NTA did not publish the JEE Main 2023 question paper, but its official answer key is hosted on this page as a PDF. For timed practice there is a clearly labelled mock paper with original questions in the reported 2023 pattern. The JEE Main exam covers Physics, Chemistry and Mathematics.
Explore: JEE Main hub · Syllabus & pattern · JEE Main 2025 · How to use PYQs
No official paper released
The conducting body did not publish the JEE Main 2023 question paper. Practise the JEE Main 2023 mock paper: original questions modelled on candidate reports, clearly labelled as a mock. The official answer key is hosted below.
Why this year matters
By 2023, you should already have an error log. Open this paper to pressure-test whether those tags still hold. 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 2023
Strong mid-recent Main for chapter-weight checks across sessions.
2023 is a mid-decade stress test. Difficulty spikes here usually come from topic mix and selection pressure, not from obsolete exam software.
What this paper stresses
- Chapters you skipped because short notes felt “enough”.
- 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.
What to extract before you move on
- One selection rule for what you attempt first in each subject.
- PCM chapters with repeated misses across sessions/years.
- Multi-correct hygiene mistakes vs pure concept gaps.
- Numerical answer re-key failures.
How 2023 sits next to other years
Compare 2023 with JEE Main 2025, JEE Main 2024, JEE Main 2022. Themes that repeat across this band deserve deliberate notes; one-off oddities do not.
Analyst note: The conducting body did not publish the JEE Main 2023 paper. The mock below is original practice in the reported pattern; treat coaching “memory-based” papers as unofficial reconstructions.
JEE Main 2023 Paper 1 mock paper (modelled on the January 2023 session)
This is a mock. NTA released JEE Main 2023 question papers only to candidates through their login; the public final key lists only question and option IDs and is hosted with this year. These original questions follow the 2023 pattern, which was unchanged from 2022.
What candidates reported about 2023
- Session 1 Paper 1 ran on 24, 25, 29, 30 and 31 January and 1 February 2023 in two shifts a day; Session 2 followed in April 2023.
- Pattern unchanged from 2022: per subject 20 MCQs plus 10 numerical questions, any 5 attempted (90 questions, 75 answered, 300 marks).
- Negative marking of −1 applied in both Section A and Section B.
- About 8.6 lakh candidates registered for Session 1 and about 9.3 lakh for Session 2.
Suggested time: 60 minutes. Each subject has 6 MCQs (Section A) and 3 integer-answer questions (Section B). In the real paper, candidates attempted any 5 of 10 Section B questions; here attempt all. Marking: +4 correct, −1 wrong in both sections, 0 if unattempted.
Physics
Section A (multiple choice)
Two forces of 3 N and 4 N act at a point at right angles to each other. The magnitude of their resultant is
- 7 N
- 1 N
- 5 N
- 12 N
Answer and explanation
Answer: C (5 N)
R = √(3² + 4²) = 5 N.
A small sphere falling through a viscous liquid reaches terminal velocity v. A sphere of the same material with twice the radius would reach a terminal velocity of
- 2v
- 4v
- v/2
- 8v
Answer and explanation
Answer: B (4v)
From Stokes' law, terminal velocity = 2r²(ρ − σ)g/(9η) ∝ r², so doubling r gives 4v.
The efficiency of a Carnot engine working between 527 °C and 127 °C is
- 50%
- 24%
- 76%
- 40%
Answer and explanation
Answer: A (50%)
In kelvin T₁ = 800 K and T₂ = 400 K, so η = 1 − T₂/T₁ = 1 − 0.5 = 50%.
A charged particle moves perpendicular to a uniform magnetic field in a circle. Its period of revolution
- is proportional to its speed
- is inversely proportional to its speed
- is proportional to the square of its speed
- does not depend on its speed
Answer and explanation
Answer: D (does not depend on its speed)
r = mv/(qB), so T = 2πr/v = 2πm/(qB), which is independent of v.
