JEE Main 24 January 2026 Shift 1 — Physics Paper with Answer Key

The morning paper of 24 January 2026, the seventh shift of Session 1. Below is the Physics section question by question, the chapter split, where the hundred marks sat, how tough it was, and the official answer key from NTA.

The shift

Date 24 January 2026
Shift Shift 1 — 9:00 AM to 12:00 PM IST
Paper Paper 1 (B.E./B.Tech)
Whole paper 75 questions, 300 marks, 180 minutes
Physics Q26 to Q50 — Section A 20 MCQ (80 marks), Section B 5 numeric (20 marks)
Marking +4 right, −1 wrong, nothing for a blank

Physics, question by question

Q What it asked Chapter
26 Ball thrown at 45°, peaks at 2 s and lands on a roof at 3 s — the roof height Motion in a Plane
27 Potential varying as ar³ + b — the charge inside unit radius Electrostatic Potential and Capacitance
28 Two electrons of different speeds in orbits of the same radius — the quantum numbers Atoms
29 Prism at minimum deviation exactly meeting the critical angle — the prism angle Ray Optics
30 Power factor of a series LCR circuit on a 220 V, 50 Hz supply Alternating Current
31 Match each kind of electromagnetic wave to how it is produced Electromagnetic Waves
32 Match magnetic induction, flux, permeability and inductance to their dimensions Units and Measurements
33 Taut wire cooled further — the temperature that raises the tension to 1.4T Mechanical Properties of Solids
34 Three concentric charged shells — the potential of each Electrostatic Potential and Capacitance
35 Reflected and refracted rays at right angles — the angle of incidence Wave Optics
36 Rise in the internal energy of water heated from 4 °C to 20 °C Thermodynamics
37 Spring cut in the ratio 1 : 3 — force constant of the shorter piece Oscillations
38 Two masses over a heavy pulley — the pulley’s rotational inertia Rotational Motion
39 Two statements on binding energy per nucleon Nuclei
40 Metre bridge null point shifted by a resistor in parallel — the unknown resistance Current Electricity
41 Microscope of 10 cm tube length — magnification written as a power of 5 Ray Optics
42 Voltmeter of finite resistance across one of two equal resistors — its reading Current Electricity
43 Three masses moved to a larger triangle about the same centre — work done Gravitation
44 Floating cylinder pushed down and let go — the period Oscillations
45 Three charges in the xy plane — the resultant dipole moment about the origin Electric Charges and Fields
46 numeric Sixty-four raindrops merging into one — the new terminal velocity Mechanical Properties of Fluids
47 numeric Stacked blocks with friction everywhere — force to pull one out steadily Laws of Motion
48 numeric Bar magnet turned from the most stable to the most unstable position Magnetism and Matter
49 numeric Protective resistance for a Zener diode of given power rating Semiconductor Electronics
50 numeric Gas in a closed cylinder heated to twice its temperature — the new pressure Kinetic Theory of Gases

Chapter by chapter

Chapter Questions Marks
Current Electricity 2 8
Electrostatic Potential and Capacitance 2 8
Oscillations 2 8
Ray Optics 2 8
Alternating Current 1 4
Atoms 1 4
Electric Charges and Fields 1 4
Electromagnetic Waves 1 4
Gravitation 1 4
Kinetic Theory of Gases 1 4
Laws of Motion 1 4
Magnetism and Matter 1 4
Mechanical Properties of Fluids 1 4
Mechanical Properties of Solids 1 4
Motion in a Plane 1 4
Nuclei 1 4
Rotational Motion 1 4
Semiconductor Electronics 1 4
Thermodynamics 1 4
Units and Measurements 1 4
Wave Optics 1 4

All 19 shifts of JEE Main 2026, counted together chapter by chapter: JEE Main Physics important chapters.

Twenty-one chapters again. Two of the three questions on electric potential and charge came close together, and the two on current electricity were both circuit-measurement questions — a metre bridge and a voltmeter that loads the circuit it is measuring. Those two are the pair of practical questions most students lose marks on.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 8 32 32%
Optics and Waves 5 20 20%
Mechanics 4 16 16%
Properties of Matter and Heat 4 16 16%
Modern Physics and Electronics 3 12 12%
Units, Dimensions and Measurement 1 4 4%

Electrodynamics 32, mechanics only 16, and units down to a single question. Of all the January shifts this one leaned hardest away from mechanics.

