JEE Main 24 January 2026 Shift 2 — Physics Paper with Answer Key

The afternoon paper of 24 January 2026, the eighth 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 2 — 3:00 PM to 6:00 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 Two coaxial loops carrying I and 4I — the field at a point on the axis Moving Charges and Magnetism
27 Flexible chain hanging at 30° at its supports — tension at the lowest point Laws of Motion
28 Detector sphere whose volume is made eight times bigger — the intensity Waves
29 Fifth harmonic of a closed pipe in unison with the first of an open pipe Waves
30 Vertical circle where the thread goes slack at 30° — speed at the bottom Laws of Motion
31 Three ways of filling a capacitor with two dielectrics — ranking them Electrostatic Potential and Capacitance
32 Object and its three times magnified real image 40 cm apart — the focal length Ray Optics
33 Intensity on the screen directly in front of one of the two slits Wave Optics
34 Rod pivoted a third of the way up, falling flat — angular velocity on landing Rotational Motion
35 Vernier callipers with 50 divisions matching 48 — the least count Units and Measurements
36 Correct truth table for the given logic circuit Semiconductor Electronics
37 Heat involved as an ideal gas goes from one state to another Thermodynamics
38 Ammeter reading in the steady state of the given circuit Current Electricity
39 Shunt needed to turn a galvanometer into a 5 mA ammeter Moving Charges and Magnetism
40 Hexagon of wires with spokes to the centre — resistance across a diagonal Current Electricity
41 Five sets of object and image distances for one lens — which reading is wrong Ray Optics
42 Velocity against distance graph — the matching acceleration against distance graph Motion in a Straight Line
43 Stopping potential at one wavelength and at twice it — the first wavelength Dual Nature of Radiation and Matter
44 Cubical block floating in a denser liquid — depth of the submerged part Mechanical Properties of Fluids
45 Two nuclear reactions with given energies — order of stability of He-3, He-4, He-5 Nuclei
46 numeric Work done in blowing a soap bubble to twice its radius Mechanical Properties of Fluids
47 numeric Point charge near the end of a uniformly charged rod — the force Electric Charges and Fields
48 numeric Metre bridge null point shifted 10 cm by a parallel resistor Current Electricity
49 numeric Small cylinder carved from a big one — ratio of their moments of inertia Rotational Motion
50 numeric 300 J given to a gas at constant volume — the mass of the gas Thermodynamics

Chapter by chapter

Chapter Questions Marks
Current Electricity 3 12
Laws of Motion 2 8
Mechanical Properties of Fluids 2 8
Moving Charges and Magnetism 2 8
Ray Optics 2 8
Rotational Motion 2 8
Thermodynamics 2 8
Waves 2 8
Dual Nature of Radiation and Matter 1 4
Electric Charges and Fields 1 4
Electrostatic Potential and Capacitance 1 4
Motion in a Straight Line 1 4
Nuclei 1 4
Semiconductor Electronics 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.

Only sixteen chapters here, the tightest of any January shift, and eight of them took two or three questions. Current Electricity alone carried 12 marks across three questions — a steady-state circuit, a hexagon network and a metre bridge. A student who had drilled circuits was handsomely paid in this shift.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 7 28 28%
Mechanics 5 20 20%
Optics and Waves 5 20 20%
Properties of Matter and Heat 4 16 16%
Modern Physics and Electronics 3 12 12%
Units, Dimensions and Measurement 1 4 4%

The same 28 / 20 / 20 / 16 / 12 / 4 that most 2026 shifts settled into. Nothing surprising in the balance; the difficulty was all in the individual questions.

How tough it was

Level Questions Share Which ones
Straight recall or one step 7 28% 28, 32, 35, 38, 39, 44, 45
Two or three steps 12 48% 26, 27, 29, 33, 36, 37, 41, 42, 43, 46, 48, 50
Long or awkward 6 24% 30, 31, 34, 40, 47, 49

The hexagon with spokes to its centre is a classic that cannot be done by inspection — it needs node equations, and the answer 4r/5 is not one anybody guesses. The rod pivoted a third of the way up is almost as awkward, because the moment of inertia and the drop of the centre of mass both have to be worked out about a pivot that is at neither the end nor the middle.

