Gravitation – Physics Questions Asked in SSC & Railway Exams (PYQs with Answers)

SSC Stenographer 2026 – 9 September 2026 Shift 2 · Acceleration due to gravity (g) and its variation

Q1. Acceleration due to gravity on Earth is approximately?

(a) 4.9 m/s²
(b) 9.8 m/s²
(c) 6.67 m/s²
(d) 10.6 m/s²

Answer & solution

Answer: (b) 9.8 m/s²

Answer: Option 2 — 9.8 m/s²
At the Earth's surface g = GM/R², and with M ≈ 6 × 10²⁴ kg and R ≈ 6,400 km this works out to about 9.8 m/s².
Every freely falling body near the surface picks up 9.8 m/s of speed every second.
Revision note — Acceleration due to gravity
• g = GM/R², where G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant. g depends on the planet; G does not.
• The standard value is 9.8 m/s² (taken as 10 m/s² in most numericals). g is a vector, always directed towards the centre of the Earth.
• g is greatest at the poles (about 9.83 m/s²) and least at the equator (about 9.78 m/s²), because the Earth is flattened at the poles and because of its rotation.
• g decreases with height above the surface and also decreases with depth, becoming zero at the centre of the Earth.
• On the Moon g is about one-sixth of the Earth's value, roughly 1.63 m/s², because the Moon's mass is far smaller.
• In free fall every body has the same acceleration g whatever its mass — a coin and a feather fall together in a vacuum; in air the feather is held back by air resistance.
Do not mix these up
• g vs G — g = 9.8 m/s² changes with place and planet; G = 6.67 × 10⁻¹¹ is a universal constant, the same everywhere. The value 6.67 in the options belongs to G, not to g.
• Mass vs weight — mass is constant and measured in kg; weight is W = mg, measured in newton, and drops to one-sixth on the Moon.
Also asked from this topic: the kind of motion a body in free fall has (uniformly accelerated motion); why gravity on the Moon is weaker (its mass is much smaller); the weight of a 70 kg astronaut on the Moon (about 117 N).

RRB NTPC UG CBT-1 2025-26 – 19 June 2026 Shift 2 · Free fall

Q2. If a ball is thrown vertically upward, which of the following describes its velocity at the highest point?

(a) Positive
(b) Negative
(c) Zero
(d) Equal to acceleration due to gravity

Answer & solution

Answer: (c) Zero

Vertical Motion of Ball
When ball is thrown upward:
• Velocity decreases continuously due to gravity
• At highest point, ball momentarily stops
• Then starts falling downward
At Highest Point:
Velocity = 0 m/s (instantaneous rest)
Acceleration = -g (downward, constant throughout)
Note: Velocity is zero, but acceleration is NOT zero
Answer: Option 3 → Zero

SSC GD 2026 – 30 May 2026 Shift 1 · Mass and weight

Q3. What is the SI unit of weight?

(a) Kilogram (Kg)
(b) Newton (N)
(c) Joule (J)
(d) Pascal (Pa)

Answer & solution

Answer: (b) Newton (N)

Correct Answer: Newton (N) — Option (b)
The SI unit of weight is Newton (N). About Weight vs. Mass: This is a very common conceptual confusion — weight and mass are different quantities; Mass: Amount of matter in an object; SI unit: Kilogram (kg); Scalar quantity; Remains constant everywhere; Measured using a balance; Weight: The gravitational force acting on an object; Weight = Mass × Acceleration due to gravity (W = mg); SI unit: Newton (N) — because weight is a FORCE; Vector quantity (has direction — towards center of Earth); Changes with location (less on Moon, zero in space); Measured using a spring balance/weighing scale; Why Newton is the unit of weight: Weight is the force of gravity on an object; Force = Mass × Acceleration (Newton's Second Law: F = ma); SI unit of force is Newton (N) = kg⋅m /s²; Therefore weight (being a force) is measured in Newtons; Numerical example: A person of mass 60 kg on Earth: Weight = 60 × 9.8 = 588 N; On the Moon (g = 1.63 m/s²): Weight = 60 × 1.63 = 97.8 N (less); Mass remains 60 kg in both cases; Common confusion: In everyday life, we say 'I weigh 60 kg' — this is technically mass (60 kg), not weight (588 N); The scientific/SI definition uses Newton for weight; About other options: Joule (J): SI unit of energy/work; Pascal (Pa): SI unit of pressure (Force per unit area); Kilogram (Kg): SI unit of MASS (NOT weight).
Option (a) Kilogram (Kg) — Incorrect: Kilogram is the SI unit of MASS, NOT weight; weight and mass are different physical quantities — mass is the amount of matter (measured in kg), while weight is the gravitational force on that mass (measured in Newton); W = mg (Weight = Mass × g), where g is acceleration due to gravity (~9.8 m/s² on Earth); while we colloquially say 'I weigh 60 kg,' scientifically weight is in Newtons. The candidate incorrectly selected this option.
Option (c) Joule (J) — Incorrect: Joule is NOT the SI unit of weight; Joule is the SI unit of ENERGY and WORK; 1 Joule = 1 Newton-metre = 1 kg⋅m²/s²; energy and weight are different physical quantities; the SI unit of weight (which is a force) is Newton (N).
Option (d) Pascal (Pa) — Incorrect: Pascal is NOT the SI unit of weight; Pascal is the SI unit of PRESSURE; 1 Pascal = 1 Newton per square metre (N/m²); pressure and weight are different quantities — pressure is force per unit area, while weight is simply a force; the SI unit of weight is Newton (N).

SSC GD 2026 – 27 May 2026 Shift 1 · Universal law of gravitation and G

Q4. Who experimentally determined the value of the Universal Gravitational Constant (G)?

(a) Isaac Newton
(b) Albert Einstein
(c) Henry Cavendish
(d) Johannes Kepler

Answer & solution

Answer: (c) Henry Cavendish

Correct Answer: Henry Cavendish — Option (c)
Henry Cavendish experimentally determined the value of the Universal Gravitational Constant (G) in 1798 through his famous torsion balance experiment. About Henry Cavendish and the Cavendish Experiment: Scientist: Henry Cavendish (1731-1810) — English physicist and chemist; Year of experiment: 1798; Experiment: The Cavendish Torsion Balance Experiment; Setup: Cavendish used a torsion balance — a wooden rod suspended horizontally from a thin wire with small lead spheres (m) at each end; Two larger fixed lead spheres (M) were placed near the small spheres; The gravitational attraction between the spheres caused the wire to twist; By measuring the angle of twist and knowing the torsion constant of the wire, Cavendish calculated the gravitational force; Result: Measured G ≈ 6.74 × 10⁻¹¹ N⋅m²/kg² (remarkably close to today's accepted value of 6.674 × 10⁻¹¹ N⋅m²/kg²); Cavendish actually described his experiment as 'weighing the Earth' rather than measuring G; From his measurement, the density and mass of Earth could be calculated; About Newton's role: Isaac Newton: Formulated the Universal Law of Gravitation: F = Gm₁m ₂/r²; Newton identified the existence and role of G in his formula, but did NOT experimentally determine its numerical value; Newton's contribution was the theoretical framework — Cavendish's was the experimental determination of G's value; About others: Albert Einstein: Formulated General Theory of Relativity (a more advanced theory of gravity, 1915); Johannes Kepler: Formulated the three Laws of Planetary Motion (early 17th century) — precursor to Newton's work.
Option (a) Isaac Newton — Incorrect: Isaac Newton formulated the Universal Law of Gravitation (F = Gm₁m ₂/r²) and identified the gravitational constant G in theory, but he did NOT experimentally determine the numerical value of G; the experimental determination was done by Henry Cavendish in 1798. The candidate selected this incorrect option.
Option (b) Albert Einstein — Incorrect: Albert Einstein developed the General Theory of Relativity (1915), which provides a more sophisticated description of gravity — but Einstein did NOT determine the value of G; that was accomplished by Henry Cavendish in 1798 using his torsion balance experiment.
Option (d) Johannes Kepler — Incorrect: Johannes Kepler (1571-1630) formulated the three laws of planetary motion (Kepler's Laws) based on Tycho Brahe's astronomical observations — these laws describe how planets move around the Sun but do NOT involve the gravitational constant G; Henry Cavendish determined G experimentally in 1798.

SSC GD 2026 – 26 May 2026 Shift 1 · Universal law of gravitation and G

Q5. Which of the following phenomena is explained by the Universal Law of Gravitation?

(a) Nuclear fusion in the Sun
(b) Tides due to Moon and Sun
(c) Magnetic attraction between poles
(d) Motion of electrons around the nucleus

Answer & solution

Answer: (b) Tides due to Moon and Sun

Correct Answer: Tides due to Moon and Sun — Option (b)
Tides due to the Moon and Sun are explained by Newton's Universal Law of Gravitation. About Newton's Universal Law of Gravitation: Every object in the universe attracts every other object with a force proportional to the product of their masses and inversely proportional to the square of the distance between them; Formula: F = G × (m1 × m2) / r²; G = Universal Gravitational Constant (6.67 × 10⁻¹¹ N⋅m²/kg²); About Tides and Gravitation: Tides are caused by the differential gravitational pull of the Moon (primarily) and the Sun on Earth's oceans; The Moon's gravitational pull creates a bulge of ocean water on the side facing the Moon (high tide) and on the opposite side (due to inertia); As Earth rotates, different areas experience high and low tides; This is directly explained by Newton's Universal Law of Gravitation ✓ ; Other phenomena explained by Universal Gravitation: Planetary orbits around the Sun; Motion of Moon around Earth; Weight of objects; Falling of objects to the Earth; Phenomena NOT explained by gravitational law: Nuclear fusion in the Sun: Explained by nuclear force (strong force) — a fundamental force at subatomic level; Magnetic attraction between poles: Explained by electromagnetic force (Maxwell's equations); Motion of electrons around the nucleus: Explained by electromagnetic force and quantum mechanics (NOT gravitational force — gravity at atomic scale is negligible).
Option (a) Nuclear fusion in the Sun — Incorrect: Nuclear fusion in the Sun is explained by the strong nuclear force — it involves hydrogen nuclei fusing to form helium at extremely high temperatures and pressures; this is a nuclear physics phenomenon, NOT explained by Newton's gravitational law.
Option (c) Magnetic attraction between poles — Incorrect: Magnetic attraction between poles is explained by electromagnetic force (one of the four fundamental forces) — this is governed by Maxwell's equations and the Biot -Savart law, NOT by the Universal Law of Gravitation.
Option (d) Motion of electrons around the nucleus — Incorrect: The motion of electrons around the nucleus is explained by electromagnetic force (attraction between negatively charged electrons and positively charged protons) and quantum mechanics (wave-particle duality) — gravitational force at atomic level is negligibly small compared to electromagnetic force.

SSC GD 2026 – 26 May 2026 Shift 2 · Mass and weight

Q6. Which physical property remains the same on both Earth and Moon?

(a) Weight
(b) Mass
(c) Gravitational acceleration
(d) Pressure

Answer & solution

Answer: (b) Mass

Correct Answer: Mass — Option (b)
Mass remains the same on both Earth and Moon (and everywhere else in the universe). About Mass: Definition: Mass is the amount of matter contained in an object; it is an intrinsic property of matter; SI Unit: Kilogram (kg); Mass is a scalar quantity; Mass does NOT depend on gravity — it remains constant everywhere (Earth, Moon, space, or any planet); An object with mass 60 kg on Earth will have the same 60 kg mass on the Moon; About Weight vs Mass: Weight = Mass × Gravitational acceleration (W = mg); Weight CHANGES with location because gravitational acceleration (g) varies; Earth's g ≈ 9.8 m/s²; Moon's g ≈ 1.67 m/s² (about 1/6th of Earth's); So an object weighing 60 N on Earth weighs only 10 N on the Moon; But its MASS remains 60/9.8 ≈ 6.12 kg on both Earth and Moon; About other options: Weight: Weight = mg; since Moon's g is 1/6th of Earth's, weight on Moon is 1/6th of weight on Earth — NOT the same; Gravitational acceleration (g): g = GM/R²; Since Moon has different mass and radius than Earth, g on Moon ≈ 1.67 m/s² vs g on Earth ≈ 9.8 m/s² — NOT the same; Pressure: Atmospheric pressure is vastly different — Earth has 1 atm atmospheric pressure while Moon has virtually no atmosphere (near vacuum) — NOT the same; Therefore, MASS is the only property that remains the same on Earth and Moon.
Option (a) Weight — Incorrect: Weight (W = mg) depends on gravitational acceleration; since Moon's gravity is approximately 1/6th of Earth's, an object weighs about 1/6th as much on the Moon compared to Earth — weight is NOT the same on Earth and Moon.
Option (c) Gravitational acceleration — Incorrect: Gravitational acceleration (g) is different on Earth (9.8 m/s²) and Moon (1.67 m/s²) because it depends on the mass and radius of the celestial body (g = GM/R²); g is NOT the same on Earth and Moon.
Option (d) Pressure — Incorrect: Earth has a substantial atmosphere with 1 atm (101,325 Pa) atmospheric pressure at sea level, while the Moon has virtually no atmosphere and near-zero atmospheric pressure; pressure is NOT the same on Earth and Moon.

SSC GD 2026 – 26 May 2026 Shift 3 · Free fall

Q7. Due to which factor does a sheet of paper fall more slowly than a stone in air?

(a) Mass difference
(b) Air resistance
(c) Difference in size/shape
(d) Gravitational difference

Answer & solution

Answer: (b) Air resistance

Correct Answer: Air resistance — Option (b)
A sheet of paper falls more slowly than a stone in air due to air resistance (drag force). About Air Resistance and Falling Objects: Newton's Law of Gravitation: Both the stone and paper are pulled toward Earth by gravity with the same gravitational acceleration g ≈ 9.8 m/s² (since gravitational acceleration is independent of mass); In vacuum: Both a stone and a paper sheet would fall at the SAME rate and reach the ground simultaneously — this was famously demonstrated by Galileo and later by Apollo 15 astronauts (who dropped a feather and a hammer on the Moon — they landed simultaneously); In air: The paper falls more slowly because: Air resistance (drag force) depends on: Surface area (larger area = more air resistance), Shape of the object, Velocity of the object; A paper sheet has a LARGE surface area relative to its mass; A stone has a SMALL surface area relative to its mass; The paper experiences proportionally much more air resistance compared to its weight (gravitational force), so it falls more slowly; Key distinction: The question asks about falling in AIR — the air resistance is the determining factor; Mass difference (Option a): Both stone and paper have different masses, but this alone doesn't explain the different falling speeds — a heavy paper and a light paper both fall slowly; Difference in size/shape (Option c): Shape affects air resistance, but the root cause is AIR RESISTANCE — shape is a secondary factor that determines HOW MUCH air resistance acts; the correct answer is 'air resistance' as the causal factor; Gravitational difference (Option d): Gravity acts equally on all objects — there is NO gravitational difference for objects near Earth's surface.
Option (a) Mass difference — Incorrect: In a vacuum, objects of different masses fall at the same rate — this proves mass difference alone does NOT cause different falling speeds; it is air resistance that causes the paper to fall more slowly than the stone in air.
Option (c) Difference in size/shape — Incorrect: While the paper's large flat shape contributes to higher air resistance, the ROOT CAUSE of the slower fall is air resistance itself — shape is what determines the magnitude of air resistance, but 'air resistance' is the correct causal factor. The candidate selected this option but 'air resistance' is more precisely correct.
Option (d) Gravitational difference — Incorrect: Gravitational acceleration (g) is the same for all objects near Earth's surface regardless of their mass or composition — there is NO gravitational difference between a stone and a paper; the correct answer is air resistance, which acts differently on them due to their different surface area-to-mass ratios.

SSC GD 2026 – 20 May 2026 Shift 3 · Universal law of gravitation and G

Q8. According to Newton's Universal Law of Gravitation, what symbol is used to denote the constant of proportionality?

