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JEE • PHYSICS • GRAVITATION

Gravitation practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 61-90 of 392 questions

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61 Medium
61 of 392
Analyzing +4 | -1
A system consists of N{N} identical particles of mass m{m} placed rigidly on the vertices of a regular polygon with each side of length ℓ\ell. If K1{K}_1 be the kinetic energy imparted to one of the particles so that it just escapes the gravitational pull of the system and there after kinetic energy K2{K}_2 is given by to the adjacent particle to escape, then the difference (K1−K2)\left({K}_1-{K}_2\right) is
View answer and explanation
Correct option: D
Gm2a\frac{G m^2}{a}
Option D (Gm2a\frac{G m^2}{a}) is correct because the difference in kinetic energies is equal to the gravitational potential energy difference between the last particle's interaction.
62 Medium
62 of 392
Analyzing +4 | -1
Assertion: On decreasing the distance between two masses potential energy increases.
Reason: Potential energy U=GMmrU=\frac{G M m}{r}.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because both the assertion and the reason are false; decreasing the distance between two masses decreases (becomes more negative) the potential energy, and the potential energy formula is actually U=−GMmrU = -\frac{{GMm}}{{r}}.
63 Medium
63 of 392
Understanding +4 | -1
Choose the incorrect statement from the following:
View answer and explanation
Correct option: C
The linear speed of a planet revolving around the sun remains constant
Option C ( The linear speed of a planet revolving around the sun remains constant) is the incorrect statement because the linear speed of a planet varies as it orbits the sun in an elliptical path.
64 Medium
64 of 392
Application +4 | -1
Self potential energy of a thin uniform shell of mass 2 kg and radius 10 cm is
View answer and explanation
Correct option: A
−20 G- 20\,G
Option A (-20G) is correct because the self-potential energy of a thin uniform shell is given by U = -Gm²/2R.
65 Medium
65 of 392
Application +4 | -1
Given mass of the moon is 1/811 / 81 of the mass of the earth and corresponding radius is 1/41 / 4 of the earth. If escape velocity on the earth surface is 11.2 km/s11.2 {~km} / {s}, the value of same on the surface of the moon is
View answer and explanation
Correct option: C
2.5 km/s2.5 {~km} / {s}
Option C (2.5 km/s2.5 {~km} / {s}) is correct because the escape velocity is proportional to the square root of (mass/radius), and the moon's mass and radius are given relative to Earth.
66 Medium
66 of 392
Application +4 | -1
A particle is projected vertically upward from the surface of the earth with a speed of 32gR,R\sqrt{\frac{3}{2} g R}, R being the radius of the earth and gg is the acceleration due to gravity on the surface of the earth. Then the maximum height ascended is (neglect cosmic dust resistance)
View answer and explanation
Correct option: B
3 R
Option B (3 R) is correct because it is derived from the conservation of energy principle, equating initial kinetic energy and potential energy to the final potential energy at maximum height.
67 Medium
67 of 392
Understanding +4 | -1
Assertion: The gravitational field of moon is much less than that of earth.
Reason: Gravitational field of a given mass (M)(M) depends upon M/r2{M} / {r}^2, which is smaller for moon.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the moon's mass and radius are smaller than Earth's, resulting in a weaker gravitational field.
68 Medium
68 of 392
Application +4 | -1
A body placed at a distance R0R_0 from the centre of earth, starts moving from rest. The velocity of the body on reaching at the earth's surface will be (Re=\left(\mathrm{R}_{\mathrm{e}}=\right. radius of earth and Me=\mathrm{M}_{\mathrm{e}}= mass of earth)
View answer and explanation
Correct option: D
2GMe(1Re−1R0)\sqrt{2 G M_e\left(\frac{1}{R_e}-\frac{1}{R_0}\right)}
Option D (2GMe(1Re−1R0)\sqrt{2 G M_e\left(\frac{1}{R_e}-\frac{1}{R_0}\right)}) is correct because it is derived from the conservation of energy principle, equating initial and final energies.
69 Medium
69 of 392
Understanding +4 | -1
The ratio of the escape velocity to the orbital velocity of a surface satellite of the earth
View answer and explanation
Correct option: A
2\sqrt 2
Option A (2\sqrt 2) is correct because the escape velocity is 2\sqrt{2} times the orbital velocity for a satellite orbiting close to the Earth's surface.
