Gravitation – NDA Physics PYQs

NDA 2026-I

Q. 1. For an object in gravitational field, the gravitational potential is the same at two points A and B, but the gravitational field is not the same at these two points. Which one of the following statements is correct?

(a)  Positive work is done by gravity in moving the object from point A to point B.

(b)  Negative work is done by gravity in moving the object from point A to point B.

(c)  Gravity must be non-conservative in such a region.

(d)  No work is done by gravity in moving the object from point A to point B.

Answer: (d) No work is done by gravity in moving the object from point A to point B.

Explanation: When gravitational potential φ is identical at A and B, the potential difference Δφ = 0. Work done by gravity = −m·Δφ = 0. No work is done regardless of the gravitational field values at A and B individually: the field can differ at the two points and yet the work is zero because the potential is the same. Gravity remains conservative in all regions. Concept Tested: Equipotential surfaces: zero work done when gravitational potential is equal at start and end

Q. 2. One year at a planet is 8 times as large as compared to the one year at the Earth. Which one of the following is correct about the planet’s orbit?

(a)  The semimajor axis of the planet’s orbit is same as that of the Earth.

(b)  The semimajor axis of the planet’s orbit is twice as compared to that of the Earth.

(c)  The semimajor axis of the planet’s orbit is thrice as compared to that of the Earth.

(d)  The semimajor axis of the planet’s orbit is four times as compared to that of the Earth.

Answer: (d) The semimajor axis of the planet’s orbit is four times as compared to that of the Earth.

Explanation: Kepler’s Third Law: T² ∝ a³. T_planet/T_Earth = 8, so (a_planet/a_Earth)³ = 8² = 64. Therefore a_planet/a_Earth = 64^(1/3) = 4. The semi-major axis of the planet’s orbit is 4 times that of Earth’s orbit. Concept Tested: Kepler’s Third Law: semi-major axis ratio from period ratio: (a₂/a₁)³ = (T₂/T₁)²

★ JOVIK Exam Insight The second Kepler’s Third Law question in this collection (see also Q. 20). The standard method: cube root of (T ratio)² = a ratio. Here: (8)^(2/3) = (64)^(1/3) = 4. Always square the period ratio first, then take the cube root.

NDA 2024-II

Q. 3. The masses of two planets are in ratio 1:7. Ratio between their diameters is 2:1. The ratio of forces which they exert on each other is:

(a)  1:7

(b)  7:1

(c)  1:1

(d)  2:1

Answer: (c) 1:1

Explanation: By Newton’s Third Law, the gravitational force planet A exerts on planet B is exactly equal and opposite to the force planet B exerts on planet A. This is true regardless of the mass or size of the planets. The ratio of the forces is always 1:1. The masses and diameters are irrelevant: they are distractors. Concept Tested: Newton’s Third Law applied to gravity: mutual gravitational forces are equal (1:1)

★ JOVIK Exam Insight A clean Newton’s Third Law trap: the force each planet exerts on the other is always equal in magnitude: regardless of mass, size, or distance. The ratio is always 1:1. The mass ratio (1:7) and diameter ratio (2:1) are distractors.

NDA 2024-I

Q. 4. Escape speed from the Earth is close to 11·2 km s⁻¹. On another planet whose radius is half of the Earth’s radius and whose mass density is four times that of the Earth, the escape speed in km s⁻¹ will be close to:

(a)  11·2

(b)  15·8

(c)  5·6

(d)  7·9

Answer: (a) 11·2

Explanation: Escape velocity v_e = √(2GM/R). New planet: radius = R/2, density = 4ρ. Mass = (4/3)π(R/2)³ × 4ρ = (4/3)πR³ρ × (1/8) × 4 = (1/2)M_Earth. So v_e(new) = √(2G × M/2 ÷ R/2) = √(2GM/R) = v_e(Earth) = 11.2 km/s. The four-times density exactly compensates the halved radius, giving the same escape speed. Concept Tested: Escape velocity: density ×4 and radius ×½ combine to give same escape speed as Earth

★ JOVIK Exam Insight A post-2023 NDA synthesis question: must substitute density into mass formula before computing escape velocity. Key: when density × R² is the same as Earth’s, escape velocity equals Earth’s. Always compute the mass from density × volume first.

