JEE Main 23 January 2026 Shift 1 — Physics Paper with Answer Key

The morning paper of 23 January 2026, the fifth 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 23 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 Pendulum length needed to double the number of oscillations in the same time Oscillations
27 Bob on a rod hits an identical bob, which then just completes a vertical circle Work, Energy and Power
28 Screw gauge with the circular scale zero above the line — corrected thickness Units and Measurements
29 de Broglie wavelength of an oxygen molecule at 27 °C Dual Nature of Radiation and Matter
30 Critical angle from the speeds of light in the two media Ray Optics
31 Four claims about the series limits of the hydrogen spectrum Atoms
32 Two spheres stick after a head-on collision — rise in temperature Thermal Properties of Matter
33 Four measurements of a rod — relative error in the average Units and Measurements
34 Copper wire falling through a horizontal field — emf after 200 m Electromagnetic Induction
35 Charge at a cube corner and another at a face centre — flux through the cube Electric Charges and Fields
36 Match four integral equations to the laws they state Electromagnetic Waves
37 Series resistor for a Zener voltage regulator Semiconductor Electronics
38 Ring of wire with a diameter wire across it — resistance between the ends Current Electricity
39 Two blocks on two springs — energy stored in the second one Oscillations
40 Square loop of four solid cylinders — moment of inertia Rotational Motion
41 Two unequal balls on a light rod — angular velocity from the angular momentum Rotational Motion
42 Four materials on a stress-strain plot — which has the largest Young’s modulus Mechanical Properties of Solids
43 Assertion and reason on domains in a ferromagnetic material Magnetism and Matter
44 Two thin prisms giving dispersion without deviation — the second angle Ray Optics
45 Projectile at 60° — kinetic energy at two later points compared with the start Motion in a Plane
46 numeric Average power of a wave from the amplitude of its electric field Electromagnetic Waves
47 numeric Capacitor gap filled with dielectric slabs — the new capacitance Electrostatic Potential and Capacitance
48 numeric Two double slit set-ups with equal fringe width — ratio of screen distances Wave Optics
49 numeric Swinging pendulum wire in a field — largest emf along the wire Electromagnetic Induction
50 numeric LCR circuit at its largest current — the capacitance Alternating Current

Chapter by chapter

Chapter Questions Marks
Electromagnetic Induction 2 8
Electromagnetic Waves 2 8
Oscillations 2 8
Ray Optics 2 8
Rotational Motion 2 8
Units and Measurements 2 8
Alternating Current 1 4
Atoms 1 4
Current Electricity 1 4
Dual Nature of Radiation and Matter 1 4
Electric Charges and Fields 1 4
Electrostatic Potential and Capacitance 1 4
Magnetism and Matter 1 4
Mechanical Properties of Solids 1 4
Motion in a Plane 1 4
Semiconductor Electronics 1 4
Thermal Properties of Matter 1 4
Wave Optics 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, and six of them took two questions each. Magnetism and Matter turned up here, in a shift where it was tested by an assertion and reason question on ferromagnetic domains.

Where the 100 marks sat

Part of Physics Questions Marks Share
Electrodynamics 9 36 36%
Optics and Waves 5 20 20%
Mechanics 4 16 16%
Modern Physics and Electronics 3 12 12%
Units, Dimensions and Measurement 2 8 8%
Properties of Matter and Heat 2 8 8%

36 marks of electrodynamics is the highest of any shift in 2026, and mechanics fell to 16, its lowest. A student strong in electricity and magnetism and weak in mechanics got the paper they would have asked for.

How tough it was

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

The square loop built out of four cylinders needed four separate moment-of-inertia terms, and the colliding spheres needed a collision, an energy loss and a calorimetry step one after the other. Those two were the clock-eaters. The four hydrogen-spectrum claims, on the other hand, could be checked on sight by anyone who knew that a series limit is the shortest wavelength, not the longest.

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  ·  Oscillations  ·  on the pattern of Q26

A simple pendulum with a string 30 cm long completes 20 oscillations in 10 s. The string length needed for it to complete 40 oscillations in the same 10 s is ______ cm. (The mass of the bob is unchanged.)