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  ·  Moving Charges and Magnetism  ·  on the pattern of Q26

Two circular loops of radii r and 2r lie in the same plane with the same centre. They carry currents I and 4I in opposite senses. The magnetic field at the common centre is ______.

Answer  μ0I/2r, in the sense of the larger loop

SolutionAt its own centre a loop gives B = μ0I/2R. The small loop gives μ0I/2r. The large loop gives μ0(4I)/(2 × 2r) = μ0I/r, which is twice as much. They oppose, so the net is μ0I/r − μ0I/2r = μ0I/2r, pointing the way the larger current takes it.

Where it goes wrong  Adding the two because the larger loop carries more current. The senses are opposite, so they subtract whatever their sizes.

Q2  ·  Laws of Motion  ·  on the pattern of Q27

A flexible chain of mass m hangs between two fixed points at the same level. At each support the chain makes 30° with the horizontal. The tension in the chain at its lowest point is ______.

Answer  √3mg/2

SolutionTake one half of the chain, of mass m/2. Three forces act on it: the tension T at the support, along the chain at 30° to the horizontal; the tension T0 at the lowest point, which is horizontal because the chain is flat there; and its weight mg/2. Vertically, T sin30° = mg/2, so T = mg. Horizontally, T cos30° = T0, giving T0 = mg × √3/2 = √3mg/2.

Where it goes wrong  Using the whole mass m for one half. Each half carries only half the weight, which is the whole reason for cutting it in two.

Q3  ·  Waves  ·  on the pattern of Q28

A point source sits at the centre of a spherical detector. If the volume enclosed by the detector is made 8 times bigger, the intensity at its surface ______.

Answer  falls to one quarter

SolutionVolume goes as R3, so eight times the volume means twice the radius. Intensity is power per unit area and the area goes as R2, so doubling the radius makes the area four times as large and the intensity one quarter of what it was. The source is putting out the same power throughout; it is just spread over more surface.

Where it goes wrong  Dividing the intensity by 8, as though it followed the volume. It follows the area, and the area is the two-thirds power of the volume.

Q4  ·  Waves  ·  on the pattern of Q29

The fifth harmonic of a closed organ pipe is in unison with the first harmonic of an open pipe. The ratio of the length of the closed pipe to that of the open pipe is 5/x. The value of x is ______.

Answer  2

SolutionA closed pipe gives f = nv/4Lc with n odd, so its fifth harmonic is 5v/4Lc. An open pipe’s first harmonic is its fundamental, v/2Lo. Setting them equal: 5/4Lc = 1/2Lo, so 10Lo = 4Lc and Lc/Lo = 5/2. Hence x = 2.

Where it goes wrong  Using v/2L for the closed pipe. A closed pipe has an antinode at one end and a node at the other, so its fundamental wavelength is 4L, not 2L.

Q5  ·  Laws of Motion  ·  on the pattern of Q30

A particle on a thread of length r is moving in a vertical circle. The tension in the thread falls to zero when the thread is 30° above the horizontal through the centre. The speed of the particle at the lowest point A of the circle is ______.

Answer  √(7gr/2)

SolutionWhere the tension is zero, the only inward force is the part of gravity along the thread, so mg sin30° = mv2/r and v2 = gr/2 at that point. That point is a height r + r sin30° = 3r/2 above the bottom. Energy conservation from the bottom gives vA2 = v2 + 2g(3r/2) = gr/2 + 3gr = 7gr/2, so vA = √(7gr/2).

Where it goes wrong  Measuring the 30° from the vertical, or forgetting that the point is above the centre so its height is r plus r sin30°, not r sin30°.

Q6  ·  Capacitance  ·  on the pattern of Q31

A parallel plate capacitor of plate area A and separation d is filled with dielectrics in three ways. A: the stronger dielectric k1 fills the whole gap. B: k1 covers half the area and k2 the other half, each over the full gap. C: k1 fills half the thickness and k2 the other half, each over the full area. With k1 > k2, rank the three capacitances.