(a) μ
(b) k
(c) G
(d) φ

Answer & solution

Answer: (c) G

Correct Answer: G — Option (c)
In Newton's Universal Law of Gravitation, the constant of proportionality is denoted by the symbol G (the Universal Gravitational Constant). Newton's Law of Universal Gravitation: Every two objects with mass attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematical expression: F = G × (m₁ × m₂) / r², where F = gravitational force between two objects, m₁ and m₂ = masses of the two objects, r = distance between their centres , G = Universal Gravitational Constant (constant of proportionality). Value of G: G = 6.674 × 10⁻¹¹ N m² kg⁻² (or m³ kg⁻¹ s⁻²); First measured by Henry Cavendish in 1798 using a torsion balance experiment; G is a universal constant — its value remains the same everywhere in the universe. Key distinction: G (universal gravitational constant) is different from g (acceleration due to gravity on Earth's surface = 9.8 m/s²).
Option (a) μ — Incorrect: μ (mu) is not the symbol for the gravitational constant; μ is used as the symbol for coefficient of friction, permeability in electromagnetism, or micro (10⁻⁶) as a prefix; the gravitational constant is denoted by G.
Option (b) k — Incorrect: k is typically used for spring constant (Hooke's Law: F = - kx ), Boltzmann constant ( k_B ), or Coulomb's constant in electrostatics; the Universal Gravitational Constant is denoted by G, not k.
Option (d) φ — Incorrect: φ (phi) is used in mathematics to represent the golden ratio (approximately 1.618), in physics for magnetic flux (Φ), or in geometry for angles; the constant of proportionality in Newton's Law of Gravitation is G, not φ. The candidate selected this incorrect option.

SSC GD 2026 – 5 May 2026 Shift 1 · Satellites, orbital velocity and escape velocity

Q9. A geostationary satellite remains stationary above a point on Earth's equator. Its orbital radius is approximately 42,200 km.
Calculate the orbital period of the satellite in hours.

(a) 24 hours
(b) 12 hours
(c) 6 hours
(d) 48 hours

Answer & solution

Answer: (a) 24 hours

Correct Answer: 24 hours — Option (a)
A geostationary satellite has an orbital period exactly equal to Earth's rotation period — 24 hours (more precisely, 23 hours 56 minutes 4 seconds or one sidereal day). This is why it appears stationary above a fixed point on the equator. The orbital radius of approximately 42,200 km (or about 35,786 km above Earth's surface) is specifically calculated so that the satellite's orbital speed matches Earth's rotational speed. This property makes geostationary satellites ideal for communications, weather monitoring, and broadcasting.
Option (b) 12 hours — Incorrect: A 12-hour orbital period corresponds to a Medium Earth Orbit (MEO) satellite like GPS satellites, not a geostationary satellite.
Option (c) 6 hours — Incorrect: A 6-hour orbital period would correspond to a much lower orbit than geostationary.
Option (d) 48 hours — Incorrect: A 48-hour period would mean the satellite orbits slower than Earth's rotation and would not remain stationary above a point.

SSC GD 2026 – 5 May 2026 Shift 2 · Satellites, orbital velocity and escape velocity

Q10. If the orbital period of a satellite is 90 minutes, what is the primary force keeping it in orbit?

(a) Earth's gravitational force
(b) Centrifugal force of the satellite
(c) Magnetic force from Earth
(d) Electromagnetic force

Answer & solution

Answer: (a) Earth's gravitational force

Correct Answer: Earth's gravitational force — Option (a)
The primary force keeping a satellite in orbit is Earth's gravitational force. Gravity pulls the satellite toward Earth's centre , providing the centripetal force required for circular orbital motion. For a Low Earth Orbit (LEO) satellite with a 90-minute orbital period (altitude ~300–400 km), Earth's gravity is the sole force maintaining the orbit. The satellite continuously 'falls' toward Earth but moves forward fast enough that Earth curves away beneath it — resulting in stable circular motion.
Option (b) Centrifugal force of the satellite — Incorrect: Centrifugal force is a pseudo-force experienced in a rotating reference frame; it is not an actual force keeping the satellite in orbit. In an inertial frame, only gravity acts on the satellite.
Option (c) Magnetic force from Earth — Incorrect: Earth's magnetic field does exert some force on charged particles, but it is not the primary force maintaining a satellite's orbit.
Option (d) Electromagnetic force — Incorrect: Electromagnetic forces act on charged particles; they are not the primary orbital force for a satellite.

SSC GD 2026 – 5 May 2026 Shift 3 · Satellites, orbital velocity and escape velocity

Q11. At an altitude of 300 km, a satellite's orbital speed is 7.8 km/s.
If the satellite's altitude increases by 200 km, what will happen to its orbital period?

(a) It will increase
(b) It will decrease
(c) No change
(d) It will change randomly

Answer & solution

Answer: (a) It will increase

Correct Answer: It will increase — Option (a)
According to Kepler's Third Law, T² ∝ r³ (where T is the orbital period and r is the orbital radius). As altitude increases, the orbital radius (r = Earth's radius + altitude) also increases. Therefore, the orbital period T increases with increasing altitude. Additionally, at a higher altitude, the gravitational force is weaker, so the satellite moves more slowly — meaning it takes longer to complete one orbit. Moving from 300 km to 500 km altitude will result in a longer orbital period.
Option (b) It will decrease — Incorrect: The orbital period increases with altitude, not decreases. A lower orbit has a shorter period (faster speed, smaller radius).
Option (c) No change — Incorrect: The orbital period definitely changes with altitude according to Kepler's Third Law.
Option (d) It will change randomly — Incorrect: The change in orbital period with altitude is deterministic and governed by Kepler's Third Law, not random.

SSC GD 2026 – 29 April 2026 Shift 2 · Numericals on gravitation

Q12. If an astronaut's mass is 70 kg, what will be their weight on the Moon (1/6th of Earth's gravity)?

(a) 70 N
(b) 117 N
(c) 420 N
(d) 686 N

Answer & solution

Answer: (b) 117 N

Correct Answer: 117 N — Option (b)
Weight on the Moon = Mass × Gravitational acceleration on Moon. Calculation: On Earth: Weight = mass × g_Earth = 70 kg × 10 m/s² = 700 N (using g = 10 m/s²); On the Moon: g_Moon = g_Earth / 6 = 10 / 6 m/s²; Weight on Moon = 70 × (10/6) = 700/6 ≈ 116.67 N ≈ 117 N. Alternatively using g = 9.8 m/s²: Weight on Earth = 70 × 9.8 = 686 N; Weight on Moon = 686 / 6 ≈ 114.3 N ≈ 114 N (closest to 117 N among the options). Key concept: Mass remains the same everywhere (70 kg on Earth, 70 kg on Moon); Weight changes with gravitational acceleration; The Moon's surface gravity is approximately 1.62 m/s² (about 1/6th of Earth's 9.8 m/s²); Astronauts on the Moon experience 'reduced weight' (not reduced mass) — they feel lighter and can jump higher.
Option (a) 70 N — Incorrect: 70 N would mean the astronaut's weight on the Moon equals their mass in newtons, which is incorrect; weight on Moon = mass × (g/6) = 70 × (10/6) ≈ 117 N, not 70 N.
Option (c) 420 N — Incorrect: 420 N would represent the weight if g_Moon = 6 m/s² (70 × 6), which is wrong; the Moon's gravity is 1/6th of Earth's, not 6 m/s². The candidate selected this incorrect option.
Option (d) 686 N — Incorrect: 686 N is the astronaut's weight on Earth (70 kg × 9.8 m/s² = 686 N); on the Moon, the weight would be 686/6 ≈ 114 N ≈ 117 N — not the same as on Earth.

SSC GD 2026 – 28 April 2026 Shift 1 · Acceleration due to gravity (g) and its variation

Q13. Why is the gravitational force experienced on the Moon
less compared to Earth?

(a) It is smaller in size
(b) Its mass is less
(c) It rotates faster
(d) It is farther from the Sun

Answer & solution

Answer: (b) Its mass is less

Correct Answer: Its mass is less — Option (b)
The gravitational force on the surface of the Moon is approximately 1/6th (about 1.62 m/s²) of that on Earth's surface (9.8 m/s²). According to Newton's Law of Universal Gravitation (g = GM/R²), the surface gravity depends on both the mass (M) and the radius (R) of the celestial body. The Moon's mass is approximately 7.35 × 10²² kg, which is about 1/81st of Earth's mass (5.97 × 10²⁴ kg). Although the Moon is also smaller in radius, the dominant factor for the lower surface gravity is its significantly lower mass. The combination of lower mass and smaller radius results in the Moon having only about 1/6th of Earth's surface gravity.
Option (a) It is smaller in size — Incorrect: While the Moon is smaller in size (radius ≈ 1,737 km vs. Earth's 6,371 km), size alone does not determine gravitational force; the primary reason for less gravity is the Moon's much lower mass.
Option (c) It rotates faster — Incorrect: The Moon actually rotates very slowly (once in about 27.3 days, same as its orbital period around Earth); the rotation speed of a body does not directly cause lower surface gravity.
Option (d) It is farther from the Sun — Incorrect: The distance from the Sun does not determine the surface gravity of a body; surface gravity is determined by the body's own mass and radius, not its distance from the Sun.

SSC GD 2026 – 28 April 2026 Shift 3 · Universal law of gravitation and G

Q14. If the distance between two objects increases six times, what will be the effect on the gravitational force?

(a) Becomes 36 times larger
(b) Becomes 6 times larger
(c) Becomes 36 times smaller
(d) Becomes 1/6 smaller

Answer & solution

Answer: (c) Becomes 36 times smaller

Correct Answer: Becomes 36 times smaller — Option (c)
According to Newton's Law of Universal Gravitation: F = Gm₁m ₂/r², where F is the gravitational force, G is the gravitational constant, m₁ and m₂ are the masses of the two objects, and r is the distance between them. The gravitational force is inversely proportional to the square of the distance (F ∝ 1/r²). If the distance is increased 6 times (r becomes 6r), then: New Force = Gm₁m ₂/(6r)² = Gm₁m ₂/36r² = F/36. So the gravitational force becomes 1/36th of the original, i.e., 36 times smaller. This is the inverse square law — doubling the distance reduces force to 1/4, tripling reduces to 1/9, and increasing by 6 times reduces to 1/36.
Option (a) Becomes 36 times larger — Incorrect: If distance increases, gravitational force DECREASES (not increases) due to the inverse square relationship; increasing distance 6 times makes the force 36 times smaller, not larger.
Option (b) Becomes 6 times larger — Incorrect: This would be true only if force were directly proportional to distance, which is not the case; gravity follows an inverse square law (F ∝ 1/r²), so increasing distance reduces the force.
Option (d) Becomes 1/6 smaller — Incorrect: This would be the result if force were inversely proportional to distance (F ∝ 1/r), not the square of distance; since gravity follows the inverse square law, the force becomes 1/36 (not 1/6) of the original.

RRB NTPC GL CBT-1 2025-26 – 17 March 2026 Shift 1 · Free fall

Q15. Free fall corresponds to which type of motion?

(a) Non-uniform motion
(b) Uniform motion
(c) Uniformly accelerated motion
(d) Periodic motion

Answer & solution

Answer: (c) Uniformly accelerated motion

Free Fall Motion
In free fall, object falls under gravity alone.
Acceleration = g = 9.8 m/s² (constant)
Since acceleration is constant and non-zero,
velocity increases uniformly with time.
Free fall = Uniformly accelerated motion
Answer: Option 3 → Uniformly accelerated motion

RRB Technician Grade III 2025-26 – 10 March 2026 Shift 2 · Acceleration due to gravity (g) and its variation

Q16. If the radius of the Earth were to double while mass remains the same, the new value of g would be ___________the original value.

(a) 2 times
(b) Half
(c) One-fourth
(d) 4 times

Answer & solution

Answer: (c) One-fourth

Acceleration due to gravity g = G × M / R²
If radius R doubles and mass M remains the same, new g' = G × M / (2R)²
= G × M / 4R²
= (1/4) × original g
So, new value of g will be one-fourth the original value.

RRB Technician Grade III 2025-26 – 10 March 2026 Shift 2 · Universal law of gravitation and G

Q17. The SI unit of the universal gravitation constant is ____________.

(a) Nm2kg-2
(b) Nm2kg-1
(c) Nm-2kg-1
(d) Nm-2kg2

Answer & solution

Answer: (a) Nm2kg-2

The universal gravitation constant (G) relates gravitational force, mass, and distance.
Its SI unit comes from Newton's law of gravitation: F = G × (m1 × m2) / r².
Force (F) unit = Newton (N).
Mass (m) unit = kilogram (kg).
Distance (r) unit = metre (m).
Rearranging for G: G = F × r² / (m1 × m2).
Substitute units: [G] = N × m² / kg² = Nm²kg⁻².
Therefore, the SI unit of the universal gravitation constant is Nm²kg⁻².

RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Universal law of gravitation and G

Q18. Which of the following is NOT explained by the Universal Law of Gravitation?

(a) Weight of objects on Earth
(b) Formation of tides
(c) Resistance offered by air to a moving object
(d) Falling of apples from a tree

Answer & solution

Answer: (c) Resistance offered by air to a moving object

The Universal Law of Gravitation explains the gravitational attraction between two masses.
It accounts for the weight of objects on Earth (gravitational force between Earth and object).
It explains the formation of tides due to gravitational pull of the Moon and Sun on Earth.
It also explains the falling of apples from a tree (gravity pulls objects towards Earth).
However, resistance offered by air to a moving object is due to air friction or drag, not gravitation.
Therefore, option 3 is NOT explained by the Universal Law of Gravitation.

RRB Technician Grade III 2025-26 – 10 March 2026 Shift 2 · Mass and weight

Q19. Which of the following statements correctly describes the difference between mass and weight?

(a) Mass changes from one planet to another, but weight stays the same.
(b) Weight and mass both remain constant everywhere.
(c) Mass stays the same everywhere, but weight changes with location.
(d) Both mass and weight change with location.

Answer & solution

Answer: (c) Mass stays the same everywhere, but weight changes with location.

Mass is the quantity of matter in a body and remains constant everywhere.
Weight is the force due to gravity acting on the mass and varies with location.
Hence, mass stays the same everywhere, but weight changes according to gravitational force.

RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Mass and weight

Q20. An object's mass remains constant everywhere (on the Earth or the Moon), whereas its weight changes because ___________.

(a) Mass is dependent on inertia, but weight is not
(b) Weight depends on the local value of g, which varies with location
(c) The SI unit of weight is Newtons, while mass is kilograms
(d) Weight is a scalar quantity, while mass is vector

Answer & solution

Answer: (b) Weight depends on the local value of g, which varies with location

Mass is the amount of matter in an object and does not change with location.
Weight is the force due to gravity acting on the mass and depends on gravitational acceleration (g).
Gravitational acceleration varies between the Earth and the Moon.
Therefore, the weight changes while mass remains constant depending on the local value of g.

RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Universal law of gravitation and G

Q21. If the gravitational force experienced by a body of mass 'm' on the surface of Earth is 'X' N. If the mass of the body is tripled, what will be the value of the gravitational force experienced by the body?

(a) X/9 N
(b) X/3 N
(c) 3X N
(d) X N

Answer & solution

Answer: (c) 3X N

The gravitational force is directly proportional to the mass of the body.
If the mass is tripled, the force also becomes three times.
Therefore, the new force = 3 × X = 3X Newtons.

RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Universal law of gravitation and G

Q22. The gravitational force between two masses increases when ________.

(a) The product of their masses increases
(b) The distance between them increases
(c) They are kept in water
(d) Their masses decrease

Answer & solution

Answer: (a) The product of their masses increases

Correct option: 1
The gravitational force (F) between two masses (m1 and m2) is given by Newton's law of gravitation:
F = G × (m1 × m2) / r², where G is the gravitational constant and r is the distance between the masses.
From the formula, force increases when the product of the masses increases or when the distance decreases.
Option 1 states that the product of their masses increases, so gravitational force increases.
Option 2 is incorrect: increasing the distance decreases the force.
Option 3 is irrelevant to gravitational force.
Option 4 is incorrect: decreasing masses decreases the force.
Therefore, the gravitational force increases when the product of their masses increases.

RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Universal law of gravitation and G

Q23. The motion of the moon around the earth requires a centripetal force, which is provided by _____________.

(a) The inertia of the moon
(b) The thrust of the sun
(c) The force of attraction of the earth
(d) The internal forces of the moon

Answer & solution

Answer: (c) The force of attraction of the earth

The moon revolves around the Earth due to centripetal force.
This force is the gravitational attraction exerted by the Earth on the moon.
Hence, it is the force of attraction of the Earth that keeps the moon in orbit.
Other options like inertia or forces of the moon itself do not provide centripetal force.
The sun's thrust is not responsible for the moon's motion around the Earth.
Therefore, the correct answer is the force of attraction of the Earth.

RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Numericals on gravitation

Q24. The weight of a body on the moon is 3 N. What will be the weight of the body on Earth?

(a) 0.5 N
(b) 3 N
(c) 18 N
(d) 2 N

Answer & solution

Answer: (c) 18 N

Weight on the moon is 1/6th of the weight on Earth.
Given weight on moon = 3 N.
Therefore, weight on Earth = 3 × 6 = 18 N.

RRB Technician Grade III 2025-26 – 6 March 2026 Shift 2 · Numericals on gravitation

Q25. A block of mass 20 kg is dropped from the top of a building of height 200 m. What will be the velocity with which the block will hit the ground? (Take the value of g = 10 m/s²)

(a) 40 m/s
(b) 20√5 m/s
(c) 20√10 m/s
(d) 20 m/s

Answer & solution

Answer: (c) 20√10 m/s

Given:
Mass, m = 20 kg (not needed in calculation)
Height, h = 200 m
Acceleration due to gravity, g = (10 m)/(s²)
Use the formula for velocity of an object under free fall:
v = √(2gh)
Substitute the values:
v = √(2 × 10 × 200)
v = √(4000)
v = √(4000) = √(400 × 10) = (20√10 m)/(s)
(Thus, velocity with which the block hits the ground is 20√10 m)/(s.)
Final answer: Option 3

RRB Technician Grade III 2025-26 – 6 March 2026 Shift 3 · Numericals on gravitation

Q26. An object is launched straight upward and reaches a maximum height of 1000 cm. What was the initial velocity of the object? (Take g = 9.8 m/s²)

(a) 10 m/s
(b) 14 m/s
(c) 2 m/s
(d) 6 m/s

Answer & solution

Answer: (b) 14 m/s

Given
Maximum height, h = 1000 cm = 10 m (since 100 cm = 1 m)
Acceleration due to gravity, g = (9.8 m)/(s²)
At maximum height, velocity = 0.
Use the formula
v² = u² - 2gh where v = 0 at max height
0 = u² - 2 × 9.8 × 10
u² = 2 × 9.8 × 10
u² = 196
u = √(196)=14 m/s
Final answer: Option 2

RRB Technician Grade III 2025-26 – 6 March 2026 Shift 2 · Mass and weight

Q27. If mass is doubled and g remains constant, weight becomes ________.

(a) Same
(b) Half
(c) Double
(d) Four times

Answer & solution

Answer: (c) Double

Weight = mass × gravitational acceleration (g).
If mass is doubled, weight becomes 2 × mass × g.
Since g remains constant, weight doubles.
Hence, weight becomes double when mass is doubled.
Final answer: Option 3

RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Universal law of gravitation and G

Q28. Which statement correctly describes the universal law of gravitation?

(a) The gravitational force increases with the product of the masses and decreases with the square of the distance between them.
(b) The gravitational force decreases when the masses of the objects increase.
(c) The gravitational force increases as the distance between objects increases.
(d) The gravitational force depends only on the distance between the objects, not their masses.

Answer & solution

Answer: (a) The gravitational force increases with the product of the masses and decreases with the square of the distance between them.

Option 1 is correct.
According to the universal law of gravitation, the force between two masses is directly proportional to the product of their masses.
The force is inversely proportional to the square of the distance between them.
Hence, gravitational force increases with the product of the masses and decreases with the square of the distance.
Other options contradict these direct proportionality and inverse proportionality relationships.
Therefore, Option 1 correctly describes the universal law of gravitation.
Final answer: Option 1

RRB JE CBT-1 2025-26 – 25 February 2026 Shift 3 · Numericals on gravitation

Q29. If a body at rest is dropped from a certain height and takes 4 seconds to reach the ground, what is the approximate height from which it was dropped? (Take g = 10 m/s²)

(a) 160 meters
(b) 100 meters
(c) 40 meters
(d) 80 meters

Answer & solution

Answer: (d) 80 meters

When an object is dropped from rest, use the equation of motion for free fall: s =(1)/(2) × g × t²
Given t = 4 seconds and g = 10 m/s², substitute values: s =(1)/(2) × 10 × (4)²
Calculate: s = 5 × 16 = 80 meters
So, the height from which the object was dropped is approximately 80 meters.
Final answer: Option 4

RRB JE CBT-1 2025-26 – 20 February 2026 Shift 2 · Numericals on gravitation

Q30. A ball initially at rest is dropped from a height and takes 5 seconds to reach the ground. What is the average speed during its fall? (g = 10 m/s²)

(a) 25 m/s
(b) 50 m/s
(c) 10 m/s
(d) 20 m/s

Answer & solution

Answer: (a) 25 m/s

Initial velocity u = 0
Time t = 5 s, g = 10 m/s²
Distance fallen:
s = ½ × g × t²
= ½ × 10 × 25
= 125 m
Average speed =(total distance)/(total time)
= 125 / 5 =25 m/s

RRB JE CBT-1 2025-26 – 19 February 2026 Shift 3 · Numericals on gravitation

Q31. A ball at rest is dropped from the top of a 20 m high building. Calculate the speed with which it hits the ground. (Take g = 10 m/s²)

(a) 40 m/s
(b) 14 m/s
(c) 10 m/s
(d) 20 m/s

Answer & solution

Answer: (d) 20 m/s

The ball is dropped from rest, so initial velocity u = 0 m/s.
Height h = 20 m, acceleration due to gravity g = 10 m/s².
Use the formula v² = u² + 2gh to find the final speed v.
v² = 0 + 2 × 10 × 20 = 400.
v = √400 = 20 m/s.
The speed with which it hits the ground is 20 m/s.
Correct option is Option 4.
Hence, final answer: Option 4

RRB JE CBT-1 2025-26 – 19 February 2026 Shift 2 · Universal law of gravitation and G

Q32. A student claims that the force exerted by the Earth on the Moon is much stronger than the force exerted by the Moon on the Earth. Based on Newton's third law and the Universal Law of Gravitation, is this claim correct?

(a) No, Earth exerts a much smaller force on the Moon
(b) No, the Moon exerts a stronger force because it is closer to Earth
(c) No, both forces are equal in magnitude but opposite in direction
(d) Yes, because Earth is more massive, it exerts more force

Answer & solution

Answer: (c) No, both forces are equal in magnitude but opposite in direction

According to Newton's third law, forces between two bodies are equal in magnitude and opposite in direction.
The Universal Law of Gravitation states that the force between two masses is mutual and equal.
Thus, the force exerted by the Earth on the Moon and that exerted by the Moon on the Earth are equal in magnitude.
The claim that the Earth exerts a much stronger force is incorrect.
Both forces are equal and opposite as per physical laws.
Final answer: Option 3

SSC MTS 2025 – 18 February 2026 Shift 3 · Satellites, orbital velocity and escape velocity

Q33. A geo-stationary satellite completes one revolution around the Earth in approximately _____ hours.

(a) 6
(b) 12
(c) 24
(d) 48

Answer & solution

Answer: (c) 24

The correct answer is option 3.
A geostationary satellite orbits the Earth at an altitude of approximately 35,786 km above the equator and completes one revolution in exactly 24 hours — the same time it takes Earth to complete one rotation. This makes the satellite appear stationary relative to a point on Earth's surface, which is why it is used for communication, weather monitoring, and broadcasting.
Option 1 is incorrect: 6 hours is too short a period for a geostationary orbit; this would correspond to a much lower orbit.
Option 2 is incorrect: A 12-hour orbit corresponds to medium Earth orbit satellites like GPS satellites, not geostationary satellites.
Option 4 is incorrect: A 48-hour orbit would require a much higher altitude than that of geostationary satellites.

SSC MTS 2025 – 13 February 2026 Shift 1 · Universal law of gravitation and G

Q34. Why do objects fall towards the Earth?

(a) Due to the Earth's rotation
(b) Due to the Earth's magnetic field
(c) Due to the force of gravity
(d) Due to air pressure

Answer & solution

Answer: (c) Due to the force of gravity

Correct Answer: Option 3 – Due to the force of gravity
Objects fall towards the Earth because of gravitational force. According to Newton's Law of Universal Gravitation, every object with mass attracts every other object with mass. The Earth, having a very large mass, exerts a strong gravitational pull on objects near it, causing them to accelerate downward (towards the Earth's centre ) at approximately 9.8 m/s².
Why other options are wrong:
Option 1 (Earth's rotation): Rotation actually creates a centrifugal effect that slightly reduces the pull at the equator; it does not cause objects to fall.
Option 2 (Earth's magnetic field): Magnetic fields act on magnetic materials; they do not cause all objects to fall.
Option 4 (Air pressure): Air pressure acts in all directions and does not cause objects to fall towards Earth.

SSC MTS 2025 – 13 February 2026 Shift 2 · Mass and weight

Q35. What is the formula to calculate the weight of an object?

(a) W = m × g
(b) W = m / g
(c) W = m / (g × g)
(d) W = m × g × g

Answer & solution

Answer: (a) W = m × g

Weight (W) is the gravitational force acting on an object. It is calculated using the formula W = m × g, where 'm' is the mass of the object (in kg) and 'g' is the acceleration due to gravity (approximately 9.8 m/s² on Earth). Weight is measured in Newtons (N).
Why other options are incorrect:
- W = m / g: This does not represent any standard physical formula for weight.
- W = m / (g × g): Incorrect; no such physical relationship exists for weight.
- W = m × g × g: Incorrect; multiplying by g twice gives units of m/s⁴, which is not weight.

SSC MTS 2025 – 11 February 2026 Shift 1 · Universal law of gravitation and G

Q36. The motion of the moon around the earth is due to the _____.

(a) centripetal force
(b) magnetic field
(c) friction force
(d) acceleration

Answer & solution

Answer: (a) centripetal force

The correct answer is centripetal force (Option 1). The Moon revolves around the Earth because of gravitational attraction, which acts as the centripetal force — a force directed toward the center (Earth). This centripetal force keeps the Moon in its circular orbit. Option 2 (magnetic field) is not the cause of orbital motion. Option 3 (friction force) does not exist in vacuum space. Option 4 (acceleration) is an effect, not a cause of the orbital motion.

SSC MTS 2025 – 10 February 2026 Shift 2 · Universal law of gravitation and G

Q37. The universal law of gravitation states that the gravitational force between two objects is proportional to _____.

(a) the difference of the masses
(b) the sum of the masses
(c) the product of the masses
(d) the square of the masses

Answer & solution

Answer: (c) the product of the masses

Correct Answer: 3 – the product of the masses.
Newton's Universal Law of Gravitation states: F = G × (m1 × m2) / r², where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. The force is directly proportional to the product of the masses (m1 × m2) and inversely proportional to the square of the distance between them.
The difference (Option 1) and sum (Option 2) of masses are not part of the law.
The square of masses (Option 4) is incorrect; it is the square of distance that appears in the denominator.

Delhi Police Head Constable (Ministerial) 2025-26 – 8 January 2026 Shift 1 · Weightlessness

Q38. A person stands on a weighing machine inside a lift. Consider the following about their apparent weight:
I. It is less than actual weight when the lift accelerates downward.
II. It becomes zero during free fall.
III. It increases if the lift moves upward with constant speed.
Which of the above statements is/are correct?

(a) I and II only
(b) I and III only
(c) II and III only
(d) I, II and III all are correct

Answer & solution

Answer: (a) I and II only

Correct Answer: Option 1 – I and II only
Statement I is CORRECT: When a lift accelerates downward, the apparent weight = m(g - a), which is less than actual weight mg. So the scale shows a lower reading.
Statement II is CORRECT: During free fall, the lift (and person) both accelerate downward at g (acceleration due to gravity). Apparent weight = m(g - g) = 0. The person feels weightless and the scale reads zero.
Statement III is INCORRECT: When the lift moves upward with CONSTANT speed, acceleration = 0. Therefore, apparent weight = m(g + 0) = mg = actual weight. The reading does NOT increase; it remains equal to the actual weight.
Why other options are wrong:
Options 2, 3, 4: All include or exclude statements incorrectly. Statement III is wrong because constant speed means zero acceleration, so apparent weight equals actual weight.

Delhi Police Constable (Exe) 2025-26 – 6 January 2026 Shift 3 · Satellites, orbital velocity and escape velocity

Q39. If a satellite in 700 km orbit completes 15 orbits/day, how many minutes per orbit?

(a) 96 min
(b) 95 min
(c) 100 min
(d) 105 min

Answer & solution

Answer: (a) 96 min

The correct answer is (a) 96 min.
Calculation: 1 day = 24 hours = 24 × 60 = 1440 minutes
Number of orbits per day = 15
Time per orbit = 1440 ÷ 15 = 96 minutes
This is consistent with actual orbital data: satellites at approximately 700 km altitude (Low Earth Orbit) complete each orbit in about 96-98 minutes.
(b) 95 min: 15 × 95 = 1425 minutes, which is less than 1440 minutes/day — incorrect.
(c) 100 min: 15 × 100 = 1500 minutes, more than 1440 — incorrect.
(d) 105 min: 15 × 105 = 1575 minutes, more than 1440 — incorrect.

Delhi Police Constable (Exe) 2025-26 – 5 January 2026 Shift 3 · Numericals on gravitation

Q40. If Mars' gravity is 3.7 m/s², what would be the weight of a 60 kg astronaut on Mars?

(a) 222 N
(b) 588 N
(c) 180 N
(d) 300 N

Answer & solution

Answer: (a) 222 N

The correct answer is (a) 222 N.
Weight = Mass × Acceleration due to gravity
Weight on Mars = 60 kg × 3.7 m/s² = 222 N
This shows that on Mars, the astronaut would weigh about 38% of their Earth weight (60 × 9.8 = 588 N on Earth).
(b) 588 N: This would be the weight on Earth (60 × 9.8 = 588 N); not on Mars.
(c) 180 N: This would correspond to gravity of 3.0 m/s² (60 × 3.0 = 180 N); not the value given.
(d) 300 N: This would correspond to gravity of 5.0 m/s² (60 × 5.0 = 300 N); not Mars' gravity.

SSC Stenographer 2025 – 8 December 2025 Shift 2 · Kepler's laws and planetary motion

Q41. Which of the following quantities remains conserved when an object moves under the gravitational pull of another object?

(a) Potential energy and linear momentum
(b) Kinetic energy and linear momentum
(c) Angular momentum and mechanical energy
(d) Only linear momentum

Answer & solution

Answer: (c) Angular momentum and mechanical energy

The correct answer is Angular momentum and mechanical energy . When a body moves under the gravitational influence of another body (like planets orbiting the Sun), the gravitational force is a conservative central force . This ensures that mechanical energy (kinetic + potential energy) remains conserved. Also, since the force always acts along the line joining the two bodies, there is no torque about the centre of force, which means angular momentum of the system remains conserved. However, linear momentum of the object is not necessarily conserved individually, as the force continuously changes its direction. A classic example is the Earth revolving around the Sun, where both angular momentum and total mechanical energy are conserved.

SSC Stenographer 2025 – 7 December 2025 Shift 1 · Kepler's laws and planetary motion

Q42. According to the Law of Periods, the square of a planet’s time period is proportional to:

(a) The cube of the semi-major axis of its orbit
(b) The product of its mass and orbital speed
(c) The square root of its orbital radius
(d) The square of its distance from the sun

Answer & solution

Answer: (a) The cube of the semi-major axis of its orbit

According to Kepler’s Third Law , also known as the Law of Periods , there is a mathematical relationship between the orbital period of a planet and the size of its orbit. The law states that the square of the orbital period (T²) of a planet is directly proportional to the cube of the semi-major axis (a³) of its orbit . This means that planets that are farther from the Sun take a much longer time to complete one revolution compared to those closer to the Sun.
Mathematically: T² ∝ a³ Here, T = orbital period of the planet, a = semi-major axis of the planet’s orbit. This law is important because it applies to all planets, moons, and even artificial satellites. For example, Earth takes 1 year to orbit the Sun, whereas Jupiter, being farther away, takes about 12 years, which is consistent with this law. Thus, the correct option is the cube of the semi-major axis of its orbit .

