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JEE • PHYSICS • GRAVITATION

Gravitation practice · Mixed difficulty

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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 450 questions

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61 Easy
61 of 450
Application +4 | -1
The acceleration due to gravity on a planet is same as that on earth and its radius is 16 times that of earth. What will be the value of escape velocity on that planet if it is Ve{V_e} on earth.
View answer and explanation
Correct option: C
4Ve4{V_e}
Option C (4Ve4{V_e}) is correct because the escape velocity is proportional to the square root of the radius, and since the radius is 16 times that of Earth, the escape velocity is 4 times that of Earth.
62 Medium
62 of 450
Analyzing +4 | -1
A clock S is based on oscillation of a spring and a clock P is based on pendulum motion. Both clocks run at the same rate on earth. On a planet having the same density as earth but twice the radius
View answer and explanation
Correct option: B
PP will run faster than SS
Option B (PP will run faster than SS) is correct because the pendulum clock's time period depends on gravity, which changes on the new planet, while the spring clock's time period is independent of gravity.
63 Medium
63 of 450
Application +4 | -1
Time period of revolution of a nearest satellite around a planet of radius RR is TT. Period of revolution around another planet, whose radius is 4R4 R but having same density is
View answer and explanation
Correct option: A
TT
Option A (TT) is correct because the time period of a satellite orbiting close to a planet depends only on the planet's density, which remains constant.
64 Medium
64 of 450
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 is correct because inside the Earth, the acceleration due to gravity is directly proportional to the distance from the center.
65 Medium
65 of 450
Application +4 | -1
The gravitational field at some point in space is g→=8i^+6j^ N/kg\overrightarrow g = 8\hat i + 6\hat j\,N/kg. The force exerted on a 1 kg mass placed at that point is
View answer and explanation
Correct option: B
10N,  37∘10N,\,\,37^\circ with x axis
Option B (10N, 37∘10N,\,37^\circ with x axis) is correct because it is the force calculated using F→=mg→\overrightarrow F = m\overrightarrow g and the angle is found using tan⁡θ=yx\tan \theta = \frac{y}{x}.
66 Medium
66 of 450
Understanding +4 | -1
If mass of a satellite is doubled and time period remain constant the ratio of orbit in the two cases will be
View answer and explanation
Correct option: B
1:11:1
Option B (1:11:1) is correct because the orbital radius of a satellite depends on its time period and the mass of the central body (Earth), but not on the mass of the satellite itself.
67 Medium
67 of 450
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 is correct because, according to Kepler's third law, the time taken to travel a path is proportional to the distance from the sun, making DAB longer than BCD.
68 Medium
68 of 450
Understanding +4 | -1
A uniform spherical shell gradually shrink maintaining its shape. The gravitational potential at its centre will (Mass of the shell remains same)
View answer and explanation
Correct option: B
Decrease
Option B (Decrease) is correct because as the radius of the shell decreases, the gravitational potential becomes more negative, thus decreasing.
69 Medium
69 of 450
Application +4 | -1
A body is projected vertically upwards from the surface of the earth with a speed equal to 14\frac{1}{4}th of escape velocity from surface of earth. Maximum height attained by the body from the surface of the earth will be (RR-Radius of earth)
View answer and explanation
Correct option: D
R15\frac{R}{15}
Option D (R15\frac{R}{15}) is correct because the maximum height is found by equating the initial kinetic and potential energy to the final potential energy using conservation of energy.
70 Medium
70 of 450
Application +4 | -1
The angular speed with which earth should spin about its axis, such that a person feels (59)th\left(\frac{5}{9}\right)^{\text {th}} times his weight at equator as compared to his weight at pole
View answer and explanation
Correct option: C
23g/R\frac{2}{3} \sqrt{g / R}
Option D (13g/R\frac{1}{3} \sqrt{g / R}) is incorrect. The correct answer is C (23g/R\frac{2}{3} \sqrt{g / R}) because it is derived from setting the effective weight at the equator to 5/9 of the weight at the pole and solving for the angular speed.
71 Medium
71 of 450
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.
72 Medium
72 of 450
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}}.
73 Medium
73 of 450
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.
74 Medium
74 of 450
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.
75 Easy
75 of 450
Understanding +4 | -1
Which of the following statement is correct? (Take, earth as a sphere of uniform density)
View answer and explanation
Correct option: D
Acceleration due to gravity decreases with increasing depth.
Option D (Acceleration due to gravity decreases with increasing depth) is correct because the effective mass pulling the object towards the Earth's center decreases as depth increases.
76 Easy
76 of 450
Understanding +4 | -1
When you move from equator to pole, the value of acceleration due to gravity ( g\mathrm{g} )
View answer and explanation
Correct option: A
increases
Option A is correct because the acceleration due to gravity increases as you move from the equator to the poles due to the Earth's shape and rotation.
77 Medium
77 of 450
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.
78 Medium
78 of 450
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.
79 Medium
79 of 450
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.
80 Medium
80 of 450
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.
81 Medium
81 of 450
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.
82 Medium
82 of 450
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.
83 Medium
83 of 450
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.
84 Medium
84 of 450
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.
85 Medium
85 of 450
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.
86 Medium
86 of 450
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.
87 Medium
87 of 450
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.
88 Medium
88 of 450
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.
89 Medium
89 of 450
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.
90 Medium
90 of 450
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.
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