Answer  7.5 cm

SolutionTwice as many oscillations in the same time means the period is halved. Since T = 2π√(L/g), the length goes as T2, so halving T divides the length by four: 30/4 = 7.5 cm. The mass of the bob never enters, which is why the question says it stays the same — it would not matter either way.

Where it goes wrong  Halving the length to 15 cm. The period goes as the square root of the length, so the length must change four-fold.

Q2  ·  Work, Energy and Power  ·  on the pattern of Q27

A bob A of mass m is fixed to a light rigid rod 1 m long pivoted at P, held at 60° to the vertical and released. At the lowest point it strikes an identical bob B resting on a smooth horizontal surface, the collision being elastic. B then runs into a vertical circular track and just manages to complete the circle. The radius R of the track is ______ m. (g = 10 m/s2.)

Answer  0.2 m

SolutionA drops through a height l(1 − cos60°) = 1 × 0.5 = 0.5 m, so it arrives with v2 = 2 × 10 × 0.5 = 10 m2/s2. In an elastic collision between equal masses the moving one stops dead and the other leaves with the whole speed, so B starts with v2 = 10. To just complete a vertical circle a body on the inside of a track needs v2 = 5gR at the bottom, so 10 = 5 × 10 × R and R = 0.2 m.

Where it goes wrong  Using v2 = 2gR, the condition for reaching the top with zero speed. On the inside of a loop the string or track cannot push inward, so the speed at the top must still be √(gR), which makes it 5gR at the bottom.

Q3  ·  Units and Measurement  ·  on the pattern of Q28

In a screw gauge of least count 0.01 mm, the zero of the circular scale sits 3 divisions above the horizontal pitch line when the studs are brought together. A sheet is then measured: the pitch scale reads 1 mm and the circular scale reads 51. The correct thickness of the sheet is ______ mm.

Answer  1.54 mm

SolutionThe observed reading is 1 + 51 × 0.01 = 1.51 mm. The zero sitting above the line is a negative zero error, −3 × 0.01 = −0.03 mm, and a correction is always subtracted from the observed reading: 1.51 − (−0.03) = 1.54 mm.

Where it goes wrong  Subtracting 0.03 and answering 1.48 mm. Above the line means the instrument reads low, so the correction is added.

Q4  ·  Dual Nature of Matter  ·  on the pattern of Q29

The de Broglie wavelength of an oxygen molecule at 27 °C is x × 10−12 m. The value of x is about ______. (h = 6.63 × 10−34 J·s, k = 1.38 × 10−23 J/K, mass of an oxygen molecule = 5.31 × 10−26 kg.)

Answer  about 26

SolutionA gas molecule at temperature T has momentum p = mvrms = √(3mkT), so λ = h/√(3mkT). Here 3mkT = 3 × 5.31 × 10−26 × 1.38 × 10−23 × 300 = 6.59 × 10−46, whose square root is 2.57 × 10−23. So λ = 6.63 × 10−34/2.57 × 10−23 = 2.58 × 10−11 m, that is about 26 × 10−12 m.

Where it goes wrong  Using the average speed or the most probable speed instead of the r.m.s. speed. The √3 in √(3mkT) is what fixes the answer.

Q5  ·  Ray Optics  ·  on the pattern of Q30

Light travels from medium A into medium B across a flat boundary and undergoes total internal reflection. The speeds of light in A and B are 2.4 × 108 m/s and 2.7 × 108 m/s. The critical angle is ______.

Answer  sin−1(8/9)

SolutionThe refractive indices are μA = 3/2.4 = 1.25 and μB = 3/2.7 = 1.111, so A is the denser of the two, which is why total internal reflection is possible going from A into B. At the critical angle, μAsin C = μBsin90°, so sin C = μB/μA = 2.4/2.7 = 8/9 and C = sin−1(8/9), about 62.7°.

Where it goes wrong  Putting the ratio the other way up and getting sin C = 9/8, which is impossible. The sine of the critical angle is always the smaller index over the larger.

