.

(ii) It suffers no acceleration.

(iii) Its angular momentum w.r.t. the earth remains conserved.

(iv) Its distance from centre must be equal to 2\sqrt{2} times of earth's radius.

The correct option is
View answer and explanation
Correct option: C
Only (iii) is true.
Option C (Only (iii) is true) is correct because a satellite in a circular orbit has constant speed but changing velocity (due to changing direction), experiences acceleration, and its angular momentum is conserved.
51 Medium
51 of 135
Understanding +4 | -1
Gravitational potential on the surface of earth is (M=mass of the earth, R= radius of earth)
View answer and explanation
Correct option: B
−gR- \mathrm{g}R
Option B (−gR- \mathrm{g}R) is correct because gravitational potential on the surface of the earth is given by V=−GM/RV = -GM/R, and since g=GM/R2g = GM/R^2, then V=−gRV = -gR.
52 Medium
52 of 135
Understanding +4 | -1
Orbital velocity of an artificial does not depend upon
View answer and explanation
Correct option: B
Mass of the satellite
Option B (Mass of the satellite) is correct because the orbital velocity formula does not include the mass of the satellite.
53 Medium
53 of 135
Understanding +4 | -1
A space ship moves from earth to moon and back. The greatest energy required for the space ship is to overcome the difficulty in
View answer and explanation
Correct option: B
Take off from earths field
Option B (Take off from earths field) is correct. Earth's gravity is much stronger than the moon's, requiring more energy to escape.
54 Medium
54 of 135
Analyzing +4 | -1
In the following four periods
(i) Time of revolution of a satellite just above the earth’s surface (Tst)\left( {{T_{st}}} \right) , (ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth (Tma)\left( {{T_{ma}}} \right) , (iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8N/kg(Tsp)9.8N/kg\left( {{T_{sp}}} \right) , (iv) Period of an infinite length simple pendulum in the earth’s real gravitational filed (Tis)\left( {{T_{is}}} \right)
View answer and explanation
Correct option: C
Tsp>Tis{T_{sp}} > {T_{is}}
Option C (Tsp>Tis{T_{sp}} > {T_{is}}) is correct because the period of a simple pendulum with length equal to Earth's radius in a uniform field is greater than that of an infinite length pendulum in Earth's real field.
55 Medium
55 of 135
Application +4 | -1
The centripetal acceleration of a body moving in a circle of radius 100 m with a time period of 2 s will be
View answer and explanation
Correct option: D
985.9  ms  −2985.9\;\mathrm{ms}{\mathrm{\;}^{ - 2}}
Option D (985.9  ms  −2985.9\;\mathrm{ms}{\mathrm{\;}^{ - 2}}) is correct because it is the centripetal acceleration calculated using the formula ac=4π2rT2a_c = \frac{{4\pi^2r}}{{T^2}}.
56 Medium
56 of 135
Understanding +4 | -1
A cylindrical vessel partially filled with water is rotated about its vertical central axis. It’s surface will
View answer and explanation
Correct option: B
Rise from the sides
Option B (Rise from the sides) is correct because when a cylindrical vessel filled with water is rotated, the water surface takes a parabolic shape, rising at the sides and dipping in the middle.
57 Medium
57 of 135
Analyzing +4 | -1
Assertion: The work done to project a body of mass mm from the earth's surface (radius RR ) to infinity is mgRm g R.

Reason: The required work done is equal to change in potential energy of the mass, which is equal to GMm/RG M m / R.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of the assertion.
Option A (If both assertion and reason are true and reason is the correct explanation of the assertion) is correct because the work done equals the change in potential energy from the surface to infinity.
58 Medium
58 of 135
Understanding +4 | -1
If ve and vo{v_e}\,and\,{v_o} represent escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R, then
View answer and explanation
Correct option: B
2vo=ve\sqrt 2 {v_o} = {v_e}
Option B (2vo=ve\sqrt 2 {v_o} = {v_e}) is correct because the escape velocity is 2\sqrt{2} times the orbital velocity.
59 Medium
59 of 135
Understanding +4 | -1
The gravitational force between two point masses m1 and m2{m_1}\,and\,{m_2} at separation rr is given by F=km1m2r2F = k\frac{{{m_1}{m_2}}}{{{r^2}}}. The constant kk
View answer and explanation
Correct option: A
Depends on system of units only
Option A is correct because the gravitational constant's numerical value depends on the system of units used (e.g., SI or CGS).
60 Medium
60 of 135
Analyzing +4 | -1
Statement 1: Value of acceleration due to gravity is same at all the points inside earth assuming it to be made up of uniform density.
Statement 2: Value of gravitational field increases as we go towards centre in a uniform spherical shell.
View answer and explanation
Correct option: D
Both statement 1 and statement 2 are false.
Option D (Both statement 1 and statement 2 are false) is correct because the acceleration due to gravity decreases linearly inside the Earth, and the gravitational field is zero inside a uniform spherical shell.
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