JEE Main 21 January 2026 Shift 2 — Physics Paper with Answer Key

The afternoon paper of 21 January 2026, the second 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 21 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 Unknown resistance in a network whose equivalent equals the unknown itself Current Electricity
27 Pulley built from a rim and two rods — acceleration of the two hanging blocks Rotational Motion
28 Charge on a capacitor in a circuit that has reached the steady state Electrostatic Potential and Capacitance
29 Oxygen and hydrogen with the same r.m.s. speed — the hydrogen temperature Kinetic Theory of Gases
30 Two charged spheres thrown at each other — least speed for them to touch Electrostatic Potential and Capacitance
31 Potentiometer wire — the balancing length for a zero galvanometer reading Current Electricity
32 Bohr orbit from the energy, then the angular momentum in units of h/2π Atoms
33 Prism with the emergent ray just grazing the second face Ray Optics
34 Work done between two instants, from a displacement written as a function of time Work, Energy and Power
35 LC loop — time for a quarter of the capacitor energy to reach the inductor Alternating Current
36 Capillary rise in two tubes of different bore, comparing heights and surface tensions Mechanical Properties of Fluids
37 Condition on the external resistance for the largest power in it Current Electricity
38 Kinetic energy of an oscillator swinging at a given angular frequency — the oscillator frequency Oscillations
39 Largest error in the viscosity worked out from a terminal velocity measurement Units and Measurements
40 Angular momentum of one car about another on a parallel track Rotational Motion
41 Sum of three lengths, written to the right number of significant figures Units and Measurements
42 Boat across a flowing river — least time for a round trip, and the drift Motion in a Plane
43 Two statements on the angular separation of fringes in a double slit Wave Optics
44 Spinning drum — least angular speed that keeps a body on the inner wall Laws of Motion
45 Long wire bent into a loop — field at the centre of the circular part Moving Charges and Magnetism
46 numeric Conduction current density against displacement current density in a conductor Electromagnetic Waves
47 numeric Refractive index of a sheet from the shift of the central fringe Wave Optics
48 numeric Heat supplied to a diatomic gas that does known work at constant pressure Thermodynamics
49 numeric Terminal velocity of a smaller ball of the same material in the same fluid Mechanical Properties of Fluids
50 numeric de Broglie wavelength of a charged particle after it is accelerated Dual Nature of Radiation and Matter

Chapter by chapter

Chapter Questions Marks
Current Electricity 3 12
Electrostatic Potential and Capacitance 2 8
Mechanical Properties of Fluids 2 8
Rotational Motion 2 8
Units and Measurements 2 8
Wave Optics 2 8
Alternating Current 1 4
Atoms 1 4
Dual Nature of Radiation and Matter 1 4
Electromagnetic Waves 1 4
Kinetic Theory of Gases 1 4
Laws of Motion 1 4
Motion in a Plane 1 4
Moving Charges and Magnetism 1 4
Oscillations 1 4
Ray Optics 1 4
Thermodynamics 1 4
Work, Energy and Power 1 4

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

Nineteen chapters in twenty-five questions. Current Electricity alone carried 12 marks here, which it did not do in any April shift, and the three questions were spread wide: a network puzzle, a potentiometer and the maximum power condition. No chapter went past three questions.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 8 32 32%
Mechanics 5 20 20%
Properties of Matter and Heat 4 16 16%
Optics and Waves 4 16 16%
Modern Physics and Electronics 2 8 8%
Units, Dimensions and Measurement 2 8 8%

32 marks of electrodynamics is the heaviest share in any 2026 shift we have on record. Modern physics dropped to two questions, and there was no semiconductor or logic gate question at all — the one place where this shift broke the pattern that held through the rest of the year.

How tough it was

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

Six long ones, the most in any January shift. The resistance network that ends in a quadratic and the pulley built out of a rim and two rods were the two that broke the clock for most students. Against that, seven questions were one line each, so a student who picked order well still had time left.