Answer  CA > CB > CC

SolutionWrite C0 = ε0A/d. Case A is simply k1C0. In case B the two halves are in parallel, so CB = [(k1 + k2)/2]C0, the arithmetic mean of the two constants. In case C they are in series, so CC = [2k1k2/(k1 + k2)]C0, the harmonic mean. For two unequal positive numbers the larger one beats the arithmetic mean, which beats the harmonic mean. So CA > CB > CC.

Where it goes wrong  Mixing up which split is series and which is parallel. Dividing by area gives parallel; stacking across the gap gives series.

Q7  ·  Ray Optics  ·  on the pattern of Q32

A mirror forms a real image three times the size of the object, and the object and image are 40 cm apart. The focal length of the mirror is ______ cm.

Answer  −15 cm

SolutionA real magnified image from a mirror means a concave mirror with the object between the focus and the centre of curvature, and the magnification is −3. With m = −v/u = −3 we get v = 3u, so if u = −x then v = −3x. Both are on the same side, so the separation is 3x − x = 2x = 40, giving x = 20. Then u = −20 cm, v = −60 cm, and 1/f = 1/v + 1/u = −1/60 − 1/20 = −1/15, so f = −15 cm.

Where it goes wrong  Adding the two distances to get 40. For a real image from a mirror the object and image are on the same side, so the gap is the difference, not the sum.

Q8  ·  Wave Optics  ·  on the pattern of Q33

In a Young’s double slit experiment each slit on its own would give intensity I0 on the screen. The slits are 2 mm apart, the screen is 10 m away and the wavelength is 6000 Å. The intensity at the point on the screen directly in front of one of the slits is ______.

Answer  I0

SolutionThe point in front of one slit is y = d/2 = 1 mm from the centre. The path difference there is yd/D = (10−3)(2 × 10−3)/10 = 2 × 10−7 m, which is one third of the wavelength 6 × 10−7 m. So the phase difference is 2π/3. Then I = 2I0(1 + cos120°) = 2I0(1 − ½) = I0.

Where it goes wrong  Expecting a bright fringe in front of a slit and answering 4I0. There is nothing special about that point; the numbers decide, and here they give a phase of 120°.

Q9  ·  Rotational Motion  ·  on the pattern of Q34

A thin uniform rod of mass M and length L is pivoted at a point L/3 from its lower end. It is held vertical and allowed to fall until it lies horizontally on the table. Its angular velocity as it hits the table is ______.

Answer  √(3g/L)

SolutionThe centre of mass is at L/2 from the lower end, so it is L/2 − L/3 = L/6 from the pivot. The moment of inertia about the pivot is ML2/12 + M(L/6)2 = ML2/12 + ML2/36 = ML2/9. Going from vertical to horizontal, the centre of mass falls by L/6. So ½(ML2/9)ω2 = Mg(L/6), which gives ω2 = 3g/L and ω = √(3g/L).

Where it goes wrong  Using ML2/3, the value about an end. The pivot is a third of the way up, so the parallel axis theorem has to be applied from the centre.

Q10  ·  Units and Measurement  ·  on the pattern of Q35

In a vernier callipers 50 vernier scale divisions coincide with 48 main scale divisions, and one main scale division is 0.05 mm. The least count is ______ mm.

Answer  0.002 mm

SolutionOne vernier division is 48/50 of a main division, that is 0.96 × 0.05 = 0.048 mm. The least count is the difference between one main division and one vernier division: 0.05 − 0.048 = 0.002 mm. The same answer comes faster from LC = (1 MSD)/50 × (50 − 48) = 0.05 × 2/50 = 0.002 mm.

Where it goes wrong  Dividing 0.05 by 50 and answering 0.001 mm. That is right only when 50 vernier divisions match 49 main ones.

Q11  ·  Semiconductor Electronics  ·  on the pattern of Q36

Inputs A and B go to an OR gate and also to a NAND gate. The two outputs are then fed into an AND gate, whose output is Y. Write the truth table for Y.

