JEE Main 28 January 2026 Shift 2 — Physics Paper with Answer Key

The afternoon paper of 28 January 2026, the last 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 28 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 Prism whose minimum deviation equals its own angle — range of the index Ray Optics
27 Photons per second from a 6 mW laser at 663 nm Dual Nature of Radiation and Matter
28 Magnetic field vector of a wave whose electric field is given Electromagnetic Waves
29 Two masses joined by a stretched spring and released — angular frequency Oscillations
30 Match viscosity, surface tension, pressure and surface energy to their dimensions Units and Measurements
31 Five claims about electrostatic field lines and Gauss’s law — which hold Electric Charges and Fields
32 Four readings in millimetres — the least count of the instrument used Units and Measurements
33 Two nuclei, one with eight times the mass number — ratio of their radii Nuclei
34 Percentage error in a spring constant from a mass and a timing measurement Units and Measurements
35 Block sliding down an incline that is itself accelerating — the time taken Laws of Motion
36 Four claims about capacitors and dielectrics — which hold Electrostatic Potential and Capacitance
37 Wheatstone bridge with one arm heated — the potential difference that appears Current Electricity
38 Which of four electrical quantities is not a measurable quantity at all Current Electricity
39 Density and Young’s modulus both raised slightly — the new wave speed Waves
40 Two diodes wired to two inputs — which gate the circuit makes Semiconductor Electronics
41 Biconvex lens built from two plano-convex lenses — the magnification Ray Optics
42 Where the field of a thick current-carrying cylinder is least Moving Charges and Magnetism
43 Which vector does not change sign when the position vector is reversed Rotational Motion
44 Five claims about a particle with x = 4t³ − 3t — which hold Motion in a Straight Line
45 Mean free path of a molecule at a given temperature and pressure Kinetic Theory of Gases
46 numeric Total work done by a system round a triangular cycle Thermodynamics
47 numeric Two wavelengths in a double slit — where their bright fringes first coincide Wave Optics
48 numeric Beats before and after loading a fork with wax — its original frequency Waves
49 numeric Flywheel and hanging mass — kinetic energy of the wheel after a 3 m drop Rotational Motion
50 numeric Inductor storing 16 J and wasting 32 W — ratio of reactance to resistance Alternating Current

Chapter by chapter

Chapter Questions Marks
Units and Measurements 3 12
Current Electricity 2 8
Ray Optics 2 8
Rotational Motion 2 8
Waves 2 8
Alternating Current 1 4
Dual Nature of Radiation and Matter 1 4
Electric Charges and Fields 1 4
Electromagnetic Waves 1 4
Electrostatic Potential and Capacitance 1 4
Kinetic Theory of Gases 1 4
Laws of Motion 1 4
Motion in a Straight Line 1 4
Moving Charges and Magnetism 1 4
Nuclei 1 4
Oscillations 1 4
Semiconductor Electronics 1 4
Thermodynamics 1 4
Wave Optics 1 4

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

Twenty chapters, with five of them taking two questions each. Three of the twenty-five questions were of the “which of these statements are correct” kind — on field lines, on capacitors, and on a particle’s motion — which is more of that format than any other shift in January.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 7 28 28%
Optics and Waves 6 24 24%
Mechanics 4 16 16%
Units, Dimensions and Measurement 3 12 12%
Modern Physics and Electronics 3 12 12%
Properties of Matter and Heat 2 8 8%

The odd one out of the whole year. Units and measurement got 12 marks here, three times what it got on 24 January morning, and optics and waves climbed to 24. Heat and properties of matter fell to 8, the lowest in any 2026 shift. A student who treats units as a throwaway chapter would have lost twelve marks in this one paper.

How tough it was

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

The block on an accelerating incline was the hardest question of Session 1: the answer has to be written with a double-angle identity before it matches any of the options, and getting the sign of the pseudo-force wrong gives one of the other three. The two-wavelength fringe coincidence is the other one worth practising — it needs the smallest whole numbers in the ratio 13 : 11 before any arithmetic can start.

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  ·  Ray Optics  ·  on the pattern of Q26

For a transparent prism the angle of minimum deviation turns out to be equal to the refracting angle of the prism itself. The refractive index n then satisfies ______.