70 Medium
70 of 392
Application +4 | -1
Two identical satellite A and B are circulating round the earth at the height of R and 2R respectively. (where R is radius of the earth). The ratio of kinetic energy of A to that of B is
View answer and explanation
Correct option: D
32\frac{3}{2}
Option D (32\frac{3}{2}) is correct because the ratio of kinetic energies is 3/2, calculated by considering the heights of the satellites and their relation to kinetic energy.
71 Medium
71 of 392
Understanding +4 | -1
The total energy of a satellite is
View answer and explanation
Correct option: B
always negative
Option B is correct because the total energy of a satellite in orbit is always negative, indicating a bound system.
72 Medium
72 of 392
Application +4 | -1
The mean distance of a planet from the sun is approximately 14\frac{1}{4} times that of the earth from the sun. Find the number of years required for the planet to make one revolution about the sun. [years].
View answer and explanation
Correct option: A
0.125
Option A (0.125) is correct because it is the result of applying Kepler's third law relating orbital period and distance from the sun.
73 Medium
73 of 392
Application +4 | -1
The height above the surface of the earth, where acceleration due to gravity is (1/64)(1 / 64) th of its value at surface of the earth is approximately
View answer and explanation
Correct option: A
45×106 m45 \times 10^6 \mathrm{~m}
Option A (45×106 m45 \times 10^6 \mathrm{~m}) is correct because it is calculated using the formula for the variation of gravity with height.
74 Medium
74 of 392
Application +4 | -1
Two bodies of masses 10 kg and 20 kg are separated by a distance 5 m. Find the gravitational force attraction between two mass [G=6.67×10−11 Nm2/kg2]\left[ {G = 6.67 \times {{10}^{ - 11}}\,N{m^2}/k{g^2}} \right] is
View answer and explanation
Correct option: B
53.36×10−1153.36 \times {10^{ - 11}}
Option B (53.36×10−1153.36 \times {10^{ - 11}}) is correct because it is the gravitational force calculated using Newton's law of gravitation.
75 Medium
75 of 392
Application +4 | -1
A body is projected vertically upwards from the surface of planet of mass M\mathrm{M} and radius R\mathrm{R} with upward speed GMR\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}. The maximum height attained by body is:
View answer and explanation
Correct option: B
R\mathrm{R}
Option B (R) is correct because it's derived by equating the change in kinetic energy to the change in potential energy, considering the gravitational potential energy formula.
76 Medium
76 of 392
Understanding +4 | -1
If earth stands still what will be its effect on man's weight
View answer and explanation
Correct option: A
Increases
Option A (Increases) is correct because if the Earth stops spinning, the centrifugal force acting on a person at the equator disappears, effectively increasing their weight.
77 Medium
77 of 392
Understanding +4 | -1
The condition for a uniform spherical mass mm of radius rr to be a black hole is [G = gravitational constant and g = acceleration due to gravity]
View answer and explanation
Correct option: C
(2Gm/r)1/2≥c{\left( {2Gm/r} \right)^{1/2}} \ge c
Option C ((2Gm/r)1/2≥c{\left( {2Gm/r} \right)^{1/2}} \ge c) is correct because the escape velocity must be greater than or equal to the speed of light for a black hole.
78 Medium
78 of 392
Understanding +4 | -1
Choose the wrong option.
View answer and explanation
Correct option: D
Gravitational mass of a particle like proton can depend on the presence of neighouring heavy objects but the inertial mass cannot.
Option D is correct because gravitational and inertial mass are considered equivalent and do not depend on neighboring heavy objects.
79 Medium
79 of 392
Application +4 | -1
A chain of 125 links is 1.25 m long and has mass of 2 kg with the ends fastened together. It is set for rotating at 50 revs−1\mathrm{rev}{\mathrm{s}^{ - 1}}, centripetal force on each links is
View answer and explanation
Correct option: C
314 N
Option D (None of these) is incorrect. Option A (3.14 N) is incorrect. Option B (0.314 N) is incorrect. Option C (314 N) is correct because it is the centripetal force on each link, calculated using the formula F=mrω2F = mr\omega^2.
80 Medium
80 of 392
Understanding +4 | -1
If the earth stops rotating, the value of gg at the equator
View answer and explanation
Correct option: A
increases
Option A (increases) is correct because the effective gravitational acceleration at the equator is reduced due to Earth's rotation; stopping the rotation would eliminate this reduction, increasing g.