SSC Stenographer 2025 – 6 December 2025 Shift 1 · Universal law of gravitation and G

Q43. Which force governs the motion of planets, stars, and galaxies?

(a) Gravitational force
(b) Nuclear force
(c) Magnetic force
(d) Electromagnetic force

Answer & solution

Answer: (a) Gravitational force

The correct answer is Gravitational force . It is the weakest of the four fundamental forces, yet it is the most dominant at astronomical scales. Gravitational force governs the motion of planets around the Sun, stars in galaxies, and even the galaxies themselves. For example, Earth revolves around the Sun due to the Sun’s gravitational pull, and the Moon revolves around the Earth for the same reason. Without gravity, the large-scale structure of the universe would not exist.

SSC CHSL Tier-1 2025 – 30 November 2025 Shift 1 · Satellites, orbital velocity and escape velocity

Q44. Generally, satellites do not fly below _____ km.

(a) 12
(b) 180
(c) 55
(d) 10000

Answer & solution

Answer: (b) 180

Correct Answer: Option 2 – 180
Satellites generally do not fly below 180 km altitude. Below this altitude, the Earth's atmosphere becomes dense enough to cause significant drag, which rapidly decays a satellite's orbit and causes it to re-enter the atmosphere. The Low Earth Orbit (LEO) range begins at approximately 180–200 km.
Why other options are wrong:
Option 1 (12): At 12 km, we are still in the troposphere; no satellite can maintain orbit here.
Option 3 (55): Rockets and high-altitude balloons can reach 55 km, but satellites cannot sustain stable orbit at this height.
Option 4 (10000): This is within the Medium Earth Orbit (MEO) range; satellites do fly at 10,000 km, but that is not the minimum. The candidate incorrectly selected this.

SSC CHSL Tier-1 2025 – 30 November 2025 Shift 3 · Satellites, orbital velocity and escape velocity

Q45. At 800 km, a satellite in Sun Synchronous Orbit will travel at a speed of approximately _____ km per second.

(a) 1.5
(b) 15
(c) 22.5
(d) 7.5

Answer & solution

Answer: (d) 7.5

Correct Answer: Option 4 – 7.5
At an altitude of 800 km, a satellite in Sun Synchronous Orbit (SSO) travels at approximately 7.5 km per second. Using the orbital velocity formula v = sqrt(GM/r), where r = Earth's radius (6371 km) + 800 km = 7171 km: v ≈ sqrt(3.986×10^14 / 7,171,000) ≈ 7,453 m/s ≈ 7.5 km/s.
Why other options are wrong:
Option 1 (1.5 km/s): Too slow; only deep space probes travel this slowly at large distances.
Option 2 (15 km/s): Too fast for LEO; this speed is associated with escape velocity contexts.
Option 3 (22.5 km/s): Incorrect; the candidate incorrectly selected this. 22.5 km/s far exceeds orbital velocity.

SSC CHSL Tier-1 2025 – 21 November 2025 Shift 1 · Universal law of gravitation and G

Q46. Which of the following statement state the importance of the universal law of gravitation?
I. The motion of planets around the Sun
II. The tides due to the moon and the Sun

(a) Only I
(b) Only II
(c) Both I and II
(d) Neither I nor II

Answer & solution

Answer: (c) Both I and II

Both statements are correct applications of Newton's Universal Law of Gravitation: Statement I - The motion of planets around the Sun: The gravitational force between the Sun and the planets keeps planets in their orbits (explained by Newton's law of gravitation). Statement II - The tides due to the moon and the Sun: Ocean tides are caused by the gravitational pull of both the Moon and the Sun on the Earth's water bodies. Both are classic examples of the importance of the universal law of gravitation. Why other options are wrong: Option 1 (Only I): Incorrect; Statement II is also a valid application. Option 2 (Only II): Incorrect; Statement I is also a valid application. Option 4 (Neither): Both statements correctly illustrate Newton's law of gravitation.

SSC CHSL Tier-1 2025 – 17 November 2025 Shift 3 · Satellites, orbital velocity and escape velocity

Q47. A communication satellite acts like a television tower placed at a height of _____ km.

(a) 1000
(b) 1500
(c) 22000
(d) 36000

Answer & solution

Answer: (d) 36000

Correct Answer: 36000
A communication satellite (geostationary satellite) orbits at approximately 36,000 km (36,000 km = 35,786 km) above the Earth's surface. At this altitude, the satellite's orbital period matches the Earth's rotation period (24 hours), making it appear stationary relative to the Earth. This is why a communication satellite can act like a television tower at a height of 36,000 km, covering a large area of Earth's surface.
Why other options are wrong:
- 1000 km: Low Earth Orbit (LEO); used for weather and spy satellites, not geostationary communication.
- 1500 km: Still in LEO; not the geostationary orbit altitude.
- 22000 km: Medium Earth Orbit; used for GPS satellites, not geostationary communication satellites.

SSC CHSL Tier-1 2025 – 15 November 2025 Shift 1 · Satellites, orbital velocity and escape velocity

Q48. Why can a geo-stationary satellite monitor the movement of clouds accurately?

(a) Because it moves in North to South direction
(b) Because it moves in South to North direction
(c) Because it is stationary relative to any place on Earth
(d) Because it rotates faster than Earth

Answer & solution

Answer: (c) Because it is stationary relative to any place on Earth

A geo-stationary satellite can accurately monitor cloud movement because it is stationary relative to any place on Earth. It orbits at approximately 36,000 km above the equator and rotates at the same angular speed as Earth, making it appear stationary. This allows continuous observation of the same region for weather monitoring. Why others are wrong: - Moving North to South or South to North: These describe polar orbits, not geostationary satellites. - Rotates faster than Earth: Geostationary satellites rotate at the SAME speed as Earth, not faster.

SSC CHSL Tier-1 2025 – 15 November 2025 Shift 1 · Numericals on gravitation

Q49. An object weighs 18 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?

(a) 1 N
(b) 3 N
(c) 7 N
(d) 9 N

Answer & solution

Answer: (b) 3 N

The acceleration due to gravity on the Moon is approximately 1/6th of that on Earth. Weight on Moon = Weight on Earth / 6 = 18 / 6 = 3 N Why others are wrong: - 1 N: Would be 1/18th of Earth weight, which is not the moon's gravity ratio. - 7 N: Would require gravity ratio of about 1/2.57, which is incorrect. - 9 N: Would mean moon's gravity is half of Earth's, which is incorrect.

SSC CHSL Tier-1 2025 – 15 November 2025 Shift 3 · Satellites, orbital velocity and escape velocity

Q50. Lagrange points are also known as _____ points.

(a) Middle
(b) Libration
(c) Centre
(d) Shortcut

Answer & solution

Answer: (b) Libration

Lagrange points are also known as Libration points. These are the five positions in an orbital configuration where a small object can maintain a stable position relative to two larger orbiting bodies. The term ' libration ' refers to the oscillatory motion near these points. Option 1 (Middle) is incorrect — not the scientific term. Option 3 (Centre) is incorrect. Option 4 (Shortcut) is incorrect.

SSC CHSL Tier-1 2025 – 12 November 2025 Shift 2 · Satellites, orbital velocity and escape velocity

Q51. If a satellite is in the plane of the equator with 0 degrees inclination and at a height of about 36,000 km, then it is said to be in ________ orbit.

(a) Polar
(b) Geo-stationary
(c) Elliptical
(d) Sun-synchronous

Answer & solution

Answer: (b) Geo-stationary

A geo-stationary orbit is located at approximately 35,786 km above the equator with 0 degrees inclination. The satellite moves in sync with Earth's rotation, appearing stationary relative to a point on Earth. Polar orbits have high inclination (near 90°); elliptical orbits are non-circular; sun-synchronous orbits maintain a consistent solar angle — all are different from geostationary.

SSC CHSL Tier-1 2025 – 12 November 2025 Shift 2 · Satellites, orbital velocity and escape velocity

Q52. If a satellite is in the plane of the equator with 0 degrees inclination and at a height of about 36000 km then it is said to be in _____ orbit.

(a) Polar
(b) Geo-stationary
(c) Elliptical
(d) Sun-synchronous

Answer & solution

Answer: (b) Geo-stationary

Correct Answer: Geo-stationary
A geostationary orbit (GEO) is a circular orbit located directly above the Earth's equator (0? inclination) at an altitude of approximately 35,786 km (~36,000 km). A satellite in this orbit rotates at the same angular speed as Earth, appearing stationary relative to the ground. It is widely used for communication and weather satellites.
Why other options are wrong:
- Polar orbit: Has an inclination of ~90?, passing over the poles.
- Elliptical orbit: Has varying altitude; the orbit is not circular.
- Sun-synchronous: A near-polar orbit (~98? inclination) that passes over any given point at the same local solar time.

SSC CHSL Tier-1 2025 – 12 November 2025 Shift 3 · Numericals on gravitation

Q53. Weight of an object on earth is 441 N. What is the mass of the object?

(a) 49 kg
(b) 41 kg
(c) 40 kg
(d) 45 kg

Answer & solution

Answer: (d) 45 kg

Correct Answer: 45 kg
Weight (W) = mass (m) x acceleration due to gravity (g)
W = mg
441 = m x 9.8
m = 441 / 9.8 = 45 kg
Why other options are wrong:
- 49 kg: 49 x 9.8 = 480.2 N, not 441 N.
- 41 kg: 41 x 9.8 = 401.8 N, not 441 N.
- 40 kg: 40 x 9.8 = 392 N, not 441 N.

SSC CGL Tier-1 2025 – 20 September 2025 Shift 3 · Satellites, orbital velocity and escape velocity

Q54. What happens if a satellite orbits Earth with 24-hour period?

(a) Spirals inward
(b) Appears stationary
(c) Escapes Earth
(d) Oscillates over longitude

Answer & solution

Answer: (b) Appears stationary

A satellite with an orbital period of 24 hours (matching Earth's rotation period), in an equatorial orbit, is geostationary — it appears stationary over a fixed point on Earth.
This is the principle used for communication and weather satellites like INSAT.
Hence, 'Appears stationary' is correct.

SSC CGL Tier-1 2025 – 15 September 2025 Shift 3 · Satellites, orbital velocity and escape velocity

Q55. Evaluate the following statements and select the option that is accurate:
Statement 1: Sun-synchronous orbits allow satellites to maintain consistent local solar time over a specific region.
Statement 2: This orbital feature is ideal for continuous solar-powered communication satellites in geostationary orbit.

(a) Only Statement 1 is correct
(b) Only Statement 2 is correct
(c) Both 1 and 2 are correct
(d) Neither 1 nor 2 are correct

Answer & solution

Answer: (a) Only Statement 1 is correct

• Statement 1: A Sun-synchronous orbit (SSO) is a nearly polar LEO where the satellite passes over any given point at the same local solar time daily. This ensures consistent lighting — ideal for earth observation and remote sensing. ✓
• Statement 2: SSO is NOT geostationary orbit. Geostationary satellites orbit at ~36,000 km. SSO satellites are in Low Earth Orbit (~600–800 km). SSO is not used for continuous communication satellites. ✗
• Answer: Only Statement 1 is correct

SSC CGL Tier-1 2025 – 12 September 2025 Shift 2 · Numericals on gravitation

Q56. If Earth's radius were reduced by 1% while its mass remained unchanged, what would be the effect on the gravitational acceleration at its surface?

(a) Increase by approximately 1%
(b) Decrease by approximately 2%
(c) Increase by approximately 2%
(d) Decrease by approximately 1%

Answer & solution

Answer: (c) Increase by approximately 2%

Surface gravitational acceleration: g = GM / R².
For small changes (with M constant): Δg / g ≈ −2 × ΔR / R.
Given ΔR / R = −1% = −0.01.
Δg / g ≈ −2 × (−0.01) = +0.02 = +2%.
So g increases by about 2%.

RRB NTPC GL CBT-1 2024-25 – 21 June 2025 Shift 3 · Universal law of gravitation and G

Q57. Which force acts without physical contact and governs celestial body motion in mechanics?

(a) Tension
(b) Friction
(c) Gravitational force
(d) Spring force

Answer & solution

Answer: (c) Gravitational force

RRB Technician Grade III 2024-25 – 30 December 2024 Shift 1 · Universal law of gravitation and G

Q58. The value of universal gravitational constant is ________.

(a) 1 × 10⁻¹¹ N m² kg⁻²
(b) 6.7 N m² kg⁻²
(c) 5 × 10⁻¹ N m² kg⁻²
(d) 6.7 × 10⁻¹¹ N m² kg⁻²

Answer & solution

Answer: (d) 6.7 × 10⁻¹¹ N m² kg⁻²

Correct Answer: Option 4 — 6.7 × 10⁻¹¹ N m² kg⁻²
By Newton's universal law of gravitation, two bodies of masses M and m separated by a distance r attract each other with a force
F = G M m / r²
The constant G in this equation is the universal gravitational constant. Its accepted value is
G = 6.67 × 10⁻¹¹ N m² kg⁻², usually rounded to 6.7 × 10⁻¹¹ N m² kg⁻².
It is called *universal* because it has the same value everywhere in the universe — it does not depend on the masses, the medium between them, or the place. Its value is extremely small, which is why gravitational attraction between ordinary objects around us is far too weak to notice; it becomes important only when at least one body is of planetary size.
❌ Why the other options are incorrect
✗ 1 × 10⁻¹¹ N m² kg⁻² — the exponent is right but the number is wrong; the accepted value is about 6.7, not 1.
✗ 6.7 N m² kg⁻² — the digits are right but the factor 10⁻¹¹ is missing, making the value about 10¹¹ times too large.
✗ 5 × 10⁻¹ N m² kg⁻² — that is 0.5, enormously bigger than G; a value like this would make everyday objects stick together by gravity.
📌 SSC & Railway Exam Facts
• F = GMm/r²; gravitational force obeys the inverse-square law and is always attractive.
• G = 6.67 × 10⁻¹¹ N m² kg⁻²; it is a scalar and a universal constant.
• g = GM/R²; on Earth g ≈ 9.8 m s⁻², and on the Moon it is about g/6.
• G was first measured experimentally by Henry Cavendish using a torsion balance.
• Do not confuse G (universal constant) with g (acceleration due to gravity, unit m s⁻²).

RRB Technician Grade III 2024-25 – 30 December 2024 Shift 1 · Numericals on gravitation

Q59. A body, starting from rest, falls from a building and reaches the ground in 2 secs. What will be the velocity of the body when it reaches the ground? (Acceleration due to gravity = 10 ms⁻² )

(a) 5 m/s
(b) 20 m/s
(c) 15 m/s
(d) 10 m/s

Answer & solution

Answer: (b) 20 m/s

Correct Answer: Option 2 — 20 m/s
📖 Given
• Initial velocity, u = 0 (the body starts from rest)
• Acceleration, a = g = 10 ms⁻² (free fall, downwards)
• Time of fall, t = 2 s
✍️ Formula
v = u + at
🔢 Solution
• v = 0 + (10 × 2)
• v = 0 + 20
• v = 20 m/s
So the body strikes the ground moving downwards at 20 metres every second. (Its height can also be found: h = ½gt² = ½ × 10 × 4 = 20 m.)
❌ Why the other options are incorrect
✗ 5 m/s — this comes from dividing instead of multiplying, 10/2 = 5; time must be multiplied by g.
✗ 15 m/s — no combination of the given data gives 15; it is only a distractor lying between 10 and 20.
✗ 10 m/s — this is the velocity after just 1 second, or the value of g mistaken for the answer.
📌 SSC & Railway Exam Facts
• Equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• In free fall put u = 0 and a = g; then v = gt, h = ½gt², v = √(2gh).
• g ≈ 9.8 m s⁻² on Earth (10 m s⁻² in exams); it is independent of the mass of the falling body.
• In the absence of air, a feather and a stone dropped together reach the ground at the same time — shown in vacuum-tube experiments.
• g is maximum at the poles and minimum at the equator; it decreases with both altitude and depth.

RRB Technician Grade III 2024-25 – 30 December 2024 Shift 1 · Universal law of gravitation and G

Q60. Which of the following phenomena CANNOT be explained using the universal law of gravitation?