NDA 2023-I

Q. 5. A sphere of volume V is made of a material with lower density than water. While on Earth, it floats on water with its volume f₁V (f₁ < 1) submerged. On the other hand, on a spaceship accelerating with acceleration a < g (g is the acceleration due to gravity on Earth) in outer space, its submerged volume in water is f₂V. Then:

(a)  f₂ = f₁

(b)  f₂ = (1 − a/g) f₁

(c)  f₂ > f₁

(d)  f₂ = (a/g) f₁

Answer: (a) f₂ = f₁

Explanation: In the accelerating spaceship, effective gravity = a (upward, felt as downward in the ship). Buoyancy condition: weight of sphere = buoyant force → ρ_sphere × V × a = ρ_water × f₂V × a. The acceleration a cancels from both sides, giving f₂ = ρ_sphere/ρ_water = f₁. The submerged fraction is the same regardless of the value of a (as long as a ≠ 0), because both weight and buoyant force scale equally with the effective acceleration. Concept Tested: Buoyancy in an accelerating spacecraft: submerged fraction unchanged (a cancels)

★ JOVIK Exam Insight A cross-chapter question (fluid mechanics + gravitation). The effective gravity a in an accelerating spaceship replaces g. Since both weight and buoyancy scale with the same a, the submerged fraction is unchanged. This is a post-2023 NDA difficulty level.

Q. 6. A mass is attached to a spring that hangs vertically. The extension produced in the spring is 6 cm on Earth. The acceleration due to gravity on the surface of the Moon is one-sixth of its value on the surface of the Earth. The extension of the spring on the Moon would be:

(a)  6 cm

(b)  1 cm

(c)  0 cm

(d)  36 cm

Answer: (b) 1 cm

Explanation: Spring extension: kx = mg. On Earth: k × 6 = mg → g = 6k/m (proportional). On the Moon: g_Moon = g/6, so x_Moon = mg_Moon/k = m(g/6)/k = (1/6)(mg/k) = 6/6 = 1 cm. The extension is directly proportional to local g: one-sixth the Earth value gives one-sixth the extension. Concept Tested: Spring extension proportional to g: Moon’s lower gravity gives 6 cm ÷ 6 = 1 cm

NDA 2020-I & II

Q. 7. Two planets orbit the Sun in circular orbits, with their radius of orbit as R₁ = R and R₂ = 4R. Ratio of their periods (T₁/T₂) around the Sun will be

(a)  1/16

(b)  1/8

(c)  1/4

(d)  ½

Answer: (b) 1/8

Explanation: Kepler’s Third Law: T² ∝ R³. So (T₁/T₂)² = (R₁/R₂)³ = (R/4R)³ = (1/4)³ = 1/64. Therefore T₁/T₂ = 1/√64 = 1/8. The inner planet (radius R) has a period 8 times shorter than the outer planet (radius 4R). Concept Tested: Kepler’s Third Law: T² ∝ R³; ratio of periods from ratio of orbital radii

★ JOVIK Exam Insight Kepler’s Third Law ratio problems appear repeatedly in NDA (2020-I & II, 2026-I). The key formula: (T₁/T₂)² = (R₁/R₂)³. Always cube the radius ratio first, then take the square root for the period ratio.

NDA 2019-II

Q. 8. Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed R/2 distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be

(a)  16 F

(b)  1 F

(c)  4 F

(d)  None of the above

Answer: (a) 16 F

Explanation: System 1: F = GM²/R². System 2: F’ = G(2M)(2M)/(R/2)² = 4GM²/(R²/4) = 16GM²/R² = 16F. Doubling both masses multiplies force by 4; halving the distance multiplies force by 4. Combined: 4 × 4 = 16F. Concept Tested: Gravitational force scaling: doubling masses and halving distance gives ×16

NDA 2019-I

Q. 9. Suppose there are two planets, 1 and 2, having the same density but their radii are R₁ and R₂ respectively, where R₁ > R₂. The accelerations due to gravity on the surface of these planets are related as

(a)  g₁ > g₂

(b)  g₁ < g₂

(c)  g₁ = g₂

(d)  Can’t say anything

Answer: (a) g₁ > g₂

Explanation: For planets of the same density ρ: mass M = (4/3)πR³ρ. So g = GM/R² = G(4/3)πR³ρ/R² = (4/3)πGρR. Surface gravity is directly proportional to radius when density is constant. Since R₁ > R₂, we get g₁ > g₂. Concept Tested: Surface gravity with same density: g ∝ R (larger planet has larger g)

★ JOVIK Exam Insight When density is the same: g ∝ R. Larger radius → larger g. This is counterintuitive: a larger planet of the same density has stronger surface gravity. Contrast: same mass × doubled radius → g halved.

Q. 10. ‘Black hole’ is a

(a)  huge black star which has zero acceleration due to gravity on its surface

(b)  star which has moderate acceleration due to gravity on its surface

(c)  star which has collapsed into itself and has large acceleration due to gravity on its surface

(d)  star which has collapsed into itself and has zero acceleration due to gravity on its surface

Answer: (c) star which has collapsed into itself and has large acceleration due to gravity on its surface

Explanation: A black hole is a stellar remnant that has collapsed under its own gravity to an extremely small volume. This produces an extraordinarily large gravitational field. The escape velocity at the event horizon equals the speed of light: nothing, not even light, can escape. Black holes do not have zero gravity; they have the highest known gravitational field strength. Concept Tested: Black hole: collapsed star with extremely large surface gravity (not zero)

Q. 11. LIGO stands for

(a)  Laser Interferometer Gravitational wave Observatory

(b)  Light Interferometer Gravitational wave Observatory

(c)  Light Induced Gravity Observatory

(d)  Laser Induced Gaseous Optics

Answer: (a) Laser Interferometer Gravitational wave Observatory

Explanation: LIGO is the Laser Interferometer Gravitational-wave Observatory. It detects gravitational waves: ripples in spacetime produced by massive accelerating objects such as merging black holes or neutron stars: using laser interferometry across kilometre-scale arms. It is a laser-based instrument (not light-intensity or gas-optics based). Concept Tested: LIGO: full form and purpose (laser-based gravitational wave detector)

NDA 2018-II

Q. 12. A planet has a mass M₁ and radius R₁. The value of acceleration due to gravity on its surface is g₁. There is another planet 2, whose mass and radius both are two times that of the first planet. Which one of the following is the acceleration due to gravity on the surface of planet 2?

(a)  g₁

(b)  2g₁

(c)  g₁/2

(d)  g₁/4

Answer: (c) g₁/2

Explanation: g₂ = G(2M₁)/(2R₁)² = 2GM₁/4R₁² = (1/2)(GM₁/R₁²) = g₁/2. Doubling both mass and radius halves the surface gravity. The mass doubles the numerator (×2) but the squared radius quadruples the denominator (×4), giving a net factor of 2/4 = 1/2. Concept Tested: Surface gravity: doubling both mass and radius gives g/2

★ JOVIK Exam Insight Planet comparison pattern: doubling mass AND radius → g halved. This appears in 2014-I (×4 mass, ×2 radius → same g) and 2018-II (×2 mass, ×2 radius → g/2). Always substitute into g = GM/R²: never guess.

NDA 2018-I

Q. 13. Which one of the following statements about gravitational force is NOT correct?

(a)  It is experienced by all bodies in the universe

(b)  It is a dominant force between celestial bodies

(c)  It is a negligible force for atoms

(d)  It is same for all pairs of bodies in our universe

Answer: (d) It is same for all pairs of bodies in our universe

Explanation: Options (a), (b), and (c) are all true: gravity acts on all bodies, dominates at celestial scales, and is negligible at atomic scales where electromagnetic and nuclear forces dominate. Option (d) is false: the gravitational force between two bodies depends on their specific masses and separation (F = Gm₁m₂/r²): it is different for every pair. Concept Tested: Properties of gravitational force: NOT the same for all pairs (depends on specific masses and distance)

NDA 2017-II

Q. 14. Which one of the following statements about a satellite orbiting around the Earth is correct?

(a)  Satellite is kept in orbit by remote control from ground station.

(b)  Satellite is kept in orbit by retro-rocket and solar energy keeps it moving around the Earth.

(c)  Satellite requires energy from solar panels and solid fuels for orbiting.

(d)  Satellite does not require any energy for orbiting.

Answer: (d) Satellite does not require any energy for orbiting.

Explanation: A satellite in a stable circular orbit is in continuous free fall. Its horizontal velocity is perfectly matched to Earth’s curvature so it perpetually misses the ground. No thrust is required to sustain the orbit: only to change it. Energy is needed only to launch the satellite into orbit or to adjust its trajectory later. Concept Tested: Satellite in stable orbit: requires no energy to maintain orbital motion

NDA 2017-I

Q. 15. Which one of the following statements is true for the relation F = Gm₁m₂/r² ? (All symbols have their usual meanings)

(a)  The quantity G depends on the local value of g, acceleration due to gravity

(b)  The quantity G is greatest at the surface of the Earth

(c)  The quantity G is used only when earth is one of the two masses

(d)  The quantity G is a universal constant

Answer: (d) The quantity G is a universal constant

Explanation: G is the universal gravitational constant: G = 6.674 × 10⁻¹¹ N m² kg⁻². It does not depend on local g, does not vary with location, and applies to any pair of masses anywhere in the universe: not only Earth calculations. g (lowercase) is the local acceleration due to gravity and varies with location; G (uppercase) is fixed throughout the universe. Concept Tested: Universal gravitational constant G: location-independent and universally applicable

★ JOVIK Exam Insight A common NDA confusion: G vs g. G is constant everywhere in the universe. g varies with location (poles vs equator, different planets). G depends on nothing: it is a fundamental constant. g = GM/R² is derived from G, M, R.

NDA 2016-II

Q. 16. The free fall acceleration g increases as one proceeds, at sea level, from the equator toward either pole. The reason is

(a)  Earth is a sphere with same density everywhere

(b)  Earth is a sphere with different density at the polar regions than in the equatorial regions

(c)  Earth is approximately an ellipsoid having its equatorial radius greater than its polar radius by 21 km

(d)  Earth is approximately an ellipsoid having its equatorial radius smaller than its polar radius by 21 km

Answer: (c) Earth is approximately an ellipsoid having its equatorial radius greater than its polar radius by 21 km

Explanation: The Earth is oblate: slightly flattened at the poles and bulging at the equator. The equatorial radius exceeds the polar radius by about 21 km. Since g = GM/R², a smaller polar radius means larger g at the poles. It is the shape (equatorial bulge), not density differences, that causes g to increase toward the poles. Concept Tested: Variation of g from equator to poles: due to Earth’s oblate shape (equatorial bulge)

NDA 2016-I

Q. 17. Suppose the force of gravitation between two bodies of equal masses is F. If each mass is doubled keeping the distance of separation between them unchanged, the force would become

(a)  F

(b)  2 F

(c)  4 F

(d)  ¼ F

Answer: (c) 4 F

Explanation: Original force F = Gm²/r². When each mass is doubled: F’ = G(2m)(2m)/r² = 4Gm²/r² = 4F. Doubling both masses quadruples the gravitational force. The distance is unchanged so the r² denominator is unaffected. Concept Tested: Newton’s law of gravitation: force scales as product of masses (doubling both → ×4)

★ JOVIK Exam Insight Force scaling with mass is a standard NDA calculation. Doubling both masses multiplies force by 4 (not 2). Halving the distance multiplies force by 4. Both together: ×16. Remember: F ∝ m₁m₂/r².

