Skip to main content
Back to GRAVITATION

Your Progress

0 / 99 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • PHYSICS • GRAVITATION

Gravitation practice · Easy

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 1-30 of 99 questions
1 Easy
1 of 99
Analyzing +4 | -1
Suppose the law of gravitational attraction suddenly changes and becomes an inverse cube law i.e. F∝1r3F \propto \frac{1}{r^3}, but still remaining a central force. Then
View answer and explanation
Correct option: A
Kepler's law of area still holds
Option A (Kepler's law of area still holds) is correct because Kepler's second law (law of areas) is a consequence of conservation of angular momentum, which holds for any central force.
2 Easy
2 of 99
Application +4 | -1
Two satellites A and B go round a planet P is circular orbits having radii 16 R and 2 R respectively. If the speed of satellite A is 6 ms−16\,m{s^{ - 1}} the speed of the satellite B will be [in m/s] nearly
View answer and explanation
Correct option: A
17 ms−117\,m{s^{ - 1}}
Option A (17 ms−117\,m{s^{ - 1}}) is correct because the orbital speed is inversely proportional to the square root of the radius.
3 Easy
3 of 99
Understanding +4 | -1
Statement-1: There is almost no effect of rotation of earth at poles.
Statement-2: Because rotation of earth is about polar axis.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true
Option A (Statement-1 is true, Statement-2 is true) is correct because the rotation of the Earth has minimal effect at the poles, and this is because the Earth rotates about its polar axis.
4 Easy
4 of 99
Application +4 | -1
The escape velocity of a body on the surface of an imaginary planet which has thrice the radius as of earth and double the mass of earth, is (ve{v_e} is the escape velocity at the surface of earth)
View answer and explanation
Correct option: C
23ve\sqrt {\frac{2}{3}} {v_e}
Option C is correct because the escape velocity is proportional to the square root of (M/R), so with the given ratios, the escape velocity becomes sqrt(2/3) times the Earth's escape velocity.
5 Easy
5 of 99
Remembering +4 | -1
If the mass of a body is MM on the surface of the earth, then mass of the same body on the surface of the moon is
View answer and explanation
Correct option: D
MM
Option D (MM) is correct because mass is an intrinsic property of an object and does not change with location.
6 Easy
6 of 99
Application +4 | -1
A planet in a distant solar system is 5 times more massive than the earth and its radius is 5 times smaller. Given that the escape velocity from the earth is 11 k ms−1,11\,k\,m{s^{ - 1}}, the escape velocity from the surface of the planet would be [in kmps]
View answer and explanation
Correct option: B
55
Option B (55) is correct because the escape velocity is proportional to M/R\sqrt{M/R}, and with the given changes, the new escape velocity is 5 times the original.
7 Easy
7 of 99
Understanding +4 | -1
Assertion: Smaller the orbit of a planet around the sun, shorter is the time it takes to complete.
Reason: According to Kepler's third law of planetary motion, square of time period is proportional to cube of mean distance from the sun.
View answer and explanation
Correct option: A
Assertion is True, Reason is True; Reason is a correct explanation for Assertion
Option A is correct because Kepler's third law directly links orbital size to orbital period.
8 Easy
8 of 99
Understanding +4 | -1
Statement-1: It is possible to put an artificial satellite into orbit in such a way that it will always remain directly over New Delhi.
Statement-2: To put an artificial satellite into orbital such that it remains directly over head at a fixed location, its time period should be related to that of earth in the equatorial plane.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is true
Statement-1 is false, and Statement-2 is true. A geostationary satellite must be above the equator, so it can't be over New Delhi.
9 Easy
9 of 99
Understanding +4 | -1
Statement-1: Gravitational potential is zero inside a shell.
Statement-2: Gravitational potential is equal to double the work done in bringing a unit mass from infinity to a point inside the gravitational field.
View answer and explanation
Correct option: D
Statement-1 is false, Statement-2 is false
Option D is correct because both statements are false: the gravitational potential inside a shell is constant and non-zero, and gravitational potential is equal to the work done, not double the work done.
10 Easy
10 of 99
Understanding +4 | -1
A planet is moving in an elliptical orbit around the sun. If TT, V,E\mathrm{V}, \mathrm{E} and L\mathrm{L} stand respectively for its kinetic energy, gravitational potential energy, total energy and magnitude of angular momentum about the centre of force, then which of the following is correct?
View answer and explanation
Correct option: C
E is always negative
Option C (E is always negative) is correct because the total energy of a planet in an elliptical orbit is always negative.
11 Easy
11 of 99
Understanding +4 | -1
Orbital velocity of earth satellite does not depend on
View answer and explanation
Correct option: B
mass of the satellite
Option B (mass of the satellite) is correct. The orbital velocity of an Earth satellite does not depend on the mass of the satellite.
12 Easy
12 of 99
Application +4 | -1
2 kg2 \mathrm{~kg} body falls from infinity to the surface of earth what will be the kinetic energy of the body on reaching the surface of Earth? (Assume escape velocity to be 1024 J10^{24} \mathrm{~J}
View answer and explanation
Correct option: A
108 J10^8 \mathrm{~J}
Option A (108 J10^8 \mathrm{~J}) is correct because the kinetic energy is equal to the potential energy difference between infinity and the Earth's surface, which can be calculated using the escape velocity.
13 Easy
13 of 99
Application +4 | -1
The radius of a planet is twice the radius of earth. Both have almost equal average mass-densities. If VPV_P and VEV_E are escape velocities of the planet and the earth, respectively, then
View answer and explanation
Correct option: B
VP=2VEV_P=2 V_E
Option B (VP=2VEV_P=2 V_E) is correct because the escape velocity is proportional to the radius when the densities are equal.
14 Easy
14 of 99
Application +4 | -1
The time period of revolution of a planet ‘A’ around the sun is 27 times that of another planet ‘B’. The ratio of distance of planet ‘B’ to the planet ‘A’ from sun
View answer and explanation
Correct option: B
0.047916667
Option B (0.047916667) is correct. This is calculated using Kepler's Third Law.
15 Easy
15 of 99
Remembering +4 | -1
The value of ' g\mathrm{g} ' on poles is
View answer and explanation
Correct option: A
Maximum
Option A (Maximum) is correct because the value of 'g' is maximum at the poles due to the Earth's shape and rotation.