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  ·  Current Electricity  ·  on the pattern of Q26

A known resistance R and an unknown resistance X are joined in parallel, and this combination is put in series with a second known resistance 2R. The equivalent resistance of the whole arrangement comes out equal to X itself. The value of X is ______.

Answer  (√3 + 1)R

SolutionThe parallel pair is RX/(R + X), so the total is 2R + RX/(R + X), and this is to equal X. Multiplying across by (R + X): 2R(R + X) + RX = X(R + X), that is 2R2 + 2RX + RX = RX + X2. Tidying up, X2 − 2RX − 2R2 = 0, so X = [2R ± √(4R2 + 8R2)]/2 = R ± √3R. A resistance cannot be negative, so X = (√3 + 1)R.

Where it goes wrong  Keeping both roots. R(1 − √3) is negative, so it is thrown away at once.

Q2  ·  Rotational Motion  ·  on the pattern of Q27

A pulley is built from a thin rim of mass M and radius r, stiffened by two thin rods, each of mass M and each as long as the diameter of the rim, crossing at the centre. It turns freely about its centre in a vertical plane. A light string over the pulley carries blocks of mass M and m at its two ends, and does not slip. The magnitude of the acceleration of the blocks is ______.

Answer  (M − m)g / [(8/3)M + m]

SolutionFirst the moment of inertia of the pulley. The rim is all at radius r, so it gives Mr2. Each rod is a thin rod of length 2r turning about its own centre, so each gives M(2r)2/12 = Mr2/3, and the two together give 2Mr2/3. So I = Mr2 + 2Mr2/3 = 5Mr2/3. For a string over a pulley with blocks M and m, a = (M − m)g/(M + m + I/r2). Here I/r2 = 5M/3, so the bracket is M + m + 5M/3 = (8/3)M + m, giving a = (M − m)g/[(8/3)M + m].

Where it goes wrong  Treating the pulley as a disc of mass 3M, which would give 1.5M in the bracket instead of 5M/3. The rim and the rods have to be added separately.

Q3  ·  Capacitance  ·  on the pattern of Q28

A 12 V battery of negligible internal resistance drives a 2 Ω and a 4 Ω resistor joined in series. A 5 μF capacitor is connected across the 4 Ω resistor. In the steady state the charge on the capacitor is ______ μC.

Answer  40 μC

SolutionIn the steady state no current flows into the capacitor branch, so the circuit is just the two resistors in series: I = 12/(2 + 4) = 2 A. The voltage across the 4 Ω resistor is 2 × 4 = 8 V, and the capacitor sits across that same 8 V. So Q = CV = 5 × 8 = 40 μC.

Where it goes wrong  Using the full 12 V. The capacitor only ever sees the voltage across the element it is joined across.

Q4  ·  Kinetic Theory of Gases  ·  on the pattern of Q29

The r.m.s. speed of oxygen molecules at 47 °C is equal to the r.m.s. speed of hydrogen molecules kept at ______ °C. (Take the mass of an oxygen molecule to be 16 times that of a hydrogen molecule.)

Answer  −253 °C

SolutionThe r.m.s. speed is √(3RT/M), so equal speeds mean T/M is the same for both: TH/MH = TO/MO. With MO = 16MH and TO = 320 K, TH = 320/16 = 20 K, which is 20 − 273 = −253 °C.

Where it goes wrong  Working in °C. The relation holds for absolute temperature only, so convert to kelvin first and back at the end.

Q5  ·  Electrostatic Potential  ·  on the pattern of Q30

Two identical metal spheres, each of radius R, charge Q and mass m, have their centres 4R apart. Each is given a speed u towards the other. The least u for which they just manage to touch is ______. (k = 1/4πε0, and kQ2 > Gm2.)