(a) Firing of a gun
(b) Motion of planets around the Sun
(c) Motion of the moon around the earth
(d) Tides

Answer & solution

Answer: (a) Firing of a gun

Correct Answer: Option 1 — Firing of a gun
Newton's universal law of gravitation states that every body attracts every other body with a force
F = G M m / r²
It explains all phenomena caused by mutual gravitational attraction between large masses.
The firing of a gun is *not* one of them. When a gun is fired, the expanding gases push the bullet forward and, by Newton's third law, push the gun backwards — the recoil. The recoil velocity follows from the law of conservation of momentum, m₁v₁ = m₂v₂. Gravity plays no part in producing the bullet's forward motion, so this phenomenon belongs to the laws of motion, not to gravitation.
❌ Why the other options are incorrect
✗ Motion of planets around the Sun — the Sun's gravitational pull supplies the centripetal force that keeps each planet in orbit; this was Newton's greatest success and it explains Kepler's laws.
✗ Motion of the moon around the earth — the Earth's gravity continuously deflects the Moon from a straight path, holding it in orbit.
✗ Tides — the rise and fall of sea water is caused by the gravitational attraction of the Moon (and to a lesser extent the Sun) on the oceans.
📌 SSC & Railway Exam Facts
• F = GMm/r²; G = 6.67 × 10⁻¹¹ N m² kg⁻²; gravitation is always attractive.
• The universal law of gravitation explains: falling bodies, g, planetary orbits, the Moon's motion, tides and the shape of galaxies.
• Gun recoil follows from conservation of momentum and Newton's third law — action and reaction are equal and opposite.
• Gravitational force is the weakest of the four fundamental forces, yet it acts over infinite range.
• Kepler's three laws of planetary motion were later derived from Newton's law of gravitation.

RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Universal law of gravitation and G

Q61. What is the SI unit of universal gravitational constant G?

(a) N m²/kgs⁻¹
(b) N m²/kg⁻¹
(c) N kgm²
(d) N m²/kg²

Answer & solution

Answer: (d) N m²/kg²

Correct Answer: Option 4 — N m²/kg²
Start from Newton's universal law of gravitation:
F = G M m / r²
Making G the subject,
G = F r² / (M m)
Now substitute the units: force F is in newtons (N), the distance squared r² is in m², and the product of the two masses Mm is in kg². Therefore
Unit of G = N m² / kg² = N m² kg⁻²
Its numerical value is G = 6.67 × 10⁻¹¹ N m² kg⁻², and it is the same everywhere in the universe — independent of the masses, of the medium between them and of the location, which is exactly why it is called the *universal* constant.
❌ Why the other options are incorrect
✗ N m²/kgs⁻¹ — this brings in a second (time) unit, which never appears in G; the derivation gives only force, length and mass.
✗ N m²/kg⁻¹ — dividing by kg⁻¹ is the same as multiplying by kg, so this has mass in the wrong place and only to the first power.
✗ N kgm² — here mass is multiplied instead of divided; mass must sit in the denominator, squared.
📌 SSC & Railway Exam Facts
• G = 6.67 × 10⁻¹¹ N m² kg⁻²; it is a scalar and a universal constant.
• F = GMm/r² — gravitational force obeys the inverse-square law and is always attractive.
• g = GM/R²; on Earth g ≈ 9.8 m s⁻², unit m s⁻² — never confuse G with g.
• G was measured experimentally by Henry Cavendish (1798) with a torsion balance.
• Gravitation is the weakest of the four fundamental forces but has infinite range.

RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Numericals on gravitation

Q62. If the mass of a football is 3 kg, then what would be its weight on the earth? (Given, g= 9.8 m s⁻²)

(a) 30.5 N
(b) 25 N
(c) 35 N
(d) 29.4 N

Answer & solution

Answer: (d) 29.4 N

Correct Answer: Option 4 — 29.4 N
📖 Given
• Mass of the football, m = 3 kg
• Acceleration due to gravity on the Earth, g = 9.8 m s⁻²
• To find: weight W on the Earth
✍️ Formula
W = mg
🔢 Solution
• W = m × g
• W = 3 × 9.8
• W = 29.4 N
• Weight is a force, so its SI unit is the newton (N), not the kilogram.
In plain language, the Earth pulls a 3 kg football with a force of 29.4 newton; the same ball taken to the Moon would weigh only about 4.9 N, because the Moon's g is nearly 1/6 of the Earth's, even though its mass stays 3 kg everywhere.
❌ Why the other options are incorrect
✗ 30.5 N — this is what you get by rounding g to about 10.17; using the given 9.8 gives exactly 29.4 N.
✗ 25 N — no combination of 3 kg with 9.8 m s⁻² gives 25 N; it is an arbitrary value.
✗ 35 N — this comes from multiplying 3 by about 11.7; again, not the given value of g.
📌 SSC & Railway Exam Facts
• W = mg — weight is a vector (a force) directed towards the centre of the Earth; mass is a scalar and stays the same everywhere.
• g ≈ 9.8 m s⁻² on the Earth, ≈ 1.63 m s⁻² on the Moon (1/6 of the Earth's), and 0 at the centre of the Earth.
• g = GM/R²; g is maximum at the poles and minimum at the equator.
• G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant, the same everywhere.
• Mass is measured with a beam balance, weight with a spring balance; 1 kgf = 9.8 N.

RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Free fall

Q63. A body of mass m and a body of mass 4m are dropped from the same height, the ratio of the time taken by the two bodies to reach the ground would be:
(Air resistance is neglected)

(a) 4:1
(b) 1:2
(c) 1:1
(d) 1:4

Answer & solution

Answer: (c) 1:1

Correct Answer: Option 3 — 1:1
📖 Given
• Mass of the first body = m, mass of the second body = 4m
• Both are dropped from the same height h, so initial velocity u = 0
• Air resistance is neglected — this is free fall
✍️ Formula
h = ut + ½gt², and with u = 0 this becomes t = √(2h/g)
🔢 Solution
• For the first body: h = 0 + ½gt₁², so t₁ = √(2h/g).
• For the second body: h = 0 + ½gt₂², so t₂ = √(2h/g).
• The expression √(2h/g) contains only h and g — the mass does not appear at all.
• Therefore t₁ = t₂, and t₁ : t₂ = 1 : 1.
In plain language, in the absence of air resistance all bodies fall at the same rate, whatever their mass. This is exactly what Galileo demonstrated, and what astronaut David Scott repeated on the Moon by dropping a hammer and a feather together — both landed at the same moment.
❌ Why the other options are incorrect
✗ 4:1 — this assumes the lighter body takes 4 times longer, i.e. time depends on mass, which free fall contradicts.
✗ 1:2 — this comes from wrongly taking t ∝ √m and using √4 = 2; mass has no place in the equation.
✗ 1:4 — this assumes t ∝ m, i.e. the heavier body takes 4 times as long, which is doubly wrong.
📌 SSC & Railway Exam Facts
• For free fall from rest: v = gt, h = ½gt², v² = 2gh — none of them contains mass.
• Galileo Galilei is said to have shown this from the Leaning Tower of Pisa.
• In a vacuum a coin and a feather fall together; in air the feather lags because of air resistance.
• g ≈ 9.8 m s⁻² on the Earth; g is independent of the mass of the falling body.
• For a body dropped from rest, the distances covered in successive seconds are in the ratio 1 : 3 : 5 : 7 …

RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Mass and weight

Q64. Which of the following statements is true?

(a) The weight of an object is the power with which it is attracted towards the earth.
(b) The mass of an object is constant and does not change from place to place.
(c) The weight of an object is constant and does not change from place to place.
(d) The weight is equal to the product of velocity and acceleration due to gravity.

Answer & solution

Answer: (b) The mass of an object is constant and does not change from place to place.

Correct Answer: Option 2 — The mass of an object is constant and does not change from place to place.
Mass is the quantity of matter contained in a body. It is a scalar quantity, measured in kilograms (kg) with a beam balance, and it is a constant property of the body. A 5 kg bag of rice is 5 kg on the Earth, 5 kg on the Moon and 5 kg in deep space.
Weight, on the other hand, is the force with which the Earth (or any planet) attracts the body, given by W = mg. It is a vector, measured in newton (N) with a spring balance. Since g changes from place to place — it is larger at the poles, smaller at the equator, smaller still at high altitude, and only about 1/6 of its Earth value on the Moon — weight is not constant.
That is why an astronaut's mass is the same on the Moon as on Earth, but his weight there is only one-sixth.
❌ Why the other options are incorrect
✗ The weight of an object is the power with which it is attracted towards the earth. — weight is a force, not power. Power is the rate of doing work, measured in watt.
✗ The weight of an object is constant and does not change from place to place. — false; W = mg changes wherever g changes, e.g. on the Moon it is about 1/6.
✗ The weight is equal to the product of velocity and acceleration due to gravity. — wrong formula; weight is mass × g, not velocity × g. That product would not even have the units of force.
📌 SSC & Railway Exam Facts
• Mass — scalar, SI unit kg, measured by a beam balance, constant everywhere.
• Weight — vector (a force), SI unit newton (N), measured by a spring balance, varies with g.
• W = mg; 1 kgf = 9.8 N; on the Moon weight is about 1/6 of that on the Earth.
• g is maximum at the poles and minimum at the equator; g = 0 at the centre of the Earth.
• In a freely falling lift a person feels weightless, because the lift and the person fall with the same acceleration g.

RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Universal law of gravitation and G

Q65. Which of the following statements is/are true for the gravitational force between two bodies:
(i) It is repulsive in nature.
(ii) It's magnitude is same throughout the universe.
(iii) It is directly proportional to the product of the masses of the bodies.
(iv) Due to this force the earth goes around the sun.

(a) Only (iii)
(b) Both (iii) and (iv)
(c) Both (i) and (ii)
(d) Both (ii) and (iii)

Answer & solution

Answer: (b) Both (iii) and (iv)

Correct Answer: Option 2 — Both (iii) and (iv)
Newton's Universal Law of Gravitation states that every body in the universe attracts every other body with a force F = G M m / r², directed along the line joining their centres.
Checking each statement:
• (i) is false — gravitational force is always attractive, never repulsive. It is the only fundamental force that is purely attractive.
• (ii) is false — it is the constant G that has the same value everywhere in the universe, not the force. The force F changes with the masses and with the distance r between the bodies.
• (iii) is true — F ∝ Mm, so doubling either mass doubles the force.
• (iv) is true — the gravitational pull of the Sun supplies the centripetal force that keeps the Earth in its orbit; the same force keeps the Moon revolving around the Earth.
Hence only (iii) and (iv) are correct.
❌ Why the other options are incorrect
✗ Only (iii) — it wrongly leaves out (iv), which is a standard true consequence of gravitation.
✗ Both (i) and (ii) — both statements are false; gravity is attractive, and the force is not the same everywhere.
✗ Both (ii) and (iii) — (iii) is true but (ii) confuses the force F with the constant G.
📌 SSC & Railway Exam Facts
• G = 6.67 × 10⁻¹¹ N m² kg⁻² — a universal constant, same everywhere; measured by Henry Cavendish.
• F ∝ 1/r² — this is called the inverse square law.
• Gravitation is the weakest of the four fundamental forces, but it acts over infinite range.
• g = GM/R²; on the Earth g ≈ 9.8 m s⁻², on the Moon g ≈ 1.63 m s⁻² (about 1/6 of the Earth's).
• Kepler's laws of planetary motion are explained by Newton's law of gravitation.

RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Universal law of gravitation and G

Q66. The value of Universal Gravitational constant is G on a certain planet of mass M, on another planet of mass 4M, the value of the Universal Gravitational constant will be:

(a) G
(b) G/2
(c) Zero
(d) 4G

Answer & solution

Answer: (a) G

Correct Answer: Option 1 — G
The key word in the question is 'Universal'. G is the Universal Gravitational Constant, and by definition its value is exactly the same everywhere in the universe — on the Earth, on the Moon, on a planet of mass M, on a planet of mass 4M, or in deep space. It does not depend on the masses of the bodies, on the distance between them, or on the medium separating them.
Its value is G = 6.67 × 10⁻¹¹ N m² kg⁻².
What *does* change from planet to planet is g, the acceleration due to gravity, because g = GM/R² depends on the planet's mass M and radius R. Students often confuse g (variable) with G (constant) — that is exactly the trap in this question. So even on a planet of mass 4M, the constant remains G.
❌ Why the other options are incorrect
✗ G/2 — assumes G weakens on a heavier planet; G has no dependence on mass at all.
✗ Zero — G can never be zero; if it were, there would be no gravitational force anywhere in the universe.
✗ 4G — the common mistake of scaling G with the mass ratio 4M/M. It is g, not G, that would change if the radius stayed the same.
📌 SSC & Railway Exam Facts
• G = 6.67 × 10⁻¹¹ N m² kg⁻²; its value was measured experimentally by Henry Cavendish (1798) using a torsion balance.
• G is a scalar and a universal constant; g is a vector and varies with place.
• g = GM/R²; on the Earth g ≈ 9.8 m s⁻², on the Moon about 1/6 of this value.
• g is maximum at the poles and minimum at the equator; it decreases with both altitude and depth, and is zero at the centre of the Earth.
• Mass is constant everywhere, but weight (W = mg) changes from planet to planet.

RRB Technician Grade III 2024-25 – 28 December 2024 Shift 1 · Free fall

Q67. Whenever an object falls toward the earth, acceleration is involved. Which of the following forces provides this acceleration?

(a) Electromagnetic force
(b) Earth's repulsive force
(c) Nuclear force
(d) Earth's gravitational force

Answer & solution

Answer: (d) Earth's gravitational force

Correct Answer: Option 4 — Earth's gravitational force
Every object near the Earth is pulled towards its centre by the Earth's gravitational force, which we call the weight of the body, W = mg. It is this force, and only this force, that gives a falling body its downward acceleration.
By Newton's Law of Universal Gravitation, the force between the Earth (mass M) and a body (mass m) at distance R is
F = GMm/R²
Putting F = ma gives a = GM/R² = g, the acceleration due to gravity, whose value near the Earth's surface is g = 9.8 m/s².
Notice that m cancels out — that is why a heavy stone and a light stone fall together in the absence of air resistance, exactly as Galileo demonstrated.
❌ Why the other options are incorrect
✗ Electromagnetic force — this acts only between charged or magnetic bodies; a falling stone is electrically neutral, so it plays no part here.
✗ Earth's repulsive force — gravitation is always attractive; there is no such thing as a gravitational repulsion.
✗ Nuclear force — the strong nuclear force acts only inside the nucleus, over a range of about 10⁻¹⁵ m; it has no effect on everyday falling bodies.
📌 SSC & Railway Exam Facts
• F = GMm/R²; G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant.
• g = GM/R² = 9.8 m/s² on Earth; on the Moon it is about 1/6 of this value.
• g is maximum at the poles, minimum at the equator, and zero at the centre of the Earth.
• Mass is constant everywhere; weight W = mg changes from place to place.
• The four fundamental forces: gravitational, electromagnetic, strong nuclear and weak nuclear — gravitation is the weakest but has infinite range.

RRB Technician Grade III 2024-25 – 28 December 2024 Shift 2 · Free fall

Q68. When a body falls freely under gravity, then which of the following quantities DOES NOT remain constant?

(a) Universal Gravitational Constant
(b) Distance
(c) Acceleration
(d) Mass

Answer & solution

Answer: (b) Distance

Correct Answer: Option 2 — Distance
In free fall, the only force acting on the body is its weight, so the body moves with the constant acceleration due to gravity, g = 9.8 m/s² (taken as 10 m/s² in easy sums).
Because the body is accelerating, its speed keeps increasing every second, and the distance it has covered keeps growing — in fact s = ½gt², so the distance is not even growing uniformly. Therefore distance is the quantity that does not stay constant.
In the first second a freely falling body covers about 4.9 m, in the next second about 14.7 m — clear proof that distance changes continuously.
❌ Why the other options are incorrect
✗ Universal Gravitational Constant — G = 6.67 × 10⁻¹¹ N m² kg⁻² is a fixed constant of nature; it is the same everywhere in the universe.
✗ Acceleration — during free fall near the Earth's surface the acceleration stays fixed at g = 9.8 m/s², independent of the body's mass.
✗ Mass — the mass of a body is the amount of matter in it and never changes with position or motion; only its weight (W = mg) can change from place to place.
📌 SSC & Railway Exam Facts
• Equations of free fall: v = gt, s = ½gt², v² = 2gs (starting from rest).
• g = 9.8 m/s² on the Earth's surface; g = 1.625 m/s² on the Moon, about 1/6 of the Earth's.
• g is maximum at the poles and minimum at the equator; it is zero at the centre of the Earth.
• G is a universal constant; g is not — it varies with height, depth and latitude.
• In free fall all bodies fall together — a feather and a coin land together in a vacuum tube.