Q6  ·  Atoms  ·  on the pattern of Q31

For the hydrogen spectrum (R is the Rydberg constant), which of these are correct? A. The longest wavelength of the Lyman series is 4/3R. B. The Balmer series lies in the visible region. C. The shortest wavelength of the Paschen series is 9/R. D. The shortest wavelength of the Lyman series is 4/5R.

Answer  A, B and C only

SolutionThe longest wavelength of a series is the smallest jump. For Lyman that is 2 → 1: 1/λ = R(1 − 1/4) = 3R/4, so λ = 4/3R — A is right. The Balmer lines do fall in the visible band, from 656 nm down to about 410 nm — B is right. The shortest wavelength is the series limit. For Paschen that is ∞ → 3: 1/λ = R/9, so λ = 9/R — C is right. For Lyman the limit is ∞ → 1, giving λ = 1/R, not 4/5R — D is wrong.

Where it goes wrong  Treating the series limit as the longest wavelength. The limit is the biggest energy jump, so it is the shortest wavelength.

Q7  ·  Thermal Properties of Matter  ·  on the pattern of Q32

Two spheres of the same material, of masses 15 kg and 25 kg, move towards each other at 10 m/s and 30 m/s, collide and stick together. If all the heat produced stays in the spheres, their temperature rises by ______ °C. (Specific heat of the material = 31 cal/kg·°C, 1 cal = 4.2 J.)

Answer  about 1.44 °C

SolutionMomentum is conserved: 15 × 10 − 25 × 30 = −600 kg·m/s, so the combined 40 kg moves at 15 m/s. The kinetic energy before is ½(15)(100) + ½(25)(900) = 750 + 11250 = 12000 J, and after it is ½(40)(225) = 4500 J. So 7500 J has become heat, which is 7500/4.2 = 1786 cal. Then ΔT = Q/mc = 1786/(40 × 31) = 1.44 °C.

Where it goes wrong  Forgetting to turn joules into calories, since the specific heat is given in calories. One of the two has to be converted before dividing.

Q8  ·  Units and Measurement  ·  on the pattern of Q33

Four people measure the length of a rod and get 20.00 cm, 19.75 cm, 17.01 cm and 18.25 cm. The relative error in the average length is ______.

Answer  about 0.06

SolutionThe mean is 75.01/4 = 18.75 cm. The four deviations from it are 1.25, 1.00, 1.74 and 0.50 cm, and their average, the mean absolute error, is 4.49/4 = 1.12 cm. The relative error is that divided by the mean: 1.12/18.75 = 0.06.

Where it goes wrong  Letting the signs of the deviations cancel. The mean absolute error uses the sizes of the deviations, never their signs.

Q9  ·  Electromagnetic Induction  ·  on the pattern of Q34

A copper wire 20 m long is held horizontally and allowed to fall freely through a uniform horizontal magnetic field of 0.5 gauss at right angles to its length. The emf across the wire after it has fallen 200 m is ______ mV. (g = 10 m/s2.)

Answer  20√10 mV, about 63 mV

SolutionAfter falling 200 m the speed is v = √(2 × 10 × 200) = √4000 = 20√10 m/s. One gauss is 10−4 T, so B = 0.5 × 10−4 T. The emf is Blv = 0.5 × 10−4 × 20 × 20√10 = 0.02√10 V = 20√10 mV, about 63 mV.

Where it goes wrong  Leaving the field in gauss. The conversion to tesla is a factor of 10−4, and missing it throws the answer out by ten thousand.

Q10  ·  Electric Charges and Fields  ·  on the pattern of Q35

A charge 2q sits at one corner of a cube and a charge q sits at the centre of one of its faces. The total electric flux through the cube is ______.

Answer  3q/4ε0

SolutionGauss’s law gives the flux from the charge enclosed, but a charge sitting exactly on the boundary is shared. Eight identical cubes meet at a corner, so only one eighth of that charge’s flux goes through this cube: 2q/8ε0 = q/4ε0. A charge on a face is shared between two cubes, so half of its flux comes through: q/2ε0. Adding, the total is q/4ε0 + q/2ε0 = 3q/4ε0.

Where it goes wrong  Counting both charges as fully enclosed and answering 3q/ε0. A charge on the surface of the closed surface only ever contributes its fair share.

Q11  ·  Electromagnetic Waves  ·  on the pattern of Q36

Name the law stated by each equation: A. ∮E·dl = −d/dt ∮B·da, B. ∮B·dl = μ0(I + ε0 dφE/dt), C. ∮E·da = (1/ε0)∫ρdv, D. ∮B·dl = μ0I.

Answer  A → Faraday’s law of induction, B → Ampere-Maxwell law, C → Gauss’s law of electrostatics, D → Ampere’s circuital law

SolutionA has a changing magnetic flux driving an electric field round a loop, with the minus sign of Lenz — that is Faraday. B is Ampere’s law with the extra displacement current term ε0dφE/dt, the piece Maxwell added, so it is the Ampere-Maxwell law. C relates the flux of E out of a closed surface to the charge inside — Gauss. D is the same as B but without the displacement term, so it is the original Ampere’s circuital law.

Where it goes wrong  Mixing up B and D, which look almost the same. The one with dφE/dt in it is the Maxwell-corrected version.

Q12  ·  Semiconductor Electronics  ·  on the pattern of Q37

A Zener diode rated at 5 V is used as a voltage regulator, with a series resistor Rs fed from an unregulated supply of 15 V. The load needs 10 mA, and the Zener can safely carry four times the load current. The value of Rs is ______ Ω.

Answer  200 Ω

SolutionThe Zener holds its own ends at 5 V, so the series resistor has to drop 15 − 5 = 10 V. The worst case is when the whole allowance of current flows: the load’s 10 mA plus the Zener’s 40 mA, that is 50 mA. So Rs = 10/0.050 = 200 Ω.

Where it goes wrong  Sizing Rs from the load current alone. The series resistor carries the load current and the Zener current together, and the Zener takes up whatever the load does not.

Q13  ·  Current Electricity  ·  on the pattern of Q38

A wire of uniform resistance λ ohm per metre is bent into a circle of radius r. A second piece of the same wire, of length 2r, is joined straight across between the two ends of a diameter AB, passing through the centre. The resistance between A and B is ______.

Answer  2πλr / (π + 4)

SolutionA and B split the circle into two semicircles, each of length πr and so of resistance πrλ. The straight piece is 2r long, so its resistance is 2rλ. All three run between the same two points, so they are in parallel: 1/R = 1/πrλ + 1/πrλ + 1/2rλ = (1/rλ)(2/π + 1/2) = (4 + π)/2πrλ. So R = 2πλr/(π + 4), which is about 0.88λr.

Where it goes wrong  Treating the two semicircles as being in series because they form a ring. Between A and B they are two separate routes, so they are in parallel.

Q14  ·  Oscillations  ·  on the pattern of Q39

Blocks of 100 g and 200 g hang at rest from the lower ends of two springs A and B. The energy stored in A is E. If the spring constants satisfy 4kA = 3kB, the energy stored in B is ______.

Answer  3E

SolutionA hanging block stretches its spring until kx = mg, so x = mg/k and the stored energy is ½kx2 = m2g2/2k. So the energy goes as m2/k. Comparing the two, EB/EA = (mB/mA)2 × (kA/kB) = (200/100)2 × (3/4) = 4 × 0.75 = 3. So EB = 3E.

Where it goes wrong  Taking the two extensions as equal. They are not: each block hangs until its own spring balances its own weight, so the extension depends on both m and k.

Q15  ·  Rotational Motion  ·  on the pattern of Q40

A square frame is built from four uniform solid cylinders, each of radius R and length L, with R much smaller than L. The total mass of the frame is M. Its moment of inertia about an axis in the plane of the square passing through the midpoints of two opposite sides is ______.