Answer  1 < n < 2

SolutionAt minimum deviation n = sin[(A + δm)/2]/sin(A/2). Putting δm = A gives n = sin A/sin(A/2) = 2cos(A/2). Now A must be more than 0 and less than 180°, so cos(A/2) lies between 0 and 1 and n lies between 0 and 2; and since a transparent prism must be denser than air, n > 1. So 1 < n < 2. A small prism angle pushes n towards 2 and a large one towards 1.

Where it goes wrong  Forgetting that n must exceed 1, and answering 0 < n < 2. Half the algebraic range is ruled out by physics.

Q2  ·  Dual Nature of Matter  ·  on the pattern of Q27

A 6 mW laser works at 663 nm. The number of photons it emits per second is ______. (h = 6.63 × 10−34 J·s, c = 3 × 108 m/s.)

Answer  2 × 1016 per second

SolutionEach photon carries E = hc/λ = (6.63 × 10−34)(3 × 108)/(663 × 10−9) = 3.0 × 10−19 J. The source puts out 6 × 10−3 J every second, so the count is 6 × 10−3/3 × 10−19 = 2 × 1016 photons per second.

Where it goes wrong  Leaving the power in milliwatts. 6 mW is 6 × 10−3 W, and the factor of a thousand changes the exponent.

Q3  ·  Electromagnetic Waves  ·  on the pattern of Q28

A plane electromagnetic wave travels through free space with its electric field E = 54 sin(kz − ωt)ĵ V/m. The magnetic field vector of the wave is ______.

Answer  B = −1.8 × 10−7 sin(kz − ωt)ĥ T

SolutionThe size comes from B0 = E0/c = 54/3 × 108 = 1.8 × 10−7 T. The direction comes from the rule that E × B points the way the wave travels, here along +k̂. With E along ĵ, we need ĵ × B̂ = k̂, and since ĵ × (−ĥ) = +k̂, the field must lie along −ĥ. So B = −1.8 × 10−7 sin(kz − ωt)ĥ T, in step with E at every instant.

Where it goes wrong  Getting the sign of the unit vector wrong by guessing. Work the cross product out: ĥ × ĵ = k̂ is the only one worth memorising, and the rest follow by cycling.

Q4  ·  Oscillations  ·  on the pattern of Q29

A spring of force constant 150 N/m is held stretched between masses of 1 kg and 0.2 kg on a frictionless horizontal surface, and then released. The angular frequency of the resulting oscillation is ______ rad/s.

Answer  30 rad/s

SolutionWhen a spring connects two free masses, the oscillation is governed by the reduced mass μ = m1m2/(m1 + m2) = (1 × 0.2)/1.2 = 1/6 kg. Then ω = √(k/μ) = √(150 × 6) = √900 = 30 rad/s. The centre of mass does not move, because the surface is frictionless and the spring’s forces are internal.

Where it goes wrong  Using the total mass 1.2 kg, or just one of the two. The reduced mass is always smaller than either, which is why the frequency comes out higher than you might expect.

Q5  ·  Units and Dimensions  ·  on the pattern of Q30

Match each quantity to its dimensions: A. coefficient of viscosity, B. surface tension, C. pressure, D. surface energy. The options are ML−1T−2, ML2T−2, ML0T−2 and ML−1T−1.

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

SolutionViscosity comes from F = ηA dv/dx, which gives η = MLT−2/(L2 × T−1) = ML−1T−1. Surface tension is force per length, MT−2, written ML0T−2. Pressure is force per area, ML−1T−2. Surface energy is an energy, ML2T−2. Note that surface tension and surface energy per unit area share dimensions, but surface energy on its own does not.

Where it goes wrong  Giving surface energy the same dimensions as surface tension. They agree only when the energy is taken per unit area.

Q6  ·  Electric Charges and Fields  ·  on the pattern of Q31

Which of these are correct? A. Electrostatic field lines form closed loops. B. The field lines point radially outward from a positive charge. C. Gauss’s law in this form holds only because the force is an inverse square one. D. The work done in taking a charge round a closed path in a static field is zero. E. Motion of a particle under the Coulomb force of a fixed charge must stay in a plane.

Answer  B, C, D and E

SolutionA is false: electrostatic field lines begin on positive charges and end on negative ones, and never close on themselves — if they did, a charge could gain energy going round a loop. B is plainly true. C is true: the r2 in the denominator is exactly what cancels the r2 in the area of a sphere, which is why the flux comes out independent of radius. D is true, and is the same fact as A seen from the energy side. E is true because a central force has no component out of the plane set by the initial position and velocity, so the motion cannot leave that plane.

Where it goes wrong  Accepting A because magnetic field lines do form closed loops. The two fields differ in exactly this respect.

Q7  ·  Units and Measurement  ·  on the pattern of Q32

A set of readings taken with one instrument comes out as 1.24 mm, 1.25 mm, 1.23 mm and 1.21 mm. The least count of that instrument is ______ mm.

Answer  0.01 mm

SolutionThe least count is the smallest division the instrument can show, and it is given away by the last digit the readings carry. Every reading here is quoted to the second decimal place in millimetres and they differ from one another in that place, so the instrument resolves 0.01 mm. A coarser instrument could not have produced 1.23 and 1.24 as separate readings, and a finer one would have shown a third decimal.

Where it goes wrong  Working out the spread of the readings, 0.04 mm, and offering that. The spread measures the scatter in the measurement, not the resolution of the scale.

Q8  ·  Nuclei  ·  on the pattern of Q33

One nucleus has mass number α and radius Rα; another has mass number β and radius Rβ, with β = 8α. The ratio Rα/Rβ is ______.

Answer  0.5

SolutionNuclear radius goes as the cube root of the mass number, R = R0A1/3. So Rα/Rβ = (α/8α)1/3 = (1/8)1/3 = 1/2 = 0.5. Eight times the nucleons makes a nucleus only twice as wide, which is another way of saying nuclear matter has a fixed density.

Where it goes wrong  Answering 1/8. The mass number is in the volume, so the radius only changes by its cube root.

Q9  ·  Units and Measurement  ·  on the pattern of Q34

A spring oscillator has T = 2π√(m/k). The mass is measured as 10 g with an accuracy of 10 mg, and the time for 50 oscillations is found to be 60 s on a watch whose resolution is 1 s. The percentage error in the spring constant k is ______ %.

Answer  3.43%

SolutionRearranging, k = 4π2m/T2, so Δk/k = Δm/m + 2ΔT/T. The mass gives 10 mg in 10 g = 0.001, that is 0.1%. For the time, the watch resolves 1 s out of the 60 s total, and because 50 oscillations were counted together that same relative error carries over to T: ΔT/T = 1/60 = 0.0167, so the time contributes 2 × 1.67 = 3.33%. Adding, 0.1 + 3.33 = 3.43%.

Where it goes wrong  Dividing 1 s by T = 1.2 s instead of by the 60 s that was actually measured. Timing many oscillations at once is exactly how that error is cut down, and the calculation has to reflect it.

Q10  ·  Laws of Motion  ·  on the pattern of Q35

A small block slides from the top of a frictionless inclined surface of angle θ and base length L, while the incline itself accelerates towards the left with a constant a0. The time the block takes to reach the lowest point is ______.

Answer  √{ 4L / [g sin2θ − a0(1 + cos2θ)] }

SolutionWork in the frame of the incline, where a pseudo-force ma0 acts on the block in the direction opposite to the incline’s acceleration. Along the slope the block then has acceleration a = g sinθ − a0cosθ. The length of the slope is L/cosθ, so from s = ½at2, t = √{2L/[cosθ(g sinθ − a0cosθ)]} = √{2L/(g sinθcosθ − a0cos2θ)}. Now use g sinθcosθ = (g/2)sin2θ and cos2θ = (1 + cos2θ)/2, which turns the denominator into half of [g sin2θ − a0(1 + cos2θ)] and gives t = √{4L/[g sin2θ − a0(1 + cos2θ)]}.

Where it goes wrong  Getting the pseudo-force the wrong way round, which flips the minus to a plus and matches one of the wrong options exactly. And then not converting to double angles, so the right answer looks like none of the choices.

Q11  ·  Capacitance  ·  on the pattern of Q36

Which of these are correct? A. The effective capacitance of capacitors in series is always smaller than the smallest one in the set. B. When a dielectric is placed between charged plates, no displacement of charge can occur because a dielectric is an insulator. C. Increasing the plate area, or reducing the thickness of the dielectric, both raise the capacitance. D. For a point charge, concentric spheres centred on it are equipotential surfaces.

Answer  A, C and D

SolutionA is true: adding reciprocals always gives something smaller than the smallest term. B is false, and that is the point of the question — a dielectric does not conduct, but its molecules do polarise, and that displacement of bound charge is exactly what weakens the field and raises the capacitance. C is true, since C = Kε0A/d rises with A and falls with d. D is true: the potential of a point charge depends only on r, so every sphere about it is an equipotential.

Where it goes wrong  Accepting B because “insulator” sounds like “nothing moves”. Charge does not flow through a dielectric, but it does shift within each molecule.

Q12  ·  Current Electricity  ·  on the pattern of Q37

A Wheatstone bridge has all four arms equal at 100 Ω and is fed from a 21 V supply, so it is balanced. One arm is then heated and its resistance rises by 10%. The potential difference that now appears across the galvanometer arm is ______ V.

Answer  0.5 V

SolutionTake the supply across A and C, with one branch A-B-C and the other A-D-C, and the galvanometer between B and D. Put VA = 21 V and VC = 0. In the untouched branch, VB = 21 × 100/(100 + 100) = 10.5 V. In the branch whose first arm has become 110 Ω, VD = 21 × 100/(110 + 100) = 10 V. So VB − VD = 0.5 V. A 10% change in one arm shows up as only half a volt — which is why a bridge is read by finding the null, not by measuring the imbalance.

Where it goes wrong  Expecting 10% of the supply. The imbalance is diluted by the whole divider, so it comes out far smaller.

Q13  ·  Current Electricity  ·  on the pattern of Q38

Which of these is not a measurable quantity: resistance, displacement current, voltage, or voltage difference?

Answer  voltage

SolutionOnly differences of potential can be measured; a potential on its own has no meaning until a zero is chosen, and a voltmeter always has two terminals. Resistance is measurable, a voltage difference is measurable, and the displacement current, though it is not a flow of charge, produces a magnetic field that can be detected and so is measurable too. “Voltage” standing alone is the odd one out.

Where it goes wrong  Picking displacement current because it sounds abstract. It is a real, measurable quantity; it is the bare word “voltage” that is not well defined.

Q14  ·  Waves  ·  on the pattern of Q39

A longitudinal wave travels at 400 m/s in a metal bar. If the density of the material is raised by 0.5% and its Young’s modulus by 1%, the speed becomes about ______ m/s.

Answer  about 401 m/s

SolutionThe speed is v = √(Y/ρ), so for small changes Δv/v = ½(ΔY/Y) − ½(Δρ/ρ) = ½(1%) − ½(0.5%) = 0.25%. A quarter of a per cent of 400 is 1 m/s, so the new speed is about 401 m/s.

Where it goes wrong  Adding the two percentages, or forgetting the factor of a half that the square root brings. The denominator also changes sign on the way out.

Q15  ·  Semiconductor Electronics  ·  on the pattern of Q40

Two p-n junction diodes have their anodes connected to inputs A and B, and their cathodes joined together at the output C, which returns to ground through a resistor. The circuit works as a ______ gate.

Answer  OR gate

SolutionA diode conducts when its anode is the more positive end. If either input is high, that diode conducts and pulls the output up through the resistor, so C is high. If both inputs are low, neither diode conducts and the resistor holds C at ground, so C is low. High whenever at least one input is high: that is an OR gate. Wiring the diodes the other way round, with the cathodes at the inputs and a resistor to the supply, gives an AND gate instead.

Where it goes wrong  Calling it an AND gate because both diodes are in the circuit. Follow one case at a time: a single high input is enough to drive the output high.

Q16  ·  Ray Optics  ·  on the pattern of Q41

A biconvex lens is made by sticking together, at their flat faces, two thin plano-convex lenses of refractive index 1.5 whose curved faces have radii 10 cm and 15 cm. An object is placed 36 cm to the left of the pair. The magnification of the image is ______.

Answer  −0.5 (half the size, inverted, real)

SolutionEach plano-convex lens has 1/f = (n − 1)/R, so the first has f = 10/0.5 = 20 cm and the second f = 15/0.5 = 30 cm. In contact their powers add: 1/F = 1/20 + 1/30 = 5/60, so F = 12 cm. With u = −36 cm, 1/v = 1/12 − 1/36 = 2/36, giving v = +18 cm. The magnification is v/u = 18/(−36) = −0.5.

Where it goes wrong  Using the lens maker’s formula once with both radii and n = 1.5, which happens to give the same answer here only because both halves share the same index. Doing the two lenses separately and adding the powers always works, even when the two glasses differ.

Q17  ·  Moving Charges and Magnetism  ·  on the pattern of Q42

A long cylindrical conductor of large cross-section carries a current spread evenly over that cross-section. Which statement about its magnetic field is correct: maximum at the axis, minimum at the surface, minimum at the axis, or the same everywhere in the cross-section?

