JEE (Advanced) · Physics, Class 11
Gravitation: JEE Advanced PYQ, 2007–2026
Gravitation (Physics, Class 11) has been asked 21 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.5 questions a paper, 3% of all Physics questions. It is the 15th most-asked of 19 Physics chapters. The two 2026 papers had 1 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 0in 2 papers |
| 2008 | 2in 2 papers |
| 2009 | 0in 2 papers |
| 2010 | 3in 2 papers |
| 2011 | 1in 2 papers |
| 2012 | 1in 2 papers |
| 2013 | 1in 2 papers |
| 2014 | 1in 2 papers |
| 2015 | 2in 2 papers |
| 2016 | 0in 2 papers |
| 2017 | 1in 2 papers |
| 2018 | 1in 2 papers |
| 2019 | 1in 2 papers |
| 2020 | 0in 2 papers |
| 2021 | 1in 2 papers |
| 2022 | 1in 2 papers |
| 2023 | 1in 2 papers |
| 2024 | 1in 2 papers |
| 2025 | 2in 2 papers |
| 2026 | 1in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Orbital motion9
- Escape velocity7
- Acceleration due to gravity2
- Gravitational potential energy1
- Kepler's laws1
- Newton's law of gravitation1
Try it
10 Gravitation questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Gravitation
Three particles, each of mass m, are placed at the corners of an equilateral triangle of side a. What is the gravitational potential energy of the system?
Answer: (A)
There are three pairs, each a apart, each contributing −Gm²/a. The total is −3Gm²/a.
Q2 · Physics · Gravitation
Using g = 9.8 m s⁻², the radius of the Earth R = 6.4 × 10⁶ m and G = 6.67 × 10⁻¹¹ N m² kg⁻², the mean density of the Earth, taken as a uniform sphere, is closest to:
Answer: (A)
g = GM/R² gives M = gR²/G. The volume is V = (4/3)πR³, so ρ = M/V = 3g/(4πGR). Substituting, ρ = (3 × 9.8)/(4π × 6.67×10⁻¹¹ × 6.4×10⁶) = 29.4/(5.36×10⁻³) ≈ 5.5×10³ kg m⁻³.
Q3 · Physics · Gravitation
A planet of period T moves round a star in an elliptical orbit of eccentricity e. The star is at the focus S nearer to the perihelion P. B is an end of the minor axis. Using Kepler's law of areas, what is the time the planet takes to travel from B to P along the arc BP (the shorter arc, which does not pass through the aphelion)?
Answer: (C)
By the law of areas, time is proportional to area swept at the Sun, so t/T = (area SPB)/(πab). Let O be the centre, so OS = ae, OP = a and OB = b. The quarter-ellipse region OPB has area πab/4. It consists of the triangle OSB, of area ½·ae·b, plus the region SPB. So area SPB = πab/4 − abe/2. Then t = T(πab/4 − abe/2)/(πab) = T(1/4 − e/2π).
Q4 · Physics · Gravitation
A geostationary satellite stays above the same point on Earth's surface. Which statement about its orbit is correct?
Answer: (A)
A geostationary satellite must have the same period as Earth's rotation, about 24 h. Only a circular orbit in the equatorial plane, moving west to east, keeps it fixed over one point. Its radius follows from T = 2π√(r³/GM), giving roughly 42,000 km from Earth's centre.
Q5 · Physics · Gravitation
A planet's mass stays the same, but the planet contracts uniformly so that its radius decreases by 2%. The acceleration due to gravity at its surface will:
Answer: (A)
With M fixed, g = GM/R² ∝ R⁻². For a small fractional change, Δg/g ≈ −2 ΔR/R = −2(−2%) = +4%. The exact value, 1/0.98² ≈ 1.041, agrees.
