Free JEE Main mock test — 75 questions, NTA pattern
One JEE Main paper in the real exam format: 75 questions across 3 timed sections, 3 hr on a server-controlled clock.
About the real exam India's national gateway to NITs, IIITs and GFTIs — and the only route into JEE Advanced for the IITs.
Free, reversible until you begin · one attempt per account · no retake, no card.
The paper, section by section
Sample questions with worked answers
8 original practice questions written by Myndaq in the JEE Main style. They are not taken from this mock paper or from official past papers. Try each one, then open the answer.
- Question 1PhysicsType a numeric answer
A polished metal plate of work function 5.0 eV is illuminated by monochromatic radiation of wavelength 155 nm. The stopping potential for the fastest photoelectrons, in volt, is ______. (Take hc = 1240 eV nm.)
Show answer and explanation
Answer: 3
Photon energy E = hc/λ = 1240 eV nm / 155 nm = 8.0 eV.
Einstein's photoelectric equation gives the maximum kinetic energy: K(max) = E - φ = 8.0 - 5.0 = 3.0 eV.
The stopping potential V(0) satisfies eV(0) = K(max), so with K(max) expressed in electronvolt the numerical value of V(0) in volt equals 3.0. Hence the stopping potential is 3 V.
Note that 8 V would be the answer only for a metal of zero work function, and 5 V is simply the work function misread as a potential.
- Question 2PhysicsOne correct answer
A square conducting loop of side 20 cm and total resistance 2 Ω lies flat in a region of uniform magnetic field directed perpendicular to its plane. The field strength increases steadily from 0.10 T to 0.50 T in 0.8 s. The induced current in the loop during this interval is
- 5 mA
- 10 mA
- 20 mA
- 40 mA
Show answer and explanation
Answer: B
Area of the loop A = (0.20 m)² = 0.04 m². Since the field is perpendicular to the plane, flux Φ = BA.
Rate of change of field = (0.50 - 0.10) T / 0.8 s = 0.50 T per second.
Induced emf magnitude = A × dB/dt = 0.04 × 0.50 = 0.02 V.
Induced current = emf / R = 0.02 / 2 = 0.01 A = 10 mA.
The distractor 20 mA comes from quoting the induced emf 0.02 V as though it were the current, i.e. forgetting to divide by the 2 Ω resistance. The value 40 mA comes from multiplying the emf by the resistance instead of dividing, 0.02 × 2 = 0.04 A. The value 5 mA corresponds to a rate of change of only 0.25 T per second, as if the 0.40 T change were spread over 1.6 s instead of 0.8 s.
- Question 3ChemistryOne correct answer
50 mL of 0.40 M solution of a weak monoprotic acid HA (pKa = 4.80) is mixed with 50 mL of 0.10 M sodium hydroxide. The pH of the resulting solution is closest to (log 3 = 0.48)
- 4.32
- 4.80
- 5.28
- 4.20
Show answer and explanation
Answer: A
Moles taken: HA = 50 × 0.40 = 20 mmol; hydroxide = 50 × 0.10 = 5 mmol. The strong base is consumed completely.
After reaction: HA left = 20 - 5 = 15 mmol, and A- formed = 5 mmol. Both are in the same 100 mL, so the volume cancels in the ratio.
Henderson-Hasselbalch: pH = pKa + log([A-]/[HA]) = 4.80 + log(5/15) = 4.80 - log 3 = 4.80 - 0.48 = 4.32.
The value 4.80 would apply only at exact half neutralisation, which needs 10 mmol of base; 5.28 comes from inverting the ratio and adding log 3; 4.20 comes from putting the original 20 mmol of acid in the denominator instead of the 15 mmol that survive, 4.80 + log(5/20) = 4.20.
- Question 4ChemistryOne correct answer
An organic compound X has the molecular formula C4H8O. It gives a positive iodoform test but does not reduce Tollens' reagent. The compound X is
- butan-2-one
- butanal
- oxolane (tetrahydrofuran)
- but-3-en-1-ol
Show answer and explanation
Answer: A
The iodoform test is positive for a methyl ketone (a CH3-CO- group) or for an alcohol bearing a CH3-CH(OH)- group, which is oxidised to that ketone by the reagent. Failure to reduce Tollens' reagent rules out an aldehyde.
