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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 31-60 of 392 questions

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31 Medium
31 of 392
Application +4 | -1
If the earth were made of lead of relative density 11.3, what then would be the value of acceleration due to gravity on the surface of the earth? Radius of the earth =6.4×106 m=6.4 \times 10^6 \mathrm{~m} and G=6.67×10−11Nm2 kg−2{G}=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}.
View answer and explanation
Correct option: A
22.21 ms−222.21 \mathrm{~ms}^{-2}
Option A (22.21 ms−222.21 \mathrm{~ms}^{-2}) is correct because it is the calculated value of acceleration due to gravity on the surface of the lead earth.
32 Medium
32 of 392
Understanding +4 | -1
The earth is an approximately 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 if the earth's density is not uniform, the acceleration due to gravity will vary and will not be zero at any point on the surface.
33 Medium
33 of 392
Understanding +4 | -1
A point mass mm is placed inside a spherical shell of radius R and mass MM. at a distance R2\frac{R}{2} from the centre of the shell. The gravitational force exerted by the shell on the point mass is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because the gravitational force exerted by a spherical shell on a point mass inside it is zero.
34 Medium
34 of 392
Application +4 | -1
Two planets have the same average density but their radii are R1 and R2{R_1}\,and\,{R_2}. If acceleration due to gravity on these planets be g1 and g2{g_1}\,and\,{g_2} respectively, then
View answer and explanation
Correct option: A
g1g2=R1R2\frac{{{g_1}}}{{{g_2}}} = \frac{{{R_1}}}{{{R_2}}}
Option A (g1g2=R1R2\frac{{{g_1}}}{{{g_2}}} = \frac{{{R_1}}}{{{R_2}}}) is correct because the acceleration due to gravity is directly proportional to the radius when the average density is the same.
35 Medium
35 of 392
Understanding +4 | -1
Kepler’s second law states that the straight line joining the planet to the sun sweeps out equal times. This statement is equivalent to saying that
View answer and explanation
Correct option: B
Tangential acceleration is zero
Option B (Tangential acceleration is zero) is correct because Kepler's second law implies conservation of angular momentum, which means the tangential acceleration is zero.
36 Medium
36 of 392
Understanding +4 | -1
Earth is flattened at poles, bulged at the equator. This is due to:
View answer and explanation
Correct option: C
The centrifugal force is more at the equator than
at the poles
Option C (The centrifugal force is more at the equator than at the poles) is correct because the Earth's rotation causes a greater centrifugal force at the equator, leading to the bulging shape.
37 Medium
37 of 392
Application +4 | -1
Two satellites A and B go round a planet P in circular orbits having radii 4R and R respectively. If the speed of the satellite A is 3V, the speed of the satellite B will be
View answer and explanation
Correct option: B
6 V
Option B (6 V) is correct because the orbital speed is inversely proportional to the square root of the radius.
38 Medium
38 of 392
Understanding +4 | -1
Centripetal acceleration is
View answer and explanation
Correct option: D
Not a constant vector
Option D (Not a constant vector) is correct because while the magnitude of centripetal acceleration is constant in uniform circular motion, its direction continuously changes, making it a non-constant vector.
39 Medium
39 of 392
Understanding +4 | -1
Weightlessness experienced while orbiting the Earth in space-ship, is the result of:
View answer and explanation
Correct option: D
Free fall towards Earth
Option D (Free fall towards Earth) is correct because weightlessness in orbit is due to the continuous free fall of the spacecraft and its occupants towards Earth, where the gravitational force provides the necessary centripetal acceleration for the orbit.
40 Medium
40 of 392
Understanding +4 | -1
Assertion: The time period of revolution of a satellite close to surface of earth is smaller than that revolving far from surface of earth.
Reason: The square of time period of revolution of a satellite is directly proportional to cube of its orbital radius.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because the time period of a satellite is related to its orbital radius, and satellites closer to Earth have smaller orbital radii and thus shorter periods.
41 Medium
41 of 392
Application +4 | -1
Force between two objects of equal masses is FF. If 25%25 \% mass of one object is transferred to the other object, then the new force will be
View answer and explanation
Correct option: C
1516F\frac{15}{16} F
Option C (1516F\frac{15}{16} F) is correct because the force is proportional to the product of the masses, and the new masses are 1.25m and 0.75m.