Explore Other Chapters in PHYSICS

View All →
UNITS & MEASUREMENTS 392 MCQs WORK-ENERGY-POWER 355 MCQs MECHANICAL PROPERTIES OF FLUIDS 317 MCQs MOTION IN PLANE 307 MCQs MAGNETIC EFFECT OF CURRENT 282 MCQs THERMAL PROPERTIES OF MATTER 269 MCQs
16 Easy
16 of 99
Analyzing +4 | -1
Assertion: Particle of mass mm dropped into a hole made along the diameter of the earth from one end to the other end executes simple harmonic motion.
Reason: Gravitational force between any two particles is inversely proportional to the square of the distance between them.
View answer and explanation
Correct option: B
Assertion is True, Reason is True; Reason is NOT a correct explanation for Assertion
Option B (Assertion is True, Reason is True; Reason is NOT a correct explanation for Assertion) is correct because the particle executes SHM, and the gravitational force is inversely proportional to the square of the distance, but this isn't the *direct* reason for SHM.
17 Easy
17 of 99
Understanding +4 | -1
I. The geo-stationary satellite always has circular orbit.
II. The time period of satellite depends on mass of the satellite, mass of the earth and height of satellite from the surface of earth.
Which of the following statement(s) is/are incorrect?
View answer and explanation
Correct option: A
Only I
Only I is incorrect. A geostationary satellite's orbit can be elliptical, though it's usually close to circular. The time period does depend on the mass of the earth and the height of the satellite.
18 Easy
18 of 99
Application +4 | -1
Two satellites of mass 400 kg and 600 kg revolve around the earth is circular orbits of radii 10 R and 20 R respectively where R is the radius of the earth the speed of the two satellites will be in the ratio
View answer and explanation
Correct option: C
2:1\sqrt 2 :1
Option C (2:1\sqrt 2 :1) is correct because the orbital speed is independent of the mass of the satellite and inversely proportional to the square root of the radius.
19 Easy
19 of 99
Application +4 | -1
Find the work done to take a particle of mass m\mathrm{m} from surface of the earth to a height equal to 2R
View answer and explanation
Correct option: D
2mgR3\frac{2 m g R}{3}
Option D (2mgR3\frac{2 m g R}{3}) is correct because the work done is equal to the change in potential energy when moving the particle from the surface to a height of 2R.
20 Easy
20 of 99
Application +4 | -1
The diameters of two planets are in the ratio of 4:1&4: 1 \& mean density have ratio 1:21: 2, then the ratio of gg on the planets will be
View answer and explanation
Correct option: C
2:12: 1
The correct answer is C (2:12:1) because the ratio of gravitational acceleration is directly proportional to the product of radius and density.
21 Easy
21 of 99
Application +4 | -1
The binding energy of a particle of mass mm with a planet, when it is on the planets surface, is 12mv02\frac{1}{2} \mathrm{mv}_0^2. A tunnel is dug along a diameter of the planet and the particle is dropped into it from the surface, when the body reaches the centre of the planet, its speed is
View answer and explanation
Correct option: B
v02\frac{v_0}{\sqrt{2}}
Option B (v02\frac{v_0}{\sqrt{2}}) is correct because it's derived from the conservation of energy, considering the potential energy difference between the surface and the center of the planet.
22 Easy
22 of 99
Analyzing +4 | -1
Assertion: The magnitude of gravitational potential at the surface of solid sphere is less than that of the centre of sphere.
Reason: Due to solid sphere, gravitational potential is same within the sphere.
View answer and explanation
Correct option: C
Assertion is True, Reason is False
Option C (Assertion is True, Reason is False) is correct because the gravitational potential *increases* as you move from the surface to the center of a solid sphere, and is constant *only* at the center.
23 Easy
23 of 99
Analyzing +4 | -1
Statement-1: Orbiting satellite or body has K.E. of always less than that of Potential energy.
Statement-2: For any bound state, the magnitude of potential energy is always twice that of kinetic energy (K.E.)