Answer  minimum at the axis

SolutionApply Ampere’s law to a circle of radius r inside the conductor. The current enclosed is I(r2/R2), so B(2πr) = μ0Ir2/R2 and B = μ0Ir/2πR2. The field therefore grows straight out from zero at the axis to its largest value at the surface, where it equals μ0I/2πR, and falls off as 1/r outside. So the field is least at the axis, exactly zero there.

Where it goes wrong  Assuming the field is strongest where the current is “most concentrated”, at the centre. On the axis there is no enclosed current at all, so there is no field.

Q18  ·  Rotational Motion  ·  on the pattern of Q43

When the position vector r is replaced by −r, which of these does not change sign: velocity, acceleration, linear momentum, or angular momentum?

Answer  angular momentum

SolutionVelocity is dr/dt, so reversing r reverses it; the same goes for acceleration and for linear momentum mv. But angular momentum is L = r × p, and both factors have flipped, so the two minus signs cancel: (−r) × (−p) = r × p. Angular momentum is unchanged. Quantities built from a cross product of two reversing vectors behave this way, which is why they are called pseudovectors.

Where it goes wrong  Reasoning that L contains r, so it must flip. It contains r once and p once, and the pair of sign changes undo each other.

Q19  ·  Motion in a Straight Line  ·  on the pattern of Q44

A particle starts at t = 0 with x(t) = 4t³ − 3t. Which of these are correct? A. It returns to the origin 0.866 units of time later. B. It is 1 unit from the origin at its turning point. C. Its acceleration is never negative. D. It is 0.5 units from the origin at its turning point. E. It never turns back, because the acceleration is never negative.

Answer  A, B and C

SolutionThe velocity is v = 12t2 − 3, which vanishes at t = 0.5, so there is one turning point. At that instant x = 4(0.125) − 1.5 = −1, so the particle is 1 unit from the origin — B is right and D is wrong. It returns to the origin when 4t3 = 3t, that is t = √3/2 = 0.866 — A is right. The acceleration is 24t, which is never negative for t ≥ 0, so C is right. But E draws the wrong conclusion from C: a positive acceleration with a negative starting velocity still allows one turn, and that is exactly what happens at t = 0.5.

Where it goes wrong  Accepting E because C is true. A body can turn round while its acceleration stays one way; it only needs the velocity to pass through zero.

Q20  ·  Kinetic Theory of Gases  ·  on the pattern of Q45

The mean free path of a molecule of diameter 5 × 10−10 m at 41 °C and a pressure of 1.38 × 105 Pa is ______ m. (kB = 1.38 × 10−23 J/K.)

Answer  2√2 × 10−8 m, about 2.8 × 10−8 m

SolutionThe mean free path is λ = kBT/(√2 πd2P). With T = 314 K the numerator is 1.38 × 10−23 × 314 = 4.33 × 10−21. The denominator is √2 × π × (5 × 10−10)2 × 1.38 × 105 = 1.414 × 3.142 × 2.5 × 10−19 × 1.38 × 105 = 1.53 × 10−13. Dividing gives 2.83 × 10−8 m, which is 2√2 × 10−8 m. That is about sixty molecular diameters — a molecule in air at this pressure travels that far between collisions.

Where it goes wrong  Leaving out the √2, which comes from the fact that the other molecules are moving too, not standing still.

Q21  ·  Thermodynamics  ·  numeric  ·  on the pattern of Q46

A system is taken round the triangular cycle ABC on a p-V diagram, with A at (1 m³, 10 Pa), B at (3 m³, 10 Pa) and C at (1 m³, 30 Pa), returning from C to A. The total work done by the system is ______ J.

Answer  −20 J (that is, 20 J is done on the system)

SolutionTake the legs in turn. A to B is an expansion at a steady 10 Pa, so W = 10 × 2 = +20 J. B to C is a straight line from (3, 10) to (1, 30): the average pressure is 20 Pa and the volume falls by 2 m³, so W = −40 J. C to A is at constant volume, so W = 0. The total is 20 − 40 = −20 J, which is just the enclosed area of 20 J taken with a minus sign because the cycle is traced anticlockwise.

Where it goes wrong  Giving +20 J, the bare area. The direction round the loop fixes the sign: clockwise means work out, anticlockwise means work in.