Q6 · Physics · Gravitation
A pendulum clock keeps correct time at the surface of the Earth. It is taken down a mine shaft to a depth of 3.2 km. Treat the Earth as a uniform sphere of radius 6400 km, and neglect temperature effects and the Earth's rotation. The time the clock loses in one day (86400 s) is closest to:
Answer: (B)
Below the surface of a uniform Earth, g' = g(1 - d/R). With d/R = 3.2/6400 = 5 × 10^-4, g falls by 0.05%. The period of a simple pendulum is T = 2π√(L/g), so T is proportional to g^(-1/2) and ΔT/T ≈ (1/2)(Δg/g) = 2.5 × 10^-4. The clock runs slow, and in 86400 s it loses about 86400 × 2.5 × 10^-4 = 21.6 s. The two common slips are skipping the factor 1/2 and using the height formula 2h/R for a depth.
Q7 · Physics · Gravitation
Treat the Earth as a sphere of radius 6.4 × 10⁶ m rotating with angular speed 7.3 × 10⁻⁵ rad s⁻¹ about its axis. Compared with its value at a pole, the acceleration due to gravity at latitude 60° is smaller by approximately:
Answer: (D)
At latitude λ a body moves in a circle of radius R cos λ. The centripetal requirement is ω²R cos λ, and its component along the line to the Earth's centre is ω²R cos²λ. So g' = g − Rω² cos²λ, and the pole (λ = 90°) has no reduction. Rω² = 6.4 × 10⁶ × (7.3 × 10⁻⁵)² ≈ 3.4 × 10⁻² m s⁻². At 60°, cos²60° = 1/4, so the reduction is ≈ 8.5 × 10⁻³ m s⁻².
Q8 · Physics · Gravitation
Suppose the force of attraction between a star and a planet of mass m were F = k/r^(5/2) instead of an inverse-square law, where k is a constant. For planets in circular orbits of radius r about the star, the period T is proportional to r^n. What is n?
Answer: (B)
The gravitational-type force supplies the centripetal force: mω²r = k/r^(5/2). So ω² = k/(m r^(7/2)), and ω ∝ r^(-7/4). Then T = 2π/ω ∝ r^(7/4). The inverse-square case gives the familiar T² ∝ r³ by the same steps.
Q9 · Physics · Gravitation
Observations show that the planets move in nearly circular orbits of radius r with period T, where T² is proportional to r³. Using only this and the centripetal force needed for circular motion, the gravitational force on a planet of mass m due to the Sun is proportional to:
Answer: (C)
For a circular orbit, the centripetal force is F = mω²r = m(2π/T)²r = 4π²mr/T². Kepler's third law gives T² = k r³, so F = 4π²m r/(k r³) = (4π²/k)·m/r². The force therefore varies as m/r², the inverse-square dependence on distance, and is proportional to the planet's mass.
Q10 · Physics · Gravitation
Satellite A of mass m moves in a circular orbit of radius r about the Earth. Satellite B of mass 3m moves in a circular orbit of radius 4r about the Earth. Taking the Earth as a point mass, what is the ratio L_A/L_B of their orbital angular momenta about the Earth's centre?
Answer: (C)
Orbital angular momentum for a circular orbit is L = mvr. Gravity supplies the centripetal force, so v = √(GM/r). Hence L = m√(GMr) ∝ m√r. For A, L ∝ m√r. For B, L ∝ 3m√(4r) = 6m√r. So L_A/L_B = 1/6. Do not take v as the same for both orbits, because speed decreases as the radius increases.
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All 21 real JEE Advanced questions from Gravitation.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Gravitation in JEE Advanced.
How many questions come from Gravitation in JEE Advanced?
Gravitation (Physics, Class 11) has been asked 21 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.5 questions a paper, 3% of all Physics questions. It is the 15th most-asked of 19 Physics chapters. The two 2026 papers had 1 from it.
Which topics of Gravitation are asked most in JEE Advanced?
Orbital motion (9), Escape velocity (7), Acceleration due to gravity (2), Gravitational potential energy (1), counted over the papers from 2007–2026.
Is Gravitation important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 15 of 19 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Gravitation JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from Gravitation with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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