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true
Option C (626\sqrt 2) is correct because it is derived from Kepler's Third Law, relating the time period to the orbital radius.
24 Easy
24 of 99
Understanding +4 | -1
If the speed of rotation of earth about its axis increases, then the weight of the body at the equator will
View answer and explanation
Correct option: B
decrease
Option B (decrease) is correct because an increase in the Earth's rotation speed increases the centrifugal force at the equator, thus reducing the effective weight of a body.
25 Easy
25 of 99
Application +4 | -1
A particle of mass 10 g10 \mathrm{~g} is kept on the surface of a uniform sphere of mass 100 kg100 \mathrm{~kg} and radius 10 cm10 \mathrm{~cm}. Find the work to be done against the gravitational force between them to take the particle far away from the sphere. (you may take G=6.67×10−11Nm2/kg2\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2 )
View answer and explanation
Correct option: B
6.67×10−10 J6.67 \times 10^{-10} \mathrm{~J}
Option B (6.67×10−10 J6.67 \times 10^{-10} \mathrm{~J}) is correct because the work done is equal to the change in potential energy, which can be calculated using the gravitational constant, masses, and radius.
26 Easy
26 of 99
Understanding +4 | -1
Assertion: If a particle projected horizontally just above the surface of earth with a speed greater than escape speed, then it will escape from gravitational influence of earth.
Reason: Escape velocity is independent of its direction.
View answer and explanation
Correct option: A
Assertion is True, Reason is True; Reason is a correct explanation for Assertion
Option A (Assertion is True, Reason is True; Reason is a correct explanation for Assertion) is correct because the assertion is true, the reason is true, and the reason correctly explains the assertion.
27 Easy
27 of 99
Application +4 | -1
The depth from the surface of the earth of radius RR at which the acceleration due to gravity will be 75%75\% of the value on the surface of the earth is
View answer and explanation
Correct option: A
R4\frac{R}{4}
Option A (R4\frac{R}{4}) is correct because at a depth of R/4, the acceleration due to gravity is 75% of its surface value.
28 Easy
28 of 99
Application +4 | -1
What is the shortest possible period of an Earth satellite in circular orbit?
View answer and explanation
Correct option: A
2πRg2 \pi \sqrt{\frac{R}{g}}
Option A (2πRg2 \pi \sqrt{\frac{R}{g}}) is correct because it represents the shortest possible period derived from equating gravitational force to centripetal force at the Earth's surface.
29 Easy
29 of 99
Application +4 | -1
ge and gp{g_e}\,and\,{g_p} denote the acceleration due to gravity on the surface of the earth and another planet whose mass and radius are twice to that of the earth, then
View answer and explanation
Correct option: A
gp=ge2{g_p} = \frac{{{g_e}}}{2}
Option A (gp=ge2{g_p} = \frac{{{g_e}}}{2}) is correct. Since the mass and radius are doubled, the acceleration due to gravity is halved.
30 Easy
30 of 99
Understanding +4 | -1
The value of acceleration due to gravity at the surface of earth
View answer and explanation
Correct option: A
is maximum at the poles
Option A is correct because the acceleration due to gravity is maximum at the poles due to the Earth's shape (being an oblate spheroid) and its rotation.
Revision notes and key takeaways

Review GRAVITATION in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 99 single-correct-answer MCQs on GRAVITATION.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
Showing 1-30 of 99 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ NEET PHYSICS - GRAVITATION MCQs (All) 255 MCQs → NEET PHYSICS - GRAVITATION MCQs (Hard) 21 MCQs → NEET PHYSICS - GRAVITATION MCQs (Medium) 135 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ UNITS & MEASUREMENTS 392 MCQs → WORK-ENERGY-POWER 355 MCQs → MECHANICAL PROPERTIES OF FLUIDS 317 MCQs → MOTION IN PLANE 307 MCQs
📚 Switch Set (EASY)
Chapter Sets
ALL 255 MCQs EASY 99 MCQs HARD 21 MCQs MEDIUM 135 MCQs
0 /99
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →