Gravitation – MCQs with Answers and Explanations
Class 9
Physics — Chapter 10: Gravitation
CBSE Class 9 Science – Chapter Wise Study Materials Based on NCERT
CBSE Board Examinations
Content Bank — Chapter 10: Gravitation (topics covered)
- Universal law of gravitation, gravitational constant (G)
- Gravitational field and acceleration due to gravity (g)
- Variation of g with altitude and depth
- Mass vs weight, apparent weight, weightlessness
- Free fall and kinematics under gravity
- Orbital motion, satellites, escape velocity, Kepler relations
- Applications: tides, pendulum, Cavendish idea
50 Multiple Choice Questions (MCQs) — Gravitation
Answers with short, concept-clearing explanations follow each question. Use these topic-wise MCQs for self-testing and classroom quizzes.
Universal Law of Gravitation (Q1–Q08)
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Q1. Newton’s Universal Law of Gravitation states that the gravitational force between two point masses is:
A. proportional to the sum of their masses and inversely proportional to distance.
B. proportional to the product of their masses and inversely proportional to the square of distance.
C. proportional to the product of their masses and inversely proportional to distance.
D. independent of distance.Answer: B.Law:F = G m₁ m₂ / r². Force ∝ product of masses and ∝ 1/r². -
Q2. The SI unit of the gravitational constant G is:
A. N/kg
B. N·m²/kg²
C. N·m/kg
D. kg·m²/s²Answer: B.FromF = G m₁ m₂ / r², G has units making F in newtons →N·m²/kg². -
Q3. If both masses in the gravitational formula are doubled and the distance between them is halved, the new force compared to original is:
A. 4 times
B. 8 times
C. 16 times
D. 32 timesAnswer: C.Force scales as (2·2)/(1/2)² = 4 / (1/4) = 16. -
Q4. Gravitational force between two point masses acts:
A. along a line perpendicular to the line joining them.
B. along a line joining their centres and is attractive.
C. only if masses are in contact.
D. repulsive if masses are similar.Answer: B.Gravity is an attractive central force along the line joining mass centres. -
Q5. Two 1 kg masses are placed 1 m apart. If gravitational constant G = 6.67×10⁻¹¹, the force between them is of order:
A. 6.67×10⁻¹¹ N
B. 6.67×10⁻⁶ N
C. 6.67×10⁻²² N
D. 6.67×10⁻¹ NAnswer: A.F = G·1·1/1² = G ≈ 6.67×10⁻¹¹ N — extremely small for everyday masses. -
Q6. Gravitational superposition principle means:
A. Forces cancel always.
B. Net gravitational force is product of individual forces.
C. Net gravitational force is vector sum of forces from each body.
D. Gravitational forces only act pairwise and cannot combine.Answer: C.Each pair contributes and net is vector sum — linear superposition. -
Q7. Two identical spheres touch each other. The distance used in gravitational formula is:
A. zero
B. sum of their radii (centre-to-centre distance)
C. difference of radii
D. their circumferencesAnswer: B.Use centre-to-centre distance; for touching spheres it's r₁ + r₂. -
Q8. Which of the following statements is correct?
A. Gravitational force depends on the medium between masses.
B. Gravitational force is independent of the medium (classical view).
C. Gravitational force vanishes in vacuum.
D. Gravitational force is repulsive in certain media.Answer: B.Classically gravity depends only on masses and separation; medium does not affect it.
Gravitational Field & Acceleration due to Gravity (Q9–Q14)
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Q9. Gravitational field intensity (g) at a point is defined as:
A. force on unit mass placed at that point.
B. mass per unit force.
C. potential energy per unit mass.
D. rate of change of force with distance.Answer: A.g = F_test / m_test; force experienced per unit mass. -
Q10. Expression for g at distance r from a spherical mass M is:
A. g = GM r²
B. g = GM / r
C. g = GM / r²
D. g = G / Mr²Answer: C.Derived from F = GMm/r² and a = F/m → g = GM/r². -
Q11. Which is true about g near Earth's surface?
A. g varies widely from place to place by orders of magnitude.
B. g is approximately 9.8 m/s² and varies slightly with latitude and altitude.
C. g is same everywhere in the universe.
D. g depends on the mass of the test object.Answer: B.g ≈ 9.8 m/s² but small variations due to Earth's shape, rotation and geology. -
Q12. Direction of gravitational field near Earth's surface is:
A. upward
B. tangential to surface
C. towards the centre of Earth (downward)
D. away from centreAnswer: C.Gravity pulls toward Earth's centre. -
Q13. If Earth’s mass were doubled but radius remained same, g at surface would:
A. halve
B. double
C. remain same
D. quadrupleAnswer: B.g ∝ M/R², doubling M doubles g. -
Q14. The value of g is independent of:
A. height above Earth (always)
B. mass of the test object
C. Earth's mass
D. Earth's radiusAnswer: B.g depends on source mass and distance, not on test mass.
Variation of g with Altitude & Depth (Q15–Q18)
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Q15. At height h above Earth's surface, g' is given by:
A. g' = g (R + h)² / R²
B. g' = g (R / (R + h))²
C. g' = g (R + h)/R
D. g' = g (R/(R − h))²Answer: B.From g = GM/r² with r = R + h → g' = g (R/(R+h))². -
Q16. For small h (h ≪ R), approximate change in g is:
A. g' ≈ g (1 + 2h/R)
B. g' ≈ g (1 − 2h/R)
C. g' ≈ g (1 − h/R)
D. g' ≈ g (1 + h/R)Answer: B.Binomial expansion: (R/(R+h))² ≈ (1 + h/R)^{-2} ≈ 1 − 2h/R. -
Q17. Inside Earth at depth d (uniform density), g behaves as:
A. g increases linearly with d
B. g decreases linearly with d: g_d = g(1 − d/R)
C. g becomes infinite at centre
D. g remains constantAnswer: B.For uniform density enclosed mass ∝ (R−d)³ leading to g_d = g(1 − d/R). -
Q18. At Earth's centre gravitational acceleration is:
A. maximum
B. zero
C. equal to surface g
D. infiniteAnswer: B.Symmetry causes net gravitational field to cancel at centre.
Mass, Weight & Apparent Weight (Q19–Q23)
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Q19. Mass of a body is measured in:
A. Newtons (N)
B. Kilograms (kg)
C. m/s²
D. Joules (J)Answer: B.Mass unit is kg; weight unit is N. -
Q20. Weight of a 10 kg object on Earth (g = 9.8 m/s²) is:
A. 9.8 N
B. 98 N
C. 10 N
D. 100 NAnswer: B.W = m g = 10 × 9.8 = 98 N. -
Q21. Apparent weight in an elevator accelerating upward with acceleration a is:
A. m(g − a)
B. mg
C. m(g + a)
D. maAnswer: C.Normal reaction N = m(g + a) when elevator accelerates upward. -
Q22. Weightlessness occurs when:
A. g = 0 always
B. normal reaction becomes zero (free fall)
C. mass becomes zero
D. speed becomes zeroAnswer: B.In free fall both object and support accelerate equally; normal reaction is zero → apparent weightlessness. -
Q23. If a person weighs 600 N on Earth, on Moon (g ≈ 1/6) their weight will be approximately:
A. 100 N
B. 600 N
C. 3600 N
D. 50 NAnswer: A.On Moon weight ≈ 600/6 ≈ 100 N (mass unchanged).
Free Fall & Kinematics under Gravity (Q24–Q29)
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Q24. A body dropped from rest falls distance s in time t. Relation is:
A. s = ut + ½ a t² with u = 0 → s = ½ g t²
B. s = ut − ½ g t²
C. s = gt + ½ u t²
D. s = utAnswer: A.Standard kinematics with u = 0 gives s = ½ g t² for free fall. -
Q25. For an object thrown upward with speed u, time to reach maximum height is:
A. u/g
B. 2u/g
C. u/2g
D. g/uAnswer: A.At top v = 0; v = u − g t → t = u/g. -
Q26. If two objects of different masses are dropped in vacuum, they reach ground at:
A. different times
B. same time
C. heavier first
D. lighter firstAnswer: B.In vacuum acceleration = g independent of mass (inertia and gravitational force scale similarly). -
Q27. Relation v² = u² + 2 g s is valid when:
A. acceleration is variable
B. acceleration is constant (here g)
C. only for horizontal motion
D. only in vacuumAnswer: B.This kinematic equation requires constant acceleration. -
Q28. A stone dropped from tower takes 5 s to reach ground. Height of tower (g = 9.8 m/s²) ≈:
A. 122.5 m
B. 49 m
C. 245 m
D. 19.6 mAnswer: A.h = ½ g t² = 0.5×9.8×25 = 122.5 m. -
Q29. In free-fall problems, air resistance is often neglected because:
A. it is always zero
B. it simplifies to uniform acceleration and is negligible for dense objects at moderate speeds
C. it acts upward only
D. gravity cancels dragAnswer: B.For small, dense objects and short distances, drag is small vs gravity; neglecting it is a useful approximation at NCERT level.
Satellites, Orbital Motion & Escape Velocity (Q30–Q39)
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Q30. For a satellite in circular orbit, centripetal force is provided by:
A. magnetic force
B. gravitational force
C. normal reaction
D. frictional forceAnswer: B.Gravity supplies required centripetal acceleration: m v² / r = GM m / r². -
Q31. Orbital speed v at radius r from Earth's centre is:
A. v = √(GM r)
B. v = GM / r²
C. v = √(GM / r)
D. v = 2π r / T onlyAnswer: C.From m v²/r = GMm/r² → v = √(GM/r). -
Q32. Orbital period T of a satellite increases when:
A. orbit radius decreases
B. orbit radius increases
C. mass of satellite increases
D. Earth's mass decreasesAnswer: B.T = 2π √(r³/GM) → larger r ⇒ larger T. Satellite mass cancels out. -
Q33. Escape velocity from Earth’s surface depends on:
A. mass of the escaping object
B. Earth's mass and radius only
C. atmospheric pressure only
D. object's shape onlyAnswer: B.v_e = √(2GM/R) depends on GM and R; independent of object mass (neglecting atmosphere). -
Q34. Numerical: If orbital speed at radius r is v, the escape speed at same radius is:
A. v/√2
B. v√2
C. 2v
D. vAnswer: B.v_orb² = GM/r; v_e² = 2GM/r = 2 v_orb² ⇒ v_e = √2 v_orb. -
Q35. Low Earth Orbit (LEO) satellites are typically placed at altitudes of:
A. 100–200 km
B. 200–2000 km
C. 36,000 km
D. 400,000 kmAnswer: B.LEO usually ranges from ~200 to 2000 km (ISS ~400 km). GEO is ~36,000 km altitude. -
Q36. A satellite in geostationary orbit has period equal to:
A. 12 hours
B. 24 hours
C. 1 hour
D. depends on satellite massAnswer: B.Geostationary satellites match Earth's rotation (24 hours) and stay over equator. -
Q37. For a satellite at orbit radius r, gravitational potential energy per unit mass is (taking zero at infinity):
A. +GM/r
B. −GM/r
C. −GM/2r
D. +GM/2rAnswer: B.Potential per unit mass V = −GM/r (zero at infinity). -
Q38. The total (kinetic + potential) specific energy of a satellite in circular orbit is:
A. −GM/r
B. 0
C. −GM/(2r)
D. +GM/(2r)Answer: C.K = ½ GM/r, U = −GM/r → total = −GM/(2r). -
Q39. If a satellite's orbit radius increases, its orbital speed:
A. increases
B. decreases
C. remains same
D. becomes zeroAnswer: B.v = √(GM/r) decreases as r increases.
Kepler's Laws, Tides & Applications (Q40–Q45)
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Q40. Kepler’s third law for circular orbits states that:
A. T ∝ r
B. T² ∝ r³
C. T² ∝ r²
D. T ∝ r³Answer: B.From Newton: T = 2π √(r³/GM) → T² ∝ r³. -
Q41. Tides on Earth are mainly caused by gravitational pull of:
A. Jupiter
B. Sun only
C. Moon (primarily) and Sun (secondarily)
D. Earth's coreAnswer: C.Moon has strongest tidal effect; Sun adds or subtracts depending on alignment (spring/neap tides). -
Q42. Kepler’s first law states that planetary orbits are:
A. circles with Sun at centre
B. ellipses with Sun at one focus
C. parabolas
D. hyperbolasAnswer: B.Planets move in elliptical orbits with Sun at a focus. -
Q43. When Sun, Earth and Moon are in straight line (new or full moon), tides are:
A. neap tides (smaller)
B. spring tides (larger)
C. no tides
D. unpredictableAnswer: B.Gravitational effects of Sun and Moon reinforce → spring tides. -
Q44. Kepler’s second law (equal areas) implies conservation of:
A. linear momentum
B. angular momentum
C. kinetic energy
D. potential energyAnswer: B.Equal areas in equal times reflect constancy of angular momentum for central force motion. -
Q45. Which practical method estimates G or Earth's mass conceptually in classrooms?
A. Mercury thermometer
B. Cavendish-like torsion balance (conceptual)
C. Pendulum only
D. BarometerAnswer: B.Cavendish experiment measures tiny gravitational forces to determine G (classroom description often given).
Concept Checks, Numericals & Common Errors (Q46–Q50)
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Q46. A student computes g at altitude equal to Earth's radius (h = R) as g/2. Correct value is:
A. g/2
B. g/3
C. g/4
D. gAnswer: C.At r = 2R, g' = g (R/(2R))² = g (1/2)² = g/4, not g/2. -
Q47. If distance between two masses is tripled, gravitational force becomes:
A. 3 times
B. 1/3
C. 1/9
D. 9 timesAnswer: C.Inverse-square law: factor = 1/3² = 1/9. -
Q48. A rock thrown up returns to same spot. Its speed just before hitting is (neglect air resistance):
A. greater than initial speed
B. equal to initial speed
C. less than initial speed
D. zeroAnswer: B.Energy conservation: kinetic energy regained → speed equal to launch speed (ignoring air resistance). -
Q49. A common exam mistake is using surface g for problems far from Earth. Best practice is:
A. always use 9.8
B. compute g using g = GM/r² at given r when required
C. ignore variation
D. use g = 0 far awayAnswer: B.When r differs significantly from R, use GM/r² or appropriate approximation. -
Q50. Which statement is correct for orbital mechanics at NCERT level?
A. Larger orbit radius → larger orbital speed.
B. Larger orbit radius → smaller orbital speed and larger period.
C. Orbit period independent of radius.
D. Satellite mass affects required orbital speed significantly.Answer: B.v = √(GM/r) decreases with r; T = 2π √(r³/GM) increases with r. Satellite mass cancels out in speed formula.
Note: These 50 MCQs are written topic-wise and strictly follow NCERT-level coverage for Class 9 Physics — Chapter 10: Gravitation. You can paste this HTML into a WordPress "Custom HTML" block. If you want, I can:
- These Multiple Choice Questions (MCQs) are designed strictly as per the NCERT syllabus, making them ideal for CBSE Class 9 board exams standard.