Answer  √{ (kQ2/4mR)(1 − Gm2/kQ2) }

Solution“Just touch” means they arrive at centre separation 2R with zero speed, so all the kinetic energy goes into the change in potential energy. The kinetic energy is 2 × ½mu2 = mu2. The electrostatic energy rises from kQ2/4R to kQ2/2R, a gain of kQ2/4R, while the gravitational energy falls from −Gm2/4R to −Gm2/2R, a loss of Gm2/4R. So mu2 = (kQ2 − Gm2)/4R, and u = √{(kQ2/4mR)(1 − Gm2/kQ2)}.

Where it goes wrong  Counting only one sphere’s kinetic energy. Both are moving, so the total is mu2, not ½mu2.

Q6  ·  Current Electricity  ·  on the pattern of Q31

A potentiometer wire AB is 50 cm long and has a total resistance of 10 Ω. A 2 V driver cell of negligible internal resistance is connected straight across AB. A cell of emf 0.8 V is connected with its positive end at A through a galvanometer. The galvanometer reads zero when the sliding contact is at P. The length AP is ______ cm.

Answer  20 cm

SolutionThe whole 2 V is dropped across the 50 cm of wire, so the potential gradient is 2/50 = 0.04 V per cm. At the null point the length AP carries exactly the emf being balanced: 0.04 × AP = 0.8, so AP = 20 cm.

Where it goes wrong  Bringing the 10 Ω into the calculation. It only matters if something else is in series with the wire; here the full cell voltage already sits across AB.

Q7  ·  Atoms  ·  on the pattern of Q32

The energy of an electron in a certain Bohr orbit of hydrogen is −0.04E0, where E0 is the magnitude of the ground state energy. If L is the angular momentum of the electron in that orbit and h is Planck’s constant, then 2πL/h is ______.

Answer  5

SolutionIn the Bohr model En = −E0/n2. Setting E0/n2 = 0.04E0 gives n2 = 25, so n = 5. The angular momentum is quantised as L = nh/2π, so 2πL/h = n = 5.

Where it goes wrong  Reading 0.04 as 1/4 and getting n = 2. It is 1/25.

Q8  ·  Ray Optics  ·  on the pattern of Q33

A prism of refracting angle 60° is made of a material of refractive index √2. Light falls on one face and the emergent ray just grazes along the second face. The angle of incidence is about ______.

Answer  about 21.5°

Solution“Grazes along the second face” means the ray leaves at 90°, so inside the glass it meets that face at exactly the critical angle: sin C = 1/√2, C = 45°. So r2 = 45°. For any prism r1 + r2 = A, giving r1 = 60 − 45 = 15°. At the first face, sin i = √2 sin15° = 1.414 × 0.2588 = 0.366, so i ≈ 21.5°.

Where it goes wrong  Assuming the path is symmetric, r1 = r2 = 30°. That is minimum deviation, a different situation entirely.

Q9  ·  Work, Energy and Power  ·  on the pattern of Q34

A body of mass 2 kg moves along the x-axis with x(t) = αt2 + βt + γ, where α = 1 m/s2, β = 1 m/s and γ = 1 m. The work done on the body between t = 2 s and t = 3 s is ______ J.

Answer  24 J

SolutionThe velocity is v = dx/dt = 2αt + β = 2t + 1. At t = 2 s, v = 5 m/s; at t = 3 s, v = 7 m/s. By the work-energy theorem the work done is the change in kinetic energy: W = ½ × 2 × (72 − 52) = 49 − 25 = 24 J.

Where it goes wrong  Working out force × displacement with the wrong displacement, or forgetting that the force here is constant so the work-energy route is much shorter.

Q10  ·  Alternating Current  ·  on the pattern of Q35

A capacitor C is charged to a potential difference V0 and then disconnected from the battery. It is then connected across an inductor of inductance L. The time taken for 25% of the initial energy to pass from the capacitor to the inductor is ______.