How tough it was

Level Questions Share Which ones
Straight recall or one step 8 32% 31, 32, 35, 37, 39, 41, 44, 50
Two or three steps 11 44% 26, 27, 29, 30, 33, 34, 42, 45, 46, 48, 49
Long or awkward 6 24% 28, 36, 38, 40, 43, 47

Two questions were unusual enough to be worth a second look even now. The water heated from 4 °C to 20 °C asks for the change in internal energy rather than the heat, so the tiny work done against the atmosphere has to be subtracted — and that work turns out to be less than a joule out of 268,800. And the two electrons in orbits of the same radius looks like a formula question but turns out to hinge on a single step: at equal radius, the speed is proportional to the quantum number itself.

All 25 Physics questions, solved

Read this first. The questions below are written by AcceleratePhysics. Each one takes the same idea, the same chapter and the same shape as the question at that place in the shift, with our own numbers. They are not NTA’s wording, and nothing on this page is copied from anyone. NTA never put the January 2026 question papers on its public site — only the final answer key, which is linked further down.

Q1  ·  Motion in a Plane  ·  on the pattern of Q26

A ball is thrown into the air at 45° to the horizontal. It reaches its greatest height 2 s after launch and lands on the roof of a building 3 s after launch. The height H of the roof is ______ m. (g = 10 m/s2.)

Answer  15 m

SolutionAt the top the vertical velocity is zero, so the vertical part of the launch velocity is uy = g × 2 = 20 m/s. The height at any time is y = uyt − ½gt2, and at t = 3 s that gives 20 × 3 − 5 × 9 = 60 − 45 = 15 m. So H = 15 m. The 45° is extra information here — useful only if the horizontal distance were also asked for.

Where it goes wrong  Working out the maximum height, 20 m, and stopping there. The ball is on its way down at 3 s, so the roof is lower than the peak.

Q2  ·  Electrostatic Potential  ·  on the pattern of Q27

In a charged spherical region the potential varies with distance as V = ar3 + b, where a and b are constants. The total charge inside a sphere of unit radius is απaε0. The value of α is ______.

Answer  −12

SolutionThe radial field is E = −dV/dr = −3ar2. By Gauss’s law the charge inside a sphere of radius r is ε0 times the flux: Q = ε0(E × 4πr2) = ε0(−3ar2)(4πr2) = −12πaε0r4. Setting r = 1 gives Q = −12πaε0, so α = −12. The constant b drops out, as any additive constant in a potential must.

Where it goes wrong  Losing the minus sign from E = −dV/dr. For a potential that rises outward, the enclosed charge is negative.

Q3  ·  Atoms  ·  on the pattern of Q28

Electrons in two different hydrogen-like atoms move at 3 × 105 m/s and 2.5 × 105 m/s, and the radii of the two orbits are nearly the same. A possible pair of quantum numbers is ______.

Answer  n = 6 and n = 5

SolutionFor a hydrogen-like atom v ∝ Z/n and r ∝ n2/Z. Equal radii means n2/Z is the same for both, so Z ∝ n2. Putting that into the speed, v ∝ n2/n = n — at equal radius the speed is simply proportional to the quantum number. So n1/n2 = 3/2.5 = 6/5, and the pair 6 and 5 fits.

Where it goes wrong  Using v ∝ 1/n straight away and looking for the smaller n with the larger speed. That holds at fixed Z, not at fixed radius, and here it is the radius that is fixed.

Q4  ·  Ray Optics  ·  on the pattern of Q29

A prism of refractive index √2 stands in air. It is set at minimum deviation, and at that setting the ray inside it exactly meets the critical angle condition at the exit face. The prism angle is ______.

Answer  90°

SolutionAt minimum deviation the path inside the prism is symmetric, so r1 = r2 = A/2. Meeting the critical angle at the exit face means r2 = C, where sin C = 1/√2 and C = 45°. So A/2 = 45° and A = 90°. That is the limiting prism for this material: any larger angle and light at minimum deviation could not leave the second face at all.

Where it goes wrong  Setting A = C instead of A/2 = C. The symmetry at minimum deviation is what halves it.

Q5  ·  Alternating Current  ·  on the pattern of Q30

A series LCR circuit on a 220 V, 50 Hz supply has R = 8 Ω, and its inductive and capacitive reactances differ by 6 Ω. The power factor is α/10. The value of α is ______.