NDA 2015-II

Q. 18. The acceleration due to gravity ‘g’ for objects on or near the surface of earth is related to the universal gravitational constant ‘G’ as (‘M’ is the mass of the earth and ‘R’ is its radius):

(a)  G = g(M/R²)

(b)  g = G(M/R²)

(c)  M = gG/R²

(d)  R = gG/M²

Answer: (b) g = G(M/R²)

Explanation: The gravitational force on a surface mass m is GMm/R². This equals mg (Newton’s second law). Cancelling m: g = GM/R². This can be written as g = G(M/R²). G is the universal gravitational constant: it is not derived from g; g is derived from G, M, and R. Concept Tested: Relation between g and G: g = GM/R² (not G as function of g)

NDA 2015-I

Q. 19. A spring can be used to determine the mass m of an object in two ways: (i) by measuring the extension in the spring due to the object; and (ii) by measuring the oscillation period for the given mass. Which of these methods can be used in a space-station orbiting Earth?

(a)  Both

(b)  Only the extension method

(c)  Only the oscillation method

(d)  Neither

Answer: (c) Only the oscillation method

Explanation: In a space station, all objects are in free fall: there is weightlessness. A hanging mass produces no extension in the spring (no weight to stretch it), so the extension method fails. The oscillation period T = 2π√(m/k) depends on mass and spring constant, not on gravity. The oscillation method works because it uses inertia, not weight. Concept Tested: Mass measurement in weightlessness: oscillation works; extension method fails

★ JOVIK Exam Insight In orbit, the spring extension method fails: no gravity means no extension. The oscillation period T = 2π√(m/k) is gravity-independent and works in space. This is a fundamental distinction between weight-dependent and inertia-based measurement.

Q. 20. The radius of the Moon is about one-fourth that of the Earth and acceleration due to gravity on the Moon is about one-sixth that on the Earth. From this, we can conclude that the ratio of the mass of Earth to the mass of the Moon is about

(a)  10

(b)  100

(c)  1,000

(d)  10,000

Answer: (b) 100

Explanation: From g = GM/R², mass M = gR²/G. M_Earth/M_Moon = (g_Earth × R_Earth²)/(g_Moon × R_Moon²) = (g_Earth/g_Moon) × (R_Earth/R_Moon)² = 6 × (4)² = 6 × 16 = 96 ≈ 100. Concept Tested: Deriving mass ratio using g = GM/R² for Earth and Moon

NDA 2014-I

Q. 21. Planet A has double the radius than that of Planet B. If the mass of Planet A is 4 times heavier than the mass of Planet B, which of the following statements regarding weight of an object is correct?

(a)  Heavier on Planet A than on Planet B

(b)  Heavier on Planet B than on Planet A

(c)  Same on both the Planets

(d)  Cannot be measured on Planet B

Answer: (c) Same on both the Planets

Explanation: Surface gravity g = GM/R². For Planet A: g_A = G(4M)/(2R)² = 4GM/4R² = GM/R². For Planet B: g_B = GM/R². So g_A = g_B. Doubling the radius quadruples R², exactly cancelling the four-times mass increase. An object weighs the same on both planets. Concept Tested: Surface gravity comparison: doubling radius and quadrupling mass gives identical g

★ JOVIK Exam Insight Planet comparison is the most tested concept in NDA Gravitation. When mass scales as R², gravity is unchanged. Always compute g = GM/R² for each planet: do not assume a larger mass means higher gravity.

