NTA did not publish the JEE Main 2021 question paper, but its 2 official answer keys are hosted on this page as PDFs. For timed practice there is a clearly labelled mock paper with original questions in the reported 2021 pattern. The JEE Main exam covers Physics, Chemistry and Mathematics.
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No official paper released
The conducting body did not publish the JEE Main 2021 question paper. Practise the JEE Main 2021 mock paper: original questions modelled on candidate reports, clearly labelled as a mock. The official answer key is hosted below.
Why this year matters
2021 will not mirror every new question type, but it still trains selection, stamina, and the chapters that keep coming back. Prefer official NTA / JEE Advanced releases. Session papers differ; note which session you sat.
Write down one selection rule you will reuse on exam day: what you attempt first, and what you abandon after a fixed time box.
Paper analysis
Snapshot of JEE Main 2021
Multi-session era firmly in place. Practise calm normalisation thinking.
2021 may feel differently paced than the newest cycle. Borrow stamina and theme lessons; do not copy obsolete section orders or marking quirks.
What this paper stresses
- Session-to-session difficulty swings (Main) or Paper 1 vs Paper 2 fatigue (Advanced).
- Silly PCM arithmetic after a correct setup.
- Chapters you skipped because short notes felt “enough”.
- Multi-correct or numerical items that need calm option hygiene.
What to extract before you move on
- Multi-correct hygiene mistakes vs pure concept gaps.
- Numerical answer re-key failures.
- One selection rule for what you attempt first in each subject.
- PCM chapters with repeated misses across sessions/years.
How 2021 sits next to other years
Use 2021 only after JEE Main 2025, JEE Main 2024, JEE Main 2023 are logged. Archive years are for frequency maps, not for predicting tomorrow’s cut-off.
Analyst note: The conducting body did not publish the JEE Main 2021 paper. The mock below is original practice in the reported pattern; treat coaching “memory-based” papers as unofficial reconstructions.
JEE Main 2021 Paper 1 mock paper (modelled on the four-session 2021 pattern)
This is a mock. In 2021 NTA displayed question papers only inside candidate logins; publicly it released only the final answer keys, which list question IDs, not questions. Those keys are hosted with this year. These original questions follow the 2021 pattern of Section A MCQs plus Section B numerical questions with internal choice.
What candidates reported about 2021
- JEE Main 2021 was held in four sessions (February, March, July and August–September); 2021 was the only year with up to four attempts.
- Each subject had 30 questions: 20 MCQs in Section A and 10 numerical-value questions in Section B, of which any 5 were to be attempted (90 questions, 75 to answer).
- Section A carried +4/−1. Section B had no negative marking in 2021.
- The February session ran on 24–26 February 2021 and the March session on 16–18 March 2021, two shifts a day.
- Sessions 3 and 4 were postponed during the second COVID-19 wave and finally held from 20 July 2021 onwards.
Suggested time: 60 minutes. Each subject has 6 MCQs (Section A) and 3 numerical-value questions (Section B). In the real paper, candidates attempted any 5 of 10 Section B questions; here attempt all. Section A: +4 correct, −1 wrong. Section B: +4 correct, no negative marking (2021 rule). Section B answers are numbers.
Physics
Section A (multiple choice)
A planet has twice the radius of the Earth and the same average density. If the escape speed from the Earth is 11.2 km/s, the escape speed from the planet is
- 11.2 km/s
- 15.8 km/s
- 22.4 km/s
- 44.8 km/s
Answer and explanation
Answer: C (22.4 km/s)
vₑ = √(2GM/R) and M = (4/3)πR³ρ, so vₑ ∝ R√ρ. With the same density and double the radius, vₑ doubles to 22.4 km/s.
The ratio of the rms speed of H₂ molecules to that of O₂ molecules at the same temperature is
- 1 : 4
- 4 : 1
- 16 : 1
- 2 : 1
Answer and explanation
Answer: B (4 : 1)
v_rms = √(3RT/M), so v(H₂)/v(O₂) = √(32/2) = √16 = 4.
The energy stored in an inductor of 0.2 H carrying a current of 3 A is
- 0.9 J
- 0.6 J
- 1.8 J
- 0.3 J
Answer and explanation
Answer: A (0.9 J)
U = ½LI² = ½ × 0.2 × 9 = 0.9 J.
In the Bohr model of hydrogen, the ratio of the radius of the third orbit to that of the first orbit is
- 3 : 1
- 1 : 9
- 27 : 1
- 9 : 1
Answer and explanation
Answer: D (9 : 1)
rₙ ∝ n², so r₃/r₁ = 9.