Answer  8

SolutionThe impedance is Z = √(R2 + (XL − XC)2) = √(64 + 36) = 10 Ω. The power factor is cosφ = R/Z = 8/10 = 0.8, so α = 8. The 220 V and 50 Hz are not needed for the power factor at all; they would only matter for the current or the power.

Where it goes wrong  Adding XL and XC instead of subtracting. In series they oppose each other, so it is the difference that enters the impedance.

Q6  ·  Electromagnetic Waves  ·  on the pattern of Q31

Match each wave to how it is produced: A. radio wave, B. microwave, C. infrared, D. X-ray. The mechanisms are: a magnetron valve; changes in the vibrational modes of atoms; inner shell electrons dropping to a lower level; rapid acceleration of electrons.

Answer  A → rapid acceleration of electrons, B → magnetron valve, C → vibrational modes, D → inner shell electrons

SolutionRadio waves come from charges driven back and forth in an aerial, that is from electrons being accelerated. Microwaves are generated in the laboratory and in ovens by a magnetron. Infrared comes from molecules and atoms vibrating, which is why it is felt as heat. X-rays come from the innermost electrons of an atom falling into a vacancy, a jump big enough to give photons of that energy.

Where it goes wrong  Swapping infrared and microwave. Both are below visible light, but the vibrational-mode origin belongs to infrared; microwaves come from an electronic valve.

Q7  ·  Units and Dimensions  ·  on the pattern of Q32

Match each quantity to its dimensional formula: A. magnetic induction, B. magnetic flux, C. magnetic permeability, D. self inductance. The formulae are MLT−2A−2, ML2T−2A−2, ML0T−2A−1 and ML2T−2A−1.

Answer  A → ML0T−2A−1, B → ML2T−2A−1, C → MLT−2A−2, D → ML2T−2A−2

SolutionStart from the force on a current-carrying wire, F = BIl, which gives B = MLT−2/(A·L) = ML0T−2A−1. Flux is B times area, so it carries two more powers of length: ML2T−2A−1. Inductance is flux per unit current, so divide by A: ML2T−2A−2. And permeability, from B = μ0nI, is B divided by current per length: MLT−2A−2.

Where it goes wrong  Confusing permeability with inductance, since both end in A−2. They differ by one power of length.

Q8  ·  Mechanical Properties of Solids  ·  on the pattern of Q33

A brass wire is held taut between two rigid supports at 27 °C with no tension. It is cooled to −43 °C, which sets up a tension T. The temperature to which it must be cooled for the tension to become 1.4T is ______ °C.

Answer  −71 °C

SolutionWith the ends fixed, the wire cannot contract, so the thermal strain is converted into tension: T = YAαΔθ, which means T is proportional to the drop in temperature below 27 °C. The first drop is 27 − (−43) = 70 °C. For 1.4T the drop must be 1.4 × 70 = 98 °C, so the temperature is 27 − 98 = −71 °C.

Where it goes wrong  Multiplying −43 by 1.4. It is the temperature drop from 27 °C that scales, not the reading on the thermometer.

Q9  ·  Electrostatic Potential  ·  on the pattern of Q34

Three concentric conducting spherical shells A, B and C of radii a, b and c, with c > b > a, carry charges q1, q2 and q3. Write down the potential of each shell.

Answer  VA = k(q1/a + q2/b + q3/c), VB = k((q1+q2)/b + q3/c), VC = k(q1+q2+q3)/c

SolutionA charged shell gives kq/R at every point inside it or on it, and kq/r outside. So at shell A, which is inside B and C, each of those two contributes at its own radius while A contributes at a: VA = k(q1/a + q2/b + q3/c). At shell B, A is now inside so it contributes q1/b, B contributes q2/b, and C still contributes q3/c. At shell C everything is inside it, so all three contribute at c.

Where it goes wrong  Using 1/a for q1 at shell B. Once a charge is inside the point you are at, its contribution is divided by the distance to the point, not by its own radius.

Q10  ·  Wave Optics  ·  on the pattern of Q35

Unpolarised light travelling in a medium of refractive index √2 meets a boundary with a medium of refractive index √6. The angle of incidence at which the reflected and refracted rays come out at right angles to each other is ______.