81 Medium
81 of 392
Remembering +4 | -1
According to Kepler's law of planetary motion, if TT represents time period and rr is orbital radius, then for two planets these are related as
View answer and explanation
Correct option: C
[T1T2]2=[r1r2]3\left[\frac{T_1}{T_2}\right]^2=\left[\frac{r_1}{r_2}\right]^3
Option C (T1T22=r1r23\frac{T_1}{T_2}^2=\frac{r_1}{r_2}^3) is correct because it represents Kepler's Third Law, which relates the period and orbital radius of planets.
82 Medium
82 of 392
Understanding +4 | -1
An iron ball and a wooden ball of the same radius are released from a beight hh in vacuum. The time taken by both to reach the ground is. The correct option is:
View answer and explanation
Correct option: A
Acceleration due to gravity in vacuum is same irrespective of size and mass of the body
Option A (Acceleration due to gravity in vacuum is same irrespective of size and mass of the body) is correct because in a vacuum, the only force acting on the balls is gravity, and the acceleration due to gravity is independent of mass.
83 Medium
83 of 392
Application +4 | -1
A mass ' m\mathrm{m} ' suspended from a spring stretches it by 5 cm5 \mathrm{~cm} when on the surface of the earth. The mass is then taken on to a height of 1600 km1600 \mathrm{~km} above earth's surface and again suspended from the same spring. At this altitude the extension of the spring is (Radius of earth = 6400 km6400 \mathrm{~km})
View answer and explanation
Correct option: A
3.2 cm3.2 \mathrm{~cm}
Option A (3.2 cm3.2 \mathrm{~cm}) is correct because the extension of the spring is proportional to the weight, which changes with altitude.
84 Medium
84 of 392
Understanding +4 | -1
Astronaut is in a stable orbit around the earth when he weighs a body of mass 5 kg. What is reading of spring balance?
View answer and explanation
Correct option: A
Spring will not be extended
Option A is correct because in a stable orbit, the astronaut and the body are in freefall, so the spring balance will not be extended.
85 Medium
85 of 392
Application +4 | -1
Gravitational potential of the body of mass mm at a height hh from surface of earth of radius RR is (Take g=g= acceleration due to gravity at earth's surface).
View answer and explanation
Correct option: B
−g(R−h)-g(R-h)
Option B (−g(R−h)-g(R-h)) is correct because it represents the gravitational potential at height h, derived from the potential energy formula.
86 Medium
86 of 392
Understanding +4 | -1
If the rotational motion of earth about its axis
increases, then the weight of a body on equator
would
View answer and explanation
Correct option: B
Decrease
Option B (Decrease) is correct because as the rotational speed increases, the effective weight decreases due to increased centrifugal force.
87 Medium
87 of 392
Understanding +4 | -1
A small satellite is in elliptical orbit around Earth as shown. If L denotes the magnitude of its angular momentum and K denotes kinetic energy
View answer and explanation
Correct option: D
L2=L1 and K2>K1{L_2} = {L_1}\,and\,{K_2} > {K_1}
Option D (L2=L1 and K2>K1{L_2} = {L_1}\,and\,{K_2} > {K_1}) is correct because angular momentum is conserved in an elliptical orbit, and kinetic energy is greater at the point closest to the Earth.
88 Medium
88 of 392
Application +4 | -1
A satellite of mass m is circulation around the earth with constant angular velocity. If radius of the orbit is R0{R_0} and mass of the earth M, the angular momentum about the centre of the earth is
View answer and explanation
Correct option: A
mGMR0m\sqrt {GM{R_0}}
Option A (mGMR0m\sqrt {GM{R_0}}) is correct because it is the angular momentum calculated using the orbital velocity and radius.
89 Medium
89 of 392
Application +4 | -1
Satellite A and B are revolving around the orbit of earth. The mass of A is 10 times of mass of B.The ratio of time period (TATB)\left( {\frac{{{T_A}}}{{{T_B}}}} \right) is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the time period of a satellite's orbit is independent of its mass.
90 Medium
90 of 392
Analyzing +4 | -1
Assertion(A): Areal velocity of a planet around sun will become two times, if mass of planet is halved.
Reason(R): Areal velocity =L2m=\frac{L}{2 m}, where LL is angular momentum of planet about centre of sun.
View answer and explanation
Correct option: D
If Assertion is incorrect but Reason is correct.
Option D is correct because the areal velocity is independent of the mass of the planet, so halving the mass does not change the areal velocity. The reason is correct as it states the formula for areal velocity.
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