Answer  Y = 0, 1, 1, 0 for (A,B) = (0,0), (0,1), (1,0), (1,1) — an exclusive OR

SolutionTake the rows. (0,0): OR = 0, NAND = 1, so Y = 0. (0,1): OR = 1, NAND = 1, so Y = 1. (1,0): the same, Y = 1. (1,1): OR = 1, NAND = 0, so Y = 0. Y is 1 exactly when the two inputs differ, which is the exclusive OR — and indeed (A + B)·(AB)′ is the standard way of building an XOR from basic gates.

Where it goes wrong  Reading the NAND as an AND, which would give a plain OR instead. The inversion in the NAND is what knocks out the (1,1) row.

Q12  ·  Thermodynamics  ·  on the pattern of Q37

10 mole of an ideal gas with CV = 21 J/K·mol is held at constant volume at 300 K. It is heated until its pressure rises from 20 Pa to 30 Pa. The heat supplied is ______ J.

Answer  31500 J

SolutionAt constant volume the pressure is proportional to the absolute temperature, so raising the pressure from 20 to 30 Pa raises the temperature from 300 K to 300 × 30/20 = 450 K, a rise of 150 K. No work is done because the volume is fixed, so all the heat goes into the internal energy: Q = nCVΔT = 10 × 21 × 150 = 31500 J.

Where it goes wrong  Using CP out of habit. At constant volume there is no expansion to pay for, so CV is the right one.

Q13  ·  Current Electricity  ·  on the pattern of Q38

A battery, a resistor and a capacitor are joined in one series loop, with an ammeter of negligible resistance in the loop. The ammeter reading in the steady state is ______ A.

Answer  zero

SolutionA capacitor in a d.c. loop charges up until the voltage across it equals the battery’s, at which point nothing more can flow. In the steady state there is no current anywhere in that loop, so the ammeter reads zero. There was a current while the capacitor was charging; “steady state” means after all of that has died away.

Where it goes wrong  Working out V/R and giving that. That is the current at the first instant, when the capacitor is still empty.

Q14  ·  Moving Charges and Magnetism  ·  on the pattern of Q39

A moving coil galvanometer of resistance 100 Ω gives a full scale deflection for 1 mA. The resistance needed to turn it into an ammeter reading 5 mA at full scale is ______ Ω.

Answer  25 Ω, connected in parallel

SolutionThe extra current must be sent round the galvanometer through a shunt. Of the 5 mA, 1 mA goes through the coil and 4 mA through the shunt, and both carry the same voltage across them: S × 4 = 100 × 1, so S = 25 Ω. The shunt goes in parallel, which is what makes an ammeter a low-resistance instrument.

Where it goes wrong  Connecting it in series, which is how a voltmeter is made. An ammeter must not add resistance to the circuit it measures, so the extra resistance goes alongside, not in line.

Q15  ·  Current Electricity  ·  on the pattern of Q40

A regular hexagon is made of six wires, each of resistance r, and each corner is also joined to the centre by a wire of the same resistance r. Current enters at one corner and leaves at the opposite corner. The resistance between those two corners is ______.

Answer  4r/5

SolutionNumber the corners 1 to 6 and call the centre O, with current in at 1 and out at 4. By the mirror symmetry about the line 1-O-4, corners 2 and 6 are at the same potential, and so are 3 and 5. By the antisymmetry of the network, O sits exactly halfway, so taking V1 = V and V4 = 0 gives VO = V/2 and V3 = V − V2. Writing Kirchhoff’s current law at corner 2, which joins to 1, to 3 and to O: (V − V2) + (V − 2V2) + (V/2 − V2) = 0, so V2 = 5V/8. The current leaving corner 1 is then 2(V − 5V/8)/r + (V/2)/r = (3V/4 + V/2)/r = 5V/4r. So R = V/I = 4r/5.

Where it goes wrong  Trying to reduce it by series and parallel alone. The spokes to the centre make the network irreducible that way, and node equations are the only honest route.

Q16  ·  Ray Optics  ·  on the pattern of Q41

Five people measure object and image distances with the same convex lens of power +5 D and record (u, v) in cm as (25, 96), (30, 62), (35, 37), (45, 35) and (50, 32). Whose reading is wrong?