A logic gate gives output 0 only when both its inputs are 1, and output 1 otherwise. The gate is
- AND
- NOR
- NAND
- XOR
Answer and explanation
Answer: C (NAND)
This is the complement of AND: AND gives 1 only for (1, 1), so NAND gives 0 only for (1, 1).
A wire of resistance R is stretched uniformly until its length doubles. Its new resistance is
- 2R
- 4R
- R/2
- 8R
Answer and explanation
Answer: B (4R)
Volume stays constant, so doubling the length halves the area. R = ρL/A becomes ρ(2L)/(A/2) = 4R.
Section B (numerical value)
A 2 kg body moving at 4 m/s collides with a 6 kg body at rest and they stick together. The loss of kinetic energy (in J) is ______. (numerical answer)
Answer and explanation
Answer: 12
Momentum: 2 × 4 = 8 v, so v = 1 m/s. KE before = ½ × 2 × 16 = 16 J. KE after = ½ × 8 × 1 = 4 J. Loss = 12 J.
The wavelength (in nm) of a photon of energy 2.48 eV is ______. (Take hc = 1240 eV nm.) (numerical answer)
Answer and explanation
Answer: 500
λ = hc/E = 1240/2.48 = 500 nm.
A parallel-plate capacitor with air between its plates has capacitance 5 pF. When the gap is completely filled with a dielectric of constant 4, its capacitance (in pF) becomes ______. (numerical answer)
Answer and explanation
Answer: 20
C = Kε₀A/d, so C becomes 4 × 5 = 20 pF.
Chemistry
Section A (multiple choice)
Which of the following contains the largest number of molecules?
- 1 g of O₂
- 1 g of H₂
- 1 g of N₂
- 1 g of CO₂
Answer and explanation
Answer: B (1 g of H₂)
Number of molecules ∝ moles = mass/molar mass. With equal masses the smallest molar mass wins: H₂ (2 g/mol) gives 0.5 mol, far more than the others.
The shape of the ClF₃ molecule is
- T-shaped
- trigonal planar
- trigonal pyramidal
- see-saw
Answer and explanation
Answer: A (T-shaped)
Cl has 3 bond pairs and 2 lone pairs (5 domains, sp³d). The lone pairs take equatorial positions, leaving a T-shaped molecule.
The solubility product of AgCl is 1.8 × 10⁻¹⁰ at 298 K. Its solubility in pure water is about
- 1.8 × 10⁻¹⁰ mol/L
- 1.34 × 10⁻¹⁰ mol/L
- 9 × 10⁻⁶ mol/L
- 1.34 × 10⁻⁵ mol/L
Answer and explanation
Answer: D (1.34 × 10⁻⁵ mol/L)
For AgCl ⇌ Ag⁺ + Cl⁻, Ksp = s², so s = √(1.8 × 10⁻¹⁰) ≈ 1.34 × 10⁻⁵ mol/L.
Acetamide treated with Br₂ and aqueous NaOH (Hofmann bromamide degradation) gives
- ethylamine
- acetic acid
- methylamine
- acetonitrile
Answer and explanation
Answer: C (methylamine)
Hofmann degradation converts RCONH₂ to RNH₂ with one carbon fewer: CH₃CONH₂ → CH₃NH₂.
Which of the following oxides is amphoteric?
- Na₂O
- Al₂O₃
- SO₃
- CaO
Answer and explanation
Answer: B (Al₂O₃)
Al₂O₃ dissolves in both acids and bases (forming Al³⁺ salts and aluminates). Na₂O and CaO are basic and SO₃ is acidic.
Glucose on oxidation with bromine water gives
- gluconic acid
- saccharic acid
- sorbitol
- n-hexane
Answer and explanation
Answer: A (gluconic acid)
Bromine water is a mild oxidant that converts only the –CHO group to –COOH, giving gluconic acid. HNO₃ would also oxidise the primary –CH₂OH (saccharic acid).
Section B (numerical value)
The mass (in g) of glucose (molar mass 180 g/mol) that must be dissolved in 500 g of water to make a 0.1 molal solution is ______. (numerical answer)
Answer and explanation
Answer: 9
Molality = moles/kg of solvent, so moles = 0.1 × 0.5 = 0.05 mol, i.e. 0.05 × 180 = 9 g.