Unpolarised light of intensity I₀ passes through a polariser and then an analyser whose axis makes 60° with the polariser’s axis. The transmitted intensity is
- I₀/2
- I₀/4
- I₀/8
- 3I₀/8
Answer and explanation
Answer: C (I₀/8)
The polariser passes I₀/2. Malus's law at the analyser: (I₀/2)cos²60° = (I₀/2)(1/4) = I₀/8.
In an AC circuit containing only an ideal inductor, the current
- leads the voltage by 90°
- lags the voltage by 90°
- is in phase with the voltage
- lags the voltage by 45°
Answer and explanation
Answer: B (lags the voltage by 90°)
V = L dI/dt. For I = I₀ sin ωt, V = ωLI₀ cos ωt, so the voltage leads the current by π/2 (the current lags by 90°).
Section B (numerical value)
A stone is dropped from rest from a height of 80 m. Taking g = 10 m/s², the time (in s) it takes to reach the ground is ______. (numerical answer)
Answer and explanation
Answer: 4
h = ½gt²: 80 = 5t², so t² = 16 and t = 4 s.
A wire 2 m long with cross-sectional area 1 mm² and Young’s modulus 2 × 10¹¹ N/m² is loaded with 100 N. Its extension (in mm) is ______. (numerical answer)
Answer and explanation
Answer: 1
ΔL = FL/(AY) = (100 × 2)/(10⁻⁶ × 2 × 10¹¹) = 10⁻³ m = 1 mm.
Point charges +2 μC and +8 μC are 30 cm apart. The distance (in cm) from the +2 μC charge, on the line joining them, where the net field is zero is ______. (numerical answer)
Answer and explanation
Answer: 10
At distance x from the 2 μC charge: 2/x² = 8/(30 − x)², so (30 − x)/x = 2 and x = 10 cm.
Chemistry
Section A (multiple choice)
Which element has the most negative electron gain enthalpy?
- F
- Cl
- Br
- O
Answer and explanation
Answer: B (Cl)
Chlorine has the most negative value. Fluorine is less negative than expected because the small, crowded 2p subshell repels the incoming electron.
The hybridisation of the central carbon atom in allene, CH₂=C=CH₂, is
- sp
- sp²
- sp³
- dsp²
Answer and explanation
Answer: A (sp)
The central carbon forms two σ bonds and two π bonds (two double bonds), so it is sp hybridised.
The oxidation state of chromium in K₂Cr₂O₇ is
- +3
- +7
- +12
- +6
Answer and explanation
Answer: D (+6)
2(+1) + 2x + 7(−2) = 0, so 2x = 12 and x = +6.
Which of the following aldehydes undergoes aldol condensation?
- HCHO
- C₆H₅CHO
- CH₃CHO
- (CH₃)₃C–CHO
Answer and explanation
Answer: C (CH₃CHO)
Aldol condensation needs an α-hydrogen. Only acetaldehyde has one; HCHO, benzaldehyde and pivaldehyde have none (they undergo the Cannizzaro reaction instead).
Zinc reacts with hot concentrated aqueous NaOH to liberate
- oxygen
- hydrogen
- chlorine
- no gas
Answer and explanation
Answer: B (hydrogen)
Zn + 2NaOH → Na₂ZnO₂ + H₂↑. Zinc is amphoteric and forms sodium zincate.
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH < 0, the yield of ammonia at equilibrium is increased by
- increasing the pressure
- increasing the temperature
- adding a catalyst
- removing N₂
Answer and explanation
Answer: A (increasing the pressure)
Higher pressure favours the side with fewer gas moles (2 versus 4). Heating favours the endothermic reverse reaction, and a catalyst does not shift the equilibrium.
Section B (numerical value)
The number of moles of KMnO₄ needed to oxidise 5 mol of Fe²⁺ to Fe³⁺ in acidic medium is ______. (numerical answer)
Answer and explanation
Answer: 1
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. One MnO₄⁻ accepts 5 electrons, so 1 mol oxidises 5 mol Fe²⁺.