RRB Technician Grade III 2024-25 – 28 December 2024 Shift 2 · Free fall

Q69. When an object is thrown upward, the force of gravity is _____.

(a) Zero
(b) the same as the direction of motion
(c) opposite to the direction of motion
(d) independent of direction of motion

Answer & solution

Answer: (c) opposite to the direction of motion

Correct Answer: Option 3 — opposite to the direction of motion
The force of gravity always acts vertically downwards, towards the centre of the Earth. When an object is thrown upward, its motion is directed upwards, so gravity is acting exactly opposite to the direction of motion.
Because the force opposes the motion, it produces a retardation of g = 9.8 m/s². The upward speed keeps falling until it becomes zero at the highest point. After that the object starts coming down, and now gravity is along the direction of motion, so it accelerates the body.
Note that during the upward journey the work done by gravity is negative, since force and displacement point in opposite directions.
❌ Why the other options are incorrect
✗ Zero — gravity never becomes zero near the Earth's surface; even at the highest point where v = 0, the acceleration is still g.
✗ the same as the direction of motion — this is true only while the body is falling down, not while it is going up.
✗ independent of direction of motion — gravity's own direction is fixed (downward), so relative to an upward motion it is definitely opposite, not unrelated.
📌 SSC & Railway Exam Facts
• Going up: v = u − gt; maximum height H = u²/2g; time of ascent t = u/g.
• Time of ascent = time of descent for a body thrown vertically up (air resistance neglected).
• The body returns to the thrower's hand with the same speed with which it was thrown.
• Weight W = mg; weight is a force measured in newton (N), mass is measured in kg.
• In free fall a body feels weightlessness, as in an orbiting spacecraft.

RRB Technician Grade III 2024-25 – 28 December 2024 Shift 3 · Acceleration due to gravity (g) and its variation

Q70. The value of acceleration due to gravity 'g' is ________.

(a) the same at the poles and equator
(b) negative at the equator and positive at the poles
(c) greater at the poles than at the equator
(d) greater at the equator than at the poles

Answer & solution

Answer: (c) greater at the poles than at the equator

Correct Answer: Option 3 — greater at the poles than at the equator
The acceleration due to gravity on the surface of the Earth is
g = GM / R²
so g is inversely proportional to the square of the Earth's radius R at that place.
The Earth is not a perfect sphere — it is an oblate spheroid, slightly flattened at the poles and bulging at the equator. Hence the polar radius (≈ 6357 km) is smaller than the equatorial radius (≈ 6378 km). A smaller R means a larger g, so g is maximum at the poles and minimum at the equator.
The Earth's rotation adds to the same conclusion: the centrifugal effect of rotation reduces the effective g most at the equator and not at all at the poles. Numerically, g ≈ 9.83 m/s² at the poles and g ≈ 9.78 m/s² at the equator.
❌ Why the other options are incorrect
✗ the same at the poles and equator — this would be true only if the Earth were a perfect non-rotating sphere; in reality g varies with latitude.
✗ negative at the equator and positive at the poles — g is an acceleration directed towards the centre of the Earth everywhere; it never becomes negative in this sense.
✗ greater at the equator than at the poles — exactly the reverse of the truth, since the equatorial radius is the larger one.
📌 SSC & Railway Exam Facts
• g = GM/R²; average value g = 9.8 m/s²; g is maximum at the poles, minimum at the equator.
• g decreases with height (g' = g(1 − 2h/R)) and with depth (g' = g(1 − d/R)), and is zero at the centre of the Earth.
• G = 6.67 × 10⁻¹¹ N m² kg⁻² is a universal constant; g is not a constant.
• Value of g on the Moon is about 1/6 of that on the Earth, i.e. 1.625 m/s².
• A body's weight W = mg is therefore slightly more at the poles than at the equator, though its mass stays the same.

RRB Technician Grade III 2024-25 – 28 December 2024 Shift 3 · Free fall

Q71. If a stone is thrown up, the value of acceleration due to gravity 'g' will be taken as ________.

(a) positive
(b) infinite
(c) negative
(d) zero

Answer & solution

Answer: (c) negative

Correct Answer: Option 3 — negative
When a stone is thrown upwards, we naturally take the upward direction as positive, because that is the direction of the initial velocity u.
But the acceleration due to gravity always acts downwards, towards the centre of the Earth. Since 'down' is the direction opposite to the chosen positive direction, g must be taken as negative (g = −9.8 m/s²) in the equations of motion.
That is why the stone decelerates as it rises: v = u − gt, and its velocity becomes zero at the highest point. The maximum height reached is h = u²/2g and the time to reach the top is t = u/g. On the way down, if we keep the same convention, the stone speeds up again because the acceleration and the (now downward) velocity point the same way.
❌ Why the other options are incorrect
✗ positive — g would be taken positive only if the downward direction were chosen as positive, as in a body falling freely from rest.
✗ infinite — g is a finite quantity, 9.8 m/s² near the Earth's surface; it can never be infinite.
✗ zero — g is zero only at the centre of the Earth or in a state of free fall inside a satellite; even at the highest point of the throw, where the velocity is zero, g is still 9.8 m/s² downwards.
📌 SSC & Railway Exam Facts
• Thrown up → g is negative (retardation); falling down → g is positive (acceleration).
• At the topmost point velocity v = 0 but acceleration g ≠ 0.
• Maximum height h = u²/2g; time of ascent = time of descent = u/g; total time of flight = 2u/g.
• A body returns to the thrower's hand with the same speed with which it was thrown (ignoring air resistance).
• g = 9.8 m/s² on Earth; g = GM/R²; astronauts feel weightless because they are in continuous free fall.

RRB Technician Grade III 2024-25 – 27 December 2024 Shift 1 · Universal law of gravitation and G

Q72. If the magnitude of the force exerted by the earth on the moon is 2.02 × 10²⁰ N, then the magnitude of the force exerted by the moon on the earth is ______.

(a) 2.02 × 10²⁰ N
(b) 6.06 × 10²⁰ N
(c) 1.01 × 10²⁰ N
(d) 4.04 × 10²⁰ N

Answer & solution

Answer: (a) 2.02 × 10²⁰ N

Correct Answer: Option 1 — 2.02 × 10²⁰ N
📖 Given
• Force exerted by the earth on the moon, F = 2.02 × 10²⁰ N
• Both bodies attract each other gravitationally
✍️ Formula
F = Gm₁m₂/r² — the same expression holds for both bodies, and by Newton's third law action and reaction are equal in magnitude and opposite in direction.
🔢 Solution
• Force on the moon due to the earth = Gm(earth)m(moon)/r²
• Force on the earth due to the moon = Gm(moon)m(earth)/r²
• The two expressions are identical, so the magnitudes are equal
• Force exerted by the moon on the earth = 2.02 × 10²⁰ N
The moon pulls the earth exactly as hard as the earth pulls the moon; the moon simply moves much more because its mass is far smaller.
❌ Why the other options are incorrect
✗ 6.06 × 10²⁰ N — this is 3 times the given force; no rule of gravitation multiplies the reaction by 3.
✗ 1.01 × 10²⁰ N — this is half the given force, wrongly assuming the smaller body feels a smaller force.
✗ 4.04 × 10²⁰ N — this is double the given force; the reaction is equal, not doubled.
📌 SSC & Railway Exam Facts
• Newton's third law: to every action there is an equal and opposite reaction, acting on two different bodies.
• Universal law of gravitation: F = Gm₁m₂/r², with G = 6.67 × 10⁻¹¹ N m² kg⁻².
• Gravitational force is always attractive and acts along the line joining the two centres.
• F ∝ 1/r² — doubling the distance makes the force one-fourth.
• The moon's gravitational pull on the earth's water causes tides.

RRB Technician Grade III 2024-25 – 27 December 2024 Shift 1 · Mass and weight

Q73. What is the SI unit of mass and weight of a body, respectively?

(a) kg, s
(b) kg, N
(c) kg, kg
(d) gm, N

Answer & solution

Answer: (b) kg, N

Correct Answer: Option 2 — kg, N
Mass is the quantity of matter contained in a body. It is a scalar quantity, it is the measure of a body's inertia, and it stays the same everywhere — on the earth, on the moon or in space. Its SI unit is the kilogram (kg).
Weight is the force with which the earth attracts a body, W = mg. Being a force, it is a vector quantity and its SI unit is the newton (N). Weight changes from place to place because the value of g changes — a body weighs about one-sixth as much on the moon as on the earth.
For example, a body of mass 10 kg has a weight of 10 × 9.8 = 98 N on the earth.
❌ Why the other options are incorrect
✗ kg, s — the second (s) is the SI unit of time, not of weight.
✗ kg, kg — the kilogram is right for mass but cannot be the unit of weight; weight is a force, measured in newton.
✗ gm, N — the unit of weight is correct here, but gram (g) is a CGS unit; the SI unit of mass is the kilogram.
📌 SSC & Railway Exam Facts
• Mass → kilogram (kg), a scalar, measured with a beam balance.
• Weight → newton (N), a vector, measured with a spring balance.
• W = mg, with g = 9.8 m s⁻² on the earth.
• Weight is zero at the centre of the earth and maximum at the poles.
• 1 kgf = 9.8 N; 1 N = 10⁵ dyne.

RRB Technician Grade III 2024-25 – 27 December 2024 Shift 2 · Universal law of gravitation and G

Q74. What is the unit of universal gravitational constant (G)?

(a) N m² kg⁻¹
(b) N m² kg⁻²
(c) N m²
(d) N m² kg

Answer & solution

Answer: (b) N m² kg⁻²

Correct Answer: Option 2 — N m² kg⁻²
📖 Given
• Two masses m₁ and m₂ separated by a distance r
• The gravitational force between them is F, measured in newton (N)
✍️ Formula
Newton's law of universal gravitation: F = Gm₁m₂/r²
🔢 Solution
• Rearranging for G, G = Fr²/(m₁m₂)
• Unit of F = N, unit of r² = m², unit of m₁m₂ = kg × kg = kg²
• So unit of G = N × m²/kg² = N m² kg⁻²
Its numerical value is G = 6.67 × 10⁻¹¹ N m² kg⁻², and because it is the same everywhere in the universe it is called the universal gravitational constant.
❌ Why the other options are incorrect
✗ N m² kg⁻¹ — this would follow if only one mass appeared in the denominator; the formula has the product of two masses.
✗ N m² — this ignores the masses altogether; it is the unit you get from Fr² alone.
✗ N m² kg — here kg is multiplied instead of divided, which is dimensionally the opposite of the correct unit.
📌 SSC & Railway Exam Facts
• G = 6.67 × 10⁻¹¹ N m² kg⁻², first measured experimentally by Henry Cavendish.
• Newton's law of gravitation: F ∝ m₁m₂/r² — the inverse square law.
• g = GM/R² ≈ 9.8 m/s² on the earth's surface; g and G are entirely different quantities.
• G is a scalar and its value does not change with place, medium or temperature.
• Dimensional formula of G is [M⁻¹L³T⁻²].

RRB Technician Grade III 2024-25 – 27 December 2024 Shift 2 · Universal law of gravitation and G

Q75. The planets go around the sun due to _______ force.

(a) Frictional
(b) Magnetic
(c) Electrical
(d) Gravitational

Answer & solution

Answer: (d) Gravitational

Correct Answer: Option 4 — Gravitational
The planets revolve around the sun because of the gravitational force of attraction exerted on them by the sun. According to Newton's universal law of gravitation, every body in the universe attracts every other body with a force
F = Gm₁m₂/r²
where G = 6.67 × 10⁻¹¹ N m² kg⁻².
This gravitational pull acts along the line joining the planet and the sun and provides exactly the centripetal force needed to keep the planet in its orbit. Without it, a planet would fly off along a straight line in accordance with Newton's first law of motion.
The same force keeps the moon revolving around the earth, holds artificial satellites in orbit and causes the tides in the sea.
❌ Why the other options are incorrect
✗ Frictional — friction acts only between surfaces in contact; there is no contact and practically no medium in the vacuum of space.
✗ Magnetic — magnetic force acts on magnets and moving charges. Planets are electrically neutral bodies and are not held by magnetism.
✗ Electrical — electrical force needs net charge; the sun and planets are overall neutral, so no net electrostatic force binds them.
📌 SSC & Railway Exam Facts
• Newton's law of gravitation: F = Gm₁m₂/r² — an inverse square law.
• Kepler's three laws: orbits are ellipses, equal areas in equal times, and T² ∝ r³.
• Centripetal force for a planet: F = mv²/r, always directed towards the centre.
• Gravitational force is the weakest of the four fundamental forces but acts over infinite range.
• Tides are caused mainly by the gravitational pull of the moon on the earth's oceans.

RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Mass and weight

Q76. The weight of an object on the Moon is ____________

(a) greater than the weight on the earth
(b) less than the weight on the earth
(c) equal to the weight on the earth
(d) may be greater than or less than depending on the density of the object

Answer & solution

Answer: (b) less than the weight on the earth

Correct Answer: Option 2 — less than the weight on the earth
📖 Given
• Weight is the gravitational pull on a body: W = mg
• Mass of the moon is about 1/100 of the earth's mass
• Radius of the moon is about 1/4 of the earth's radius
✍️ Formula
g = GM/R² and W = mg
🔢 Solution
• g(moon)/g(earth) = (M(moon)/M(earth)) × (R(earth)/R(moon))²
• = (1/100) × (4)² = 16/100
• So g(moon) ≈ 1/6 of g(earth), about 1.6 m/s²
• Since mass m stays the same, W(moon) = m × g(moon) = 1/6 of W(earth)
So a 60 kg astronaut weighing 600 N on the earth weighs only about 100 N on the moon — which is why astronauts can take those long, floating leaps.
❌ Why the other options are incorrect
✗ greater than the weight on the earth — impossible, because the moon's gravitational acceleration is much smaller than the earth's.
✗ equal to the weight on the earth — only the mass stays the same everywhere; weight changes with the value of g.
✗ may be greater than or less than depending on the density of the object — the density of the object plays no part; the ratio g(moon)/g(earth) = 1/6 is the same for every body.
📌 SSC & Railway Exam Facts
• Weight on the moon = 1/6 of the weight on the earth; g(moon) ≈ 1.6 m/s².
• Mass is constant and measured in kg; weight = mg is a force, measured in newton (N).
• g is maximum at the poles, minimum at the equator, and zero at the centre of the earth.
• Weight becomes zero in a freely falling lift — the state of weightlessness.
• The moon has no atmosphere because its escape velocity (2.4 km/s) is too small to hold gases.

RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Universal law of gravitation and G

Q77. Which of the following equations is true, when two objects A and B of masses M and m are separated by a distance d, F is the force of attraction between them and G is the universal gravitational constant?

(a) F = G 1/d²
(b) F = G (M × m)/d
(c) F = G (M × m)/d²
(d) F = G M/d²

Answer & solution

Answer: (c) F = G (M × m)/d²

Correct Answer: Option 3 — F = G (M × m)/d²
Newton's universal law of gravitation states that every object in the universe attracts every other object with a force that is
• directly proportional to the product of their masses: F ∝ M × m
• inversely proportional to the square of the distance between them: F ∝ 1/d²
Combining the two, F ∝ Mm/d², and introducing the constant of proportionality G,
F = G (M × m)/d²
Here G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant, so called because its value is the same for every pair of bodies anywhere in the universe. The force acts along the line joining the two bodies and is always attractive.
❌ Why the other options are incorrect
✗ F = G 1/d² — the masses are missing; the force must depend on both M and m, otherwise a heavy and a light body would be attracted equally.
✗ F = G (M × m)/d — here d appears to the first power; gravitation is an inverse square law, so d must be squared.
✗ F = G M/d² — only one mass appears; the product of the two masses is required by the law.
📌 SSC & Railway Exam Facts
• F = GMm/d² — an inverse square law; doubling the distance makes the force one-fourth.
• G = 6.67 × 10⁻¹¹ N m² kg⁻², measured experimentally by Henry Cavendish.
• On the earth's surface, g = GM/R² ≈ 9.8 m/s²; g and G are different quantities.
• The gravitational forces on the two bodies are equal and opposite — an action–reaction pair.
• Gravitation explains the motion of planets, tides, satellites and the free fall of bodies.

RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Acceleration due to gravity (g) and its variation

Q78. The acceleration due to gravity on an object of mass m placed near the surface of the earth is:

(a) independent of m
(b) linearly proportional to m
(c) inversely proportional to m
(d) linearly proportional to m²

Answer & solution

Answer: (a) independent of m

Correct Answer: Option 1 — independent of m
📖 Given
• Mass of the object, m
• Mass of the earth, M, radius of the earth, R
• Universal gravitational constant, G
✍️ Formula
Gravitational force, F = GMm/R², and from Newton's second law F = mg
🔢 Solution
• mg = GMm/R²
• Cancel m from both sides
• g = GM/R²
The mass m of the object has completely disappeared from the expression, so g does not depend on the mass of the falling body. That is why a feather and a stone dropped together in a vacuum tube reach the bottom at the same time.
❌ Why the other options are incorrect
✗ linearly proportional to m — the weight (W = mg) is proportional to m, but g itself is not.
✗ inversely proportional to m — nothing in g = GM/R² contains the object's mass, so no inverse relation exists.
✗ linearly proportional to m² — again, the object's mass cancels out completely; only the earth's mass M matters.
📌 SSC & Railway Exam Facts
• g = GM/R² ≈ 9.8 m/s² on the earth's surface; G = 6.67 × 10⁻¹¹ N m² kg⁻².
• g is maximum at the poles and minimum at the equator, because R is least at the poles.
• g becomes zero at the centre of the earth.
• On the moon, g is about 1/6 of its value on the earth.
• Mass is constant everywhere and measured in kg; weight = mg changes with place and is measured in newton (N).

RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Free fall

Q79. If u, g, and h are the initial velocity, gravitational acceleration, and height, respectively, of an object falling under gravitational force, then its final velocity v can be obtained from:

(a) v² + u² = 2gh
(b) v²/u² = 2gh
(c) v²u² = 2gh
(d) v² − u² = 2gh

Answer & solution

Answer: (d) v² − u² = 2gh

Correct Answer: Option 4 — v² − u² = 2gh
📖 Given
• Initial velocity = u
• Acceleration = g (the body is falling freely under gravity)
• Distance fallen = h
• Final velocity = v
✍️ Formula
The third equation of motion, v² = u² + 2as
🔢 Solution
• For vertical fall put a = g and s = h
• v² = u² + 2gh
• Shifting u² to the left side, v² − u² = 2gh
This equation connects velocity with distance directly, without needing the time of fall — which is why it is the quickest route whenever time is not given.
❌ Why the other options are incorrect
✗ v² + u² = 2gh — the sign is wrong; u² must be subtracted, since v² = u² + 2gh.
✗ v²/u² = 2gh — dividing the squares is meaningless here and is also dimensionally incorrect.
✗ v²u² = 2gh — multiplying the squares gives units of m⁴/s⁴, which cannot equal 2gh (m²/s²).
📌 SSC & Railway Exam Facts
• The three equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• For free fall from rest (u = 0): v = gt, h = ½gt², v² = 2gh.
• Taking g = 9.8 m/s² downward; for a body thrown upward a = −g.
• At the highest point of an upward throw the velocity is zero but the acceleration is still g.
• Equations of motion apply only for uniform acceleration in a straight line.

RRB Technician Grade III 2024-25 – 26 December 2024 Shift 3 · Numericals on gravitation

Q80. If an object on the surface of the moon weighs 3 N, the weight of the same object on surface of the Earth is ______.

(a) 18 N
(b) 3 N
(c) 12 N
(d) 24 N

Answer & solution

Answer: (a) 18 N

Correct Answer: Option 1 — 18 N
📖 Given
• Weight of the object on the moon, W(moon) = 3 N
• Acceleration due to gravity on the moon is one-sixth of that on the Earth, g(moon) = g(earth)/6
• The mass m of the object stays the same everywhere
✍️ Formula
W = mg, so W(earth)/W(moon) = g(earth)/g(moon) = 6
🔢 Solution
• W(earth) = 6 × W(moon)
• W(earth) = 6 × 3
• W(earth) = 18 N
So the same object that presses down with only 3 N on the moon presses down with 18 N on the Earth, because the Earth pulls six times as strongly.
❌ Why the other options are incorrect
✗ 3 N — this assumes weight is the same everywhere; that is true of mass, not of weight.
✗ 12 N — this comes from multiplying by 4 instead of 6.
✗ 24 N — this comes from multiplying by 8 instead of 6.
📌 SSC & Railway Exam Facts
• g(earth) = 9.8 m s⁻²; g(moon) ≈ 1.63 m s⁻², about 1/6 of the Earth's value.
• Mass is measured in kg and is constant; weight is a force measured in newton (N).
• Newton's law of gravitation: F = Gm₁m₂/r², with G = 6.67 × 10⁻¹¹ N m² kg⁻².
• The value of g is maximum at the poles and minimum at the equator.
• In free fall (and inside an orbiting satellite) a body feels weightlessness.

RRB Technician Grade III 2024-25 – 24 December 2024 Shift 1 · Free fall

Q81. The motion of a freely falling body is an example of a:

(a) uniform velocity motion
(b) non-uniformly accelerated motion
(c) uniformly accelerated motion
(d) uniform speed motion

Answer & solution

Answer: (c) uniformly accelerated motion

Correct Answer: Option 3 — uniformly accelerated motion
A body in free fall moves only under the pull of gravity, and near the Earth's surface this gives it a constant acceleration g = 9.8 m/s², directed vertically downwards. Because the acceleration stays the same in both magnitude and direction, the motion is called uniformly accelerated motion.
In such a motion the velocity increases by an equal amount in equal intervals of time — 9.8 m/s in every second. Its velocity–time graph is a straight sloping line and the distance–time graph is a parabola. Also, g does not depend on the mass of the falling body, so in a vacuum a coin and a feather fall side by side.
❌ Why the other options are incorrect
✗ uniform velocity motion — uniform velocity means the velocity does not change at all, so the acceleration would be zero; but a falling body clearly speeds up.
✗ uniform speed motion — for the same reason, the speed of a falling body keeps increasing every second, so it is not uniform.
✗ non-uniformly accelerated motion — this would require the acceleration itself to change; near the Earth's surface g is practically constant, so the acceleration is uniform.
📌 SSC & Railway Exam Facts
• Free fall — motion under gravity alone; g = 9.8 m/s² on the Earth, 1.63 m/s² on the Moon.
• Free-fall equations (u = 0): v = gt, h = ½gt², v² = 2gh.
• g is independent of the mass of the body — the famous Galileo result.
• g = GM/R²; g is maximum at the poles, minimum at the equator, and zero at the centre of the Earth.
• In uniformly accelerated motion the velocity–time graph is a straight line, and the area under it gives the displacement.

RRB Technician Grade III 2024-25 – 24 December 2024 Shift 1 · Universal law of gravitation and G

Q82. Which of the following constant values given in the options is equal to the universal gravitational constant 'G'?

(a) 6.7 × 10⁻¹¹ Nm²kg⁻²
(b) 9.1 × 10⁻³¹ kg
(c) 1.6 × 10⁻¹⁹ C
(d) 9.8 ms⁻²

Answer & solution

Answer: (a) 6.7 × 10⁻¹¹ Nm²kg⁻²

Correct Answer: Option 1 — 6.7 × 10⁻¹¹ Nm²kg⁻²
Newton's law of universal gravitation states that two bodies of masses m₁ and m₂ whose centres are a distance r apart attract each other with a force
F = Gm₁m₂/r²
Here G is the universal gravitational constant. Its accepted value is G = 6.67 × 10⁻¹¹ N m² kg⁻², rounded in the option to 6.7 × 10⁻¹¹ Nm²kg⁻². It is called *universal* because it has the same value everywhere in the universe, for every pair of bodies and in every medium. Its value was first measured experimentally by Henry Cavendish using a torsion balance. The unit itself confirms the choice: rearranging the law gives G = Fr²/m₁m₂, whose unit is N m²/kg².
❌ Why the other options are incorrect
✗ 9.1 × 10⁻³¹ kg — this is the mass of an electron; the unit kg shows at once that it is a mass, not a gravitational constant.
✗ 1.6 × 10⁻¹⁹ C — this is the charge on an electron (or proton), measured in coulomb (C), an electrical quantity.
✗ 9.8 ms⁻² — this is g, the acceleration due to gravity on the Earth's surface; g changes from place to place, whereas G never changes.
📌 SSC & Railway Exam Facts
• F = Gm₁m₂/r²; G = 6.67 × 10⁻¹¹ N m² kg⁻², a universal constant measured by Henry Cavendish.
• g = GM/R² = 9.8 m/s² — g is not a constant; it varies with height, depth and latitude.
• Gravitational force follows the inverse-square law: double the distance and the force becomes one-fourth.
• Mass of an electron = 9.1 × 10⁻³¹ kg; charge of an electron = 1.6 × 10⁻¹⁹ C.
• Other standard constants: speed of light c = 3 × 10⁸ m/s, Planck's constant h = 6.63 × 10⁻³⁴ J s.

RRB Technician Grade III 2024-25 – 24 December 2024 Shift 2 · Universal law of gravitation and G

Q83. The gravitational force between two objects:

(a) only depends on the masses of the objects
(b) depends on the masses of the objects and the distance between them
(c) does not depend on either the masses of the objects or the distance between them
(d) only depends on the distance between the objects

Answer & solution

Answer: (b) depends on the masses of the objects and the distance between them

Correct Answer: Option 2 — depends on the masses of the objects and the distance between them
Newton's universal law of gravitation states that every two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres:
F = G M m / r²
Here G is the universal gravitational constant, whose value is 6.67 × 10⁻¹¹ N m² kg⁻². So the force clearly depends on both the masses and the separation. Doubling either mass doubles the force, while doubling the distance cuts the force to one-fourth.
❌ Why the other options are incorrect
✗ only depends on the masses of the objects — incomplete; it ignores the inverse-square dependence on r, the reason gravity weakens with distance.
✗ does not depend on either the masses of the objects or the distance between them — completely wrong; it would make gravity the same everywhere, which contradicts orbits and free fall.
✗ only depends on the distance between the objects — incomplete; the product Mm is very much part of the formula, which is why the Sun pulls far harder than the Moon.
📌 SSC & Railway Exam Facts
• F = GMm/r²; G = 6.67 × 10⁻¹¹ N m² kg⁻², first measured by Henry Cavendish.
• G is a universal constant; g is not — it changes with place, g = GM/R².
• Average value of g on Earth = 9.8 m/s²; on the Moon it is about 1/6 of this.
• Gravitational force is the weakest of the four fundamental forces but always attractive.

RRB Technician Grade III 2024-25 – 24 December 2024 Shift 3 · Numericals on gravitation

Q84. The weight of an object was 60 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?

(a) 30 N
(b) 60 N
(c) 39 N
(d) 10 N

Answer & solution

Answer: (d) 10 N

Correct Answer: Option 4 — 10 N
📖 Given
• Weight on the earth, W(earth) = 60 N
• Acceleration due to gravity on the moon is 1/6 of that on the earth: g(moon) = g(earth)/6
✍️ Formula
W = mg, so W(moon)/W(earth) = g(moon)/g(earth) = 1/6
🔢 Solution
• W(moon) = W(earth) × 1/6
• W(moon) = 60 × 1/6
• W(moon) = 10 N
The mass stays the same everywhere; only the weight falls, because the moon is much smaller and less massive, so it pulls the object with only one-sixth of the earth's force. This is why astronauts can take long, floating leaps on the moon.
❌ Why the other options are incorrect
✗ 30 N — this is half of 60 N, obtained by using a factor of 1/2 instead of the correct 1/6.
✗ 60 N — this assumes the weight does not change; that is true of mass, not of weight.
✗ 39 N — no step of the calculation gives 39; it is a distractor with no physical meaning here.
📌 SSC & Railway Exam Facts
• Weight W = mg; SI unit newton (N) — weight is a force and a vector.
• g(earth) = 9.8 m/s², g(moon) ≈ 1.63 m/s² ≈ g/6.
• Mass is constant everywhere; SI unit kilogram (kg), a scalar.
• g = GM/R² — it depends on the mass and radius of the planet, not on the object.
• Weight is zero at the centre of the earth and maximum at the poles, least at the equator.

RRB Technician Grade III 2024-25 – 24 December 2024 Shift 3 · Universal law of gravitation and G

Q85. Which of the following forces is responsible for the tides, due to the Moon and the Sun?

(a) Nuclear force
(b) Electromagnetic force
(c) Gravitation force
(d) Electrostatic force

Answer & solution

Answer: (c) Gravitation force

Correct Answer: Option 3 — Gravitation force
Tides — the regular rise and fall of sea water — are caused by the gravitational pull of the Moon and, to a smaller extent, of the Sun on the water of the oceans. By Newton's universal law of gravitation, F = GMm/r², so the side of the Earth facing the Moon is pulled a little more strongly than the Earth's centre, and the far side a little less. This difference in pull across the Earth raises a bulge of water on both sides, producing high tide there and low tide in between.
The Moon's effect is about twice that of the Sun even though the Sun is far more massive, because the Moon is very much nearer and the force falls off as 1/r². When the Sun, Earth and Moon are in a straight line (new moon and full moon) their pulls add and we get the very high spring tides; at quadrature we get the weak neap tides.
❌ Why the other options are incorrect
✗ Nuclear force — the strong force that binds protons and neutrons inside the nucleus; it acts only over about 10⁻¹⁵ m, so it cannot reach across space.
✗ Electromagnetic force — acts between charges and magnets; sea water and the Moon are electrically neutral overall, so this force plays no role in tides.
✗ Electrostatic force — the force between stationary charges, given by Coulomb's law; again, the bodies concerned carry no net charge.
📌 SSC & Railway Exam Facts
• F = GMm/r²; G = 6.67 × 10⁻¹¹ N m² kg⁻², measured by Henry Cavendish.
• Gravity is the weakest of the four fundamental forces but has infinite range and is always attractive.
• Four fundamental forces: gravitational, electromagnetic, strong nuclear, weak nuclear.
• Spring tides — new moon and full moon; neap tides — first and last quarter.
• The Moon is the chief cause of tides; the Sun's tidal effect is roughly half that of the Moon.

RRB Technician Grade III 2024-25 – 23 December 2024 Shift 1 · Universal law of gravitation and G

Q86. Which of the following is/are NOT related to the phenomenon of the universal law of gravitation?
(A) The motion of the moon around the earth.
(B) The tides caused by the moon.
(C) The motion of electron around the nucleus.

(a) (A) and (B)
(b) (B) and (C)
(c) Only (C)
(d) Only (B)

Answer & solution

Answer: (c) Only (C)

Correct Answer: Option 3 — Only (C)
Newton's universal law of gravitation states that every body in the universe attracts every other body with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them:
F = G m₁m₂/r², where G = 6.67 × 10⁻¹¹ N m² kg⁻².
(A) is related — the Moon revolves around the Earth because the Earth's gravitational pull supplies the centripetal force that keeps it in orbit.
(B) is related — the tides in the oceans are produced by the gravitational attraction of the Moon (helped by the Sun) on the water of the Earth.
(C) is NOT related — an electron revolves around the nucleus because of the electrostatic force of attraction between the negative electron and the positive nucleus, given by Coulomb's law. Gravitational force between them is about 10³⁹ times weaker and is completely negligible. Hence only (C) is unrelated.
❌ Why the other options are incorrect
✗ (A) and (B) — both of these are classic consequences of gravitation, so neither can be "not related".
✗ (B) and (C) — the tides in (B) are directly caused by the Moon's gravitational pull, so (B) cannot be included.
✗ Only (B) — (B) is related to gravitation; the statement that is unrelated is (C).
📌 SSC & Railway Exam Facts
• F = G m₁m₂/r²; G = 6.67 × 10⁻¹¹ N m² kg⁻², a universal constant measured by Henry Cavendish.
• g = GM/R²; g = 9.8 m/s² on the Earth, 1.63 m/s² on the Moon (about 1/6 of the Earth's).
• Gravitational force is the weakest of the four fundamental forces; the strong nuclear force is the strongest.
• Gravitation explains the motion of planets (Kepler's laws), the orbits of satellites and the tides.
• The electrostatic force follows Coulomb's law, F = kq₁q₂/r² — the same inverse-square form, but enormously stronger.