Answer  the third one, (35, 37)

SolutionA power of +5 D means f = 100/5 = 20 cm, so every honest pair should satisfy 1/u + 1/v = 1/20 = 0.05. Checking: (25, 96) gives 0.0504; (30, 62) gives 0.0495; (35, 37) gives 0.0556; (45, 35) gives 0.0508; (50, 32) gives 0.0513. Four of the five sit within a per cent or two of 0.05, but the third is out by more than ten per cent, so that is the faulty reading.

Where it goes wrong  Suspecting (25, 96) because the numbers look extreme. An object just outside the focus genuinely throws its image a long way off; the arithmetic is what decides.

Q17  ·  Motion in a Straight Line  ·  on the pattern of Q42

The velocity of a particle grows in proportion to the distance covered, v = kx. The graph of acceleration against distance is ______.

Answer  a straight line through the origin, a = k2x

SolutionWhen velocity is given as a function of position, use a = v dv/dx. Here dv/dx = k, so a = (kx)(k) = k2x. So the acceleration also rises in proportion to x, and the a-against-x graph is a straight line through the origin, steeper than the v-against-x line by a factor k.

Where it goes wrong  Differentiating with respect to time and getting stuck. a = dv/dt needs x as a function of t; a = v dv/dx does the job in one line from the graph you already have.

Q18  ·  Dual Nature of Matter  ·  on the pattern of Q43

Light of a certain wavelength falling on a metal surface gives a stopping potential of 3.2 V. With light of twice that wavelength the stopping potential drops to 0.7 V. The first wavelength is about ______ m. (h = 6.63 × 10−34 J·s, e = 1.6 × 10−19 C, c = 3 × 108 m/s.)

Answer  about 2.5 × 10−7 m

SolutionEinstein’s equation gives hc/λ = φ + eVs. For the two cases, hc/λ = φ + 3.2e and hc/2λ = φ + 0.7e. Subtracting removes the work function: hc/λ − hc/2λ = 2.5e, that is hc/2λ = 2.5e. So λ = hc/5e = (6.63 × 10−34)(3 × 108)/(5 × 1.6 × 10−19) = 2.5 × 10−7 m.

Where it goes wrong  Trying to find φ first. Subtracting the two equations kills it off straight away and leaves a single step.

Q19  ·  Mechanical Properties of Fluids  ·  on the pattern of Q44

A cubical block of density 600 kg/m3 and height 8.0 cm floats in a liquid of density 900 kg/m3. The height of the submerged part is ______ cm.

Answer  about 5.3 cm

SolutionFor a floating body the weight equals the upthrust, so ρblockV = ρliquidVsubmerged. Since the block is a cube of uniform cross-section, the volumes are in the ratio of the heights: hsub/H = 600/900 = 2/3. So hsub = 8 × 2/3 = 5.33 cm, and about 2.7 cm stands above the liquid.

Where it goes wrong  Putting the ratio the other way up and getting 12 cm, which is taller than the block. A floating body is always less dense than the liquid, so the submerged fraction is always less than one.

Q20  ·  Nuclei  ·  on the pattern of Q45

Two reactions: 3He + n → 4He releasing 20 MeV, and 4He + n → 5He absorbing 0.9 MeV. If X3, X4 and X5 are the stabilities of 3He, 4He and 5He, the correct order is ______.

Answer  X4 > X5 > X3

SolutionA reaction that gives out energy ends in something more tightly bound, and one that must be fed energy ends in something less tightly bound. The first reaction releases 20 MeV, so 4He is far more stable than 3He. The second needs 0.9 MeV, so 5He is less stable than 4He — but only by 0.9 MeV, nothing like the 20 MeV gained in the first step, so 5He is still well above 3He. Hence X4 > X5 > X3.

Where it goes wrong  Putting 5He at the bottom because its reaction absorbed energy. The two energies have to be compared with each other, not just read for their signs.

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

A soap bubble of surface tension 0.04 N/m is blown to a radius of 7 cm. The work done in blowing it further to a radius of 14 cm is (15000 − x) μJ. The value of x is ______. (π = 22/7.)