NDA 2013-I

Q. 22. Gravitational force shares a common feature with electromagnetic force. In both cases, the force is

(a)  between massive and neutral objects

(b)  between charged objects

(c)  a short range

(d)  a long range

Answer: (d) a long range

Explanation: Both gravitational and electromagnetic forces obey inverse-square laws and act over unlimited distances: they are long-range forces. Gravitational force acts between all massive objects (not just charged or neutral ones). Electromagnetic force acts between charged particles. Their shared feature is long-range action across vast distances without a material medium. Concept Tested: Gravitational and electromagnetic forces: both are long-range (inverse-square law)

NDA 2012-II

Q. 23. A body weighs 5 kg on equator. At the poles it is likely to weigh

(a)  5 kg

(b)  less than 5 kg but not zero

(c)  0 kg

(d)  more than 5 kg

Answer: (d) more than 5 kg

Explanation: The Earth is oblate: its polar radius is smaller than its equatorial radius. Since g = GM/R², a smaller radius at the poles means larger g at the poles. A body therefore weighs more at the poles than at the equator. It is a common misconception that the poles are lighter: the opposite is true. Concept Tested: Variation of weight with latitude: weight is greater at poles than at equator

★ JOVIK Exam Insight g is larger at the poles than at the equator because the polar radius is smaller (oblate Earth). NDA tested this in 2012-II and 2016-II. A body always weighs more at the poles than at the equator.

NDA 2012-I

Q. 24. A body attached to a spring balance weighs 10 kg on the Earth. The body attached to the same spring balance is taken to a planet where gravity is half that of the Earth. The balance will read

(a)  20 kg

(b)  10 kg

(c)  5 kg

(d)  2.5 kg

Answer: (c) 5 kg

Explanation: A spring balance measures weight (gravitational force), not mass. On Earth, weight = mg = 10g. On the other planet, gravity = g/2, so weight = m(g/2) = 5g-force. The balance is calibrated for Earth gravity, so it reads 5 kg-force: half the Earth reading. The actual mass of the body remains 10 kg throughout. Concept Tested: Spring balance: reads weight (not mass); halved gravity gives half the reading

★ JOVIK Exam Insight A recurring NDA trap: mass is constant everywhere; weight and spring balance reading change with gravity. Mass = 10 kg always. Spring balance on a half-g planet reads 5 kg-force (half the Earth weight).

NDA 2011-II

Q. 25. Which one among the following is the correct value of the gravitational force of the Earth acting on a body of mass 1 kg?

(a)  8.9 N

(b)  9.8 N

(c)  89 N

(d)  98 N

Answer: (b) 9.8 N

Explanation: Gravitational force = mg = 1 kg × 9.8 m/s² = 9.8 N. The standard value of g at the Earth’s surface is 9.8 m/s². Option (a) 8.9 N uses an incorrect value of g. Options (c) and (d) are off by a factor of 10: a decimal-place error. Concept Tested: Gravitational force on 1 kg body: F = mg = 9.8 N

NDA 2010-II

Q. 26. An annular solar eclipse occurred during January 2010 with duration of annularity around 12 minutes. It is predicted that such long annular duration will not occur till the year 3043. Such prediction is possible due to:

(a)  Einstein’s theory of relativity

(b)  Darwin’s theory of natural selection

(c)  Newton’s theory of gravitation

(d)  Hawking’s theory of black hole

Answer: (c) Newton’s theory of gravitation

Explanation: Long-term prediction of celestial events: including solar eclipses, orbital positions, and their recurrence: is made possible by Newton’s theory of gravitation. By calculating gravitational interactions between the Earth, Moon, and Sun with high precision, astronomers can predict orbital configurations centuries into the future. Concept Tested: Predictive power of Newton’s theory of gravitation: celestial event prediction

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