Answer  (3/8)MR2 + (1/6)ML2

SolutionEach cylinder has mass M/4. The two cylinders that the axis cuts through are at right angles to it, with their centres on it, so each contributes (M/4)(R2/4 + L2/12); the pair gives MR2/8 + ML2/24. The other two lie parallel to the axis at a distance L/2 from it, so each contributes its own (M/4)(R2/2) about its long axis plus (M/4)(L/2)2 from the parallel axis theorem; the pair gives MR2/4 + ML2/8. Adding all four, I = (1/8 + 1/4)MR2 + (1/24 + 1/8)ML2 = (3/8)MR2 + (1/6)ML2.

Where it goes wrong  Using the same formula for all four cylinders. Two are crossed by the axis and two run parallel to it, and those are different cases.

Q16  ·  Rotational Motion  ·  on the pattern of Q41

Two small balls of masses m and 2m are fixed to the two ends of a rigid rod of length d and negligible mass. The angular momentum of the system about an axis A through its centre of mass, at right angles to the rod, is L. The angular velocity about A is ______.

Answer  3L / 2md2

SolutionThe centre of mass is nearer the heavier ball: m sits 2d/3 from it and 2m sits d/3 from it. So I = m(2d/3)2 + 2m(d/3)2 = 4md2/9 + 2md2/9 = 2md2/3. Then ω = L/I = 3L/2md2.

Where it goes wrong  Putting the axis at the middle of the rod. The centre of mass of m and 2m is one third of the way along, not halfway.

Q17  ·  Mechanical Properties of Solids  ·  on the pattern of Q42

Four wires A, B, C and D give stress-strain graphs that are straight lines through the origin, with A the steepest and D the shallowest. Which wire has the largest Young’s modulus?

Answer  A

SolutionYoung’s modulus is stress divided by strain, so on a graph of stress against strain it is the slope. The steepest line, A, has the largest modulus — it needs the most stress to produce the same strain, which is another way of saying it is the stiffest.

Where it goes wrong  Not checking which quantity is on which axis. If the plot is strain against stress, the modulus is one over the slope, and then the shallowest line wins instead.

Q18  ·  Magnetism and Matter  ·  on the pattern of Q43

Assertion: the individual atoms of a ferromagnetic material have magnetic dipole moments and interact so as to line themselves up spontaneously, forming domains. Reason: at a high enough temperature the domain structure breaks up and the magnetisation disappears, at what is called the Curie temperature. Choose the correct option.

Answer  Both are true, but the reason is not the explanation of the assertion

SolutionBoth statements are correct physics. Ferromagnetic atoms do have dipole moments and the exchange interaction between them does line them up into domains; and above the Curie temperature the thermal agitation does destroy that order. But the second explains how the domains are lost, not why they form. So both are true and the reason is not the explanation.

Where it goes wrong  Marking the reason as the explanation because both sentences are about domains. A correct statement on the same topic is not automatically the cause of the first one.

Q19  ·  Ray Optics  ·  on the pattern of Q44

A thin prism of angle 5° made of material of refractive index 1.72 is combined with a second thin prism of refractive index 1.9, so that the pair gives dispersion without deviation. The angle of the second prism is ______.

Answer  4°

SolutionA thin prism deviates light by (μ − 1)A. For the two deviations to cancel, with the prisms set opposite ways round, (μ1 − 1)A1 = (μ2 − 1)A2. So 0.72 × 5 = 0.9 × A2, giving A2 = 3.6/0.9 = 4°. The colours do not cancel, because the two materials disperse by different amounts — which is the point of the arrangement.

Where it goes wrong  Using μ in place of (μ − 1). It is the excess over 1 that produces the bending.

Q20  ·  Motion in a Plane  ·  on the pattern of Q45

An object is thrown with kinetic energy K at 60° to the horizontal. Ignoring air resistance, the ratio of its kinetic energy at the highest point of the flight to K is ______.

Answer  1 : 4

SolutionAt the top the vertical part of the velocity is zero and only the horizontal part is left, and that part never changed: u cos60° = u/2. Kinetic energy goes as the square of the speed, so at the top it is K(cos260°) = K/4. The ratio is 1 : 4, and the other three quarters of the energy is sitting as potential energy at the peak.

Where it goes wrong  Answering zero, as though the object stopped at the top. Only the vertical motion stops; the horizontal motion carries on untouched.