Butan-2-one, CH3-CO-CH2-CH3, has the required methyl ketone group, is C4H8O, and being a ketone it does not reduce Tollens' reagent. It fits both clues.
Butanal is C4H8O but is an aldehyde, so it would give a silver mirror with Tollens' reagent. Oxolane is a cyclic ether with no carbonyl and no CH3-CH(OH)- unit, so it fails the iodoform test. But-3-en-1-ol, CH2=CH-CH2-CH2-OH, is also C4H8O but its hydroxyl carbon carries no methyl group, so it too gives no iodoform.
- Question 5ChemistryType a numeric answer
Among the species O2^-, N2, NO, CN^-, B2 and F2, the number of species that are paramagnetic in their ground state is ______.
Show answer and explanation
Answer: 3
Paramagnetism requires at least one unpaired electron in the molecular orbital configuration.
O2^- (superoxide) has 17 electrons in all, one more than O2; the extra electron pairs up in one antibonding pi orbital, leaving a single unpaired electron, so it is paramagnetic.
NO has 15 electrons, an odd number, so one electron must remain unpaired in the antibonding pi level: paramagnetic.
B2 has 10 electrons; because the s-p mixing places the bonding pi orbitals below the sigma orbital, the last two electrons occupy the two degenerate pi orbitals singly with parallel spins: paramagnetic.
N2 (14 electrons) and CN^-, which is isoelectronic with N2, have every orbital doubly filled, and F2 likewise fills all bonding and antibonding pi orbitals completely; all three are diamagnetic.
Count of paramagnetic species = 3.
- Question 6MathematicsOne correct answer
A drawer holds 6 blue pens and 3 black pens that are alike except in colour. Three pens are taken out together at random. The probability that exactly two of them are blue is
- 15/28
- 3/14
- 5/21
- 5/28
Show answer and explanation
Answer: A
Total number of ways of choosing 3 pens from 9 is C(9,3) = 84.
Favourable ways: choose 2 of the 6 blue pens and 1 of the 3 black pens, that is C(6,2) × C(3,1) = 15 × 3 = 45.
Probability = 45/84 = 15/28.
The distractor 3/14 equals 18/84, which is the probability of exactly two black pens, C(3,2) × C(6,1) = 18. The value 5/21 equals 20/84, the probability that all three are blue, C(6,3) = 20. The value 5/28 comes from computing the sequential probability of the single order blue, blue, black, (6/9)(5/8)(3/7) = 5/28, and forgetting that the black pen can be drawn in any of the three positions; multiplying by 3 restores 15/28.
- Question 7MathematicsType a numeric answer
The absolute value of the coefficient of x^5 in the expansion of (2x² - 1/x)^7 is ______.
Show answer and explanation
Answer: 560
The general term is T(r+1) = C(7,r) (2x²)^(7-r) (-1/x)^r = C(7,r) 2^(7-r) (-1)^r x^(14-2r-r) = C(7,r) 2^(7-r) (-1)^r x^(14-3r).
For the power of x to be 5, set 14 - 3r = 5, giving r = 3, which is a valid index since 0 ≤ 3 ≤ 7.
Coefficient = C(7,3) × 2^4 × (-1)³ = 35 × 16 × (-1) = -560.
The absolute value asked for is therefore 560. A common slip is to write the exponent as 14 - 2r, which gives no integer solution for the power 5.
- Question 8MathematicsOne correct answer
Let f(x) = x³ - 9x² + 15x + 7 on the closed interval [0, 7]. The difference between the greatest and the least value attained by f on this interval is
- 7
- 14
- 18
- 32
Show answer and explanation
Answer: D
f'(x) = 3x² - 18x + 15 = 3(x - 1)(x - 5), so the critical points inside the interval are x = 1 and x = 5.
Evaluate f at the critical points and at both endpoints: f(0) = 7, f(1) = 1 - 9 + 15 + 7 = 14, f(5) = 125 - 225 + 75 + 7 = -18, f(7) = 343 - 441 + 105 + 7 = 14.
Greatest value = 14, attained at both x = 1 and x = 7; least value = -18 at x = 5. The required difference is 14 - (-18) = 32.
The option 14 is the greatest value alone rather than the difference, 18 is the size of the least value with its sign dropped, and 7 is the gap between the two endpoint values f(7) - f(0), which ignores the interior minimum at x = 5.