Answer  60°

SolutionThe reflected and refracted rays are perpendicular at the Brewster angle, where tan iB equals the ratio of the two indices taken in the direction of travel: tan iB = √6/√2 = √3, so iB = 60°. Checking with Snell’s law, the refracted ray is then at 30°, and 60 + 30 + 90 = 180 as it should be.

Where it goes wrong  Putting the ratio upside down. The index of the second medium goes on top when light travels from the first into the second.

Q11  ·  Thermodynamics  ·  on the pattern of Q36

The density of water is 1000 kg/m3 at 4 °C and 998 kg/m3 at 20 °C. The increase in internal energy of 4 kg of water when it is heated from 4 °C to 20 °C is ______ J. (Specific heat capacity of water = 4200 J/kg·°C, atmospheric pressure = 105 Pa.)

Answer  268799.2 J

SolutionThe heat supplied is Q = mcΔT = 4 × 4200 × 16 = 268800 J. The water also expands slightly against the atmosphere, doing work W = PΔV. Its volume goes from 4/1000 = 4.000 × 10−3 m3 to 4/998 = 4.008 × 10−3 m3, so ΔV = 8.0 × 10−6 m3 and W = 105 × 8.0 × 10−6 = 0.8 J. By the first law, ΔU = Q − W = 268800 − 0.8 = 268799.2 J.

Where it goes wrong  Answering 268800 J. That is the heat, not the change in internal energy — and the point of giving the two densities is to make you find the work, however tiny it turns out to be.

Q12  ·  Oscillations  ·  on the pattern of Q37

A spring of force constant 15 N/m is cut into two pieces whose lengths are in the ratio 1 : 3. The force constant of the shorter piece is ______ N/m.

Answer  60 N/m

SolutionThe force constant of a spring is inversely proportional to its length, because a longer spring has more turns to share the same stretch. The shorter piece is one quarter of the original length, so its constant is four times as large: 4 × 15 = 60 N/m. (The other piece, three quarters as long, has 20 N/m.)

Where it goes wrong  Reading “ratio 1 : 3” as halves or thirds. One part in four and three parts in four is what it means.

Q13  ·  Rotational Motion  ·  on the pattern of Q38

Masses of 400 g and 350 g hang from the ends of a light string over a heavy pulley of radius 2 cm. Released from rest, the heavier mass is seen to fall 81 cm in 9 s. The rotational inertia of the pulley is ______ kg·m2. (g = 9.8 m/s2.)

Answer  9.5 × 10−3 kg·m2

SolutionFrom rest, s = ½at2 gives a = 2 × 0.81/81 = 0.02 m/s2. For masses over a pulley, a = (m1 − m2)g/(m1 + m2 + I/r2), so 0.02 = (0.05)(9.8)/(0.75 + I/r2). That gives 0.75 + I/r2 = 24.5, so I/r2 = 23.75 and I = 23.75 × (0.02)2 = 9.5 × 10−3 kg·m2.

Where it goes wrong  Forgetting to square the radius when converting I/r2 back to I. 2 cm squared is 4 × 10−4 m2, not 2 × 10−2.

Q14  ·  Nuclei  ·  on the pattern of Q39

Two statements. Statement I: for all elements, the heavier the nucleus the greater its binding energy per nucleon. Statement II: for all elements, nuclei with a lower binding energy per nucleon tend to change into nuclei with a higher one. Which is true?

Answer  Statement I is false, Statement II is true

SolutionThe binding energy per nucleon rises steeply up to about iron, A ≈ 56, where it peaks near 8.8 MeV, and then falls slowly all the way to uranium. So it is not true that heavier always means more tightly bound — Statement I is false. Statement II is the whole reason nuclear energy exists: light nuclei fuse and heavy ones split, both moving towards the peak and releasing the difference.

Where it goes wrong  Confusing total binding energy with binding energy per nucleon. The total does keep rising with mass; it is the per-nucleon figure that peaks and turns over.

Q15  ·  Current Electricity  ·  on the pattern of Q40

A metre bridge has 2 Ω and 3 Ω in its two gaps and balances at some point of the wire XY. When an unknown resistor is joined in parallel with the 3 Ω, the balance point shifts 22.5 cm towards Y. The unknown resistance is ______ Ω.