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

A plane electromagnetic wave travelling through free space has electric field E = 377 sin(6.27 × 103t − 2.09 × 10−5x) N/C. The average power carried per unit area is ______ W/m2. (Take √(μ0/ε0) = 377 SI units.)

Answer  188.5 W/m2

SolutionThe intensity is the average of the Poynting vector, I = E02/2μ0c, and μ0c is the same as √(μ0/ε0) = 377 Ω, the impedance of free space. So I = 3772/(2 × 377) = 377/2 = 188.5 W/m2. The numbers in the bracket are there only to tell you it is a free-space wave; the amplitude alone settles the power.

Where it goes wrong  Leaving out the factor of 2. The average of sin2 over a cycle is a half, and that is where the 2 comes from.

Q22  ·  Capacitance  ·  numeric  ·  on the pattern of Q47

A parallel plate capacitor of capacitance C has air between its plates. The gap is now filled with two slabs, each of half the gap thickness and of the full plate area, with dielectric constants 2 and 3. The new capacitance is ______.

Answer  12C/5, that is 2.4C

SolutionSlabs stacked one above the other across the gap are in series, each with half the thickness. So 1/C′ = (d/2)/(2ε0A) + (d/2)/(3ε0A) = (d/2ε0A)(1/2 + 1/3) = (d/ε0A)(5/12). Since C = ε0A/d, this gives C′ = 12C/5 = 2.4C.

Where it goes wrong  Treating the slabs as being in parallel and adding. They would be in parallel only if each covered part of the area over the whole gap; stacked across the gap they are in series.

Q23  ·  Wave Optics  ·  numeric  ·  on the pattern of Q48

Two separate Young’s double slit set-ups, using light of different wavelengths, give fringes of exactly the same width. The ratio of their slit separations is 2 : 1 and the ratio of their wavelengths is 1 : 2. The ratio of the screen distances D1/D2 is ______.

Answer  4

SolutionThe fringe width is β = λD/d. Setting the two equal, λ1D1/d1 = λ2D2/d2, so D1/D2 = (λ2/λ1)(d1/d2) = 2 × 2 = 4. The first set-up is working with half the wavelength and twice the slit separation, so its screen has to be four times further away to open the fringes out to the same width.

Where it goes wrong  Inverting one of the two ratios. Write β out for each set-up and divide; do not try to see the answer in one step.

Q24  ·  Electromagnetic Induction  ·  numeric  ·  on the pattern of Q49

A simple pendulum has a bob of mass 10 g on a metallic wire 10 cm long, hanging in a uniform magnetic field of 2 T at right angles to the plane in which it swings. It is released from 60° to the vertical. The largest emf between the point of suspension and the bob is ______ mV. (g = 10 m/s2.)

Answer  100 mV

SolutionThe bob is fastest at the bottom, where v = √(2gl(1 − cos60°)) = √(2 × 10 × 0.1 × 0.5) = 1 m/s, so the wire’s angular velocity there is ω = v/l = 10 rad/s. A rod turning about one end in a perpendicular field has an emf ½Bωl2 across it: ½ × 2 × 10 × (0.1)2 = 0.1 V = 100 mV. The mass of the bob does not appear, since it cancels out of the energy equation.

Where it goes wrong  Using Blv with the bob’s speed, which gives 0.2 V. Different points of the wire move at different speeds, and the integral along it brings the factor of a half.

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

A series LCR circuit with L = 1 H is driven by a source whose frequency can be varied. The largest current measured is 50 mA, and it occurs for V = 5 sin(100t). The capacitance is ______ μF.

Answer  100 μF

SolutionThe current is largest at resonance, where the two reactances cancel and the circuit behaves like R alone. That the peak occurs at ω = 100 rad/s tells us the resonant frequency is 100 rad/s, so 1/√(LC) = 100, giving C = 1/(1002 × 1) = 10−4 F = 100 μF. As a by-product, R = 5/0.050 = 100 Ω.

Where it goes wrong  Trying to use the impedance formula with a guessed R. The phrase “largest current” is the whole clue: it means resonance, and at resonance only L and C matter for the frequency.

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, 24 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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