Q22  ·  Wave Optics  ·  numeric  ·  on the pattern of Q47

A beam containing wavelengths 650 nm and 550 nm lights up a double slit of separation 2 mm, with the screen 1.2 m away. The least distance from the central maximum at which bright fringes of both wavelengths fall together is ______ × 10−5 m.

Answer  429 × 10−5 m, that is 4.29 mm

SolutionTwo bright fringes land together when n1λ1 = n2λ2, that is 650n1 = 550n2, or 13n1 = 11n2. The smallest whole numbers are n1 = 11 and n2 = 13. So the position is y = 11 × 650 × 10−9 × 1.2/(2 × 10−3) = 4.29 × 10−3 m = 429 × 10−5 m. Checking the other way round: 13 × 550 × 10−9 × 600 gives the same 4.29 mm.

Where it goes wrong  Pairing the larger order with the larger wavelength. It is the other way about: the longer wavelength needs fewer fringes to reach the same place.

Q23  ·  Waves  ·  numeric  ·  on the pattern of Q48

Two tuning forks A and B sounded together give 8 beats in 2 s. When A is loaded with wax the beats fall to 4 in 2 s. If B has a frequency of 380 Hz, the original frequency of A was ______ Hz.

Answer  384 Hz

SolutionEight beats in 2 s is 4 beats per second, so A was either 384 Hz or 376 Hz. Loading a fork with wax always lowers its frequency. If A were 384, lowering it brings it closer to 380 and the beat rate falls — which is what happened, from 4 Hz to 2 Hz, so A was at 382 Hz after loading. If A were 376, lowering it would take it further from 380 and the beats would increase. So A was 384 Hz.

Where it goes wrong  Stopping at “384 or 376”. The wax is there precisely to settle which, and the rule is that it always lowers the frequency.

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

A flywheel of mass 3 kg and radius 5 m, free to turn about a horizontal axis, has a light string wound round it with a 3 kg mass tied to the loose end. The mass is released from rest. The kinetic energy of the wheel when the mass has descended 3 m is ______ J. (g = 10 m/s2; treat the flywheel as a uniform disc.)

Answer  30 J

SolutionThe wheel has I = ½MR2 = ½ × 3 × 25 = 37.5 kg·m2, and the string does not slip, so v = ωR. Energy conservation gives mgh = ½mv2 + ½Iω2: 3 × 10 × 3 = ½(3)v2 + ½(37.5)(v2/25), that is 90 = 1.5v2 + 0.75v2 = 2.25v2, so v2 = 40. The wheel’s share is ½Iω2 = 0.75v2 = 30 J, and the falling mass keeps the other 60 J.

Where it goes wrong  Giving the whole 90 J to the wheel. The hanging mass is moving too, and here it takes twice as much of the energy as the wheel does.

Q25  ·  Alternating Current  ·  numeric  ·  on the pattern of Q50

An inductor stores 16 J of magnetic energy and wastes 32 W as heat in its own resistance when an alternating current of 2 A r.m.s. at 50 Hz flows through it. The ratio of its inductive reactance to its resistance is ______. (π = 3.14.)

Answer  314

SolutionThe average stored energy is ½LIrms2 = ½L(4) = 2L, so 2L = 16 and L = 8 H. The heat gives Irms2R = 4R = 32, so R = 8 Ω. The reactance is XL = 2πfL = 2 × 3.14 × 50 × 8 = 2512 Ω. The ratio is 2512/8 = 314, which is just 100π — a coil with a very large reactance beside a small resistance, as a good inductor should have.

Where it goes wrong  Using the peak current for the stored energy. The 2 A is already an r.m.s. value, and ½LIrms2 is the average energy, which is what “stores 16 J” means here.

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, 28 January 2026, is on its own page, and the other January dates are 21, 22, 23 and 24 January.

Official final answer key →

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

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Mathematical Tools · Units and Measurements · Motion in a Straight Line · Motion in a Plane · Laws of Motion · Work Energy and Power · Rotational Motion · Centre of Mass · Gravitation · Mechanical Properties of Solids · Mechanical Properties of Fluids · Thermal Properties of Matter · Thermodynamics · Kinetic Theory of Gases · Oscillations · Waves · Electric Charges and Fields · Electrostatic Potential and Capacitance · Current Electricity · Moving Charges and Magnetism · Magnetism and Matter · Electromagnetic Induction · Alternating Current · Electromagnetic Waves · Ray Optics · Wave Optics · Dual Nature of Radiation and Matter · Atoms · Nuclei · Semiconductor Electronics

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