Answer  216

SolutionA soap bubble has two surfaces, so the energy stored is 2T × 4πR2 and the work needed is 2T × 4π(R22 − R12). Here R22 − R12 = 0.142 − 0.072 = 0.0196 − 0.0049 = 0.0147 m2. So W = 2 × 0.04 × 4 × (22/7) × 0.0147 = 0.014784 J = 14784 μJ. Then 15000 − x = 14784, so x = 216.

Where it goes wrong  Leaving out the factor of 2 for the two surfaces. A soap film has an inside and an outside; a liquid drop would have only one.

Q22  ·  Electric Charges and Fields  ·  numeric  ·  on the pattern of Q47

A point charge q = 1 μC sits on the line of a thin insulating rod, 2 cm from one end. The rod is 10 cm long and carries Q = 24 μC spread evenly along it. The force between the charge and the rod is ______ N. (1/4πε0 = 9 × 109 N·m2/C2.)

Answer  90 N

SolutionThe line charge density is λ = 24 × 10−6/0.10 = 2.4 × 10−4 C/m. An element dx at distance x from the charge pulls with kqλdx/x2, and everything is along one line so the contributions simply add: F = kqλ∫dx/x2 from 0.02 to 0.12 = kqλ(1/0.02 − 1/0.12) = (9 × 109)(10−6)(2.4 × 10−4)(50 − 8.33) = 2.16 × 41.67 = 90 N.

Where it goes wrong  Treating the rod as a point charge at its centre, 7 cm away, which gives about 44 N. The near end is only 2 cm away and dominates, so the integral cannot be shortcut.

Q23  ·  Current Electricity  ·  numeric  ·  on the pattern of Q48

In a metre bridge, 2 Ω is in the left gap and 3 Ω in the right, and the null point is found at l cm from the left. When an unknown resistance x is joined in parallel with the 3 Ω, the null point moves 10 cm to the right. The value of x is ______ Ω.

Answer  6 Ω

SolutionAt first 2/3 = l/(100 − l), giving l = 40 cm. Afterwards the balance is at 50 cm, which means the two gaps are now equal: the right-hand resistance has become 2 Ω. That is the 3 Ω in parallel with x: 3x/(3 + x) = 2, so 3x = 6 + 2x and x = 6 Ω.

Where it goes wrong  Expecting x to be smaller than 3 Ω because the parallel combination drops to 2. A 6 Ω resistor across 3 Ω gives exactly 2 Ω; the parallel value only has to fall below 3, not the resistor itself.

Q24  ·  Rotational Motion  ·  numeric  ·  on the pattern of Q49

A uniform solid cylinder of length L and radius R has moment of inertia I1 about its own axis. A smaller coaxial cylinder of length L/2 and radius R/3 is carved out of it, and its moment of inertia about the same axis is I2. The ratio I1/I2 is ______.

Answer  162

SolutionFor a cylinder about its own axis, I = ½MR2 with M = ρπR2L, so I = ½ρπR4L — the fourth power of the radius times the length. The small cylinder has one third the radius and half the length, so I2/I1 = (1/3)4 × (1/2) = 1/162. Hence I1/I2 = 162.

Where it goes wrong  Using R2 and forgetting that the mass also carries R2. Together they make the fourth power, which is where the big factor comes from.

Q25  ·  Thermodynamics  ·  numeric  ·  on the pattern of Q50

300 J of heat is given to oxygen, for which CP = (7/2)R, and its temperature rises from 20 °C to 50 °C at constant volume. The mass of the gas is about ______ g. (R = 8.314 J/mol·K, molar mass of oxygen = 32 g/mol.)

Answer  about 15.4 g

SolutionAt constant volume the heat goes entirely into internal energy, Q = nCVΔT, and CV = CP − R = (5/2)R = 20.79 J/mol·K. The rise is 30 K, so n = 300/(20.79 × 30) = 0.481 mole. At 32 g per mole that is 0.481 × 32 = 15.4 g.

Where it goes wrong  Using the given CP directly. It is handed over only so that CV can be found from it, since the volume is fixed.

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. Shift 1 of the same day, 24 January 2026, is on its own page, and the other January dates are 21, 22, 23 and 28 January.

Official final answer key →

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

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