The number of σ bonds in one molecule of ethene (C₂H₄) is ______. (numerical answer)
Answer and explanation
Answer: 5
Four C–H σ bonds plus one σ component of the C=C double bond: 5 σ bonds (and 1 π bond).
A first-order reaction has rate constant 0.0693 min⁻¹. Its half-life (in min) is ______. (numerical answer)
Answer and explanation
Answer: 10
t½ = 0.693/k = 0.693/0.0693 = 10 min.
Mathematics
Section A (multiple choice)
If z = (1 + i)/(1 − i), then z⁴ equals
- 1
- −1
- i
- −i
Answer and explanation
Answer: A (1)
z = (1 + i)²/((1 − i)(1 + i)) = 2i/2 = i, so z⁴ = i⁴ = 1.
The variance of the first 10 natural numbers is
- 5.5
- 9.17
- 30.25
- 8.25
Answer and explanation
Answer: D (8.25)
For 1, 2, …, n the variance is (n² − 1)/12 = 99/12 = 8.25.
The value of lim (x→∞) (1 + 2/x)^x is
- e
- 2e
- e²
- 1
Answer and explanation
Answer: C (e²)
lim (1 + a/x)^x = eᵃ, here with a = 2.
If A is a 3 × 3 matrix with det A = 4, then det(2A) equals
- 8
- 32
- 16
- 64
Answer and explanation
Answer: B (32)
det(kA) = kⁿ det A for an n × n matrix: 2³ × 4 = 32.
The slope of the normal to the curve y = x² at x = 1 is
- −1/2
- 2
- 1/2
- −2
Answer and explanation
Answer: A (−1/2)
The tangent slope is 2x = 2, so the normal slope is −1/2.
The statement p → q is logically equivalent to
- p ∨ ~q
- ~p ∧ q
- q → p
- ~p ∨ q
Answer and explanation
Answer: D (~p ∨ q)
p → q is false only when p is true and q is false, which matches ~p ∨ q. (q → p is the converse, not equivalent.)
Section B (numerical value)
A committee of 3 is to be chosen from 6 men and 4 women so that it has exactly 2 women. The number of ways is ______. (numerical answer)
Answer and explanation
Answer: 36
Choose 2 women in C(4, 2) = 6 ways and 1 man in C(6, 1) = 6 ways: 6 × 6 = 36.
The sum of the first 20 terms of the arithmetic progression 3, 7, 11, … is ______. (numerical answer)
Answer and explanation
Answer: 820
a = 3, d = 4: S₂₀ = (20/2)[2 × 3 + 19 × 4] = 10 × 82 = 820.
The value of ∫₀¹ 6x(1 − x) dx is ______. (numerical answer)
Answer and explanation
Answer: 1
∫₀¹ (6x − 6x²) dx = 3 − 2 = 1.
Reports this mock is modelled on
How to sit JEE Main 2021
- Work through the mock paper above under exam timing. It follows the reported pattern, but the questions are our own.
- Sit the paper under current brochure rules (timers, negatives, NAT/MSQ norms), even if 2021’s interface looked different.
- When the clock ends, stop. Review should take at least as long as the attempt.
- Fill the error log the same day: section, topic, and miss type (concept, silly, time, or misread).
- Next session, repair only logged items. Then pick another year from the decade list.
Other years in this decade
- JEE Main 2025
- JEE Main 2024
- JEE Main 2023
- JEE Main 2022
- JEE Main 2020
- JEE Main 2019
- JEE Main 2018
- JEE Main 2017
- JEE Main 2016
FAQ
Where can I download the official JEE Main 2021 question paper?
The conducting body did not release it publicly. This page has a clearly labelled mock paper modelled on candidate reports instead.
What does the JEE Main 2021 paper analysis include?
A year snapshot, what the paper stresses, what to extract into your error log, and how 2021 compares with neighbouring years on this hub.
How should I practise the JEE Main 2021 previous year paper?
Open the official PDF, sit it timed under current brochure rules, fill an error log the same day, read the paper analysis, then move to another year in the decade list.
Is JEE Main 2021 still useful if I finished newer papers?
Yes, for theme and weightage mining after your newest papers are logged. Skip deleted syllabus pockets.
Are memory-based JEE Main 2021 papers reliable?
Treat memory-based sets as interim. Reconcile contested items with official keys or response sheets when the conducting body publishes them.
Where is the JEE Main syllabus and pattern for this paper?
See the JEE Main syllabus & pattern page on this hub, and always re-check the latest official brochure before exam day.