42 Medium
42 of 392
Analyzing +4 | -1
Assertion: If gravitational force is proportional to 1R3\frac{1}{\mathrm{R}^3} then Keplar's second law would not be affected.
Reason: For Keplar's second law, central nature of force is the only condition.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because Kepler's second law depends only on the central nature of the force, not the specific radial dependence.
43 Medium
43 of 392
Application +4 | -1
The gravitational force between two bodies of masses 2 kg and 8 kg is 6.67×10−11N.6.67 \times {10^{ - 11}}N. Find the distance between two bodies [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: D
4 m
Option D (4 m) is correct because it is the result of solving the gravitational force equation for the distance between the two masses.
44 Medium
44 of 392
Understanding +4 | -1
The centripetal force acting on a satellite orbiting round the earth and the gravitational force of earth acting on the satellite both equal FF. The net force on the satellite is
View answer and explanation
Correct option: B
FF
Option B (FF) is correct because the gravitational force provides the necessary centripetal force for the satellite's orbit, so the net force is simply the centripetal force.
45 Medium
45 of 392
Analyzing +4 | -1
Both earth and moon are subject to the gravitational force of the sun. As observed from the sun, the orbit of the moon
View answer and explanation
Correct option: B
will not be strictly elliptical because the total gravitational force on it is not central.
Option B is correct because the moon's orbit is not strictly elliptical due to the combined gravitational forces of the Sun and Earth, resulting in a non-central net force.
46 Medium
46 of 392
Analyzing +4 | -1
If gravitational constant is decreasing with time, what will remain unchanged in case of a satellite orbiting around earth
View answer and explanation
Correct option: C
tangential velocity
Tangential velocity remains unchanged because the gravitational constant affects the orbital period and radius, but not directly the tangential velocity if angular momentum is conserved.
47 Medium
47 of 392
Application +4 | -1
The height from the surface of earth at which the gravitational potential energy of a ball of mass mm is half of that at the centre of earth is (where RR is the radius of earth)
View answer and explanation
Correct option: B
R3\frac{R}{3}
Option B (R3\frac{R}{3}) is correct because it is calculated by equating half the gravitational potential energy at the center of the Earth to the gravitational potential energy at a height h from the surface.
48 Medium
48 of 392
Application +4 | -1
When a body is lifted from surface of earth at a height equal to radius of earth, then the change in its potential energy is
View answer and explanation
Correct option: C
12mgR\frac{1}{2} m g R
Option C (12mgR\frac{1}{2} m g R) is correct because the change in potential energy when lifting a body to a height equal to the Earth's radius is half of mgRmgR.
49 Medium
49 of 392
Understanding +4 | -1
At the surface of a certain planet, acceleration due to gravity is one-quarter of that on earth. If a brass ball is transported to this planet, then which one of the following statements is not correct
View answer and explanation
Correct option: A
The mass of the brass ball on this planet is a quarter of its mass as measured on earth
Option A (The mass of the brass ball on this planet is a quarter of its mass as measured on earth) is not correct because mass is an intrinsic property and does not change with location or gravitational field.
50 Medium
50 of 392
Understanding +4 | -1
Assertion: Space ship while entering the Earth’s atmosphere is likely to catch fire.
Reason: Temperature of upper atmosphere is very high.
View answer and explanation
Correct option: C
If assertion is true but reason is false
Option C is correct because while the assertion is true, the reason is false; the spaceship catches fire due to air friction, not the temperature of the upper atmosphere.
51 Medium
51 of 392
Application +4 | -1
The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant =6.7×10−11Nm2/kg2,=6.7 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2, mass of an electron =9.1×10−31 kg,=9.1 \times 10^{-31} \mathrm{~kg}, charge on an electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C})
View answer and explanation
Correct option: C
24×10−4424 \times 10^{-44}
Option C (24×10−4424 \times 10^{-44}) is correct because it is calculated by dividing the gravitational force by the electrostatic force between two electrons.
52 Medium
52 of 392
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.
53 Medium
53 of 392
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.
54 Medium
54 of 392
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.
55 Medium
55 of 392
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}.
56 Medium
56 of 392
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.
57 Medium
57 of 392
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.
58 Medium
58 of 392
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.
59 Medium
59 of 392
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.
60 Medium
60 of 392
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.
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