Answer  2 Ω

SolutionAt first 2/3 = l/(100 − l), which gives l = 40 cm. After the shunt the balance is at 40 + 22.5 = 62.5 cm, so 2/R′ = 62.5/37.5 = 5/3 and R′ = 1.2 Ω. That R′ is the 3 Ω in parallel with the unknown X: 3X/(3 + X) = 1.2, so 3X = 3.6 + 1.2X, 1.8X = 3.6 and X = 2 Ω.

Where it goes wrong  Forgetting that the shunt must lower the right-hand resistance, which is why the balance moves away from the 2 Ω side. If the shift comes out the other way, the two gaps have been read the wrong way round.

Q16  ·  Ray Optics  ·  on the pattern of Q41

A compound microscope of tube length 10 cm uses two convex lenses of focal lengths 2 cm and 5 cm. Its total magnification for normal adjustment is 5k. The value of k is ______. (Least distance of distinct vision = 25 cm.)

Answer  2

SolutionFor normal adjustment the magnification is (L/fo)(D/fe) = (10/2)(25/5) = 5 × 5 = 25. Writing 25 as a power of 5 gives 52, so k = 2. The shorter focal length is naturally the objective, since a microscope objective must be the stronger lens.

Where it goes wrong  Putting the 5 cm lens as the objective, which gives (10/5)(25/2) = 25 as well — the same number by accident here, but the wrong reasoning, and it fails whenever the figures are less kind.

Q17  ·  Current Electricity  ·  on the pattern of Q42

Two resistors of 100 Ω each are joined in series across a 9 V battery. A voltmeter of resistance 400 Ω is connected across one of them. The voltmeter reads ______ V.

Answer  4 V

SolutionThe voltmeter is not ideal, so it draws current and changes the circuit. In parallel with 100 Ω it makes 100 × 400/500 = 80 Ω. The circuit is then 100 + 80 = 180 Ω, so the current is 9/180 = 0.05 A and the voltage across the 80 Ω part is 0.05 × 80 = 4 V. Without the voltmeter the true reading would have been 4.5 V, so the instrument has pulled it down by half a volt.

Where it goes wrong  Answering 4.5 V. That is the voltage before the meter was attached; a real voltmeter always reads a little low.

Q18  ·  Gravitation  ·  on the pattern of Q43

Masses of 200 kg, 300 kg and 400 kg sit at the corners of an equilateral triangle of side 20 m. They are moved to the corners of a larger equilateral triangle of side 25 m with the same centre. The work done in the process is ______ J. (G = 6.7 × 10−11 N·m2/kg2.)

Answer  1.74 × 10−7 J

SolutionThe potential energy of three masses at mutual distance a is U = −G(m1m2 + m2m3 + m3m1)/a. The sum of the products is 200 × 300 + 300 × 400 + 400 × 200 = 260000. So U goes from −260000G/20 = −13000G to −260000G/25 = −10400G. The work done is the rise in energy, 2600G = 2600 × 6.7 × 10−11 = 1.74 × 10−7 J.

Where it goes wrong  Forgetting that all three sides of an equilateral triangle are the same, so one denominator covers all three pairs. And the answer is positive: pulling masses apart against gravity needs work.

Q19  ·  Oscillations  ·  on the pattern of Q44

A cylindrical block of mass M and cross-sectional area A floats with its axis vertical in a liquid of density ρ. It is pushed down a little and released, and it oscillates. The period of the oscillation is ______.

Answer  2π√(M/ρAg)

SolutionPush the block down by x and an extra volume Ax is submerged, so the extra upthrust is ρAxg, directed back towards the floating position. That is a restoring force proportional to the displacement, with effective spring constant k = ρAg. So the motion is simple harmonic with T = 2π√(M/k) = 2π√(M/ρAg).

Where it goes wrong  Putting ρAg on top instead of underneath. It plays the part of the spring constant, so it belongs in the denominator with M above it.

Q20  ·  Electric Charges and Fields  ·  on the pattern of Q45

Charges +2q, +3q and −4q sit at (0, −3a), (2a, 0) and (−2a, 0) in the xy plane. The resultant dipole moment about the origin is ______.

Answer  2qa(7ĥ − 3ĵ)

SolutionThe dipole moment of a set of charges about a point is p = Σqiri, taking each charge times its position vector. Here: 2q(0, −3a) = (0, −6qa); 3q(2a, 0) = (6qa, 0); and (−4q)(−2a, 0) = (8qa, 0). Adding component by component gives (14qa, −6qa), which is 2qa(7ĥ − 3ĵ).