The spin-only magnetic moment of Mn²⁺ in BM, to the nearest integer, is ______. (numerical answer)
Answer and explanation
Answer: 6
Mn²⁺ is 3d⁵ with 5 unpaired electrons: μ = √(n(n + 2)) = √35 ≈ 5.92 BM ≈ 6.
The number of structurally isomeric alcohols with the formula C₄H₁₀O is ______. (numerical answer)
Answer and explanation
Answer: 4
Butan-1-ol, butan-2-ol, 2-methylpropan-1-ol and 2-methylpropan-2-ol: 4.
Mathematics
Section A (multiple choice)
If |a| = 3, |b| = 4 and |a × b| = 6, and the angle between a and b is acute, then a · b equals
- 6√3
- 6
- 12
- 6√2
Answer and explanation
Answer: A (6√3)
sin θ = 6/(3 × 4) = 1/2, so θ = 30°. Then a · b = 12 cos 30° = 6√3.
The value of sin⁻¹(1/2) + cos⁻¹(1/2) is
- π/3
- π
- 2π/3
- π/2
Answer and explanation
Answer: D (π/2)
sin⁻¹x + cos⁻¹x = π/2 for every x in [−1, 1]. Here it is π/6 + π/3 = π/2.
The number of onto functions from {1, 2, 3} to {a, b} is
- 8
- 2
- 6
- 9
Answer and explanation
Answer: C (6)
There are 2³ = 8 functions in all. Removing the 2 constant (non-onto) ones leaves 6.
The focus of the parabola y² = 12x is
- (0, 3)
- (3, 0)
- (12, 0)
- (6, 0)
Answer and explanation
Answer: B ((3, 0))
y² = 4ax with 4a = 12 gives a = 3, so the focus is (3, 0).
The median of the data 2, 8, 5, 11, 7, 4 is
- 6
- 5
- 7
- 6.5
Answer and explanation
Answer: A (6)
Sorted: 2, 4, 5, 7, 8, 11. For six values the median is the mean of the 3rd and 4th: (5 + 7)/2 = 6.
∫ eˣ (sin x + cos x) dx equals
- eˣ cos x + C
- eˣ (sin x − cos x) + C
- −eˣ sin x + C
- eˣ sin x + C
Answer and explanation
Answer: D (eˣ sin x + C)
∫ eˣ (f + f') dx = eˣ f + C. With f = sin x, f' = cos x, the answer is eˣ sin x + C.
Section B (numerical value)
The coefficient of x² in the expansion of (1 + x)⁶(1 − 2x) is ______. (numerical answer)
Answer and explanation
Answer: 3
Coefficient = C(6, 2) × 1 + C(6, 1) × (−2) = 15 − 12 = 3.
The determinant of the matrix with rows (1, 2, 3), (0, 4, 5), (0, 0, 6) is ______. (numerical answer)
Answer and explanation
Answer: 24
The matrix is upper triangular, so the determinant is the product of the diagonal entries: 1 × 4 × 6 = 24.
The number of diagonals of a decagon is ______. (numerical answer)
Answer and explanation
Answer: 35
n(n − 3)/2 = 10 × 7/2 = 35.
Reports this mock is modelled on
- NTA – JEE (Main) 2023 Session 1 final answer keys (exam dates and shifts)
- NTA – JEE (Main) 2022 Information Bulletin (pattern carried into 2023)
- NTA – Public Notice (17 Oct 2024): discontinuation of optional questions in Section B
- Wikipedia – Joint Entrance Examination – Main (year-wise sessions, shifts and events)
How to sit JEE Main 2023
- 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 2023’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 2022
- JEE Main 2021
- JEE Main 2020
- JEE Main 2019
- JEE Main 2018
- JEE Main 2017
- JEE Main 2016
FAQ
Where can I download the official JEE Main 2023 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 2023 paper analysis include?
A year snapshot, what the paper stresses, what to extract into your error log, and how 2023 compares with neighbouring years on this hub.
How should I practise the JEE Main 2023 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 2023 still useful if I finished newer papers?
Yes. Recent years sharpen timing; stack them after a same-day review of each attempt.
Are memory-based JEE Main 2023 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.