What you get back
The JEE Main at a glance
About the real JEE Main
JEE Main is the national entrance examination for undergraduate engineering, architecture and planning admissions, conducted by the National Testing Agency (NTA) twice a year (Session 1 in January, Session 2 in April). Paper 1 covers Physics, Chemistry and Mathematics — 75 compulsory questions across three subjects in a three-hour computer-based test, offered in 13 languages. JEE Main scores feed two outcomes: direct B.E. / B.Tech / B.Arch admission to the 31 NITs, 26 IIITs and 32 other Government-Funded Technical Institutes (GFTIs) through JoSAA counselling, AND eligibility for JEE Advanced — the IIT entrance — for the top 2.5 lakh qualified candidates.
What it covers
The full official NTA JEE Main 2026 syllabus for Paper 1 — Physics, Chemistry and Mathematics — drawn from the Class 11 and 12 NCERT curriculum. Each subject has Section A (20 multiple-choice questions) and Section B (5 numerical-entry questions), all compulsory. Numeric Section B is treated as a first-class question type with no negative marking, just like the real paper.
Who sits it
- Class 11 and 12 science students with Physics, Chemistry and Mathematics aiming for B.E. / B.Tech, B.Arch or B.Planning at NITs, IIITs or GFTIs.
- Aspirants for the IITs — JEE Main is the mandatory qualifying gate before you can sit JEE Advanced.
- Anyone who wants disciplined, NCERT-anchored practice with the real +4 / −1 / 0 marking and 20 MCQ + 5 numeric per subject format.
How people prepare for it
- NCERT first — every JEE Main question is rooted in NCERT theory. Read both volumes of Physics, Chemistry and Mathematics before any reference book.
- Layer reference books on top: HC Verma and DC Pandey for Physics; NCERT plus MS Chouhan / OP Tandon for Chemistry; RD Sharma / Cengage / Arihant for Mathematics.
- Practise the 20 MCQ + 5 numeric format from day one. Section B numericals are scored differently (no negative) and reward calculated risks.
- Take full 3-hour mocks weekly under exam conditions. Time management across the three subjects is the single biggest differentiator at the top end.
- Use both sessions strategically. Session 1 is your dry run; the better of your two NTA scores is what counts for JoSAA counselling.
What it opens up
- B.E. / B.Tech at the 31 NITs (~24,000 seats) across Computer Science, Electronics, Mechanical, Civil and other branches.
- B.Tech at the 26 IIITs (~7,500 seats) with a strong tilt towards Computer Science and IT specialisations.
- B.E. / B.Tech / B.Arch at the 32 GFTIs (~5,500 seats) including IIEST Shibpur, the SPAs, and institutes like ISM Dhanbad's affiliates.
- Eligibility to sit JEE Advanced — the IIT entrance — for the top 2.5 lakh JEE Main qualifiers.
- B.Arch (5-year) and B.Planning (4-year) at the SPAs and several NITs through Paper 2 of JEE Main.
Figures about the real exam, as recorded in the Myndaq course. They describe the real sitting, not this free mock paper - check current details with the exam body before you book.
Figures above are as reported for the cycle named in each row and are refreshed each cycle. Sourced, dated facts for this exam (official links on every row) are on the JEE Main course page.
How this mirrors the real format
- The paper structure is the real format: 3 sections in exam order, 75 questions in total.
- Timing is server-controlled for the full 3 hr -- closing the tab does not pause the clock, exactly like our paid mock series.
- Marking follows the published paper: the per-section marks shown in the paper map are the marks the scorer applies.
- Sections run strictly in order and a completed section cannot be reopened.
- Break windows fall where the paper schedules them, and each one is shown in the paper map.
"Real format" means the paper shape, timing, marking and navigation above - not the official exam software, its visual identity, or any affiliation with the exam body.
What differs or is limited
- This free mock follows the same paper shape, timing, marking scheme and navigation as our paid JEE Main 2026 mock series. It is scored automatically. It does not include tutor explanations, analysis dashboards or practice content.
Device and setup checklist
- A stable internet connection for the full duration.
- A laptop or desktop is recommended for the closest exam experience.
One attempt, explicitly
Adding this mock to your account is free and reversible - you can change the exam any time before you begin. Only the confirmed Begin action uses your one attempt. No retake, no reset, no card.