Answer  (π/6)√(LC)

SolutionIn an LC loop the charge runs as q = q0cosωt with ω = 1/√(LC), so the energy left in the capacitor is U0cos2ωt. If a quarter has gone across, three quarters remain: cos2ωt = 3/4, so cosωt = √3/2 and ωt = π/6. Therefore t = (π/6)/ω = (π/6)√(LC).

Where it goes wrong  Putting cos2ωt = 1/4 because the question says 25%. The 25% is what left, so three quarters is what is still there.

Q11  ·  Mechanical Properties of Fluids  ·  on the pattern of Q36

Two liquids of the same density have the same surface tension. Their rise is measured by the capillary method in two tubes whose inner radii are r1 and r2, with r1 > r2. The measured heights h1 and h2 then satisfy ______.

Answer  h1 < h2

SolutionThe rise is h = 2Tcosθ/(ρgr). With T, θ and ρ the same for both, h goes as 1/r. Since the first tube is the wider one, r1 > r2 gives h1 < h2: the narrower tube lifts the liquid higher.

Where it goes wrong  Reading “wider tube” as “more liquid, so higher”. More liquid is lifted in the wide tube, but to a smaller height.

Q12  ·  Current Electricity  ·  on the pattern of Q37

A battery of emf E and internal resistance r is connected across an external resistance R. The power used up in R is largest when ______.

Answer  R = r

SolutionThe current is I = E/(R + r), so the power in R is P = E2R/(R + r)2. Write the denominator as (R − r)2 + 4Rr, so P = E2/[(R − r)2/R + 4r]. The bracket is smallest, and P therefore largest, when (R − r)2 = 0, that is R = r. The power delivered there is E2/4r.

Where it goes wrong  Answering R = 0 or R very large, from looking at only the numerator or only the denominator. The product has a peak in between.

Q13  ·  Oscillations  ·  on the pattern of Q38

The kinetic energy of a simple harmonic oscillator is itself oscillating with an angular frequency of 176 rad/s. The frequency of the oscillator is ______ Hz. (Take π = 22/7.)

Answer  14 Hz

SolutionFor x = A sinωt the kinetic energy is ½mω2A2cos2ωt, and cos2ωt = (1 + cos2ωt)/2. So the energy swings at 2ω, twice the oscillator’s own rate. Here 2ω = 176, so ω = 88 rad/s and f = ω/2π = 88 × 7/(2 × 22) = 14 Hz.

Where it goes wrong  Dividing 176 by 2π straight away. The factor of 2 between the energy and the motion has to come out first.

Q14  ·  Units and Measurement  ·  on the pattern of Q39

A small sphere of radius r and density σ falls through a viscous liquid of density ρ and viscosity η and settles to a terminal velocity v0. Only r and v0 are measured, with errors Δr and Δv0. The largest possible error in the value of η worked out from them is ______.

Answer  2(Δr/r) + (Δv0/v0)

SolutionStokes’ law at terminal velocity gives v0 = 2r2(σ − ρ)g/9η, so η = 2r2(σ − ρ)g/9v0. Taking relative errors, a squared quantity in the numerator contributes twice its own relative error and a quantity in the denominator contributes its relative error, and for the largest error the two are added, never subtracted: Δη/η = 2(Δr/r) + (Δv0/v0).

Where it goes wrong  Subtracting the second term because v0 sits in the denominator. For the maximum error every term is added.

Q15  ·  Rotational Motion  ·  on the pattern of Q40

Two cars A and B, each of mass 103 kg, travel in the same direction along parallel tracks 10 m apart, at 72 km/h and 36 km/h. The magnitude of the angular momentum of A about B is ______ J·s.

Answer  105 J·s

SolutionAngular momentum about B is worked out with the velocity of A relative to B. That relative speed is 72 − 36 = 36 km/h = 10 m/s, along the track. The perpendicular distance between the tracks is 10 m, so L = mvreld = 103 × 10 × 10 = 105 J·s. It stays the same at every instant, because the perpendicular distance never changes.