RRB Technician Grade III 2024-25 – 23 December 2024 Shift 1 · Free fall

Q87. Which of the following statements is true about the free fall of an object under gravitational force?

(a) The velocity of the object remains constant during free fall.
(b) The acceleration experienced by an object is independent of its mass.
(c) The acceleration experienced by an object is opposite to the direction of motion.
(d) The acceleration experienced by an object depends on its mass.

Answer & solution

Answer: (b) The acceleration experienced by an object is independent of its mass.

Correct Answer: Option 2 — The acceleration experienced by an object is independent of its mass.
Free fall is the motion of a body under the action of gravity alone, with air resistance neglected. The gravitational force on the body is F = mg, and by Newton's second law this force also equals F = ma. Equating them:
ma = mg, so a = g — the mass m cancels out completely.
Therefore every freely falling body, heavy or light, falls with the same acceleration g = 9.8 m/s² (often taken as 10 m/s²), directed vertically downwards. This is why a coin and a feather dropped together in an evacuated tube reach the bottom at the same instant — the famous experiment that confirms Galileo's conclusion.
❌ Why the other options are incorrect
✗ The velocity of the object remains constant during free fall — false; the velocity increases steadily at 9.8 m/s every second, since the motion is accelerated.
✗ The acceleration experienced by an object is opposite to the direction of motion — false for a falling body; g acts downwards, the same direction as the motion. It is opposite only for a body thrown upwards.
✗ The acceleration experienced by an object depends on its mass — false; the mass cancels in ma = mg, so a = g for every body.
📌 SSC & Railway Exam Facts
• In free fall: v = gt, h = ½gt², v² = 2gh (taking u = 0).
• g = 9.8 m/s² on the Earth; g = GM/R², so it is independent of the falling body's mass.
• g is maximum at the poles, minimum at the equator, and becomes zero at the centre of the Earth.
• In a freely falling lift a person feels weightlessness, because the apparent weight becomes zero.
• Galileo is said to have demonstrated at the Leaning Tower of Pisa that bodies of different masses fall together.

RRB Technician Grade III 2024-25 – 23 December 2024 Shift 2 · Numericals on gravitation

Q88. An object weighs 60 N when measured on the Earth. What would be its weight when measured on the moon? (Given, acceleration due to gravity (g) of the moon is 1/6 th of the g of the earth)

(a) 40 N
(b) 60 N
(c) 10 N
(d) 70 N

Answer & solution

Answer: (c) 10 N

Correct Answer: Option 3 — 10 N
📖 Given
• Weight on the Earth, W(earth) = 60 N
• g(moon) = (1/6) × g(earth)
• Mass m stays the same everywhere
✍️ Formula
W = mg, so W(moon) = m × g(moon) = (1/6) × W(earth)
🔢 Solution
• W(moon) = (1/6) × 60
• W(moon) = 60/6
• W(moon) = 10 N
The object becomes one-sixth as heavy on the moon even though it still contains exactly the same amount of matter.
❌ Why the other options are incorrect
✗ 60 N — this is the Earth weight; it would be right only for mass, which does not change, but weight does change.
✗ 40 N — this is 60 × (2/3); the fraction 1/6 has not been used correctly.
✗ 70 N — weight on the moon must be less than on the Earth, because the moon's gravity is weaker, so a larger value is impossible.
📌 SSC & Railway Exam Facts
• Mass is constant everywhere and is measured in kg; weight = mg is a force measured in newton (N).
• On the Earth g = 9.8 m/s² (often taken as 10 m/s²); on the moon g ≈ 1.63 m/s².
• The moon's gravity is weak because its mass is about 1/81 and its radius about 1/4 of the Earth's.
• A body's weight is zero at the centre of the Earth and maximum at the poles.
• g = GM/R², so g decreases with height and with depth below the surface.

RRB Technician Grade III 2024-25 – 23 December 2024 Shift 3 · Free fall

Q89. What is the initial speed of any object if it is dropped from a height of 50 m?

(a) 20 m/s
(b) 50 m/s
(c) 0
(d) 25 m/s

Answer & solution

Answer: (c) 0

Correct Answer: Option 3 — 0
The word "dropped" is the key to this question. When a body is simply dropped, it is released from rest — nobody throws it or pushes it downwards. Therefore its initial speed u = 0, whatever the height may be. The figure 50 m is extra data put there only to distract the candidate.
After release, gravity gives the body a constant downward acceleration g = 9.8 m/s², and its speed keeps increasing. Using v² = u² + 2gh = 0 + 2 × 9.8 × 50 = 980, it would hit the ground at v ≈ 31.3 m/s — but that is the final speed, not the initial one.
❌ Why the other options are incorrect
✗ 20 m/s — a dropped body has no starting speed; this value would apply only if the object were thrown downwards at 20 m/s.
✗ 50 m/s — this is simply the number 50 copied from the height; 50 m is a distance, not a speed, and the two cannot be interchanged.
✗ 25 m/s — again a thrown-down starting speed; nothing in the question says the object was projected with any speed.
📌 SSC & Railway Exam Facts
• "Dropped / released / let fall" ⇒ u = 0; "thrown / projected" ⇒ u ≠ 0.
• Free-fall equations (u = 0): v = gt, h = ½gt², v² = 2gh.
• g = 9.8 m/s² on the Earth (often taken as 10 m/s²) and is independent of the mass of the falling body.
• In a vacuum a feather and a coin fall together — the classic Galileo demonstration.
• Time of fall from height h: t = √(2h/g); here t = √(100/9.8) ≈ 3.2 s.

RRB Technician Grade III 2024-25 – 20 December 2024 Shift 3 · Numericals on gravitation

Q90. With an initial velocity of 10 m/s, an object is thrown vertically upwards. The maximum height attained by the object is ______. (Assume g = 10 m/s² )

(a) 20 m
(b) 5 m
(c) 100 m
(d) 1 m

Answer & solution

Answer: (b) 5 m

Correct Answer: Option 2 — 5 m
📖 Given
• Initial velocity, u = 10 m/s (upwards)
• Final velocity at the highest point, v = 0 m/s
• Acceleration due to gravity, g = 10 m/s² (acting downwards, so a = −10 m/s²)
• To find: maximum height, h
✍️ Formula
v² = u² + 2as, which for vertical upward motion becomes v² = u² − 2gh
🔢 Solution
• 0² = (10)² − 2 × 10 × h
• 0 = 100 − 20h
• 20h = 100
• h = 100/20
• h = 5 m
In plain language: at the topmost point the object stops for an instant, and the whole of its kinetic energy has been converted into potential energy, which happens 5 m above the throwing point.
❌ Why the other options are incorrect
✗ 20 m — this comes from using h = u²/(2g) with g = 2.5, or from doubling the correct answer; it is not supported by the data.
✗ 100 m — this is simply u² = 100 written as a height, forgetting to divide by 2g.
✗ 1 m — this comes from dividing u by 2g (10/20 = 0.5) or similar mis-substitution; the velocity must be squared first.
📌 SSC & Railway Exam Facts
• Maximum height for vertical projection: h = u²/2g; time of ascent t = u/g.
• Here time of ascent = 10/10 = 1 s, and total time of flight = 2 s.
• The body returns to the thrower's hand with the same speed 10 m/s (ignoring air resistance).
• g = 9.8 m/s² on the Earth (often taken as 10), 1.63 m/s² on the Moon, and zero at the centre of the Earth.
• g is maximum at the poles and minimum at the equator, and it decreases with height as well as with depth.

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Q1. Acceleration due to gravity on Earth is approximately?

Q2. If a ball is thrown vertically upward, which of the following describes its velocity at the highest point?

Q3. What is the SI unit of weight?

Q4. Who experimentally determined the value of the Universal Gravitational Constant (G)?

Q5. Which of the following phenomena is explained by the Universal Law of Gravitation?

Q6. Which physical property remains the same on both Earth and Moon?

Q7. Due to which factor does a sheet of paper fall more slowly than a stone in air?

Q8. According to Newton's Universal Law of Gravitation, what symbol is used to denote the constant of proportionality?

Q9. A geostationary satellite remains stationary above a point on Earth's equator. Its orbital radius is approximately 42,200 km.Calculate the orbital period of the satellite in hours.

Q10. If the orbital period of a satellite is 90 minutes, what is the primary force keeping it in orbit?

Q11. At an altitude of 300 km, a satellite's orbital speed is 7.8 km/s.If the satellite's altitude increases by 200 km, what will happen to its orbital period?

Q12. If an astronaut's mass is 70 kg, what will be their weight on the Moon (1/6th of Earth's gravity)?

Q13. Why is the gravitational force experienced on the Moonless compared to Earth?

Q14. If the distance between two objects increases six times, what will be the effect on the gravitational force?

Q15. Free fall corresponds to which type of motion?

Q16. If the radius of the Earth were to double while mass remains the same, the new value of g would be ___________the original value.

Q17. The SI unit of the universal gravitation constant is ____________.

Q18. Which of the following is NOT explained by the Universal Law of Gravitation?

Q19. Which of the following statements correctly describes the difference between mass and weight?

Q20. An object's mass remains constant everywhere (on the Earth or the Moon), whereas its weight changes because ___________.

Q21. If the gravitational force experienced by a body of mass 'm' on the surface of Earth is 'X' N. If the mass of the body is tripled, what will be the value of the gravitational force experienced by the body?

Q22. The gravitational force between two masses increases when ________.

Q23. The motion of the moon around the earth requires a centripetal force, which is provided by _____________.

Q24. The weight of a body on the moon is 3 N. What will be the weight of the body on Earth?

Q25. A block of mass 20 kg is dropped from the top of a building of height 200 m. What will be the velocity with which the block will hit the ground? (Take the value of g = 10 m/s²)

Q26. An object is launched straight upward and reaches a maximum height of 1000 cm. What was the initial velocity of the object? (Take g = 9.8 m/s²)

Q27. If mass is doubled and g remains constant, weight becomes ________.

Q28. Which statement correctly describes the universal law of gravitation?

Q29. If a body at rest is dropped from a certain height and takes 4 seconds to reach the ground, what is the approximate height from which it was dropped? (Take g = 10 m/s²)

Q30. A ball initially at rest is dropped from a height and takes 5 seconds to reach the ground. What is the average speed during its fall? (g = 10 m/s²)

Q31. A ball at rest is dropped from the top of a 20 m high building. Calculate the speed with which it hits the ground. (Take g = 10 m/s²)

Q32. A student claims that the force exerted by the Earth on the Moon is much stronger than the force exerted by the Moon on the Earth. Based on Newton's third law and the Universal Law of Gravitation, is this claim correct?

Q33. A geo-stationary satellite completes one revolution around the Earth in approximately _____ hours.

Q34. Why do objects fall towards the Earth?

Q35. What is the formula to calculate the weight of an object?

Q36. The motion of the moon around the earth is due to the _____.

Q37. The universal law of gravitation states that the gravitational force between two objects is proportional to _____.

Q39. If a satellite in 700 km orbit completes 15 orbits/day, how many minutes per orbit?

Q40. If Mars' gravity is 3.7 m/s², what would be the weight of a 60 kg astronaut on Mars?

Q41. Which of the following quantities remains conserved when an object moves under the gravitational pull of another object?

Q42. According to the Law of Periods, the square of a planet’s time period is proportional to:

Q43. Which force governs the motion of planets, stars, and galaxies?

Q44. Generally, satellites do not fly below _____ km.

Q45. At 800 km, a satellite in Sun Synchronous Orbit will travel at a speed of approximately _____ km per second.

Q46. Which of the following statement state the importance of the universal law of gravitation?I. The motion of planets around the SunII. The tides due to the moon and the Sun

Q47. A communication satellite acts like a television tower placed at a height of _____ km.

Q48. Why can a geo-stationary satellite monitor the movement of clouds accurately?

Q49. An object weighs 18 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?

Q50. Lagrange points are also known as _____ points.

Q51. If a satellite is in the plane of the equator with 0 degrees inclination and at a height of about 36,000 km, then it is said to be in ________ orbit.

Q52. If a satellite is in the plane of the equator with 0 degrees inclination and at a height of about 36000 km then it is said to be in _____ orbit.

Q53. Weight of an object on earth is 441 N. What is the mass of the object?

Q54. What happens if a satellite orbits Earth with 24-hour period?

Q56. If Earth's radius were reduced by 1% while its mass remained unchanged, what would be the effect on the gravitational acceleration at its surface?

Q57. Which force acts without physical contact and governs celestial body motion in mechanics?

Q58. The value of universal gravitational constant is ________.

Q59. A body, starting from rest, falls from a building and reaches the ground in 2 secs. What will be the velocity of the body when it reaches the ground? (Acceleration due to gravity = 10 ms⁻² )

Q60. Which of the following phenomena CANNOT be explained using the universal law of gravitation?

Q61. What is the SI unit of universal gravitational constant G?

Q62. If the mass of a football is 3 kg, then what would be its weight on the earth? (Given, g= 9.8 m s⁻²)

Q63. A body of mass m and a body of mass 4m are dropped from the same height, the ratio of the time taken by the two bodies to reach the ground would be:(Air resistance is neglected)

Q64. Which of the following statements is true?

Q66. The value of Universal Gravitational constant is G on a certain planet of mass M, on another planet of mass 4M, the value of the Universal Gravitational constant will be:

Q67. Whenever an object falls toward the earth, acceleration is involved. Which of the following forces provides this acceleration?

Q68. When a body falls freely under gravity, then which of the following quantities DOES NOT remain constant?

Q69. When an object is thrown upward, the force of gravity is _____.

Q70. The value of acceleration due to gravity 'g' is ________.

Q71. If a stone is thrown up, the value of acceleration due to gravity 'g' will be taken as ________.

Q72. If the magnitude of the force exerted by the earth on the moon is 2.02 × 10²⁰ N, then the magnitude of the force exerted by the moon on the earth is ______.

Q73. What is the SI unit of mass and weight of a body, respectively?

Q74. What is the unit of universal gravitational constant (G)?

Q75. The planets go around the sun due to _______ force.

Q76. The weight of an object on the Moon is ____________

Q77. Which of the following equations is true, when two objects A and B of masses M and m are separated by a distance d, F is the force of attraction between them and G is the universal gravitational constant?

Q78. The acceleration due to gravity on an object of mass m placed near the surface of the earth is:

Q79. If u, g, and h are the initial velocity, gravitational acceleration, and height, respectively, of an object falling under gravitational force, then its final velocity v can be obtained from:

Q80. If an object on the surface of the moon weighs 3 N, the weight of the same object on surface of the Earth is ______.

Q81. The motion of a freely falling body is an example of a:

Q82. Which of the following constant values given in the options is equal to the universal gravitational constant 'G'?

Q83. The gravitational force between two objects:

Q9. A geostationary satellite remains stationary above a point on Earth's equator. Its orbital radius is approximately 42,200 km.
Calculate the orbital period of the satellite in hours.

Q11. At an altitude of 300 km, a satellite's orbital speed is 7.8 km/s.
If the satellite's altitude increases by 200 km, what will happen to its orbital period?

Q13. Why is the gravitational force experienced on the Moon
less compared to Earth?

Q46. Which of the following statement state the importance of the universal law of gravitation?
I. The motion of planets around the Sun
II. The tides due to the moon and the Sun

Q63. A body of mass m and a body of mass 4m are dropped from the same height, the ratio of the time taken by the two bodies to reach the ground would be:
(Air resistance is neglected)

Q86. Which of the following is/are NOT related to the phenomenon of the universal law of gravitation?
(A) The motion of the moon around the earth.
(B) The tides caused by the moon.
(C) The motion of electron around the nucleus.