The JEE Main 2020 question paper was not officially published, so this page does not host it. Instead it has a clearly labelled mock paper with original questions modelled on the reported 2020 pattern, plus a timed attempt plan and an error-log routine. 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 2020 question paper. Practise the JEE Main 2020 mock paper: original questions modelled on candidate reports, clearly labelled as a mock.
Why this year matters
Older JEE Main papers like 2020 are for mining, not nostalgia. Keep current marking rules, and skip topics the latest brochure dropped. 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 2020
Disrupted calendar year. Keep current NTA norms.
2020 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
- Multi-correct or numerical items that need calm option hygiene.
- 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”.
What to extract before you move on
- PCM chapters with repeated misses across sessions/years.
- Multi-correct hygiene mistakes vs pure concept gaps.
- Numerical answer re-key failures.
- One selection rule for what you attempt first in each subject.
How 2020 sits next to other years
Use 2020 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 2020 paper. The mock below is original practice in the reported pattern; treat coaching “memory-based” papers as unofficial reconstructions.
JEE Main 2020 Paper 1 mock paper (modelled on the January 2020 pattern)
This is a mock. NTA displayed JEE Main 2020 question papers only inside candidate logins during the answer-key challenge; its public "question papers" page covered only 2019. These original questions follow the 2020 pattern, the first with numerical-value questions: 20 MCQs and 5 numerical questions per subject.
What candidates reported about 2020
- JEE Main 2020 introduced a new Paper 1 pattern of 75 questions: 25 per subject, of which 20 were MCQs and 5 needed a numerical answer.
- MCQs carried +4/−1. Numerical-value questions carried +4 with no negative marking.
- The January session was held on 7–9 January 2020 for B.E./B.Tech, in computer-based mode with two shifts a day.
- The second session, scheduled for April, was postponed because of the COVID-19 pandemic and held in September 2020.
- An impersonation case in Assam (a candidate used a proxy test-taker) led to several arrests that year.
Suggested time: 60 minutes. Each subject has 6 MCQs and 2 numerical-value questions (the real 2020 paper had 20 + 5 per subject, 3 hours). MCQs: +4 correct, −1 wrong. Numerical-value questions: +4 correct, no negative marking. Enter numerical answers as numbers.
Physics
Section A (multiple choice)
The dimensional formula of Planck’s constant is
- [M L² T⁻²]
- [M L T⁻¹]
- [M L² T⁻¹]
- [M L² T]
Answer and explanation
Answer: C ([M L² T⁻¹])
h = E/ν. Energy is [M L² T⁻²] and frequency is [T⁻¹], so h = [M L² T⁻¹] (the same as angular momentum).
A projectile is launched with speed 40 m/s at 30° to the horizontal. Its horizontal range is (g = 10 m/s²)
- 80 m
- 80√3 m
- 160 m
- 40√3 m
Answer and explanation
Answer: B (80√3 m)
R = u² sin 2θ / g = 1600 × sin 60° / 10 = 160 × (√3/2) = 80√3 m ≈ 138.6 m.
A 0.5 kg block attached to a spring of force constant 200 N/m oscillates on a smooth floor. The period of oscillation is
- π/10 s
- π/5 s
- 2π s
- π/20 s
Answer and explanation
Answer: A (π/10 s)
ω = √(k/m) = √(200/0.5) = 20 rad/s, so T = 2π/ω = 2π/20 = π/10 s.
The magnetic field at the centre of a circular loop of radius 10 cm carrying 5 A is (μ₀ = 4π × 10⁻⁷ T m/A)
- 2π × 10⁻⁵ T
- π × 10⁻⁶ T
- 2 × 10⁻⁵ T
- π × 10⁻⁵ T
Answer and explanation
Answer: D (π × 10⁻⁵ T)
B = μ₀I/(2R) = (4π × 10⁻⁷ × 5)/(2 × 0.1) = π × 10⁻⁵ T.
In Young’s double-slit experiment, λ = 600 nm, slit separation 0.2 mm and screen distance 1 m. The fringe width is
- 0.3 mm
- 1.2 mm
- 3 mm
- 6 mm
Answer and explanation
Answer: C (3 mm)
β = λD/d = (600 × 10⁻⁹ × 1)/(0.2 × 10⁻³) = 3 × 10⁻³ m = 3 mm.
A radioactive sample has a half-life of 5 days. The fraction of the sample that has decayed after 15 days is
- 1/8
- 7/8
- 3/4
- 1/3
Answer and explanation
Answer: B (7/8)
15 days = 3 half-lives, so the fraction remaining is 1/8 and the fraction decayed is 1 − 1/8 = 7/8.
Section B (numerical value)
A car starts from rest and accelerates uniformly to 72 km/h in 10 s. The distance (in m) it covers in this time is ______. (numerical answer)
Answer and explanation
Answer: 100
72 km/h = 20 m/s. Average speed = (0 + 20)/2 = 10 m/s, so distance = 10 × 10 = 100 m.
Resistors of 4 Ω and 12 Ω are connected in parallel across an ideal 9 V battery. The total power (in W) dissipated is ______. (numerical answer)
Answer and explanation
Answer: 27
Equivalent resistance = 4 × 12/16 = 3 Ω. P = V²/R = 81/3 = 27 W.
Chemistry
Section A (multiple choice)
The number of moles in 11.2 L of O₂ gas at STP (molar volume 22.4 L) is
- 1
- 0.25
- 0.5
- 2
Answer and explanation
Answer: C (0.5)
n = 11.2/22.4 = 0.5 mol.
The bond order of the superoxide ion O₂⁻ is
- 2
- 1.5
- 2.5
- 1
Answer and explanation
Answer: B (1.5)
O₂⁻ has 17 electrons. Bonding electrons = 10 and antibonding = 7 (MO theory), so bond order = (10 − 7)/2 = 1.5.
The pH of a 0.01 M NaOH solution at 25 °C is
- 12
- 2
- 10
- 13
Answer and explanation
Answer: A (12)
[OH⁻] = 10⁻² M, so pOH = 2 and pH = 14 − 2 = 12.
Which of the following species is aromatic?
- Cyclopentadienyl cation
- Cycloheptatrienyl anion
- Cyclooctatetraene (planar form assumed)
- Cyclopentadienyl anion
Answer and explanation
Answer: D (Cyclopentadienyl anion)
The cyclopentadienyl anion is cyclic, planar and fully conjugated with 6 π electrons (4n + 2, n = 1). The cation has 4, the cycloheptatrienyl anion 8 and cyclooctatetraene 8 π electrons.
A consequence of the lanthanoid contraction is that
- La has a smaller radius than Lu
- all lanthanoids show only the +2 state
- Zr and Hf have almost identical atomic radii
- Hf is much larger than Zr
Answer and explanation
Answer: C (Zr and Hf have almost identical atomic radii)
The poor shielding by 4f electrons makes the 5d series atoms contract, so Hf (after the lanthanoids) ends up nearly the same size as Zr above it.
The major product of the addition of HBr to propene in the absence of peroxide is
- 1-bromopropane
- 2-bromopropane
- 1,2-dibromopropane
- propan-2-ol
Answer and explanation
Answer: B (2-bromopropane)
Markovnikov addition proceeds through the more stable secondary carbocation CH₃–CH⁺–CH₃, so Br attaches to C-2.
Section B (numerical value)
The mass (in g) of NaOH (molar mass 40 g/mol) needed to prepare 500 mL of a 0.2 M solution is ______. (numerical answer)
Answer and explanation
Answer: 4
Moles = 0.2 × 0.5 = 0.1 mol. Mass = 0.1 × 40 = 4 g.
The number of unpaired electrons in a free Fe³⁺ ion is ______. (numerical answer)
Answer and explanation
Answer: 5
Fe (Z = 26) is [Ar]3d⁶4s². Fe³⁺ is [Ar]3d⁵, and by Hund’s rule all five d electrons are unpaired.
Mathematics
Section A (multiple choice)
If the vectors a = i + 2j + k and b = 2i + λj − 4k are perpendicular, then λ equals
- 2
- −1
- 1
- 3
Answer and explanation
Answer: C (1)
a · b = 2 + 2λ − 4 = 0 gives λ = 1.
The term independent of x in the expansion of (x + 1/x)¹⁰ is
- 210
- 252
- 120
- 45
Answer and explanation
Answer: B (252)
General term C(10, r) x^(10 − 2r). It is independent of x when r = 5, giving C(10, 5) = 252.
The equation of the tangent to the curve y = x³ at the point (1, 1) is
- y = 3x − 2
- y = 3x + 2
- y = x
- y = 2x − 1
Answer and explanation
Answer: A (y = 3x − 2)
dy/dx = 3x² = 3 at x = 1. Tangent: y − 1 = 3(x − 1), i.e. y = 3x − 2.
The sum of the infinite geometric series 8 + 4 + 2 + 1 + … is
- 15
- 32
- 12
- 16
Answer and explanation
Answer: D (16)
a = 8, r = 1/2, so S = a/(1 − r) = 8/(1/2) = 16.
The function f(x) = x² + 2x is strictly increasing on
- (−∞, −1)
- (0, ∞) only
- (−1, ∞)
- (−∞, ∞)
Answer and explanation
Answer: C ((−1, ∞))
f'(x) = 2x + 2 > 0 exactly when x > −1.
The eccentricity of the ellipse x²/25 + y²/16 = 1 is
- 4/5
- 3/5
- 5/3
- 9/25
Answer and explanation
Answer: B (3/5)
e = √(1 − b²/a²) = √(1 − 16/25) = 3/5.
Section B (numerical value)
The value of ∫₀² (3x² + 1) dx is ______. (numerical answer)
Answer and explanation
Answer: 10
∫ (3x² + 1) dx = x³ + x. From 0 to 2: 8 + 2 = 10.
The number of three-digit numbers with distinct digits that can be formed using the digits 1, 2, 3, 4, 5 is ______. (numerical answer)
Answer and explanation
Answer: 60
Choices: 5 for the hundreds place, 4 for tens, 3 for units: 5 × 4 × 3 = 60.
Reports this mock is modelled on
How to sit JEE Main 2020
- 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 2020’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 2021
- JEE Main 2019
- JEE Main 2018
- JEE Main 2017
- JEE Main 2016
FAQ
Where can I download the official JEE Main 2020 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 2020 paper analysis include?
A year snapshot, what the paper stresses, what to extract into your error log, and how 2020 compares with neighbouring years on this hub.
How should I practise the JEE Main 2020 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 2020 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 2020 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.