Where it goes wrong  Using 20 m/s, A’s own speed. About B it is A’s speed relative to B that counts.

Q16  ·  Units and Measurement  ·  on the pattern of Q41

Keeping significant figures in view, the sum of the lengths 52.01 m, 153.2 m and 0.123 m is ______.

Answer  205.3 m

SolutionThe plain sum is 205.333 m. For addition it is the number of decimal places that decides, not the count of significant figures, and the worst of the three is 153.2 m with one decimal place. So the answer is rounded to one decimal place: 205.3 m.

Where it goes wrong  Counting significant figures instead of decimal places. That rule is for multiplication and division.

Q17  ·  Motion in a Plane  ·  on the pattern of Q42

A river 200 m wide flows from west to east at 18 km/h. A boat that does 36 km/h in still water makes one round trip, bank to bank and back. The least time for the journey, and the displacement along the bank at the end of it, are ______ and ______.

Answer  40 s and 200 m

SolutionThe crossing time is shortest when the boat is pointed straight at the far bank, because then its whole 10 m/s is used across the stream: t = 200/10 = 20 s each way, so 40 s for the round trip. The current carries the boat downstream at 5 m/s the whole time, and it does so in the same direction on the way back as on the way out, so the drift adds up: 5 × 40 = 200 m along the bank.

Where it goes wrong  Expecting the drift to cancel on the return leg. The river keeps flowing one way, so the two drifts add.

Q18  ·  Wave Optics  ·  on the pattern of Q43

Two statements about Young’s double slit experiment. Statement I: the angular separation of the fringes increases as the screen is moved further from the slits. Statement II: the angular separation increases when the source is replaced by one of longer wavelength. Which is true?

Answer  Statement I is false, Statement II is true

SolutionThe angular separation of consecutive fringes is θ = λ/d. The screen distance D does not appear in it at all, so moving the screen back widens the fringes on the screen (β = λD/d) but leaves the angle untouched — Statement I is false. The wavelength does appear, in the numerator, so a longer wavelength does widen the angle — Statement II is true.

Where it goes wrong  Mixing up fringe width with angular separation. One depends on D, the other does not.

Q19  ·  Laws of Motion  ·  on the pattern of Q44

A large drum of radius R spins about its own vertical axis with angular velocity ω. A body of mass M rests against the inner wall, where the coefficient of friction is μ. The least ω that keeps the body from sliding down is ______.

Answer  √(g/μR)

SolutionThe wall pushes inward on the body with the normal force that supplies the centripetal force: N = Mω2R. Along the wall, friction has to hold the weight up, and at the least ω it is at its limit: μN = Mg. Putting the two together, μMω2R = Mg, so ω = √(g/μR). Note the mass cancels out, so a heavy body and a light one need the same spin.

Where it goes wrong  Setting friction along the direction of the centre. Friction acts along the surface, which here is vertical, so it is the weight it fights, not the centripetal need.

Q20  ·  Moving Charges and Magnetism  ·  on the pattern of Q45

A long straight wire carrying current I is bent so that a complete circular loop of radius r is formed in the middle, the two straight portions lying along one and the same straight line through the centre of the loop. At the centre the field of the loop and the field of the straight portions point the same way. The magnitude of the field at the centre O is ______.

Answer  (μ0I/2πr)(π + 1)

SolutionTreat the two pieces separately and add. The two straight portions together make one infinitely long straight wire at a perpendicular distance r from O, giving μ0I/2πr. The circular loop gives μ0I/2r at its own centre, which can be written πμ0I/2πr. Since they point the same way they add: B = (μ0I/2πr)(1 + π).

Where it goes wrong  Forgetting to put both terms over the same denominator before adding, or subtracting when the two fields actually agree in direction. Always fix the direction with the right-hand rule first.