Where it goes wrong  Dropping one of the two minus signs in the third term. A negative charge at a negative coordinate gives a positive contribution.

Q21  ·  Mechanical Properties of Fluids  ·  numeric  ·  on the pattern of Q46

Sixty-four raindrops, each of radius 1 mm, are falling at a terminal velocity of 10 cm/s. They merge into one bigger drop. The terminal velocity of the bigger drop is ______ cm/s.

Answer  160 cm/s

SolutionVolume is conserved, so 64r3 = R3 and R = 4r. The terminal velocity goes as the square of the radius, so it is multiplied by 16: 10 × 16 = 160 cm/s.

Where it goes wrong  Scaling by 64 or by 4 instead of by 16. The radius grows as the cube root of the number of drops, and the velocity then goes as its square.

Q22  ·  Laws of Motion  ·  numeric  ·  on the pattern of Q47

A block of 4 kg rests on top of a block of 6 kg, which rests on the ground. The coefficient of sliding friction is 0.5 between every pair of surfaces, including block and ground. The top block is held still by a string tied to a wall. The force needed to pull the lower block out at a constant speed is ______ N. (g = 10 m/s2.)

Answer  70 N

SolutionThe lower block has two surfaces rubbing. On top of it the 4 kg block presses with 40 N, so that friction is 0.5 × 40 = 20 N. Underneath, the ground carries both blocks, 100 N, so that friction is 0.5 × 100 = 50 N. At constant speed the applied force just matches the two together: 20 + 50 = 70 N.

Where it goes wrong  Using only 6 kg for the normal force at the ground. The ground holds up both blocks, so the full 10 kg presses on it.

Q23  ·  Magnetism and Matter  ·  numeric  ·  on the pattern of Q48

A short bar magnet set with its axis at 30° to an external field of 800 gauss feels a torque of 0.016 N·m. The work done in turning it from its most stable to its most unstable position is α × 10−3 J. The value of α is ______.

Answer  64

SolutionFirst find the magnetic moment. B = 800 gauss = 0.08 T, and τ = mB sinθ, so 0.016 = m × 0.08 × sin30° = 0.04m, giving m = 0.4 A·m2. The most stable position is aligned with the field, U = −mB, and the most unstable is opposite it, U = +mB. So the work is 2mB = 2 × 0.4 × 0.08 = 0.064 J = 64 × 10−3 J, and α = 64.

Where it goes wrong  Answering mB instead of 2mB. The magnet is turned the whole way from aligned to anti-aligned, which costs twice as much.

Q24  ·  Semiconductor Electronics  ·  numeric  ·  on the pattern of Q49

A voltage regulating circuit uses a Zener diode of breakdown voltage 10 V, able to dissipate at most 0.4 W, run from a 15 V supply. The protective series resistance needed is about ______ Ω.

Answer  about 125 Ω

SolutionThe diode holds 10 V across itself, so the largest current it can take is 0.4/10 = 0.04 A. The series resistor has to drop 15 − 10 = 5 V at that current, so R = 5/0.04 = 125 Ω. Any smaller and the diode would burn out in the worst case, with no load drawing current.

Where it goes wrong  Dividing the power by 15 V instead of 10 V. The diode’s own power is its own voltage times its own current, and its voltage is the breakdown value.

Q25  ·  Kinetic Theory of Gases  ·  numeric  ·  on the pattern of Q50

A fixed mass of gas in a closed cylinder is at 3.23 kPa and 50 °C. It is heated until its temperature is doubled. The new pressure is ______ Pa.

Answer  6460 Pa

Solution“Closed cylinder” means the volume is fixed, so P/T stays constant. Doubling the temperature means doubling the absolute temperature: 50 °C is 323 K, so the new temperature is 646 K and the pressure doubles too, from 3.23 kPa to 6.46 kPa = 6460 Pa.

Where it goes wrong  Reading “double the temperature” as 100 °C. In gas problems a doubling is always of the kelvin value, and 100 °C is only 373 K.

The official answer key

NTA did not publish the Session 1 question papers. They were shown only inside each candidate’s own login during the answer key challenge window in February 2026, and that window is closed. The one official file for this shift is the final answer key, which covers all ten January shifts. The other January 2026 dates were 21, 22, 23 and 28 January, two shifts on each.

Official final answer key →

All shifts, date by date: JEE Main Previous Years Papers.

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