Q21  ·  Electromagnetic Waves  ·  numeric  ·  on the pattern of Q46

An electromagnetic wave of frequency 100 MHz travels through a medium of conductivity σ = 10 mho/m. The ratio of the largest conduction current density to the largest displacement current density is ______. (Take 1/4πε0 = 9 × 109 N·m2/C2.)

Answer  1800

SolutionThe conduction current density is Jc = σE and the displacement current density is Jd = ε0dE/dt, whose peak is ε0ωE0. So the ratio is σ/ε0ω. Here ε0 = 1/(4π × 9 × 109) and ω = 2π × 108, so the ratio is 10 × 4π × 9 × 109/(2π × 108) = 10 × 180 = 1800.

Where it goes wrong  Using the frequency 108 in place of ω. The derivative brings in ω = 2πf, so the 2π must be there.

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

In a double slit set-up the slits are 0.4 mm apart and the screen is 1 m away. A thin transparent sheet 20 μm thick is placed in front of one slit, and the central bright fringe shifts by 20 mm. If the refractive index of the sheet is written as α/10, the value of α is ______.

Answer  14

SolutionThe sheet adds an extra path (μ − 1)t on one side, and the centre moves across by S = D(μ − 1)t/d. Putting the numbers in: 0.020 = 1 × (μ − 1) × 20 × 10−6/(0.4 × 10−3) = (μ − 1) × 0.05. So μ − 1 = 0.4, μ = 1.4 = 14/10, and α = 14.

Where it goes wrong  Using μt instead of (μ − 1)t. The sheet replaces 20 μm of air, so only the extra path counts.

Q23  ·  Thermodynamics  ·  numeric  ·  on the pattern of Q48

A diatomic gas (γ = 1.4) does 100 J of work while expanding at constant pressure. The heat given to the gas is ______ J.

Answer  350 J

SolutionAt constant pressure W = nRΔT and Q = nCPΔT, so Q/W = CP/R. Since CP = γR/(γ − 1), the ratio is γ/(γ − 1) = 1.4/0.4 = 3.5. So Q = 3.5 × 100 = 350 J. Of this, 250 J goes into the internal energy and 100 J comes out as work.

Where it goes wrong  Answering 100 J, as though all the heat became work. At constant pressure the gas warms up as well, so most of the heat stays inside.

Q24  ·  Mechanical Properties of Fluids  ·  numeric  ·  on the pattern of Q49

A metal ball of radius 6 mm has a terminal velocity of 20 cm/s in a viscous fluid. A second ball of the same material and radius 3 mm, in the same fluid, has a terminal velocity of ______ cm/s.

Answer  5 cm/s

SolutionThe terminal velocity is v = 2r2(σ − ρ)g/9η. Same material and same fluid means everything except r2 is unchanged, so v goes as r2. Halving the radius divides the speed by four: 20/4 = 5 cm/s.

Where it goes wrong  Halving the velocity along with the radius. The dependence is on r2, not r.

Q25  ·  Dual Nature of Matter  ·  numeric  ·  on the pattern of Q50

A particle of charge 3 × 10−19 C and mass 6 × 10−27 kg is accelerated through a potential difference of 1.21 V. The de Broglie wavelength of the particle is α × 10−12 m. The value of α is ______. (Planck’s constant = 6.6 × 10−34 J·s.)

Answer  10

SolutionThe work done by the field becomes kinetic energy, qV = p2/2m, so p = √(2mqV) and λ = h/√(2mqV). Here 2mqV = 2 × 6 × 10−27 × 3 × 10−19 × 1.21 = 43.56 × 10−46, whose square root is 6.6 × 10−23. So λ = 6.6 × 10−34/6.6 × 10−23 = 10−11 m = 10 × 10−12 m, giving α = 10.

Where it goes wrong  Writing λ = h/mv and then using v from qV = ½mv2 in two separate steps, which is the same thing but invites an arithmetic slip. Going straight to h/√(2mqV) is safer.

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

Official final answer key →

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

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